Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$ Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x... | Noticing that $x=1$ and $x=-1$ are roots and by long division, you'll get $x^6-2x^5+3x^4-3x^2+2x-1=(x-1)(x+1)(x^4-2x^3+4x^2-2x+1)$
Considering $x^4-2x^3+4x^2-2x+1=0$, you now divide by $x^2$ and you'll get $x^2+\frac{1}{x^2}-2(x+\frac{1}{x})+4=0$ which, using the substitution $y=x+\frac{1}{x}$, gives $y^2-2y+2=0$.
I be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Inscribed circle in a trapezoid BA = 3. The radius of the circle is 2. What is DC?
I know that AG = AF = 1. And i know that FC = CE. And DC = DE + EC. I drew a line from point A to line DC to create a right triangle, but I can’t figure out where to go from there
| Attach a coordinate system to the figure, and let its origin be at point $D$.
Then $Z = (2, 2)$, and $A = (3, 4) $, and $C = (x_1, 0) $
When want to solve for $x_1$ such that the distance between $Z$ and $AC$ is $2$.
The equation of $AC $ is $ y - 4 = \dfrac{-4}{x_1-3} (x - 3) $
We want to make sure the coefficients of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation. The given question is:
The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation.
I am confused about how to work out this question, but I feel l... | We know that the equation has a triple root, so if let $\alpha$ be the triple root and $\beta$ be the single root, we get that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = (x-\alpha)^3(x-\beta)
\end{equation}
By expanding the RHS, we see that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = x^4 - (3\alpha+\beta)x^3 + (3\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-... | In this answer I use Lagrange multipliers to find the critical points of the function within the region $x,y,z\geq0$, show that the only critical point is a minimum at $x=y=z=1$, and then argue it must achieve a maximum on the boundary and find that maximum.
Let $f=xy+yz+zx-xyz$ then $$\nabla f=(1-(y-1)(z-1),1-(z-1)(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Two different answers to $\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$ Let us evaluate
$$\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$$
Dividing the numerator and denominator by $x^3$ we will be left with
$$\lim_{x \to -\infty} \frac{\frac{8}{x}-2+\frac{1}{x^3}}{\frac{6}{x}+\frac{13}{x^2}+\frac{4}{x^3}}$... | Note that
*
*for $x\to -\infty$
$$\frac{6}{x}+\frac{13}{x^2}+\frac{4}{x^3} = \frac{1}{x}\left(6+\frac{13}{x}+\frac{4}{x^2}\right)\to 0 \cdot6=0$$
with $\frac 1x <0$, then $0$ is reached from the left side ($0^-$).
*
*for $t=-x\to \infty$
$$\frac{6}{t}-\frac{13}{t^2}+\frac{4}{t^3} = \frac{1}{t}\left(6-\frac{13}{t}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | I've revised your work as follows.
The first step is fine
$$ 0<x<2 \implies -\frac a2<2x-\frac a2<4-\frac a2$$
As a minor issue we should use $\iff$ instead of $\implies$.
*
*Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$
The condition corresponds to $a\ge8$.
The inequality you reach $\left(4-\frac a2\right)^2+\frac {3a^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\i... | It's easy to show that the given integral comes down to evaluating:
$$I = \int_{0}^{1} \sqrt{\sqrt{x+1}-\sqrt{x}} \ dx$$
Define the following function on $[0,1]$:
$$f(x) = -\sqrt{\sqrt{x+1}-\sqrt{x}}$$
I claim that that this is invertible. Let's just find an explicit inverse for this. So:
$$y^2 = \sqrt{x+1}-\sqrt{x}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 5
} |
Rationalize the denominator of $\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$ Rationalize the denominator of $$\dfrac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$$ Usually we are supposed to use one of the formulas $$x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$$ I don't think they will work here. We can say $\sqrt[3]{3}=t\Rightarrow t^3=3$ and the gi... | You can proceed by identification
$(1+t-t^2)(a+bt+ct^2)=(b+c-a)t^2+(a+b-3c)t+(3c-3b+a)$
To get rid of the surds, solve $\begin{cases}b+c-a=0\\a+b-3c=0\end{cases}\implies\begin{cases}a=2c\\b=c\end{cases}$
We can set $c=1$ which correspond to the $(2+t+t^2)$ indicated in the other answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integration Issue with Finding a Centroid I've been brushing up on my understanding of centroids in 2-dimensions, and I chose to try to find $M_x$ of the region bounded by $$f(x) = \sin(x-\frac{\pi}{2})+3$$ and the x-axis, $g(x)=0$, from $x=\frac{\pi}{2}$ to $x=\frac{7\pi}{2}$.
This gives:
$$M_x=\int_\frac{\pi}{2}^\fr... | The integral $$M_x = \frac{1}{2} \int_{x=\pi/2}^{7\pi/2} \left(\sin \left(x - \frac{\pi}{2}\right) + 3 \right)^2 \, dx$$ is more easily evaluated by first performing the substitution $u = x - \pi/2$, $du = dx$ to obtain
$$\begin{align}
M_x &= \frac{1}{2} \int_{u=0}^{3\pi} \sin^2 u + 6 \sin u + 9 \, du \\
&= \frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4533038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all real solutions of the equation $x^{10} - x^8 + 8x^6 - 24x^4 + 32x^2 - 48 = 0$ I have been able to factorize the polynomial as follows: $$(x^2 - 2)(x^8 + x^6 + 10x^4 - 4x^2 + 24)$$ from which $\sqrt2$ and $-\sqrt2$ are obvious solutions. My guess is that $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ does not have any rea... | Try to write $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ as sum of squares and positive constant.
There is only one negative term, and else are all square already. So I try to factorize the negative term into a square, like
$x^8 + x^6 + 9x^4 + x^4 - 4x^2 + 4 + 20 =...>0$
You can try to finish the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Inequality with trigonometric relations performing a Von Neumann analysis, I arrived to the relation
\begin{equation}
E_m(t+\Delta t)=\underbrace{\left[1-\frac{r}{4}\left(3-4e^{-i\theta}+e^{-2i\theta}\right)\right]}_{G}E_m(t),
\end{equation}
where $\theta\in[-\pi,\pi]$ and $r>0$. The question is to find the largest $r$... | Note that for $z=e^{-i\theta}$ one has $G(z)=1-\frac{r}{4}(z^2-4z+3)$ so by the polar C-R the maximum of $|G(z)|$ when $|z|=1$ is attained at points where $\frac{zG'(z)}{G(z)} \ge 0$ (which of course means $\frac{zG'(z)}{G(z)}$ real)
Now $G'(z)=\frac{r}{2}(2-z)$ is non zero on $|z|=1$ so the above is equivalent to $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $|z| \leq 1$, then $|\text{Im}(\bar{z} + 1 )| \leq \sqrt{3-x^2}$ If $|z| \leq 1$, then $|\text{Im}(\bar{z} + 1 )| \leq \sqrt{3-x^2}$ $\quad $ for $0 \leq x \leq1.$
Attempt:
Let $z = x + iy$ such that $\bar{z} = x -iy$.
Thus,
$$|z|^2 = x^2 + y^2 \leq 1 \quad \quad \quad (1)$$ and
$$\bar{z} = x - iy$$
$$\bar{z} + 1 = ... | Your way is fine and correct, more simply, as an alternative, from here $|y| \leq \sqrt{3-x^2}$ we can square both side to obtain
$$y^2\le 3-x^2 \iff x^2+y^2 \le 3$$
which is true since $x^2 + y^2 \leq 1$.
Another way, by polar coordinates with $|r|\le 1$
$$|y| \leq \sqrt{3-x^2} \iff r|\sin \theta|\le\sqrt{3-r^2\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what positive integers $A$ is the sequence eventually constant?
For each integer $n \ge 0$ let $S(n)=n-m^2$, where $m$ is the greatest integer such that $m^2\le n$. Define $(a_k)_{k=0}^\infty$ as $a_0=A$ and $a_{k+1}=a_k+S(a_k)$ for $k \ge0$. For what positive integers $A$ is the sequence eventually constant?
I'm... | Suppose $a_k=n=m^2+e$ is a nonsquare term, where $m$ is the largest integer such that $m^2\le n$. Then $1\le e\le 2m$ since the next square is $(m+1)^2=m^2+2m+1$, and $S(n)=e$ so $a_{k+1}=m^2+2e$. We have
$$m^2<m^2+2e\le m^2+4m<m^2+4m+4=(m+2)^2$$
and $m^2+2e\ne(m+1)^2$ since the parities of both sides differ (subtract ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find the second derivative of $y=\left(1-2\sqrt{x}\right)^3$ Find the second derivative of $$y=\left(1-2\sqrt{x}\right)^3$$ Let's find the first derivative: $$y'=3(1-2\sqrt{x})^2(1-2\sqrt{x})'=3\left(0-2\dfrac{1}{2\sqrt{x}}\right)(1-2\sqrt{x})^2=-3\dfrac{1}{\sqrt{x}}(1-2\sqrt{x})^2$$ The second derivative of a function... | Your calculations are correct. After simplification, the result is $$y'' = \frac{3(1-4x)}{2x^{3/2}}.$$
A simpler approach that works in this case is to expand the cube:
$$y = (1 - 2x^{1/2})^3 = 1 - 6x^{1/2} + 12x - 8x^{3/2}.$$ Then no product or chain rule is required:
$$y' = -3x^{-1/2} + 12 - 12x^{1/2},$$
and
$$y'' ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\ln(2) > \frac{2}{3}$ I'm trying to show that $\ln(2) > \frac{2}{3}$. However, I'm pretty stuck on how to proceed.
I tried to show that this is the case using a similar technique to
$$
e < 4 \iff \sqrt{e} < 2 \iff \frac{1}{2} < \ln(2)
$$
However, while I can show that $\sqrt{e} < 2$ (because $2^2 = 4 ... | $\ln(2) > \frac{2}{3}$ is equivalent to $e^2 < 8$, so we need an upper bound for $e$. A standard trick is to replace the “tail” of the exponential series by a geometric series, which can be computed explicitly:
$$
e = \sum_{n=0}^\infty \frac{1}{n!}= 1 + 1 + \frac 12 \left(
1 + \frac{1}{3} + \frac{1}{3\cdot 4} + \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\cos(105°)$ using sum and difference identities I was trying to solve $\cos(105°)$ using sum and difference identities.
My solution:
$\cos(105°) = \cos(60°+45°)$
$\cos(60°)\cos(45°) - \sin(60°)\sin(45°)$
so,
$\frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ - $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$
then,
$\frac{\sqrt{2}}{4... | $a - b = a + (-b) = (-b) + a = -b + a$.
Thus,
$$\frac{\sqrt{2} - \sqrt{6}}{4} = \frac{-\sqrt{6} + \sqrt{2}}{4}$$
and both your answers are correct (they are equivalent).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4544050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification
Solve the quartic polynomial :
$$x^4+x^3-2x+1=0$$
where $x\in\Bbb C$.
Algebraic, trigonometric and all possible methods are allowed.
I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, t... | You can easily observe that, the expression $(x-1)^2$ is almost included in the polynomial $P(x):=x^4+x^3-2x+1$.
Let's rewrite your polynomial as follows:
$$
\begin{align}P(x)=x^4+x^3-\color{red}{x^2}+\color{red}{x^2}-2x+1\end{align}
$$
Based on this observation, we will represent the polynomial $P(x)$ as a "quadratic"... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4545364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 1
} |
2022 Gr11 Fermat math contest question #20 Question: A sequence of numbers $t_1, t_2, t_3$,... has its terms defined by $t_n = \frac{1}{n} - \frac{1}{n+2}$ for every integer $n \geq 1.$ For example, $t_4 = \frac{1}{4} - \frac{1}{6}$. What is the largest positive integer k for which the sum of the first k terms (that is... | Given, $$t_n=\frac1n-\frac{1}{n+2}$$ we get $$S_k=\sum_{n=1}^k t_n=\sum_{n=1}^k \frac1n-\frac{1}{n+2}$$$$= \sum_{n=1}^k\frac1n-\sum_{n=1}^k\frac{1}{n+2}$$
Reindexing the second sum and replacing $n+2=i$, we get $$S_k= \sum_{n=1}^k\frac1n- \sum_{i=3}^{k+2}\frac1k$$$$=1+\frac12-\frac{1}{k+1}-\frac{1}{k+2}=\frac{k(3k+5)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$
Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$.
I think it might be useful to prove that f is injective and t... | Let $P(x,y)$ be the claim $$f(x+f(y+xy)) = (y+1)f(x+1)-1.$$
Notice that $P(x,\frac{y}{x+1})$ gives
$$
f(x+f(y))=(y+x+1)\frac{f(x+1)}{x+1}-1.
$$
To prove injectivity, assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$. Then comparing $P(x,\frac{a}{x+1})$ and $P(x,\frac{b}{x+1})$ we get
$$
(a+x+1)\frac{f(x+1)}{x+1}-1=(b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $
\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
=
\begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
-1
... | One way to instantly see that $0$ is an eigenvalue is that the three columns are clearly not linearly independent --- in fact, they're all constant multiples of each other! Next, any dependency vector of those columns gives you an eigenvector for the eigenvalue $0$. For example, clearly the expression
$$\text{column 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Creating a complex number so that its norm equals to 1 I would like to create a complex number c so that its norm is equal to some number a (for the purpose of this question let's assume a = 1) if I already have either its real or imaginary part. I know that:
$$\lVert \mathbf{c} \rVert = \sqrt{\sum_{i=1} ^{n} c_i \over... | Take any nonzero complex number $c=a+bi$. Divide by it's length, which is $\sqrt{a^2+b^2}$ to get a complex number of length $1$.
$$z = \frac{c}{||c||}= \frac{a}{\sqrt{a^2+b^2}} + \frac{b}{\sqrt{a^2+b^2}}i.$$
Compute the length of $z$ to see how it works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The solution of differential equation $2xy+6x+(x^2-4)y'=0$ When I solved the above DE, I reached to the step and confused how to continue
$2x(y+3)+(x^2-4)dy/dx$
$(x^2-4) dy/dx=-2x(y+3)$
$dy/(y+3)=-2x/(x^2-4) dx$
$∫dy/(y+3)=∫-2x/(x^2-4) dx$
$\ln|y+3|=-ln|x^2-4|+C$
$\ln|y+3|+ln|x^2-4|=C$
$\ln|(y+3)(x^2-4)|=C$
$|(y+... | From this step you use exponential
$$\ln\vert y+3\vert=-\ln\vert x^2-4\vert +C$$
$$\Rightarrow e^{\ln\vert y+3\vert}=e^{-\ln\vert x^2-4\vert +C}$$
$$\Rightarrow \vert y+3\vert=\frac{e^C}{e^{\ln\vert x^2-4\vert}}$$
$$\Rightarrow \vert y+3\vert=\frac{K}{\vert x^2-4\vert}$$
For some positive number $K$
$$\Rightarrow y+3=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the closed formula for the following recusive sequence. Find the closed formula for the recursive sequence defined by $a_0 = 4$, $a_1 = 12$, $a_n = 6a_{n-1}-a_{n-2}$ for $n>1$. This question is stumping me. I don't know any methods besides guess and check.
Current progress:
The first five terms are $4, 12, 68, 396... |
Using the following fact,
$$\begin{cases}a_{n}=ba_{n-1}+ca_{n-2},\\a_{0}=c,\quad a_{1}=d \end{cases}$$ then $$\begin{cases} a_{n}=\alpha r_{1}^{2}+\beta r_{2}^{2}, \quad \text{if} \quad r_{1}\not=r_{2},\quad r^{2}-br-c=0,\\a_{n}=br_{1}^{n}+dnr_{2}^{n},\quad \text{if}\quad r_{1}=r_{2},\quad r^{2}-br-c=0,\end{cases}$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Can we make the inequality $A-B \leq \frac{A^2}{4}$ strict? I have an interesting problem:
Given that $$
A=\frac{1}{2022}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)$$
$$B=\frac{1}{2023}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2023}\right)
$$
I got an upper bound on $A-B$.The Lower bound is zero,... | If $\;A-B=\dfrac{A^2}4\;,\;$ then , by solving the quadratic equation in $H$, we get that $\;H=2-A\;,\;$ hence ,
$H=\dfrac{4044}{2023}<2<1+\dfrac12+\dfrac68<H_8<H_{2022}=H\;.$
So, $\;A-B\neq\dfrac{A^2}4\;,\;$ consequently ,$\;\;A-B<\dfrac{A^2}4\,.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\sum_{n=1}^\infty \frac{n^5}{e^{2\pi n}-1}$? I recently found the following result on Twitter.
$$\sum_{n=1}^\infty \frac{n^5}{e^{2\pi n}-1}=\frac{1}{504}$$
I know that $\int_0^\infty \frac{x^5}{e^{2 \pi x}-1} dx = \frac{5!}{(2\pi)^6}\zeta(6)=\frac{1}{504}$
How to show that the sum is also equal to the ... | I'm not sure if it's more appropriate to post this here or elsewhere, but we can integrate the function $$f(z) = \frac{z^{5}e^{iz}}{\cosh(z) - \cos(z)} $$ around the contour $$\left[-\frac{(2N+1) \pi \sqrt{2}}{2}, \frac{(2N+1) \pi \sqrt{2}}{2}\right] \cup \frac{(2N+1) \pi \sqrt{2}}{2} e^{i[0, \pi]}. $$
(For any positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)? Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)?
My answer is not possible because for $a^2$ to be equal to $b$ means that the argument of $b$ is t... | The arguments $\arg(a)$ and $\arg(b)$ are only defined up to a multiple of $2\pi$. Or, if we require that $\arg(z) \in (-\pi,\pi]$ for all $z$, then it's not necessarily the case that $\arg$ is multiplicative: we might end up having to add or subtract $2\pi$.
If we go with the first option, we know that $\arg(a) \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Prove that $a+b+c+d$ is a composite number Let $a,b,c,d$ be positive integers such that $a^2+3ab+b^2=c^2+3cd+d^2$. Prove that $a+b+c+d$ is a composite number.
I think this statement is correct, at least using the program I came across only composite numbers. An approximate solution plan is as follows: The equality $(2a... | Suppose $p:= a+b+c+d$ is a prime. Since $d=p-a-b-c$ we have:
$$a^2+3ab+b^2 = c^2+3pc-3ac-3bc-3c^2+p^2-2p(a+b+c)+a^2+b^2+c^2+2ab+2bc+2ca$$
i.e.
$$ab+ ac+bc+c^2=p^2-p(2a+2b-c)$$
or $$(a+c)(b+c) = p(p-2a-2b+c)$$
Last equation implies $p\mid a+c$ or $p\mid b+c$. In each case we have a contradiction, since $p$ is greater th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$.
Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$.
Firstl... | Your analysis is good.
Starting in the middle:
*
*$u = \sqrt{4+x}.$
*$v = \sqrt{4-x}.$
*$\dfrac{u}{2+u} = \dfrac{v}{2-v}.$
So,
$$2u - uv = 2v + uv \implies 2u - 2v = 2uv \implies $$
$$2\sqrt{4+x} - 2\sqrt{4-x} = 2\sqrt{16 - x^2} \implies $$
$$4(4 + x) + 4(4-x) - 8\sqrt{16 - x^2} = 4(16 - x^2) \implies $$
$$32 - 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=2\pi$. Probably equation $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=\int_0^{2\pi} \frac{ab}{a^2\sin^2t+b^2\cos^2t}\mathrm dt$ is useful. Double integration is also a tool, but I don't know next step.
| Substitute $y = \tan t$
\begin{align}
\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}dt=&
\ 4ab \int_0^{\pi/2} \frac{ \sec^2 t}{a^2+b^2\tan^2t}dt\\
=& \ 4ab\int_0^\infty \frac {1}{a^2+b^2y^2}dy\\
=& \ 4\arctan \left(\frac{b\tan y}a\right) \bigg|_0^\infty = 2\pi
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Trick of selecting fundamental matrix in nonhomogeneous differential equations to simplify calculations Problem:
I came across the following differential equation set:
$$
\begin{cases}
\dot{x}&=3x-2y+t\\
\dot{y}&=4x-y+t^2\\
\end{cases}
$$
My solution
I followed normal practices and found the general solution to the h... | We want to solve for $x,y$ such that
$$
\frac{d}{dt}\left[\begin{array}{c}x \\ y\end{array}\right]
= \left[\begin{array}{cc}3 & -2 \\ 4 & -1\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right] +\left[\begin{array}{c}t \\ t^2\end{array}\right].
$$
Let $C$ be the constant coefficient matrix on the ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
I have a circle with $n$ equal tangent circles inscribed, what is the diameter of the largest circles that could be inscribed in the space left? Hello I know that the question is very long and maybe confused, I'll do my best to explain it step by step.
The problem is relevant in the cable design industry. The first ste... | Let's put the diagram on the coordinate plane, with the supercircle centered at the origin, and the $+x$-axis going through the tangent point of two blue circles and the center of one red circle. Call $S$ the radius of the supercircle, $R$ the radius of the blue circles, and $r$ the radius of the red circle.
All the bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proportions of ornamental gables in gothic architecture I'm trying to figure out mathematically precise proportions for gothic architecture-style gables.
The gable has the shape of a triangle. There is a central incircle, as well as a smaller incircle in the remaining space at the top. The legs of the triangle are tang... |
We have isosceles $\triangle ABC$ with $AB = AC$. $O$ is the midpoint of $BC$, and angle $ t = \angle OAC $. Further the height is $h = AO$
From this, it follows that
$OB = OC = h \ \tan t $
The semi-perimeter is given by
$ s = OB + AB = h ( \tan t + \sec t ) $
The radius of the incircle (shown in red) is
$ R = \dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find a, real number for the following limit Find a, real number, such that $\displaystyle{\lim_{n\to\infty}{(n^3+an^2)^\frac{1}{3}-(n^2-an)^\frac{1}{2}}}=1$
I noted x=$(n^3+an^2)^\frac{1}{3}$
and y=$(n^2-an)^\frac{1}{2}$
I applied with the conjugate ($x^2-xy+y^2$) but I do not know how to continue. Any ideas?
Thank you... | Let $\alpha=n^3+an^2$ and $\beta=n^2-an$. Introduce a difference of $6^{\rm th}$ powers:
$$\begin{align*}
\alpha^{1/3} - \beta^{1/2} &= (\alpha^2)^{1/6} - (\beta^3)^{1/6} \\[1ex]
&= \frac{\left((\alpha^2)^{1/6}\right)^6 - \left((\beta^3)^{1/6}\right)^6}{(\alpha^2)^{5/6} + (\alpha^2)^{4/6}(\beta^3)^{1/6} + (\alpha^2)^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[... | The integral admits elementary close-form as evaluated below
\begin{align}
I_n=&\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} \overset{x\to 1/x}{dx}\\
=& \ \frac12 \int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx-\frac n2 \int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx
\end{align}
where $\int_0^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Pairs of eigenvectors that are not multiple of each other I have the following matrix
\begin{align}
J
&= \begin{bmatrix}
0 & \lambda a \\
a & \delta \\
\end{bmatrix}
\end{align}
For which I find two eigenvalues ($e_-$ and $e_+$):
\begin{align}
\det \left( J - e I \right) &= 0 \\
\det \left(
\begin{bmatrix}
-e & \lambd... | They are scalar multiples of each other:
$$
\left(\frac{e_+-\delta}{a},1\right)
=\frac{e_+-\delta}{a}\left(1,\frac{e_+}{\lambda a}\right).
$$
Note that $\dfrac{e_+-\delta}{a}\dfrac{e_+}{\lambda a}=1$ because $e_+$ is a root of the characteristic equation $x^2-\delta x-\lambda a^2=0$ of $J$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s... | For even cases, apply
\begin{align}
&1-x^{4m}=(1-x^4) \prod_{k=1}^{m-1} \left(1+2x^2\cos\frac{k\pi}{m}+x^4 \right)\\
& 1-x^{4m+2}= (1-x^2)\prod_{k=0}^{m-1} \left(1+2x^2\cos\frac{(2k+1)\pi}{2m+1}+x^4\right)
\end{align}
and
$$\int_0^1 \frac{\ln(1+2x^2\cos \theta +x^4)}{1+x^2}dx
=\pi \ln\left(2\cos\frac{\theta}4\right)-2G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
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A lower bound for upper Darboux sums of $\int_0^1 x^3 \, dx$
Find the largest $\lambda$ such that $$\sum_{k=1}^n x_k^3\left(x_{k}-x_{k-1}\right)\ge \frac{1}{4}+\frac{\lambda}{n}$$ holds for any $n \in \mathbb{N}$ and any $x_0,x_1,x_2,\cdots, x_n\in\mathbb{R}$ satisfying $$0=x_0\le x_1\le x_2\le \cdots\le x_n=1.$$
It'... | For every partition $0 = x_0 \le x_1 \le x_2 \le \cdots \le x_n = 1$ of $[0, 1]$ is
$$
\sum_{k=1}^n x_k^3(x_k - x_{k-1}) \ge \frac 14 + \frac 38 \frac 1n
$$
and the factor $\boxed{\lambda = \frac 38}$ is best possible.
Proof: The identity
$$
a^3 (a-b) = \frac 14 (a^4 - b^4) + \frac 38 (a^2-b^2)^2 + \frac 18 (a-b)^3(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find all the real solutions of $(1+x^2)(1+x^4)=4x^3$. As title suggests, the question is to find all the real roots to the polynomial:
$$(1+x^2)(1+x^4)=4x^3$$
This problem was asked in the Kettering University Math Olympiad a few years back, it's an interesting problem with many different ways to approach it. I'm going... | Nice answer.
An amusing way is to use AM/GM. First, note that any solution is positive, since the left side is positive, and thus $4x^3$ must be positive, so $x$ must be positive.
Then AM/GM says $$\frac{1+x^2}{2}\geq x\\\frac{1+x^4}{2}\geq x^2$$
So $$\frac{(1+x^2)(1+x^4)}{4}\geq x^3$$ for $x>0,$ with equality only whe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Solution for $\Delta u = 0$ for which $u(\cos(t), \sin(t)) = t(2\pi - t) \ , \ 0 \leq t \leq 2\pi$ I have the following problem: Find the solution for $\Delta u = 0$ for which $u(\cos(t), \sin(t)) = t(2\pi - t), \\ 0 \leq t \leq 2\pi$.
My attempt:
I rewrite $u(\cos(t), \sin(t)) = u(x(t), y(t))$ with $x(t) = \cos(t), \ ... | I’ve made some mistakes in the original answer and is easier to redo everything. I can’t delete the answer so I’m compelled to at another one.
As I understand the problem is this: solve $\Delta u=0$ for $u(\cos t, \sin t)=t(2\pi-t), 0\le t\lt 2\pi$
There is a lot of differentiating from now on.
$$\frac{du}{dx}=\frac{\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Source: https://www.hkage.edu.hk/uploads/file/202207/6cda89c718b674f6ac3aa2c19049abe5.pdf
I tried substituting $y = 2x+3$ and ended up with $\frac{y^2}{2}-\frac{7}{2}y^{\frac 3 2} ... | Notice that
\begin{align*}
2x^2+6x+9-7x\sqrt{2x+3}&=2x^2-7x\sqrt{2x+3}+3(2x+3)\\
&=\left(x-3\sqrt{2x+3}\right)\left(2x-\sqrt{2x+3}\right).
\end{align*}
The real solution of $x-3\sqrt{2x+3}=0$ is $x=9+6\sqrt 3$ and the real solution of $2x-\sqrt{2x+3}=0$ is $x=\frac{1+\sqrt{13}}4$. Finally,
$$9+6\sqrt 3>9>\frac{1+4}4>\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Number of walks of length k between 2 vertices in path graph Let $P_n$ define a path graph with $n$ vertices (example $P_5$ shown on image). Let $f_{n, k}(a, b)$ define a number of walks of length $k$ in $P_n$ starting from vertex $a$ and ending in vertex $b$.
Is there a formula for $f_{n, k}(a, b)$? Even after some re... | We can use the adjacency matrix: $$f_{n,k}(a,b) = (A(P_n)^k)_{ab}$$ where $A(P_n)$ is the adjacency matrix of $P_n$. That adjacency matrix has the form $$\begin{bmatrix}0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How might I have anticipated that $\frac14(\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}})$ simplifies to a single surd (namely, $\frac14\sqrt{25+10\sqrt{5}}$)? This is perhaps a silly question related to calculating with surds. I was working out the area of a regular pentagon ABCDE of side length 1 today and I ended up with the... | $10 + 2 \sqrt 5$ has norm $100 - 5 \cdot 4 = 80.$ $5 + 2 \sqrt 5$ has norm $25 - 5 \cdot 4 = 5.$ The ratio of the norms is $\frac{80}{5} = 16,$ which is an integer and a square, so the ratio might be very nice.
$$ \frac{10+2 \sqrt 5}{5 + 2 \sqrt 5} \cdot \frac{5-2 \sqrt 5}{5 - 2 \sqrt 5}
= \frac{30-10 \sqrt 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[... | The exact algebraic answer, which expressible in terms of real quantities by real-valued radicals, is:
$$\bbox[5px,border:2px solid #C0A000]{\begin{align}&x=-\alpha_k\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\\
&y=-\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\end{align}}$$
where,
$$\begin{align}\alpha_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
$(a^n-1)(b^n-1)$ can't be a square for all $n$ unless $ab$ is a square Let $a,b$ be positive integers $>1$ such that $(a^n-1)(b^n-1)$ is a square for all $n\ge 1$. Prove that $ab$ is a perfect square.
I'm not asking for a solution to this problem because I already know one. What I'm asking is would the following approa... | The limit of $d_n$ is not zero except if $a=1$ or $b=1$. Assume that $b \ge a \ge 2$. Then,
\begin{align*}
\sqrt{\Big(a^n-\frac{1}{a^2}\Big)\Big(b^n-\frac{1}{b^2}\Big)} & \ge \sqrt{\Big(a^n-\frac{1}{4}\Big)\Big(b^n-1\Big)} \\
& \ge \sqrt{(a^n-1)(b^n-1) + \frac{3}{4}(b^n-1)} \\
& \ge \sqrt{(a^n-1)(b^n-1)} + \frac{3(b^n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$?
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ?
The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$
How to proceed?
| If the question is asking you to factor you need not necessarily solve for the roots.
Use the FOIL (First, Out, In, Last) method to expand $(ax + b)(cx + d)$ and simultaneously solve for the coefficients $a, b, c, d$.
We would have $$\begin{cases} ac = 20 \\ ad + bc = 13 \\ bd = -15 \end{cases}$$
Furthermore, once you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$
My attempt:
$$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$
$$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1... | Your answer is correct.... Sort of!
The answer you arrived at, can be simplified further to arrive at the answer you wish to achieve, since your attempt is completely valid. So, we have:
$$\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2}$$
$$\frac{x^3-5x^2+7x-8+6}{x^3-6x^2+9x-8+6}$$
$$\frac{(x^3-8)-(5x^2-7x-6)}{(x^3-8)-(6x^2-9x-6)}... | {
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Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \i... | Also using complex analysis for $\theta\ne \frac{\pi}{2}$ we have
$$
\int_0^{2 \pi} \frac{dz}{1+\cos \theta \cos x}
{=
\frac{1}{i}\oint_{|z|=1} \frac{2dz}{2z+\cos \theta (z^2+1)}
\\=
\frac{1}{i}2\pi i\text{Res}\left\{\frac{2}{2z+\cos \theta (z^2+1)}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
2\pi\text{Res}\... | {
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Where is error in this solution to $ \sqrt{x} + \sqrt{-x} = 2 $? Given the equation:
$$ \sqrt{x} + \sqrt{-x} = 2 $$
The solutions are $x = \pm 2i$. This can be seen via Wolfram Alpha
$$
\left( \sqrt{x} + \sqrt{-x} \right)^2 = 2^2
$$
$$
\sqrt{x}^2 + 2\sqrt{x}\sqrt{-x} + \sqrt{-x}^2 = 4
$$
$$
x + 2\sqrt{x}\sqrt{-x} - x =... | For the function $ \ f(x) \ = \ \sqrt{x} + \sqrt{-x} \ \ , \ $ if we are restricted to $ \ x \ \in \ \mathbb{R} \ \ , \ $ the first term is only defined for $ \ x \ \ge \ 0 \ \ $ and the second term, for $ \ x \ \le \ 0 \ \ , \ $ so the domain for $ \ f(x) \ $ is just the intersection of these intervals, $ \ x \ = \ 0 ... | {
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Is there other method to evaluate $\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x, \textrm{ where }n\in N?$ Letting $x\mapsto \frac{1}{x}$ transforms the integral into
$\displaystyle I=\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x=-\int_0^1 \frac{x^{2 n-1} \ln x}{x^n+1} d x \tag*{} $
Splittin... | Substitute $t=\frac1{x^n}$
$$\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x
=-\frac1{n^2}\int_0^1 \frac{t\ln t}{1+t}dt
= \frac{1}{n^2}\left(1-\frac{\pi^2}{12}\right)
$$
| {
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I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me. $x,y,z>0$. Prove that:$$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$ My solution:$$\frac{x+y}{2}≥\frac{2xy}{x+y}$$ $$\frac{y+z}{2}≥\frac{2yz}{y+z}$$ $$\frac{x+z}{2}≥\frac{2xz}{x+z}$$... | Well I think using means inequalities $HM\le GM\le AM$ is pretty much the way to go, and it all boils down in the end to $(x-y)^2\ge 0$ if you decide to prove it directly.
Indeed the $AM\ge HM$ inequality is just
$AM-HM=\dfrac{x+y}2-\dfrac 2{\frac 1x+\frac 1y}=\dfrac{x+y}2-\dfrac{2xy}{x+y}=\dfrac{(x+y)^2-4xy}{x+y}=\dfr... | {
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Integrating $\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}$
I'm struggling with the integral, $$\int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx.$$
I tried it as follows:
Substituting $x-2 = \frac1t \implies dx = \frac{-dt}{t^2}.$
$$\therefore \int\frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} = \int \frac{- dt}{\frac{t^2}{t^4} \sqrt{(\f... | Continuing where you left off,
$$\begin{align*}
I &= - \int \frac{t^3}{\sqrt{18t^2+10t+1}} \, dt \\[1ex]
&= -16 \int \frac{(s-5)^3}{(s^2-18)^4} \, ds
\end{align*}$$
by employing the substitution,
$$s = \frac{\sqrt{18t^2+10t+1}-1}t \implies t = \frac{2s-10}{18-s^2} \implies dt = \frac{2(18-10s+s^2)}{(18-s^2)^2} \, ds$$
... | {
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Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any. Given the curve $y=\frac{5x}{x-3}$. To examine its asymptotes if any.
We are only taking rectilinear asymptotes in our consideration
My solution goes like this:
We know that, a straight line $x=a$, parallel to $y$ axis can be a vertical asymptote of a br... | A homographic function is the quotient of two first-degree polynomial functions, through an expression in the form, with $c\ne 0$,
$$f \left( x \right)=\dfrac {ax+b} {cx+d}=\frac{5x+0}{x-3}$$
$a=5, b=0, c=1, d=-3$.
*
*If $c=0$ then $y =\frac{a \cdot x}{d}+\frac{b}{d}$, which is the equation of a line of angular coeff... | {
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"source": "stackexchange",
"question_score": "3",
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Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$ Though $n\geq 2$ is a real number, which is not necessarily an integer, we can still resolve the integrand into two fractions,
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^... | We have
\begin{align*}
&\int_{0}^{\infty} \frac{x^n - 2x + 1}{x^{2n} - 1} \, \mathrm{d}x \\
&= \int_{0}^{\infty} \left( \frac{1}{x^n - 1} - \frac{2x}{x^{2n} - 1} \right) \, \mathrm{d}x \\
&= \mathrm{PV}\!\int_{0}^{\infty} \frac{1}{x^{n} - 1} \, \mathrm{d}x - \mathrm{PV}\!\int_{0}^{\infty} \frac{2x}{x^{2n} - 1} \, \math... | {
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"source": "stackexchange",
"question_score": "14",
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"answer_id": 2
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Convert a boolean expression to just NOR operatons How do we convert this expression to just $\text{NOR}$ operation?
$$A \cdot B + B \cdot C + C \cdot D$$
My attempt:
$A \cdot B + B \cdot C + C \cdot D = \overline{\overline{A \cdot B + B \cdot C + C \cdot D}} $
Using De Morgan's law
$ \to \overline{\overline{(A \cdot B... | WolframAlpha returns
(a NOR c) NOR (b NOR c) NOR (b NOR d)
You can derive it by using the steps here.
| {
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Find the equation of the circle touching the line $(x-2)\cos\theta+(y-2)\sin\theta=1$ for all values of $\theta$
Find the equation of the circle touching the line $(x-2)\cos\theta+(y-2)\sin\theta=1$ for all values of $\theta$
The answer is given on toppr website.
It says, $(x-2)\cos\theta+(y-2)\sin\theta=\cos^2\theta... | Here's a simpler and correct (verified by GeoGebra) solution.
The circle center is $(m,n)$, then its distance to the line must be constant as $\theta$ varies. If we write the line's equation as
$$x\cos\theta+y\sin\theta-2\cos\theta-2\sin\theta-1=0,$$
apply the distance formula to have
$$d=\frac{|m\cos\theta+n\sin\theta... | {
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How do we find the value of $\tan\biggr(\sum_{1\leq i\leq 21}f(i)\biggr)$ if $f(x) = \arctan\left(\frac{1}{x^{2 }+ x + 1}\right)$? Let's assume that $f : \mathbb{R}\rightarrow (-\frac{\pi}{2}, \frac{\pi}{2})$ such that for every $x\in \mathbb{R}$, $f(x) = \arctan\biggr(\frac{1}{x^2+x+1}\biggr)$. How can we find $\tan\b... | As @TheBestMagician has mentioned, notice that:
\begin{align*}
\arctan\left(\frac{1}{x^{2} + x + 1}\right) = \arctan\left(\frac{(x + 1) - x}{1 + (x + 1)x}\right) = \arctan(x + 1) - \arctan(x)
\end{align*}
Based on such identity, the proposed sum telescopes.
Hopefully this helps!
| {
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Finding the angle $\angle BDC$
Let's assume that $\angle DBC = 50^{\circ}$, $[DC]$ bisects $\angle ACB$, and that $|AC| = |BC|-|AD|$. How could we find the angle $\angle BDC$?
Applying angle bisector theorem:
$$\frac{|AD|}{|BD|} = \frac{|AC|}{|BC|}$$
Since $|AC| = |BC|-|AD|$,
$$\frac{|AD|}{|BD|} = \frac{|BC|-|AD|}{|BC... | Let $\displaystyle x = \frac{∠A}{2} \;,\; y = \frac{∠C}{2}\;$
Given $∠B=50° \quad → 2x+2y = 130°$
Law of Tangent, on ΔACD, given $AD = (a-b)$
$\displaystyle \frac{b-(a-b)}{b+(a-b)}
= \frac{2b}{a} - 1
= \frac{\tan \left(\frac{(180-2x-y)-y}{2}\right)}{\tan \left(\frac{(180-2x-y)+y}{2}\right)}
= \frac{\tan 25°}{\tan (90°... | {
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"source": "stackexchange",
"question_score": "8",
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Sum of primes in the interval $(\delta n,n)$ Let $0<\delta<1$. What is an asymptotic formula for $$\sum_{\delta n\leqslant p\leqslant n}p,$$where $p$ is a prime number?
| Let $s(n) = \sum_{p\leqslant n, p \text{ is prime}}p$. Then, as detailed in this MathOverflow answer, $s(n) \sim \frac{n^2}{2\log{n}}$.
Thus,
$$
\begin{align}
2s(n) - 2s(\delta n) &= \frac{n^2}{\log{n}} - \frac{\delta^2n^2}{\log{\delta n}} + o\left(\frac{n^2}{\log{n}}\right)\\
&= \frac{n^2 \log{\delta n} - \delta^2n^2\... | {
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Can $7^n+1$ be a perfect cube when $n>1$? $7^n+1$ (when $n>1$) can in principle be a perfect cube, since it sometimes is divisible by $8$, and sometimes leaves $8$ as a remainder when dividied by $9$. But when I tried to find perfect cubes that is form of $7^n+1$, I did not suceed.
Can $7^n+1$ be a perfect cube?
| This solution is based on a solution to a similar problem about $2^n + 1$ found here.
No.
Suppose that $7^n + 1 = m^3$ for some positive integers $n$ and $m$. Rearranging, we get $$7^n = m^3 - 1 = (m-1)(m^2 + m + 1).$$
Because $7$ is prime, we know that both $m-1$ and $m^2 + m + 1$ must both be powers of $7$ (includin... | {
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Proving $n(n-2)(n-1)^2$ is divisible by 12 I want to show that $n^4 - 4n^3 + 5n^2 -2n$ is divisible by $12$ whenever $n>0$.
I reduced this to $n(n-2)(n-1)^2$. My approach has been to check divisibility by $3$ and $4$.
In both cases, squares are always congruent to $0$ or $1$. So
\begin{align*}n^2 &\equiv_3 0, 1 \\ \imp... | Suppose $(n,12)=1.$ We use Euler and lil' Fermat.
$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 2(n-1)\pmod {2^2},$$
and
$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 0\pmod 3.$$
So by CRT, using Bezout, we have $1×4×2×0-1×3×2(n-1)\equiv6(n-1)\equiv0\pmod{12}.$
Suppose $(n,12)\neq1.$ Then either $2\mid n$ or $3\mid n.$
If $2\mid n,$ th... | {
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Turn a real number (with a complex closed form) into its trigonometric form This post outlines 'fake' complex numbers (real numbers with complex closed form that usually come from the roots of unfactorable cubics (the example I need right now), or they can come from things like $i^i = e^{-\frac{\pi}{2}}$), in his own a... | Recall DeMoivre's formula:
$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
$$\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$$
This can trivially be derived from Euler's formula.
Also note that in polar coordinates,
$$1 + i\sqrt{7} = \sqrt{8} \operatorname{cis}(\tan^{-1}\sqrt{7})$$
So by De... | {
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Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. Proof that ${\left(1-\dfrac{2}{x^2}\right)}^x\!< \dfrac{x-1}{x+1}$ for any $x > 2$.
any ideas?
| Hi @max We can use Binomial series as we have :
$$\left(1-\frac{2}{x^{2}}\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}=\left(\frac{\left(x-1\right)}{x+1}\right)^{1+\frac{1}{x}-1}$$
So at order two we have :
$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-... | {
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How is $\frac{\cot10^\circ-\cot20^\circ}{\tan40^\circ(\cot10^\circ+\cot20^\circ)-2}=\frac{\sqrt3}{3}$ Can someone show me how $$\dfrac{\cot10^\circ-\cot20^\circ}{\tan40^\circ(\cot10^\circ+\cot20^\circ)-2}=\dfrac{\sqrt3}{3}$$
Is this an obvious simplification, as I really cannot see it. The LHS is equivalent to $$\dfrac... | You need factorisation and defactorisation formulas to solve the problem.
\begin{align}...&=\dfrac{\sin10^\circ \cos40^\circ}{\sin 20^\circ(\cos20^\circ-2\sin10^\circ\cos40^\circ)}\\
&=\dfrac{ \cos40^\circ}{2\cos 10^\circ(\cos20^\circ-2\sin10^\circ\cos40^\circ)}\\
&=\dfrac{ \cos40^\circ}{2\cos 10^\circ\cos20^\circ-2\si... | {
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Show that if $x,y>0$, $\left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$ Through some rearrangement of the inequality and expansion, I have been able to show that the inequality is equivalent to
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6≥0$$
However, I am not sure how to prove the above or if expansion and rear... | Note that your inequality is a trivial equality for $x=y$ i.e. $x^6=x^6$ as the factors $2$ cancel out.
This indicates that you can factorize $(x-y)$ out of your final equation (i.e. it's value is $0$ when $x=y$)
You get:
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6
=(x-y)(x^5+x^4y-2x^3y^2+2x^2y^3-xy^4-y^5)$$
The degree $5$ RHS e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the Maclaurin expansion of exact ODE general solution has the same form as the power series solution So, given the ODE
$$
y''+2y'+y=0
$$
I have found a power series solution, coefficient recurrence relation and the general solution in terms of elementary functions:
$$
\begin{aligned}
a_{n+2} &= -\frac{2a_{n+... | Just expand $(C_1+C_2x)e^{-x}$ to get
$$y=C_1+(C_1+C_2)x+(C_1/2+C_2)x^2+\cdots$$
from which we get $a_0=C_1$ and $C_1+C_2=a_1$, i.e. $C_2=a_1-a_0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Expressing complex roots in a trigonometric form
Given that
$$(z+2)^{12}=z^{12}$$
Show that its roots may be expressed in the form
$$-1-i\cot\left(\frac{1}{12}k\pi\right)$$
where $k=\pm1,\pm2,\pm3,\pm4,\pm5.$
My attempt at solving this:
$$\frac{(z+2)^{12}}{z^{12}}=1 \rightarrow \left({\frac{z+2}{z}}\right)^{12}=1$$
$... | Your calculations are correct so far, you can continue with
$$
\frac{2}{-1+e^{i\frac{1}{6}k\pi}} = -1 + \frac{e^{i\frac{1}{6}k\pi}+1}{e^{i\frac{1}{6}k\pi}-1}
= -1 + \frac{e^{i\frac{1}{12}k\pi}+e^{-i\frac{1}{12}k\pi}}{e^{i\frac{1}{12}k\pi}-e^{-i\frac{1}{12}k\pi}}
= -1 +\frac{2\cos(\frac{1}{12}k\pi)}{2i\sin(\frac{1}{12}k... | {
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"url": "https://math.stackexchange.com/questions/4642052",
"timestamp": "2023-03-29T00:00:00",
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Range of $f(x)=x \sqrt{1-x^2}$ I have to find the range of $f(x)=x\sqrt{1-x^2}$ on the interval $[-1,1]$. I have done so by setting $x=\sin\theta$ and thus finding it to be $[-0.5,0.5]$.
Let $x=\sinθ$. Then, for $x\in[-1,1]$ we get that $θ \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, $f(x)$ becomes:
$f(\theta)=\sin\thet... | Alternatively, observe that the function is odd over $[-1,1]$. Hence, if the max value occurs at $x_0 > 0$, then the min occurs at $-x_0$. But if $x > 0$, then by AM-GM inequality: $f(x) =x\sqrt{1-x^2} \le \dfrac{x^2+ (1-x^2)}{2}=\dfrac{1}{2}$ with $=$ occurs when $x=\sqrt{1-x^2}\implies x^2 = 1-x^2\implies x=\dfrac{1}... | {
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Evaluate $\lim_{x \to \infty} (e^{\frac{1}{x}}(x^3 - x^2 + \frac{x}{2}) - \sqrt{x^6 +1})$ using Taylor $\lim\limits_{x \to \infty}\left[e^{\frac{1}{x}}\left(x^3 - x^2 + \dfrac{x}{2}\right) - \sqrt{x^6 +1}\right]$
(I assume that we need to use $t = \frac{1}{x}$ in order to get lim, where argument goes to $0$. But for n... | We have that
$$e^{\frac{1}{x}} = 1+\frac1x+\frac1{2x^2}+\frac1{6x^3} +o\left(\frac1{x^3}\right)$$
$$\sqrt{x^6 +1} = x^3\left(1+\frac1{x^6}\right)^\frac12= x^3+\frac12\frac1{x^3}+o\left(\frac1{x^3}\right)$$
and then
$$e^{\frac{1}{x}}\left(x^3 - x^2 + \frac{x}{2}\right) - \sqrt{x^6 +1}=$$
$$=x^3 - x^2 + \frac{x}{2}+x^2 -... | {
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"answer_id": 0
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Does $a c>b^2$ if and only if $(1-a) (1-c)<(1-b)^2$, where $0Does $$a c>b^2$$ if and only if $$(1-a) (1-c)<(1-b)^2,$$ where $0<a<1, 0<c<1$, and $0<b<1$ If it is correct, then how to prove it?
| I finally find a easy way to prove it.
First, $ac>b^2$ implies $b^2<(\frac{a+c}{2})^2$. And note that $$(1-a)(1-c)-(1-b)^2 \leq (1-\frac{a+c}{2})^2-(1-b)^2=(2-\frac{a+c}{2}-b)(b-\frac{a+c}{2})<0$$
The last inequality is from $b^2\leq(\frac{a+c}{2})^2$ combined with $a,b,c,\in (0,1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this series $f(n) = f(n/2) + n$? Can I solve this as:
$f(n) = f(n/2) + n$
let find, $$f(n/2) = f(n/2/2) + n/2\\
f(n/2) = f(n/4) + n/2$$
Now,
$$\begin{split}
f(n) &= f(n/4) + n/2 + n\\
f(n) &= f(n/8) + n/4 + n/2 + n
\end{split}$$
hence so on.
$$\vdots$$
Now, $n = 2^i$.
$$\begin{split}
f(2^i) &= f(2^i/2^i)... | Given $$f(n) = f(n/2) + n$$ for all even integers $n\geq 2$, substitute the linear ansatz $f(n)= A n + B$ to yield
$$A n + B = A/2 n + B/2 + n.$$ Comparing coefficients of $n^1=n$ and of $n^0=1$ yields
the constraint equations $$A = A/2 + 1$$ and $$B=B/2$$ which are simultaneously solved only by $A=2$ and $B=0$. So
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| For example,
$$X = 1+2+3+4+5+6$$
Then twice $X$ is
$$2X = (1+2+3+4+5+6) + (1+2+3+4+5+6)$$
which we can rearrange as
$$2X = (1+2+3+4+5+6) + (6+5+4+3+2+1)$$
and add term by term to get
$$2X = (1+6)+(2+5)+(3+4)+(4+3)+(5+2)+(6+1)$$
to get
$$2X = 7+7+7+7+7+7 = 6*7 = 42$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 27
} |
Perfect numbers, the pattern continues The well known formula for perfect numbers is
$$
P_n=2^{n-1}(2^{n}-1).
$$
This formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$:
$$
496=1+2+4+8+16+31+62+124+248
$$
one can see that the first pattern is a sequence o... | The patterns you're after tell us nothing about the perfect(ness) of a number.
Since they hold even if the number is not perfect, for example, take n = 6. $2^5 (2^6-1) = 2016 = 2^{10} + 2^9 + ... + 2^5$. Which is valid for any n, since in binary, 2^6-1 is 6-1=5 1's from left, and multiplication by 2^5 is adding 5 zeros... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Exploring the quadratic equation $x^2 + \lvert x\rvert - 6 = 0$ This question and the described solution are copied from a test-paper :
For the equation $x^2$ + |x| - 6 = 0 analyze the four statements below for correctness.
*
*there is only one root
*sum of the roots is + 1
*sum of the roots is zero
*the product ... | The statement is saying:
$|x| = -x$ (for $x < 0$)
and divide both sides by $x$ to give us:
$\frac{|x|}{x} = -1$
Let us test a few values to make sure this holds.
$x = -3$ and $|x| = 3$:
$\frac{|-3|}{-3} = -1$
$\frac{3}{-3} = -1$
$-1 = -1$
True.
$x = -5$ and $|x| = 5$
$\frac{5}{-5} = -1$
$-1 = -1$
True.
We can verify th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
| for any $\theta \in \mathbb{R}$ the transformation $\psi: x \rightarrow 2x^2 -1$ sends $cos \theta$ to $cos 2\theta$ . hence $\psi^2 $ sends $cos \theta$ to $cos4\theta$
if $\alpha = \frac{2\pi}5$ then $cos 4\alpha = cos \alpha$ so that $cos \alpha$ is a fixed point for $\psi$ and if $c=cos \alpha$ we have
$$
\psi^2(c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/7695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 11,
"answer_id": 4
} |
Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I have some trouble to do that and I'd glad with any help I may get.
| Once someone gives the answer, it becomes easier!
I presume you are only looking for real roots and that $\displaystyle (5x-x^3)^{1/3}$ is the unique real number whose cube is $\displaystyle 5x - x^3$.
First observe that
$\displaystyle (x-2)^3 + 5x - x^3 = -6x^2 + 17x -8$
So if $\displaystyle a = (5x-x^3)^{1/3}$ and $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?
| In general, in $\triangle{ABC}$
If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.
If $\sin^2 A + \sin^2 B + \sin^2 C = 2$ then $\triangle{ABC}$ is right angled.
If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.
Assume $A \le B \lt \pi/2$ and $ A \le B \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A Vector Arithmetic Function, Special Integer Partitions and Residue Sequences Consider the following arithmetic function, $f(a,b) = (a+1, \lceil a- \frac{a}{b} \rceil , \lceil a- \frac{2a}{b} \rceil, \dots, \lceil \frac{a}{b} \rceil, 1)$, where all terms except the first and last are computed by the application of a c... | Here's an answer to the second part of Problem 1: $$k = \frac{(a+1)(b+1) - \gcd(a,b) + 3}{2}.$$
According to your definition,
$$k = 1 + (a+1) + \sum_{j=1}^{b-1} \left\lceil \frac{ja}{b} \right\rceil.$$
Now, $\left\lceil \frac{ja}{b} \right\rceil = \left\lfloor \frac{ja + b - 1}{b} \right\rfloor.$ (See, for example, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/10726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem:
$$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$
| $\textbf{HINT:}$ Actually this is the answer in itself.
$\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$.
This is all that you will need when you do induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Does the sum of reciprocals of primes converge? Is this series known to converge, and if so, what does it converge to (if known)?
Where $p_n$ is prime number $n$, and $p_1 = 2$,
$$\sum\limits_{n=1}^\infty \frac{1}{p_n}$$
| Let's start with three lemmas:
*
*Suppose $A\subseteq\{1,2,3,\ldots\}$ and $\sum\limits_{n\in A} \dfrac 1 n < \infty$. Then $\sum\limits_{n\in B} \dfrac 1 n <\infty$ where $B$ is the closure of $A$ under multiplication.
*The closure of the set of primes under multiplication is all of $\{1,2,3,\ldots\}$.
*$\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/15946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 0
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Maximize the product of two integer variables with given sum Let's say we have a value $K$. I want to find the value where
$A+B = K$ (all $A,B,K$ are integers) where $AB$ is the highest possible value.
I've found out that it's:
*
*$a = K/2$; $b = K/2$; when $K$ is even
*$a = (K-1)/2$; $b = ((K-1)/2)+1$; when $K$ i... | In the even case, assuming $A\leq B$, the numbers will be of the form $A=\frac{K}{2}-m$ and $B=\frac{K}{2}+m$ in order to get $A+B=K$, with $m=0,1,\ldots,\frac{K}{2}$. Then multiplying gives $AB=\left(\frac{K}{2}\right)^2 - m^2 \leq \left(\frac{K}{2}\right)^2$.
In the odd case, you'll have $A=\frac{K}{2}-m-\frac{1}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Integration by parts: $\int e^{ax}\cos(bx)\,dx$ I need to evaluate the following function and then check my answer by taking the derivative:
$$\int e^{ax}\cos(bx)\,dx$$
where $a$ is any real number and $b$ is any positive real number.
I know that you set $u=\cos(bx)$ and $dv=e^{ax} dx$,
and the second time you need t... | So you have
$$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$
Multiplying out you get
$$\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,dx.$$
At this poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
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Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ Using congruences, show that the following is always an integer for every integer
value of $n$:
$$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$
| Taking the lcm we have $\displaystyle \frac{1}{15} \cdot \Bigl[ 3n^{5}+5n^{3} + 7n\Bigr]$ Now show that the quantity $3n^{5} + 5n^{3}+7n$ is always divisible by $15$. Induction may be useful.
Clearly for $n=1$, $3+5+7=15$ is divisible by $15$. Assume that it is true for $n=k$. That is assume that $3k^{5}+5k^{3}+7k$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/21548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 0
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What are the steps to solve this simple algebraic equation? This is the equation that I use to calculate a percentage margin between cost and sales prices, where $x$ = sales price and $y$ = cost price:
\begin{equation}
z=\frac{x-y}{x}*100
\end{equation}
This can be solved for $x$ to give the following equation, which c... | First, clear the denominator by multiplying both sides by $x$:
\begin{align*}
z &= \frac{100(x-y)}{x}\\
zx &= 100(x-y)
\end{align*}
Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$:
\begin{align*}
zx &= 100x - 100y\\
zx - 100x &= -100y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $a^{k} \equiv b^{k} \pmod m $ and $a^{k+1} \equiv b^{k+1} \pmod m $ and $\gcd( a, m ) = 1$ then $a \equiv b \pmod m $ My attempt:
Since $a^{k} \equiv b^{k}( \text{mod}\ \ m ) \implies m|( a^{k} - b^{k} )$ and $a^{k+1} \equiv b^{k+1}( \text{mod}\ \ m ) \implies m|( a^{k+1} - b^{k+1} ) $
Using binomial id... | This is related to what Bill Dubuque said but is more elementary:
Since $a^k = b^k$ mod $m$, one has $m | (a^k - b^k)$, and similarly $m | (a^{k+1} - b^{k+1})$. Thus by the first equation $m$ divides $b(a^k - b^k) = (a^kb - b^{k+1})$. Subtracting, we have that $m | (a^{k+1} - b^{k+1}) - (a^kb - b^{k+1}) = a^{k+1} - a^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 3
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Resolution of a non-homogeneous heat equation I'm looking for solution to this non-homogeneus problem.
$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$ for $0<x<\pi$, $t>0$
$u(x,0)=0$
$u(0,t)=\frac{\partial u}{\partial x}(x=\pi)=0$
Does anyone know how to proceed?
| Let $u(x,t)=\sum\limits_{s=0}^\infty C(s,t)\sin\dfrac{(2s+1)x}{2}$ so that it automatically satisfies $u(0,t)=\dfrac{\partial u}{\partial x}(x=\pi)=0$ ,
Then $\sum\limits_{s=0}^\infty\dfrac{\partial C(s,t)}{\partial t}\sin\dfrac{(2s+1)x}{2}+\sum\limits_{s=0}^\infty\dfrac{(2s+1)^2}{4}C(s,t)\sin\dfrac{(2s+1)x}{2}=F(x,t)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/24480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus:
$|z-1| = 2|z+1|$
that is, where the modulus of $z -1$ is twice the modulus of $z+1$
I've solved this problem algebraically (by letting $z=x+iy$) as follows:
$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2... | A perhaps easier way to see that it is a circle is to rewrite as
$\displaystyle |z'| = 2|z' + 2|, \text{where } z' = z -1$
And so
$\displaystyle 1/4 = |1/2 + 1/z'|$
Thus $\displaystyle \frac{1}{z'}$ gives a circle centred at $(-1/2,0)$ and radius $1/4$.
As we can see from the answers and comments to this recent questi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/27199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 1
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Integer multiplication using FFT I've some trouble in understanding integer multiplication using FFT.
I'm using the algorithm described on wikipedia.
Here is an example of how I understand this algorithm:
$$a=173$$
$$b=95$$
Lets take $w=4$, then we have
$$a=13*2^{4\cdot 0}+10*2^{4\cdot1}$$
$$b=15*2^{4\cdot 0}+5*2^{4\cd... | -- answer first posted as comment --
At this point I think you're supposed to reinterpret this result as a natural number that's the product of the numbers you started with. Wait, $\sqrt{2}$? Have you double-checked the calculations? Also I think your vectors should have 4 positions because now the result will be crop... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Simple 4-cycle permutation I call a 4-cycle permutation simple if I can write it as $(i,i+1,i+2,i+3)$ so $(2,3,4,5)$ is a simple 4-cycle but $(1,3,4,5)$ is not. I want to write $(1,2,3,5)$ as a product of simple 4-cycles. So this is what I did:
$$
(1,2,3,5)=(1,2)(1,3)(1,5)
$$
but
$$\begin{align}
(1,3)&=(2,3)(1,2)(2,... | (2,3,4,5) is (1,2,3,5) conjugated by (1,2,3,4) so (1,2,3,5) = (1,2,3,4)(2,3,4,5)(1,2,3,4)−1.
While this may appear to be coincidence, I think you'll find it works quite well in general as suggested by Carl Brannen.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$ To prove the convergence of the p-series
$$\sum_{n=1}^{\infty} \frac1{n^p}$$
for $p > 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test.
I am wondering if there is a self-contained proof that th... | This is the most direct and elementary way I know how to prove the result, although it only works for powers in the range $[0,1] \cup [2,\infty)$ which is exactly the uninteresting set, and Joriki's answer is much better regardless. I had already written this, and perhaps somebody finds it useful.
First we consider $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/29450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "184",
"answer_count": 9,
"answer_id": 6
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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates.
What methods can be used to evaluate the limit $$\lim_{x\righta... | Alternatively, rewrite this limit as
$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right).$$
Consider the Taylor expansion around $0$ of $\sqrt[n]{1+z}$. We have
$$\sqrt[n]{1+z}=1+\frac{1}{n}z+O\left(z^{2}\right).$$ Setting $z=\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/30040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 6,
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How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$? How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$?
The book solution
$$\left(\dfrac{27}{101}\right) = \left(\dfrac{3}{101}\right)^3 = \left(\dfrac{101}{3}\right)^3 = (-1)^3 = -1$$
My solution
$$\left(\dfrac{27}{101}\right) = \left(\dfrac{101}{27... | Your mistake is calculating $\left(\dfrac{2^2}{27}\right)\:.\:$ But since you wish to use only $2$'s it's much simpler, viz.
$$ \left(\dfrac{27}{101}\right)\ =\ \left(\dfrac{128}{101}\right)\ =\ \left(\dfrac{2}{101}\right)^{7}\ =\ -1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/31200",
"timestamp": "2023-03-29T00:00:00",
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Find equation of quadratic when given tangents? I know the equations of 4 lines which are tangents to a quadratic:
$y=2x-10$
$y=x-4$
$y=-x-4$
$y=-2x-10$
If I know that all of these equations are tangents, how do I find the equation of the quadratic?
Normally I would be told where the tangents touch the curve, but that ... | Since the two pairs of tangents are symmetric with respect to the $y$-axe,
the quadratic function $f(x)=ax^{2}+bx+c$ must be even ($f(x)=f(-x)$),
which implies that $b=0$. The equations of the tangents to the graph of $f(x)=ax^{2}+c$ at points $%
\left( x_{1},f(x_{1})\right) $ and $\left( x_{2},f(x_{2})\right) $ are
$$... | {
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"url": "https://math.stackexchange.com/questions/31321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$ Problem
Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$
My attempt was,
Since $p$ divides $n^4 + 1 \implies n^4 + 1 \equiv 0 \pmod{p} \Leftrightarrow n^4 \... | As you have noticed, $p \equiv 1 \bmod 4$. Then $1 \equiv n^{p-1} = (n^4)^{(p-1)/4} \equiv (-1)^{(p-1)/4} \bmod p$. This means that $(p-1)/4$ is even, i.e., $p\equiv 1 \bmod 8$.
By induction, this argument generalizes to: if an odd prime $p$ divides a number of the form $n^k+1$, where $k$ is a power of $2$, then $p \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/33392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
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How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series
$$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$
My attempt at a solution - I saw that I co... | You use this formula $$(1+x)^{-p/q} = 1 - \frac{\displaystyle\frac{p}{q}}{1!}\cdot x + \frac{ \displaystyle\frac{p}{q} \cdot \Bigl(\frac{p}{q}+1\Bigr)}{2!} x^{2} - \cdots$$
So your series can be rewritten as $$ \frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/34671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
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Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is
Write each of the following as a
decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%
For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{... | It happens to be well known that a number over 9 (other than 0 or 9) has a repeating decimal. That is: $\dfrac{1}{9} = 0.11 \bar{1}$, $\dfrac{4}{9} = 0.44 \bar{4} $, and so on. Why is this true?
It comes from the fact that if we have the equation $10x = 1.\bar{1}$, then we have $x = 0.\bar{1}$ by division. Subtracting,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/35631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Residue of $f(z) = \frac{z-2}{z^2}\sin(\frac{1}{1-z})$ at $z = 1$ Given the complex function $f(z) = \frac{z-2}{z^2}\sin(\frac{1}{1-z})$, how can we calculate the residue at the essential singularity at $z = 1$?
| Let us define $\xi=z-1$ (such that we can calculate the residue at $\xi=0$). The function is given by
$$\frac{1-\xi}{(1+\xi)^2} \sin(\xi^{-1}).$$ For the residue, we need to obtain the coefficient in front of $\xi^{-1}$ in the Laurent expansion around $\xi=0$. We have the Laurent expansions
$$\sin(\xi^{-1}) = \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/37612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that a repeating decimal equals a fraction I'm having some trouble with this question:
Prove that 0.1636363636...=9/55, using infinite series. I'd appreciate any help you can give me. Thanks!
| You can do this: \begin{align*} 0.1636363\cdots &= \frac{1}{10} + \biggl[ 63 \times 10^{-3} + 63\times 10^{-5} + 63 \times 10^{-7} + \cdots\biggr] \\ &= \frac{1}{10} + 63 \cdot \Bigl[ 10^{-3} + 10^{-5} + 10^{-7} + \cdots \Bigr] \\ &= \frac{1}{10} + 63 \cdot \frac{10^{-3}}{1-10^{-2}} \\ &= \frac{1}{10} + \frac{7}{110}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/38838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using Horner's Method I'm trying to evaluate a polynomial recursively using Horner's method.
It's rather simple when I have every value of $x$ (like: $x+x^2+x^3...$), but what if I'm missing some of those? Example: $-6+20x-10x^2+2x^4-7x^5+6x^7$.
I would also appreciate it if someone could explain the method in more det... | Horner's method or form is also sometimes called nested form. You can think of it as starting with the whole polynomial $$6x^7-7x^5+2x^4-10x^2+20x-6,$$ setting aside the constant term (if it's zero, you can set aside a zero here) and factoring out an $x$ from the remaining terms $$(6x^6-7x^4+2x^3-10x+20)x-6,$$ and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Is it possible to solve a separable equation a different way and still arrive at the same answer? I have the following equation
$$(xy^2 + x)dx + (yx^2 + y)dy=0$$ and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.
Let $M = xy^2 + x $ and $N = yx^2 + y$
$$My =... | Taking the other route you get $y'y(x^2+1)=-xy^2-x$. Denote $z=y^2$, hence $z'=2yy'$. So you get $z'(x^2+1)=-2xz-x$. Hence you have $z'+\frac{2x}{x^2+1}z=-\frac{x}{x^2+1}$. This gives you $z=\frac{k(x)}{x^2+1}$. Substituting, you have $k'(x)=-x$, which implies $k(x)=-\frac{x^2}{2}+C$. Hence $z=-\frac{x^2}{2(x^2+1)}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/50887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to add compound fractions? How to add two compound fractions with fractions in numerator like this one:
$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$
or fractions with fractions in denominator like this one:
$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$
| Here is a start for the first one:
$$\frac{\frac{1}{x}}{2} + \frac{\frac{2}{3x}}{x} = \frac{x\frac{1}{x}}{2x} + \frac{2\frac{2}{3x}}{2x}
= \frac{1}{2x} + \frac{\frac{4}{3x}}{2x} = \frac{1}{2x} + \frac{4}{3x}\frac{1}{2x} = \frac{1}{2x} + \frac{4}{6x^2} = \frac{1}{2x} + \frac{2}{3x^2}$$
Now try to derive $\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/51410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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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.