Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
why $x^2 - 25y^2 = 1$ cannot have integer solutions other than $( \pm 1,0)$ pell equations must have n to be nonsquared,so $x^2 - 25y^2 = 1$ is not a pell equation, but the thing is how can i show $x^2 - 25y^2 = 1$ have no integer solutions? or it has integer solutions.
Solve : $x^2-25y^2=1$
| Given $x^2-25y^2=1$
$(x-5y)(x+5y)=1$
$1=-1 \cdot -1$ and $1=1 \cdot 1$ are the only ways to write $1$ as the product of two integers.
Hence we must have $x-5y=x+5y=1$ or $x-5y=x+5y=-1$ and so it gives the only solutions as $x= \pm 1$ and $y=0$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Argument and modules of complex numbers Can you help me to find the argument and modules of this complex number please?
$$z=\frac{2-i}{-\sqrt {3}-2i}$$
| $$z=\frac{2-i}{-\sqrt {3}-2i}\cdot\frac{-\sqrt {3}+2i}{-\sqrt {3}+2i}=\frac{-2\sqrt3+2i-i\sqrt3-2i^2}{3+4}=$$
$$=\frac{2-2\sqrt3+(2-\sqrt3)i}{7}=\frac{2-2\sqrt3}{7}+\frac{2-\sqrt3}{7}i=a+bi$$
then
$$|z|=\sqrt{a^2+b^2}$$
$$\arg z=\arctan\frac{a}{b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\sum_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}$ absolute convergent,conditional convergent or divergent? Is $\sum_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}$ absolute convergent,conditional convergent or divergent ?
$\sum_{n=2}^{\infty} |\frac{(-1)^n }{n^2+(-1)^n}| =\frac{1}{|n^2+(-1)^n|}$
now $|n^2+(-1)^n| > n^2 - ... | $\sum\limits_{n=2}^{\infty} \frac{(-1)^n }{n^2+(-1)^n}=\sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2+1}-\sum\limits_{n=1}^{\infty} \frac{1}{(2n+1)^2-1}$
with
$\sum\limits_{n=1}^{\infty} \frac{1}{(2n)^2+1}\leq \frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{24}$
and
$\sum\limits_{n=1}^{\infty} \frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279208",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Algebra: Prove inequality $\sum_{n=1}^{2015} \frac1{n^3} < \frac 54$ Can someone prove inequality (n is natural):$$\sum_{n=1}^{2015} \frac{1}{n^3} < \frac 5 4$$
I have tried some predictions like $a^3 > a(a - 1)(a - 2) $ but couldn't get anything out of them.
| Note that for $n>1$,
$$\frac{1}{n^3}<\frac{1}{n^3-n}=\frac{1}{2(n-1)}-\frac{1}{n}+\frac{1}{2(n+1)}$$
$$\sum_{n=2}^{2015}\frac{1}{n^3-1}=\frac{1}{2(1)}-\frac{1}{2(2)}-\frac{1}{2(2015)}+\frac{1}{2(2016)}=\frac{2031119}{8124480}<\frac{1}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\frac{2}{\pi}\ln\frac{\pi + 2}{2} < \int_0^{\pi/2} \frac{\sin x}{x(x+1)} \, \mathrm{d}x < \ln\frac{\pi+2}{2}$ How to prove that $$\frac{2}{\pi}\ln\frac{\pi + 2}{2} < \int\limits_0^{\frac{\pi}{2}} \frac{\sin x}{x(x+1)}dx < \ln\frac{\pi+2}{2}$$
How prove tasks like this?
| We have $\sin x < x$ for $x > 0$ hence $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \mathrm{d}x < \int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{x+1} = \ln (x+1)\big]_0^{\pi/2} = \ln \left(\frac{\pi}{2} + 1\right).$$
Also $\sin x > \frac{2}{\pi}x$ for $x > 0$ (and $x < \pi/2$) hence $$\int_0^{\pi/2} \frac{\sin x }{x(x+1)} \, \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex eigenvectors... I can't get the right answer even though I'm using software I'm using software to calculate my eigenvectors, and I can't get the correct answers...
I have $$A = \begin{pmatrix} 1 & 2+i \\ 2+i & 1 \end{pmatrix}.$$ I solved the eigenvalues to be $3+i, -1-i$ (which is correct according to my softwa... | We all agree that given
$$
\mathbf{A} = \left[
\begin{array}{cc}
1 & 2+i \\
2+i & 1 \\
\end{array}
\right]
$$
the eigenvalues are
$$
\lambda \left( \mathbf{A} \right) = \left\{ 3 + i, -1 - i \right\}
$$
The eigenvectors are
$$
v_{1} =
\color{blue}{\left[
\begin{array}{c}
1 \\
1 \\
\end{array}
\right]}, \qquad... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find partial sum of this series: $\sum_{n=1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$? $\sum_{1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$
How to find partial sum, sum and prove convergence by definition? Thanks a lot.
| $$\sum_{n=1}^{\infty } \frac{2n+3}{n(n+1)(n+2)(n+3)}$$
Resolve it into partial fractions,
$$=\frac{1}{2}\sum_{n=1}^{\infty }[\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}]$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find the locus of $|z-2i|=3|z+3|$ I got as far as
$$|z-2i|=3|z+3| \Leftrightarrow \\
(\ldots) \Leftrightarrow \\
x^2-y^2+4y-4=9x^2+54x+81-9y^2 \Leftrightarrow \\
x^2-9x^2-y^2+9y^2+4y-4-54x-81=0\Leftrightarrow \\
-8x^2+8y^2+4y-85-54x=0 \Leftrightarrow \\
8y^2+4y-8x^2-54x-85=0\Leftrightarrow \\
y^2+\frac{1}{2}y-x^2-\fr... | Problem is way up at the top. You wrote $x^2 - y^2 + 4y-4$, when it should have been $x^2 + y^2 - 4y+4$. Similarly, $9x^2 - 9y^2 + 54 x + 81$ should have been $9x^2 + 9y^2 + 54 x + 81$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Let $ABCD$ be a square with sidelength 1 and $AKL$ is an equilateral triangle where $K$ lies on BC and L lies on CD. What is the area for AKL? I tried to use the Pythagorean Theorem to find one side of $AKL$, but there isn't enough information. What am I missing?
| Put it on a cartesian plane $A = (0,0); B= (1,0); C=(1,1); D=(0,1)$ and $K = (1,k)$ and $L=(j, 1)$.
Then ... it all comes out in the wash.
$AK = \sqrt{k^2 + 1} = KL = \sqrt{(1-k)^2 + (1-j)^2} = AL = \sqrt{1 + j^2} ; 1>k > 0; 1>j> 0$
So $k^2 + 1 = k^2 + j^2 - 2(k+j) + 2 = 1+ j^2$
So $j = k$ and $k^2 - 4k + 1 = 0$ so $k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding solutions to polynomial modular equations like $(x+1)^6 - x^6 \equiv 0\ (\textrm{mod}\ 19)$ I'm more or less okay with linear modulo equations, and was wondering how to solve polynomial modulo equations like $(x+1)^6 - x^6 \equiv 0\ (\textrm{mod}\ 19)$.
The possible solutions are 2, 7, 9, 11 and 16 (found on an... | there are six sixth roots of one $\pmod {19},$ those being the three cube roots of $1$ and the three cube roots of $-1.$ For each of these, solve $\frac{x+1}{x} \equiv t \pmod {19}$ where $t^6 \equiv 1 \pmod {19}.$ Or, for $t \neq 1,$ you see
$$ \frac{1}{x} \equiv t - 1 \pmod {19}, $$
$$ x \equiv \frac{1}{t-1} \pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$ uniformly continuous or not Find out if function $$f(x,y)=\frac{\sin\sqrt[3]{x^3+y^3}}{\sqrt[5]{x^5+y^5}}$$ is uniformly continous or not in area $D=\{0<x^2+y^2<1\}$.
I found out that we have no $\lim\limits_{x,y\to 0}f(x,y)$, because $$\lim\limits_{x,y\to 0}\... | It's not uniformly continuous. Take $x = \rho\cos{\alpha},y= \rho\sin{\alpha}$. Let $\alpha =\pi/4,$, and let $\rho \rightarrow 0^+$, then $f(x,y) \rightarrow \frac{2^{1/3}}{2^{1/5}}$, but take $\alpha =0$, then $f(x,y) \rightarrow 1$.
So let $\epsilon = \frac{1}{2}|\frac{2^{1/3}}{2^{1/5}} - 1|$, then $\forall \delta >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to take $\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$? The integral:
$$\int_0^{+\infty} \frac{x^2+1}{x^4+1}dx$$
If num were greater than denum I would just devide it normally with long division, but it is not, how should I handle it then?
| By substituting $x\mapsto\frac1x$, we get
$$
\int_0^1\frac{x^2+1}{x^4+1}\,\mathrm{d}x=\int_1^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
$$
Therefore, using formula $(6)$ from this answer, we get
$$
\begin{align}
\int_0^\infty\frac{x^2+1}{x^4+1}\,\mathrm{d}x
&=2\int_0^1\frac{x^2+1}{x^4+1}\,\mathrm{d}x\\
&=2\int_0^1\left(1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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Large powers of sine appear Gaussian -- why? As part of approximating an integral, I have noticed that $\sin^k(x), x\in[0, \pi]$ look almost identical to $\exp\left(-\frac{k}{2}(x-\frac{\pi}{2})^2\right)$ once $k$ is large enough (in practice, the two equations are visually identical for $k\geq 15$). As an example, see... | This is too long for a comment.
Semiclassical made a very good comment.
Let us use Taylor series around $x=\frac \pi 2$
$$\sin(x)=1-\frac{1}{2} \left(x-\frac{\pi }{2}\right)^2+\frac{1}{24} \left(x-\frac{\pi
}{2}\right)^4+O\left(\left(x-\frac{\pi }{2}\right)^6\right)$$ $$\sin^k(x)=1-\frac{1}{2} k \left(x-\frac{\pi }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
} |
Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| Note that if $z^3=1, z\neq 1$ so that $z^3-1=(z-1)(z^2+z+1)=0$ then $z^5+z+1=z^2+z+1=0$. A key to this observation is just seeing whether an appropriate root of unity may be a root.
This tells you that $x^2+x+1$ is a factor of $x^5+x+1$, and the question then is how you do the division. The factorisation method you hav... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Why is $x^2+x$ the same as writing $(x+0.5)^2-0.25$? I find it extremely weird that a translation vertically by a variable should cause the graph to move both vertically and horizontally. Also, why 0.5 and 0.25?
| Okay so $0.5 = \frac{1}{2}$ and $0.25 = \frac{1}{4}$.
Let's start from $$\left ( x + \frac{1}{2} \right )^{2} - \frac{1}{4}$$
$$= \left (\frac{2x+1}{2} \right )^{2} - \frac{1}{4}$$
$$= \frac{(2x+1)^{2}}{4} - \frac{1}{4}$$
$$= \frac{(2x+1)^{2}-1}{4}$$
$$= \frac{4x^{2}+4x}{4}$$
Then factor out $4x$ in the numerator to o... | {
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"url": "https://math.stackexchange.com/questions/2293553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
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What can be the values of $m$ in $(m-3)x^2 + (m+2)x + 2m + 1$? What can be the values of $m$, if $(m-3)x^2 + (m+2)x + 2m + 1$ should be always greater than or equal to zero?
I think we should be using Delta, but I've got no idea how.
| Let,
\begin{align}
y&=ax^2+bx+c\\
&=a\left(x+\dfrac{b}{2a}\right)^2+\left(c-\dfrac{b^2}{4a}\right)\\
&=a\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b^2-4ac}{4a}\right)\\
&=a\left[\left(x+\dfrac{b}{2a}\right)^2-\dfrac{D}{4a^2}\right]
\end{align}
From this we can conclude that, when $D\le0$ and $a\ge0$, then only $y\ge0... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Primes where $(p-1) ^ 2 + 1$ and $(p+1) ^ 2 + 1$ are also prime I'm following a math course about basic number theory. The course contains an Open Problems section with Landau’s conjecture, that states: "There are infinitely many primes of the form $n^2 + 1$.". Examples include:
$2 = 1^2 + 1$
$5 = 2^2 + 1$
$17 = 4^2 + ... | Have you thought about comparing $(p - 1)^2 + 1$ to $(p + 1)^2 + 1$ for just a few values of $p$ and seeing what happens?
*
*$p = 2$ gives us $(p - 1)^2 + 1 = 2$ and $(p + 1)^2 + 1 = 10 = 2 \times 5$.
*$p = 3$ gives us $(p - 1)^2 + 1 = 5$ and $(p + 1)^2 + 1 = 17$.
*$p = 5$ gives us $(p - 1)^2 + 1 = 17$ and $(p + 1... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Find values of constants a and b such that the given improper integral converges Find values of constants a and b such that:$$\int_{3}^\infty \left(\frac{ax+2}{x^2+3x}-\frac{b}{3x-2}\right) dx=k$$ then by partial fractions we get:
$$\lim_{N \to \infty} \int_{3}^N \left(\frac{2}{3}\frac{1}{x}+\frac{a-\frac{2}{3}}{x+3}-\... | Hint: It converges when
$$ \lim _{ n\to \infty } \ln { \left[ \sqrt [ 3 ]{ \frac { { n }^{ 2 }{ \left(
n+3 \right) }^{ 3a-2 } }{ { \left( 3n-2 \right) }^{ b } } } \right] } \\ \\ 2+3a-2=b\quad \Rightarrow b=3a $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Complex Analysis - Laurent series of $\frac{z}{z^2+1}$ in $|z-3|>2$ I'm supposed to find the Laurent series for $\frac{z}{z^2+1}$ in $D$ : $|z-3|>2$
the nature of the domain $D$ tells me that the analytic part of the Laurent series expansion must be $0$
which means the series I'm looking for should look like this :
$... | Let $f(z)=\frac{z}{z^2+1}$. There are pole singularities of $f$ at $z=\pm i$. Hence, if we expand $f$ in a Laurent series around $z=3$, that series will have only a principal part for $|z-3|<\sqrt{10}$ and will have no principal part for $|z-3|>\sqrt{10}$.
For $|z-3|>\sqrt{10}$, we can write
$$\begin{align}
\frac{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Count permutations of [1, 1, 1, 2, 2, 3, 3, 4, 5] with length 6 How can I count number of permutations such as:
[1, 1, 1, 2, 2, 3]
[1, 2, 1, 3, 2, 4]
[1, 1, 1, 2, 2, 5]
...
So I have:
'1' x 3
'2' x 2
'3' x 2
'4' x 1
'5' x 1
So permutation valid if it contains $\le3$ of '1' and $\le2$ of '2' and and $\le2$ of '3' and $\... | I believe you are forced to sum mutually exclusive and exhaustive possibilities.
I suggest defining $N(a,b)$ as the number of permutations having $a$ "1"'s and $b$ other doubles. Then ...
$$N(a,b) = \binom 2 b \cdot \frac{6!}{a!(2!)^b}$$
and
$$ N_{tot} = N(0,2)+N(1,2)+N(1,1)+N(2,2)+N(2,1)+N(2,0)+N(3,1)+N(3,0) $$
* EDIT... | {
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"timestamp": "2023-03-29T00:00:00",
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Egyptian fraction for $\varphi- {F(2n+2) \over F(2n+1)}$ The sum of the reciprocals of the ${2^n}$th Fibonacci numbers is known to be $\dfrac{3-\sqrt{5}}{2}$.
https://math.stackexchange.com/a/746678/134791
This may be written as the following closed form for an Egyptian fraction.
$$\varphi=2-\sum_{k=0}^\infty \frac{1}{... | Let $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2} = -\alpha^{-1}$.
In terms of $\alpha, \beta$, we have the Binet's formula:
$$F(n) = \frac{\alpha^n - \beta^n}{\alpha - \beta}$$
For any odd integer $m$, this leads to
$$\varphi - \frac{F(m+1)}{F(m)} = \alpha - \frac{\alpha^{m+1} - \beta^{m+1}}{\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove that $\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$? Is there a trick to show that
$$\sqrt{-3-2i}+\sqrt{-3+2i} = \sqrt{2(\sqrt{13}-3)}$$
is true ?
| Consider the following.
\begin{align}
\sqrt{-3-2i}+\sqrt{-3+2i} &= \sqrt{a} \\
(\sqrt{-3 - 2 i} + \sqrt{-3 + 2 i} )^2 &= a \\
-6 + \sqrt{(-3-2i)(-3+2i)} &= a \\
(-3-2i)(-3+2i) &= (a+6)^2 \\
(a+6)^2 - 13 &=0 \\
a^2 + 12 a + 23 &= 0 \\
\end{align}
from here it is determined that $a = -6 \pm \sqrt{13}$, which yields the d... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Evaluate series $\sum\limits_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$
Determine the value of
$$\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}}{1-x^{2^n}}$$
or $$\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8}+\cdots$$
for $x\in\mathbb{R}$.
The answer is $\dfrac{x}{1-x}$ for $x\in(0,1)$. To prove this, notice
$$\f... | Let $x\in(0,1)$ and let $\varepsilon>0$. I shall prove that there is a natural number $p$ such that$$n\geqslant p\Longrightarrow\left|\frac x{1-x}-\sum_{k=1}^n\frac{x^{2^{k-1}}}{1-x^{2^x}}\right|<\varepsilon.$$Take $p'\in\mathbb N$ such that $\left|\frac x{1-x}-(x+x^2+\cdots+x^{p'})\right|<\varepsilon$. Take $p\in\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
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Regarding a number theory proof Before I begin I should mention that I have only recently started studying number theory.
It is required to prove that there are no positive integers $a,b,n>1$ such that
$$a^n - b^n\mid a^n+b^n$$
that is $k$ can never be an integer if
$$\frac{a^n+b^n}{a^n-b^n} = k$$
Now if $b|a$ then ... | Notice if $a^n - b^n|a^n + b^n$ then $a^n - b^n|a^n + b^n + a^n - b^n = 2a^n$ and $a^n - b^n| a^n + b^n -(a^n- b^n) = 2b^n$.
So if $p|a^n - b^n$ and $p$ is prime either $p = 2$ or $p|a$ and $p|b$.
Let $d = \gcd (a,b)$ and let $a = a'd; b = b'd$ then $\frac{a^n - b^n}{a^n + b^n}= \frac {a'^n - b'^n}{a'^n + b'^n}$ and $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
3-Variable Constrained Inequality I had a chance to read a newly published problem-solving journal very recently and saw this inequality:
Statement: If $a, b, c \in [1,\infty)$ with $a+b+c = 6$, then show that:
$\sqrt{a^2-1}+\sqrt{b^2-1} + \sqrt{c^2-1} < \dfrac{3\sqrt{3}abc}{2}$. Just curious about your techniques,etc... | Let's look at $f(x)= \sqrt{x^2-1}$ for $x\gt1$
$f''(x)=\dfrac {-1}{(x^2-1)^{3/2}}\lt0$ so it's concave and by jensen
$\sqrt{a^2-1}+\sqrt{b^2-1} + \sqrt{c^2-1} \leq 3 \sqrt{(\dfrac {a+b+c}{3})^2-1}=3\sqrt{3}$
By symmetry we can assume $a\geq2 \Rightarrow abc\geq2 \Rightarrow \dfrac{3\sqrt{3}abc}{2}\geq3\sqrt{3}$
(I didn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to compute the series $\sum_{k\geq 1}\frac{1}{k(k+m)}$ I tried to do it like this:
$$\sum_{k\geq 1} \frac{1}{k(k+m)}=$$ $$\ = \sum_{k\geq 1} \frac{1}{km}-\frac{1}{m(k+m)}=$$ $$=\sum_{k\geq 1} \frac{1}{m}(\frac{1}{k}-\frac{1}{k+m})=$$ $$=\frac{1}{m}\sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} $$
I do know that this sum... | $$\require{cancel} \sum_{k\geq 1}\frac{1}{k}-\frac{1}{k+m} = \left(\frac{1}{1} - \cancel{\frac{1}{m+1}}\right) + \left(\frac{1}{2} - \cancel{\frac{1}{m+2}}\right) + \ldots + \left(\cancel{\frac{1}{m+1}} - \frac{1}{2m+1}\right) + \left(\cancel{\frac{1}{m+2}} - \frac{1}{2m+2} \right) + ...$$
So, all the second terms will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2319430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric integrals. Integrate $\int\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}\mathrm{d}x$ I tried factoring the denominator but does not seem to suit the factorization of quadratic. Is there another way?
| \begin{align*}
J &= \int \frac{1}{\cos^4 - \cos^2 x \sin^2 x + \sin^4 x} dx \\
&= \int \frac{\sec^4}{1 - \tan^2 x + \tan^4 x} dx
\end{align*}
Let $u = \tan x$ and $\sec^2 x = \tan^2 x + 1$
$$ J = \int \frac{(1+\tan^2 x) \sec^2 x}{1- \tan^2 x +\tan^4 x} dx $$
again $ u = \tan x$
$$ J = \int \frac{u^2 + 1}{u^4 - u^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2322595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as ... | Green triangle has 2x area of red triangle. Heights of both triangles are equal because all sides of square are equal. so base of green triangle = 2x base of red triangle.So r=length of base of red triangle and 2r=base of green triangle. Let s= length of side of square also height of both triangles. for red triangle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 11
} |
Show that $2222^{5555} + 5555^{2222}$ is divisible by 7 (without modular arithmetic) I tried using the following approach:
$$x=2222^{5555}+5555^{2222} = (2222^5)^{1111}+(5555^2)^{1111}$$
Now we know $(x^n+y^n)$ is divisible by $(x+y)$ for odd natural number $n$. So,
$$x=(2222^5+5555^2)k,\ k\in N$$
$$x=(1111^2)(32\cdot1... | Because $2226$ divided by $7$, $5551$ divided by $7$,
$a^n-b^n=(a-b)(a^{n-1}+...+b^{n-1})$, for odd $n$ we have $a^n+b^n=(a+b)(a^{n-1}-...+b^{n-1})$ and
$$2222^{5555}+5555^{2222}=$$
$$=(2226-4)^{5555}+4^{5555}+$$
$$+(5551+4)^{2222}-4^{2222}+$$
$$-\left(4^{5555}-4^{2222}\right)$$
and $$4^{5555}-4^{2222}=4^{2222}\left(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving a relation related to quadratic equation Question:If $α$ and $β$ be the roots of $ax^2+2bx+c=0$ and $α+δ$, $β+δ$ be those of $Ax^2+2Bx+C=0$, prove that, $\frac{b^2-ac}{a^2}=\frac{B^2-AC}{A^2}$.
My Attempt: Finding the sum of roots and product of roots for both the equations we get,
$α+β=\frac{-2b}{a}$
$αβ=\... | Instead doing all the hard work you did, you can notice that the difference of roots $(\vert x_1-x_2\vert )$ is same for both the equations. Hence :
$$|\alpha-\beta|=|(\alpha+\delta)-(\beta+\delta)|=\sqrt{(\alpha+\beta)^2-4\alpha \beta}=\sqrt{(\alpha+\delta +\beta+ \delta)^2-4(\alpha+\delta)( \beta+\delta)}$$
$$\implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation:
\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t) \hat k \\ y(t... | $\mathbf u = (\frac {\sqrt 2}{2} \mathbf i - \frac {\sqrt 2}{2} \mathbf j)\\
\mathbf v = (\frac {\sqrt 6}{6} \mathbf i + \frac {\sqrt 6}{6} \mathbf j -\frac {\sqrt 6}{3} \mathbf k)$
$(x,y,z) = (1,1,-1) + 3\mathbf u \cos t + 3\mathbf v \sin t\\
x = 1 + 3\frac {\sqrt 2}{2} \cos t + \frac{\sqrt {6}}{2} \sin t\\
y = 1 - 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
The volume of the water in the hemispherical bowl is given by $V=\frac{\pi}{3}y^2(3R-y)$ when the water is $y$ m deep. Water is flowing out at the rate of $6m^3$/min from a reservoir shaped like a hemispherical bowl of radius $R=13m.$The volume of the water in the hemispherical bowl is given by $V=\frac{\pi}{3}y^2(3R-y... | Since your hemisphere is the lower half of a circle, you can express the radius as the x-coordinate (see image added in edit below),
$$ R^2 = x^2 + y^2 \implies x = \sqrt{R^2 - y^2} $$
now differentiate to find $\frac{dx}{dt}$,
$$ \frac{dx}{dt} = \frac{1}{2} (R^2-y^2)^{-\frac{1}{2}} (-2y) \frac{dy}{dt} = \frac{-y}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $\int \frac{1}{6+(x+4)^2} dx$. $\int \frac{1}{6+(x+4)^2} dx = 6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx$
Now $u=\frac{(x+4)}{\sqrt{6}}$ and $du=\frac{1}{\sqrt{6}}dx$.
$6\int \frac{1}{1+\frac{(x+4)^2}{6}} dx=\frac{6}{\sqrt{6}}\int\ \frac{1}{1+u^2}=\frac{6}{\sqrt{6}}$ arctan$(u)$=$\frac{6}{\sqrt{6}}$ arctan$(\frac{... | You can write it shortly as $$\int \frac { 1 }{ 6+\left( x+4 \right) ^{ 2 } } dx=\frac { 1 }{ 6 } \int \frac { \sqrt { 6 } d\left( \frac { x+4 }{ \sqrt { 6 } } \right) }{ 1+{ \left( \frac { x+4 }{ \sqrt { 6 } } \right) }^{ 2 } } =\frac { \sqrt { 6 } }{ 6 } \arctan { \left( \frac { x+4 }{ \sqrt { 6 } } \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\int_0^{2\pi} \frac{d\theta}{(A+B\cos\theta)^2}$? How to evaluate $\displaystyle\int_0^{2\pi} \frac{d\theta}{(A+B\cos\theta)^2}$?
I know that I can evaluate the integral using residue theorem, but which is the result with passages? Please
| With the substitution $z(\theta) = e^{i \theta}$ so that $d\theta = \frac{dz}{iz}$, we can rewrite the integral as the counterclockwise contour integral
$$
\oint_{|z| = 1} \frac{1}{(A + \frac 12 B(z + z^{-1}))^2} \frac 1{iz}\,dz =
\frac 4i \oint_{|z| = 1} \frac{z}{(Bz^2 + 2Az + B)^2}\,dz
$$
Evaluate this integral usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}?$
$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\left(\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}\right)\tag1$$
Where $\b... | Here is a fairly elementary proof of this result. I derive a closed form equivalent to the OP's.
We begin with the following Lemma:
Lemma $1$: $\sum_{k=1}^p \ln(r*k+n) = \ln\left[r^n\left(\frac{n+r}{r}\right)_p\right] \quad \forall p\in\mathbb{N}$
Proof of Lemma $1$: $\sum _{k=1}^p\ln \left(rk+n\right) = \ln \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
How do I solve $\int\frac{dx}{\sin x+\cos x-1}$? Please help me find the following indefinite integral:
$$\int\dfrac{dx}{\sin x+\cos x-1}$$
I have tried many different substitutions to no avail. Any help is appreciated.
| $$I = \int \frac{1}{\sin x + \cos x - 1}\, dx$$
Apply integral Subtitution
$$u = \tan\left(\frac{x}{2}\right)$$
$$\sin(x) = \frac{2u}{1+u^2},\quad \cos(x) = \frac{1 - u^2}{1 + u^2}\, \quad dx = \frac{2}{1 + u^2}$$
It follows
$$\int\frac{1}{u(-u + 1)}\, du = -\int\frac{1}{u(u - 1)}\, du $$
by partial fraction
$$-\int\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2338996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
$1-x+x^2-x^3+..(-1)^nx^n$ I have the following sum:
$1-x+x^2-x^3+..(-1)^nx^n, x\neq -1$
So what I thought was separating it in two cases like this:
Case 1. n is even
$$
1+x^2+x^4+...+x^n-x(1+x^2+...+x^n)
$$
Which I can turn into $\frac{1-x^{n+2}}{1-x^2}-\frac{x(1-x^{n+2})}{1-x^2}=\frac{1-x^{n+2}}{1-x^2}(1-x)$
Case 2. n... | Call $S(x)$ your sum and note that
$$
S(x)(1+x)=1+(-1)^{n}x^{n+1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$| \cos (x) -\cos (y)| < \frac{y-x}{2}$ if $0\le x < y \le \pi/6$ If $0\leq{x<y<\frac{\pi}{6}}$
I am trying to prove that
$$| \cos (x) - \cos (y)| < \frac{y-x}{2}$$
Using the mean value theorem, defining $g(x)=\frac{y-x}{2}$ and $f(x)=| \cos (x) - \cos (y)|$. Then if show that $h(x)=g(x)-f(x)>0$, I would have proved i... | If $0<y<x<\frac {\pi}{6}$
then $\cos y>\cos x$ since cosine is a decreasing function in this interval.
$|\cos x - \cos y| = -\cos x + \cos y$
Proposition:
$\frac {(x-y)}{2} > - \cos x + \cos y\\
\frac 12 > \frac {- \cos x + \cos y}{x-y}$
Mean value theorem: If $f(x)$ is differentiable in the interval $[x,y]$ there exi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to calculate divergence of the given function? The vector function is:
$$\mathbf{v}=\frac{1}{r^2}\hat{\mathbf{r}}$$
$r$ is the magnitude of position vector and
$\hat{\mathbf{r}}$ is the unit vector along the position vector
Now divergence will be
$$\nabla \cdot \mathbf{v}={\left(\frac{\partial}{\partial x},\frac{\p... | Without switching coordinate systems, this is my favorite method, since it breaks down the identity into small pieces.
Let $\mathbf{r} = x\mathbf{i} + y \mathbf{j} + z \mathbf{k}$, and $r = \sqrt{x^2 + y^2 + z^2}$. Notice that
\begin{align*}
\mathbf{v} &= \frac{\mathbf{r}}{r^3}\\
\mathbf{r}\cdot\mathbf{r} &= r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is it possible factor out $(x - b)$ from $(x^n-b^n)$ when $n$ is a fraction? Factoring out $(x-b)$
But how to factor out $(x-1)$ when the polynomial has the power of a negative integer where $n=-4$
$x^{-4}-1$
And even worst how to perform the factorization when the polynomial is to the power of an fraction? $n=-4/7$
$x... | For your example :
You could call $x^{-1/7}=\frac{1}{x^{1/7}}=y$
So
$$y^4-1=(y-1)(y+1)(y^2+1)=\left(\frac{1}{x^{1/7}}-1\right)(y^3+y^2+y+1)$$
$$=-\frac{1}{x^{1/7}}\left(x^{1/7}-1\right)(y^3+y^2+y+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integration by parts
I have done the 1st part of question and the answer I got is $\frac{5e^4 - 1}{32} $which I verified from the calculator too. But I am confused how to approach to the deducing part using previous result(since it is stated HENCE ). Any help is greatly appreciated.
| $$\int _{ 1 }^{ e }{ { x }^{ 3 }{ \left( \ln { x } \right) }^{ 2 } } dx=\frac { 1 }{ 4 } \int _{ 1 }^{ e }{ { \left( \ln { x } \right) }^{ 2 } } d\left( { x }^{ 4 } \right) ={ \frac { { x }^{ 4 }{ \left( \ln { x } \right) }^{ 2 } }{ 4 } }_{ 1 }^{ e }-\frac { 2 }{ 4 } \int _{ 1 }^{ e }{ { x }^{ 3 } } \ln { x } dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Transform curve into canonical form and determine type I have the curve:
$$9x^2-6xy+y^2+6x-2y-3=0$$
I have to transform it into the nnormal form and determine its type.
Well, we all know that such a curves are given by equation:
$$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F = 0$$
Which is purely the case.
$\textbf{Thoughts:}$
1.) I dete... | Hint:
Use Gauß' algorithm to write this polynomial as a sum of squares:
\begin{align}
9x^2-6xy+y^2+6x-2y-3&=(9x^2-6xy+6x)+y^2-2y-3\\
&=\bigl[(3x-y+1)^2-y^2+2y-1\bigr]+y^2-2y-3\\
&=(3x-y+1)^2-4\\
&=(3x-y-1)(3x-y+3).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Consider the sequence $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2 $ . Does this converge? Consider the sequence $ b_{n}=2+\frac{1}{b_{n-1}} , \ \ b_{0}=2 $ . Does this converge ? If then find the limit . Also the order of convergence $ \eta $.
Answer: \begin{align} b_{0 }=2 \\ b_{1}=2+\frac{1}{2} \\ b_{2}=2+\frac{1}{2+\f... | It's obvious that $b_n>2$.
Thus, $$|b_n-1-\sqrt2|=\left|\frac{1}{b_{n-1}}-\frac{1}{1+\sqrt2}\right|=|\frac{|b_{n-1}-1-\sqrt2|}{(1+\sqrt2)b_{n-1}}<\frac{|b_{n-1}-1-\sqrt2|}{2(1+\sqrt2)},$$
which says that $\{b_n\}$ converges to $1+\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A double-asymptote function I am facing a calculus problem which returns me a function which looks like this:
It seems to me, that it behaves half like a logarithm, half like an exponential. It is, of course, neither of both. It is different.
Some conditions are:
*
*$f\left(\frac{1}{2e}\right) = \frac{1}{2e}$
*$f(... | Here is a possible answer for your question.
As Lambert W function is used in the question, it inspired me to think of the following algebraic expression as the form:
$$(a-bx)^{a-cy}=(a-by)^{a-cx},\tag{1}$$
where $a>e$ and $b,c>0$.
The above is the graph of the expression by substituting $a=\frac 1 e+e$ and $b=c=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Why is $\sum\limits_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \Delta D$? In a book on economics, I have read that $\sum_{t = 1}^{\infty} \frac{r\Delta D}{(1 + r)^t} = \frac{r\Delta D}{r} = \Delta D$.
$t$ is the index of the current time period; $r$ represents the interest rate; $\Delta D$ is the change of debts.
W... | The main theme here is the geometric series expansion
\begin{align*}
\sum_{t=0}^\infty q^t=\frac{1}{1-q}\qquad\qquad |q|<1
\end{align*}
Since OPs series starts with index $t=1$ we consider
\begin{align*}
\sum_{t=1}^\infty q^t&=\left(\sum_{t=0}^\infty q^t\right)-1\\
&=\frac{1}{1-q}-1\\
&=\frac{q}{1-q}\qquad\qquad |q|<1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve $\frac{x+12}{x-3} \gt |x-1| + 1$ Solve $\frac{x+12}{x-3} \gt |x-1| + 1$
I finally got $\frac{(x-6)(x+2)}{x-3} \lt 0$
That means the solutions are $3\lt x \le 6$ and $x\lt -2$ Am I right?
| at first we note that $$x\ne 3$$ and then we do case work:
$$x\geq 1$$ then we have $$\frac{x+12}{x-3}>x$$ then we have to solve
$$\frac{-x^2+4x+12}{x-3}>0$$
solving this we obatin $$3<x<6$$
in the second case we assume
2) $$x<1$$ and this is equivalent to
$$\frac{x+12}{x-3}>2-x$$
can you finish?
the last inequation is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving Pinter 7.B.4 with a program Here is exercise 7.B.4 from 'A Book of Abstract Algebra' by Charles C. Pinter.
A solution to this using a C# program is posted.
Is there another good approach using a computer program?
Any language is welcome.
The subgroup of $S_5$ generated by
$
f = \begin{pmatrix}1 & 2 & 3 & 4 & 5... | You don't need to use a computer program for the computation. The two given maps act nontrivially on disjoint subsets of $\{1, \dots, 5\}$, so the map $\langle{f, g\rangle} \to \mathbb{Z}_2 \oplus \mathbb{Z}_3$ given by $f \to (1,0)$ and $g\to (0, 1)$ is a well-defined isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding the tangent line that intersects a quartic at two points. I'm trying to find a straightforward calc solution to this question. I found an algebraic solution but I don't think it's the quickest way to do it. Thanks in advance!
Find the linear function $g(x)=mx+b$ whose graph is tangent to the graph of $f(x)=x^4-... | Let $h(x)=f(x)-g(x)$ be the distance between the functions. Note that $h(x)$ is a monic quartic polynomial. If the tangent line $g(x)$ intersects the graph of $f(x)$ at $x=c$ and $x=d$, then $h(x)$ must vanish $x=c$ and $x=d$, so the monomials $x-c$ and $x-d$ divide $h(x)$.
Additionally, since the derivatives of $f(x)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Let $a,b,c$ be roots of $x^3+px+r=0$. Find the cubic whose roots are $(a-b)^2$,$ (b-c)^2$ and $(c-a)^2$ Question
Let $a,b,c$ be roots of $x^3+px+r=0$. Then find the cubic whose roots are $(a-b)^2, (b-c)^2$ and $ (c-a)^2$
Attempt
I have tried using Vieta's formulas to compute coefficients of the sought cubic.
For sum of... | HINT.-$$a+b+c=0\\ab+ac+bc=p\\abc=r$$ This gives
$$(a-b)^2+(b-c)^2+(c-a)^2=(c^2+4bc+4b^2)+(a^2+4ac+4c^2)+(b^2+4ab+4b^2)
=5(a^2+b^2+c^2)+4p$$
Since $a^2+b^2+c^2+2(ab+ac+bc)=0$ we have $$(a-b)^2+(b-c)^2+(c-a)^2=-10p+4p=-6p$$ Let $A$ and $B$ the other Vieta coefficients we need so$$X^3+6pX^2+AX+B=0$$ Taking into account w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\text{If } x(x+y+z)=20,y(x+y+z)=30 \text{ and } z(x+y+z)=50 \text{ then the value of } 2(x+y+z) is:$
If $x(x+y+z)=20$, $y(x+y+z)=30$ and $z(x+y+z)=50$ then what is the value of $2(x+y+z)$?
Ans. $20$
I have tried the following:
$$ \frac{20}{x}=\frac{30}{y}=\frac{50}{z}$$
From which I get:
$$x:y:z=2:3:5$$
Now,
$$2(x+y... | Simple, follow the equations you get:
$${y(x+y+z)\over x(x+y+z)}={30\over20} \implies {y\over x}={3\over2}\implies y={3\over2}x\\{z(x+y+z)\over x(x+y+z)}={50\over20} \implies {z\over x}={5\over2}\implies z={5\over 2}x\\x(x+y+z)=20\implies x\left(x+{3\over2}x+{5\over2}x\right)=20\implies{10\over 2}x^2=20\implies x^2=4\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find 3 vectors in $\mathbb{R}^3$ such that the angle between each two is $\pi/3$ Find $3$ vectors $\vec{a},\vec{b},\vec{c}$ in $\mathbb{R}^3$ such that the angle between each two is $\pi/3$
I know that the dot-product of two vectors in $\mathbb{R}^3$ is represented like this $(\vec{a},\vec{b})=|\vec{a}||\vec{b}|\cos(\a... | Let see if we can find the most symmetric triple of vectors all in the positive octant. Let the $3$ unit vectors be
\begin{eqnarray*}
( \alpha, \alpha , \beta) , ( \alpha,\beta ,\alpha , )( \beta,\alpha, \alpha )
\end{eqnarray*}
We require that they are of unit length & their dot products are $1/2$. So we have
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Rotate $xy=1$ by $\frac{\pi}{4}$ in a negative (clockwise) direction. I was studying hyporbolae for the first time and noticed that $y=\frac{1}{x}$ is a rotated hyperbola. I had seen equations like $y=\frac{1}{x}$ before but never noticed they where hyperbolae.
Anyway using geometry and the general form of a hyperbola,... | I think its easier if we keep the variable $y$ around. Then
$$
\begin{pmatrix}
X\\
Y
\end{pmatrix}
=
\begin{pmatrix}
\cos(\pi/4) & -\sin(\pi/4)\\
\sin(\pi/4) & \cos(\pi/4)
\end{pmatrix}
\begin{pmatrix}
x\\
y
\end{pmatrix}
$$
so $X = \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{2} y$ and $Y = \frac{\sqrt{2}}{2} x + \frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:
| At $x=2$ we have
$$y=\frac{4 c}{(2-a) (2-b) (2-c)}+\frac{2 b}{(2-a) (2-b)}+\frac{a}{2-a}+1=-\frac{8}{(a-2) (b-2) (c-2)}$$
Which is
$$y=\frac{8}{-a b c-4 (a+b+c)+2 (a b+a c+b c)+8}$$
In the given equation $x^3 -3x^2 + 2x +4 = 0$ we know the roots $a,b,c$ so we can write $(x-a)(x-b)(x-c)=0$ that is $$x^3-x^2 (a+b+c)+x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Need help in limits that contain arithmetic progression Let $a_n$ be the $n$-th term of an arithmetic progression with the initial term $a_1=2$ and with the common difference 5. That is, $a_n=2+5(n-1).$
Evaluate $$\lim_{n\rightarrow\infty}n(\sqrt{a_n^2+3}-\sqrt{a_n^2-3}).$$
Well I consider that arithmetic progression h... | $$\sqrt{a_n^2+3}-\sqrt{a_n^2-3} = (\sqrt{a_n^2+3}-\sqrt{a_n^2-3})\cdot\frac{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}} = \\ = \frac{a_n^2+3 - (a_n^2-3)}{\sqrt{a_n^2+3}+\sqrt{a_n^2-3}}$$
Simplify this expression, and then also use the fact that
$$\sqrt{an^2 + bn + c} =n\cdot\sqrt{1 + \frac bn +\frac c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How to calculate $\sum_{r=1}^\infty\frac{8r}{4r^4+1}$?
Calculate the following sum: $$\frac{8(1)}{4(1)^4+1} + \frac{8(2)}{4(2)^4+1} +\cdots+ \frac{8(r)}{4(r)^4+1} +\cdots+ \text{up to infinity}$$
MY TRY:- I took $4$ common from the denominator. and used $a^2+b^2=(a+b)^2-2ab$. It gave me two brackets, whose subtractio... | We can use partial fraction decomposition to write this as \begin{align}\sum_{n=1}^\infty \frac{8n}{4n^4+1}&=\sum_{n=1}^\infty\left(\frac{2}{2 n^2 - 2 n + 1} - \frac{2}{2 n^2 + 2 n + 1}\right)\end{align}
If we start writing out terms of this sequence, we get \begin{array}{ccccccccccc}(n=1)&2&-&\frac{2}{5}\\
(n=2)&&+&\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
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An expression involving the roots of a quadratic polynomial I have the equation $$ 9x^2 - 11x + 1 = 0 $$ whose two roots are $ \alpha $ and $ \beta $ .
I need to evaluate $$ \frac 1 {(9\alpha-11)^2} + \frac{11\beta - 1} 9$$
What I've tried
*
*Expanded the denominator and add them, but nothing simplifies and I get e... | Just to show another way:
$$
\eqalign{
& 9\alpha ^{\,2} - 11\alpha + 1 = 0\quad \Rightarrow \quad \alpha \left( {9\alpha - 11} \right) = - 1\quad \Rightarrow \quad \left( {9\alpha - 11} \right) = - 1/\alpha \cr
& 9\beta ^{\,2} - 11\beta + 1 = 0\quad \Rightarrow \quad 11\beta - 1 = 9\beta ^{\,2} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Transition from exponential form to logarithmic form in an inequality This is my first question on math stack exchange and I am happy to join the community!
I would like to ask if someone could explain me in detail, how this transition is made (because I could not find any reference in the book):
$$\left(1-\frac{2}{n}\... | It may be more complicated, as there are counter-examples to your statement, such as
*
*$n=1,k=3$ in which case $\left(1-\frac{2}{n}\right)^k = -1 \lt \frac13$
*$n=2, k=2$ in which case $\left(1-\frac{2}{n}\right)^k = 0 \lt \frac13$
*$n=-1, k=-2$ in which case $\left(1-\frac{2}{n}\right)^k = \frac19 \lt \frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2357885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the minimum of the value $3x^2-2xy$ if $\frac{x^2}{4}-y^2=1$ Let $x,y\in R$,such
$$\dfrac{x^2}{4}-y^2=1$$
find the minium of the
$$3x^2-2xy$$
I think $x=2\sec{t},y=\tan{t}$,then
$$3x^2-2xy=12(\sec{t})^2-4\sec{t}\tan{t}$$
| This sure looks like a Lagrange multiplier problem to me. Let $f(x,y) = 3x^2-2xy$ and $g(x,y) = \frac{x^2}{4}-y^2.$ Then $\nabla f = (6x-2y,-2x)$ and $\nabla g = (\frac{x}{2}, -2y)$. We get the system of 3 equations in three variables:
$$6x-2y = \lambda \frac{x}{2}$$
$$-2x = \lambda (-2y)$$
$$\frac{x^2}{4}-y^2 = 1$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
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$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:
$$\sqrt{9x^2-16}>3x+1$$
Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\lo... | First the domain of the inequation is $(-\infty,-4/3]\cup[4/3,+\infty)$.
Next, you should know that, when $A\ge 0$,
$$\sqrt A>B\iff A >B^2\quad\text{or}\quad B<0.$$
In the present case, one gets
\begin{cases}
9x^2-16>(3x+1)^2\iff x<-\dfrac{17}6 \\
\quad \text{or}\\
x<-\dfrac13.
\end{cases}
Both conditions yield $x\in (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Sum of square root of $n$ primes as a nested square root $$\sqrt{2}+\sqrt{3}=\sqrt{5+\sqrt{24}}$$
$$\sqrt{2}+\sqrt{3}+\sqrt{5}=\sqrt{14+\sqrt{140+\sqrt{4096+\sqrt{8847360}}}}$$
These are two examples of how sum of k square root of primes (not necessarily consecutive) can be represented as nested square roots.
Is there... | The first it's just
$$\sqrt{p}+\sqrt{q}=\sqrt{p+q+2\sqrt{pq}}=\sqrt{p+q+\sqrt{4pq}}.$$
For the second and for the rest maybe the following would help.
$$\sqrt2+\sqrt3+\sqrt5=\sqrt{10+2(\sqrt{6}+\sqrt{10}+\sqrt{15})}=$$
$$=\sqrt{14+2(\sqrt{6}+\sqrt{10}+\sqrt{15}-2)}=$$
$$=\sqrt{14+2\sqrt{35+2\sqrt{10}+6\sqrt6}}=$$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving that $ \sum_{i=1}^{b-1} \left\lfloor \frac{a}{b}i \right\rfloor = \sum_{j=1}^{a-1} \left\lfloor \frac{b}{a}j \right\rfloor $ I am trying to solve the following problem:
Let $a$ and $b$ be integers greater than one which have no common divisors. Prove that:
$$ \sum_{i=1}^{b-1} \left\lfloor \frac{a}{b}i \right\r... | Since $a$ and $b$ are coprime, the numbers $0,a,2a,\ldots,(b-1)a$ form a complete set of residues modulo $b.$ Hence, if $\{ia/b\}$ denotes the fractional part of $ia/b,$ and if $i\neq j$ are any two distinct elments of $\{0,1,\ldots,b-1\},$ then $b^{-1}\{ia/b\}\neq b^{-1}\{ja/b\},$ which implies that each $b^{-1}\{ia/b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A generalized Ahmed's integral Let $\vec{A}:=(A_1,A_2,A_3)$ be a vector where all its components are positive real numbers.
In the context of this question An integral involving error functions and a Gaussian we came across a following integral.
\begin{equation}
I(\vec{A}) := A_1\int\limits_0^{A_3} \frac{ \arctan\left... | Define $\phi:= \arccos(A_2/\sqrt{1+A_1^2+A_2^2})$ and $\alpha := \sqrt{1+A_1^2}-A_1$ and $\beta:=\sqrt{1+A_1^2}+A_1$ and
\begin{eqnarray}
&&{\mathcal F}^{(a,b)}(t):=\int \arctan(\frac{t}{a}) \frac{1}{t-b} dt = \log(t-b) \arctan(\frac{t}{a})\\
&&-\frac{1}{2 \imath} \left( \log(t-b) \left[ \log(\frac{t-\imath a}{b-\imath... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Functional equation: $ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) $
Exercise: Find all of functions $f: \mathbb{R} \to \mathbb{R}$ such that $$ f \left( x + y ^ 2 + z ^ 3 \right) = f ( x ) + f ^ 2 ( y ) + f ^ 3 ( z ) \text , \forall x , y , z \in \mathbb R \text . $$
In fact, I solved th... | The case $ f ( 0 ) = 0 $ is solved by @TobEnrack. For the case $ f ( 0 ) = -1 $, let $ x = y = 0 $ in
$$ f \big( x + y ^ 2 + z ^ 3 \big) = f ( x ) + f ( y ) ^ 2 + f ( z ) ^ 3 \tag 0 \label 0 $$
and you get
$$ f \big( z ^ 3 \big) = f ( z ) ^ 3 \tag 1 \label 1 $$
Again, letting $ x = z = 0 $ in \eqref{0} you have
$$ f \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find angles in a triangle, with two similar triangles with scale factor $\sqrt{3}$
Triangle ABC has point D on BC which creates triangle ABD and ACD. They differ with the scale factor $\sqrt{3}$. What are the angles?
I know ADB and ADC cannot be right, as it shares the side AD and cannot have a scale factor of $\sqrt... | A solution is given by dropping an altitude to the hypotenuse of a 30-60-90 right triangle. Is this the only one? To find out we define the six quantities $a = BC$, $b = CA$, $c = AB$, $d = AD$, $m = BD$, and $n = DC$. There are four ways in which the triangles $ABD$ and $ACD$ could be similar:
\begin{align}
1)& \;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Show that $\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
I want to prove that $\displaystyle\int_0^\pi\frac{2\cos(2\theta)+\cos(3\theta)}{5+4\cos(\theta)}d\theta=\frac{\pi}{8}$
My ideas, I don't know if they lead anywhere:
Let's substitute $\cos(\theta)=\frac{e^{i\theta}+e^{-i\t... | Looks like there should be a partial fraction decomposition:
$\frac{z^2+z^{-2}+\frac{1}{2}z^3+\frac{1}{2}z^{-3}}{5z+2z^2+2} = Az + B + \frac C{z} + \frac D{z^2} + \frac E{z^3} + \frac {F}{z+2} + \frac {G}{2z + 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understa... | An alternative solution ...
We have $n^4-n^2+64=A^2$. Multiply this by $4$ and complete the square $(2n^2-1)^2+255=4A^2$. So
\begin{eqnarray*}
(2A+2n^2-1)(2A-2n^2+1)= 3 \times 5 \times 17.
\end{eqnarray*}
This gives
\begin{eqnarray*}
2A+2n^2-1 = x \\
2A-2n^2+1= y
\end{eqnarray*}
There are $4$ possible values for $(x,y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 0
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Peculiar Sum regarding the Reciprocal Binomial Coefficients Whilst playing around on Wolfram Alpha, I typed in the sum
$$\sum_{x=0}^\infty \frac{1}{\binom{2x}{x}}=\frac{2}{27}(18+\pi\sqrt 3)$$
I'm not sure how to derive the answer. My first instinct was to expand the binomial coefficient to get
$$\sum_{x=0}^\infty \fra... | \begin{eqnarray*}
\binom{2n}{n} ^{-1} = \frac{2n+1}{2^{2n+1}} \int_{-1}^{1} (1-x^2)^n dx
\end{eqnarray*}
Substitute this for summand and inerchange the order of the integral and sum.
\begin{eqnarray*}
\int_{-1}^{1} \sum_{n=0}^{ \infty} \frac{2n+1}{2^{2n+1}} (1-x^2)^n dx &=& \frac{1}{2} \int_{-1}^{1} \left(2\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
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Find the value of $\Biggl(\Biggl(\tan^2\frac{7\pi}{24}-\tan^2\frac{\pi}{24}\Biggr):\Biggl(1-\tan^2\frac{7\pi}{24}\tan^2\frac{\pi}{24}\Biggr)\Biggr)^2$ This problem reminds me of the formula $\tan(x\pm y)=\frac{\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}.$ But two minus signs in the problem interfere in applying the formul... | Hint: $A^{2}-B^{2} = (A-B)(A+B)$
Further hint:
Use the hint to rewrite the expression as $\left(\frac{\tan{(7\pi/24)}-\tan{(\pi/24)}}{1+\tan{(7\pi/24)}\tan{(\pi/24)}}\cdot\frac{\tan{(7\pi/24)}+\tan{(\pi/24)}}{1-\tan{(7\pi/24)}\tan{(\pi/24)}}\right)^{2}$
Answer:
The addition formula for $\tan$ then gives you the an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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If $\;\log_m mn + \log_n mn = 49$, find $(\log_m n)^{3/4} + (\log_n m)^{3/4}.$ Given that $\log_m mn + \log_n mn = 49,\;$ find the value of $$(\log_m n)^{3/4} + (\log_n m)^{3/4}.$$
I have rewritten this expression but have hit a dead end.
| Let $\sqrt[4]{\log_nm}=x$. Hence, $x>0$ and $$ x^4+\frac{1}{x^4}=47 \iff
\left(x^2+\frac{1}{x^2}\right)^2=49$$
$$\iff x^2+\frac{1}{x^2}=7$$
$$\iff\left(x+\frac{1}{x}\right)^2=9$$
$$\iff x+\frac{1}{x}=3.$$
Thus, $$\left(\log_m n\right)^{3/4} + \left(\log_n m\right)^{3/4}=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Region of Convergence for Laurent Expansion Find the Laurent expansion of $\frac{z}{(z+1)(z+2)}$ about the singularity $z=-2$. Specify the region of convergence and the nature of singularity at $z = -2$.
The Laurent expansion I get is
$1+(z+2)+(z+2)^2+ \ldots + \frac{2}{z+2}$
The singularity is a simple pole.
But how... | The Laurent series can be obtained as
$$\frac{z}{(z+1)(z+2)} = \frac{1}{z+2}\frac{z+2 - 2}{z+2 - 1} \\ = \left(\frac{2}{z+2} - 1\right)\frac{1}{1 - (z+2)} \\ = \left(\frac{2}{z+2} - 1\right)\sum_{n=0}^\infty (z+2)^n$$
and the geometric series on the RHS converges when $|z+2| < 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Prove $3^n$ divides any number formed from $3^n$ identical digits I know this is a repost of this, however I had some trouble understanding the answers as I am new to number theory. I asked my doubt in a comment but got no reply.
One of the answers says:
A number whose digits are all equal and of length $3^n$ is thus ... | We have
$$\frac{10^{3^{n+1}}-1}9=\frac{a^3-1}9=\frac{10^{3^n}-1}9(a^2+a+1)$$
where $a=10^{3^n}$. The induction step boils down to confirming that
$3$ divides $a^2+a+1$. If we could prove that $a\equiv1\pmod 3$ we could
confirm that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do we know that these roots are non-real? The roots, x and y, of a quadratic equation $$ax^2 + bx + c=0$$ satisfy the following inequality:
$$x^2 + y^2 < 0$$
What can you conclude about the nature of the roots of the quadratic equation?
$$x^2 + y^2 = (x+y)^2 - 2xy = \frac{b^2 - 2ac}{a^2}$$
since $$x^2 + y^2 <0$$, ... | Alternatively:
$$x^2+y^2=\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2=\frac{b^2-4ac}{a^2}<0 \Rightarrow$$
$$b^2-4ac<0 \Rightarrow x,y\in \mathbb{C}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$. Question:
Let $\ a,b,c$ be integers such that $ \ a^{2} + b^{2} = c^{2}$. If $c$ is divisible by $3$, prove that $a$ and $b$ are both divisible by $3$.
My attempt:
Proof by cont... | We have $a^2+b^2=c^2\equiv 0\pmod 3$. Since $a^2,b^2\equiv 0$ or $1\pmod 3$ we must to have $a\equiv b\equiv 0\pmod 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the value of $x^2 + y^2 + z^2$ if $x^2 y + y^2 z + z^2 x=2186$ and $xy^2 + y z^2 + z x^2=2188$
What is the value of $x^2 + y^2 + z^2$ if $x^2 y + y^2 z + z^2 x=2186$ and $xy^2 + y z^2 + z x^2=2188$, where $x,y,z$ are integers.
My Attempt
$(x^2 y + y^2 z + z^2 x) +(x^2 y + y^2 z + z^2 x) =2186+2188=4374$.
... | Subtracted the first equation from the second:
$$\left(x^2 z+x y^2+y z^2\right)-\left(x^2 y+x z^2+y^2 z\right)=2$$
Simplify and factor:
$$-x^2 y+x^2 z+x y^2-x z^2-y^2 z+y z^2=(x-z)(-x y+x z+y^2-y z)= (x - z)(y - x) (y - z)$$
Because $x$, $y$ and $z$ are integers, $x-z$, $x-y$ and $z-y$ must be integers too, and the onl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Suppose that you have a set of 12 colored balls, two each of six different colors $C_{1}$ through $C_{6}$... Suppose that you have a set of 12 colored balls, two each of six different colors $C_{1}$ through $C_{6}$. Find the number of six-ball combinations if balls of the same color are considered identical.
Using the... | Generating Function Approach
$$
\begin{align}
\left[x^6\right]\left(1+x+x^2\right)^6
&=\left[x^6\right]\left(\frac{1-x^3}{1-x}\right)^6\\
&=\left[x^6\right]\sum_{k=0}^6(-1)^k\binom{6}{k}x^{3k}\sum_{j=0}^\infty(-1)^j\binom{-6}{j}x^j\\
&=\sum_{k=0}^2(-1)^k\binom{6}{k}(-1)^{6-3k}\binom{-6}{6-3k}\\
&=\sum_{k=0}^2(-1)^k\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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$\lim\limits_{n \rightarrow \infty} \sqrt{n^2+n} -n$?
Calculate
$\displaystyle\lim_{n \to \infty}
\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right)$.
$\displaystyle\lim_{n \to \infty}\left(\,{\sqrt{\,{n^{2} + n}\,} - n}\,\right) =
\infty - \infty$ We have an indeterminate form
So I proceeded to factorize $$\sqrt{n^2+n... | Trick is to write $\sqrt{n^2 + n} - n$ in a nice form.Below we use rationalization
$\lim _{n\rightarrow \infty}\sqrt{n^2 + n} - n = \lim_{n \rightarrow \infty} \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \lim_{n \rightarrow \infty} \sqrt{n}(\frac{1}{\sqrt{n+1}+\sqrt{n}}) = \lim_{n \rightarrow \infty} \frac{\sqrt{n}(1)}{\sqrt{n}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c)$ I am stuck with the following problem.
If $a>0$, $b>0$, $c >0$ and not all equal then prove that:
$$b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c).$$
Additional info:I'm looking for solutions using AM-GM .
I don't know how to progress .
I will be grateful if som... | From C-S:
$$\left(\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\ge \left(a+b+c\right)^2 \Rightarrow $$
$$\left(a^2b^2+b^2c^2+c^2a^2\right)^2\ge a^2b^2c^2(a+b+c)^2.$$
equality occurs when $a=b=c.$
Taking square root and $a\ne b\ne c$ will do.
| {
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"url": "https://math.stackexchange.com/questions/2384503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating a Discontinuous Integral How do you evaluate this integral? $$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$
My work:
The expression $x^2 - 2x - 3$ is discontinuous at $x = 3$ in the interval $x = 0$ to $x = 4$, so I got to integrate the expression like this:
$$\int _0 ^4\frac{dx}{x^2 - 2x - 3}$$ is equal to
$$\int _0 ... | $$I=\int _0 ^4\frac{dx}{x^2 - 2x - 3} =\frac{1}{4} \int _0 ^4\left(\frac{1}{x - 3}- \frac{1}{x+1}\right)dx$$
As already pointed out, there is a singularity at $x=3$.
So, at the common sens, the integral isn't convergent. But this is not true in the sens of Cauchy integration (Cauchy Principal Value : http://mathworld.w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $P(a)=0 \Rightarrow P(a+1)=1$ then $P(x)$ has no repeated roots.
Let $P(x) \in \mathbb{R}[x]$ be polynomial with all real roots and has the property that $P(a)=0 \Rightarrow P(a+1)=1$ for all $a \in \mathbb{R}$. Prove that $P(x)$ has a repeated root.
I think this problem statement is not true because if we suppose... | For $n>1$ we have $P(x)=(x-a)^n$ then $P(a+1)=(a+1-a)^n=1$
If $P(x)$ has no repeated root then it doesn't happen in general
Anyway carat, the original poster, is right: for $n=1$ it is false. It even makes no sense talking of multiple root for $n=1$
Suppose the root are $3$ then we have
$P(x)=(x - a) (x - b) (x - c)$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
is there any analytical way to konw if $\frac{1}{2x}+\frac{x}{2} >1$ for $(1,\infty)$ or $(0,\infty)$? I was solving a physics problem and I got
$$\frac{a^2+b^2}{2ab}$$
or equivalents
$$\frac{1+ \left( \frac{b}{a}\right)^2}{2\cdot\frac{b}{a}}.$$
If $x=\frac{b}{a}$, we obtain
$$\frac{1+x^2}{2x}$$ or $$\frac{1}{2x}+\fra... | For $x > 0$ (and $x$ must be greater than zero):
$\frac {1}{2x} + \frac x2 > 1 \iff$
$\frac 1x + x > 2 \iff$
$1 + x^2 > 2x \iff $
$x^2 - 2x + 1 > 0 \iff$
$(x - 1)^2 > 0 \iff$
$x - 1 \ne 0 \iff$
$x \ne 1$.
That's it. If $x = 1$ then $\frac 1{2x}+ \frac x2 = 1$ other wise $\frac 1{2x}+ \frac x2 > 1$
If $x < 0$ well ...... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Closed-form for Floor Sum 1 Does a closed form exist for the following sum?
$$\sum_{k=0}^n \lfloor \sqrt{k} + \sqrt{k + n} \rfloor$$
If not, why is this sum so radically different than the sums below?
Closed forms do exist for the following sums*:
$$\sum_{k=0}^n \lfloor \sqrt{k + n} \rfloor$$
$$\sum_{k=0}^n \lfloor \sq... | Hint:
Such sums are made of runs of equal values, that are delimited by the indexes such that the general term crosses an integer.
$$\sqrt k+\sqrt{k+n}=m$$ when
$$k+2\sqrt k\sqrt{k+n}+k+n=m^2$$
or
$$4k(k+n)=(m^2-n-2k)^2$$
or
$$k=\frac{(m^2-n)^2}{m^2}=m^2-2n+\frac1{m^2}.$$
As $k$ is an integer, the final fraction can be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Convergence of a series depending on a parameter. I was studying the convergence of a series with a parameter and I want to ask you if my conclusion is correct and if there is a better method to do it.
$$\sum_{n=1}^{\infty}\left(\frac{\pi}{2}-\arcsin\frac{n}{n+4} \right)^{\alpha}$$
I'm asked to say for which $\alpha$ t... | $\sin(x) = \sin(\frac{\pi}{2}) + \cos\left(\frac{\pi}{2}\right)\left(x - \frac{\pi}{2}\right) - \frac{1}{2}\sin\left(\frac{\pi}{2}\right)\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow \frac{\pi}{2}}\left(\left(\frac{\pi}{2} - x\right)^2\right) \\ = 1 - \frac{1}{2}\left(\frac{\pi}{2} - x\right)^2 + o_{x \rightarrow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
Find the curve of intersection between $x^2 + y^2 + z^2 = 1$ and $x+y+z = 0$
My attempt:
*
*$x^2 + y^2 + z^2 = 1$
*$x+y+z=0$
$$(2) \implies z = -(x+y)$$
$$(1) \implies x^2+y^2+(x+y)^2 = 1$$
$$2x^2 + 2y^2 + 2xy = 1$$
This is the curve i... | HINT: it is $$y_{1,2}=-\frac{x}{2}\pm\sqrt{\frac{1}{2}-\frac{3}{4}x^2}$$ and with that $y$ we can compue $z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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The largest integer that divides $p^4-1$, in which p is a prime greater than $5$ The question is to find the largest integer that divides all $p^4-1$, where p is a prime greater than 5. Being asked this question, I just assume this number exists. Set $p = 7$, then $p^4-1=2400$. I don't have any background in number the... | Let $n$ be the largest integer that divides $p^4-1$ for all prime $p\geq 7.$
We have $11^4-1=14640$ and $7^4-1=2400.$ The $gcd$ of $14640$ and $2400$ is $240.$ So $$n\leq 240.$$ If $p$ is odd then modulo $16$ we have $p^4\in \{(\pm 1)^4, (\pm 3)^4,(\pm 5)^4,(\pm 7)^4\}=\{1^2,9^2, 25^2, 49^2\}=\{1^2,9^2,9^2,1^2\}=$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Solve $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}+\frac{1}{z}$
Solve $$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}+\frac{1}{z},$$ where $x, y, z$ are positive integers.
So far I have:
If $x=y=z \Rightarrow(x, y, z) = (2, 2, 2)$.
If $x=z \Rightarrow (x, y, z) = (k, 2, k)$, where $k$ is natural.
If $y=z \Rightarrow (x, y, z) = (2, ... | By symmetry, we assume $x\leq y$ throughout this discussion.
Cases:
If $x=1$, then the equation simplifies to $\frac{1}{2}+\frac{1}{y}=\frac{1}{z}$. Then $z=1$ since otherwise the LHS is larger than the RHS. When $z=1$, we have $y=2$.
If $x=2$, then the equation simplifies to $\frac{1}{y}=\frac{1}{z}$, which can be s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2397190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluate the given limit: Evaluate the given limit:
$$\lim_{x\to 1} \dfrac {1+\cos \pi x}{\tan^2 \pi x}$$
My Attempt:
$$=\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \times \dfrac {\cos^2 \pi x}{\sin^2 \pi x}$$
$$=\lim_{x\to 1} (1+\cos \pi x) \cos^2 \pi x (\... | $\lim_{x\to 1} \dfrac {1+\cos \pi x}{\dfrac {\sin^2 \pi x}{\cos^2 \pi x}}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{\sin^2 \pi x}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{1-\cos^2 \pi x}=$
$=\lim_{x\to 1} \dfrac {(1+\cos \pi x)(\cos^2 \pi x)}{(1-\cos \pi x)(1+\cos \pi x)}=$
$=\lim_{x\to 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Is $f(x)=\frac{x^2}{1+x^2}$ bijective? $f: \mathbb{R} \to \mathbb{R}$ via $f(x) = \frac {x^2}{1+x^2}$
Injective:
Suppose f(a) = f(b)
$$ \implies \frac {a^2}{1+a^2} = \frac {b^2}{1+b^2} \implies a^2 +a^2b^2 = b^2 +b^2a^2$$
$$ \implies a^2 = b^2 \implies a \neq b$$
Surjective: If $x<0$, the image of $f$ is $[0, \infty)$
... | $f $ is an even function
$$\implies f (1)=f (-1)=\frac {1}{2} $$
$\implies f $ is not injective
$\implies f $ is not bijective.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If one root of the equation $ax^2+bx+c=0$ be the square of the other. If $a \neq 0$ and if one root of the equation $ax^2+bx+c=0$ is the square of the other, prove that: $$b^3+a^2c+ac^2=3abc.$$
My Attempt:
Given:
$$ax^2 + bx + c=0$$
Let $\alpha $ and $\beta $ be the roots of the equation.
$$\alpha + \beta = \dfrac {-b}... | $$\alpha + \beta = \dfrac {-b}{a}$$
$$\alpha . \beta = \dfrac {c}{a}$$
Substitute
$$\alpha = \beta^2$$
into previous equations,
$$a(\beta^2+\beta) = a\beta(\beta+1) = - b\tag{1}$$
$$a\beta^3 = c\tag{2}$$
Cube equation $(1)$,
$$a^3 \beta^3 (\beta^3 + 3 \beta^2 + 3 \beta + 1) = -b^3$$
Using equation $(2)$,
$$a^2 c (\bet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2398823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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solving $\cos z + \sin z = i$ I found a solution but when testing it, it does not work....?
$$\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\righ... | You may have applied the quadratic form incorrectly as I can see.
You have:
$$e^{2iz}(1+i)+2e^{iz}-1+i=0$$
with $a=1+i,b=2,c=-1+i$.
$$\implies e^{iz}=\frac{-2\pm \sqrt{2^2-4(i+1)(i-1)}}{2(1+i)}=\frac{-2\pm\sqrt{4-4(i^2-1)}}{2(i+1)}=\frac{-2\pm\sqrt{4-4(-2)}}{2(i+1)}$$
$$=\frac{-2\pm\sqrt{4+8}}{2(i+1)}=\frac{-2\pm\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2399959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\triangle BAC$ is a right-angled isosceles triangle Suppose in the figure below, $AD\ \text{//} \ BC$, $BD=BC$, $CD=CE$,
and $ABCD$ is a trapezoid; $\measuredangle ABD=15°$
Prove that: $\triangle BAC$ is a right-angled isosceles triangle.
| Let $F$ the intersection of lines $BA$ and $CD$, $\angle BCD= x$ and $BC=L$.
Then
$$\angle BFC= x-\frac{\pi}{12}. \tag{1}$$
Applying some trig we get:
$$CD=2L\cos x \tag{2}$$
$$DE =4L(\cos x)^2 \tag{3} $$.
Applying sine rule in triangle $DAE$ we get:
$$DA=\frac{4L(\cos x)^2\sin x}{-\sin(3x)} \tag{4} $$
The similarit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2402637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.
I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{... | \begin{eqnarray*}
p_2(x) &=& (x^6+1)(x^5+x^4+x^3+x^2+x+1)\\
&=& (x^6+1)(x^3+1)(x^2+x+1)\\
&=& (x^6+1)(x+1)(x^2-x+1)(x^2+x+1)\\
&=& (x+1)(x^6+1)(x^4+x^2+1)
\end{eqnarray*}
\begin{eqnarray*}
p_1(x) &=& (x^{12}+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\
&=& (x^{12}+1)(x^6+1)(x^4+x^2+1)
\end{eqnarray*}
Writ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2404083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer. Given
$$(1+x)^n= \binom {n}{0} + \binom{n}{1} x+ \binom{n}{2} x^2+ \cdots + \binom {n}{n} x^n.$$
Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer.
I tried to u... | Hint: What is $(1+1)^n+(1-1)^n+(1+i)^n+(1-i)^n$?
| {
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"url": "https://math.stackexchange.com/questions/2404878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a... | hint AM-GM
$$(a+b)^2-a^2-b^2=2ab\le a^2+b^2$$
$$\implies 1-a^2-b^2\le (a^2+b^2) $$
$$\implies 1\le 2 (a^2+b^2). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$ or $a^3-b^3$ will be divisible by $7$. This is a very interesting word problem that I came across in an old textbook of mine:
Show that, for any two integers $a$ and $b$, at least one of the expressions $a^3, b^3, a^3+b^3$... | Hint:
Note that, $a^3\equiv 0,1^3,2^3,3^3,4^3,5^3,6^3(\mod 7)\Rightarrow a^3\equiv 0,1,1,-1,1,-1,-1(\mod 7)$
More explanation :
Then $7\mid a\Rightarrow 7\mid a^3$ or $7\mid b\Rightarrow 7\mid b^3$.
If $7$ divides none of $a,b$, then either $a^3+b^3\equiv 0(\mod 7)$ or $a^3-b^3\equiv 0(\mod 7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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What are some mathematically interesting computations involving matrices? I am helping designing a course module that teaches basic python programming to applied math undergraduates. As a result, I'm looking for examples of mathematically interesting computations involving matrices.
Preferably these examples would be ... | Here's the polynomial derivative matrix (where $n$ is the degree of the polynomial):
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}\newcommand{\i}{\mathrm{i}}\newcommand{\text}[1]{\mathrm{#1}}\newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 26,
"answer_id": 7
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Factorization and GCD for Rijndael's finite field I'm factoring polynomials in $GF(2^8)$ with modulo polynomial $m = 2^8 + 2^4 + 2^3 + 2^1 + 2^0$
In particular, I factored
a = 0x49 = $$2^6 + 2^3 + 2^0 = 2^1 * (2^1 + 2^0)^6 * (2^2 + 2^1 + 2^0) * (2^4 + 2^1 + 2^0) * (2^3 + 2^1 + 2^0) \bmod{m}$$
b = 0x64 = $$2^6 + 2^5 + 2... | A fundamental problem is that while there are primes in the ring of polynomials $GF(2)[x]$ (where you can also run the (extended) Euclidean algorithm, there are no primes in the field $GF(2^8)=GF(2)[x]/\langle m(x)\rangle$. The same holds in all fields. Technically, all the non-zero elements of a field are units, and h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number of real solutions of $f(f(x))$ Given $f(x)=x^3-12x+3$, find number of real solutions of $f(f(x))=0$.
My Try: Since $f(x)$ is having three real toots say $\alpha$, $\beta$ and $\gamma$, we have $$f(f(\alpha))=f(f(\beta))=f(f(\gamma)) =f(3).$$
Hence by Rolles theorem $\exists$ atleast one $c $ such that $f'(c)=0$
... | $f(x)=x^3 - 12 x + 3$ has $3$ real solutions, as you stated: $\alpha,\beta,\gamma\in\mathbb{R}$
$f(f(x))=0 $ means $\left(x^3-12 x+3\right)^3-12 \left(x^3-12 x+3\right)+3=0$
translates in three $3-$rd degree equations
$\alpha ^3-12 \alpha +3=\alpha;\;\beta ^3-12 \beta +3=\beta;\;\gamma^3-12 \gamma +3=\gamma$
that is
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to solve a matrix system using Gauss elimination $$\left(\begin{array}{ccc|c}
-1 & 2 & 1 & 3\\
3 & \alpha & -2 & \beta\\
-1 & 5 & 2 & 9
\end{array}\right)$$
I am struggling to solve this system $Ax=b$. I understand the basics of Gauss elimination but am not sure how to handle it with the alpha and beta. It needs to... | $$\det\left(
\begin{array}{rrr}
-1 & 2 & 1 \\
3 & a & -2 \\
-1 & 5 & 2 \\
\end{array}
\right)=-3-\alpha$$
if $-3-\alpha\ne 0$ that is $\alpha\ne -3$ there exists one and only one solution
$$\left\{\frac{3 a-b+6}{a+3},\frac{b+3}{a+3},\frac{3 (2 a-b+3)}{a+3}\right\}$$
if $\alpha=-3$ the determinant of the matrix $A$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maclaurin Series Expansion $\ln(8+x^3)$.
Find Maclaurin Series Expansion of function $\ln(8+x^3)$. For which $x$ series converges? For which $ x $ ,$ f(x)$ equals sum of this series?
$$ (\ln(8+x^3))'=\frac{3x^2}{8+x^3} =\frac{\frac{3}{8}x^2}{1-(-\frac{x^3}{8})}=$$
$$ =\sum_{n=1}^{\infty}\frac{3x^2}{8}(-1)^n\frac{x^{... | HINT:
$$\ln(8+x^3)=\ln8+\ln\left(1+\dfrac{x^3}8\right)$$
See Why does the taylor series of $\ln (1 + x)$ only approximate it for $-1<x \le 1$?
OR
Taylor series for $\log(1+x)$ and its convergence
OR
What is the correct radius of convergence for $\ln(1+x)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2412952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the minimum value of $a^2+b^2+c^2$
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$.
I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me $a+b+c$, wh... | Scalar product:
Let $\vec A : = (a,b,c)$, $\vec B: = (b,c,a)$.
$|\vec A \cdot \vec B| \le |A| |B| $.
$\Rightarrow$ :
$2 \le ab +bc + ca \le a^2 +b^2 + c^2$.
Equality:
$a=b=c = \sqrt{\frac{2}{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2413134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.