Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find all integer $n$ so that both $n+2008$ divides $n+2008^2$ and $n+2009$ divides $n+2009^2$ Find all integer $n$ so that both $n+2008$ divides $n+2008^2$ and $n+2009$ divides $n+2009^2$
Obviously $n$=$0$ is a solution, but are there any others? Thank you
| $$a|b \iff \gcd(a, b) = a$$
$$\gcd(a, b) = \gcd(a, b-a)$$
So $\gcd(n + 2008, n+2008^2) = \gcd(n+2008, 2008^2 - 2008) = n + 2008$, similarly for 2009.
$$\begin{cases} n + 2008 ~|~ 2008 \cdot 2007 \\
n + 2009 ~|~ 2009 \cdot 2008\end{cases}$$
So $n=0$ is the only nonnegative solution. For negative solutions, brute force ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the inverse of a polynomial in a quotient ring Consider $\mathbb Z_5[x]/I$ with $I$ as ideal generated by $b=x^3+3x+2$. If $(x+2) + I$ is element of $\mathbb Z_5[x]/I$ that has an inverse. Find the inverse of $(x+2) + I$.
I stuck to get the inverse because the gcd of that is not $1$.
| Hint: $$(x+2)(ax^2+bx+c) + I =ax^3+(2a+b)x^2+(2b+c)x+2c+I\\ = (2a+b)x^2 +(2b+c-3a)x+2(c-a) + I$$
Now solve
\begin{align}
2a+b &=0\\
2b+c-3a &= 0\\
2(c-a) &= 1
\end{align}
in $\Bbb Z_5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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$1!+2!+3!+4!+5!+...+n!$ is a square of an integer At what n is the number $1!+2!+3!+4!+...+n!$ a square of another integer number?
All in all I have found $2$ values: $1$ and $3$. I think they are the only ones, but the only thing left is to prove it. How?
| Let $1!+2!+3!+\cdots \cdots +n! = y^2$
$\star$ if $n=1\;,$ then $1=y^2$ (True)
$\star$ if $n=2\;,$ then $1+2=3=y^2$ (False)
$\star$ if $n=3,$ then $1+2+6=9=y^2$ (True)
$\star$ if $n=4,$ then $1+2+6+24=33=y^2$(False)
for $n\geq 5,$ then L .H .S end with $3$ and we now that square of any integer does not have last digit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Using the binomial theorem I'd like some help with proving the next equation:
$$\sqrt{1+x}=\sum_{0}^{\infty }\frac{(-1)^{n-1}}{2^{2n-1}\cdot n}\binom{2n-2}{n-1}\cdot x^{n}$$
| Let's rewrite the expression in terms of the Gamma function
$$
\sqrt {1 + x} = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n - 1} }}
{{2^{\,2n - 1} n}}\left( \begin{gathered}
2(n - 1) \\
n - 1 \\
\end{gathered} \right)x^{\,n} } = \sum\limits_{0 \leqslant n} {\frac{{\left( { - 1} \right)^{\,n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Difficult definite integration $\int_{-2}^{0}\frac{x^2 + x - 5}{(x-1)^2}e^x\,\mathrm dx$ I found an integral in a contest that seems very difficult to compute. The answer is $-2$, however, I do not know how to arrive at this answer.
$$\int_{-2}^{0}\frac{x^2 + x - 5}{(x-1)^2}e^x\,\mathrm dx$$
At first I tried to make th... | Integration by part we get
\begin{align*}
\int\frac{x^2 + x - 5}{(x-1)^2}e^x \, \mathrm{d}x&=-\frac{x^2 + x - 5}{x-1}e^x +\int e^{x}\left ( x+4 \right )\, \mathrm{d}x \\
&=-\frac{x^2 + x - 5}{x-1}e^x+e^{x}\left ( x+3 \right )\\
&=\frac{x+2}{x-1}e^{x}
\end{align*}
Then you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Number of pairs of prime numbers $p$ and $q$ such that $p$ divides $q-1$ and $q$ divides $p^3-1$
Determine the number of pairs of prime numbers $p$ and $q$, both of which are less than $2017,$ such that $p$ divides $q-1$ and $q$ divides $p^3-1$.
If $p = 2$, then $q = 7$ and if $q = 2$, there are no solutions. Now sup... | If $p\mid q-1$ then certainly $p<q$. Then from $q\mid p^3-1=(p-1)(p^2+p+1)$, we conclude $q\mid p^2+p+1$.
As you wrote, $p=2$ leads to $q$ odd (wow!) and $q\mid p^-1=7$, so $q=7$.
Next, $p=3$ leads to $q\mid 9+3+1=13$, so $q=13$, which gives a solution.
Then, $p=5$ leads to $q\mid 25+5+1=31$, so $q=31$, which gives a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Generating function for counting compositions of $n$ I was given the next question: I was asked to find the generating function for the number of divisions of $n$ (a given natural number) with exactly 3 elements.
For example: if $n$ equals to 5 then 3,1,1 are a set of one of the options (3 + 1 + 1 = 5)
I tried thinki... | We derive the wanted generating function for the number of partitions of $n$ with exactly three parts by starting with a seemingly different generating function.
The generating function for the number of partitions which consist of zero or more of $1,2,$ and $3$ is
\begin{align*}
&(1+x+x^2+x^3+\cdots)(1+x^2+x^4+x^6+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
if $x$ is rational and $x^2$ is natural, prove that $x$ is integer If $x$ is rational: There exists $\frac{a}{b}$ such that $a$, $b$ are integers.
If $x^2$ is natural:
$\left(\frac{a}{b}\right)^2$ is natural => $\frac{a^2}{b^2}$ is natural
Then $a^2$ divides $b^2$ => $a$ divides $b$
If $a$ divides $b$ and $a$, $b$ are ... | Let $x^2 = n \in \mathbb{N}\,$, then $x$ is a root of $x^2-n=0$. Since $x^2-n$ is a monic polynomial, any rational root must be an integer by the rational root theorem, so $x$ is an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show the recursive sequence is increasing
How do I show that the recursive sequence
$$a_n = a_{\lfloor n/2 \rfloor} + a_{\lceil n/2 \rceil} +3n+1, \quad
n\geq 2, \phantom{x} a_1 = 3$$
is an increasing sequence?
1. attempt:
If I can show that $a_{n+1}-a_n>0$, I would be able to show it is increasing.
\begin{align*}
a... | For an even index we have
$$a_{2n}=2a_n+6n+1$$
and for an odd one,
$$a_{2n+1}=a_n+a_{n+1}+6n+4$$
so if we assume (using induction) that $a_{n+1}\ge a_n$ then
$$a_{2n+2}=2a_{n+1}+3(2n+2)+1=2a_{n+1}+6n+7>a_n+a_{n+1}+6n+4=a_{2n+1}$$
and
$$a_{2n+1}=a_n+a_{n+1}+6n+4>2a_n+6n+1=a_{2n}$$
Remark: Perhaps one of two more base ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\leq2$ Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{3\sqrt[3]{abc}}{2(a+b+c)}\leq2$$
I tried C-S, BW, uvw and more, but without any success.
| Here is my proof. Dedicated to dear Dr. Sonnhard Graubner.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$,$abc=w^3$ and $u=xw$.
Hence, $x\geq1$ and we need to prove that $$\sum_{cyc}\frac{a}{a+b}+\frac{w}{2u}\leq2$$ or
$$\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)\leq\frac{1}{2}-\frac{w}{2u}$$ or
$$(u-w)(9uv^2-w^3)\geq u\sum_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding a closed form expression for $\sum_{i=1}^{n-1}\csc{\frac{i\pi}{n}}$ Consider
$\displaystyle{S =
\sum_{k = 1}^{n - 1}\csc\left(k\,{\pi \over n}\right) =
\frac{1}{\sin\left(\pi/n\right)} +
\frac{1}{\sin\left(2\pi/n\right)} +
\frac{1}{\sin\left(3\pi/n\right)} + \cdots +
\frac{1}{\sin\left(\left[n - 1\right]\pi/n\r... | Let we assume for first that $n$ is an odd number, $n=2N+1$. In such a case the given sum is
$$ \sum_{k=1}^{2N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)} = 2\sum_{k=1}^{N}\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}=\frac{4N+2}{\pi}H_N+2\sum_{k=1}^{N}\left[\frac{1}{\sin\left(\frac{\pi k}{2N+1}\right)}-\frac{1}{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I ha... |
I thought it would be instructive to present a way forward that relies on elementary analysis, including knowledge of the sum $\sum_{k=1}^\infty \frac1{k^2}=\pi^2/6$, partial fraction expansion, and telescoping series. It is to that end we proceed.
The Harmonic number, $H_{n+2}$, can be written as
$$H_{n+2}=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Double summation equation involving a parameter While trying to solve a statistics probability density function question, I have reached a point where I unable to proceed with solving for the value of $c$ in the equation:
$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{c}{(i+j-1)(i+j)(i+j+1)} = 1$, where $c>0$
I went throug... | The starting point is partial fraction decomposition :
$$\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2(i+j-1)}-\frac{1}{i+j}+\frac{1}{2(i+j+1)}$$
Now, let's compute a partial sum (indexed by $j$) :
$$\sum_{j=1}^q\frac{1}{(i+j-1)(i+j)(i+j+1)}=\frac{1}{2}\sum_{j=1}^q\left(\frac{1}{i+j-1}-\frac{1}{i+j}\right)-\frac{1}{2}\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $f(x)=(x-2)q(x)-8$ for polynomial $q$, and $x+2$ is a factor of $f(x)$, find the remainder when $f(x)$ is divided by $x^2-4$
Given:
$f(x)=(x-2)q(x) -8$, where $q(x)$ is a polynomial
$(x+2)$ is a factor of $f(x)$
Find the remainder when $f(x)$ is divided by $(x^2-4)$.
I know the answer is $-2x-4$, but I do not know... | We know that when $f(x)$ is divided by $x-2$ the remainder is $-8$, so
$$f(x) = (x-2)q(x) - 8$$
for some polynomial $q(x)$.
We also know that $x+2$ is a factor of of $f(x)$. This means that when we plug in $x=-2$ to $f(x)$ we should get $0$. So
$$0 = (-2 -2)q(-2) - 8$$
$$8 = -4q(-2)$$
$$-2 = q(-2)$$
This last result ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding the matrix associated with a linear map
Find the matrix associated with the linear map $f:R^2 \rightarrow R^2$ defined by $f(x,y)=(3x-y,y-x)$ with respect to the ordered basis ${(1,0),(1,1)}$
Let the matrix be $A$ and let $f(x)=AX$ where
$$X=\begin{bmatrix}
x \\
y
\end{bmatrix}$$
and
$$A=\begin{bmatrix}
... | Regarding to the standard basis $e_i$ the map has the matrix
$$
A =
\begin{pmatrix}
3 & -1 \\
-1 & 1
\end{pmatrix}
$$
because
$$
f(x)
= x A
= (x_1 e_1 + x_2 e_2) A
= x_1 e_1 A + x_2 e_2 A
= (x_1, x_2)
\begin{pmatrix}
e_1 A \\
e_2 A
\end{pmatrix}
\\
e_1 = (1,0):\quad f(1,0) = (3,-1) = e_1 A = (1,0) A = (a_{11} a_{12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Evaluating $\iint_V x^4+y^4+z^4$ with the divergence theorem
Use Gauss divergence theorem to evaluate
$$\iint_S \left(x^{4} + y^{4} + z^{4}\right)$$ over sphere S of radius $a$.
So I wrote this as
$$
\begin{align}
&a\iint_{\partial V} \Big(x^3 \hat{i}+y^3\hat{j}+y^3\hat{k}\Big)\cdot\Big(\frac{x\hat{i}+y\hat{j}+z\hat{... | The normal vector $\mathrm{n}$ at at point $(x,y,z)$ on a sphere is given by $\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$, hence it you want to write $x^4+y^4+z^4$ as the dot product $\mathrm{F}\cdot\mathrm{n}$ you have to take $\mathrm{F}$ as $a(x^3+y^3+z^3)$, then $\text{div }\mathrm{F}=3a(x^2+y^2+z^2)$ and the original inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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What is the minimum value of $x^2+y^2$ subject to $x^3+y^3+xy=1$?
What is the minimum value of $x^2+y^2$ under the constraint $x^3+y^3+xy=1$? Please do not use partial differentials (multivariable calculus) or Lagrange multipliers. You can use elementary algebra or single variable calculus.
I plotted the graph of $x^... | Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Hence, $x^3+y^3+xy=1$ gives $2u(4u^2-3v^2)+v^2=1$ or $v^2=\frac{8u^3-1}{6u-1}$.
But $u^2\geq v^2$.
Hence, $u^2\geq\frac{8u^3-1}{6u-1}$ or $\frac{2u^3+u^2-1}{6u-1}\leq0,$ which gives $\frac{1}{6}<u\leq u_1$,
where $u_1$ is a real root of the equation $2u^3+u^2-1=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\sum_{k=1}^\infty\left \lfloor{\frac{n}{p^k}}\right\rfloor \leq 2\sum_{k=1}^\infty\left \lfloor{\frac{n-1}{p^k}}\right\rfloor$ I'm trying to prove that:
$$\sum_{k=1}^\infty\left \lfloor{\frac{n}{p^k}}\right\rfloor \leq 2\sum_{k=1}^\infty\left \lfloor{\frac{n-1}{p^k}}\right\rfloor $$ for composite $n > 4$ where $... | We want to prove that
$$\sum_{k=1}^{\infty}\left(2\left\lfloor\frac{n-1}{p^k}\right\rfloor-\left\lfloor\frac{n}{p^k}\right\rfloor\right)\ge 0$$
Let $n=p^kM_k+R_k$ and $n-1=p^km_k+r_k$ where $M_k,R_k,m_k,r_k$ are integers such that $M_k\ge 0,m_k\ge 0,0\le R_k\lt p^k$ and $0\le r_k\lt p^k$.
Let us separate it into two ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If cosB=3/4, how do I find cos 2B and cos(B/2)? I got $\cos 2B$;
$\cos 2B=2 \cos^2 B - 1
=9/8 -1
=1/8$
but when I tried cos1/2B, I got: COS1/2B=cos^2*1/4B-1, then I solved it, and didn't get the answer which was in the book, why?
| $\cos 2B = 2 \cos^2 B - 1
= \frac9{8} - 1 = \frac{1}{8}$
Second answer -
$\cos B = 2 \cos^2(\frac B2) - 1$
$\frac34 = 2 \cos^2(\frac B2) - 1$
$\frac 74 = 2 \cos^2(\frac B2)$
$\frac 78 = \cos^2(\frac B2)$
$\sqrt{\frac78} = \cos (\frac B2)$
$\frac12 \sqrt{\frac72} = \cos (\frac B2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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For all triangle prove that $\sum\limits_{cyc}\frac{a}{(a+b)^2}\geq\frac{9}{4(a+b+c)}$
Let $a$, $b$ and $c$ be a sides-lengths of the triangle. Prove that:
$$\frac{a}{(a+b)^2}+\frac{b}{(b+c)^2}+\frac{c}{(c+a)^2}\geq\frac{9}{4(a+b+c)}$$
The Buffalo way kills it, but I am looking for a nice proof for this nice inequa... | Have: $\Leftrightarrow \left(\sum _{cyc}a\right)\left(\sum _{cyc}\frac{a}{\left(b+c\right)^2}\right)\ge \frac{9}{4}$
By C-S and Nesbitt : $LHS\ge \left(\frac{a}{b+c}+\frac{b}{b+c}+\frac{c}{c+a}\right)^2=\frac{9}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Sum of $\frac{z^{k+1}}{(1-z)^2}$ over the roots of $x^n+1=0$ equals $\frac{n}{2}(k-\frac{n}{2})$ The problem is to show that for every $k \in \{1,2,...,n\}$ if we sum $\frac{z^{k+1}}{(1-z)^2}$ over the roots $z$ of $x^n+1=0$, we obtain the result $\frac{n}{2}(k-\frac{n}{2})$.
One idea I had is to consider the polynomia... | We have with
$$f(z) = \frac{z^{k+1}}{(1-z)^2} \frac{nz^{n-1}}{z^n+1}
= \frac{1}{(z-1)^2} \frac{nz^{n+k}}{z^n+1} $$
and $\zeta_{n,q} = \exp(\pi i/n+ 2\pi i q/n)$
$$\sum_{x^n+1=0} \frac{x^{k+1}}{(1-x)^2}
= \sum_{q=0}^{n-1}
\mathrm{Res}_{\large z=\zeta_{n,q}} f(z).$$
This is also given by (residues sum to zero)
$$-\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2109493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove the inequality ${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $ given that $abc = 1$ Prove the inequality $${1 \over a^3 (b + c)} + {1 \over b^3 (a+c)} + {1 \over c^3 (a+b)} \ge \frac 32 $$ So, I know a proof for this, but I basically memorized it without understanding. It's
$$ {(... | By Holder and AM-GM
$$\sum\limits_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(b+c)}\geq\frac{3abc(a+b+c)(ab+ac+bc)}{3\sum\limits_{cyc}(b+c)}\geq\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$... | HINT:
Take the sine of both sides, apply the double angle formula for the sine function, and use $\cos(\arcsin(x))=\sqrt{1-x^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $P=log_{x}xy $ and $Q=log_{y}xy$, then how is $P+Q=PQ$? If $P$ and $Q$ have different bases for the log, how do we prove $P+Q=PQ$?
| Consider $P + Q$ , it is $log_{x}{xy}+log_{y}{xy}$ which is equal to
$ log_{x}{x} + log_{x}{y} + log_{y}{x} + log_{y}{y}= 2 + log_{x}{y} + log_{y}{x} $
now consider rightside $P.Q$ which is $log_{x}{xy} .log_{y}{xy} = (1 + log_{x}{y}).(1+log_{y}{x}) $
which is equal to $1 + log_{x}{y} + log_{y}{x} + log_{x}{y}.log_{y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Prove that if the sides $a,b,c$ of a triangle satisfy $a^2+b^2=kc^2$, then $k>\frac {1}{2}$.
Prove that if the sides $a,b,c$ of a triangle satisfy $a^2+b^2=kc^2$, then
$k>\frac {1}{2}$.
Source : CRUX (Page Number $1$;Question Number $74$)
Obviously, for $k=1$, a right angled triangle exists. I tried assuming $k<\fr... | By triangle inequality we can write:
$$c<a+b \to c^2<a^2+b^2+2ab$$
Multiply both sides by $k$ and use $kc^2=a^2+b^2$:
$$kc^2<k(a^2+b^2)+(2ab)k \to (1-k)(a^2+b^2)<(2ab)k$$
If $k \le 1/2$ then $1-k \ge 1/2$, so we have:
$$\frac{a^2+b^2}{2}\le (1-k)(a^2+b^2)<(2ab)k \le ab$$
what give us
$$\frac{a^2+b^2}{2}< ab \to (a-b)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$... | You don't to begin using equality!! You must to arrive in this...
$1+2+⋯+k+k+1= $(here you use the induction hipoteses) Than...We start correct:
$$1+2+⋯+k+k+1=**induction hipoteses(ih)**\frac{k(k+1)}{2}**finish(ih)**+**hereyourepeatlikeabove**k+1$$
I hope I've helped.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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System of two cubic equations: $x+y^2 = y^3$, $y+x^2=x^3$ I got stuck on this system of equations. Could you help and tell me how should I approach this problem?
\begin{align*}
x+y^2 &= y^3\\
y+x^2 &= x^3
\end{align*}
These are the solutions:
\begin{align*}
(0, 0); \\((1+\sqrt{5})/2, (1+\sqrt{5})/2); \\((1-\sqrt... | Notice that $x=0 \implies y=0$, hence $(0,0)$ is a trivial solution. We will consider the case where $x \neq 0$.
$$x=y^2(y-1)$$
$$y=x^2(x-1)$$
Suppose on the contrary that $x=1$, then from the second equation $y=0$ which contradicts the first equation $x=0$. Hence $x \neq 1$.
Suppose on the contrary that $x \in (0,1)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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Find the summation of the following sequence Please give me an idea on how to proceed as I am really stuck with this, I have not encountered this type of question before yet my friend gave this summation to me and I am stuck.
$${4\choose 1} + \frac{5\choose2}{2}+\frac{6\choose3}{4}+\dots$$
| Let $\displaystyle\binom{r+3}r\left(\dfrac12\right)^{r-1}=\binom{r+3}3\left(\dfrac12\right)^{r-1}=f(r+1)-f(r)$
where $f(m)=\left(\dfrac12\right)^m\sum_{r=0}^na_rm^r$ where $a_r$ are arbitrary constants
$\dfrac16(r+3)(r+2)(r+1)\left(\dfrac12\right)^{r-1}$
$=\left(\dfrac12\right)^{r+1}\left(a_0+a_1(r+1)+a_2(r+1)^2+a_3(r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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$a,b,c,d>0$ and $abcd=1$ prove $\sum \frac {1}{(1+a)(1+a^2)} \ge 1$ If $a,b,c,d>0$ and $abcd=1$ prove:
$$\sum \frac{1}{(1+a)(1+a^2)}\ge 1$$
Here's my solution:
I try to prove it by reverse:
$$3\ge\sum\frac{a^3+a^2+a}{(a+1)(a^2+1)}$$Then by AM-GM:
$$3\ge\sum\frac{a(1+a+a^2)}{2a(a+1)}$$$$$6\ge\sum\frac{1+a+a^2}{a+1}$$
T... | The inequality $\sum\limits_{cyc}\frac{a^2}{a+1}\leq2$ is wrong.
Try $b=c=d\rightarrow0^+$ and $a\rightarrow+\infty$.
The starting inequality we can prove also by the following two inequalities.
$$\frac{1}{(a^2+1)(a+1)}+\frac{1}{(b^2+1)(b+1)}\geq\frac{1}{1+\sqrt{a^3b^3}}$$ and
$$\frac{1}{(c^2+1)(c+1)}+\frac{1}{(d^2+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can I solve $x^2+2xy+y^2+3x-3y-18=0$ $$x^2+2xy+y^2+3x-3y-18=0$$
I don't know how to solve it.
| When a general conic equation
$$ Ax^2+Bxy+Cy^2+Dx +Ey +F=0$$
has $B\ne0$ and $A=C$ then rotating it $45^\circ$ will eliminate the $xy$ term.
This will be accomplished by the substitution
\begin{eqnarray}
x&=&\frac{X-Y}{\sqrt{2}}\\
y&=&\frac{(X+Y)}{\sqrt{2}}
\end{eqnarray}
\begin{equation}
x^2+2xy+y^2+3x-3y-18=0
\end{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ wher $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find value of $\tan(a+2b)$ and $\tan(2a+b)$ If $\sin(a+b)=1$ and $\sin(a-b)=\frac{1}{2}$ where $a \geq 0$ and $b \leq \frac{\pi}{2}$ , find the value of $\tan(a+2b)$ and $\tan(2a+b)$.
I have tried to apply the formula for $\tan(a+... | Hint:
$$\sin(a + b) = 1$$
$$\sin(a + b) = \sin \frac π2$$
$$a + b = \frac π2 = 90^\circ$$
Also,
$$\sin(a - b) = \frac 12$$
$$\sin(a - b) = \sin \frac π6$$
$$a - b = \frac π6 = 30^\circ$$
Solve equations to find $a$, $b$.
$$a = \frac π3 = 60^\circ$$
$$b = \frac π6 = 30^\circ$$
Put values of $a$ and $b$ to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate: $\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^2}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$ The problem is to evaluate the following sum:
$$\frac{1}{2^1}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\ldots-\frac{100}{2^{100}}$$
My approach was to find the common denominator ($2^{100}$), then the series becomes:
$$ \fra... | Notice that
$$\sum_{n=1}^{100}x^n=\frac{x-x^{101}}{1-x}$$
which is just a geometric series. Differentiate once to get
$$\sum_{n=1}^{100}nx^{n-1}=\frac d{dx}\frac{x-x^{101}}{1-x}$$
Multiply both sides by $x$ and set $x=-\frac12$ to get
$$\frac1{2^1}-\frac2{2^2}+\dots-\frac{100}{2^{100}}=-\frac12\left(\frac d{dx}\bigg|_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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how to solve the following recurrence? $t(n)=[4-t(n-1)]^{-1}$ I am trying to solve the following recurrence :
$T_n=\frac{1}{4-T_{n-1}}$
I tried various methods using range transformation but still can't figure it out.
| If it has a limit $L$,
$T_n=\frac{1}{4-T_{n-1}}
$
becomes
$L=\frac{1}{4-L}
$
or
$L^2-4L+1 = 0$
so that
$L
=\dfrac{4\pm\sqrt{16-4}}{2}
=2\pm\sqrt{3}
$.
Let
$L_1
=2+\sqrt{3}
$
and
$L_2
=2-\sqrt{3}
$.
If
$0 < T_{n-1} < 1$,
then
$\frac14 < T_n
< \frac13
$.
After this,
$\frac1{4-1/3}
< T_{n+1}
< \frac1{4-1/4}
$
or
$\frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many triples $(x,y,n)$ are there such that $x^n - y^n = 2^{100}$
How many triples $(x,y) \in \mathbb{N^+}^2$ and $n \gt 1$ are there such that $x^n - y^n = 2^{100}$
I dont know how to start. Any hint will be helpful.
I know the identity $x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})$.
I... | Since the question is being asked rather often and the solution in the comments seems to have a flaw I propose the following answer.
We claim: the equation
$$
a^n-b^n=2^k,\tag1
$$
where $a,b,n,k$ are positive integers ($n>1$), has exactly $\left\lfloor\frac{k-1}2\right\rfloor$ solutions $(a,b)$ for $n=2$ and no soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Is there easy way to factor polynomials for calculus? factor of $$4y^9-4y$$
comes out to be $$4y(y^4+1)(y^2+1)(y+1)(y-1)$$
How would you approach to factor?
PS: My math is very rusty.
| \begin{align*}
4y^9-4y&=4y(y^8-1)\\
&=4y((y^4)^2-1)\\
&=4y(y^4-1)(y^4+1)\\
&=4y((y^2)^2-1)(y^4+1)\\
&=4y(y^2-1)(y^2+1)(y^4+1)\\
&=4y(y-1)(y+1)(y^2+1)(y^4+1).
\end{align*}
I am simply using $a^2-b^2=(a-b)(a+b)$. For example, at the second equation we took $a=y^4,\ b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Combinatorics and limit problem For every natural number $n$ let us consider $a_n$ the greatest natural non-zero number such that:
$$\binom {a_n}{n-1} \gt \binom {a_n-1}{n}.$$
Compute $$\lim_{n \to \infty} \frac {a_n}{n}.$$
I started by using the formula for the binomial coefficient, I obtained a second degree inequati... | $$
\begin{align}
&\, \binom{a_{\small n}-1}{n}\lt\binom{a_{\small n}}{n-1} \,\Rightarrow \frac{(a_{\small n}-1)!}{(n)!\,(a_{\small n}-n-1)!}\lt\frac{(a_{\small n})!}{(n-1)!\,(a_{\small n}-n+1)!} \\[4mm]
&\, \Rightarrow 1\lt\frac{a_{\small n}\,n}{(a_{\small n}-n)\,(a_{\small n}-n+1)} \,\Rightarrow\, a_{\small n}^2-(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx$ How do i evaluate $$\int \frac{1-7\cos^2x}{\sin^7x\cos^2x}dx$$. I tried using integration by parts and here is my approach
$\int \frac{ sinx}{(1-cos^2x)^4\cos^2x} dx$ and then put $cos x=t$ and then tried to use partial fractions.I applied similar logic for the oth... | Well, we know that:
$$\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}=\csc^7\left(x\right)\left(\sec^2\left(x\right)-7\right)\tag1$$
So, for the integral we get:
$$\int\frac{1-7\cos^2\left(x\right)}{\sin^7\left(x\right)\cos^2\left(x\right)}\space\text{d}x=\int\csc^7\left(x\right)\sec^2\left(x\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2... | Step 5. onwards is not ok
$$\int_a^bf(x) dx = F(b)-F(a) $$
$$\int_a^bf(x) dx = F(b-a) $$
The first result is correct but second one is in general wrong.
EDIT1:
The constant term in $ (x^2-3 x +2) $ is missing for integration. Factorization is not beneficial,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Triple Integral $\iiint x^{2n}+y^{2n}+z^{2n}dV$ Evaluate:
$$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV $$
I have tried to convert to spherical polars and then compute the integral, but it gets really messy because of the 2n power. Any tips?
| First observation: it is symmetric in $x,y,z$, so by linearity we have
$$\iiint_{x^2+y^2+z^2 \leqslant 1} x^{2n}+y^{2n}+z^{2n} dV =3\iiint_{x^2+y^2+z^2 \leqslant 1} z^{2n} dV.$$
Choosing spherical coordinates it becomes
$$3\iiint_{x^2+y^2+z^2 \leqslant 1} (r\cos \theta)^{2n} dV$$
where $dV= r^2 \sin \theta \ \text{d}r ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Euler's Method and induction Euler's Method for series associates with a given series $\sum\limits_{j=0}^\infty(-1)^ja_j$ the transformed series $\sum\limits_{n=0}^\infty\frac{\Delta^n a_0}{2^{n+1}}$ where $\Delta^0a_j=a_j$, $\Delta^na_j=\Delta^{n-1}a_j-\Delta^{n-1}a_{j+1}$, $j=1,2,\cdots.$
I am given that $\ln 2 = 1-\... | A formula for $ \Delta ^n a_j$ is given by
\begin{eqnarray*}
\Delta ^n a_j = \frac{n! \ j!}{(n+j+1)!}
\end{eqnarray*}
We now need to use double induction ... it suffice to show that the formula $\Delta^na_j=\Delta^{n-1}a_j-\Delta^{n-1}a_{j+1}$ holds, which is easy
\begin{eqnarray*}
\Delta ^n a_j = \Delta^{n-1}a_j-\Del... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2132581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Which integration techniques I should use for $\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$
$$\int{\frac{\sqrt{x^2-3}-3\sqrt{x^2+3}}{\sqrt{x^4-9}}}dx$$
I can simplify it to:
$$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}$$
but I can't go from here.
| Hint :
$$\int{\frac{dx}{\sqrt{x^2+3}}} - 3\int{\frac{dx}{\sqrt{x^2-3}}}\\=\int{\frac{dx}{\sqrt{3}\sqrt{(\frac{x}{\sqrt3})^2+1}}} - 3\int{\frac{dx}{\sqrt{3}\sqrt{(\frac{x}{\sqrt3})^2-1}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What are the solutions to $a^2+ab+b^2$ $=$ $3^n$? What are all known integer solutions ($a, b, n$) to $a^2+ab+b^2$ $=$ $3^n$ besides ($1, 1, 1$) and ($-2, 1, 1$)? Do any others even exist?
This question comes from the identity that ($a^3±b^3$)/($a±b$) $=$ $0, 1$ $\pmod 3$. If ($a^3±b^3$)/($a±b$) $=$ $0$ $\pmod 3$, then... | Let $a=u+v$ and $b=u-v$. It follows:
$$a^2+ab+b^2=(u+v)^2+(u^2-v^2)+(u-v)^2$$
Hence
$$3u^2+v^2=3^n$$
If $n=0$, then, $u=0$ and $v=\pm 1$
If $n=1$, then $v^2=3(1-u)(1+u)$ $\implies$ $(u,v)=(\pm 1,0),(\pm 1, \pm 1 )$
If $n>1$, then $u,v$ are both divisible by 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find the Derivative of $f(x)=\frac{7}{\sqrt {x}}$ using the definition. I get that $$\frac{d}{dx}\left(7\times\dfrac{1}{\sqrt{x}}\right)=\frac{d}{dx}(7x^{.5})=\dfrac{7}{2}x^{-.5}$$ is the derivative, but I can't ever use $\lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.
If someone or anyone could go step by step and do the prob... | \begin{align*}
\dfrac{\frac{7}{\sqrt{x+h}} - \frac{7}{\sqrt{x}}}{h} &= 7\left(\dfrac{\sqrt x - \sqrt{x+h}}{h\sqrt{x(x+h)}}\right) \\
&= 7\left(\dfrac{\sqrt x - \sqrt{x+h}}{h\sqrt{x(x+h)}}\right)\left(\dfrac{\sqrt x + \sqrt{x+h}}{\sqrt x + \sqrt{x+h}}\right) \\
&= 7\left(\dfrac{x-(x+h)}{h\sqrt{x(x+h)}}\right)\left(\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove for all real numbers $a, \ b,$ and $c$, $bc + ac + ab \leq a^2 + b^2 + c^2$. I've been having some trouble with this particular problem. Could someone provide a hint of some sort? My original attempt that turned out to be useless is to start off with
$$(b+c)^2 + (a+c)^2 + (a+b)^2 = (b+c)^2 + (a+c)^2 + (a+b)^2$$
a... | $
a^2 + b^2 + c^2 \ge ab + bc + ca\\
\Longleftrightarrow \frac12(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca) \ge 0\\
\Longleftrightarrow (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ac + a^2) \ge 0 \\
\Longleftrightarrow (a-b)^2 +(b-c)^2 +(c-a)^2 \ge 0
$
which is obviously true. Giving a hint for this problem is as good a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Homogeneous D.E show that $ x\frac{dy}{dx} = y + \frac{x}{lny-lnx} $ is a homogeneous D.E.
Find the general solution in the form $ye^{y/x} = f(x) $
The answer given is $ye^{y/x} = Ax^2 $ but I am getting $ (\frac{y}{x})^{y/x} = Axe^{y/x} $
Is the given answer correct?
| Consider the given answer $\quad ye^{y/x} = Ax^2 \quad\to\quad \ln(y)+\frac{y}{x}=\ln(A)+2\ln(x) \quad$ and differentiate it :
$$\frac{dy}{y}+\frac{dy}{x}-\frac{ydx}{x^2}=2\frac{dx}{x}$$
$$\left( \frac{1}{y}+\frac{1}{x}\right)dy=\left(\frac{2}{x}+\frac{y}{x^2} \right)dx$$
$$\frac{dy}{dx}=\frac{\frac{2}{x}+\frac{y}{x^2}... | {
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"url": "https://math.stackexchange.com/questions/2140407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $47^x = 8$ and $376^y = 128$ , find $\frac{3}{x}-\frac{7}{y}$ What I know:
$x={\log_{47}8}$ and $y=\log_{376}128$
How do I do this without using a calculator?
| No logs, no calculator. :)
$$47^x=8=2^3\Longrightarrow 47=2^{3/x}\\
376^y=128=2^7\Longrightarrow 376=2^{7/y}$$
Dividing,
$$\frac {2^{3/x}} {2^{7/y}}=\frac {47}{376}=\frac 18\\
2^{3/x-7/y}=2^{-3}\\
\frac 3x-\frac 7y=\color{red}{-3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Evaluate the given limit Given a function $f : R → R$ for which $|f(x) − 3| ≤ x^2$. Find
$$\lim_{ x\to0}\frac{f(x) - \sqrt{x^2 + 9}}{x}$$
Can the function $f(x)$ be considered as $x^2 + 3$ and go about evaluating the limit using the Limit laws?
| We have $|f(x)-\sqrt{x^2 +9}| = |f(x)-3 + 3 - \sqrt{x^2 + 9}| \le |f(x)-3| + |3 - \sqrt{x^2+9}| \le x^2 + |3 - \sqrt{x^2 + 9}|$. So:
$$\left|\frac{f(x)-\sqrt{x^2+9}}{x}\right| \le |x| + \left| \frac{\sqrt{x^2+9}-3}{x}\right|$$
Now squeeze.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What values for a prime $p$ are possible if $3p+1$ is a perfect cube? There is a similar case like this already on the site, but it deals with perfect squares and is relatively easy to solve.
But what about perfect cubes? Thus $3p+1= n^3$ ?
any help?
Thanks!
| This is not the solution, I tried all four cases and didn't get anywhere.... sigh. Can someone help?
Case 1: assume p=n-1 and 3=(n^2)+n+1
Then n=p+1, so
3=(p+1)^2 + (p+1) +1
3= p^2 +2p +1 +p +1 +1
3= p^2 +3p +3
but then what?
0=p^2 +3p
0= p(p +3) so p is 0 or -3? would 3 be the answer? i thought it wou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2149048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's I tried:
$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\
\frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\
\frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\
\frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\
\frac{\ln{x^3}}{x-1} - 1-x = \\
???$$
What do I d... | Recall that
$$\ln(x^3)=3\ln(x)$$
and
$$\lim_{x\to1}\frac{\ln(x)}{x-1}=1$$
which should give you
$$\lim_{x\to1}\frac{3\ln(x)}{x-1}-1-x=3-1-1=\boxed1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Probability that $x^2+y^2$ is divisible by $5$ and $7$ Two natural numbers $x$ and $y$ are chosen at random with a uniform distribution.Then the question is to find out the probability that $x^2+y^2$ is divisible by $5$ and $7$.
I understand that divisible by $5$ and $7$ is the same thing as divisible by $35$.But I am ... | $1^2≡1, 2^2≡4, 3^2≡2, 4^2≡2, 5^2≡4, 6^2≡1 \mod 7$.
Then $x,y$ are multiple of $7$. Since about one's place $1+7^2≡1+9,4^2+7^2≡6+9, 4^2+8^2≡6+4,1^2+8^2≡1+4≡0\pmod5$, these are multiple of 5. Therefore they are 9 patterns of $(0,0)(7,14)(7,21)(14,7)(14,28)(21,7)(21,28)(28,14)(28,21) \pmod{35}$$$\dfrac9{35^2}=0.735\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Epsilon-Delta: Prove $\frac{1}{x} \rightarrow 7$ as $x \rightarrow \frac{1}{7}$
Prove that $\displaystyle\frac{1}{x} \rightarrow 7$ as $\displaystyle x \rightarrow \frac{1}{7}$.
I need to show this with an $\epsilon-\delta$ argument. Still figuring these types of proofs out though, so I could use some tips/critiques ... | The only logical error I could find is related to the definition of $\delta$. If you want to use strict inequalities, you should have $\delta<\min\left\{\frac{1}{14},\frac{\varepsilon}{98}\right\}$. Otherwise, at least one of the strict inequalities should be changed (depending on the value of $\varepsilon$).
Regardle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
... | You can even prove more: if $a,b,c$ are the sides and $\alpha,\beta,\gamma$ the respectively opposite angles, summing up the three identities of the law of cosine, we get
$$
a^2+b^2+c^2=2ab\cos\gamma+2bc\cos\alpha+2ca\cos\beta
$$
which implies the statement, as all the cosines are less than $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Keep getting wrong answer for showing $(a_n)=(\frac{2n^2-1}{n^2},\frac{1}{n})$ is $d^{(2)}$-convergent? A sequence in $\mathbb{R}^2$ is given by:
$a_n=(\frac{2n^2-1}{n^2},\frac{1}{n})$, for each $n∈\mathbb{N}$.
I must show that $(a_n)$ is $d^{(2)}$-convergent.
My textbook says that I can show $a_n$ is convergent by sho... | The sequence does converge, but not to $(0,0)$. Observe that
\begin{align*}
d^{(2)}(a_n,(2,0))&=\sqrt{\left(\frac{2n^2-1}{n^2}-2\right)^2+\left(\frac{1}{n}-0\right)^2} \\
&=\sqrt{\left(\frac{1}{n^2}\right)^2+\left(\frac{1}{n}\right)^2} \\
&=\sqrt{\frac{1+n^2}{n^4}}.
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
find $g^{-1}(x)$ in terms of $f^{-1}$ Consider $g(2x-3) = \frac{2f(x-2) + 3}{5 - f(x-2)}$ . Also $f$ and $g$ are invertible . Now find $g^{-1}(x)$ in terms of $f^{-1}$.
My try : Because in the parentheses we have $x-2$ and $2x-3$ instead of $x$ , I don't know how we can solve this problem.
| HInt:
first $$x \to x+2 \\\to
g(2(x+2)-3)=\dfrac{2f(x+2-2) + 3}{5 - f(x+2-2)}\to \\
g(2x+1)=\dfrac{2f(x) + 3}{5 - f(x)}$$
now find $f(x)$ then apply $f^{-1}$
$$y=\dfrac{2f(x) + 3}{5 - f(x)} \to \\
5y-yf(x)=2f(x)+3 \to \\f(x)(y+2)=5y-3 \\ \to \\
f(x)=\dfrac{5y-3}{y+2}$$
apply $f^{-1}$to both sides
$$f^{-1}f(x)=f^{-1}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2156683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Producing circles with $r^2=x^2+y^2$ $r^2=x^2+y^2\ +\pi $ produces graph with two circles :
$r = 2$
This graph was produced using Desmos, is this correct or a quirk of the graph software ?
Ive noticed $\forall n \epsilon N [r^2=x^2+y^2+n\pi]$ appears to produce same circle, so $r^2=x^2+y^2\ +3\pi $ produces same circl... | Your first picture is correct if you are graphing
\begin{align*}
2^2 &= x^2 + y^2\\
- \pi + 2^2 &= x^2 + y^2
\end{align*}
for red and yellow respectively. If we have the equation
$$
r^2 = x^2 + y^2 + c
$$
where $c$ is a constant, that is the same as
$$
r^2 - c = x^2 + y^2
$$
and so our new radius is $r' = \sqrt{r^2 - c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divi... | For $7$: $(n^3-1)(n^3+1)=n^6-1$ then apply Fermat's little theorem.
For $3^2$ : First, $(k-1)k(k+1)$ has to be divisible by $3$.
Second, $(n^3-1)\cdot(n^3+1)=(n-1)(n^2+n+1)\cdot (n+1)(n^2-n+1)$ which has the factor $(n^2-1)$ which is divisible by 3 by Fermat's LT.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
What operations can be performed on three variables $x$, $y$, and $z$ and produce one unique sum? I'm having a programming problem, but I feel this would be more relevant here. I have three unique variables, $x$, $y$, and $z$. What operations can I perform on these variables to produce a unique value? Are there any app... | $f(x,y,z) = x\sqrt{2} + y\sqrt{5} + z\sqrt{7}$
This function would increase by a similar order of magnitude to its inputs. Moreover it's possible to prove that for $x,y,z \in \mathbb{Z}, f(x,y,z)$ must be unique $\forall x,y,z$.
Edit: Proof:
Suppose for contradiction that $\exists x_{1},y_{1},z_{1},x_{2},y_{2},z_{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
find extremum of $y = x(x-1)^{\frac{1}{3}}$ given function:
$$y = x(x-1)^{\frac{1}{3}}$$
steps:
$$y'=\frac{(x^2-x)^{\frac{-2}{3}}2x-1}{3}$$
After simplifying:
$$ y' = \frac{2x-1}{(3x^2-3x)^{\frac{2}{3}}}$$
therefore
$$ x = \frac{1}{2}$$ $$x \ne 0$$
Am I right with calculus?
| The function is
$$
f(x)=x\sqrt[3]{1-x}
$$
not
$$
\sqrt[3]{x(1-x)}
$$
at least according to common notation. Thus the derivative is
$$
f'(x)=(1-x)^{1/3}+x\cdot\frac{1}{3}(1-x)^{-2/3}\cdot(-1)=
\frac{1}{3(1-x)^{2/3}}(3-3x-x)=\frac{3-4x}{3(1-x)^{2/3}}
$$
which vanishes at $x=3/4$ and is undefined at $x=1$.
Note that the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $CE=2AD$ I already have a proof, but if you can please give another:
Let $ABC$ be an isosceles triangle with $AB=AC$ and point $D$ be on segment $AB$. The line perpendicular to $AB$ which passes $D$ intersects $BC$ (extended) and $AC$ at $E$ and $F$ respectively. $C$ is on segment $BE$, between $B$ and $E$.... | Let $\measuredangle A=\alpha$, $AB=a$ and $AD=xa$.
Hence, $$FC=a-\frac{ax}{\cos\alpha}=\frac{a(\cos\alpha-x)}{\cos\alpha}$$ and
$$CE=BE-BC=\frac{a(1-x)}{\sin\frac{\alpha}{2}}-2a\sin\frac{\alpha}{2}=\frac{a(\cos\alpha-x)}{\sin\frac{\alpha}{2}}.$$
Thus, $$S_{\Delta{FCE}}=\frac{FC\cdot CE\sin\left(90^{\circ}+\frac{\alpha}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$5x^2−10x+7$ in completed square form? I was studying quadratic equations and practicing to solve them using the technique of completing the squares.
My answer was as follows:
$$5x^2−10x=−7$$
$$5x^2−10x+25=18$$
$$x(x-5)^2−5(x-5)=18$$
$$\boldsymbol{(x-5)^2-18=0}$$
The answer in the book is:
$$\boldsymbol{5(x-1)^2+2=... | Usually, if it's an equation, then we can divide both sides to get a quadratic term of the form $x^2$ instead of $ax^2$:
\begin{eqnarray*}
5x^2 -10x &=& -7 \\
x^2 - 2x &=& - \frac{7}{5} \\ \\
\left(x-1\right)^2-1 &=& -\frac{7}{5} \\ \\
\left(x-1\right)^2 &=& - \frac{2}{5}
\end{eqnarray*}
This has no real roots. You are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Beta function-like integral $$\int\limits_0^1 \frac{x^{1-\alpha} (1-x)^\alpha}{(1+x)^3} \, dx $$
After the substitution $z=\frac{1}{x} - 1$,
I've got this:
$$\int\limits_0^\infty \left(\frac{1}{z}-1\right)^\alpha\left(\frac{1}{z}+1\right)^{-3} dz $$
But it is still far from the Beta function.
| The substitution $u = \frac{1}{x}-1$ is actually useful because it transforms it into an integral that can be evaluated fairly easily using contour integration.
Specifically, for $-1 <a <2$,
$$\int_{0}^{1} \frac{x^{1-a}(1-x)^{a}}{(1+x)^{3}} \, dx = -\int^{0}_{\infty} \frac{\left(\frac{1}{1+u} \right)^{1-a} \left(1- \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Prove that if LCMs are equal, then the numbers are equal too. For $a,b \in \mathbb{N}$, how do I prove:
$$lcm(a,a+5)=lcm(b,b+5) \implies a=b$$
| I assumed $a$ and $b$ were positive integers.
We have $\mathrm{lcm}(a,a+5) = \frac {a(a+5)}{\gcd(a,a+5)}$, so that the given equality implies
$$\frac {a(a+5)}{\gcd(a,a+5)} = \frac {b(b+5)}{\gcd(b,b+5)}.$$
But $\gcd(x,y) = \gcd(x,y+kx)$ for all $k \in \mathbb{Z}$ so that $\gcd(a,a+5) = \gcd(a,5)$ and $\gcd(b,b+5) = \gcd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2169008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Proving: $\left(a+\frac1a\right)^2+\left(b+\frac1b\right)^2\ge\frac{25}2$
For $a>0,b>0$, $a+b=1$ prove:$$\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\ge\dfrac{25}{2}$$
My try don't do much, tough
$a+b=1\implies\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{ab}$
Expanding: $\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfra... | Using Cauchy-Schwarz Inequality we have that:
$$\left(\bigg(a+\dfrac{1}{a}\bigg)^2+\bigg(b+\dfrac{1}{b}\bigg)^2\right)(1 + 1) \ge \left( a + b + \frac 1a + \frac 1b \right)^2 \ge 5^2 = 25$$
The last inequality follows as:
$$\frac 1a + \frac 1b = \frac{b + a}{ab} = \frac 1{ab} \ge \frac{4}{(a+b)^2} = 4$$
Another way to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Evaluating surface integral. The question is:
Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$. Given $\vec{V}=x^3j+z^3k$, I want to evaluate the surface intergal
$$\iint_s\nabla\times V.\hat n dS$$
would the paramatisation be something like $x=r\cos\theta,y=r\sin\... | alternative
Green's / Stokes theorem:
$\iint_s\nabla\times V.\hat n dS = \oint_c V\cdot dr_2 - \oint_c V \cdot dr_1$
$r_1 = (\cos\theta, \sin\theta, 0), r_2 = (2\cos\theta, 2\sin\theta, 0)$
$\int_0^{2pi} 2^4\cos^4\theta\ d\theta - \int_0^{2pi} \cos^4\theta\ d\theta\\
15\int cos^4 \theta\ d\theta\\
\frac {15}{2}\int (\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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On the Diophantine equation $2(a^3+b^3+c^3)=abc(abc+6)$
I am sure that $(3, 2, 1)$ is the only natural triple satisfying the following Diophantine equation $$2(a^3+b^3+c^3)=abc(abc+6)\quad\text{for}\quad a\ge b\ge c.$$
But I can not prove or refute that. Help me, please.
What I've tried so far. Rearrange the equati... | HINT
Using the identity
$$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca),$$
one can get the original equation in the form of
$$2(a+b+c)((a+b+c)^2-3(ab+bc+ac)) = (abc)^2,$$
or
$$r=\frac{2s^3-p^2}{6s},$$
where
$$s=a+b+c,\quad r=ab+bc+ac,\quad p=abc,$$
$$a>0,\quad b>0,\quad c>0,\quad s^3-27p\ge 0.$$
That allows to write... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2177372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Probability of product isn't greater than a specific number I have a joint distribution of 2 random variables $X$ and $Y$:
$$ f(x, y) = \frac{3}{2}x + \frac{y}{2} $$ if $x,y \in [0;1]$ and $0$ otherwise.
What's the following probability:$$ P\left(XY < \dfrac{1}{2}\right) $$
I thought about sum of probabilities: $P(X < ... | Actually, if $X>\frac{1}{2}$ and $Y>\frac{1}{2}$, you may still have $XY<\frac{1}{2}$. For instance, if $X=Y=\frac{5}{8}$, $XY=\frac{25}{64}<\frac{1}{2}$.
What you really need is $Y<\frac{1}{2X}$. So you can integrate the joint density of $X$ and $Y$ over the region $R$ where $0 \leq x \leq 1$, $0 \leq y \leq 1$, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove the inequality, power. $\{ x,y \in\Bbb R\ \}$ If $x+y = 2$ then prove the inequality:
$x^4 + y^4 \ge 2$
How I started
*
*$(x+y)^2 = 4$
*$x^2 + y^2 = 4 - 2xy$
*$(x^2+y^2)^2 - 2(xy)^2 \ge 2$
*$(4-2xy)^2 - 2(xy)^2 \ge 2$
*$16-16xy + 4(xy)^2 -2(xy)^2 - 2 \ge 0$
*$2(xy)^2 - 16xy + 14 \ge 0$
*for $t=xy$
*$2t^... | We need to prove that $\frac{x^4+y^4}{2}\geq\left(\frac{x+y}{2}\right)^4,$
which enough to prove for non-negatives $x$ and $y$.
Let $x^2+y^2=2uxy$.
Hence, $u\geq1$ and we need to prove that $2u^2-1\geq\left(\frac{u+1}{2}\right)^2,$
which is $(u-1)(7u+5)\geq0$.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rank of the $n \times n$ matrix with ones on the main diagonal and $a$ off the main diagonal I want to find the rank of this $n\times n$ matrix
\begin{pmatrix}
1 & a & a & \cdots & \cdots & a \\
a & 1 & a & \cdots & \cdots & a \\
a & a & 1 & a & \cdots & a \\
\vdots & \vdots & a& \ddots & & \vdots\\
\vdots & \vdots & \... | Let
$$\mathrm M_n (a) := \begin{bmatrix} 1 & a & a & \dots & a & a\\ a & 1 & a & \dots & a & a\\ a & a & 1 & \dots & a & a\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ a & a & a & \dots & 1 & a\\ a & a & a & \dots & a & 1\end{bmatrix} = (1-a) \mathrm I_n + a 1_n 1_n^{\top}$$
The eigenvalues of rank-$1$ matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2181367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find real $P$ s.t. $B=P^{-1}AP$ TASK:
$\frac{dx}{dt}=Ax$
Given $A=\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 & -3 \\
1 & 3 &2
\end{bmatrix}$
Find real matrix $P$ s.t change of coordinates $x=Py$ transforms the system to
and
$\frac{dy}{dt}=By$
$B=\begin{bmatrix}
1 & 0 & 0 \\
0 & 2 &... | Besides the hint given in my comment to a related post, consider that the similarity relation is not unique, because in fact:
$$
\mathbf{A = R}\;\mathbf{C}\;\mathbf{R}^{\,\mathbf{ - 1}} \quad \quad \Leftrightarrow \quad \left\{ {\begin{array}{*{20}c}
{\mathbf{A} = \mathbf{A}\;\mathbf{A}\;\mathbf{A}^{\,\mathbf{ - 1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help with a combinatoric formula Anyone can help solve th following? $\left( {\begin{array}{*{20}{c}}
{2n} \\
n
\end{array}} \right)$ means $2n$ chooses $n$. Thanks!
$\sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}{c}}
{2k} \\
k
\end{array}} \right){2^{ - 2k}}} = \sum\limits_{k = 0}^n {\frac{{(2k)!}}{{... | How about induction? The base case $n = 0$ is easily verified. Then
\begin{align}
\binom{2n}{n}2^{-2n}(2n+1) + \binom{2n+2}{n+1}2^{-2n-2} &= \binom{2n+2}{n+1}2^{-2n-2}\biggl[\frac{4(n+1)^2}{2n+2} + 1\biggr] \\
&= \binom{2n+2}{n+1}2^{-2n-2}\bigl(2n+2+1\bigr)
\end{align}
is the induction step.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof: $n^2 - 2$ is not divisible by 4 I tried to prove that $n^2 - 2$ is not divisible by 4 via proof by contradiction. Does this look right?
Suppose $n^2 - 2$ is divisible by $4$. Then:
$n^2 - 2 = 4g$, $g \in \mathbb{Z}$.
$n^2 = 4g + 2$.
Consider the case where $n$ is even.
$(2x)^2 = 4g + 2$, $x \in \mathbb{Z}$.
... | For odd $n$, $n^2-2$ is odd.
For even $n$, $n^2$ is divisible by $4$, so that $n^2-2$ is not.
By contradiction:
Let $n$ be odd. Then $n^2-2$ is both odd and a multiple of four.
Let $n$ be even. Then $n^2-2$ and $n^2$ are both multiples of $4$, so that $2$ is a multiple of $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2184937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
$\int_{-\infty}^{\infty }\frac{x^{2} \, dx}{x^{4}+a^{4}}$ I'm considering the following integral:
$$\int_{-\infty}^{\infty }\frac{x^{2} \, dx}{x^{4}+a^{4}}$$
I do not know how to factor the denominator to calculate the residues.
I'm stuck here:
$$(x^{2}-|a|^{2}i)(x^{2}+|a|^{2}i)$$
Thank you so much.
| Find the indefinite integral.
First, determine constants $p, q, r, s$ such that
$$
\frac{px+q}{x^2+\sqrt{2}ax+a^2}+\frac{rx+s}{x^2-\sqrt{2}ax+a^2} = \frac{x^2}{x^4+a^4}
$$
By comparing the coefficients of $x^3, x^2, x, 1$, we have the following relational expression:
$$
p+r=0, \\
-\sqrt{2}a(p-r)+q+s=1, \\
a^2(p+r)-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Does the sequence $x_{n+1}=x_n+\frac{1}{x_n}$ converge or diverge? Set $x_1=a$, where $a > 0$ and let $x_{n+1}= x_n + \frac{1}{x_n}$. Determine if the sequence $ \lbrace X_n \rbrace $ converges or diverges.
I think this sequence diverges since Let $x_1= a > 0$ and $x_{n+1}= x_n + (1/x_n)$ be given. Let $x_1=2$ since $a... | We have $$x_{N+1}-x_1=\sum_{n=1}^N x_{n+1}-x_n =\sum_{n=1}^N \frac1{x_n}$$ so $x_n$ converges if and only if the rightmost series converges.
But the series converges only if $x_n$ diverges to $\pm\infty $, impossible: both the sequence and the series must diverge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2190592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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Prove that $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
Prove that
$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}=\frac{\sqrt{7}}{8}$.
What I've tried doing :
If $\theta=\frac{\pi}{7}:$
$$
3\theta+4\theta=\pi
$$
This allowed me to prove that :
$$
\tan^2\frac{\pi}{7}+\tan^2\fra... | Since $\sin\frac{k\pi}{7}=\sin\frac{(7-k)\pi}{7}$ for k=1,2,3 we get:
$$(\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7})^2=
\prod_{k=1}^6 {\sin\frac{k\pi}{7}}=\frac{7}{2^6}$$
where the last equality is the case n=7 in this question:
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$
So we get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $ I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x... | Here is an efficient way forward. We write $\log(1+x^3)=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^{3n}}{n}$, which is valid for $|x|<1$.
The series converges uniformly on $[0,1/2]$ and we may exchange the series and integration to obtain
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)\,dx&=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Maximum value of $x^2+y^2$ Question
Find the maximum value of $x^2+y^2$ if
$4x^4+9y^4=64$
Now I really don't understand how to proceed or whether I should change my approach all together. Any help is appreciated.Thanks :)
| $$4x^4+9y^4=64$$
$$\frac{(x^2)^2}{4^2}+\frac{(y^2)^2}{(8/3)^2}=1$$
Let $x=\pm \sqrt{4\cos(\theta)}$ and $y=\pm \sqrt{\frac{8}{3} \sin (\theta)}$. With $\theta \in [0,\frac{\pi}{2}]$.
Then,
$x^2+y^2=4\cos (\theta)+\frac{8}{3} \sin (\theta)$
$$=\langle 4, \frac{8}{3} \rangle \cdot \langle \cos (\theta), \sin (\theta) \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $a>b>0$ and $a^3+b^3+27ab=729$ then $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.
If $a>b>0$ and $a^3+b^3+27ab=729$ then the quadratic equation $ax^2+bx-9=0$ has roots $\alpha,\beta,(\alpha<\beta)$. Find the value of $4\beta -a\alpha$.
By looking at the equation... | Hint ( I haven't completely verified)
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
If this equal zero then
$$(a+b+c)=0$$ or
$$(a^2+b^2+c^2-ab-bc-ca)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a... | Repeatedly squaring mod $19$ and noting that $3\cdot13\equiv1\pmod{19}$:
$$
\begin{align}
10^2&\equiv5&\pmod{19}\\
10^4&\equiv6&\pmod{19}\\
10^8&\equiv17&\pmod{19}\\
10^{16}&\equiv4&\pmod{19}\\
10^{32}&\equiv16&\pmod{19}\\
10^{33}&\equiv8&\pmod{19}\\
10^{33}-1&\equiv7&\pmod{19}\\
\frac{10^{33}-1}3&\equiv7\cdot13&\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
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Suppose a is a real number for which all the roots of the equation $ x^4 - 2ax^2 + x +a^2 -a =0$ are real.Find the range of values of a? I have tried the question with $Rolle's$ theorem...
Let the equation have $2$ unequal real roots
$f'(x)= 4ax^3-4ax+1=0$
But cannot proceed after this.
| we can write above equation as $x^4-2ax^2+x+a^2-a=0$ as $a^2-(2x^2+1)a+x^4+x=0$
So $$\displaystyle a= \frac{(2x^2+1)\pm\sqrt{(2x^2+1)^2-4(x^4+x)}}{2}=\frac{(2x^2+1)\pm (2x-1)}{2}$$
So $x^2+x-a=0$ or $x^2-x+1-a=0$
So for real roots , $1+4a\geq 0$ and $1-4(1-a)\geq 0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ I am supposed to proof by induction that $3^{2n-1} + 2^{2n-1}$ is divisible by $5$ for $n$ being a natural number (integer, $n > 0$).
What I have so far:
Basis: $n = 1$
\begin{align}
3^{2 \cdot 1-1} + 2^{2 \cdot 1-1} & = 3^1 + 2^1\\
... | From where you got to:
\begin{align}
3^{2 \cdot (k+1)-1} + 2^{2 \cdot (k+1)-1} & = 3^{2k+2-1} + 2^{2k+2-1}\\
& = 9\cdot3^{2k-1} + 4\cdot 2^{2k-1}\\
& = 5\cdot3^{2k-1} + 4(3^{2k-1}+ 2^{2k-1})\\
& =5\cdot3^{2k-1} + 4\cdot 5m \tag {from hypothesis}\\
& =5(3^{2k-1} + 4m)\\
\end{align}
... so divisible by $5$ as required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Find the singular values of $T: p(x)\mapsto xp'(x)+2x^2p''(x)$ I would appreciate help with this problem I am self-studying this problem:
Find the singular values of the operator $T\in \mathcal P_2(\mathbb{C}): p(x)\mapsto xp'(x)+2x^2p''(x)$. The inner-product on $\mathcal P_2(\mathbb{C})$ is defined as $\langle p,q ... | Start with the basis $\{1, x, x^2\}$ for $\mathcal{P}_2(\mathbb{C})$. Performing Gram-Schmidt yields the following orthonormal basis for $\mathcal{P}_2(\mathbb{C})$: $$\mathcal{B} = \left\{ \frac{1}{\sqrt{2}}, \sqrt{\frac{3}{2}}x, \sqrt{\frac{45}{8}}\left(x^2-\frac{1}{3}\right)\right\}_.$$
Let $A$ denote the matrix of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int \frac{x^2-1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx$
The question is to evaluate $$\int \frac{x^2-1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx$$
I tried to rewrite the integral as $$\frac{ab}{b-a}\int \frac{1/a(x^2+ax+1)-1/b(x^2+bx+1)}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx-2\int \frac{1}{x\sqrt{x^2+ax+1}\sqrt{x... | Suppose $x>0$. Letting $x+\frac{1}{x}\to u$ and then $du=(1-\frac{1}{x^2})dx$, one has
\begin{eqnarray}
&&\int \frac{x^2-1}{x\sqrt{x^2+ax+1}\sqrt{x^2+bx+1}} dx\\
&=&\int\frac{1-\frac1{x^2}}{\sqrt{x+a+\frac{1}{x}}\sqrt{x+b+\frac{1}{x}}} dx\\
&=&\int\frac{1}{\sqrt{u+a}\sqrt{u+b}}du\\
&=&\int\frac{1}{\sqrt{(u+\frac{a+b}2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Approaching Tricky Combinatorial Proofs - Tough Example I'm struggling with finding effective ways to approach questions (especially proofs) that use summations and Taylor Series. I've worked through several simpler examples, but always get stuck once a non-trivial question arises. In particular, I'm hoping to get some... | Here is an approach based on the observation that
\begin{align*}
\sum_{i=0}^{n-1}\binom{2i}{i}\frac{1}{i+1}\cdot\frac{1}{2^{2i}}
\end{align*}
is a Cauchy-product which can be interpreted as the coefficient of the product of two (ordinary) generating functions. We recall the generating function of the Catalan numbers $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the inequality $(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$
Solve the inequality $$(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)<0$$
My work so far:
1) $a>0$
2) Let $f(x)=(x^2-2\sqrt{a}\cdot x+1)(2^x+\lg a)$.
$f(x)=0$ if and only if $x^2-2\sqrt{a}\cdot x+1=0$ or $2^x+\lg a=0$
$$x^2-2\sqrt{a}\cdot x+1=0$$
$$x_{1,2}=\sqrt a \p... | $1)$ If $0<a<1$ then we have $x^2-2\sqrt{a}\cdot x+1>0$ for all real number $x$.
Then we must have
$$2^x+\log a<0\to 2^x<\log a^{-1}\to x<\log_2(\log a^{-1})$$
$2)$ If $a>1$ then we have $2^x+\log a>0$ for all real number $x$.
Then we must have
$$x^2-2\sqrt{a}\cdot x+1<0\to \sqrt{a}-\sqrt{a-1}<x<\sqrt{a}+\sqrt{a-1}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+... | Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$ ......(1)
Using property -
$\int_a^{b} f(x) dx = \int_a^{b} f(a + b - x) dx$
We have,
$I = \int_0^{\frac{\pi}{2}} \frac{\sin^3(\frac{\pi}{2}-x)}{\cos^3(\frac{\pi}{2}-x)+\sin^3(\frac{\pi}{2}-x)} dx$
$I = \int_0^{\frac{\pi}{2}} \frac{\cos^3(x)}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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If $\dfrac {1}{a+b} +\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ $ If $\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$, prove that $\angle B=60^\circ$.
My Attempt
$$\dfrac {1}{a+b}+\dfrac {1}{b+c}=\dfrac {3}{a+b+c}$$
$$\dfrac {a+2b+c}{(a+b)(b+c)}=\dfrac {3}{a+b+c}$$
$$a^2-ac-b^2+c^2=0$$.
How t... | from your condition we get $$a^2+c^2-ac=b^2$$ apllying the theorem of cosines we get
$$a^2+c^2-ac=a^2+c^2-2ac\cos(\beta)$$
thus we get $$\cos(\beta)=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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System of Equation. (Complicated) Given that $x,y \in \mathbb{R}$. Solve the system of equation.
$$5x(1+\frac{1}{x^2+y^2})=12$$
$$5y(1-\frac{1}{x^2+y^2})=4$$
My attempt,
I made them become $x+\frac{x}{x^2+y^2}=\frac{12}{5}$ and $y-\frac{y}{x^2+y^2}=\frac{4}{5}$.
I don't know how to continue from here. Hope anyone woul... | $$5x(1+\frac{1}{x^2+y^2})=12\cdots \cdots (1)$$
$$5y(1-\frac{1}{x^2+y^2})=4\cdots \cdots \cdots \cdots (2) \times i$$
$$5(x+iy)+5\cdot \frac{x-iy}{(x+iy)(x+iy)}=12+4i$$
Now adding these two equations
put $z=x+iy$ and $\bar{z} = x-iy$ and $|z|^2= z \cdot \bar{z} = x^2+y^2$
So $$5z+\frac{5\bar{z}}{z \cdot \bar{z}}=12+4i$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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If $c$ is equal to $\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$ then find $\log_2{(-c)}$. If $c$ is equal to $$\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$$then find $\log_2{(-c)}$.
I think the binomial theorem might h... | Since
$$
\begin{gathered}
\left( {1 + i\,} \right)^{\,n} = \sqrt 2 ^{\,n} \left( {\cos \left( {\frac{{n\,\pi }}
{4}} \right) + i\sin \left( {\frac{{n\,\pi }}
{4}} \right)} \right) = \sum\limits_{0\, \leqslant \,k\left( { \leqslant \;n} \right)} {\left( \begin{gathered}
n \\
k \\
\end{gathered} \right)i^{\,k} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Partial sum back into series. If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n−1}{n+1}$
then find $a_n$ and find
the sum of this series. Fully justify your answer.
Using the definition.
$a_n = S_{n} - S_{n-1} $
$$a_n = \frac{n-1}{n+1} - \frac{n-2}{n}$$
$$= \frac{n(n-1) - (n-2)(n+1)}{n(n+1)}$$
$$= \f... | Obviously, the sequence can be defined as
$$\begin{cases}a_0=S_0=-1,\\a_n=S_n-S_{n-1}=\dfrac{n-1}{n+1}-\dfrac{n-2}{n}=\dfrac2{n(n+1)},n>0.\end{cases}$$
(In particular, $a_1=1$, ensuring $S_1=a_0+a_1=0$.)
The sum is indeed $\lim_{n\to\infty}\dfrac{n+1-2}{n+1}=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solution for continued fraction To solve the finite continued fraction $[2,....,2]$, can I use this solution.
Let $y$ be the value of $[2,....,2]$.
Then $y=[2;y]=2+\frac{1}{y}$.
So $y-2=\frac{1}{y}$.
Thus $y^2-2y-1=0$
My question is is my method correct. Is this approach suitable to solve $[2,....,2]$
|
Is this approach suitable to solve $[2,....,2]$
No, because your continued fraction is finite. Moreover, it is rational and the solutions of $y^2-2y-1=0$ are irrational.
Notation.
\begin{equation*}
x_{n}=a_0+\cfrac{1}{a_1+\cfrac{1}{\begin{array}{ccc}a_{2}+ & & \\& \ddots & \\& & +\cfrac{1}{a_{n-1}+\cfrac{1}{a_{n}}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
If the roots of $9x^2-2x+7=0$ are $2$ more than the roots of $ax^2+bx+c=0$, then value of $4a-2b+c=$?
My approach is: Let the roots of $ax^2+bx+c=0$ are $\alpha$ and $\beta$. So, $$\alpha+\beta=-\frac{b}{a}\\ \alpha\... | Hint: Find the roots of equation $9x^2-2x+7=0$ by solving this let say the roots be $\alpha$ and $\beta$ (You know the value of $\alpha$ and $\beta$.) then the roots of other equation must be $\alpha-2$ and $\beta-2$ form a equation using $$(x-(\alpha-2))(x-(\beta-2))$$ compare the coefficient. then put the value in $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2209958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Method to prove limit in $\mathbb{R}^2$ Given is the following limit $\lim_{(x,y)\to(0,0)}\frac{x^3y}{x^4+y^2}$.
Now, it holds that
$0\leq\left\vert\frac{x^3y}{x^4+y^2}\right\vert\leq \frac{x^3y}{y^2}=\left\vert\frac{x^3}{y}\right\vert$.
It is clear that $(x,y)\mapsto\frac{x^3}{y}$ is homogeneous on $\mathbb{R^2}\setmi... | Let $S=\frac {|x^3y|}{x^4+y^2}.$ We have $x^2|y|\leq \frac {(x^2+|y|)^2}{4}$ so $$S=\frac {|x|\cdot x^2|y|}{x^4+y^2} \leq \frac {|x|(x^2+|y|)^2}{4(x^4+y^2)}=$$ $$=\frac {|x|(x^4+y^2)+2|x^3y|}{4(x^4+y^2)}=$$ $$ =\frac {|x|}{4}+\frac {S}{2}.$$ $$ \text {Therefore }\quad S\leq \frac {|x|}{2}.$$
Not as slick as the answe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Eigen value and eigen vectors I have given a matrix $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -3 & 3 \end{pmatrix}. $$
We have to find 3 linearly independent eigen vector. I have calculated the eigen values that is 1 with multiplicity 3 . I have found the eigen vector $ \begin{pmatrix}1 & 1 & 1 \end{pmatrix}^T... | The characteristic polynomial is
\begin{align}
\det\begin{pmatrix}
0-X & 1 & 0 \\
0 & 0-X & 1 \\
1 & -3 & 3-X
\end{pmatrix}
&=
-X\det\begin{pmatrix}-X & 1 \\ -3 & 3-X\end{pmatrix}
-\det\begin{pmatrix}0 & 1 \\ 1 & 3-X\end{pmatrix}
\\
&=-X(-3X+X^2+3)+1
\\[6px]
&=-X^3+3X^2-3X+1=(1-X)^3
\end{align}
There cannot exist three... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $(a,b)=1$, how to find the suitable $u$ and $v$ such that $(a+b)u+(a^2-ab+b^2)v=3$? Suppose $(a,b)=1$. I want to prove that $(a+b,a^2-ab+b^2)=1$ or $3$.
Since $(a,b)=1$, so $(a^2,b^2)=1$ and therefore there are $x,y\in\mathbb Z$ such that $a^2x+b^2y=1$ or $3a^2x+3b^2y=3$ and by adding and subtracting $3abx+3aby+3a^2... | \begin{align}a^{2} - a b + b^{2} &= (a + b)(a - 2 b) + 3 b^{2} \\&= (a + b) (b - 2 a) + 3 a^{2}.\end{align}
This you can see by first regarding $a^{2} - a b + b^{2}$ and $a + b$ as polynomials in $a$, and dividing $a^{2} - a b + b^{2}$ by $a + b$. And then doing the same regarding them as polynomials in $b$, or simply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Matrix and a Linear Function I'm a student of electrical engineering. At our math class we were given following assignment: given a matrix
$A = \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}$ and a function $f(x) = x^5 - x^3 + 5x^2 + 8$, compute $f(A)$.
I've checked out this PDF document, but I don'... | I claim that
$$A^n = \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}^n=\begin{pmatrix}
1 & 0 & 2n\\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}.\tag 1$$
By the principle of mathematical induction, first, for $n=2$
$$A^2 = \begin{pmatrix}
1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1
\end{pmatrix}^2=\begin{pmatrix}
1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real? Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real?
Let $f(z)=z^5+2z^2+7z+1$, and $g(z)=z^5$. For $|z|=2,$ we note that
$$
|f(z)-g(z)|\leq2|z|^2+7|z|+1=23<32=2^5=|g(z)|.
$$
Since $g$ has $5$ zeros in $|z|\leq 2$,... | I'd never seen Sturm's theorem before, but it was interesting enough that I'll add the computation:
\begin{align}
p_0(x) &= x^5+2x^2+7x+1\\
p_0'(x) = p_1(x) &= 5x^4+4x+7\\
\text{remainder of $-p_0/p_1$:}\quad p_2(x) &= -\tfrac{6}{5}x^2 - \tfrac{28}{5}x - 1\\
\text{remainder of $-p_1/p_2$:}\quad p_3(x) &= \tfrac{12562}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int {t+ 1\over t^2 + t - 1}dt$
Find : $$\int {t+ 1\over t^2 + t - 1}dt$$
Let $-w, -w_2$ be the roots of $t^2 + t - 1$.
$${A \over t + w} + {B \over t+ w_2} = {t+ 1\over t^2 + t - 1}$$
I got $$A = {w - 1\over w - w_2} \qquad B = {1- w_2\over w - w_2}$$
$$\int {t+ 1\over t^2 + t - 1}dt = A\int {1\over t+ w} ... | To get to their answer more directly, you can write the integral as$$\frac 12\int \frac{2t+1}{t^2+t-1}dt+\frac 12\int\frac{1}{(t+\frac 12)^2-\frac 54}dt$$
$$=\frac 12 \ln|t^2+t-1|+\frac{1}{2\sqrt{5}}\ln\left|\frac{t+\frac 12-\frac{\sqrt{5}}{2}}{t+\frac 12+\frac{\sqrt{5}}{2}}\right|+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Vector magnitude subtraction Please help! This was in a textbook and I cannot seem to make sense of it. Would it not be 2?
The two vectors a and b are perpendicular. If a has magnitude 8 and b has magnitude 3, what is |a−2b|??
I
| Since your vectors are perpendicular, they don't directly add and subtract like scalars would.
Denoting them in R$^2$:
$\textbf{a} - 2\textbf{b} = \begin{pmatrix}
8 \\
0
\end{pmatrix} -
2 \begin{pmatrix}
0 \\
3
\end{pmatrix} =
\begin{pmatrix}
8 \\
-6
\end{pmatrix}.$
Thus we need to consider what the value of $\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Knowing that $\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3}$ and $\tan \alpha > 0$, find the exact value of $\tan \alpha$ First I tried to solve for $\alpha$:
$$\cos(\frac{3\pi}{2}+\alpha)=-\frac{1}{3} \Leftrightarrow \alpha = \arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{2} \lor \alpha = -\arccos(-\frac{1}{3})+2k\pi-\frac{3\pi}{... | Hint: your first equation gives $\sin (\alpha)\,=\,\frac{1}{3}$ which can be easily solved to get $\tan(\alpha)\,=\,\frac{\sqrt2}{4}$
$$\cos(\frac{3\pi}{2}+\alpha)\,=\,-\sin(\alpha)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing that $f(x)=\frac x{x+1}$ is the unique function satisfying $f(x)+f\left(\frac1x\right)=1$ and $f(2x)=2f\big(f(x)\big)$ We are given a function $ f : \mathbb Q ^ + \to \mathbb Q ^ + $ such that $$ f ( x ) + f \left( \frac 1 x \right) = 1 $$
and
$$ f ( 2 x ) = 2 f \big( f ( x ) \big) \text . $$
Find, with proof, ... | A bit too long for a comment, but here's a start:
You have shown $f(1) = 1/2$, $f(2) = 2/3$ and hence $f(1/2) = 1/3$.
Plug $x = 1/2$ into your second identity to get $f(1) = 2f(f(1/2))$, i.e $1/2 = 2 f(1/3)$, i.e $\color{blue}{f(1/3) = 1/4}$. Thus, $\color{blue}{f(3) = 3/4}$ as well.
To get $f(4)$ and $f(1/4)$, we must... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"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.