Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge 5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge 5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\... | Viewed geometrically, $x+y+z=5$ is a plane and $x^2+y^2+z^2=a^2$ is a sphere with radius $a$ at the origin; the smallest (positive) $a$ for which the sphere touches the plane corresponds to the shortest distance between the plane and the origin, and the point of tangency represents the smallest possible value of $x^2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\sum_{n=1}^{\infty} a_n$ is a series and {$s_n$} is the sequence of partial sums. If $s_n = \frac{n^2+1}{4n^2-3}$, find $\sum_{n=1}^{\infty} a_n$ Only idea I have is to use $a_n = s_n-s_{n-1}$, which would mean $a_n = \frac{n^2 + 1}{4n^2 - 3}-\frac{(n-1)^2 + 1}{4(n-1)^2 -3}$. But I'm not sure if I'm thinking in th... | Thanks to a commenter @MartinR, I was able to find $a_n$. Originally I was confused as to what the question was asking to find $a_n$ as a rule or a single value/sum, and Martin's comment helped me realize that we're looking for a single value.
$$\sum_{n=1}^{\infty} a_n = \lim_{n \rightarrow \infty} s_n = \lim_{n \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$.
The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$.
This is what I tried below.
I am not... | Following your work with $F(2) = 4 \implies c = 8a^2$. Thus: $F(x) = \dfrac{x^4}{4} -2a^2x^2+8a^2= \left(\dfrac{x^2}{2} - 2a^2\right)^2+8a^2 - 4a^4\ge 8a^2-4a^4$. This minimum value equals to $0$. So $8a^2 - 4a^4 = 0\implies 4a^2(2-a^2)=0\implies a = 0,\pm \sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}.$ I want to evaluate the following integral ($\theta_0>0$)
\begin{equation*}
\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta
\end{equation*}
So I thought about $d(\cos\theta)=-\si... | Spectree’s step is correct and can be carried on as follows:
$$\begin{aligned}\int_{\theta_0}^{\pi} \sqrt{\frac{1-\cos\theta}{\cos\theta_0-\cos\theta}} d\theta &= - \int_{\theta_0}^{\pi} \frac{d(\cos\theta)}{\sqrt{(\cos\theta_0-\cos\theta)(1+\cos\theta)}}\\ &\stackrel{y=\cos \theta}{=} -\int_{a}^{-1} \frac{d y}{\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4340894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
A point $(X, Y )$ is chosen randomly from the unit circle. Let $R = \sqrt{(X^2 +Y^2)}$. Find the joint density function of the random vector $(X, R)$ A point $(X, Y )$ is chosen randomly from the unit circle.
Let $R = \sqrt{(X^2 +Y^2)}$.
Find the joint density function of the random vector $(X, R)$
My try:
$R=\sqrt{X^{... | $R = \sqrt{X^2 + Y^2}$ where $X$ and $Y$ are uniformly distributed on a unit disk i.e. $x^2 + y^2 \leq 1$. So you should have $0 \leq r \leq 1$ and not $0 \leq r \leq \sqrt2$.
I will use the Jacobian method.
Use the fact that $ \displaystyle f_{XY}(x,y) = \frac{1}{\pi}~$ over unit circle.
As $r = \sqrt{x^2 + y^2}$, $y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
When is the integer between twin primes a perfect square? Explain why there exists an integer $n$ such that $4n^2 − 1 = p$ Suppose that $$ and $ + 2$ are both prime numbers.
a) Is the integer between $$ and $ + 2$ odd or even? Explain your answer.
All prime numbers are odd except for $\pm 2.$
So the integer between the... | Given$\quad A^2+B^2=C^2\quad $ the most common means of generating Pythagorean triples is Euclid's formula
$A=m^2-k^2,\quad B=2mk, \quad C=m^2+k^2.\quad$ We can restrict this formula to generating only primitive triples and odd square multiples of primitives if we let
$\quad m=(2n-1+k),\quad $ i.e.
$A=(2n-1+k)^2-k^2,\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How many ways to deal with the integral $\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}$? I tackle the integral by rationalization on the integrand first.
$$
\frac{1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{2 x}
$$
Then splitting into two simpler integrals yields $$
\int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{1}{... | Seek $f$ so$$\frac{\sqrt{1+x}+\sqrt{1-x}}{2x}=f(1+x)-f(1-x),$$e.g.$$f(y):=\frac{\sqrt{y}}{2(y-1)}.$$You want to evaluate$$\int(f(1+x)-f(1-x))dx=F(1+x)+F(1-x)+C$$with $F^\prime(y)=F(y)$.
Just about any concise solution technique will exploit the above facts. Your approach is equivalent to next taking $y=z^2$, so$$F(y)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How many squares with vertices on $x^2y^2 =1$ are possible? We can form a square with vertices on the curve $x^2 y^2 = 1$ and which does not intersect this curve.
Can we get any other square with vertices on $x^2 y^2 = 1$ and which does not intersect this curve? Can anyone please tell me elaborately?
| Yes. Rotating and scaling so as to keep the vertices on the curve, we get a square with vertices at $\pm(a,1/a)$ and $\pm(-1/a,a)$ with $a>0$. We can rotate clockwise until there is an intersection when the slope of the side from $(a,1/a)$ to $(-1/a,a)$ is less than the slope at $(a,1/a)$. This condition on $a$ may be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$\binom{n}{r}$ where $r$ can only be chosen in groups of $k$ I just started my learning in combinatorics, and I know that number of ways to choose $r$ items from $n$ items without order is given by $\binom{n}{r}= \frac{n!}{r!(n-r)!}$. However, I couldn't figure out the number of ways to choose $r$ items from $n$ items,... |
The number of valid paths of length $n$ consisting of $r$ horizontal $(1,0)$-steps and $n-r$ vertical $(0,1)$-steps with at most $k$ consecutive horizontal steps is
\begin{align*}
\color{blue}{\sum_{q=0}^{\left\lfloor\frac{r}{k+1}\right\rfloor}\binom{n-(k+1)q}{r-(k+1)q}\binom{n-r+1}{q}(-1)^q}\tag{1}
\end{align*}
where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Simplify $A(t)=\frac{1-t}{1-\sqrt[3]{t}}+\frac{1+t}{1+\sqrt[3]{t}}$ Simplify $$A(t)=\dfrac{1-t}{1-\sqrt[3]{t}}+\dfrac{1+t}{1+\sqrt[3]{t}}$$ and calculate $A(3\sqrt3).$ For $t\ne\pm1$ we have, $$A=\dfrac{(1-t)(1+\sqrt[3]{t})+(1+t)(1-\sqrt[3]{t})}{1-\sqrt[3]{t^2}}=\\=\dfrac{2-2t\sqrt[3]{t}}{1-\sqrt[3]{t^2}}$$ What to do ... | As a general principle, sometimes things look simpler if we perform a suitable substitution: let $u = \sqrt[3]{t}$, so that $t = u^3$ and $$A(t) = A(u^3) = \frac{1 - u^3}{1 - u} + \frac{1 + u^3}{1 + u}.$$ Now it becomes obvious that we either need to factor the numerators, or put everything over a common denominator.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Surface delimited by an elliptical cylinder and z+y = 9. We're covering Stokes' Theorem in class right now, and I can't understand anything. I'm struggling to solve even the most trivial examples. Here's one I can't solve: Calculate the surface integral of $\mathbf{F}(x,y,z) = (z,x,y)$ where the surface $S$ is the surf... | I am going to interpret the question as computing the line integral of $\mathbf{F}(x, y, z)=(z, x, y)$ over the closed curve given by the intersection of the elliptical cylinder $\frac{x^2}{4}+\frac{y^2}{9} = 1$ and the plane $y+z=9$. The reason I interpret it like this is twofold: the elliptical cylinder and the plane... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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if a+b+c+d=5m, what are the sets of values of a,b,c,d for which 5 divides none of them? Let's assume $a+b+c+d$ is an integer multiple of $5$, that is
$$a+b+c+d=5m-------(1)$$
For all integer $m,$
where $5$ does not divide any of $a,b,c$ and $d$. What are all the sets of values of $a,b,c,d$?
The sets I could find are:
$... | You missed $5w+3,5x+4,5y+4,5z+4$. Also your first set can be rendered with the last term as $5z+2$.
We may render each term as $\in\{1,2,3,4\}\bmod 5$ and require four such numbers to add up to a multiple of $5$. The sum must then be $\in\{5,10,15\}$.
Thus we now have a restricted partitioning problem. Clearly only $1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof verification: Given $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, show that $\lim _{x\to 0}\frac {f(x)}{x}=0$. Given that $\lim_{x\to 0}\frac{f(2x)-f(x)}{x}=0$ and $\lim_{x\to 0} f(x)=0$, it is to be proven that $\lim _{x\to 0}\frac {f(x)}{x}=0$.
Proof: Let $\epsilon\gt 0$ be fixed.
\begin{ali... | Let $\epsilon > 0$ be arbitrary, and let $x_\epsilon$ be small enough so that
$
\Bigg| \dfrac{f \big( x \big) - f \big(\frac {x} {2} \big)}{\frac {x} {2}} \Bigg|
\le \epsilon
$ for all $|x| \le x_\epsilon$.
Then,
$$
\Bigg| \dfrac{f \big(x \big)}{x} \Bigg| =
$$
$$
\dfrac{1}{2} \Bigg|
\dfrac{f \big(x \big) - f \big(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to find the formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}}$, where $n\in N$? By the generalization in my post,we are going to evaluate the integral $$\int_{0}^{\infty} \frac{d x}{\left(x^{2}+1\right)^{n}},$$
where $n\in N.$
First of all, let us define the integral $$I_n(a)=\int_{0}... | If you use complex analysis, then there is a pole at $x = i$
If we take the contour of the semi-circle in the upper half-plane, then when $n\ge 1$ the integral along the semi-circle vanishes. That means that we can focus on the residual at $i.$
$\int_0^\infty \frac {1}{(x^2 + 1)^n} \ dx = \frac {\pi i}{(n-1)!} \frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
$S_n = \frac{12}{(4-3)(4^2-3^2)} + \frac{12^2}{(4^2-3^2)(4^3-3^3)} + ... + \frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})}$
So I know that in order to find the series we need to change the form because we can cancel the terms $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} = \frac{A}{4^n-3^n} + \frac{B}{4^{n+1}-3^{n+1}}$$
But my q... | $$\frac{12^n}{(4^n-3^n)(4^{n+1}-3^{n+1})} =\frac {4^n\cdot3^n}{(4^n-3^n)(4\cdot4^n-3\cdot3^n)}$$
Divide top and bottom by $3^{2n}$ and write $x=\frac{4^n}{3^n}$ and we get
$$\frac{x}{(x-1)(4x-3)}$$
In partial fractions, this is $$\frac{1}{x-1}-\frac{3}{4x-3}=\frac{1}{x-1}-\frac{1}{\frac43x-1}$$
So the expression is $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating $\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)} $ I used all the trig and log tricks but still can't compute this limit
$$\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}.$$
I tried the following:
\begin{align*}\lim_{x\to0}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}
&= \lim_... | Just using the standard limit $\lim_{t\to 0}\frac{\ln (1+t)}{t} = 1$ you get
\begin{eqnarray*}\frac{\ln\left(1+\sin^2(2x)\right)}{1-\cos^2(x)}
& = & \frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(2x)}\cdot \frac{\sin^2(2x)}{\sin^2 x} \\
& = & 4\cos^2 x \cdot \frac{\ln\left(1+\sin^2(2x)\right)}{\sin^2(2x)} \\
& \stackrel{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? How to calculate integral $\int_0^\infty \frac{x}{(e^{2\pi x}-1)(x^2+1)^2}dx$? I got this integral by using Abel-Plana formula on series $\sum_{n=0}^\infty \frac{1}{(n+1)^2}$. This integral can be splitted into two integrals with bounds from 0 to 1 and from 1 to infi... | Here is a brutal-force computation using contour integral.
Let $\operatorname{Log}(\cdot)$ be the logarithm with the branch cut $[0, \infty)$ so that $\operatorname{Arg}(z) \in (0, 2\pi)$. Also, for $s > 0$ we define
$$ I(s) = \int_{0^+ i}^{\infty + 0^+ i} f_s(z) \, \mathrm{d}z, \quad \text{where} \quad f_s(z) = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find power series solution for $y''-xy=0$ close to $x=0$. Find power series solution for $y''-xy=0$ close to $x=0$.
$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}-x\sum_{n=0}^\infty a_nx^n=0$
Then,
$a_{n+2}(n+2)(n+1)-a_{n-1}=0 \implies a_{n+2}(n^2+3n+2)-a_{n-1}=0 \implies a_{n+3}=a_n\cdot \frac{1}{(n+3)(n+2)}$
I got two problems:... | If you look at the pattern (not very simple, I agree), there is a closed form for the coefficients
$$a_n=\frac{3^{-2 n/3}\,\, \Gamma \left(\frac{2}{3}\right)}{ \Gamma \left(\frac{n+2}{3}\right) \Gamma \left(\frac{n+3}{3}\right)}\left(a_0+a_1 \cos \left(\frac{2 \pi n}{3}\right)+a_2\sin \left(\frac{2 \pi n}{3}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of the square harmonic series I stumbled across the following series reviewing some HW from a few years ago
$\sum_{i=1}^{n}\left(\sum_{j=i}^{n}\frac{1}{j}\right)^2$
i.e.
$(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n})^2+(\frac{1}{2}+\ldots+\frac{1}{n})^2+\ldots+(\frac{1}{n})^2$
This series equals $2n-\sum_{i=1}^{n}\f... | Proof by induction. For $n = 1$ it is easy to show that this holds. Assume the induction hypothesis for some $n$. Then we want to show that
$$
\sum_{p = 1}^{n + 1} \biggl( \sum_{q = p}^{n + 1} \frac{1}{q} \biggr)^2
\;=\;
2 (n + 1) - \sum_{p = 1}^{n + 1} \frac{1}{p}
\text{.}
$$
We have
$$
\biggl( \sum_{q = p}^{n + 1} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Finding extreme values of $ f :\mathbb R^2\rightarrow \mathbb R$, when the determinant $\Delta = AC - B^2 = 0$. We generally rely on the result:
[Let $ f$ be a real valued function from $\mathbb R^2$ with continuous partial derivatives at a stationary point $\vec a$ in $\mathbb R^2$. Let
$A = D_{11}f(\vec a)$,
$ B = D_... | Consider the problem:
Find and classify the extreme values (if any) of the function: $$f(x,y) = y^2+x^2y+x^4$$
Here we have,
$D_1f = \partial f/\partial x = 2xy +4x^3$
$D_2f = \partial f/\partial y = 2y +x^2$
$D_{11}f = \partial^2 f/\partial x^2 =2y +12x^2$
$D_{12}f = \partial^2 f/\partial x \partial y =2x$
$D_{22}f = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4373797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
n-th root in limits I'm having trouble with this kind of series: $$a_n = \sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})$$
I tried to make something like perfect square to simplify those $7$th root junk which is obviously impossible, and as $n$ tends to infinity L'Hopital's rule also can't solve the problem in this form.
The mul... | One more, multistorey, without series, way
$$\sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})=$$
$$=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-\sqrt[7]{1-\frac{4}{n}}\right)
=\\
=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-1-\sqrt[7]{1-\frac{4}{n}}+1 \right) =\\
=\sqrt{n}\sqrt[7]{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$
My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead ... | Using complex numbers:
$$
\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})
= \arg((5+\sqrt{3}i)(2+\sqrt{3}i))
= \arg(7+7\sqrt{3}i)
= \arctan(\sqrt{3})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$
I found the similar question in Quora. There was slightly a mistake.
$$(\tan\theta+\sin\theta)^2-n^2=4\sqrt{mn}$$
$$(\tan\thet... | Assume that $\theta\in (0,\pi/2)$ and hence $\tan\theta,\,\sin\theta>0$.
If $w=\sin\theta$, then $\tan\theta=\frac{w}{\sqrt{1-w^2}}$ and
$$
m=w+\frac{w}{\sqrt{1-w^2}}=\frac{w+w\sqrt{1-w^2}}{\sqrt{1-w^2}}
$$
Now
$$
F(n)=m^2-n^2-4\sqrt{mn}, \quad n\ge 0,
$$
is strictly decreasing and continuous $F(0)=m^2>0$ and $F(\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Why is $13^n\cdot 14 + 13\cdot 2^{2n}$ divisible by $9$ for any nonnegative integer $n$? Prove that statement
$13^n\cdot14+13\cdot2^{2n}$ is divisible by $9$ for any natural $n$
So, the first step is to represent $13^n$ as sum of
$(9+4)^n$ and what the next one?
Give me a hint, please.
I've no idea where to go for now.... | One can always do it by induction.
$13^{n+1} \cdot 14 + 13\cdot 2^{2(n+1)}=$
$13^n\cdot 14\times 13 + 13\cdot 2^{2n}\times 4=$
$13^n\cdot 14\times(9+4) + 13 \cdot2^{2n}\times 4=$
$13^n\cdot 14\times 9 + 4\times (13^n\cdot 14+ 13 \cdot2^{2n})$
which is a multiple of $9$ if $13^n\cdot 14+ 13 \cdot2^{2n}$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$ I am trying to evaluate this integral:
$$\displaystyle\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}\arctan{\sqrt{x^2+y^2+z^2}}dxdydz$$
I coverted the region by using polar coordinate system for xy-plane and it be... | A simpler way using symmetry.
Firstly denote $r=\sqrt{x^2+y^2+z^2}$ and split the integrand
$$
I=\int_0^1\int_0^1\int_0^1r\arctan r~dV=\frac\pi2\underbrace{\int_0^1\int_0^1\int_0^1r~dV}_{I_1}-\underbrace{\int_0^1\int_0^1\int_0^1r\operatorname{arccot} r ~dV}_{I_2}
$$
The former is the average distance from a point in a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$
Evaluate $$\lim_{x\to 0} \frac{\sin(x+\tan x)-2x\cos x}{x(\sin x^2)^2}$$
Wolfram alpha gives $\dfrac{-7}{20}$.
Here is my work
For $x \to 0 $
$\tan x \sim x$
$\sin(x+\tan x) \sim \sin2x$
$\sin2x \sim 2x$ (I wasn't sure about this but I evaluated the limit ... | You have messed up with asymptotic relations. You must be really really careful when you use them.
In this limit, we know from the denominator that we have to search for a $5$ degree expansion. Namely:
$$x\cdot(\sin(x^2))^2\,\,\sim\,\, x^5$$
Now, we set things ready for Tyalor-MacLaurin series:
$$\sin(t)=t-\frac{1}{6}t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In how many ways we can place 9 different balls in 3 different boxes such that in every box at least 2 balls are placed?
In how many ways we can place $9$ different balls in $3$ different boxes such that in every box at least $2$ balls are placed?
Approach 1:
Let $x_1$, $x_2$, $x_3$ denote the number of balls in boxe... | There are two small mistakes otherwise your work using the second approach is also correct. It should be -
$ \displaystyle \small 3^9 - \underbrace{\left[{3 \choose 1}2^9 - {3 \choose 2}1^9 \right]}_{\Large {(a)}} - \underbrace{\left[{3 \choose 1}{9 \choose 1} (2^8 - \color { blue } {2}) - \color { blue } {2} \cdot {3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Domain of solution of differential equation $y’=\frac{1}{(y+7)(t-3)}$ If we are given the initial-value problem $$\frac{dy}{dt}=\frac{1}{(y+7)(t-3)}, y(0)=0$$
I want to solve said initial value problem & state the domain of the solution. I also want to observe what happens when $t$ approaches the limits of the solution... | The equation in question is $$y'(t)=\frac1{[y(t)+7](t-3)}.$$ Thus, there is a singularity at $t=3,$ which suggests that the solutions will have a domain that is a subset of $(-\infty,3)\cup(3,\infty).$ To solve the equation, we consider $$[y(t)+7]y'(t)=\frac1{t-3}.$$ Notice that $$\left(\frac{y^2}2+7y\right)'(t)=[y(t)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4383197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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show this equation $3x^4-y^2=3$ has no integer solution show this diophantine equation
$$3x^4-y^2=3$$ has no integer $(y\neq 0)$ solution?
My try: WLOG Assmue $(x,y)$ is postive integer solution,then $3|y$,let $y=3y'$,then we have
$$x^4-1=3y'^2\tag{1}$$
and following I want $\pmod5$,since $x^4\equiv 0,1\pmod 5$
(1):if... | I tried the following.
$3x^4-y^2=3$
$3x^4-3=y^2$
$3(x^2-1)(x^2+1)=y^2 \tag{1}$
If $y^2$ is divisible by $3$ then one of the factors of $(x^2-1)(x^2+1)$ should be divided by $3$ to make the right hand side a perfect square. So $x$ can't be $0 \pmod{3}$. Now if we divide 3 from the both hand side we can see the following... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$e^x= x^2+x$ has no roots for $x>0$ I need to prove using elementary calculus that:
$$e^x=x^2+x$$
has no roots for $x>0$.
I could easily observe it graphically but how can I prove it.
Please suggest.
| Fix a positive integer $n$ and define
$$
f_n(x):=e^x-1-x-x^2/2-\cdots-x^n/n!.
$$
Then $f^\prime_n(x)=e^x-1-x-\cdots-\frac{x^{n-1}}{(n-1)!}$, which is positive (by induction) on $(0,\infty)$.
Hence $f_n$ in strictly increasing on $(0,\infty)$. In particular, this gives the known inequality
$$
\forall x>0, \quad
e^x>1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$
Evaluate
$$\sqrt[3]{40+11\sqrt{13}}+\sqrt[3]{40-11\sqrt{13}}$$
The solution is $5$.
Suppose $\sqrt[3]{40+11\sqrt{13}}=A, \sqrt[3]{40-11\sqrt{13}}=B$
We have
$$A^3+B^3=80, A^3-B^3=22\sqrt{13}$$
Two unknowns, two equations, so we should be able to solve ... | Note that
$$\sqrt[3]{40+11\sqrt{13}}=\frac{5+\sqrt{13}}{2}$$
$$\sqrt[3]{40-11\sqrt{13}}=\frac{5-\sqrt{13}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4391660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Differential Equation by substitution How can I solve:
$$
yy'+x=\sqrt{x^2+y^2}
$$
I tried:
$$
y\frac{dy}{dx}+x=\sqrt{x^2+y^2}$$
let $v = x^2+y^2$,
$dv=2xdx+2ydy$
then I don't know what's next
I tried isolating the dy
$$dy=\dfrac {(dv-2xdx)}{2\sqrt {(v-x^2)}}$$
and substituting back to the equation
$$\sqrt {(v-x^2)}\dfr... | Notice that the equation only requires that $y$ be differentiable for real numbers $x$ such that $y(x)\neq0.$ Therefore, $x^2+y(x)^2\gt0.$ Let $D=\{x\in\mathbb{R}:y(x)\neq0\}.$ Let $w:D\to\mathbb{R}$ with $w(x)=x^2+y(x)^2.$ Hence $y(x)y'(x)+x=\sqrt{x^2+y(x)^2}$ is equivalent to $w'(x)=2\sqrt{w(x)},$ where $w(x)\gt0.$ T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2$ for $0 \le a, b, c, d \le 1$
Let $0\le a,b,c,d\le 1$, prove that $$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+d}+\frac{d}{1+a}\le2.$$
I’ve solved it, but I want a solution without derivatives. My solution is posted below.
The part of the expression ... | Note that, for all $x\in [0, 1]$,
$$1 - x/2 - \frac{1}{1 + x}
= \frac{x(1 - x)}{2(1 + x)}\ge 0.$$
We have
\begin{align*}
&\frac{a}{1 + b} + \frac{b}{1 + c}
+ \frac{c}{1 + d} + \frac{d}{1 + a}\\
\le\, & a (1 - b/2) + b(1 - c/2) + c(1 - d/2) + d(1 - a/2)\\
=\, & a + b + c + d - \frac12(ab + bc + cd + da)\\
=\,& a + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Proving $2\cos(k\theta) = \sum_{r=0}^{\lfloor k/2\rfloor} c(k,r) (2\cos \theta)^{k-2r}$ for $c(k,r)$ as defined
Let $k$ be a positive integer. Define for $n\ge 1, 0\leq r\leq \lfloor n/2\rfloor$, the integers $c(n,r)$ so that $c(1,0) = 1, c(2,0) = 1, c(2,1) = -2$ and for $n\ge 3$,
$$\begin{cases}
c(n,0)\phantom{/2} = ... | Remarks: The proof for even $m$ is similar. So I omit it.
The proof for odd $m$:
By the inductive hypothesis, we have
$$2\cos (m - 1)\theta = \sum_{r=0}^{(m - 1)/2} c(m - 1,r)(2\cos \theta)^{m - 1 - 2r}$$
and
$$2\cos (m - 2)\theta = \sum_{r=0}^{(m - 3)/2} c(m - 2,r)(2\cos \theta)^{m - 2 - 2r}.$$
Using the identity $\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Generalization of the law of tangents for a cyclic quadrilateral The law of tangents is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides.
Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle, and $\alpha$, $\beta$ and $\gamma$ be ... | The Usual Law of Tangents
Applying the formulae for the Sum of Sines and the Sum of Cosines, we get
$$
\begin{align}
\frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)}
&=\frac{2\sin\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}{2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)}\\[6pt]
&=\tan\left(\tfrac{A+B}2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$. Prove that if $ z$ is a complex number such that $ |z| \leq 1 $ then $|z^2 -1|\cdot |z-1|^2 \leq 3 \sqrt{3}$.
I tried geometric solution or using triangle inequality but doesn't work because I lose equality c... | There exists a very beautiful graphic solution.
We have $ |z| \le 1$ the unit disk bounded by the green circle, with centre $C_1$.
Let's assume: $|z-1| = r$, drawn as the purple circle $(C_2,r)$.
Note that we must have $0 \le r \le 2 $.
Now our inequality is $|z^2-1||z-1|^2 \le 3\sqrt{3} $ which is equivalent to: $|z+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \... | The line $y = m x + c$ has the parametric equation
$ (x, y) = (0, c) + t (\cos \theta, \sin \theta) $
where $\theta$ is the angle between the line and the positive $x$-axis.
and $m = \tan \theta $
Intersecting this line with the parabola $y = x^2$ yields
$ c + t \sin \theta = t^2 \cos^2 \theta $
which has two solutions... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find equation of the plane that this $x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}$ curve lies
Prove that all points of the given curve lie in one plane, and find
the equation of that plane:
$$x=\frac{1+t}{1-t}, y=\frac{1}{1-t^2}, z=\frac{1}{1+t}.$$
If the given curve lies in one plane, then
$$a\left(\frac{... | We have the following parameteric curve
$x(t) = \dfrac{1 + t}{1 - t} , y(t) = \dfrac{1}{1 - t^2} , z(t) = \dfrac{1}{1 + t} $
As the OP pointed out, we can simply assume that the curve lies on the plane
$$ a x + b y + c z + d = 0 $$
And then try and find $a,b,c,d$. Using this method, and multiplying through by $(1 - t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Common region between an icosahedron and a dodecahedron This is admittedly one of the hard problems I've come across. It involves the common region (intersection) between two dual platonic solids: icosahedron, and dodecahedron.
The question is as follows:
A dodecahedron and an icosahedron intersect as shown below. The... | As @David K observed, by considering the edge of the icosahedron, it is segmented into 3 equal parts, each being of length $1$. Hence the edge length of the icosahedron is $3$.
The second concern is the edge length of the dodecahedron. Consider a face of the dodecahedron. Suppose the edge length is $a$, and the smal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove $a^3-b^3 \geq a^2b - b^2a$ Given that $a>b>0$, prove that $a^3-b^3 \geq a^2b - b^2a$.
I have considered difference of cubes, where $a^3-b^3 = (a-b)(a^2+ab+b^2)$. However this doesn't seem to get me that far - especially when working backwards from the statement I need to prove - where I factorised the ri... | Alternative approach:
$0 < (a - b)^3 = (a^3 - b^3) - (3)(a^2b - ab^2) \implies$
$(a^3 - b^3) > 3(a^2b - ab^2).$
Further, since $a > b,$
you know that $a^2b > ab^2 \implies (a^2b - ab^2) > 0.$
Therefore $3(a^2b - ab^2) > (a^2b - ab^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Generate an explicit formula for: $a_n=a_{n-1}+n^2-5n+7, a_3=2$ So I'm currently solving this math homework problem and I've set up a recursive formula:
$a_n=a_{n-1}+n^2-5n+7; a_3=2$
Right now, I'm having trouble as to where to start. I was listing out the cases but they didn't get me very far: $a_3=2,a_4=5,a_5=12,a_... | We may observe that:
$$
a_n-a_{n-1}=n^2-5n+7.
$$
Then we could have:
$$
\begin{align}
&a_n-a_3\\
=&\sum_{k=4}^n{a_k-a_{k-1}}\\
=&\sum_{k=4}^n{k^2-5k+7}\\
=&\left[ \left( \sum_{k=1}^n{k^2} \right) -1^2-2^2-3^2 \right] -\frac{5\left( 4+n \right) \left( n-3 \right)}{2}+7\left( n-3 \right)\\
=&\left[ \frac{n\left( n+1 \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding $\sum_{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$ I want to find the closed form of:
$\displaystyle \tag*{} \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n2^n \binom{2n}{n}}$
Where $H_{k}$ is $k^{\text{th}}$ harmonic number
I tried to expand the numerator (Harmonic numbers) in ter... | Simplify the integral
\begin{align}
&\int_0^1\frac{x\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,dx\\
=& \>\sqrt2 \left(\sinh^{-1}\frac x{\sqrt8}\right)^2\bigg|_0^1
- \int_0^1\frac{\sinh^{-1}\frac x{\sqrt8}}{(1+x)\sqrt{1+\frac{x^2}8}}\,\overset{x=\sqrt2(y-\frac1y)}{dx}\\
=&\>\frac1{2\sqrt2}\ln^22-2\sqrt2 \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4422112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Maximize $z$ over $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$ Suppose that $x$, $y$, and $z$ are real numbers such that $x + y + z = 3$ and $x^2 + y^2 + z^2 = 6$. What is the largest possible value of $z$?
I tried applying Cauchy-Schwarz to get $(x^2+y^2+z^2)(1+1+1)\geq (x+y+z)^2$, but this doesn't say anything. I also ... | Here is a solution avoiding calculus.
Use cylindrical coordinates.
$x = r \cos\theta, y = r \sin\theta, z = z, r \geq 0$
From $x^2 + y^2 + z^2 = 6$, we get
$$z^2 = 6 - r^2 \tag1$$
From $x + y + z = 3$, we get
$ \displaystyle r = \frac {3-z}{\cos\theta + \sin\theta} \tag2$
From $(1), z \leq \sqrt6~$ and so we know $(3 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$... | First way: Using the Extended Euclidean Algorithm in $\Bbb Q[x]$.
Notice that
\begin{equation*}
15 = 4(3n^2+4n+5) - (6n+5)(2n+1).
\end{equation*}
Hence, if $2n+1$ divides $3n^2+4n+5$, then it also divides $15$.
Thus, $2n+1 \in \{\pm 1, \pm 3, \pm 5, \pm 15\}$, i.e.,
\begin{equation*}
n \in \{-8,-3,-2,-1,0,1,2,7\}.
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum possible value of $x+y-2xy$ over reals. Find the minimum possible value of $x+y-2xy$, if $x+y\ge 3xy,~ 2(x+y)\ge 1+3xy~$ and $~0\le x\le y\le 1.$
My guess is $x=y=\frac{2}{3},$ gives the minimum value of $\frac{4}{9}$.
I thought this manipulation might be useful,
$$x+y-2xy=\frac{3(x+y)-6xy}{3}\ge \fr... | I think that your second idea works.
Let $k:=s-2p$, and $f : p=\frac 14s^2, g : p=\frac 23s-\frac 13$ and $h : p=\frac s3$.
Then, $f$ and $g$ intersect at $A(2/3,1/9)$ and $C(2,1)$, and $f$ and $h$ intersect at $O(0,0)$ and $B(4/3,4/9)$.
So, one can see that
*
*for $s\lt\frac 23$, $p\leqslant \frac 23s-\frac 13$
*f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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prove that the sum of the elements in two subsets is the same
Eight consecutive positive integers are partitioned into two subsets such that the sum of the squares in each subset is the same. Prove that the sum of the elements in each subset is also the same, assuming that the smallest element is at least 56.
The met... | I have a simple proof if the smallest number, name it $a$, is greater than $56$. We have $$(a+a_1)^2+...+(a+a_4)^2=(a+a_5)^2+...+(a+a_8)^2,$$ where $a_1,...a_8\in \{0,...,7\}$. Hence $$2a(S_1-S_2)=a_5^2+...a_8^2-a_4^2-...-a_1^2,$$ where $S_1=a_1+...+a_4, S_2=a_5+...+a_8$. So $$|S_1-S_2|< (7^2+6^2+5^2+4^2-3^2-2^2-1)/112... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2... | Your analysis is spot on, up until the point where you concluded that
$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}. \tag1 $$
I didn't bother reading after that, because there is no point. All that is necessary from the point of (1) above is to consider that any equation of the form
$$[f(x)]^2 = C$$
may b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove $ (3\Bbb N + 1) \cap (5\Bbb N + 3) = 15\Bbb N + 13$ I'm trying to prove that: $$ (3\Bbb N + 1) \cap (5\Bbb N + 3) = 15\Bbb N + 13$$ What have I done so far is, I'm getting an x from left hand side of equation:
$$
x \in (3\Bbb N + 1) \cap (5\Bbb N + 3)
$$
Then I'm simplifying it, like this:
$$
x \in (3\Bbb ... | Let $n=3x+1$ and assume $x=5y+z$, then $n=15y+3z+1$. If $n\equiv 3 \pmod{5}$, then $z\equiv -1 \pmod5$. Thus we can take $z=-1$ and obtain the result you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluating $\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$ I want to find the closed form of:
$\displaystyle \tag*{}\sum \limits_{k=1}^{\infty} \frac{\binom{4k}{2k}}{k^2 16^k}$
I tried to use the taylor expansion of $\frac{1}{\sqrt{1-x}}$ and $\frac{1}{\sqrt{1-4x}}$ but both of them had $\binom{2n}{n}$ in... | Using @RobPratt's elegant solution
$$S_1(x)=\sum_{k=1}^\infty \frac{\binom{2k}{k}}{k^2 }x^k=2 x \, _4F_3\left(1,1,1,\frac{3}{2};2,2,2;4 x\right)$$ Simplifying
$$S_1(x)=2 x \left(\frac{\text{Li}_2\left(\frac{1}{2}-\frac{1}{2} \sqrt{1-4
x}\right)}{x}-\frac{\left(\log \left(1+\sqrt{1-4 x}\right)-\log (2)\right)^2}{2
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Evaluating $\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\tanh(x)}-\frac{1}{x}\bigg)$ using L'Hôpital's rule I need to evaluate the following limit (using L'Hôpital's rule).
$$\lim_{x \rightarrow 0}\frac{3}{x}\bigg(\frac{1}{\tanh(x)}-\frac{1}{x}\bigg)$$
I expressed $\tanh(x)$ as $\frac{\sinh(x)}{\cosh(x)}$:
$$\lim_{... | Solution :
$$\lim _{x\to \:0}\left(\frac{3}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1}{x}\right)\right)$$
Taking out constant :
$$\lim _{x\to a}\left[c\cdot f\left(x\right)\right]=c\cdot \lim _{x\to a}f\left(x\right)$$
We get :
$$=3\cdot \lim _{x\to \:0}\left(\frac{1}{x}\left(\frac{1}{\tanh \left(x\right)}-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving system of three equations in three unknowns for electric circuit analysis problem My electrical engineering circuit analysis textbook presents the following problem:
Find the currents and voltages in the circuit shown in Fig. 2.28.
Answer: $v_1 = 6 \ \text{V}$, $v_2 = 4 \ \text{V}$, $v_3 = 10 \ \text{V}$, $i_... | It is easier to write the three equations and use Gaussian Elimination, the equations being
$$ 2i_1 + 8i_2 + 0 i_3 = 10 \ \text{V}\\ 0 i_1 - 8i_2 +4i_3 = 6 \ \text{V} \\ i_1 - i_2 - i_3 = 0 \text{V}\\ $$
Method 1: We have the augmented matrix
$$ \left(
\begin{array}{ccc|c}
2 & 8 & 0 & 10 \\
0 & -8 & 4 & 6 \\
1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$x^4+y^4+z^4=\frac{m}{n}$, find $m+n$. $x^4+y^4+z^4$=$m\over n$
x, y, z are all real numbers,
satisfying $xy+yz+zx=1$ and $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$
m, n are positive integers and their greatest common divisor is 1. Calculate m+n.
My thinkings so far are as f... | Notice if $(x,y,z)$ satisfies the two conditions, so does $(-x,-y,-z)$. Furthermore, it is clear $x,y,z$ have the same sign. Since $(x,y,z)$ and $(-x,-y,-z)$ gives the same $\frac{m}{n}$, we only need to consider the case $x,y,z > 0$.
Take three numbers $\alpha,\beta,\gamma \in (0,\frac{\pi}{2})$ such that
$$x = \tan\a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\lim_{n \to \infty}a_n$ where $a_1=1$ and $a_{n+1}=a_n+\frac{1}{2^na_n}$. There are some attempts as follows.
Obviously, $\{a_n\}$ is increasing. Thus $a_n\ge a_1=1$. Therefore
$$a_{n+1}-a_n=\frac{1}{2^na_n}\le \frac{1}{2^n},$$
which gives that
$$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)\le a_1+\sum_{k=1}^{n-1}\frac{... | Here is a short proof that the limit is between 1.810 and 1.828. This can be improved to get better precision, I believe.
Squaring both sides of the recurrence, we get
$$
a_{n+1}^2 = a_n^2 + \frac{2}{2^n} + \frac{1}{4^n a_n^2} \tag{1}
$$
and so, summing,
$$
a_n^2 = a_1^2 + \sum_{k=1}^{n-1} (a_{k+1}^2 - a_k^2) = 1+\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Is there a solution for $(\frac{3}{x+y+z})^n+(\frac{3}{x+y+z})^{5-n}<2$? Is there a solution for
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n}<2$$
where $n\in\mathbb Z$ and $x,y,z>0, xyz=1$.
This is the part of my attempts for my homework, that I stucked.
This is obvious $x+y+z≥3\implies \frac {3}... | Hint
By AM-GM inequality:
$$x+y+y\ge3(xyz)^{1/3}\Rightarrow \frac{3}{x+y+z}\le1.$$
The equality holds only when $x=y=z=1$.
It means that, for $0\le n \le 5$, we have
$$\left(\frac{3}{x+y+z}\right)^n+\left(\frac{3}{x+y+z}\right)^{5-n} \le 2.$$
You want to know if there is a solution for
$$\left(\frac{3}{x+y+z}\right)^n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tough integral $\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3) $ How to prove
$$\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx =6\pi\zeta(3), $$
and does there even exist a closed form of $$\int_0^{ \pi }\frac{x^3(\pi-x)^3}{\sin^3 x} dx \ ? $$
(Note that the easier one $$\int_0^{ \pi }\frac{x (\pi-x) }{... | Integrate by parts twice
\begin{align}
\int_0^{ \pi }\frac{x^2(\pi-x)^2}{\sin^2 x} dx
=& \>2 \int_0^{ \frac\pi2}\frac{x^2(\pi-x)^2}{\sin^2 x} dx \\
=&\>4 \int_0^{\frac\pi2}(\pi^2x-3\pi x^2+2x^3)\cot x\> dx\\ =&\> 4\int_0^{\frac\pi2} (6\pi x-6x^2 -\pi^2)\ln(2\sin x)dx
\end{align}
Then, utilize the known results $\int_0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Integrate $(1-\sqrt{x})/(1+\sqrt{x})$
Integrate $(1-\sqrt{x})/(1+\sqrt{x})$.
So, this was on my test, and I got
$$4(1+\sqrt{x})-4\ln(1+\sqrt{x})-x+c$$
as a result using symbolab. But my teacher gave the result shown below (his solution is shown here, and transcribed below).
$$\int \frac{1-\sqrt{x}}{1+\sqrt{x}} \, dx ... | Your teacher made 2 separate mistakes. First when splitting the integrals:
$$2\int\frac{3u-u^2-2}{u}du=\int6\,du-2\int u \,du-4\int \frac1udu$$
They wrote a 2 instead of 4 in the final term. Then when resubstituting the expression for $u$ they used $1-\sqrt{x}$ instead of $1+\sqrt{x}$:
$$6(1+\sqrt{x})-(1+\sqrt{x})^{2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the maximum of $x+\frac{1}{2y}$ under the condition $(2xy-1)^2=(5y+2)(y-2)$ Assume $x,y \in \mathbb{R}^+$ and satisfy $$\left(2xy-1\right)^2 = (5y+2)(y-2)$$, find the maximum of the expression $x+\frac{1}{2y}$.
$\because (5y+2)(y-2) \geq 0$ , $\therefore y\geq 2$ or $y \leq -\frac{2}{5}$, and we have $$2xy = \sqrt... | Here is another way. Change variables to replace $x$ with $z$ under the relation $z=2xy\iff x=\frac{z}{2y}$. So now you are maximizing $\frac{z+1}{2y}$ subject to $(z-1)^2=(5y+2)(y-2)$.
Lagrange multipliers can handle this. You have a system of three equations in three variables $y,z,k$:
$$\left\{\begin{align}
(z-1)^2&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2... | Hint:
\begin{align}
S & = \sum_{n=1}^\infty \frac{1}{n^2} \\
& = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \sum_{n=1}^\infty \frac{1}{(2n)^2} \\
& = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \frac{1}{4}\sum_{n=1}^\infty \frac{1}{n^2} \\
& = \sum_{n=1}^\infty \frac{1}{(2n-1)^2} + \frac{1}{4}S
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the value of expression $Q = \frac{x + 1}{y}$ when $xy > 1$ and expression $P = x + 2y + \frac{5x + 5y}{xy - 1}$ reaches its maximum value.
Consider two positives $x$ and $y$ where $xy > 1$. The maximum value of the expression $P = x + 2y + \dfrac{5x + 5y}{xy - 1}$ is achieved when $x = x_0$ and $y = y_0$. C... | Just note that since the points $xy=1$ are excluded, the maximizers/minimizers will be a stationary points in the open set defined by the condition $x y>1$. Once you determine the stationary points (some are complex numbers, rule them out), you'll see that the only stationary point with $x,y>0$ and $xy>1$ is $(x,y)=(3,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\sum_{1\le iLet $n \ge 2$ be a an integer and $x_1,...,x_n$ are positive reals such that $$\sum_{i=1}^nx_i=\frac{1}{2}$$
Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}$$
Here is the source of the problem (in french) here
Edit:
I'll present my best bound yet on $$\... | You should be able to look for the maximum, by the method of Lagrange. We set
$$F(x_1,...,x_n) \equiv \sum_{i\neq j} \frac{x_i x_j}{(1-x_i)(1-x_j)} \stackrel{\text{wts.}}{\leq} \frac{n(n-1)}{(2n-1)^2} \, ,$$
and define the Lagrange-function as $$L(x_1,...,x_n) = F(x_1,...,x_n) - \lambda \left(x_1+...+x_n-1/2\right) \,.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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In a school of $1025$ students, $400$ of them ($\frac{1}{5}$ of the boys and $\frac{4}{7}$ of the girls) cannot swim. How many boys are there? Not sure how to set up an equation for this problem.
A school has $1025$ students. A total of $400$ students cannot swim. This consists of $\frac{1}{5}$ of the boys and $\frac{... | What the equation is saying is that the number of girls who cannot swim is $3/4$ of the number of girls who can swim. This is true since the fraction of girls who can swim is
$$\frac{3}{7} = \frac{3}{4} \cdot \frac{4}{7}$$
As you observed, $625 - x$ is the number of girls who can swim. Since $x$ is the number of boys ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450861",
"timestamp": "2023-03-29T00:00:00",
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How to find the explicit solution of $y'(x)= |y(x)|(1-y(x)) \frac{x^3}{1+x^4}$ Given,
$$y'(x)= |y(x)|(1-y(x)) \frac{x^3}{1+x^4}$$ with initial conditions $y(0) = 2$
I know that the y(x) can be found using Bernoulli's method such that
$$\frac{y'}{y^2} -\frac{x^3}{y(1+x^4)} = -\frac{x^3}{1+x^4} $$
Assuming $\frac{1}{y} =... | The derivative of $(1+x^4)^{1/4}$ is $x^3(1+x^4)^{-3/4}$, which is exactly the integrand you are in doubt of.
The equation is separable, so a better substitution would use and preserve that pattern by setting $v=y^{-1}-1$. Then
$$
v'=-y^{-2}y'=-sv\frac{x^3}{1+x^4}, ~~~ s=sign(y(0)).
$$
This is obviously again separabl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4452277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that for positive reals $x,y,z$, $x^6+y^6+z^6 + 6x^2y^2z^2 \geq 3xyz(x^3+y^3+z^3)$. I am not sure if the inequality is true. My first attempt was to try AM-GM inequality in clever ways. I also tried Schur's inequality which gives
$$
x^6+y^6+z^6 + 6x^2y^2z^2 \geq (x^2+y^2+z^2)(x^2y^2+y^2z^2+z^2x^2)
$$
but it is no... | The problem is equivalent to prove that
$a^2+b^2+c^2+6 \ge 3(a+b+c)$, $abc=1$
But the last inequality is true because
$$LHS - RHS = \frac{(a+b+c-3)^2+a^2+b^2+c^2+2abc+1 -2(ab+bc+ca)}{2} \ge 0$$
Note that the inequality $a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$ is well-known.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the minimum value of the function $f(x)= \frac{x^2+3x-6}{x^2+3x+6}$? I was trying to use the differentiation method to find the minimum value of the person but it did not give any result, I mean when I differentiated this function and equated to zero for finding the value of $x$ when the function value would be... | Another didactic way, we prove that $-11/5 \le y=f(x)<1, x \in \Re$.
Let $y=\frac{x^2+3x-6}{x^2+3x+6}$ and check that $x^2+3x+6 >0$
Then we can write $(y-1)x^2+3(y-1)x+6(y+1)=0, x \in \Re$
$\implies B^2 \ge 4AC \implies 9(y-1)^2\ge 24(y+1)(y-1).$
Case 1: $y<1$ we get $9y-9 \le 24y+24 \implies y \ge -11/5$
and hence $-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{... | The equation $x^3 - x^2 + 2 = 0$ has three solutions. Therefore we can write it as $(x-a)(x-b)(x-c)=0$. Expanding the second expression and equating terms we get:
$a+b+c = 1$; $ab+bc+ac=0$; $abc=-2$
Now substitute $a = 1+i$. We get $b+c = -i$ and $bc=-1+i$. Which is easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
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Why does my solution of the closed form formula for the Fibonacci sequence differ from the actual solution by a sign? I have $a_{n}=a_{n-1}+a_{n-2}$ with $a_0=0$ and $a_1=1$.
I've found the generating function $G(x)=\frac{x}{1-x-x^2}$. Solving $1-x-x^2=0$ gives the solutions $\alpha_1=\frac{-1+\sqrt{5}}{2}$ and $\alpha... | Hint: It seems we have
\begin{align*}
\frac{x}{1-x-x^2}=\color{blue}{-}\frac{\frac{1}{\sqrt5}\alpha_1}{x-\alpha_1}\color{blue}{+}\frac{\frac{1}{\sqrt{5}}\alpha_2}{x-\alpha_2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $ I want to prove that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $, I started proving but I got stuck.
Let $ x, y \in \mathbb{R} $. Therefore:
$ \lfloor{x}\rfloor \le x $
$ \lfloor{y}\rfloor \le y $
$ \Rightarrow $
$ \lfloor... | $ \lfloor{x}\rfloor $ is an integer $ \leq x;\ \lfloor{y}\rfloor $ is an integer $ \leq y.$
Therefore, $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y.$
Since $\lfloor{x+y}\rfloor$ is the greatest integer $ \leq x+y,$ and $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You... | Since $x$ and $y$ are positive integers, we can use AM-GM inequality. Also, notice that $3y = 5(20 -x)$, therefore $y$ is a multiple of $5$ or $xy$ is a multiple of $5$.
So, $$\frac{5x + 3y}{2} \ge \sqrt{15xy} $$
or, $$50 \ge \sqrt{15xy}$$
By squaring both sides, we ge:$$2500 \ge 15xy$$
or $$166.67 \ge xy$$
Since $xy$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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How to prove this summation of floor function I am supposed to show that
$$
\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor + \cdots = n
$$
For all postive integers $n \geq 1 ... | The statement is trivially true for $n=1$ and $n=2$. Suppose it is true for any positive integer less than $n$, and let's prove it is true for $n$. We have two cases:
$n=2m$ is even. Then:
$$
\begin{array}{cll}&\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+... | {
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"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Series solution for $x^2y''-x(x+6)y'+10y=0$ I have to solve this differential equation:
$$x^2y''-x(x+6)y'+10y=0$$
by using this method and I am stuck at this step.
Please help me to solve it.
Here is my attempt:
https://i.stack.imgur.com/XglDm.jpg
| The given differential equation is a secondorder linear ordinary differential equation. We are required to find a series solution to this problem.
Step 1: Assume that an infinite series solution exists such that
$$
y = \sum_{n = 0}^{\infty}c_{n}x^{n}
$$
Step 2: Calculate $y''$, $y'$:
$$
y' = \sum_{n=1}^{\infty}nc_{n}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $S=\{\overline{0},\overline{5},\overline{10}\}\subset\mathbb{Z}_{15}$ isomorphic to$\mathbb{Z}_{3}$ as rings? My book asked me to prove that it is in fact. I generated the tables for addition
\begin{array}{c|ccc}
+ & 0 & 5 & 10 \\
\hline 0 & 0 & 5 & 10 \\
5 & 5 & 10 & 0 \\
10 & 10 & 0 & 5
\end{array} \begin{array}{c... | You are almost right.
$f(x+y)=f(x)+f(y)$ also holds as
$$f(5+5)=f(10)=1 =2+2=f(5)+f(5)$$
The remaining argument to prove the isomorphism is to note that $f$ has inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find all positive integer solutions for $3^x-2^y=1$.
Find all positive integer solutions for $3^x-2^y=1$.
Quickly we can find the solutions $(1,1)$ and $(2,3)$. Now the claim is that for $x \ge 2$ and $y \ge 3$ there are no solutions.
The equation can be expressed as $3^x=2^y+1$ from where we can get to $3^x-3 = 2^y-... | You used two times the following false statement:
$ax \equiv 0 \pmod m$ and $a$ is not multiple of $m$ so $x \equiv 0 \pmod m$
The correct statement is
$ax \equiv 0 \pmod m$ and $a$ is coprime to $m$ so $x \equiv 0 \pmod m$
The first time you used it was with $a=2$ and $m=3$ so you got a correct conclusion, but the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the provided question of AM-GM inequality in the description, I'm not able to figure out which manipulation was used here, in the solution?
Let $x, y, z \in \Bbb R_{> 0}$ such that $xyz=1$. Prove that $$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}$$
Now, in the first st... | The expression wasn't rewritten, those other two terms were more or less poofed into existence. Implicitly we're saying
$$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)}$$ $$ = \frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} + \frac{(1+y)}{8} + \frac{(1+z)}{8} - \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$ integer. Find the smallest $n\in \mathbb Z^+$ that makes $\sqrt{100+\sqrt n}+\sqrt{100-\sqrt n}$
Clearly if $n=0$ then we will have $20$ but I couldn't decide that how can I find the other integers. Any hint?
If I say that $x=\sqrt{... | If
$$
x=\sqrt{100-\sqrt n}+\sqrt{100+\sqrt n}\tag{1}
$$
then squaring both sides gives us
$$
x^2=200+2\sqrt{100^2-n}
$$
Rearranging and squaring again gives
$$
n=100x^2-\frac14x^4\tag{2}
$$
For this to be an integer greater than zero, $x$ must be non-zero and even.
From (1) we have that $\sqrt n\leq 100$. This gives us... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find symmetric equations for the line of intersection of the two planes $x + 2y + 5z = 3$ and $2x + 3y = 1$. The question asks to find symmetric equations for the line of intersection of the two planes $x + 2y + 5z = 3$ and $2x + 3y = 1$. I have done the following work below. Could you please provide feedback on whethe... | Another way to do it is to find 2 points which should lie on this line;
Let $z=0$
$$x+2y=3\\
2x+3y=1\\x=-7,y=5,z=0 $$
Let $z=1$ $$x+2y=-2\\
2x+3y=1\\x=13,y=-5,z=1$$
So line through these 2 points is:
$$\frac{x+7}{20}=\frac{y-5}{-10}=\frac{z}{1}$$
where $<13,-5,1>-<-7,5,0>=<20,-10,1>$ represents the direction ratios of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the maximum value of the multiple variable function Show that the maximum value of$$xy(z-h)\left (\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right )$$is$$\left (\frac{2}{5}h\right )^5\frac{ab}{c^4}.$$
| Take the partial derivatives with respect to $x$, $y$, and $z$, then set them to $0$. For $x$,
$$y(z-h)\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)+xy(z-h)\frac{2x}{a^2}=0\\y(z-h)\left(\frac{3x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}\right)=0$$
Similarly for $y$:
$$x(z-h)\left(\frac{x^2}{a^2}+\frac{3y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?
I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$... | $$20!=2^{18}\cdot3^8\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$$
$$19\cdot 17\cdot 13=4199>2^{12}, \\3\cdot 11>2^5$$
So it suffices to show:
$$3^7\cdot 5^4\cdot 7^2>2^{25}.$$
Now, $5\cdot 7>2^5,$ and $3^2>2^3,$ so it suffices to showing:
$$75=3\cdot 5^2>2^{6}=64$$
This shows: $$\frac{20!}{2^{60}}=\left(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Finding the roots of $x^5+x^2-9x+3$ I have to find all the roots of the polynomial $x^5+x^2-9x+3$ over the complex. To start, I used Wolfram to look for a factorization and it is $$x^5+x^2-9x+3=(x^2 + 3) (x^3 - 3 x + 1)$$
I can take it from here employing the quadratic and cubic formula, but how can I get the same answ... | It can be factorize by adding and subtracting $3x^3$
$$x^5+x^2-9x+3=x^5-3x^3+x^2+3x^3-9x+3=x^2(x^3-3x+1)+3(x^3-3x+1)=(x^2 + 3) (x^3 - 3 x + 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Closed form for $\int_ 0^{1/2} {\frac {x} {3 +4 x^2}\ln\frac {\ln\left (1/2 +x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $ Find the integral $$\int_ 0^{1/2} {\frac {x} {3 + 4 x^2}\ln\frac {\ln\left (1/2 + x \right)} {\ln\left (1/2 - x \right)}\mathrm {d} x} $$
Wolfram | Alpha tells me that the answer is $ - 0.... | After integration by parts we have
\begin{gather*}
I=\int_{0}^{1/2}\frac{x}{3+4x^2}\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\,\mathrm{d}x= \underbrace{\left[\dfrac{1}{8}\ln\left(\frac{3+4x^2}{4}\right)\ln\dfrac{\ln(1/2+x)}{\ln(1/2-x)}\right]_{0}^{1/2}}_{=0}-\\[2ex]
\dfrac{1}{8}\int_{0}^{1/2}\ln\left(\frac{3+4x^2}{4}\right)\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I c... | Continue with
\begin{align}
A= &\ 4\pi \int_{1}^2 r \sqrt{\frac{r^2}{r^2 - 1} + 1} \ \overset{t=r^2-1}{dr }\\
=& \ 2\pi \int_0^3 \sqrt{\frac{2t+1}t}\ dt\overset{ibp}=
2\pi\bigg(\sqrt{t(2t+1)}\bigg|_0^3+\frac12\int_0^3 \frac1{\sqrt{t(2t+1}}dt\bigg)\\
=& \ 2\pi\bigg( \sqrt{21}+\frac1{\sqrt2}\sinh^{-1} \sqrt{2t}\bigg|_0^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What is the average of rolling two dice and only taking the value of the lower dice roll? My question is related to this question, with the exception that I'm looking for the average when taking only the lower of two dice rolls.
My question is:
What is the average of rolling two dice and only taking the value of the l... | Let the random variable $X$ be defined as the maximum value when two fair six-sided dice are rolled. We wish to find $E[X]$.
Since there are $k$ possible values which are at most $k$, there are $k^2$ ordered pairs of values in which both dice display a number which is at most $k$. To ensure that the maximum value is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left... | Putting another answer here, a simpler one. We start by multiplying both sides of the equation by $x\sqrt{x^2-1}$. This gives $$(x^4-3x^2+x+1)\sqrt{x^2-1}+(x^5-4x^3+x^2+3x-1)=0$$
Subtracting this from the original equation, we get directly:
$$\sqrt{x^2-1}=x^3-2x+1$$
We can see that $x=1$ is a solution and other solutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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parametrization of the result of $\sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1}\sum_{k=j+1}^N$ as a function of $N$ As stated in the title, what's the result as a function of N for
$$\sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1}\sum_{k=j+1}^N 1$$
| Summing from $a$ to $b$ where $b\geqslant a-1$ has $b-a+1$ summands
$$\sum_a^b 1 = \underbrace{1+1+\cdots 1}_{\textstyle b-a+1 \text{ times}} \tag0$$
where the empty sum equals $0$ by definition. Hence
$$\begin{align}
\sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1}\sum_{k=j+1}^N 1
&= \sum_{i=1}^{N-2}\sum_{j=i+1}^{N-1} (N-j)\\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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What does passing the base 3 test mean here? I have got this one:
Let $n$ be a positive odd integer passing the base 3 test, i.e. $3^n\equiv 3\pmod n$. Prove that $\frac{3^n-1}{2}$ also passes the base 3 test.
My attempt:
$3^n\equiv 3\pmod n$
$\Rightarrow 3^n-1\equiv 2\pmod n$
$\Rightarrow \frac{3^n-1}{2}\equiv 1 \pmod... | For simpler algebra, let
$$r = \frac{3^n-1}{2} \tag{1}\label{eq1A}$$
Note the problem is actually asking to prove that, given $n$ passes the test, then $r$ also passes the "base $3$ test", i.e., $r$ is a positive, odd integer and $3^r \equiv 3 \pmod{r}$.
Since $n$ is a positive, odd integer, then $r \ge 1$ and $3^n \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492510",
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"source": "stackexchange",
"question_score": "1",
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How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+... | Personally I'm somewhat inclined to use the fifth row of Pascal's triangle: $$
1\\
1\quad 1 \\
1\quad 2\quad 1\\
1\quad 3\quad 3\quad 1\\
1 \quad 4\quad 6\quad 4 \quad 1\\
1\quad 5\quad 10\quad 10\quad 5 \quad 1$$
Thus $$(1-x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
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A golden question $\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$ How would you solve this problem for real $x$?
$$\sqrt{2+\sqrt{2-x}}=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
It can be easily shown that both equations
$$x=\sqrt{2+\sqrt{2-x}}\tag{1}$$
and $$x=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}\tag{2}$$
h... | Let $$f(x)=\sqrt{2+\sqrt{2-x}}$$ and
$$g(x)=\sqrt{x-\frac 1x} + \sqrt{1-\frac 1x}$$
See the fig below where $y=f(x)$ blue and $y=g(x)$ green and $y=x$ (red).
The solution of $x=f(x)$ is given by the intersection of red and blue lines at $x=1.6180...=\frac{1+\sqrt{5}}{2}$ (the same golden number)
The root of $x=g(x)$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4495058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to make this proof rigorous by introducing partition of numbers? Question: Let $n$ be a positive integer and $ H_n=\{A=(a_{ij})_{n×n}\in M_n(K) : a_{ij}=a_{rs} \text{ whenever }
i +j=r+s\}$. Then what is $\dim H_n$?
Proof:
For $n=2$
$H_2=\begin{pmatrix} a_{11}& a_{12}\\a_{12}&a_{22}\end{pmatrix}$
Matrix of sum ... | Sums of indices can go from $2$ to $2n$. Define, for $2\le k\le n$ the matrix $M_k$ that has $1$ where $i+j=k$ and $0$ elsewhere.
It should be easy to prove that $\{M_k:2\le k\le 2n\}$ is a basis of your subspace. That's all, you don't need to count the number of nonzero entries in $M_k$.
Let's look at a typical matrix... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to prove that $\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$? $$\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$$
Equality seems to be when x = 1
I have managed to show that the derivative is 0 at x = 1, and that this is a minimum (by the second der... | Apply Cauchy-Schwarz inequality twice followed by an application of AM-GM inequality:
$\sqrt{\dfrac{x^2+1}{x+1}}+\dfrac{2}{\sqrt{x}+1}\ge \dfrac{\sqrt{x+1}}{\sqrt{2}}+\dfrac{2}{\sqrt{x}+1}\ge\dfrac{\sqrt{x+1}}{\sqrt{2}}+\dfrac{2}{\sqrt{2(x+1)}}\ge 2\sqrt{\dfrac{\sqrt{x+1}}{\sqrt{2}}\cdot \dfrac{2}{\sqrt{2(x+1)}}}=2$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$
Let $a,b,c>0$ and $ab+bc+ca=3$. Prove that $abc(a+b)(b+c)(c+a)\leq8$
My try: $a^3b^3c^3(a+b)(b+c)(c+a)\leq8a^2b^2c^2$ so $(a^2bc+ab^2c)(ab^2c+abc^2)(abc^2+a^2bc)\leq 8ab.bc.ca$. Then choose $ab=x,bc=y,ca=z$. Problem become $x,y,z>0$, $x+y+z=3$. Prov... | Your idea is briliant indeed. Put $x = ab, y = bc, z = ac$, then $x+y+z = 3$, and we prove: $(x+y)(y+z)(z+x)\le 8$. By AM-GM inequality: $(x+y)(y+z)(z+x) \le \left(\dfrac{x+y+y+z+z+x}{3}\right)^3= 2^3 = 8$. Equality occurs when $x+y=y+z=z+x \implies x = y = z = 1\implies a = b =c = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Regular heptagon area formula Be a regular heptagon. Can anyone demonstrate this formula?
I know:
Let apothem = ap
side = l
$ S=\frac{p.ap}{2}\implies S = \frac{7l.ap}{2}\\
\triangle OBC: OC^2 = a^2+l^2\\
\triangle PAB: b^2+l^2 = PB^2\\
OC^2-a^2=PB^2-b^2$
but i can't go on
|
Hints: You can use this formula to find area of regular polygon:
$$A=\frac {n\cdot l^2}{4 tan \frac {\pi}n}$$
where n is number of sides, $l$. is length of side. For $l=1$ and $n=7$ we get $A=3.63..$
Now you can find a as :
$\angle KBC=\angle OFH$
because their rays are perpendicular on each other. So we have:
$\trian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Product of $i^4+4$ is a perfect square The following is the Problem $50$ (Folklore) from this pdf:
Determine all positive integers $k$ such that $\prod_{i=1}^k (i^4+4)$ is a perfect square.
I realised that $i^4+4$ is just the Sophie-Germain identity, so the problem resolves to proving $2(k^2+2k+2)(k^2+1)$ is square, ... | I suppose this question requires to find all integers $k$ satisfying the statement than proving that statement is true ($k=1$ is false since it evaluates to 5)
So we can start from $2(k^2+1)(k^2+2k+2)$ which you found. For perfect squares we need two factors of the number to be equal:
$2(k^2+1)=k^2+2k+2_{---(1)}\\OR\\2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to ... | A proof based on the definition. Let $\epsilon>0$ such that $\epsilon<1$. We are seeking a $\delta$ such that if $|x|<\delta$ we get:
$\dfrac{x^{2}}{x^{2}+2x+1}<\epsilon^{2}$ which is equivalent to,
$\dfrac{\epsilon^{2}-\epsilon}{1-\epsilon^{2}}<x<\dfrac{\epsilon^{2}+\epsilon}{1-\epsilon^{2}}$. Now take $\delta=\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to use Mittag Leffler expansion (if possible) to solve the following integrals? In a physics related context I've been trying to solve the following two integrals:
(i)
$$
\text{int}_1 = \int_{-\infty}^\infty \csc\left( \frac{\pi+2 i z}{2 \sqrt{2} } \right) \text{sech}^2(z) \,dz
$$
(ii)
$$
\text{int}_2 = \int_{-\... | To solve $int_2$, I tried the following using Mittag-Lefflers' theorem and the corresponding expansions of $\csc(x)$ and $\text{sech}^2(x)$ given by
$$
\text{csc}(x) = \underbrace{\frac{1}{x}}_{A_1} + \underbrace{2 x \sum_{k=1}^\infty \frac{(-1)^k}{x^2 - (\pi k)^2}}_{A_2}, \forall \frac{x}{\pi} \notin \mathbb{Z} \\
\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to show $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$, with $n$=odd? In previous post, I got the answer that $\gcd \left(\frac{p^2+1}{2}, \frac{p^5-1}{2} \right)=1$, where $p$ is prime number.
I am looking for more general case, that is for $p$ prime,
When is $\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2}... | Suppose that
$$\gcd\left(\frac{p^m + 1}{2}, \frac{p^n - 1}{2}\right) \neq 1.$$
There are two cases:
*
*Both $p^m + 1$ and $p^n - 1$ are multiples of $4$.
*Both $p^m + 1$ and $p^n - 1$ are multiples of the same odd prime $q$.
In case 1, $p^m + 1 \equiv 0 \pmod 4$ implies that $p^m \equiv 3 \pmod 4$, so that $p \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $|ax^2+bx+c|\le100$ for all $|x|\le 1$, What is the maxima for $|a|+|b|+|c|$ Here is a similar problem posted before. I try to use this method to solve this problem.
$$f(-1)=a-b+c, f(0)=c, f(1)=a+b+c$$
So we have $|c|=|f(0)|\le 100$
$$|2a|=|f(-1)-2f(0)+f(1)|\le|f(-1)|+2|f(0)|+|f(1)|\le400$$
So we have $|a|\le200$
Bu... | Let $|ax^2+bx+c|\le100$ for all $|x|\le 1$
Firstly if $x=0, |c|\le 100(1)$
And if $x = \dfrac{1}{2}, then \left| \dfrac{a}{4}+\dfrac{b}{2}+c \right| < 100\Longrightarrow| a + 2b + 4c | < 400$
$| a | = | 2a + 2b + 2c - (a + 2b + 4c) + 2c | < 2 | a + b + c | + | a + 2b + 4c | + 2 | c | < 200+ 400 + 200< 800$ (if $x=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2... | Your answers for $x=(1\pm i)/\sqrt2$, $1/x=(1\mp i)/\sqrt2$ are correct, now notice that
$$
{1\over x}={\bar x},
$$
the "complex conjugate", and
$$
x^2+{\bar x}^2={1+2i-1+1-2i-1\over 2}=0
$$
and $x^4={\bar x}^4=-1$,
so
$$
\begin{align}
x^{2022}+1/x^{2022}&=x^{4\times505}x^2+{\bar x}^{4\times505}{\bar x}^2\\
&=(-1)^{505... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2} \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) ... | If $S = 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2021}} + \frac{1}{2^{2022}}$, then just add $\color{blue}{\frac{1}{2^{2022}}}$ to both sides to get the result:
\begin{align}
S + \color{blue}{\frac{1}{2^{2022}}} &= 1 + \frac{1}{2} + \cdots + \frac{1}{2^{2020}} + \frac{1}{2^{2021}} + \underbrace{\frac{1}{2^{2022}} + \colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplifying indefinite integral (possibly in terms of elliptic integrals) I would like to simplify the following integral
$$\int ds \frac{s^n}{\sqrt{1-s^2}}\frac{1}{\sqrt{-(s-s_1)(s-s_2)(s-s_3)}}$$.
for $n=0, 1$, and $2$. Here, $s_1$, $s_2$, and $s_3$ are constants (they come from the roots of a complicated cubic polyn... | I will write here a (very) partial answer to your problem.
First of all, I am going to consider only one case, that is $n = 0$.
Even in this case, it seems like no expression exists for a "primitive" of that integral.
However if we make a deal (a bad one) we can manage to find an expression. The deal is: one of the thr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| If we take mod 3 and $y=2z+1$ then $2^y\equiv -1$ but $$x^2+7x+4 \equiv 0,1$$ a contradiction, so $y=2z$. Let $2^z = t$, then we have
*
*If $x>0$ then
$$(x+2)^2\leq \underbrace{x^2+7x+4}_{t^2} \leq (x+4)^2 \implies t = x+3\implies x=5$$
*
*If $x<0$ then $x=-n$ with positive $n$. So $$(n-4)^2\leq \underbrace{n^2-7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as fol... | First of all, clearly $z = 0$ works, so let's just assume $z \neq 0$.
Let $z = r e^{i \theta}$. We have:
\begin{align}
\bar{z} - z^2 &= i(\bar{z} + z^2) \\
(1 - i)\bar{z} &= (1 + i)z^2 \\
\frac{1 - i}{1 + i} &= \frac{z^2}{\bar{z}} \\
e^{-i \pi/2} &= r e^{3i\theta}
\end{align}
This shows that $r = 1$ and $\theta = -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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