Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find real parts of the complex roots of this $9^{th}$ order polynomial in explicit form I have a following polynomial. (See WolframAlpha ):
$$x^9-6x^8+14x^7-16x^6+36x^5-56x^4+ 24x^3-320x+\frac{640}{9}=0 \tag{1}$$
Wolfram says that $(1)$ has three real roots and three pairs of complex conjugate roots.
I know one of the ... | The other two real roots, quite expectedly, are $\tan \left({\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$ and $\tan \left({2\pi\over3}+\frac{1}{3} \arctan \frac{1}{3} \right)+\tanh \left(\frac{1}{3} \text{arctanh} \frac{1}{3} \right)$. The complex roots ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve in integers the equation $\sqrt{x^3-3xy^2+2y^3}=\sqrt[3]{13x+8}$
Solve in integers the equation
$$\sqrt{x^3-3xy^2+2y^3}=\sqrt[3]{13x+8}$$
My work so far:
I used www.wolframalpha.com. Then $x=9,y=8 -$ solution.
My attempt:
1) Let $\sqrt{x^3-3xy^2+2y^3}=a, \sqrt[3]{13x+8}=b$. Then
$$\begin{cases}
a=b\\
a^2-b^... | There are no integer solutions with $x<0$, since $13x+8<0$ for such $x$. When $0\leq x\leq 8$ then $13x+8$ is not an integer cubic. Since the latter is easily seen to be a necessary condition it follows that there are no solutions with $x$ in this range. On the other hand you already have found the solution $(9,8)$. In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displa... | Expanding $$A= \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$$ we get $$A=\frac{a^3-a^2 (b+c)-a \left(b^2-3 b c+c^2\right)+(b-c)^2 (b+c)}{a b c}$$ Replace $c$ by $-(a+b)$, expand and simplify.
The result is a whole number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and ... | When $y> 0$, you have
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{y}.$$
When $y<0$, you have
$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{-y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Factoring $x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$ For a quintic polynomial to be reducible to the following form over $\mathbb{Q}$:
$$x^5+B x^4+C x^3+D x^2+E x+F=(x^2+a x+b)(x^3+p x+q)$$
we need to match the coefficients ($a=B$ obviously, so we write the system of equations for the rest):... | The statement is true.
You already proved the following :
If a quintic polynomial in the form:
$$x^5+B x^4+C x^3+D x^2+E x+F,~~~~~B,C,D,E,F \in \mathbb{Q}$$
can be factored into $(x^2+a x+b)(x^3+p x+q)$ over $\mathbb{Q}$, then
$$E=BD+\frac{1}{4}(B^2-C)^2-n^2,~~~~~n \in \mathbb{Q}$$
$$F=\frac{1}{2} \left(B^5-B^3C+3B^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Eliminate the parameter given $x = \tan^{2}\theta$ and $y=\sec\theta$ $x = \tan^{2} (\theta)$ and $y = \sec (\theta)$
knowing that $\tan^{2} (\theta) = (\tan (\theta))^2 = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}$
and that $\sec(\theta) = \dfrac{1}{\cos(\theta)}$ $\to$ $y=\dfrac{1}{\cos(\theta)}$
For $y$ we can get an e... | You can save some passages:
$$
x=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}=
\frac{1}{\cos^2\theta}-\frac{\cos^2\theta}{\cos^2\theta}=y^2-1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Baricenter of a region bounded by a closed parametric curve I've always known how to get the center of mass of any region, but now i met a new question with a region bounded by a parametric curve and the question is to get its baricenter! My question is what changes in this case than the normal way of getting the baric... | So we are looking for the baricenter (centroid) of the region enclosed by the curve
$$
C:\left\{ \begin{gathered}
0 \leqslant t \leqslant 1 \hfill \\
x(t) = \cos \left( {2\pi t} \right) \hfill \\
y(t) = t - t^{\,3} \hfill \\
\end{gathered} \right.
$$
Among the methods exposed in answering to
Baricenter of a re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| Squaring and adding,we get
$1+cosec^2\theta+\sec^2\theta=m^2+n^2.....(1)$
We need to find $cosec\theta$ and $\sec\theta$ from the given equations
First equation becomes $\frac{1}{\csc\theta}+\csc\theta=m$
$cosec^2\theta-m$ $cosec\theta+1=0$
Similarly second equation becomes
$\sec^2\theta-n\sec\theta-1=0$
Solving these ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Finding the parameters of an ellipsoid given its quadratic form Suppose we have the quadratic form of an ellipsoid of the form
$$ax^2 + by^2+cz^2+dxy+eyz+fxz+gx+hy+iz+j=0$$
I want to find centroid of the arbitrarily oriented ellipsoid, its semi-axes, and the angles of rotation.
For the 2D case I found an answer here.... | The equation of an ellipsoid centered at the origin has the form $q_A(\mathbf{v}) = \mathbf{v}^T A \mathbf{v} = C$ where $C > 0$ and $A$ is a symmetric matrix whose eigenvalues are all positive. If the eigenvalues of $A$ are $\lambda_i$ with a corresponding orthonormal basis of eigenvectors $v_i$ (so $Av_i = \lambda_i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is my solution wrong? Question: "Find all x such that ${\frac{x-a}{b}}+{\frac{x-b}{a}}={\frac{b}{x-a}}+{\frac{a}{x-b}}$ , where a and b are constants."
My attempt:
${\frac{x-a}{b}}+{\frac{x-b}{a}}={\frac{b}{x-a}}+{\frac{a}{x-b}}$
Let $m=x-a$ ,Let $w=x-b$ . Therefore:
${\frac{m}{b}}+{\frac{w}{a}}={\frac{b}{m}}+{\fr... | When you divided by $x-a-b$ you threw away the root $x=a+b$. Just like in the equation $t(t-1)=t(2t-5)$, if we cancel the $t$ we are losing the root $t=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{\mathrm{d}y}{\mathrm{d}x} = (x-y)/(x+y)$ Solve
$$\frac { { d }y }{ { d }x } =\frac { x-y }{ x+y } $$
It is homogeneous, thus let $y = vx$. From this, $\frac{\mathrm{d}y}{\mathrm{d}x} = x\frac{\mathrm{d}v}{\mathrm{d}x} + v$
Thus,
$v'x + v = (1-v)/(1+v)$ thus,
$\frac{2}{1-v} + \ln(v - 1) = \ln(x) + C$.
W... | $$\frac { dy }{ dx } =\frac { x-y }{ x+y } \\ \frac { dy }{ dx } =\frac { 1-\frac { y }{ x } }{ 1+\frac { y }{ x } } \\ y=zx\\ z^{ \prime }x+z=\frac { 1-z }{ 1+z } \\ z^{ \prime }x=\frac { 1-z }{ 1+z } -z=\frac { 1-2z-z^{ 2 } }{ 1+z } \\ \int { \frac { 1+z }{ 1-2z-z^{ 2 } } dz } =\int { \frac { dx }{ x } } \\ -\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving the Inequality $\frac{14x}{x+1}<\frac{9x-30}{x-4}$ The question says to find all the integral values of x for which the inequality holds.
the question is
$$\frac{14x}{x+1}<\frac{9x-30}{x-4}$$
My Solution
\begin{align}
& \frac{14x}{x+1} < \frac{9x-30}{x-4} \\[6pt]
& \frac{14x(x-4)-(9x-30)(x+1)}{(x+1)(x-4)}<0 \\... | Basic approach. Perhaps easier would be to rewrite the original inequality as
$$
14 - \frac{14}{x+1} < 9 + \frac{6}{x-4}
$$
This leads to
$$
\frac{14}{x+1} + \frac{6}{x-4} > 5
$$
which becomes
$$
\frac{4x-10}{(x+1)(x-4)} > 1
$$
You can now (a) consider the cases $x = 0, 1, 2, 3$ separately, and then otherwise, (b) for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Solving an Itô Integral Can someone please show me how to solve this Itô Integral?
$$\begin{align}\int_{1}^{t}\frac{dB_s}{B_s^2 + B_s^4} && \end{align} $$
| Set $f(x)=-\frac{1}{x}-\tan^{-1}x$. we have
$$f'(x)=\frac{1}{x^2}-\frac{1}{1+x^2}=\frac{1}{x^2+x^4}$$
and
$$f''(x)=-\frac{2x+4x^3}{(x^2+x^4)^2}$$
By application of Ito's lemma we have
$$f(B_t)=f(B_1)+\int_{1}^{t}f'(B_s)dB_s+\frac{1}{2}\int_{1}^{t}f''(B_s)ds$$
therefore
$$-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)=-\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)} = $ ? $m$, $n$ integers > 0 The integral
$$
A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)}
$$
popped up when I was playing around with the integral representatio... | Using the identities $$\frac{\sin (2m \theta)}{\sin (\theta)} = 2 \sum_{k=0}^{m-1} \cos[(2k+1)\theta]$$
and
$$\cos^{2n+1}(\theta) = \frac{1}{4^{n}}\sum_{j=0}^{n} \binom{2n+1}{j} \cos [2n+1-2j)\theta], $$ we get
$$ \begin{align} A_{mn} &= \frac{1}{\pi} \int_{0}^{\pi} \frac{\sin (2 m \theta)}{\sin (\theta)}\left(\cos (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integral value of $n$ that makes $n^2+n+1$ a perfect square. Find all integers $n$ for which $n^2+n+1$ is a perfect square.
By hit and trial we get $n=-1,0$ but could someone suggest any genuine approach as how to approach this problem?
| As other solutions stated , one of your values can be obtained as like ,
$ n^2+n+1=k^2 \equiv 1 (mod 3) \implies n(n+1)\equiv 0 (mod 3)$
So $3|n$ or $3| (n+1)$ , put $n=3k$ , $f(k)=9k^2+3k+1=(3k+1)^2 - 3k$ which implies $n=k=0$ .
Secondly if $3|(n+1)$ put $n=3k-1$ & we get $ f(k)=9k^2-3k+1=(3k-1)^2+3k$ which forces $k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Find the coefficient of $ x^{12}$ in $(1-x^2)^{-5}$ Find the coefficient of $x^{12}$ in $(1-x^2)^{-5}$
What can be said for $x^{17}$
Tried $\frac{1}{(1-x^2)^{5}}$=$\sum_{n=0}^\infty \binom{n+5-1}{n}x^n$
not sure that i can do that with $x^2$
| $$y=\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
the first derivative is
$$\frac{-1}{(1-x)^2}=\sum_{n=0}^{\infty }nx^{n-1}$$
the fourth derivative
$$\frac{24}{(1-x)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{n-4}$$
let $x\rightarrow x^2$
$$\frac{24}{(1-x^2)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{2n-8}$$
to find coefficie... | {
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"url": "https://math.stackexchange.com/questions/1861551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to find $\tan x $ from $(a+1)\cos x + (a-1)\sin x=2a+1$? How do I find $\tan x$ from this equation?
$$(a+1)\cos x + (a-1)\sin x=2a+1$$
Thanks for any help!!
| using the so-called Weierstrass substitution we get this here
$${\frac { \left( a+1 \right) \left( 1- \left( \tan \left( x/2 \right)
\right) ^{2} \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2
}}}+2\,{\frac { \left( a-1 \right) \tan \left( x/2 \right) }{1+
\left( \tan \left( x/2 \right) \right) ^{2}}}=2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
integrate $\int \frac{x^4-16}{x^3+4x^2+8x}dx$
$$\int \frac{x^4-16}{x^3+4x^2+8x}dx$$
So I first started with be dividing $p(x)$ with $q(x)$ and got:
$$\int x-4+\frac{8x^2+32x-16}{x^3+4x^2+8x}dx=\frac{x^2}{2}-4x+\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx$$
Using partial sum I have received:
$$\int \frac{8x^2+32x-16}{x^3+4... | HINT:
As $x^2+4x+8=(x+2)^2+2^2$ and $\dfrac{d(x^2+4x+8)}{dx}=2(x+2)$
for $\dfrac{Ax+B}{x^2+4x+8},$ express it as $$a\cdot\dfrac{2(x+2)}{x^2+4x+8}+b\cdot\dfrac1{(x+2)^2+2^2}$$
Choose $x+2=2\tan y$ for the second term
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Determine how many roots are real, and finding all roots of a quintic: $-2y^5 +4y^4-2y^3-y=0$ Using a computer we can see that the only real root of $f(y)=-2y^5 +4y^4-2y^3-y=0$ is $0$. Furthermore, we know from algebra that since this polynomial lives in $\Bbb R[y]$ that the roots come in complex conjugate pairs. I.e. ... | Note that$$-2y^4+4y^3-2y^2-1=-2y^2(y-1)^2-1<0$$
for real $y$, so the remaining roots are not real. We need
$$
y^2(y-1)^2=-\frac{1}{2},
$$
or
$$
y(y-1)=\pm\frac{i\sqrt{2}}{2}.
$$
For each choice of sign, this is a quadratic that you can solve with the usual formula. Specifically,
$$
y^2-y(\pm)_1\frac{i\sqrt{2}}{2}=0\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$ $$\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}$$
$$\sin^2 x =\frac{1}{2}{(1- \cos2x)}$$
$$\cos^2 x =\frac{1}{2}{(1+\cos2x)}$$
$$\int \frac{(1- \cos2x)^2}{2.(1+\cos^2 2x)}{dx}$$
$$\frac{1}{2} \int \left[1-\frac{2 \cos2x}{1+\cos^22x}\right] dx$$
What should I do nex... | $$I=\int \frac{\sin^4 x}{\sin^4 x +\cos^4 x}{dx}=\int \frac{1-2\cos^2(x)+\cos^4(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$
$$=\int\frac{1}{2}-\frac 12\frac{-1+2\cos^2(x)}{1-2\cos^2(x)+2\cos^4(x)}dx$$
$$=\int\frac{1}{2}-\frac 12\frac{\cos(2x)}{1-2\cos^2(x)\sin^2(x)}dx$$
$\color{blue}{\text{This last step above is a few manipulati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following
If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.
| $ x^{14} - x^{12} + \ldots + x^2 -1 = \frac{7}{x}, $ I am considering $ x \neq 0 $ multiply both side by $ x^2 $ and add you get $ x^{16} = 1+ 7x + \frac{7}{x}. $ Take minimum of R.H.S.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \... | The right hand side of the implies sign above can be studied, viz;
\begin{align}
\frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} &= \frac{2^{2n}+(-1)^{2n}2^{2n}}{(-1)^{2n}2^{4n}}\\
&= \frac{2.2^{2n}}{2^{4n}} \qquad (-1)^{2n}=1 \quad \forall n\\
&= \frac{2}{2^{2n}}
\end{align}
Thus,
\begin{align}
\frac{1}{(-2^n)^2} + \frac{1}{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Monotonicity of the sequence $(a_n)$, where $a_n=\left ( 1+\frac{1}{n} \right )^n$ Define $a_n=\left ( 1+\frac{1}{n} \right )^n$ for $n\geq 1$. I want to show that it is increasing. First, we have
$$\frac{a_{n+1}}{a_n}=\left ( \frac{1+\frac{1}{n+1}}{1+\frac{1}{n}} \right )^n\left ( 1+\frac{1}{n+1} \right )=\left ( 1-\f... | The left hand side is
$$\frac1{(n+1)^3}\cdot((n+1)^2-n)(n+2)=\frac{n^3+3n^2+3n+2}{(n+1)^3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Show $\cos\left( \pi n^{2}\ln\left(\frac{n}{n-1} \right) \right)=(-1)^{n+1}\frac{\pi}{3n}+\mathcal{O}\left( \frac{1}{n^2}\right) $
I would like to show :
$$\cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)=(-1)^{n+1}\dfrac{\pi}{3n}+\mathcal{O}\left( \dfrac{1}{n^2}\right) $$
by starting from the left side... | You may write
$$
\begin{align*}
\cos\left( \pi n^{2}\ln\left(\dfrac{n}{n-1} \right) \right)&=\cos\left( -\pi n^{2}\ln\left(\dfrac{n-1}{n} \right) \right)\\
&=\cos\left( -\pi n^{2}\ln\left(1-\frac1n \right) \right)\\
&=\cos\left( \pi n^{2}\left(\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{3 n^3}+\mathcal{O}\left(\dfrac{1}{n^4} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Use lagrange multipliers to calculate the maximum and minimum $f(x,y,z)=x^2y^2z^2$ constrained by $x^2+y^2+z^2=1$
$\nabla f_x$ $=$ $2xy^{2}z^{2}$,
$\nabla f_y$ $=$ $2yx^{2}z^{2}$,
$\nabla f_z$ $=$ $2zx^{2}y^{2}$
$\nabla g_x$ $=$ $2x$,
$\nabla g_y$ $=$ $2y$,
$\nabla g_z$ $=$ $2z$
setting the sets equal to each other... | You have three equations from the first order conditions:
$$2xy^2z^2=2\lambda x \qquad 2x^2yz^2=2\lambda y\qquad 2x^2y^2z=2\lambda z $$
Suppose $x=0$. Then the first equation is satisfied, and the other two equations imply that $\lambda=0$ (since we cannot have $x=y=z=0$ as that does not satisfy the constraint). This g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve the equation $xy = 1, x^{2x-y} = y^{2(x-y)}$ I have the following equation that I don't know how to solve:
$$ \begin{cases} xy = 1 \\ x^{2x-y} = y^{2(x-y)} \end{cases} $$
Here's what I've tried (but my mathematical instinct tells me that I didn't solve it right):
$$ \begin{cases} xy = 1 \\ x^{2x-y} = y^{2(... | Your attempted solution is correct. But you missed the trivial solution $(x,y)=(1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer
How to evaluate the sums $\sum\limits_{n=1}^\infty\frac{1}{n \binom{kn}{n}}$ with $k$ a positive integer?
For $k=1$, the series does not converge.
When $k=2$, I can prove that:
$$\sum_{n=1}^\infty\frac{1}{n \binom{2n}{... | I'm late for this party but the appearance of the plastic constant's minpoly, or $x^3-x-1=0$, got me interested.
The binomial sum can be expressed as a concise finite sum of logarithms. For $k>1$,
$$\sum_{n=1}^\infty\frac1{n\binom{kn}n} =\sum_{n=1}^k \frac{\ln(1-x_n)}{k-(k-1)x_n}=\int_1^\infty\frac1{x(x^k-x+1)}$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.... | More generally,
if
$t
=\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}
$,
$\begin{array}\\
t^2
&=(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}})^2\\
&=a+\sqrt{b}+2(\sqrt{a+\sqrt{b}}\sqrt{a-\sqrt{b}})+a-\sqrt{b}\\
&=2a+2\sqrt{(a+\sqrt{b})(a-\sqrt{b})}\\
&=2a+2\sqrt{a^2-b}\\
\text{so}\\
t
&=\sqrt{2a+2\sqrt{a^2-b}}\\
\end{array}
$
In this cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Proof that $P(x)=x-\frac{1}3 x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has radius of convergence $1$
Proof that
$P(x)=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+\cdots$ has
radius of convergence $1$
First of all, I need to convert this to a series:
$$\sum_{k=1}^\infty \frac{x^{2k-1}(-1)^k}{1-2k}$$
(I hope thi... | Taking the absolute value of your algebra gives $x^2$. Since the ratio test looks for when this is $<1$, we ask ourselves when is $|x^2|<1$. The answer is when $-1<x<1$, and if $|x|>1$ it does not converge. This suffices to find the radius; it is $1$.
If you want the interval of convergence as well, then you need ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
properties of ratio and rule of cross multiplication If $$\frac{l}{\sqrt a-\sqrt b}+\frac{m}{\sqrt b-\sqrt c}+\frac{n}{\sqrt c-\sqrt a} =0$$
$$\frac{l}{\sqrt a+\sqrt b}+\frac{m}{\sqrt b+\sqrt c}+\frac{n}{\sqrt c+\sqrt a} =0$$
Show that $$ \frac{l}{(a-b)(c-\sqrt ab)}=\frac{m}{(b-c)(a-\sqrt bc)}=\frac{n}{(c-a)(b-\sqrt ac... | For $a,b,c>0,$ let $a =A^2$ etc.
As $(B+C)(C-A)-(B-C)(C+A)=2(C^2-AB),$
$$\dfrac l{\dfrac n{(B-C)(C+A)}-\dfrac n{(B+C)(C-A)}}=\dfrac m{\cdots}=\dfrac1{\dfrac1{(A-B)(B+C)}-\dfrac1{(A+B)(B-C)}}$$
$$\implies\dfrac {l(B^2-C^2)(C^2-A^2)}{2(C^2-AB)}=\dfrac m{\cdots}=\dfrac{n(A^2-B^2)(B^2-C^2)}{2(B^2-CA)}$$
$$\iff\dfrac {l(C^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the real and imaginary parts of an equation
Find the real and imaginary parts of $\frac{1}{3z+2}$
So I have expanded it out to get $\frac{1}{3x+3iy+2}$
Thus giving $Re(\frac{1}{3z+2})=\frac{1}{3x+2}$ and $Im(\frac{1}{3z+2})=\frac{1}{3y}$
However in my answer book it says: $Re(\frac{1}{3z+2})=\frac{3x+2}{(3x+2)^... | Note quite, remember that $\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}$. In general, the real and imaginary part of $z$ is $x$ and $y$ where $z = x+iy$. That is, you must seek to turn your expression for $z$ into the form $x+iy$. You can then multiply $\frac{1}{3x + 3iy + 2}$ with a fraction that consists of its conju... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to ... | Write $f(x)=g(x)+h(x)$. Now $g(x)$ has minima at $x=1/2$ and $h(x)$ is increasing function with minima at $x=0$.Then $f(x)$ has minima at ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
For which $a$, equation $4^x-a2^x-a+3=0$ has at least one solution.
Find all values of $a$ for which the equation $4^x-a2^x-a+3=0$ has at least one solution.
$\bf{My\; Try::}$ We can write it as $$2^{x}-a-\frac{a}{2^x}+\frac{3}{2^x}=0$$
So $\displaystyle \left(2^x+\frac{3}{2^x}\right)=a\left(1+\frac{1}{2^x}\right).$
... | You found a necessary condition, but not a sufficient one. You've concluded that
$$a \ge \frac{2^x + \frac{3}{2^x}}{1 + \frac 1 {2^x}}$$
Your application of AM-GM is all about minimizing the numerator, but it doesn't minimize the entire fraction. In fact, AM-GM is sharp when $2^x = \sqrt{3}$, in which case the fraction... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational... | $$
\begin{align}
n^4 - 4n^3 + 6n^2 - 4n + 1 = 0\\
n^4 - n^3 - 3n^3 + 3n^2 + 3n^2 - 3n - n + 1 = 0\\
n^3(n - 1) - 3n^2(n - 1) + 3n(n - 1) - 1(n-1) = 0\\
(n - 1)(n^3 - 3n^2 + 3n - 1) = 0\\
(n - 1)(n^3 - n^2 -2n^2 + 2n + n - 1) = 0\\
(n - 1)[n^2(n - 1) - 2n(n - 1) + 1(n - 1)] = 0\\
(n - 1)(n - 1)(n^2 - 2n + 1) = 0\\
(n - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur. The question is two tangents to the hyperbola $4x^2-y^2=36$ intersect at the point $(0,4)$. Find the coordinates for the points on the hyperbola for this to occur.
My atte... | The hyperbola can be written as $\frac{x^2}{9} - \frac{y^2}{36} = 1$ and let the tangent at $(3\sec\theta, 6\tan \theta)$ pass through $(0,4)$. The equation of the tangent is $\frac{x}{3}\sec\theta - \frac{y}{6}\tan\theta = 1$ and since $(0,4)$ lies on this, we have $\tan\theta = -\frac{3}{2}$. Thus the points are give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplifying Ramanujan-type Nested Radicals Ramanujan found many awe-inspiring nested radicals, such as...
$$\sqrt{\frac {1+\sqrt[5]{4}}{\sqrt[5]{5}}}=\frac {\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{2}-1}{\sqrt[5]{125}}\tag{1}$$$$\sqrt[4]{\frac {3+2\sqrt[4]{5}}{3-2\sqrt[4]{5}}}=\frac {\sqrt[4]{5}+1}{\sqrt[4]{5}-1}\tag{2}$$$$\s... | Landau's algorithm: http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=63496 $\qquad\qquad$
https://en.wikipedia.org/wiki/Susan_Landau
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
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Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:... | Set, provisionally, $x=\tan\theta$ and $y=\tan2\theta$. Then the equation becomes
$$
x+y+\frac{x+y}{1-xy}=xy\frac{x+y}{1-xy}
$$
Note that the expression only makes sense for $1-xy\ne0$.
After removing the denominator, we get
$$
(x+y)(2-xy)=xy(x+y)
$$
so this becomes
$$
2(x+y)(1-xy)=0
$$
Since $1-xy\ne0$, we remain with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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What is the $\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$? What is the limit of
$\lim_\limits{x \to 0} \frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$?
I attempted the problem via L^Hopital's Rule so I rewrote it as
$$y=\frac{\cos(x)-1+\frac{x^2}{2}}{x^4}$$
then took the natural of both sides
$$\ln(y)=\ln(\frac{\cos... | I believe the problem should read $$\lim_{x \to 0} \frac{\cos(x) - 1 + \tfrac {x^2}2}{x^4};$$ i.e., the sign of the $1$ should be flipped. Otherwise the top goes to $2$ and the bottom to $0$ so the limit is $+\infty$. As I have written it, you can simply use l'Hopital's rule several times: \begin{align*}\lim_{x \to 0} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to prove that $\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$? As stated in the question. Thank you!
$$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$
| Just using calculus.
If you use Stirling approximation, that is to say $$n! \sim \sqrt{2\pi n}\left(\frac n e\right)^n$$ the expression $$\frac{a^ab^b}{a!b!}\le\frac{(a+b)^{(a+b)}}{(a+b)!}$$ becomes $$\frac{e^{a+b}}{2 \pi \sqrt{a} \sqrt{b}}\le \frac{e^{a+b}}{\sqrt{2 \pi } \sqrt{a+b}}$$ which becomes $$1 \le \frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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find the number of pairs $(a,b)$ where $a,b,n\in\mathbb N$ and $1\le aGiven a positive integer $n$, find the number of pairs $(a,b)$ where $a,b\in\mathbb N$ such that $(a^2-1)$ is divisible by $b$ and $(b^2-1)$ is divisible by $a$, and where $1 \le a < b \le n$.
Please help me how to solve this problem in constant time... | Clearly $(1,b)$ works always, so we concentrate on the other solutions.
Suppose we have $a$ and $b$ such that $a|(b+1)(b-1)$ and $b|(a+1)(a-1$).
Lemma $1$: Since $(a+1,a-1)=2$ or $1$ we have $a|2(b+1)$ or $a|2(b-1)$.
Lemma $2$: If $a$ is odd then $a|(b+1)$ or $a|(b-1)$. In particular $a\leq b+1$
Clearly $a$ and $b$ can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 ... | $$x = 59 + 64k$$
$$x = 32 + 81j$$
$\text{lcm}(81, 64) = 5184$
so
$$81x = 4779 + 5184k$$
$$64x = 2048 + 5184j$$
Subtract equations
$$17x = 2731 + 5184(k-j)$$
or
$$17x \equiv 2731 \bmod 5184$$
$\gcd(17, 5184)=1$, so:
$$x \equiv 2731 \cdot 17^{-1} \bmod 5184$$
$$x \equiv 2731 \cdot 305 \equiv 832955 \equiv 3515 \bmod 5184... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Range of function involving fractional part and integer part of $x$
The range of function $$f(x) = \left\{\frac{x}{4}\right\}+\cos \left(\frac{1-2\lfloor x \rfloor }{2}\right)+\sin \left(\frac{\pi \lfloor x \rfloor }{2}\right)$$
Where $\{x\} = x-\lfloor x \rfloor$ and $\lfloor x \rfloor $ represent floor function of $... | We consider four cases.
i) Let $x=4k+t$ with $k\in \mathbb{Z}$ and $t\in [0,1)$ then
$$f(x) = \frac{t}{4}+\cos \left(\frac{1-8k}{2}\right)=\frac{t}{4}+\cos \left(\frac{8k-1}{2}\right).$$
Hence if $A_0:=\cup_{k\in\mathbb{Z}}[4k,4k+1)$ then
$$f(A_0)=\bigcup_{k\in\mathbb{Z}}\left[\cos \left(\frac{8k-1}{2}\right),\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Counting even and odd numbers in the columns of a triangular arrangement of the integers I write the positive numbers starting at $1$ in a triangle:$$\mathbb{N}_\triangle = \begin{matrix}
&&&&&21&\ldots \\
&&&&15&20&\ldots \\
&&&10&14&19&\ldots \\
&&6&9&13&18&\ldots \\
&3&5&8&12&... | Note that each column forms an interval of integers, and so the numbers in the column alternate between being odd and even.
Hence if $n$ is even, so that there are an even number of integers in the $n$th column, exactly half of them will be even, so $e(2k) = k$ for every $k \in \mathbb{N}$.
If $n$ is odd, there will ei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Analytic Geometry Question: Computing the equation of a circle given 2 points and center I want to derive the equation of a circle passing through the intersection of the two circles. My method gave me a nonsensical solution. I would love if someone could tell me how my method went wrong, although since my solution is ... | First of all, it would be slightly easier if we write the resultant circle as:-
$((x-11)^2+(y-4)^2-9) + j(x^2+y^2-100) = 0$
because only one variable is involved.
Secondly, the painful completing square process can be avoided.
Therefore the equation of the required circle is $x^2 + y^2 + 2(\dfrac {-11}{1 + j})x + 2(\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ wh... | Let
$$P=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}$$
$$Q=\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}+\frac{a}{a+b}$$
$$R=\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}+\frac{b}{a+b}$$
We have $$Q+R=4\tag{1}$$
$$P+Q=\frac{a+b}{b+c}+\frac{b+c}{c+d}+\frac{c+d}{d+a}+\frac{d+a}{a+b} \overbrace{\geq}^{\color{red}{\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
| The answer by Doug M solves your problem. I just want to expand on it to the general ellipse.
Assume $$ax^2+2bxy+cy^2+dx+ey+f=0$$ is an ellipse ($ac-b^2>0, a>0$). It has center $x=\frac{eb-cd}{2(ac-b^2)}, y=\frac{bd-ae}{2(ac-b^2)}$, which means that when the equation transforms to $$ax'^2+2bx'y'+cy'^2+f-\frac{ae^2-2bd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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Challenge: Prove $\displaystyle \sum_{n \in \mathbb{N}} \frac{n!}{(2n+1)!}=e^{1/4}\sqrt{\pi}\ \ \mathrm{erf}(\frac{1}{2})$ I stumbled upon this cute sum while messing about, and I want to see what other solutions people propose before I put forward my own (which may be unnecessarily complicated).
You can use any maths ... | Outline:
*
*We know that $$e^{1/4} = \sum_{n=0}^\infty \frac{1}{2^{2n} n!}$$
*and that $$\begin{align}
\sqrt{\pi}\operatorname{erf}\left(\frac{1}{2}\right) &= \int_0^{1/2} e^{-t^2}dt
= \int_0^{1/2} \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{n!} dt
= \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\int_0^{1/2} t^{2n} dt
\\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$ I'm reading a introductory book on mathematical proofs and I am stuck on a question.
Let $a, b, c, d$ be positive real numbers, prove that if $\frac{a}{b} < \frac{c}{d}$, then $\frac{a}{b} < \frac{a+c}{b+d} <\frac{ c}{d}$.
| $$\frac{a}{b}=\frac{a(1+\frac{d}{b})}{b(1+\frac{d}{b})}=\frac{a+d(\frac{a}{b})}{b+d}<\frac{a+d(\frac{c}{d})}{b+d}=\frac{a+c}{b+d}$$
and similarly for the other inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$ if $x^2+y^2=1$. Show that $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}≥\frac{3}{1+(\frac{x+y}{2})^2}$. It is given that $x^2+y^2=1$. $x,y$ are positive real numbers.
[From a Regional Mathematical Olympiad, 2013 in India]
| $$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+xy}\ge \frac{4}{1+x^2+1+y^2} + \frac{1}{1+xy}$$
Use of AM-HM Inequality for $n=2$
$$=\frac{4}{3}+\frac{1}{1+xy} \ge \frac{1}{\frac{5}{4}+\frac{xy}{2}}$$
Expanding out
$$\frac{7+4xy}{3(1+xy)} \ge \frac{4}{5+2xy}$$
Common Denominators
$$(7+4xy)(5+2xy) \ge 12(1+xy)$$
Justified ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A quadratic equation Find all values of a for which the quadratic $$\cos^2x - (a^2 + a + 5) |\cos x| + (a^3 + 3a^2 + 2a + 6) = 0$$ has real solution(s)
A. $a=-3$, B. $a=-2$, C. $a=-1$, D. $a=0$
I have solved this question by putting the values of the given options :
Let $\cos x = t$
then it will look like $$t^2 - (a^2... | You have to check if for at least one root $t$ of the equation
$$t^2 - (a^2 + a + 5) t + (a^3 + 3a^2 + 2a + 6)=
t^2 - (a(a+1)+ 5) t + (a(a+1)(a+2) + 6)
=0$$
there is a real $x$ such that $|\cos(x)|=t$.
Since $|\cos(x)|=t$ is solvable as soon as $t\in [0,1]$, just see if there is a root $t\in [0,1]$.
i) If $a=-3$ then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A difficult functional equation Is it possible to solve the following functional equation-
Determine all functions $f: \mathcal{R} \rightarrow \mathcal{R}$ such that $f(f(x)-f(y))=f(f(x))-2x^2f(y)+f(y^2)$ for all reals $x,y$
Here , $\mathcal{R}$ denotes the set of all reals.
| I suppose that $f$ is not $0$.
1) I put first $a=f(0)$. Then $x=y=0$ gives $f(a)=0$. Putting $y=a$ gives $f(a^2)=0$, and replacing $x$ by $a$ gives $$f(-f(y))=a-2a^2f(y)+f(y^2)$$
2) Let $b$ such that $f(b)=0$. Replacing $x$ by $b$ gives
$$f(-f(y))=a-2b^2f(y)+f(y^2)$$. By the above, this is true $b=a^2$. If $a^4\not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What is the remainder when the product of the primes between 1 and 100 is divided by 16? The product of all the prime numbers between 1 and 100 is equal to $P$. What is the remainder when $P$ is divided by 16?
I have no idea how to solve this, any answers?
| If the product of the odd primes is $8k+r$ then the product of all of them is $16+2r$. So we only need to work $\bmod 8$. We only need to find primes with residues $3,5$ and $7$. We do it as follows:
$$\begin{pmatrix}
3 && 5 && 7\\
11 && 13 && 15 \\
19 && 21 && 23\\
27 && 29 && 31\\
35 && 37 && 39\\
43 && 45 && 47\\
51... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Constant when integrating I was integrating $(t+2)^2$ by using the substitution method and by expanding the function. By expanding it then integrating I got the same answer as the book ($\frac{1}{3}(t^3+6t^2+12t)+C$), but when I substituted I got $\frac{1}{3}\cdot(t+2)^3 + C = \frac{1}{3} \cdot (t^3+6t^2+12t+8) + C$.
A... | Your answers differ by a constant factor, so the values of $C$ will be different in these 2 methods.
In other words,
$$
\frac{t^3+6t^2+12t}{3} + C= \frac{t^3 + 6t^2 + 12t +8}{3} + K
$$
just $C = K + 8/3$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ I need to show that $(a+2)^3+(b+2)^3+(c+2)^3 \ge 81$ while $a+b+c=3$ and $a,b,c > 0$ using means of univariate Analysis.
It is intuitively clear that $(a+2)^3+(b+2)^3+(c+2)^3$ is at its minimum (when $a+b+c=3$) if $a,b,c$ have "equal weights", i.e. $a=b=c=1$.
To show t... | Let $f(x) = x^3 $ is a convex function. Using Jenson we get :
$$ \frac{f(a+2)+f(b+2)+f(c+2)}{3} \geq f(\frac{a+b+c+6}{3})=f(3)=27$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Basic Algebra combining exponent fractions/simplifying (George F Simmons "Precalculus in Nutshell") From George F. Simmons 'Precalculus' book, Algebra section, 5(d);
Combine and Simplify
$$\frac{x}{xy^2} + \frac{y}{x^2y}$$
Combine:
= $$\frac{x(x^2y) + y(xy^2)}{(xy^2)(x^2y)}$$
Simplify: = $$\frac{x(x^2y) + y(xy^2)}{xy(... | $$\frac{x}{xy^2}+\frac{y}{x^2y}=\frac{x}{x}\cdot\frac{x}{xy^2}+\frac{y}{y}\cdot\frac{y}{x^2y}=\frac{x\cdot x}{xy^2\cdot x}+\frac{y\cdot y}{x^2y\cdot y}=\frac{x^2}{x^2y^2}+\frac{y^2}{x^2y^2}=\frac{x^2+y^2}{x^2y^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is minimizing least squares equivalent to finding the projection matrix $\hat{x}=A^Tb(A^TA)^{-1}$? I understand the derivation for $\hat{x}=A^Tb(A^TA)^{-1}$, but I'm having trouble explicitly connecting it to least squares regression.
So suppose we have a system of equations: $A=\begin{bmatrix}1 & 1\\1 & 2\\1 &3\e... | The matrix has full column rank; we are guaranteed a unique solution.
Problem statement
$$
\begin{align}
\mathbf{A} x &= b \\
\left[
\begin{array}{cc}
1 & 1 \\
1 & 2 \\
1 & 3 \\
\end{array}
\right]
%
\left[
\begin{array}{cc}
x_{1} \\
x_{2} \\
\end{array}
\right]
&=
\left[
\begin{array}{cc}
1 \\
2 \\
2 \\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
find integers a and b such $x^2-x-1$ divides $ax^{17}+bx^{16}+1 = 0$
By really long division i got :-
$$Q=ax^{15} + (a+b)x^{14} + \dots +(610a + 377b) $$
$$R = x(987a+610b)+1+610a+377b$$
since remainder is $0 $,
$$987a+610b = 0$$
$$1+610a+377b =... | $$x^2-x-1=
\left( x-\frac{1+\sqrt{5}}{2} \right)\left( x-\frac{1-\sqrt{5}}{2} \right)$$
By factor theorem,
\begin{align*}
a\left( \frac{1+\sqrt{5}}{2} \right)^{17}+
b\left( \frac{1+\sqrt{5}}{2} \right)^{16}+1 &=0 \quad \cdots \cdots \: (1) \\
a\left( \frac{1-\sqrt{5}}{2} \right)^{17}+
b\left( \frac{1-\sqrt{5}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan ... | Changing in terms of $$\sin \text {and} \cos ,3=2+1$$
$$L.H.S=\frac {2\sin 80^{\circ}.\sin20^{\circ} +\cos (80^{\circ}-20^{\circ})}{\sin(80^{\circ}+20^{\circ})} $$
$$=\frac {\cos(80^{\circ}-20^{\circ}) -\cos(80^{\circ}+20^{\circ})+\cos 60^{\circ}}{\sin 100^{\circ}} $$
$$=\frac {1-\cos 100^{\circ}}{\sin100^{\circ}}=\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove this inequality $\sum\limits_\text{cyc}\sqrt{1-xy}\ge 2$ Let $x,y,z\ge 0$, and $x+y+z=2$, show that
$$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$
Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$
or
$$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$
| After squaring of the both sides we need to prove that
$$2\sum\limits_{cyc}\sqrt{(1-xy)(1-xz)}\geq1+xy+xz+yz$$ or
$$2\sum\limits_{cyc}\sqrt{((x+y+z)^2-4xy)((x+y+z)^2-4xz)}\geq(x+y+z)^2+4(xy+xz+yz)$$ or
$$2\sum\limits_{cyc}\sqrt{((x-y)^2+z^2+2xz+2yz)((x-z)^2+y^2+2xy+2yz)}\geq\sum\limits_{cyc}(x^2+6xy)$$
and after using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
vectorial equation i need help to Calculate the vectorial equation of the line $$ g = E_{1}\cap E_{2} $$ you get by intersecting the planes$$ E_{1}:\overrightarrow{x}=\left(\begin{array}{c}1\\ 0\\1\end{array}\right)+r.\left(\begin{array}{c}1\\ 1\\0\end{array}\right)+s.\left(\begin{array}{c}0\\ 1\\1\end{array}\right);r,... | The normal vectors to planes $E_1$ and $E_2$ are:
$$\vec{N_1}=\begin{pmatrix}1\\1\\0\end{pmatrix} \times \begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix} \ \text{and} \ \vec{N_2}=\begin{pmatrix}1\\1\\1\end{pmatrix} \times \begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\1\\-1\end{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$ Find the sum of $n$ terms of following series:
$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$
I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way... | We have
$$ n(n+1)^2(n+2) = 24\binom{n+2}{4}+12\binom{n+2}{3}\tag{1}$$
hence
$$ \sum_{n=1}^{N}n(n+1)^2(n+2) = 24\binom{N+3}{5}+12\binom{N+3}{4} \tag{2}$$
is a consequence of the Hockey Stick identity, leading to:
$$ \sum_{n=1}^{N}n(n+1)^2(n+2) = \color{red}{\frac{1}{10} N (1+N) (2+N) (3+N) (3+2 N)}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Does this system of Diophantine equations have a solution?
Are there natural numbers $a,b,c,d,e,f$ such that we have $a \neq b$ and $a \neq c$ and $b \neq c$ and that they are solution of this system of equations:
$9ab-3a-3b+1=d^2$
$9ac-3a-3c+1=e^2$
$9bc-3b-3c+1=f^2$
| There are infinitely many. Let,
$$a = (2p+1)^2+2p^2\\
b = (2q+1)^2+2q^2\\
c = (2r+1)^2+2r^2$$
then,
$$d =4(3p+1)^2(3q+1)^2\\
e =4(3p+1)^2(3r+1)^2\\
f =4(3q+1)^2(3r+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
cardinality of cartesian product of the infinite set of natural numbers One of the problems in my discrete math course states that we need to prove that $\mathcal{N}\times\mathcal{N}$ is countable specifically when there's a function $f:\mathcal{N}\times\mathcal{N}\to\mathcal{N}$ defined as follows:
$$f(a,b)=\frac{1}{2... | $\frac{1}{2}(a+b+1+1)(a+b+1)+a=\frac{a^2+ab+a+ba+b^2+b+a+b+1+a+b+1}{2}+a=\frac{a^2+b^2+2ab+3a+3b+2}{2}+a=\frac{a^2+b^2+2ab+5a+3b+2}{2}$
$\frac{1}{2}(a+b+1)(a+b)+a+\frac{1}{2}(a+b+1)\cdot 2=\frac{a^2+ab+ba+b^2+a+b}{2}+a+\frac{2a+2b+2}{1}=\frac{a^2+b^2+2ab+5a+3b+2}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Contradictory result when testing Linear independence using Gaussian elimination Consider a set of vectors - (2,3,1) , (1,-1,2) and (7,3,8).
I want to find if its linearly dependent or independent.
Putting it as:
\begin{equation}
2a + b + 7c = 0 \\
3a - b + 3c = 0 \\
a + 2b + 8c = 0
\end{equation}
If I use Gaussian ... | First note that
$$
\left[\begin{array}{ccc|c}
2&1&7&0 \\
3&-1&3&0 \\
1&2&8&0 \\
\end{array}\right]
\Rightarrow
\left[\begin{array}{ccc|c}
1&0&2&0 \\
0&1&3&0 \\
0&0&0&0 \\
\end{array}\right]
$$
Using matrix equations
We have
$$
a\begin{bmatrix}
1\\
0\\
\end{bmatrix}
+b\begin{bmatrix}
0\\
1\\
\end{bmatrix}
+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How should I have derived the extra solution my book lists? I am working through some geometry exercises, one of which is to find the point(s) where the circle defined by $x^2 + y^2 = 16$ touches the line defined by $x - 2y = 4$.
I defined $x$ in terms of $y$ starting from the equation for the line: $x = 4 + 2y$.
Subst... | plugging $$x=4+2y$$ in $$x^2+y^2=16$$ we get $$(4+2y)^2+y^2=16$$ from here we get
$$16+4y^2+16y+y^2=16$$ or equivalent to $$y(5y+16)=0$$ can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the angles using the given conditions
I got stuck in last step. What should I do next? Please help me
| $$
\text{Let } a = \frac{\alpha}{2}, b = \frac{\beta}{2}, c = \frac{\gamma}{2} \\
\sin{\frac{\alpha-\beta}{2}}+\sin{\frac{\alpha-\gamma}{2}}+\sin{\frac{3\alpha}{2}} = \frac{3}{2} \\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin{3a} = \frac{3}{2}
\\
\sin\left(a-b\right)+\sin\left(a-c\right)+\sin\left(a-b+a-c+a-a+a+b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying a rational function with infinite series in numerator and denominator We're working with Taylor Series and I have to simplify the rational expression
$$ \frac{x-2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2\frac{x^6}{6!} + \cdots}{x- \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}.$$
I'm ... | The long division mentioned in the comments is likely the way you are 'supposed' to answer the question. Here's the $\operatorname{csc}$ method:
$$\begin{align}R & = \frac{x - 2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2 \frac{x^6}{6!} + \ldots}{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determine the diameter of a circle given its length is a two digit number and when the number is reversed the length of a chord is obtained. Problem:
The length of diameter AB is a two digit integer. Reversing the digits gives the length of a perpendicular chord CD.The distance from their intersection point H to the ce... | Yes, your attempt to the problem is correct. here is another way of solving it.
The equation of a circle is $$x^2 + y^2 = R^2\Rightarrow x^2 + y^2 = (\frac{d}2)^2$$
where, R and d are the radius and the diameter of the circle respectively.
So, the equation for the top semi circle is $$y = \sqrt{(\frac{d}2)^2 - x^2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Inequality with $a+b+c=1$ Let $a,b,c\in\mathbb{R^+}$ such that $a+b+c=1$. Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\geq\frac{11}{2}.$$
I am trying to resolve this problem but actually i found some issues:
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=3+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge... | $$\sum\frac{1}{a+b}+3(ab+bc+ca)\geq\frac{11}{2}$$
$$\sum\frac{(a+b+c)^2}{a+b}+\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$\sum\frac{a^2}{b+c} + 4 +\frac{3}{2}(1-\sum a^2)\geq\frac{11}{2}$$
$$2\sum\frac{a^2(a+b+c)}{b+c}\geq3\sum a^2$$
$$2\sum\frac{a^3}{b+c}\geq\sum a^2$$
Using CSB we get:
$$(\sum\frac{a^3}{b+c})(\sum a(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
linear transformation eigenvectors and eigenvalues
Consider the operator defined by $T(x, y, z) = (-x+2y, 3y, 0)$.
*
*Find the eigenvalues of T and all corresponding eigenvectors.
*Find each generalised eigenvector corresponding to each eigenvalue.
So for 1, I found the matrix with respect to the standar... | Operator
$$
\mathbf{T} =
\left[
\begin{array}{rrr}
-1 & 2 & 0 \\
0 & 3 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
$$
Eigenvalues
To find the eigenvalues, compute $p(\lambda)$, the characteristic polynomial.
$$
p (\lambda ) = \det \left( \mathbf{T} - \lambda \mathbf{I}_{3} \right) = \det
\left[
\begin{array}{ccr}
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1915180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?
For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\fra... | Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Hence, by C-S $\sum\limits_{cyc}\tan^2\frac{\alpha}{2}=\sum\limits_{cyc}\frac{yz}{x(x+y+z)}=\sum\limits_{cyc}\frac{y^2z^2}{xyz(x+y+z)}\geq\frac{(xy+xz+yz)^2}{3xyz(x+y+z)}\geq1$.
The equality occurs for $x=y=z$, id est, for $a=b=c$.
In another hand, our sum $\rightarrow+\infty$ fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1915376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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proof using the mathematical induction Prove $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer for every positive integer k.
In proving for (n+1) integer,the expression is integer,I found $(n+1)^7$ term.I use binomial theorem to expand but finally it won't work(some term become integer While so... | Based on other similar questions, you might think to first prove that
$$ \frac{k^7 - k}{7} $$
is always an integer. Once you have done so, you can simplify
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=
\frac{k^7 - k}{7}
+\frac{k^5}{5}+\frac{2k^3}{3}+\frac{2k}{15}
$$
You can do the same with $5$ and $3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Three digit number such that $A^2+B^2+C^2$ is divisible by $26$. Find all three digit natural numbers $ABC,(A \neq 0)$ such that $A^2+B^2+C^2$ is divisible by $26$.
Could someone give me some hint as how to approach this question as I am not able to initiate?
| As $26=2\cdot13$
and $x^2\equiv x\pmod2\implies A^2+B^2+C^2\equiv A+B+C\pmod2$
So, there should be even number of odd values among $\{A,B,C\}\ \ \ \ (1)$
Again, as $A,B,C$ are decimal digits, $$0\le A,B,C\le9\ \ \ \ (2)$$
Now for any integer $y, y^2\equiv0,\pm1,\pm4,\pm3\pmod{13}$
For $13\mid(A^2+B^2+C^2),$
we need $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots
The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.
How would I find the minimum value of $a^2 + b^2$ ?
| We get the solutions
$$
0 = x^2 + a x + b + 2 = (x + a/2)^2 - a^2/4 + b + 2 \iff \\
x = \frac{-a \pm \sqrt{a^2 - 4b - 8}}{2}
$$
which are real for
$$
a^2 - 4b - 8 \ge 0
$$
We now want to solve the optimization problem
\begin{matrix}
\min & a^2 + b^2 \\
\text{w.r.t} & a^2 - 4b - 8 \ge 0
\end{matrix}
The above image sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Stars and Bars with bounds This question is related to Error solving “stars and bars” type problem
I have what I thought is a fairly simple problem: Count non-negative
integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$
such that $0 \leq x_1 \leq 9$.
The difference here is on the constraint. It b... | We cannot have $3$ $x_i= > 9$ because the result might be $>= 30$
We can have two $x_i > 9$ because the result may be in the bounds , example $10+10+1+1+1=23$ and $11+10+1+1+0=23$
1 ) let's see the case where there is one or two $x_i > 9$
We can have one $x_i > 9$ because the result may be in the bounds , example $10+9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1922819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Find all functions satisfying the functional equation $ xf(x) + f(1-x) = x^3 - x $
Find all functions, for all real x, that satisfy the following functional equation:
$$ xf(x) + f(1-x) = x^3 - x $$
| We have
\begin{align*}
xf(x) + f(1-x) &= x^3-x\\
(1-x)f(1-x) + f(1-(1-x)) &= (1-x)^3 - (1-x)
\end{align*}
and hence
\begin{align*}
f(x) + (1-x)f(1-x) = (1-x)^3 - (1-x)
\end{align*}
Multiplying the first equation by $1-x$ and subtracting from the third equation, we get
\begin{align*}
(x(1-x) - 1)f(x) &= (x^3-x)(1-x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected Run before Absorption Consider a $4$ state Markov Chain that includes the row-stochastic
transition matrix
$$P = \left(\begin{array}{cccc} 1/4&1/4&1/4&1/4 \\ 1/2&0&0&1/2 \\
1/3&1/3&1/3&0 \\ 0&0&0&1 \end{array}
\right).$$
Assume that we start the chain in state $1$. I would like ... | Considering this,
assume $$Q = \left(\begin{array}{cccc} 1/4&1/4&1/4 \\ 1/2&0&0 \\
1/3&1/3&1/3 \end{array}
\right)$$
Then the expected number of transitions from state $i$ to $j$ is
$$N=(I-Q)^{-1}=\left(\begin{array}{cccc} 2.2857 &0.8571&0.8571 \\ 1.1429 & 1.4286 & 0.4286\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$
We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.
| Note that $0 \le x\le a(a-1)$,
\begin{align*}
x &= \sqrt{a-\sqrt{a+x}} \\
x^2 &= a-\sqrt{a+x} \\
a+x &= (a-x^2)^2 \\
a+x &= a^2-2ax^2+x^4 \\
x^4-2ax^2-x+a^2-a &= 0 \\
(x^2-x-a)(x^2+x-a+1) &= 0 \\
\end{align*}
$$x=\frac{1 \pm \sqrt{4a+1}}{2} \quad \text{or} \quad \frac{-1 \pm \sqrt{4a-3}}{2}$$
For $a\ge 1$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
$\int \frac{\sqrt{1-36 x^2}}{x^2}\,dx$ I need help integrating
$$\int \frac{\sqrt{1-36x^2}}{x^2} \ dx$$
using trigonometric substitution.
My first step was simplifying the integral down to $$\int \frac{\sqrt{36(\frac{1}{36}-x^2)}}{x^2} \ dx$$
and use $x=\frac{1}{6} \sin \theta$ to perform trigonometric substitution. I ... | Consider the following substitution $x=\frac{\sin (u)}{6}$ and $dx=\frac{\cos(u)}{6}$. Then, $\sqrt{1-36x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\sin^{-1}(6x)$. Hence,
$$
\int \frac{\sqrt{1-36x^2}}{x^2} \,dx=6\int \cot^2(u) \,du=6\int (\csc^2(u)-1) \,du=-6\cot(u)-6u.
$$
Substituting back, we get
$$
-6\cot (\sin^{-1}(6x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integration of $\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$ Integral of $\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$
This is the answer, but I'm not sure how to get to it...
http://www.wolframalpha.com/input/?i=Integral+of+(x-1)%2F((sqrt(1%2Bsqrt(1-x%5E2))))dx
| $$I=\int\frac{x-1}{\sqrt{1+\sqrt{1-x^2}}}dx$$
Start from substitution $t^2=1-x^2,\quad x=\sqrt{1-t^2},\quad dx=-\frac{t}{\sqrt{1-t^2}}dt$
$$
I=\int\frac{t(1-\sqrt{1-t^2})}{\sqrt{1+t}\sqrt{1-t^2}}dt=\int\frac{t}{\sqrt{1+t}\sqrt{1-t^2}}dt-\int\frac{t}{\sqrt{1+t}}dt=I_1-I_2+const
$$
$I_2$ is trivial and equal to $\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
| Lazy people like me will do this as follows. The function
$$f(x) = \frac{1}{(x+3)(x+4)^2(x+5)^3}$$
has a simple pole at $x = -3$. The principal part of the Laurent expansion around this point (i.e. the sum of the singular terms in the expansion) is:
$$s_1(x) = \frac{1}{8}\frac{1}{x+3}$$
If you calculate the principal p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
CDF for Negative Binomial Distribution I am trying to show that the following statement is true.
$$
\sum_{x = r}^{X}\binom{x-1}{r-1}p^r(1-p)^{x-r} =
\sum_{x = r}^{X}\binom{X}{x}p^x(1-p)^{X-x}
$$
Where $X$ and $r$ and $p$ are constants, with $X \geq r$, and $ 0 \leq p \leq 1.$
How did I get there? Well, this is the stor... | Suppose we seek to prove that
$$\sum_{k=0}^m {n+k-1\choose k} q^k
= \sum_{k=0}^m {m+n\choose m-k} q^{m-k} (1-q)^k.$$
The RHS is
$$\sum_{k=0}^m {m+n\choose k} q^{k} (1-q)^{m-k}.$$
Extracting the coefficient on $q$ on the RHS we get
$$[q^p] \sum_{k=0}^m {m+n\choose k} q^{k} (1-q)^{m-k}
= \sum_{k=0}^m {m+n\choose k} [q^p]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to solve $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
Solve the equation $z^2 + \left( 2i - 3 \right)z + 5-i = 0$ in $\mathbb{C}$
I was thinking on it for a few minutes and came up with a few ideas (none of them worked).
My first idea: Use the quadratic formula. Is that allowed? If so, I got to this:
... | By solving $(a+ib)^2=8i+15$ we get
$a^2-b^2=15$, $2ab=8$ which are satisfied by $a=4$ and $b=1$.
Hence we can go on with the quadratic formula:
$$z = \frac{-(2i-3) \pm \sqrt{(2i-3)^2-4(5-i)}}{2} = \frac{3-2i \pm \sqrt{-8i-15}}{2}\\
= \frac{3-2i \pm i\sqrt{8i+15}}{2}=\frac{3-2i \pm i(4+i)}{2}=\frac{(3\mp 1)+ i(-2\pm 4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I don't know what is wrong with my graph for this problem - at most how many students take none of these courses? Is my graph wrong? If it is, what is the correct one? I used this graph to solve the problem and got a wrong answer. It is highly appreciated if one can show me the correct graph.
Of the 60 students at Hop... |
Total of all the classes
$$
a+b+c+2d+2e+2f=133-66=67\tag{1}
$$
Algebra
$$
a+d+e=40-22=18\tag{2}
$$
Trigonometry
$$
b+d+f=48-22=26\tag{3}
$$
Geometry
$$
c+e+f=45-22=23\tag{4}
$$
Number of students in one of these classes
$$
t=a+b+c+d+e+f+22\tag{5}
$$
Number of students not in any class
$$
\begin{align}
60-t
&=38-a-b-c-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions to ceiling equation system
Prove that there does not exist an $x$ with $1000 \leq x \leq 1990$ that can be expressed in the forms $$\dfrac{10000}{x} = \left \lceil\dfrac{10000}{x} \right \rceil -\dfrac{1}{m} \quad \text{and} \quad \dfrac{10000}{x-1} = \left \lceil\dfrac{10000}{x-1} \right \rceil-\dfrac{1}{k}... | Some necessary conditons can be found, for example both $x \cdot 10000 \bmod x$ and $(x-1) \cdot 10000 \bmod (x-1)$ must be even multiples of triangular numbers. Unfortunately I did not see how to use those for a systematic approach to find solutions. (Not that it matters, but I am curious what the source or backgroun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Generalizing proof of AM-GM inequality? For two positive real numbers $x$ and $y$, we can prove the AM-GM inequality between them.
$$\left(\sqrt{x}-\sqrt{y}\right)^2=x+y-2\sqrt{xy}\geq 0$$
$$\implies \frac{x+y}{2}\geq\sqrt{xy}$$
Can this be generalized easily to the case of more numbers?
| If $n=2^k$, where $k\in\mathbb N$, so we can use $\frac{x+y}{2}\geq\sqrt{xy}$ for non-negatives $x$ and $y$.
If $n>2^k$ we can use a similar to the following reasoning.
We'll prove that $\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$ for non-negatives $x$, $y$ and $z$.
Indeed, by AM-GM for four variables we obtain:
$$\frac{x+y+z+\s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $ \frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}} $ . Prove
$$
\frac 1 2 \cdot \frac 3 4 \cdot \frac 5 6 \cdots \frac{2n-1}{2n} < \frac 1 {(2n+1)^{0.5}}
$$
Can this be done by induction using the pi function.
If no, why not.
| Hint. If you want to prove it by induction, you may observe that, for $n\ge1$,
$$
(2n+2)^2=4n^2+8n+4>4n^2+5n+3=(2n+1)(2n+3)
$$ giving
$$
\frac1{2n+2}<\frac1{\sqrt{2n+1}\cdot \sqrt{2n+3}}, \qquad n\ge1,
$$ which is equivalent to the inductive step:
$$
\frac1{\sqrt{2n+1}}\cdot \frac{2n+1}{2n+2}<\frac1{\sqrt{2n+3}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Taylor Series of $ \cfrac{x}{1+x} $ at $ x = -2 $ I am trying to find the taylor series for the following:
$$ \cfrac{x}{1+x} \ \text{ at } \ x = -2 $$
The following are my steps:
$$ \cfrac{x}{1+x} = 1 - \cfrac{1}{1+x} \ \text{ (by long division) } $$
Taylor series of $ \cfrac{1}{1+x} $ at $ x = 0, $
$$ \begin{align} \c... | Hint. Note that
$$\frac{x}{1+x}=\frac{x}{-1+x+2}=\frac{-(x+2)+2}{1-(x+2)}=1+\frac{1}{1-(x+2)}.$$
Now use the result: for $|z|<1$ then
$\displaystyle \cfrac{1}{1-z} = \displaystyle\sum_{n=0}^{\infty}{ z^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1942071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Showing that $\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6$ How does one evaluate the following limit?
$$\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)$$
The answer is $6$.
How does one justify this answer?
Edit: So it really was just combine the fraction and use L'hopita... | $\lim_\limits{x\to1}\frac {23(1-x^{11})-11(1-x^{23})}{(1-x^{11})(1-x^{23})}$
$\lim_\limits{x\to1}\frac {12-23x^{11}+ 11x^{23}}{(1-x^{11})(1-x^{23})}$
Now we could apply L'Hopitals at this point, or we can use algebra.
Using algebra, numerator and denominator both divide by $(x-1)^2$
$1-x^{11} = (1-x)\sum_\limits{i=0}^{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 7
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Finding $m$ so that the roots of $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$
Find $m$ so that the roots of the equation $x^2 - (3m - 1)x + m^2 - 2 = 0$ lie inside the interval $(1, 5)$.
Having $x_1, x_2 \in (1, 5)$, we get $x_1 + x_2 \in (2, 10)$ and $x_1x_2 \in (1, 25)$. By solving these two ine... | Notice that $\Delta = 5m^2 -6m +9 > 0, \forall m \in \mathbb{R}.$ Hence,
$x = \dfrac{3m-1 \pm \sqrt{5m^2 -6m +9}}{2} \in \mathbb{R}.$
$(1):~~~\dfrac{3m-1 - \sqrt{5m^2 -6m +9}}{2} > 1 \implies (3m-3)^2 > 5m^2 -6m +9 \implies 4m(m -3) > 0 \implies m < 0 ~~\text{or}~~ m > 3.$ But, only $ m > 3$ satisfies the condition ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1946714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Does this trig equation have no solution? $\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$
Solve the following trig. equation for $x$
$$\sqrt 2\sin \left(\sqrt 2x\right)+\sin (x)=0$$
My try:
I divided by $\sqrt{\left(\sqrt2\right)^2+1^2}=\sqrt 3$,
$$\frac{\sqrt 2}{\sqrt 3}\sin\left(\sqrt 2x\right)+\frac{1}{\sqrt 3}\... | $\sqrt{2} \sin(\sqrt{2}x)$ goes from $-1$ to $1$ on the interval $[(2n-1/4) \pi/\sqrt{2}, (2n+1/4)\pi/\sqrt{2}]$ and from $1$ to $-1$ on the interval $[(2n+3/4)\pi/\sqrt{2}, (2n+5/4)\pi/\sqrt{2}]$. On each of these intervals
the derivative of $\sqrt{2}\sin(\sqrt{2}x)$ has absolute value at least $\sqrt{2}$. Thus $\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Find the set of $n\in\Bbb Z^+$ with $M=\{n,n+1,n+2,n+3,n+4,n+5\}$ partitionable into two sets
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numb... | Hint 1: Start by proving that each set of the partition must have 3 elements. To do this just argue that, for $n \geq 2$, the product of the smallest 4 elements is higher than the product of the largest 2.
Hint 2: For all $n >0 $ we have
$$(n+3)(n+4)(n+1)>(n+5)(n+2)n$$
Hint 3 one of the sets of your partition must con... | {
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"url": "https://math.stackexchange.com/questions/1948239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove by induction that $\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $2^{n}$ Prove by induction that :
$\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $ 2^{n}$
My proof is :
At $n=1$
$$\frac{\left ( 3+\sqrt{5} \right ... | An idea for you assuming the claim's true for exponents up to $\;k\;$ . Observe that I begin from the middle of your work:
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} = \left [ \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k} \right ]\left [ \left ( 3+\sqrt{5} \right )+\left ( 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving $E(X)$ and $Var(X)$ for a random variable that has a discrete uniform distribution problem, what am I doing wrong I am having a problem proving $E(X)$ and $Var(X)$ for a random variable that has a discrete uniform distribution. The question and my current workings are below, can someone please indicate what I a... | $$E(X^2) = \sum_{k=a}^{a+s-1} k^2P(X=k) = \frac 1{a+s-1}\sum_{k=a}^{a+s-1}k^2 = \frac{ (s-1)(6a^2 + 6as-6a + 2s^2 + 1 - 3s)}{6(a+s-1)}$$
$$Var(X) = E(X^2) - E(X)^2 = \frac{ (s-1)(6a^2 + 6as-6a + 2s^2 + 1 - 3s)}{6(a+s-1)} - a^2 - \frac {s^2 - 2s + 1}4 - a(s-1)$$
Now you could go ahead and simplify this, or using argumen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why do some divisibility rules work only in base 10? Aside from the zero rule (a number ending with zero means it is divisible by the base number and its factors), why is it that other rules cannot apply in different bases?
In particular for 3 one can just sum all digits and see if it is divisible. What relation exists... | There are other rules for other bases.
To understand why the rule for $3$ works in base $10$, you need to write your number $x=d_n\times 10^n+\dots+d_1\times 10^1+d_0$ where the $d_i$ are the digits of $x$ in base $10$.
You can notice that $10=3\times 3+1$, $10^2=3\times 33+1$, $10^n=3...3\times 3+1$ (where $3...3$ has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 4
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showing $\frac{x}{e^{x}-1} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $ I want to show
$\frac{x}{e^{x}-1} = -\sum_{k=0}^{\infty} x e^{kx} = \frac{\ln(e^x)}{e^x-1} = -\frac{\ln[1-(1-e^x)]}{1-e^x} = \sum_{n=1}^{\infty} \frac{(1-e^x)^{n-1}}{n} $
The first step seems trivial, what i am confused is the process of
\begin... |
In order to show the identity
\begin{align*}
\frac{x}{e^x-1}=\sum_{n=1}^\infty \frac{\left(1-e^x\right)^{n-1}}{n}\qquad\qquad |1-e^x|<1\tag{1}
\end{align*}
we recall that
\begin{align*}
\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\qquad\qquad\qquad |x|<1
\end{align*}
We obtain
\begin{align*}
x&=\ln(e^x)=\ln(1-(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Challenging Integral [indefinite] Integrate: $$\int \frac{x^2+n(n-1)}{(x\sin x+n\cos x )^2}dx$$
I've been beating my head around this problem for quite some time now, but I've got nowhere. I'd request the person writing the solution to please explain his thought process because I would like to learn how to appraoch suc... | Using the identities $n^2+x^2=(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2$ and
$n=\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)$, we can rewrite the integral as
$\displaystyle \int\frac{(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2-[\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)]}{(n\cos x+x\sin x)^2} dx$
$\displaystyle=\int\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Roots of $x^2-2ax+b=0,~a^2>>b$, stable algorithm
Find a stable algorithm for the computation of the roots of the equation $x^2-2ax+b=0$ if $a,~b>0$ and $a^2>>b$.
Attempt. The roots are $x_1=a-\sqrt{a^2-b}$ and $x_2=a+\sqrt{a^2-b}$, where we have a cancelation error for the computation of $x_1$, since $a^2-b\cong a^2$... | *
*About one part of your question, no purely mathematical answer can be given because $\displaystyle x_1=\frac{b}{a+\sqrt{a^2-b}}$ is strictly the same as $a-\sqrt{a^2-b}.$ It is a floating point issue.
*About the other part, yes, $\frac{b}{2a}$ is a good approximation.
This can be attested by applying Taylor expan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Binomial identity in a Change of Basis Matrix I need the following identity in my research and have a rather ugly proof but cannot find anything more elegant.
If $L\colon\mathbb{Q}[m]\to\mathbb{Q}[m]$ is a linear map which satisfies
$L\left(\binom{m}{c+d}\right)=L\left(\binom{m}{c}\right)L\left(\binom{m}{d}\right)$
fo... | Vandermonde's identity and diagonal summation can be used to give a non-inductive proof:
\begin{align*}
L\left(\binom{m+a}{c}\right)L\left(\binom{m+b}{d}\right)&=L\left(\sum_{i=0}^c\binom{m}{i}\binom{a}{c-i}\right)L\left(\sum_{j=0}^d\binom{m}{j}\binom{b}{d-j}\right)\\
&=\sum_{i=0}^c\sum_{j=0}^d\binom{a}{c-i}\binom{b}{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the volume of the solid obtained by rotating the region bounded by the given curves about the $x$-axis. $y=x^{3/2}$, $y=8$, $x=0$ Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the $x$-axis.
$y=x^{3/2}$, $y=8$, $x=0$
What I did... | Since the revolution is around the $x$-axis, each cylindrical shell at $x$ for $x\in [0,4]$ (where $4=8^{3/2}$) is actually an annulus with inner radius $r(x)=x^{3/2}$ and outer radius $R(x)=8$.
The volume of this shell is $(\pi R^2(x) -\pi r^2(x))\cdot dx$. Hence the total volume can be obtained by "summing" the volum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.