Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculating conditional expectation and variance of multivariate normal
Suppose $\mathbf{Y} = \begin{bmatrix} Y_1 \\ Y_2 \\ Y_3 \end{bmatrix}
\sim N(\mu, \Sigma)$ where $\mu = \begin{bmatrix} 1 \\ 2 \\ 3
\end{bmatrix}$ and $\Sigma = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 5 & 1
\\ 1 & 1 & 3 \end{bmatrix}$. Calculate $\m... | Partition the multivariate normal random vector
$\mathbf{Y}$
consisting of two subvectors as
$\mathbf{Y}=\left(\begin{array}{c}
\mathbf{Y^{(1)}}\\
\mathbf{Y^{(2)}}\\
\end{array} \right)$
where
$\mathbf{Y^{(1)}}=\left(\begin{array}{c}
Y_{1}\\
Y_{2}\\
\end{array} \right) $
and
$\mathbf{Y^{(2)}}=\left(\begin{array}{c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2485500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Use induction to prove following sum identity Using induction, find an explicit formula for the sum $\displaystyle\sum_{i=1}^n\frac{3}{i^2+3i}.$
My attempt: After writing the the first few partial sums, I was able to tell that the denominators of the sum will have the form of $(i+3)!/6$ but I'm not sure how to proceed ... | It looks like a telescoping series to me. $\frac{3}{i(i+3)}=\frac{1}{i}-\frac{1}{i+3}$, so write out some terms and see what cancels. It looks like you are left with $$1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}- \frac{1}{n+2}- \frac{1}{n+3}$$
To do induction, claim $S_N=1 + \frac{1}{2}+ \frac{1}{3} - \frac{1}{n+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the name of this theorem of Jakob Steiner's, and why is it true? In The Secrets of Triangles a remarkable theorem is attributed to Jakob Steiner.
Each side of a triangle is cut into two segments by an altitude. Build squares on each of those segments, and the alternating squares sum to each other.
The book do... | Label the squares' side lengths $a, b, c, d, e, f $ (clockwise from $A$). The claim is that $$a^2+c^2+e^2=b^2+d^2+f^2$$
Let $x$ be the altitude from $A$.
Let $y$ be the altitude from $B$.
Let $z$ be the altitude from $C$.
By the Pythagorean theorem applied to the two right triangles that include the altitude from $A$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2486718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 7,
"answer_id": 3
} |
How to calculate the limit $\lim_{x \to 1} \frac{x + x^2 +\cdots+ x^n - n}{x - 1}$? How can I find the limit of the following function:
$$\begin{equation*}
\lim_{x \to 1}
\frac{x + x^2 +\cdots+ x^n - n}{x - 1}
\end{equation*}$$
Any help will be appreciated.
Thanks!
| Remember that $$\sum_{i=1}^n x^i=\frac{x \left(x^n-1\right)}{x-1}$$ Let $x=y+1$ and consider
$$S_n=\frac {-n+\sum_{i=1}^n x^i }{x-1}=\frac{\frac{(y+1) \left((y+1)^n-1\right)}{y}-n}{y}$$ Now, using the binomial theorem or Taylor series $$(y+1)^n=1+n y+\frac{1}{2} (n-1) n y^2+\frac{1}{6} (n-2) (n-1) n y^3+O\left(y^4\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2487214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Different ways to tackle the integral $\int_0^1\sqrt\frac x{1-x}\,dx$ $$\int_0^1\sqrt\frac x{1-x}\,dx$$
I saw in my book that the solution is $x=\cos^2u$ and $dx=-2\cos u\sin u\ du$.
I would like to see different approaches, can you provide them?
| Here is another way that involves rationalising the numerator first.
For $0 \leqslant x < 1$ we can write
\begin{align*}
\int_0^1 \sqrt{\frac{x}{1 - x}} \, dx &= \int_0^1 \sqrt{\frac{x}{1 - x}} \cdot \frac{\sqrt{x}}{\sqrt{x}} \, dx\\
&= \int^1_0 \frac{x}{\sqrt{x - x^2}} \, dx
\end{align*}
Now rewriting the numerator as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 1
} |
How to prove $|z|\cdot |w|=|z \cdot w|$ form complex numbers? I am trying to prove $|z|\cdot |w|=|z \cdot w|$
$|z\cdot w|=\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}$
and
$|z|\cdot|w|=\sqrt{a^{2}+b^{2}}\cdot\sqrt{c^{2}+d^{2}}=\sqrt{(a^{2}+b^{2})\cdot(c^{2}+d^{2})}=\sqrt{(a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2})}$
And I'm stuck
... | Now, $$(ac-bd)^2+(ad+bc)^2=a^2c^2-2abcd+b^2d^2+a^2d^2+2abcd+b^2c^2=$$
$$=a^2c^2+a^2d^2+b^2c^2+b^2d^2=a^2(c^2+d^2)+b^2(c^2+d^2)=(a^2+b^2)(c^2+d^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Invariant factors So I was asked the following question: Let $ G = \{ 1 + a_1x + a_2x^2 +a_3x^3 : a_i \in \mathbb{Z}/2\mathbb{Z} \} $, and define a binary operation on $ G $ by $ p(x) * q(x) = p(x)q(x) \bmod{x^4} $. This makes $ G $ a $ \mathbb{Z} $-module. Find the invariant factors of $ G $.
I understand that the op... | The elements we have are:
$$\{1,1+x,1+x^2,1+x+x^2,1+x^3,1+x+x^3,1+x^2+x^3,1+x+x^2+x^3\}$$
It can be useful to see what the subgroups $\langle g\rangle$ look like for all $g\in G$, as this gives us a good idea what to pick as generators.
Additionally, this is an abelian group of order $8$, so it's isomorphic to one of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Summation over roots of unity Find the value of $\displaystyle\sum_{r=1}^{4} \frac{1}{2-\alpha^r} $
where $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity.
My approach:-
As we know that $ \alpha^k (k=0,1,2,3,4,5) $ are fifth roots of unity, then $ \alpha^k - 1$ should be equal to zero. Therefore the final answer ... | If $\alpha^5 = 1$, $\frac{1}{2-\alpha} = c_0 + c_1 \alpha + \ldots + c_4 \alpha^4$ where
$$\eqalign{ 2 c_0 - c_4 &= 1\cr
2 c_1 - c_0 &= 0\cr
2 c_2 - c_1 &= 0\cr
2 c_3 - c_2 &= 0\cr
2 c_4 - c_3 &= 0\cr}$$
The solution of this is $$c_0 = 16/31,\; c_1 = 8/31,\; c_2 = 4/31,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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Probability a bit in a bit string is 1 after swapping Stuck on a homework question, so I could use all the help I could get.
Let $x = x(1), \dots , x(n)$ be a bit string containing exactly $m$ occurrences of 1. Consider the following operation on $x$: we choose a random pair of indices $(i,j),$ and we swap $x(i)$ and $... | Let's start with this:
\begin{align*}
P(X_N(i)=1|X_{N-1}(i)=0) &= \frac{2m}{n^2}
\end{align*}
Because if $X_{N-1}(i)=0$, then we get $X_N(i)=1$ if the first index chosen is $i$ and the second index is one of the $m$ out of $n$ spots that have a $1$, or if the first index is one of the $m$ out of $n$ spots that have a $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$ $$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$
I tried to solve this equation.
First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2}... | You can use inequalities to simplify your problem.
Since $\lfloor x \rfloor \le x$. Therefore we've
\begin{align}
x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\
&\le \sqrt x+\sqrt {x+1}+\sqrt{x+2} \\
&\le 3\sqrt {x+2}\\
\end{align}
$$\implies x^2 \le 9(x+2) \; ; x \in \mathbb Z$$
This gives us the range $x \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2499746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 1
} |
$\iint_A\frac{y}{x^2+y^2}dxdy, \quad A: [\space {(x,y): \space x^2+y^2\ge 4, \quad x^2+y^2\le2 \sqrt{x^2+y^2}+2x}]$ I've been given a "fun" integral and I just want to know if I'm doing it good or I missed something.
$$\iint_A\frac{y}{x^2+y^2}dxdy, \quad A: [\space {(x,y): \space x^2+y^2\ge 4, \quad x^2+y^2\le2 \sqrt{x... | The integrand is odd function respect to $y$ and the area is symmetric about $x$-axis, so the integral is $0$. On the other hand, you can write
$$\iint_A\frac{y}{x^2+y^2}dxdy=\int_{-\pi/2}^{\pi/2}\int_2^{2(1+\cos\theta)}\dfrac{r\sin\theta}{r^2}r\,dr\,d\theta=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction that $\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$. Help finding my mistake
Prove by induction:
$$\sum_{i=1}^{n} i \cdot 2^i = (n-1) \cdot 2^{n+1} +2$$
Basis: let $p(n)$ be the predicate.
Let $n=1$ this gives $(1-1) \cdot 2^1+1+2 = 2$ and $1 \cdot 2^1 = 2$ so its true for $p(1)$
Induction... | Assuming $P(n)$ is true, we have
$$\begin{align*}
\sum_{i=1}^{n+1} i \cdot 2^i
&= \sum_{i=1}^n i\cdot2^i + (n+1)\cdot2^{n+1} \\\\
&= (n-1)\cdot2^{n+1}+ 2 +(n+1)\cdot2^{n+1} \\\\
&= 2^{n+1}\cdot((n-1)+(n+1))+2 \\\\
&=2^{n+1}\cdot(2n)+2 \\\\
&=n\cdot2^{n+2}+2 \\\\
&=((n+1)-1)\cdot2^{((n+1)+1)}+2
\end{align*}$$ as desir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the limit of a function . How can I calculate the following limit:
\begin{equation*}
\lim_{x \rightarrow a}
\frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a} }{\sqrt{x^2 - a^2}}
\end{equation*}
I feel that I should multiply by the conjugate, but which conjugate?
| Hint: Write $y=\sqrt{x}$ and $b=\sqrt{a}$. Then you get:$${y-b+\sqrt{y^2-b^2}\over \sqrt{y^4-b^4}}={\sqrt{y-b}+\sqrt{y+b}\over \sqrt{(y+b)(y^2+b^2)}}$$ So the limit is $1\over \sqrt{2a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Counting the number of words containing a specific subword, and not another subword Consider the $10$ letter word
$$PRRAAAATTM$$
and all the words formed by rearranging its letters. How many of these words contain the subword $RAT$ but do not contain the subword $MAP$?
This is what I tried:
Let $R$ be the set of wo... | @mathlove has provided you with a nice answer. Here is an alternative approach:
Number of words containing RAT: We have eight objects to arrange: RAT, A, A, A, M, P, R, T. We can choose three positions for the As, then arrange the remaining five distinct objects in the remaining five places in
$$\binom{8}{3}5!$$
wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It s... | it must be $$a\geq |x|$$ and $$b\geq x$$
after squaring we get the equation
$$0=2x^2-2bx+b^2-a^2=0$$
solving this equation we get
$$x_1=\frac{b}{2}+\frac{1}{2}\sqrt{2a^2-b^2}$$
$$x_2=\frac{b}{2}-\frac{1}{2}\sqrt{2a^2-b^2}$$
so we get the following solution set
$$x=b$$ and $a=b$
$$x=\frac{1}{2}(b\pm i\sqrt{b^2-2a^2}$$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
When $f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$ is an integer $a,b,m,x$ are positive integers.
For which $x>0$ is $f(x)$ an integer?
$$f(x)=\frac{-b^2m-ba+ax}{-mx-bm-a}$$
I been trying to play with it, I changed it to:
$$\frac{b^2m-a\left(b+x\right)}{a+m\left(b+x\right)}$$
And then I been trying to say:
$$a+m\left(b+x\right)|... | Here is a partial answer: Note that
\begin{eqnarray*}
f(x)&=&\frac{b^2m+ba-ax}{bm+a+mx}=\frac{b^2m+ba+bmx-bmx-ax-mx^2+mx^2}{bm+a+mx}\\
&=&\frac{(b-x)(bm+a+mx)+mx^2}{bm+a+mx}=b-x+\frac{mx^2}{bm+a+mx}.
\end{eqnarray*}
So $f(x)$ is an integer if and only if $\frac{mx^2}{bm+a+mx}$ is. This means there is a positive integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Limit of $f(x)$ when $x$ goes to zero Let $f(x) = \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}$ . Find value of $\lim_{x \to 0} f(x)$ if it exists . I can solve it using L'Hospital's Rule and Taylor series but I'm looking for another way suing trigonometric identities .
| As $x\to 0$ we have
$\tan x\sim x;\;\sin x \sim x;\;1-\cos x\sim \dfrac{x^2}{2}$
So the limit becomes
$$\lim_{x\to 0} \frac{1 + \tan x + \sin x - \cos x}{\sin^2 x + x^3}=\lim_{x\to 0}\frac{x+x+\frac{x^2}{2}}{x^2+x^3}=\lim_{x\to 0}\frac{4x+x^2}{2(x^2+x^3)}=\lim_{x\to 0}\frac{x(4+x)}{2x^2(1+x)}=$$
$$=\lim_{x\to 0}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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GCD of $n^3+3n+1$ and $7n^3+18n^2-n-2$ is always $1$ A problem I found yesterday says to prove $\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=1$ for all integers $n\ge 1$. To begin, I used the Euclidean algorithm to observe that $7n^3+18n^2-n-2=7\left(n^3+3n+1\right)+\left(18n^2-22n-9\right)$, so $$\gcd(n^3+3n+1, 7n^3+18n^2-n-2)=\gcd... | It appears the problem fails for $n=309$. WolframAlpha says so here. See discussion on AoPS here on how to find $n=309$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How would one calculate absolute error under root operations? I know absolutely nothing about error analysis and I was wondering if someone could help me out. I want to approximate the value of $t$ that satisfies this equation: $$\sqrt{(\frac{1}{200}e^{2t}-\frac{1}{200}e^{-2t})^{2}+(\frac{1}{50}e^{2t})^{2}}=1.$$ I want... | I'll not answer the error propagation part, but show you how to calculate the $t$ value explicitly: By squaring the inital equation and setting $X = e^{2t}$, you obtain
$$\frac{1}{200^2}(X-\frac{1}{X})^2 + \frac{1}{50^2}X^2 = 1$$
$$\frac{1}{200}(X^2 - 2 + \frac{1}{X^2}) + \frac{4}{50}X^2 = 200$$
$$\frac{1}{200}(X^4 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Graph the set of all complex numbers $z_n=\frac{2n+1}{n-i},n\in\mathbb R$ I'm a bit rusty on my complex numbers, how would you solve the following problem on paper?
Determine and sketch (graph) the set of all complex numbers of form:
$$z_n=\frac{2n+1}{n-i},n\in\mathbb R$$
Rationalizing yields $$S=\left\{z_n\in\math... | I'll use $x$ instead of $n$.
Note that$$\frac{2x^2+x}{x^2+1}=1+\frac{x^2+x-1}{x^2+1}=1+\frac{2x^2+2x-2}{2(x^2+1)}$$and that$$\frac{2x+1}{x^2+1}=\frac12+\frac{-x^2+4x+1}{2(x^2+1)}.$$So$$\frac{2x+1}{x-i}=1+\frac i2+\frac{2x^2+2x-2}{2(x^2+1)}+\frac{-x^2+4x+1}{2(x^2+1)}i.$$Now, observe that$$\left(\frac{2x^2+2x-2}{2(x^2+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proof explanation: finding the coefficient of $(r+1)$th term in the expansion of $\left(1-6x\right)^{-\frac{1}{2}}$? Here is the answer of the math..
$\displaystyle\left(1-6x\right)^{-\frac{1}{2}}$
$\displaystyle=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)....\left(-\frac{1}{2}-r+1_{ }\right)}{r!}\left(-... | In line 4, $(n+1/2) = (2n+1)/2$ for all n running from 0 to r-1 and in denominator there become 2's r times which give $2^r$ in denominator. In next line they have multiplied $1.2...r$ and they distribute 2's for each n running from 1 to r which becomes $2.4....2r$ ($2^r$ is in numerator too). In next line they take b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
recurrence relation, all terms of the sequence positive Let $a_1=a$, $a_2=\frac{1}{a}-a$, $a_{n+1}=\frac{n}{a_n}-a_n-a_{n-1}$ for $n=2,3,4,...$.
Find all $a$ such that $(a_n)$ is a sequence of positive reals.
My attempt was to look at $a_3=\frac{3a^2-1}{a-a^3}$, $a_4=\frac{8a^3-4a}{3a^4-4a^2+1}$ and a few more, $a_1>... | Some observations...Posting it as answer since it is too long as comment. With $a_1 = a, a_2 = \frac{1}{a} - a$ and $a_n(a_{n+1} + a_n + a_{n-1}) = n$, we get
$$S = a_2a_3 + (a_3 + a_4)^2+(a_4 + a_5)^2 + ... + (a_{n-1} + a_n)^2 -
(a_4^2 + a_5^2 + ...a_{n-1}^2) + a_na_{n+1}$$, where $S = \frac{n(n+1)}{2} - 3$.
So, $(a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 3
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A positive real number $x$ with the property $x^3=3$ is irrational. I have the following problems:
1) There exists a positive real number $x$ such that $x^3=3$.
2) A positive real number $x$ with the property $x^3=3$ is irrational.
My Idea for 1) would be (there might be a few mistakes here):
Let $S = \{ x \geq 0 | ... | To prove it's irrational, proceed just like in the proof that $\sqrt{2}$ is irrational. Assume there are integers, in lowest terms, such that $\frac{a^3}{b^3} = 3$. So, $a^3 = 3b^3$. Show that $3$ must divide both $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $ \frac{1}{r}+\frac{1}{s}=a; \frac{1}{r}\times\frac{1}{s}=b; a+b=r; a\times b=s$, find $a$. (Brazilian Math Olympics, 2016)
Given: $\{a,b,r,s\}\subset \mathbb R$, $a>0$,
$\frac{1}{r}$, $\frac{1}{s}$ are roots for $x^2-ax+b=0$, and
$a$,$~b$ are roots for $x^2-rx+s$.
Find: the numeric value of $a$.
This is qu... | $
\left\{
\begin{array}{l}
\frac{1}{r}+\frac{1}{s}=a \\
\frac{1}{r}\cdot\frac{1}{s}=b\\
a+b=r\\
a b=s
\end{array}
\right.
$
$
\left\{
\begin{array}{l}
\frac{1}{a+b}+\frac{1}{a b}=a \\
\frac{1}{(a+b) (a b)}=b
\end{array}
\right.
$
$$a^5+a^4-2 a^3-2 a^2-2 a-1=0\to \left(a^2-a-1\right) \left(a^3+2 a^2+a+1\right)=0$$
... | {
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Number of Partitions of $n$ No Part Appears Exactly Once Equals to $n$ Partitioned into 0,2 ,3 or 4 (Mod 6 ) I want to prove that the number of partitions of $n$ in which no part appears exactly once is equal to the numbers of partitions of $n$ into parts that are congruent to one of $0,2,3,$ or $4$ mod $6$.
My approa... | Let
\begin{align}
&A(n):= \text{no. of partitions of $n$ in which no part appears exactly once,}
\\&B(n):= \text{no. of partitions of $n$ into parts that are congruent to one of 0, 2, 3, or 4 mod 6.}
\end{align}
Now, for some indeterminate $q$, we have
\begin{align}
\sum_{n=0}^\infty A(n)q^n&=(1+q^2+q^3+\cdots)(1+q^4+q... | {
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How to find the Laplace inversion of $\frac{p}{p^4+4}$? How to calculate $$\mathscr{L}^{-1}\left\{\frac{p}{p^4+4}\right\}$$ where $\mathscr{L}$ is the Laplace transform operator? I thing need to apply some partial fraction first but am unable to work it out. Any idea or further help will be very good to me.
| $p^4+4$ has 4 roots: $p=\pm 1 \pm i$, so it can be written as: $(p-1-i)(p-1+i)(p+1-i)(p+1+i)=(p^2-2p+2)(p+2p+2)$, so:
$$\frac{p}{p^4+4}=\frac{p}{(p^2-2p+2)(p+2p+2)}=\frac{Ap+B}{p^2-2p+2}+\frac{Cp+D}{p^2+2p+2}$$
The solution is $A=0, B=\frac{1}{4}, C=0, D=-\frac{1}{4}$ (I can show more detail of the computation if you n... | {
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General formula for the partial sums I m having trouble figuring out how to find the general formula for partial sums of the following two series.
*
*$\sum_{i=2}^n \frac{1}{i^2-1} = \frac{3}{4}-\frac{1}{2n}-\frac{1}{2(n+1)}$
*$\lim_{n\to\infty} \frac{4n^2 -n^3}{10+2n^3} = -\frac{1}{2} $
Both are in convergence an... | Make it telescopic. For the first one, notice that:$$\frac{1}{i^2-1}=\frac{1}{2}(\frac{1}{i-1}-\frac{1}{i}+\frac{1}{i}-\frac{1}{i+1})$$
This is obtained by noting that $i^2-1$ in the denominator can be factored into two linear terms, namely $i+1$ and $i-1$. Then we assume we have a factorization in the form of:
$$\fra... | {
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uncomplicated basis for $\mathbb Q(d)$ over $\mathbb Q$ where $d$ is a root of $x^4-14x^2+9$
Find an uncomplicated basis for $\mathbb Q(d)$ over $\mathbb Q$ where $d$ is a root of $x^4-14x^2+9$. Then a basis for $\mathbb Q(\sqrt{7+2\sqrt{10}})$ over $\mathbb Q$.
I thought the way to make a basis was to take the root ... | You should note that, by denesting (justified by $7^2-40=3^2$),
$$
\sqrt{7+2\sqrt{10}}=\sqrt{\frac{7+3}{2}}+\sqrt{\frac{7-3}{2}}=\sqrt{5}+\sqrt{2}
$$
Since $(\sqrt{5}+\sqrt{2})^2=7+2\sqrt{10}$ and
$$
(\sqrt{5}+\sqrt{2})^3=
5\sqrt{5}+15\sqrt{2}+6\sqrt{5}+2\sqrt{2}=11\sqrt{5}+17\sqrt{2}
$$
we see that $\{1,\sqrt{2},\sqrt... | {
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Finding the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?
Q. What is the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?
My approach : Let the matrix be denoted by $A$. Then $$A=
\left(\begin{matrix}
0 & 2 & 0 & 2 & 0 & ... | Hint: Apply your matrix to the vector
$$\begin{pmatrix}1 \\ -1 \\ 1 \\ -1 \\ 1\\-1\\1\\-1\\1\\-1\end{pmatrix}$$
| {
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Find the number of solutions of $a_0+a_1+a_2=17$ if $2\le a_0\le 5$, $3\le a_1\le 6$, $4\le a_2\le 7$.
Find the number of solutions of $a_0+a_1+a_2=17$ if $2\le a_0\le 5$,
$3\le a_1\le 6$, $4\le a_2\le 7$.
This is an exmaple given by my professor, and his solution is:
The number of solutions to this equation sat... | generating function.
This is another method that you may be interested.
$2\le a_0\le 5$, $3\le a_1\le 6$, $4\le a_2\le 7$.
$0\le a_0-2\le 5-2$, $0\le a_1-3\le 6-3$, $0\le a_2-4\le 7-4$.
$0\le b_0\le 3$, $0\le b_1\le 3$, $0\le b_2\le 3$.
$ a_0-2 = b_0, a_1-3 = b_1, a_2-4 = b_1$
$ a_0 =b_0+2, a_1 = b_1+3, a_2 = b_... | {
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solution using synthetic geometry I managed to solve this problem only using complex numbers but I'd like to solve it using synthetic geometry and I can't.
Can someone help me to solve this problem using synthetic geometry?
Let $ABC$ an acute triangle with $AB > AC$ . Let $O$ its circumcenter and let $D$ the midpoint... |
Considering two cases where only one of the two conditions is satisfied,
first let $ABC$ be an acute triangle but with $AB=AC$, as in the figure above. Since $MD$ and $AO$ are collinear, they are parallel in that they do not intersect.
Next, suppose $AB>AC$, but that the angle at $C$ is right. Then since $AD$ is the ... | {
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Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
Let $\angle BAC =90 ^{\circ} AB=15 ,CD=10 ,AD=5$ Then $OA=?$
| $A = (0,0)\\
B = (15,0)\\
C = (0,15)
D = (0,5)\\
O = (x,y)\\
(x,y)\cdot(x-15,y) = 0\\
x^2 + y^2 = 15x\\
(x,y-5)\cdot(x,y-15) = 0\\
x^2 + y^2 - 20y + 75 = 0\\
20y = 15 x + 75\\
y= \frac 34 x + \frac {15}{4}\\
x^2 + (\frac 34 x + \frac {15}{4})^2 = 15x\\
16x^2 + (3 x + 15)^2 = 240x\\
25x^2 + (90-240) x + 225 = 0\\
x^2 -6... | {
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"source": "stackexchange",
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Find the real solutions for the system: $x^2+4xy-2y^2=5(x+y)$, $ 5x^2-xy-y^2=7(x+y)$
Find all real solutions for the system:
$$\left\{
\begin{array}{l}
x^2+4xy-2y^2=5(x+y)\\
5x^2-xy-y^2=7(x+y)\\
\end{array}
\right.
$$
This is from a math olympiad training book. No answer provided.
It is easy to spot that $(x,... | Note that $$\begin{align*}5\times\text{Eq.(2)}-7\times\text{Eq.(1)} \equiv 18x^2-33xy+9y^2=0 \\\equiv (y-3x)(3y-2x)=0\\\implies y=\frac{2x}{3} \text{ and } y=3x\end{align*}$$
Substituting $y=3x$ into $\text{Eq.(1)}$, we have, $$x^2+12x^2-2(9x^2)=5(4x) \implies x^2+4x=0 \\\implies \boxed{(x=0, y=0) \text{ and } (x=-4, y... | {
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How to find matrix exponential $e^A$ I have the matrix $$A =\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}$$
and I have to find $e^A$
I've found two complex-conjugate eigenvalues $\lambda_{1,2} = \pm i$
so substracting $\lambda_1 = i$ from the matrix's diagonal I got:
$$A_1 = \begin{pmatrix} -i & 1 \\-1 & i \end{pmatri... | Like here For any $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^{n}.$ Employing the Identification $$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{... | {
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Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$
My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+... | we have to prove that $$a^2+b^2\geq \frac{15}{2}+\frac{ab}{4}$$
if $$\frac{15}{2}+\frac{ab}{4}<0$$ then is nothing to prove, in the other case we have
$$a^4+b^4+2a^2b^2\geq \frac{225}{4}+\frac{a^2b^2}{16}+\frac{15}{4}ab$$ for $$a^4+b^4$$ we Substitute $60-a^2b^2$ and we have to prove
$$60-a^2b^2+2a^2b^2\geq \frac{225}{... | {
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Evaluate $\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$. Evaluate following in terms of Gamma function:
$$\int_0^a \sqrt{\frac{x^3}{a^3-x^3}} dx$$
I don't know how to proceed. So, please tell the intuition behind the solution.
| Assuming $a>0$,
$$\int_{0}^{a}\sqrt{\frac{x^3}{a^3-x^3}}\,dx \stackrel{x \mapsto az}{=} a\int_{0}^{1}z^{3/2}(1-z^3)^{-1/2}\,dz \stackrel{z \mapsto u^{1/3}}{=}\frac{a}{3}\int_{0}^{1}u^{-1/6}(1-u)^{-1/2}\,du $$
equals
$$\tfrac{a}{3}\,B\left(\tfrac{5}{6},\tfrac{1}{2}\right)=a \cdot\frac{\Gamma\left(\tfrac{5}{6}\right)\sqr... | {
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Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 ... | If $f(m)=5^m+2\cdot3^m-3,$
Let use eliminate either of $5^n$ or $3^n$
Method$\#1:$
$$f(n+1)-5f(n)=3^n(6-10)-(3-3\cdot5)=4(3-3^n)$$
Now as $3^n$ is odd, for integer $n\ge0,3-3^n$ is even
$\implies f(n+1)-5f(n)$ is divisible by $8$
$\implies8|f(n+1)\iff8|f(n)$ as $8\nmid5$
Now establish the base case $f(0)$
Method$\#2:$
... | {
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Finding: $\lim\limits_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$
I'm running into problems with this limit:
$$\lim_{x\to -\infty} \frac{6x^2+5\cos{\pi x}}{\sqrt{x^4+\sin{5\pi x}}}$$
I've tried using l'Hospitals rule, however we will alway keep the $\cos(\pi x)$ expression, as well for $\sin(5... | Hint: The functions sine and cosine are bounded. Therefore, even though they can oscillate wildly at infinity, they can be controlled.
This implies that in limit operations, we have the following rules of thumb that are correct if used appropriately:
$$1.\lim_{x\to \infty}(0 \times \sin(x)) = 0$$
$$2.\lim_{x\to \infty... | {
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Why is it the case that if $a$ is a primitive root of $x^p=1$, then $\frac{x^p-1}{x-1}=(x-a^2)(x-a^4)...(x-a^{2(p-1)})=1+x+x^2+...+x^{p-1}$? If $p$ is an odd prime. Why is it the case that if $a$ is a primitive root of $x^p=1$, then $\frac{x^p-1}{x-1}=(x-a^2)(x-a^4)...(x-a^{2(p-1)})=1+x+x^2+...+x^{p-1}$? I can see why ... | Note that as $ a^p= 1$. So in the factorisation after $ p-1/2$ terms you have $ a^{p+1} ,a^{p+3} .... a^{2p-2} $ which are $ a^3,a^5,...a^{p-2} $ and thus the expression on R.H.S turns out to be $$ \prod_{i=1}^{p-1} ( x -a^i) $$
which then gives you the factorisation that comes from fact that it is a primitve root
| {
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Sequence : $a_{n+1}=2a_n-a_{n-1}+2$ Let $c$ be a positive integer. The sequence $a_1, a_2, \ldots$ is defined by $a_1=1, a_2=c$ and
$a_{n+1}=2a_n-a_{n-1}+2$ for all $n \geq 2$.
Prove that for each $n \in \mathbb{N}$ there exists $k \in \mathbb{N}$ such that $a_na_{n+1} = a_k$.
My attempt :
Trying with small numbers,... | @Ian, the solutions to the quadratic are:
$-cn-n^2+3n$ and $cn+n^2-c-3n+4$ which are both integers. The second solution gives, for $n=1,2,\ldots$, the assignment $k(n)= 2, c+2, 2c+4$, which are all positive indices.
| {
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Wrong Wolfram Alpha result for $\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}$?
I'm trying to solve this limit:
$$
\lim_{(x,y)\to(0,0)}\frac{xy^4}{x^4+x^2+y^4}
$$
Here's my attempt:
$$0 \le |\frac{xy^4}{x^4+x^2+y^4} - 0| = \frac{|x|y^4}{x^4+x^2+y^4},$$ and since $x^4+x^2 \ge0$ then $\frac{y^4}{x^4+x^2+y^4} \le 1$ s... | You can also use polar coordinate by letting $x = r\cos \theta$ and $y = r\sin\theta$ and the limit becomes $$\lim_{r\to 0} \frac{r^5 \cos \theta \sin^4\theta}{r^4\cos^4\theta + r^2\cos^2\theta + r^4\sin^4\theta} = \lim_{r\to 0} r^3\left(\frac{\cos\theta\sin^4\theta}{r^2\cos^4\theta+\cos^2\theta + r^2\sin^4\theta}\righ... | {
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Simplify $\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$ I know that the result of this expression is 16 but how do I get to that result?
$$\frac{7^{ \log_{5} 15 }+3^{2+\log_{5}7}}{7^{\log_{5}3}}$$
| We simplify the sum in two parts. First notice that $\displaystyle{\frac{7^{\log_5 15}}{7^{\log_5 5}} = 7^{\log_5 15 - \log_5 3} = 7^{\log_5 5} = 7}$. Next rewrite $7^{\log_5 3} = 3^{\log_3 7 \log_5 3}$, so the second part becomes $3^{2 + \log_5 7 - \log_3 7 \log_5 3}$, but $\log_3 7 \log_5 3 = \log_5 3^{\log_3 7} = \l... | {
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Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \... | Note that $\frac{1-2x}{2x+3} \geq 0$ implies $1-2x \geq 0$ only if $2x+3>0$.
You will get $1-2x\leq 0$ for $2x+3<0$, because what you're doing basically is multiplying $\frac{1-2x}{2x+3} \geq 0$ by a negative number.
| {
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How many even numbers of four distinct digits greater than 5000 are possible How many even numbers of four distinct digits greater than 5000 are possible? Please help me
The only thousand digit that are possible 5,6,7,8,9.
The only hundred digit that are possible are 1, 2, 3, 4, 5, 6, 7, 8, 9
The only ten digit that ... | THe first digit can be $5,6,7,8,9$. Those are four posibilities.
The second digit can be any of the ten $0,1,2..., 9$ (for some inexplicable reason you didn't include $0$) but the second digit must be different from the first. So there are $9$ options.
The third digit must be different from the first two so there are... | {
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Find an angle of a triangle on a larger triangle which cut through its midpoint
In triangle $\triangle BAC$ with $\angle ABC = 30\deg$. $D$ is the midpoint of $BC$. We join $A$ and $D$ and $\angle CDA = 45 \deg$. Find $\angle BAC$.
On applying Sine rule,
$$\frac{2x}{\sin {(15+\theta)}}=\frac{AC}{\sin 30}$$
and also
... | Your reasoning looks good to me. Using your second equation, $$AC=\frac{x}{\sqrt{2}\sin{\theta}}$$ Now substituting $AC$ in the first equation, $$\frac{2x}{\sin{(15+\theta)}}=\frac{2x}{\sqrt{2}\sin{\theta}}$$ or $$\sin{(15+\theta)}=\sqrt{2}\sin{\theta}$$
Using trig identity, $$\cos15\sin\theta+\sin15\cos\theta=\sqrt{2}... | {
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Find all $\tau$ in $S_4$ such that $\tau (12)(34) \tau^{-1} = (12)(34)$ I know there should be 8 but I have only found these 6 so far, so I am missing 2:
$(1)$
$(1 2)$
$(3 4)$
$(12)(34)$
$(13)(24)$
$(14)(23)$
I know that if $\sigma$ is a $k$-cycle and $\tau\sigma\tau^{-1} = \sigma$ then $\tau$ is a power of $\sigma$. ... | In extended notation, the permutations $\tau\in S^4$ such that
$$\tau(1,2)(3,4)\tau^{-1} = (\tau(1),\tau(2))(\tau(3),\tau(4))=(1,2)(3,4)$$ clearly are
$$ \left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 3 & 4\end{smallmatrix}\right),\left(\begin{smallmatrix}1 & 2 & 3 & 4 \\ 1 & 2 & 4 & 3\end{smallmatrix}\right),\left... | {
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How to prove non negativeness of $f(x)=1-\cos x-\frac{x^2}{20}$ I'm trying to prove that $f(x)=1-\cos x -\frac{x^2}{20}$, defined on $[-\pi, \pi]$, is a non negative function.
How do I prove that $f(x)\ge 0 $ for all $x \in [-\pi, \pi]$?
Hints?
| HINT
Since
$$1-\frac{x^2}{2}\le \cos(x)\le1-\frac{x^2}{2}+\frac{x^4}{24}$$
Thus
$$f(x)=1-\cos x -\frac{x^2}{20}\geq 1-1+\frac{x^2}{2}-\frac{x^4}{24}-\frac{x^2}{20}=\frac{9x^2}{20}-\frac{x^4}{24}\geq 0 \quad x\in[-\pi,\pi]$$
Indeed
$$\frac{9x^2}{20}-\frac{x^4}{24}=x^2\left(\frac{9}{20}-\frac{x^2}{24}\right)\geq0 \iff x^... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Solving a Matrix Equation with Maple I am attempting to use Maple to solve a matrix equation of the form $aB^2 + bB + cI = B^{-2}$, where $B$ is a $3 \times 3$ matrix, and I is the $3 \times 3$ identity matrix (i.e. to find the values a, b and c which satisfy the equation).
My idea was to use a piece of code like
A:=M... | using new letters, there is the characteristic polynomial for $B.$ As $B$ is 3 by 3 and invertible, we have $r \neq 0$ in
$$ B^3 + p B^2 + q B + r I = 0. $$
When needed, this gives us the useful $$ B^3 = -p B^2 - q B - r I . $$
Next, we get
$$ B^4 = -p B^3 - q B^2 - r B = -p (-p B^2 - q B - r I ) - q B^2 - r B ,... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
using a base 3 decimal to express as a base 10 fraction using geometric series
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.
So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:
A quick way to get the answer by u... | If you are not sure on how to compute $(0.2121\ldots)_{3}$, we have: $$\begin{align} E = (0.\color{red}{2}\color{green}{1}\color{red}{2}\color{green}{1}\ldots)_3 = \frac{1}{3}\times \color{red}{2} + \frac{1}{3^2}\times \color{green}{1} + \frac{1}{3^3}\times \color{red}{2} + \frac{1}{3^4}\times \color{green}{1} + \ldots... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $x^3 +y^3 + z^3 =57$ How can we solve $x^3 + y^3 + z^3 =57$ efficiently in a shorter way. $x$ $y$ and $z$ are integers. Given that modulus of $x$ $y$ and $z$ is less than or equal to five. We can of course do by hit and trial but what is the method of solving such questions. I actually stumbled upon this equation... | Since $57$ is odd, we need $x+y+z$ odd.
Since $3\mid 57$ and $n\equiv n^3\bmod 3$, we need $3\mid x+y+z$.
Given that $-5\le x,y,z\le5$, it's relatively quick to eliminate $x+y+z=\pm 9$ and identify that we need $x+y+z=\pm 3$.
Then taking $x\ge y\ge z$ we must have $x\ge 3$ initially and then after considering a couple ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Property of Medians and Cirumcircle Let $ABC$ be a non-isosceles triangle. Medians of $\triangle ABC$ intersect the circumcircle in points $L,M,N$. If $L$ lies on the median of $BC$ and $LM=LN$, then prove that $2a^2=b^2+c^2$.
My Attempt:
Let $G$ be the centroid of $\triangle ABC$ and $D$ be the mid-point of $BC$.
Sin... | Let $\measuredangle LAC=\alpha_1$, $\measuredangle LAB=\alpha_2$, $\measuredangle MBA=\beta_1$, $\measuredangle MBC=\beta_2$, $\measuredangle NCB=\gamma_1$ and $\measuredangle NCA=\gamma_2$.
Thus,
$$\gamma_1+\alpha_2=\measuredangle NBM+\measuredangle BML=\measuredangle NML=\measuredangle MNL=\measuredangle MNC+\measure... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Two fair coins are tossed until both turn up heads A penny and a dime are tossed together until both turn up heads, after which no more tosses are made. Find the expected number of times the penny comes up heads.
What I've tried:
Let $X$ and $Y$ be the number of times the penny and dime come up heads, respectively. The... | Imagine a clock that "ticks" only when the penny comes up heads. We need he expected number of ticks $T$ until the dime comes up heads. This follows a geometric distribution with success probability $\dfrac{1}{2}$. So $\mathbb{E}[T] = 2$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Why is my alternate method of calculating scalar products not working? The exercise is such: Given that $|\vec{a}| = 3$, $|\vec{b}| = 2$ and $\varphi = 60^{\circ}$ (the angle between vectors $\vec{a}$ and $\vec{b}$), calcluate scalar product $(\vec{a}+2\vec{b}) \cdot (2\vec{a} - \vec{b})$.
My initial thought was to sol... | Note $$2\vec {b}\cdot 2\vec {b}=4b^2$$
Compare this with what you have in your first approach when calculating
$$
|\vec {a}+2\vec {b}|
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Show that $f(x,y,z)=6x^2+4y^2+2z^2+4xz-4yz \geq 0$ for all $x, y, z$ except for $x=y=z=0$ When I try to factor the quadratic form, I end up with
$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$
which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should the... | $$6x^2+4y^2+2z^2+4xz−4yz=(4x^2+z^2+4xz)+(4y^2+z^2−4yz)+2x^2=(2x+z)^2+(2y-z)^2+2x^2\ge 0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Factoring $x^7+3x^6+9x^5+27x^4+81x^3+243x^2+729x+2187$
Question: How would you factor$$P(x)=x^7+3x^6+9x^5+27x^4+81x^3+243x^2+729x+2187$$
I thought for a while and realized that the coefficients are in powers of $3$, so $x=-3$ is a factor. Taking that factor out, we see that the septic is equal to$$P=(x+3)(x^2+9)(x^4+... | $$P(X)=x^7+3x^6+3^2x^5+3^3x^4+3^4x^3+3^5x^2+3^6x+3^7=\frac{x^8-3^8}{x-3}\\=\frac{(x^4-3^4)(x^4+3^4)}{x-3}=\frac{(x^2-3^2)(x^2+3^2)(x^4+3^4)}{x-3}=\frac{(x-3)(x+3)(x^2+3^2)(x^4+3^4)}{x-3}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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How can I find the inverse of this infinite triangular matrix? I want to find the inverse of the following matrix
\begin{bmatrix}
1&0&0&0&0&\cdots&0&0&\cdots\\
0&1&0&0&0&\cdots&0&0&\cdots\\
\binom{2}{0}&0&1&0&0&\cdots&0&0&\cdots\\
\binom{4}{1}&\binom{2}{0}&0&1&0&\cdots&0&0&\cdots\\
\binom{6}{2}&\binom{4}{1}&\binom{2}{0... | Hmm, I'm not sure where your problem is.
Let's consider $L$ as the empty-matrix with all entries in the first principal subdiagonal set to $1$.
Then your matrix $M$ can be written as evaluation of the power series
$$ f(x)= x^0 + 0 \cdot x + \sum_{k=2}^\infty \binom{2k-2}{k-2}x^k \tag {1.1}$$
writing
$$ M= f(L) ... | {
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"source": "stackexchange",
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"answer_count": 2,
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Is 2018 special because of these properties? I discovered that: $$2018=(6^2)^2+(5^2)^2+(3^2)^2+(2^2)^2$$
We also have: $$13^2+43^2=2018$$
And we have:
$$2018=44^2+9^2+1^2$$
I somehow tend to believe that there could be a finite number of these numbers that are sum of two squares, three squares and four fourth powers.
... | Sum of 4 distinct fourth powers:
$$2018 =2^4 + 3^4 + 5^4 + 6^4$$
as well as sum of 12 consecutive squares:
$$2018 = 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2 + 13^2 + 14^2 + 15^2 + 16^2 + 17^2 + 18^2$$
and "smallest number equal to the product of two primes which is also equal to the sum of 33 distinct primes": http://oei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Evaluate lim$_{n\rightarrow\infty}$ $\frac{1-2+3-4+5-...............+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}$
Evaluate $$\lim_{n\rightarrow\infty}\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}.$$
My Approach
Let $A_{n}=1+3+5+\cdots\left(2n-1\right)$ and $B_{n}=2+4+6+\cdots\left(2n\r... | One may write
$$
1-2+3-4+5-\cdots+\left(-2n\right)=\underbrace{-\left(2-1\right)}_{\color{red}{-1}}\:\underbrace{-\left(4-3\right)}_{\color{red}{-1}}-\cdots\underbrace{-\left(2n-(2n-1)\right)}_{\color{red}{-1}}=\color{red}{-n}
$$ giving
$$
\frac{1-2+3-4+5-\cdots+\left(-2n\right)}{\sqrt{n^{2}+1}+\sqrt{n^{2}-1}}=\frac{-n... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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A Question About A Calculation Of A Determinant.
Calculate $\det(A)$.
$$A =\begin{bmatrix}
a&b&c&d\\
-b&a&-d&c\\
-c&d&a&-b\\
-d&-c&b&a\\
\end{bmatrix}.$$
This is an answer on a book:
$$A A^T = (a^2+b^2+c^2+d^2) I.$$
$$\det(A) = \det(A^T).$$
$$\det(A)^2 = (a^2+b^2+c^2+d^2)^4.$$
The coefficient of $a^4$ in $\det(A)$ is... | The book tells you that $A A^\top = x I$ with $x = (a^2+b^2+c^2+ d^2)$. It follows that
$$\det(A A^\top)= \det(x I) = x^4$$
But one has
$$\det(A A^\top) = \det(A)\det(A^\top) = (\det(A))^2 $$
Hence $\det(A)=\pm x^2$. The coefficient of $a^4$ shows that it is $+x^2$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Parametric equation with a trigonometric function in exponent I'm having trouble thinking of any solution or idea to solve this math problem. Any help will be appreciated.
$(a^2 - 1)*2^{-\sin^2 x} = a^2 - 4a + 3 $,
$a = ?$
The equation should have real solutions.
| For $a=1$ the equation is satisfied for every $x$. Let's look for solutions different from $1$, so we can divide both sides by $a-1$, getting
$$
(a+1)2^{-{\sin^2x}}=a-3
$$
so
$$
2^{-{\sin^2x}}=\frac{a-3}{a+1} \tag{*}
$$
(note that $a=-1$ would lead to a contradiction).
Since $-1\le-{\sin^2x}\le0$, we have $1/2\le 2^{-{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Roots of $6z^5+5z^4+4z^3+3z^2+2z+1$ By All the zeroes of $p(z)$ lie inside the unit disk, I know there all roots are inside the unit disk. How can I remove the boundary. Wolfram alpha tells me there all there are no solutions on the unit circle. Furthermore, how can I prove the e.g. all roots are inside the circle with... | Multiply the given polynomial by $(1-z)^2$ to get $f(z)=1-7z^6+6z^7$. This is the characteristic polynomial of the matrix
$$P=\begin{pmatrix}0&0&0&0&0&0&-\tfrac{1}{6}\\
1&0&0&0&0&0&0\\
0&1&0&0&0&0&0\\
0&0&1&0&0&0&0\\
0&0&0&1&0&0&0\\
0&0&0&0&1&0&0\\
0&0&0&0&0&1&\tfrac{7}{6}\\\end{pmatrix}$$
Let $M$ be the maximum modulu... | {
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"source": "stackexchange",
"question_score": "1",
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Seems Simple, forgetting some fundamentals - $\frac{1.6}{3.01} = \frac{x}{1000+x}$ I am having trouble remembering/finding simple source to review a few fundamentals.
I know I need to factor and try getting x alone. Though not able to recreate answer. Seems it would have multiple roots. Would be great to see the prope... | It's much easier than you think.
$\frac ab = \frac cd \iff ad = bc; b\ne 0; d\ne 0$.
So
$\frac{1.6}{3.01} = \frac{x}{1000+x}\implies$
$(1000 + x)\frac{1.6}{3.01} = (1000 + x) \frac{x}{1000+x}\implies$
$(1000+x)\frac{1.6}{3.01}= x \implies$
$(1000+x)\frac{1.6}{3.01}*3.01 = x*3.01 \implies$
$1.6(1000 + x) = 3.01 x$
Just... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$ I'm trying to solve this problem.
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$
I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.
I factored out the $2\cos x$ and rewrote... | Rewrite as following
$$\begin{align}
f(x) & = 8 + 6 \sin(x ) \cos(x ) + 8\cos^2(x) - 4 + 4 \\
&= 12 + 3 \sin(2x) + 4 \cos(2x) \\
&= 12 + 5 \left(\frac{3}{5} \sin(2x) + \frac{4}{5} \cos(2x)\right) \\
&= 12 + 5 \sin(2x + \arctan{\tfrac{4}{3}})
\end{align}$$
Now its easy since $\sin(...)$ always lies in $[-1,1]$, max/m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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How did they get this equation comparing three ratios? I was reading from an old maths textbook. It was giving some examples on how to solve ratios. I stumbled upon this example and felt perplexed after reading only part of it.
We're given this equation.
$$\frac{x}{l(mb+nc-la)} = \frac{y}{m(nc+la-mb)} = \frac{z}{n(la +... | The trick is that
$$\frac{a}{b}=\frac{c}{d} \implies \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$$
$$A=\frac{\frac{x}{l}}{mb + nc - la} = \frac{\frac{y}{m}}{nc + la - mb} = \frac{\frac{z}{n}}{la + mb -nc}$$
Use each two terms
$$A = \frac{\frac{x}{l} + \frac{y}{m}}{2nc} = \frac{\frac{x}{l} + \frac{z}{n}}{2mb} = \frac{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Solve $y^{(4)}-2y^{(3)}+2y'-y=xe^x.$ Solve $y^{(4)}-2y^{(3)}+2y'-y=xe^x.$
The characteristic equation is $(r-1)^3(r+1)\Rightarrow y_h=(C_1+C_2x+C_3x^2)e^x+C_4e^{-x}.$
The problem is the particular equation. Why doesn't it work with the ansatz
$$y=(ax+b)e^x?$$
I get
\begin{array}{lcl}
y & = & e^x(ax+b) \\
y' ... | The particular solution is $$y*=\frac{1}{(D-1)^3(D+1)}xe^x=e^x\frac{1}{D^3(D+2)}x=e^x\frac{1}{2D^3(1+\frac{D}{2})}x=e^x\frac{1}{2D^3}(1-\frac{D}{2}+...)x$$
$$=e^x\frac{1}{2D^3}(x-\frac{1}{2})=e^x\frac{1}{2}\int{\int{\int{(x-\frac{1}{2})dx}dx}dx}=\frac{e^x}{2}(\frac{x^4}{24}-\frac{x^3}{12})=e^x(\frac{x^4}{48}-\frac{x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Geometric proof for $\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$ Are there geometric proofs for the identitity
$$\tan{(3x)}=\tan{(x)}\tan{\left(\frac{\pi}{3}-x\right)}\tan{\left(\frac{\pi}{3}+x\right)}$$
My try:
Thank in advances.
|
I have another simple proof.
Assume $\angle AFD = \frac{\pi}{3}$, $\angle FAD = 4x$, and $\angle FDA = 4y$. Obviously, $y = \frac{\pi}{6}-x$, $\textit{i.e.}$, $x+y = \frac{\pi}{6}$.
$AQ$ bisects $\angle FAD$, and $AC$ bisects $\angle FAQ$.
Also, $DP$ bisects $\angle FDA$, and $DB$ bisects $\angle FDP$.
And $R = AQ ... | {
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Product of length of segments in Ellipse.
If the normal at any point $P$ on the Ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2} {b^2}=1$ meets the axis in $G$ and $g$ respectively, then find $PG\cdot Pg$, in terms of $a$ and $b$.
I tried considering the parametric point as $(a\cos\theta,b\sin\theta)$ on the ellipse, then cons... | Let $P(u,v)$.
Thus, $$\frac{xu}{a^2}+\frac{yv}{b^2}=1$$ is an equation of the tangent to ellipse.
Thus the slope of the normal it's $\frac{va^2}{b^2u}$ and the equation of the normal it's
$$y-v=\frac{va^2}{b^2u}(x-u),$$ which gives
$$g\left(0,v\left(1-\frac{a^2}{b^2}\right)\right)$$ and
$$G\left(\left(1-\frac{b^2}{a^2}... | {
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"source": "stackexchange",
"question_score": "1",
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When are these eigenvalues non-negative? I'm trying to find a pair of real numbers $(a,b)$ which ensure that some matrix is strictly positive semi-definite.The eigenvalues of this matrix are $$\lambda=1 + a \pm \sqrt{(c+b)^2+2(x'-ax)^2}$$ and $$\lambda=1 - a \pm \sqrt{(c-b)^2+2(x'-ax)^2}.$$
I therefore need one of the... | I will use the notations $X=\sqrt{2}x$ and $X^\prime=\sqrt{2}x^\prime$ as it makes every expression simpler.
I will prove the following:
If $c^2 +{X^\prime}^2\leq 1$, then:
*
*Pick any $b$ in the interval $\left[0, \sqrt{1-{X^\prime}^2}-c\right]$
*If $X\neq 1$ put $a=\dfrac{-\left(1+XX^\prime\right) +\sq... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving a formula for distributing $n$ objects into $r$ non-empty boxes / formula for number of onto functions Distribution of $n$ distinct objects into $r$ different boxes if empty boxes are not allowed or in each box at least one object is put is:
$$r^n - \dbinom{r}{1}(r-1)^n+ \dbinom{r}{2}(r-2)^n- \dbinom{r}{3}(r-... | $\def\peq{\mathrel{\phantom{=}}{}}$For each $1 \leqslant k \leqslant r$, denote by $A_k$ the set of scenarios in which the $k$-th box is empty. The quantity to be found is$$
|\overline{A_1} \cap \cdots \cap \overline{A_r}| = |S| - |A_1 \cup \cdots \cup A_r|,
$$
where $S$ is the set of all scenarios.
Now, by inclusion a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Can't find error at completing the square I am desperatly looking for the mistake I did when completing the square.
I have a function $f(x)=-4.905x^2+5x+6$
Nothing special. So when I was trying to find the peak of the curve I ran into a problem and couldn't figure out why this happens, since I have repeated the task a... | It seems that you are confusing finding the solutions of $ax^2+bx+c=0$ where $a\not=0$ with finding the vertex of the parabola $y=ax^2+bx+c$.
In order to find the solutions, we have
$$\begin{align}ax^2+bx+c=0&\implies x^2+\frac bax=-\frac ca\\\\&\implies x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\\\\&\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things... | We have $$(x+y+z)(x^2-xy+y^2+z^2)=x^3+y^3+z^3+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$x^2-xy+y^2+z^2=1+xz^2+yz^2+x^2z-xyz+y^2z$$ so $$-xy=xz(x+z)+yz(y+z)-xyz$$ giving $$xyz=xy+xz-xyz+yz-xyz\implies 3xyz=xy+xz+yz$$ Now $$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\implies xy=xz+yz=0$$ Hence $$\boxed{3xyz=0\implies xyz=0.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
} |
Find two different latin squares of order $5$ In the following question I am trying to find two different Latin squares of order $5$
Latin Square #1
\begin{array}
& &1 &2 &3 &4 &5 \\
&5 &1 &2 &3 &4 \\
&4 &5 &1 &2 &3 \\
&3 &4 &5 &1 &2 \\
&2 &3 &4 &5 &1 \\
\end{array}
This Latin square was made by filling in the first ... | They are different, in the sense that they're unequal.
They are the same, in the sense that they are isotopic, meaning that there is a way to permute the rows, columns, and symbols of one to obtain the other. In this case, we can permute the rows of the first one to obtain the second one.
The Cayley table of $\mathbb{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many strings of digits of length $n$ contains a substring of $k$ identical consecutive digits? Let $f(n, k)$ be the number of strings of digits of length $n$ that contain a substring of $k$ identical consecutive digits. What would be a recursive or closed-form way to compute $f(n, k)$?
For example, $f(n, 2)=10^n -... |
We consider the alphabet $V=\{0,1,\ldots,9\}$. We are looking for the number $g(n,k)$ of strings of length $n$ having runs at most length $k-1$. The wanted number is $$f(n,k)=10^n-g(n,k)$$
Strings with no consecutive equal characters at all are called Smirnov or Carlitz words. See example III.24 Smirnov words from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving in different ways that $n^{n-1}-1$ is divisible by $(n-1)^2$. I have this amazing exercise which explicitly says prove in at least six different way that
$n^{n-1}-1$ is divisible by $(n-1)^2$ where $n$ is an integer.
So far I have only prove it as follows: By geometric sum we have
$$\sum_{k=0}^{n-2}n^k=\frac{n... | $$
\begin{align}
n^{n-1}-1 &= (n-1) ( n^{n-2} \color{red}{- 1} + n^{n-3} \color{red}{- 1} +\ldots + n^2 \color{red}{- 1} + n \color{red}{- 1} + 1 \color{red}{+n-2}) \\
&= (n-1) \big(\,(n-1)(\ldots) + (n-1)(\ldots)+ \ldots + (n-1)(n+1)+ (n-1)+ (n-1))\,\big)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Find the range of $x$ for the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$
Question: Find the range of $x$ for the convergence of the series$$\sum_{n=1}^{\infty} \frac{1}{\left(-3\right)^{n+2}} \frac{\left(4x-12\right)^{n}}{n^{2}+1}$$
MY App... | $$\sum_{n=1}^{\infty}\frac{1}{\left(-3\right)^{n+2}}\frac{\left(4x-12\right)^{n}}{n^{2}+1}=\sum_{n=1}^{\infty}\frac{1}{\left(9\right)}\frac{\left(4-\frac{4x}{3}\right)^{n}}{n^{2}+1}$$
according to Cauchy root test , we get
$$|4-\frac{4x}{3}|\leqslant 1$$
so
$$\frac{9}{4}\leqslant x\leqslant \frac{15}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How can one prove that this polynomial is non-negative? How one can prove the following inequality?
$$58x^{10}-42x^9+11x^8+42x^7+53x^6-160x^5+118x^4+22x^3-56x^2-20x+74\geq 0$$
I plotted the graph on Wolfram Alpha and found that the inequality seems to hold. I was unable to represent the polynomial as a sum of squares.... | For $x<0$ it's obvious.
But for $x\geq0$ we obtain:
$$58x^{10}-42x^9+11x^8+42x^7+53x^6-160x^5+118x^4+22x^3-56x^2-20x+74=$$
$$=(x^3-x^2-x+1)(58x^7+16x^6+85x^5+85x^4+207x^3+47x^2)+$$
$$+287x^4-138x^3-103x^2-20x+74>0,$$
where $$287x^4-138x^3-103x^2-20x+74=$$
$$=(16x^2-4x-5)^2+(31x^4-10x^3+x^2)+(40x^2-60x+49)>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
System of equations involving complex numbers I'm getting confused to figure this work out. The only thing that came into my head was using AM-GM inequality, but i just get stuck. Here's the problem:
Let $a,b,c,$ be the complex numbers such that $abc=1$. and
\begin{cases}
a^{20}+b^{20} + c^{20} &= \frac{1}{a^{20}} + \... | Hint. If $abc=1$ then, for any integer $n$, the equation
$$a^{n}+b^{n} + c^{n}= \frac{1}{a^{n}} + \frac{1}{b^{n}} + \frac{1}{c^{n}}$$
is equivalent to
$$(1-a^n)(1-b^n)(1-c^n)=0.$$
Moreover note that $\gcd(20,17)=1$, $\gcd(2017,17)=1$, and $\gcd(2017,20)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Computing the definite integral $\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$
Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$
This is what I did:
$u = x^2 + a^2 $
$du/dx = 2x$
$du = 2xdx$
$1/2 du = x dx$
$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{... | You calculated the antiderivative correctly.
$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$$ from $0$ to $a$ is $$\frac{1}{3}\cdot \left(a^2+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\cdot \left(0^2+a^2\right)^{\frac{3}{2}}=\\\frac{1}{3}\cdot((2a^2)^\frac{3}{2}-(a^2)^\frac{3}{2})=\frac{1}{3}\cdot(2^\frac{3}{2}a^3-a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
The number of ways in which one can choose three distinct numbers from the set so that the product of the chosen numbers is divisible by $9$
Consider the set $A=\{1,2,3,...,30\}$
. The number of ways in which one can choose three
distinct numbers from $A$ so that the product of the chosen numbers is divisible b... | We solve the opposite: let $Y$ denote the number of ways one can choose three distinct numbers for which their multiple not divisible by 9 i.e. divisible by 3 and not 9 or not divisible by 3 at all. First note that there are $10$ multiples of 3 among those $3$ are also a multiple of 9 in $\{1,2,...,30\}$ so the number ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Partial Fraction problem solution deviates from the Rule Question:
Compute $\displaystyle \int\frac{x^2+1}{(x^2+2)(x+1)} \, dx$
My Approach:
As per my knowledge this integral can be divided in partial Fraction of form $\dfrac{Ax+B}{x^2+px+q}$ and then do the following as per to integrate it.
Solution:
Taking $\dfrac{x^... | you must write $$\frac{x^2+1}{(x^2+2)(x+1)}=\frac{Ax+B}{x^2+2}+\frac{C}{x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) The Arithmetic Mean (A.M) between two numbers exceeds their Geometric Mean (G.M.) by $2$ and the GM exceeds the Harmonic Mean (H.M) by $1.6$. Find the numbers.
My Attempt:
Let the numbers be $a$ and $b$. Then,
$$A.M=\dfrac {a+b}{2}$$
$$G.... | $$AM=GM+2$$
$$GM=HM+1.6$$
Since $$GM^2=AM\cdot HM,$$
$$GM^2=(GM+2)(GM-1.6)$$
$$GM^2=GM^2+0.4GM-3.2$$
$$GM=8$$
$$AM=10$$
$$\sqrt{ab}=8, \frac{a+b}{2}=10$$
$$ab=64, a+b = 20$$
The numbers are $16$ and $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Subgroups of General Linear Group I have a question about the order of the subgroups of $GL_2(\mathbb{R})$ generated by the following matrix:
$\begin{pmatrix}
1&-1 \\
-1&0
\end{pmatrix}$
So I computed several powers of this matrix and realized that the order of the subgroup generated by this matrix is infinity. Ho... | Hint: When you diagonalize this matrix, you get
$
S\begin{pmatrix} \frac{-1 + \sqrt{5}}{2} & 0 \\ 0 & \frac{1-\sqrt{5}}{2} \end{pmatrix}S^{-1}$
for some ugly matrices $S$ and $S^{-1}$. These numbers are larger in norm than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Equation System with 4 real variables I need to solve the next equation system:
Find all real numbers $a,b,c,d$ such that:
$$
\left\{
\begin{array}{c}
a+b+c+d=20 \\
ab+ac+ad+bc+bd+cd=150 \\
\end{array}
\right.
$$
I tried something like this:
$b+c+d=20-a$
And i put the second equation like this
$a(b+c+d) + bc+bd+cd=... | By $(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$, we find $$ a^2+b^2+c^2+d^2 = 100 $$
Also by Cauchy-Schwarz inequality, $(a+b+c+d)^2 \leq (1^2 + 1^2 + 1^2 +1^2)(a^2+b^2+c^2+d^2)$ and therefore we yields: $$ 400 \leq 400 $$
Then, the equality condition occurs if and only if $a=b=c=d=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\gcd(a,b)=2$, $3a+b^2 =3388$, $a>0$, $b>49$ I have this problem and I can't do it.
$$\begin{cases}
\gcd(a,b)=2 \\
3a+b^2 =3388 \\
a>0,b>49
\end{cases}
$$
I've tried writing $a=2a'$ and $b=2b'$, but then I have $3a'+2b'^2=1694$ and I don´t know what to do
| Assuming $a$ needs to be positive, the equation $3a + b^2 = 3388$ shows that $b \leq 58$. Since $gcd(b,a) =2 $ implies that $b$ is divisible by $2$, so the only options are $b = 50,52,54,56,58$. Note that $4$ divides $3388$, so if $4$ divides $b$, then $4$ divides $3a$ and hence $4$ divides $a$. This is not acceptab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
equations of triangle's sides, given equations of two bisectors and one point of triangle? The equations two angles bisectors are $x-3y-6=0$ and $x+y-2=0$. We also know that one point of the triangle is $A(2,-4)$. Clearly, this point doesn't satisfy those two equations, and by finding the intersection of the bisectors ... | Let $D$ be the incentre. $D=(3,-1)$.
The slope of $AD$ is $3$.
Let $\angle BAC=2a$, $\angle ABC=2b$ and $\angle ACB=2c$. Then $a+b+c=\frac{\pi}{2}$.
Note that $\angle BDC=\pi-b-c=\frac{\pi}{2}+a$.
If $\theta$ is the acute angle between the two given angle bisectors, then
$$\tan\theta=\frac{\frac{1}{3}-(-1)}{1+(-1)(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Formatting Functions To Avoid Loss Of Significant
Rewrite the following to avoid loss of significant
*
*$\ln(x+1)-\ln(x)$ where $x>>1$
*$\cos^2(x)-\sin^2(x)$ where $x\approx \frac{\pi}{4}$
*$\sqrt{x^2+1}-x$ where $x>>1$
*$\sqrt{\frac{1+\cos x}{2}}$
*
*Using taylor expansion we get $$x-\frac{x^2}{2}+\frac{x^... | You shouldn't use Taylor series for these.
*
*$\log(x+1)-\log(x) = \log(1+\frac{1}{x})$
*$\cos^2 x - \sin^2 x = \cos 2x$
*$\sqrt{x^2+1} - x = (\sqrt{x^2+1} - x) \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$
and simplify.
*$1+\cos x = 2\cos^2 \frac{x}{2}$. This identity should help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What's the expected number of times I have to roll two die until they both sum $7$? Here is my guess: the probability of summing $7$ on two rolls is $\frac 16$. This means if I repeat the experiment many times I'll roll $7$ one sixth of them (approximately). Hence,
$$N \cdot \bigg(\cfrac 16\bigg) \cdot 7 = 7$$
where ... | The probability of doing it after one roll is $1/6$, in two is $5/6 \times 1/6$, in three $(5/6)^2 1/6$ and so on ... we get
\begin{eqnarray*}
E(7)=1 \times \frac{1}{6} + 2 \times \frac{5}{6} \times\frac{1}{6} + 3 \times \left(\frac{5}{6}\right)^2 \times\frac{1}{6}+\cdots = \frac{1}{6} \sum_{i=1}^{\infty} i \left( \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Help proving $\forall x \geq 2: \cos(2 \pi/x) \leq 1 - 1/x^2$ I want to prove the following inequality:
$$\forall x \geq 2: \cos(2 \pi/x) \leq 1 - 1/x^2
$$
I believe it holds: numerical evidence.
I tried to prove it using taylor series, but it didn't work so well. Any ideas?
| Note that by Taylor's series $\forall x$
$$\cos x\le 1-\frac{x^2}2+\frac{x^4}{24}$$
thus
$$\cos \frac{2\pi}{x}\le 1-\frac{2\pi^2}{x^2}+\frac{16\pi^4}{24x^4}\le 1-\frac{1}{x^2}$$
indeed
$$1-\frac{2\pi^2}{x^2}+\frac{2\pi^4}{3x^4}\le 1-\frac{1}{x^2}\iff\frac{1}{x^2}-\frac{2\pi^2}{x^2}+\frac{2\pi^4}{3x^4}\le0$$
$$\iff 3x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Determine the whether the sequence $a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$ is convergent or divergent. Consider the recursively defined sequence $a_n = 1$
$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$$
Is the sequence convergent?
This is my attempt:
First, we prove that the sequence is positive, and monotoni... | In your proof of $a_{n+1}>a_n$ you used that $a_n>0$, but it is not proven.
I like the following reasoning.
$$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3-\sqrt5}{2}}-\frac{1}{a_n}>0$$ by induction because $a_1=1>\frac{3-\sqrt5}{2}.$
$$a_{n+1}-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Want to understand how a fraction is simplified The fraction is used to determine the sum of a telescopic series.
$$\sum_{k=1}^\infty\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$
This is the solved fraction.
$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}} = \frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1)^2k-k^2(k+1)} = \frac{(k+1)\sqrt{k}-k\sqrt... | *
*First equal: multiply and divide by $(k+1)\sqrt k-k\sqrt{k+1}$
*Second equal: Expand the denominator
*Third equal: cancel obvious terms in the denominator and write $k^2+k=k(k+1)$.
*Fourth equal: distribute along the minus in the numerator, and cancel the obvious factors $(k+1)$ in the first summand, and $k$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Expected number and probability of a series of 1s in a bit string. Suppose the sequence 1001111011011 has a total of $4$ "blocks" of one, because number of contiguous sequences made of ones (1, 1111,11 and 11) count to 4. Given a random bit array of length $N$, what's the expected value of the "number of 1 blocks"? Rel... | Using $z$ for zeros and $w$ for ones and $u$ for runs we get the
generating function
$$(1+z+z^2+\cdots)
\\ \times \sum_{q\ge 0} u^{q} (w+w^2+\cdots)^q (z+z^2+\cdots)^q
\\ \times (1+uw+uw^2+\cdots)$$
which is
$$\frac{1}{1-z}
\\ \times \sum_{q\ge 0} u^{q} \frac{w^q}{(1-w)^q} \frac{z^q}{(1-z)^q}
\\ \times \left(1+u\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How many trees on $\{1,2,3,4,5,6,7\}$ have a vertex of degree 2?
How many trees on $\{1,2,3,4,5,6,7\}$ have a vertex of degree 2 ?
Attempt -
It feels like an inclusion exclusion problem (using kailey's code) , let's define $|A_i| \Rightarrow $ vertex $i$ is of degree $1$.
$|Ai\, \cup A_j$| - vertices $i,j$ are of deg... | I think that your formula is correct. It gives $16380$. Below there is an alternative approach with the same final result.
For a labeled tree with $7$ vertices we have that
$$\sum_{v \in \{1,2,3,4,5,6,7\}} \deg v = 2(7-1)=12.$$
So if a tree has NOT vertices of degree $2$ and it has $m$ leaves of degree $1$ then
$$12\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Closed algebraic form of $\cos(\frac{\pi}{7})$ We all know that $\cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}$, $\cos(\frac{\pi}{4})=\frac{\sqrt 2}{2}$ and $\cos(\frac{\pi}{3})=\frac{1}{2}$. One can also prove that $\cos(\frac{\pi}{5})=\frac{\sqrt 5+1}{4}$. But it seems that $\cos(\frac{\pi}{7})$ cannot be put in a closed fo... | For any $n\geq 3$, $\cos\left(\frac{2\pi}{n}\right)$ is al algebraic number over $\mathbb{Q}$ with degree $\frac{\varphi(n)}{2}$, since its minimal polynomial can be computed from $\Phi_n(x)$ in a straightforward way. Our case is given by $n=14$, where the minimal polynomial of $\cos\left(\frac{\pi}{7}\right)$ is given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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When is $(p^2-1)/8$ even? I am trying to find for what values of p $\frac{p^2-1}{8}$ will be even number $\frac{p^2-1}{8}=2m\implies (p-1)(p+1)=8(2m)$ then can I write $p\equiv 1\pmod8$ or $p\equiv -1\pmod8$ ?
Also trying to find for what values of p it will be an odd number
$\frac{p^2-1}{8}=(2m+1)\\\implies ... | I think you want to look at congruences modulo $16$, not $8$. After all, $\frac{p^2-1}8=2m$ if and only if $p^2-1=16m$, that is $p^2\equiv1\pmod{16}$. Similarly $\frac{p^2-1}8$ is odd if and only if $p^2-1=8+16m$, that is $p^2\equiv9\pmod{16}$.
By writing down all eight possibilities, you see that $n^2\equiv1\pmod{16}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$ I have a question regarding the sum of this series:
$$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$
My approach:
I found that this sum is equal to:
$$\sum_{n=3}^{2008}\frac{... | What we want is $$\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!}$$
\begin{align}
\dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \dfrac{n+2}{n! \left( 1 + (n+1) + (n+1)(n+2) \right)}\\
& = \dfrac{n+2}{n! \left( n^2 + 4n + 4 \right)}\\
& = \dfrac1{n! \left( n+2 \right)}\\
& = \dfrac{n+1}{(n+2)!}\\
& = \dfrac{n+2}{(n+2)!} - \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Inverse trigonometic functions Question:
Prove that $\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)=0$
MyProblem
We can just use the formula of $\tan^{-1} A +\tan^{-1} B$ but I think it would be a waste of time. Is there any other shorter and simpler method to solve it.
| The hint:
Prove that
$$\left(\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)\right)'=0$$ and check $A=\frac{\pi}{2}.$
Indeed, $$\left(\arctan\left(\frac{1}{2}\tan2x\right)+\arctan\cot{x}+\arctan\cot^3x\right)'=$$
$$=\frac{\frac{1}{\cos^22x}}{1+\frac{1}{4}\tan^22x}+\frac{-\frac{1}{\sin^2x}}{1+\cot^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove by induction that $1+2^{2^n}+2^{2^{n+1}}$ is divisible by $7$ As said in the title you have to show by induction that 1+$2^{2^n}+2^{2^{n+1}}$ is divisible by 7. So you start with n=0, that gives 1+2+4=7. So the start is shown.
Let 1+$2^{2^n}+2^{2^{n+1}}$ be divisble by 7 for a n.
I've tried several attempts but I... | Starting with $n = 0$, we have $1+2^{2^0}+2^{2^{0+1}} = 7$, which is clearly divisible by $7$. Now suppose inductively that $n \ge 1$ and argument holds for all $n$. Then for $n+1$, we have
$$1+2^{2^{n+1}}+2^{2^{n+2}} = 1+2^{2\cdot2^n}+2^{2\cdot2^{n+1}} = 1+(2^{2^n})^2+(2^{2^{n+1}})^2$$
Now and and subtract $(2^{2^n})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Three summations
Given
$$S_n=\sum^{n-1}_{i=0}\sum^{i-1}_{j=0}\sum^{j-1}_{k=0} (i+j+k), $$
there are positive integers $A$ and $B$ such that
$$\frac{1}{S_3}+\frac{1}{S_4}+\frac{1}{S_5}+\dots=A-\frac{2\pi^2}{B}$$
Find $A+B$.
MyApproach:
I need to solve the innermost summation first and then proceed to the last... | Expanding the triple summation:
$$\begin{align} S_n=&\sum^{n-1}_{i=0}\sum^{i-1}_{j=0}\sum^{j-1}_{k=0} (i+j+k)=\\ &\sum^{n-1}_{i=0}\sum^{i-1}_{j=0} \left(ij+j^2+\frac{j(j-1)}{2}\right)=\\ &\sum^{n-1}_{i=0}\left(i\cdot \frac{i(i-1)}{2}+\frac{3}{2}\cdot \frac{(i-1)i(2i-1)}{6}-\frac{1}{2}\cdot \frac{(j-1)j}{2}\right)=\\
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Given $x = \sqrt[4]{x^{3}+6x^{2}}$, determine whether the sum of all possible roots of $x$ is equal to $1$. Given $$x = \sqrt[4]{x^{3}+6x^{2}}$$
Quantity $A$: Sum of all possible roots of $x$
Quantity $B$: $1$
My solution: I've taken fourth power on both sides of the equation to get $$x^{4} = x^{3} + 6x^{2}.$$
Rearrang... | $-2$ is not a solution of $x = \sqrt[4]{x^{3}+6x^{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to calculate the integral $\int\frac{1}{\sqrt{(x^2+8)^3}}dx$? I need to solve something like this
$$\int\frac{1}{\sqrt{(x^2+8)^3}}dx$$
Wolfram alpha says the solution is $$\frac{x}{8\sqrt{x^2+8}} + c$$
The problem is that the integrand is obtained by the quotient rule:
$$\bigg(\frac{g(x)}{h(x)}\bigg)'=\frac{g'(x)h(... | I don't know of any general formula for the quotient rule.
But for an integral of the form
$$\int \frac{dx}{\left(ax^n+b\right)^{k}}$$
You can try the following substitutions:
*
*If $n=1$, put $u=ax+b$. ($n=0$ is trivial.)
*If $n=2$, put $x=\left(\frac{b}{a}\right)^\frac{1}{n} tan u$.
*If $n\ge 3$, put $x=\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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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.