Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding difference of angles in triangle . If $\sin A+\sin B+\sin C=0$, $\cos A+\cos B+\cos C=0$, then prove that $$A-B=B-C=C-A=\dfrac{2\pi}3$$
| Squaring and adding we get $$\cos(A-B)+\cos(B-C)+\cos(C-A)=-\frac32$$
Again, we have $$\sin A+\sin B=-\sin C\text{ and }\cos A+\cos B=-\cos C$$
Squaring and adding we get $$2+2\cos(A-B)=1\implies \cos(A-B)=-\frac12=\cos\frac{2\pi}3$$
$$\implies A-B=2n\pi\pm\frac{2\pi}3$$ where $n$ is any integer
Case $A:$ Taking the '... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/623765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solution verification: solving $\sqrt{x-4}-\sqrt{x-5}+1=0$ I solved the following equation, and I just want to be sure I did it right.
This is the procedure:
$$
\sqrt{x-4}-\sqrt{x-5}+1=0\\
\sqrt{x-4}=\sqrt{x-5}-1\\
\text{squaring both sides gives me:}\\
(\sqrt{x-4})^2=(\sqrt{x-5}-1)²\\
x-4=(\sqrt{x-5})²-\sqrt{x-5}+1\\
... | The square root function is increasing on $[0,\infty)$, so
$$\sqrt{x-4} - \sqrt{x-5} \ge 0 $$
for all $x\ge 5.$ Hence the expression can never equal -1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Compute $\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\cdots+\left (\frac{1}{n-1}+\frac{1}{n} \right )^2+\left (\frac{1}{n} \right )^2$ Compute the value of the following expression
$$\left (1+\frac{1}{2}+\cdots+\frac{1}{n} \right )^2+\left ( \frac{1}{2}+\cdots + \frac{1}{n}\right )^2+\cdots+\left (\frac{1}{n-1}... | $$S_n - S_{n-1} = n\times \left(\frac{1}{n}\right)^2 + 2\frac{1}{n}\left(1 \times \frac11 +2 \times \frac12+\cdots + (n-1)\times \frac{1}{n-1} \right) = 2-\frac1n$$ and $S_0$ can be taken to be $0$, so just add up $(S_n - S_{n-1})+(S_{n-1} - S_{n-2})+\cdots + (S_{1} - S_{0})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
| Here's an alternative argument using the fact that a set of $n$ elements has $2^n$ subsets. Consider subsets $S$ of the set $\{0,1,2,\dots,9,10\}$. Since this set has 11 elements, it has $2^{11}$ subsets. Now ask, for each $n$ in the range $0\leq n\leq 10$, how many of these subsets $S$ have $n$ as their largest elem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
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Is $x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n})$ , $n \ge 1$ convergent? $$x_n =\sqrt{n} (\sqrt{n+1}− \sqrt{n}) ,\quad n \ge 1$$
My work is:
First i analyzed the convergence of this sequence in 2 parts:
a) $\lim \sqrt{n+1}− \sqrt{n} = \lim \frac{(\sqrt{n+1}− \sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} = \lim \frac{n+... | Is it correct this?
$\begin{align}
x_n
&=\sqrt{n} (\sqrt{n+1}− \sqrt{n})\\
&=\sqrt{n} (\sqrt{n+1}− \sqrt{n})\dfrac{\sqrt{n+1}+ \sqrt{n}}{\sqrt{n+1}+ \sqrt{n}}\\
&=\sqrt{n} \dfrac{(\sqrt{n+1}+ \sqrt{n})(\sqrt{n+1}− \sqrt{n})}{\sqrt{n+1}+ \sqrt{n}}\\
&=\sqrt{n} \dfrac{(n+1)-n}{\sqrt{n+1}+ \sqrt{n}}\\
&= \dfrac{\sqrt{n}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Calculate the eigenvectors We calculate the eigenvectors for the matrix
$$
\begin{equation*}
\mathbf{A} = \left(
\begin{array}{ccc}
2 & 0 & 0 \\
1 & 1 & 1 \\
1 & -1 & 3 \\
\end{array}
\right)
\end{equation*}
$$
First, I calculted the eigenvalue polynomial $\det(\mathbf{A}-\lambda \mathbf{... | Your results are correct, but that does not mean that $A$ is a degenerate matrix, just that $A$ is not diagonalizable.
Just one little thing:
$x_1=\left(
\begin{array}{ccc}
1 \\
1 \\
0 \\
\end{array} \right)$
and $x_2=\left(
\begin{array}{ccc}
0 \\
1 \\
1 \\
\end{array} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim\limits_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1} \,\right)=0$
Prove that:
$$\lim_{n\to\infty}\left(\sqrt{n^4+n^2+20n+7}\,-\,\sqrt{n^4+n^2+1}
\,\right)=0$$
Epsilon>0. According to the Archimedean Property of reals, we have n1 element N with epsilon*n1>11. (Why 11? Seems so random...)... | Extarct n^4 from each radical. So, what is left inside the remaining radicals is (1 + 1/ n^2 + ...) and, since "n" is large, each remaining radical radical is almost 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove that: $\sin{\frac{\pi}{n}} \sin{\frac{2\pi}{n}} ...\sin{\frac{(n-1)\pi}{n}} =\frac{n}{2^{n-1}}$ Using that:
$$
x^{n - 1} + x^{n - 2} + \cdots + x + 1
=
\left(x - w\right)\left(x - w^{2}\right)\ldots\left(x - w^{n - 1}\right)
$$
Prove that:
$$
\sin\left(\pi \over n\right)\sin\left(2\pi \over n\right)\ldots
\sin\le... | Notice that
$$
w=e^{i\frac{2\pi}{n}}=z^2,
$$
with
$$
z=e^{i\frac{\pi}{n}}
$$
and
\begin{eqnarray}
n&=&(1-w)\ldots(1-w^{n-1})=(|z|^2-z^2)\ldots(|z|^{2n-2}-z^{2n-2})\\
&=&(-1)^{n-1}z^{1+2+\ldots+(n-1)}(z-\bar{z})\ldots(z^{n-1}-\bar{z}^{n-1})\\
&=&(-1)^{n-1}z^{\frac{n(n-1)}{2}}2i\left(\sin\frac{\pi}{n}\right)\ldots2i\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
To find $R\circ R^{-1}$ in Discrete mathematics Today I came across a question in DMS which says:
If $R$ is the relation “Less Than” from $A = \{1, 2, 3, 4\}$ to $B = \{1,3,5\}$ then find $R\circ R^{-1}$.
Now what is $R\circ R^{-1}$?
I know how to find $R\circ R$,
like in this question firstly we will find $R$ where... | I wanna introduce the domain and the range of $R$, which are defined in Elements of Set Theory by Herbert B. Enderton:
\begin{align}
x \in \mathrm{dom} R &\Leftrightarrow \exists y :\: \left( x,\, y \right) \in R,\\
x \in \mathrm{ran} R &\Leftrightarrow \exists t :\: \left( t,\, y \right) \in R.
\end{align}
It'll help... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/634983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit of $x^x$ as $x$ tends to $0$ I am trying to solve the following limit:
$$\lim \limits_{x\to0} x^x$$
The only thing that comes to mind is to write $x^x$ as $e^{x\ln{x}}$ and getting the right sided limit would be easy but I don't see how I could get the left sided one seeing that the $\ln$ is not defined for negat... | L'Hospital's rule is quickest. I show that other approaches are possible:
For $\ x\in \left[\ \frac{1}{7},\ \frac{1}{6}\ \right),$
$$ \left( \frac{1}{7} \right)^{\frac{1}{6}}<x^x<1.$$
Now using Newton's Binomial expansion,
\begin{align} \left(1-\alpha\right)^\frac{1}{6} = 1 + \frac{1}{6}(-\alpha) + \frac{\left(\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/637401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
mixed numbers subtraction vertically In the following subtraction we are subtracting $2$ mixed numbers vertically. I know how it works except the last step.
$$ 7 \frac{1}{3} - 4 \frac{1}{2} = 3 + \frac{-1}{6} = 2 + \frac{5}{6} = 2 \frac{5}{6}$$
I am confused about this step: $ 3 + \frac{-1}{6} = 2 + \frac{5}{6}$
How do... | For "normal" subtraction, we all learned in elementary school to "borrow" a 1 from the next highest place if we needed to.
In this case, we're sort of doing the same thing. When we have $3 + -1/6$, we "borrow" a 1 to make $2 + 6/6 + -1/6$, which then becomes $2 + 5/6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/638931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $4^{2n+1}+3^{n+2} : \forall n\in\mathbb{N}$ is a multiple of $13$ How to prove that $\forall n\in\mathbb{N},\exists k\in\mathbb{Z}:4^{2n+1}+3^{n+2}=13\cdot k$
I've tried to do it by induction. For $n=0$ it's trivial.
Now for the general case, I decided to throw the idea of working with $13\cdot k$ and try to... | $$\begin{gather}4^{2n+3}+3^{n+3}=4^{2n+1}\cdot{4^2}+3^{n+2}\cdot{3}=\\
=4^{2n+1}\cdot{4^2}+3^{n+2}\cdot{4^2}-3^{n+2}\cdot{4^2}+3^{n+2}\cdot{3}=\\
=16\cdot(4^{2n+1}+3^{n+2})-3^{n+2}\cdot(16-3)=\\
=16\cdot(4^{2n+1}+3^{n+2})-13\cdot3^{n+2}\end{gather}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/640859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Simple Question on Triangles... What times the sum of the squares of the sides of a triangle is equal to the sum of the squares of the medians of the triangle.
| Hint:
$\Large 2\cdot m_a^2+\frac{a^2}{2}=b^2+c^2$
$\Large 2\cdot m_b^2+\frac{b^2}{2}=c^2+a^2$
$\Large 2\cdot m_c^2+\frac{c^2}{2}=a^2+b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of function with $\frac00$ I am unable to solve following limit:
$$\lim_{x \rightarrow 0_+}\frac{\sin \sqrt{x}}{x^2}\left(\sqrt{x+2x^2}-\sqrt{2\sqrt{1+x}-2}\right)$$
I keep getting $ \frac{0}{0}$. I admit I haven't tried to use l'Hospital rule multiple times as the square root is not so nice to derive more than... | Note that $\sqrt{2\sqrt{1+x}-2}\sim\sqrt{x+2x^2}\sim\sqrt x$, while
\begin{align}
(x+2x^2)-(2\sqrt{1+x}-2)
&=2x^2+x+2-2\sqrt{1+x}\\
&=\frac{(2x^2+x+2)^2-4(1+x)}{2x^2+x+2+2\sqrt{1+x}}\\
&\sim\frac 94x^2
\end{align}
thus
\begin{align}
\frac{\sin \sqrt{x}}{x^2}(\sqrt{x+2x^2}-\sqrt{2\sqrt{1+x}-2})
&=\frac{\sin \sqrt{x}}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/645074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to integrate this integral? I had a problem with this specific integral:
$$
\int{\sqrt{\frac{x^3 -3}{x^{11}}} \ dx}
$$
Thank you ahead of time.
| $$ \int \sqrt{\frac{x^3-3}{x^{11}}}\,dx = \int \frac{1}{x^4} \sqrt{\frac{x^3-3}{x^3}} \,dx
= \int \frac{1}{x^4} \sqrt{1-\frac{3}{x^3}} \,dx $$
Let
$$ u = \frac{1}{x^3}, \; du = -\frac{3}{x^4} dx $$
The above integral becomes
$$ -\frac{1}{3} \int \sqrt{1-3u}\:du = -\frac{1}{3} \cdot -\frac{1}{3} \cdot \frac{2}{3}
(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Complex Numbers and exponential form and roots The roots of $z^7 = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}$ are $\text{cis } \theta_1, \text{cis } \theta_2, \dots, \text{cis } \theta_7,$ where $ 0^\circ \le \theta_k < 360^\circ $for all $ 1 \le k \le 7$. Find $\theta_1 + \theta_2 + \dots + \theta_7$. Give your answer ... | $\def\cis{\operatorname{cis}}$Let's consider a more general problem. We have $a=r\cis\alpha$ and we write its $n$th roots as
\begin{gather}
\sqrt[n]{r}\cis\frac{\beta}{n}\\
\sqrt[n]{r}\cis\left(\frac{\beta}{n}+\frac{2\pi}{n}\right)\\
\sqrt[n]{r}\cis\left(\frac{\beta}{n}+2\frac{2\pi}{n}\right)\\
\dots\\
\sqrt[n]{r}\cis\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the error in the simultaneous equation. The question is:
$$\begin{align}
\tag{1} 2x - 3y &= 3\\
\tag{2} 4x^2 - 9y^2 &= 3
\end{align}$$
From equation (1):
$$\tag{3} 2x = 3 - 3y.$$
Substitute equation (3) in (2):
$$\begin{align}
4x^2 - 9y^2 &= 3\\
(2x)^2 - 9y^2 &= 3\\
(3 - 3y)^2 - 9y^2 &= 3\\
[(3)^2 - 2\cdot 3\c... | From $(3)$ you get $x$, so your solution is
$$ \begin{eqnarray}
x&=&1 \\
y&=&\frac{1}{3}
\end{eqnarray}
$$
Substitute these value in your equation then you see that all the equalities hold except the equality $(1)$. Here we get $1$ for the left-hand side and $3$ for the right-hand side. So apparently $(3)$ is erroneo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate $\iint_D\sin(xy)dA$ where $D$ is bounded by $y=\frac 1x, y=\frac2x, y=x, y=2x$ in the first quadrant. Evaluate $\iint_D\sin(xy)dA$ where $D$ is bounded by $y=\frac 1x, y=\frac2x, y=x, y=2x$ in the first quadrant.
By subbing numbers into the equation, I see that $1\leq x\leq 2, 1\leq y\leq 2.$
Without solving t... | Okay, here's what I got. Evaluating this double integral seems to be much harder on usual cartesian coordinates, since the region is pretty ugly.
Courtesy of Mathematica. If we perform the coordinate change
$$
\begin{cases}
xy & = u \\
\frac{y}{x} & = v,
\end{cases}
$$
we get the much neater region that is a square.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limits at infinity I'm working with limits at infinity and stumbled upon this exercise where I want to evaluate the indicated limit:
$$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x}$$
I tried to solve it by doing the following:
$$\lim_{x \to \infty} \frac{1}{\sqrt{x^2-2x}-x} = \lim_{x \to \infty} \frac{1}{\sqrt{x^2} \s... | The points that we can add to Adi's post are:
$$\frac{1}{\sqrt{x^2-2x}-x}\times\frac{\sqrt{x^2-2x}+x}{\sqrt{x^2-2x}+x}=\frac{\sqrt{x^2-2x}+x}{(x^2-2x)-x^2}=\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}$$ Now if $x\to+\infty$ so $|x|=x$ and so $$\color{blue}{\frac{|x|\sqrt{1-2/x}+x}{-2x}}=\frac{x\sqrt{1-2/x}+x}{-2x}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/651969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How find this value $A+B+C+AB+BC+AC+ABC$ let
$$A=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{\pi}{6}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
$$B=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\left(\dfrac{5\pi}{6}+\dfrac{i}{3}\text{arcsinh}{(\dfrac{3\sqrt{3}}{2})}\right)}$$
$$C=\dfrac{2\sqrt{3}}{3}i\cdot\cos{\lef... | Observe that $$\cos3\left(\frac\pi6+\frac i3\text{arcsinh} \frac{3\sqrt3}2
\right)=\cos\left(\frac\pi2+i\text{arcsinh} \frac{3\sqrt3}2\right)$$
$$=-\sin \left(i\cdot\text{arcsinh} \frac{3\sqrt3}2\right)$$
Now as, $\displaystyle \sin(ix)=i\sinh(x)$
$$-\sin \left(i\cdot\text{arcsinh} \frac{3\sqrt3}2\right)=-i\sinh \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving Integral that contain exponential and Power I have an integral of this form:
$$\int_0^\infty e^{-\frac{x}{a}-\frac{z^2}{bx}-\frac{z}{bx}}\left(\frac{c}{c+x+z}\right)^K~dx$$
where $K$ is a positive integer.
$a$ , $b$ and $c$ are reals and $>0$
finally,$z$ is also reals.
My question is that; is it possible to fin... | Special case of $z=-c$ :
$\int_0^\infty e^{-\frac{x}{a}-\frac{c^2}{bx}+\frac{c}{bx}}\left(\dfrac{c}{x}\right)^K~dx$
$=c^K\int_0^\infty\dfrac{e^{-\frac{x}{a}-\frac{c^2-c}{bx}}}{x^K}dx$
$=c^K\int_0^\infty\dfrac{e^{-\frac{\sqrt{\frac{a(c^2-c)}{b}}x}{a}-\frac{c^2-c}{b\sqrt{\frac{a(c^2-c)}{b}}x}}}{\left(\sqrt{\dfrac{a(c^2-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.
\begin{align} \int_{0}^{\infty} \fr... | I like to calculate this integral as follows:
Let us note that
$$\frac{1}{x^5}=\frac{1}{4!}\int_0^\infty t^4e^{-xt}dt$$ So
$$I=\frac{1}{4!}\int_{0}^{\infty}(x^{3}-\sin^{3}x)\int_0^\infty t^4e^{-xt}\;dt\;dx$$
$$=\frac{1}{4!}\int_{0}^{\infty}t^4\int_{0}^{\infty}(x^{3}-\sin^{3}x)e^{-xt}\;dx\;dt$$
$$=\frac{1}{4!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/656757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 2
} |
Linear Algebra Complex Numbers The solutions to the equation $z^2-2z+2=0$ are $(a+i)$ and $(b-i)$ where $a$ and $b$ are integers. What is $a+b$?
I simplified and got $(z+1)(z+1) = -1$ and now I'm not sure where to go from there. I did this but I'm not sure.
$(a+i)^2 = a^2 - 1$
$(b-i)^2 = b^2 + 1$
$a+b=(a+i)+(b-i)=(a^2... | Since this is a polynomial with real coefficients, $z$ is a root if and only if $\overline z$, the complex conjugate of $z$, is. Since $\overline{a + i} = a - i$, we see that $a = b$ necessarily.
Then using that $a + i$ and $a - i$ are roots, we can factor the polynomial as
\begin{align*}
z^2 + 2z + 2 &= \Big(z - (a + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/657927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If$ a+b-1=1+\frac{ln(2^a-1)}{ln4}+\frac{ln(2^b-1)}{ln4}$ then $a=b$? If $$a+b-1=1+\frac{ln(2^a-1)}{ln4}+\frac{ln(2^b-1)}{ln4}$$ where $a,b>0$ are real numbers and ln is $log_e$, then is a=b?
| Let $f(x) = x - \log_4 \left(2^{x+2}-4\right)$. Then the given relation is $f(a)+f(b) = 0$.
Now $f(x)$ has a minimum at $x=1$, which is $f(1) = 0$. So $f(x) > 0$ for all other values of $x>0$. Hence for the relation to be satisfied, $a=b=1$.
Updated: to show the minimum without calculus, note that
$f(x) \ge 0 \iff x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/658459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$(3^x-2^x)^2+3^x-2^x\leq 5^x$ Solve the following inequality
$$(3^x-2^x)^2+3^x-2^x\leq 5^x$$
I'm having difficulty in solving this inequality because there are multiple exponential with different basis. I thought, however, the replacement:
$$y=3^x-2^x$$
and so:
$$y^2+y\leq 5^x$$
treating $5^x$ as the known term of the ... | Your approach will flesh out as follows:
$$\boxed{y^2+y\le5^x}\iff y^2+y-5^x\le0\iff y\in\Big[y_{_1},y_{_2}\Big],\qquad y_{_{1,2}}=\frac{-1\pm\sqrt{5^x+1}}2$$
$$-\sqrt{5^x+1}\le2y+1\le\sqrt{5^x+1}\iff|2y+1|\le|\sqrt{5^x+1}|\iff(2y+1)^2\le5^x+1$$
$$4y^2+4y\le5^x\iff\boxed{y^2+y\le\dfrac{5^x}{4\ \ }}\iff y\in\Big[y'_{_1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/660636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How does $\frac1{n^4} =\frac1{(2n-1)^4} + \frac1{(2n)^4}$ I need to find the sum of $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4}.$
In the solution document it written:
$$\displaystyle\sum_{n=1}^\infty \frac1{n^4} = \sum_{n=1}^\infty \frac1{(2n-1)^4} + \sum_{n=1}^\infty \frac1{(2n)^4}.$$
My question is, how did he get ... | \begin{align*}
\sum_{n=1}^\infty \frac{1}{n^4}
&= \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \frac{1}{5^4} + \frac{1}{6^4} + \cdots \\
&= \left( \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \right)
+ \left( \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots \right) \\
&= \sum_{n=1}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/664188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $\left(\frac12(x+y)\right)^2 \le \frac12(x^2 + y^2)$ Prove that $$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
I've gotten that
$$\left(\frac12(x+y)\right)^2 \ge 0 $$
but stumped on where to go from here...
| This substitution is a nice way to do this:
$$x=r\sin\theta , y=r\cos\theta$$
So, our identity:
$$\left(\frac12(x+y)\right)^2 \leq \frac12(x^2 + y^2)$$
$$\frac{1}{4} (x^2+y^2+2xy) \leq \frac12(x^2 + y^2)$$
Applying our substitutions:
$$\frac{1}{4} (r^2+2\cdot r\sin\theta\cdot r\cos\theta) \leq \frac12(r^2)$$
$$\frac{r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/665206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 1
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Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
| Hint Use Cauchy-Schwarz.
Second solution
$$(a+b)^2=a^2+b^2+2ab \leq 2+2ab$$
You got that far, you are almost there:
By AM-GM
$$\sqrt{a^2b^2} \leq \frac{a^2+b^2}{2}$$
which implies $$2ab \leq a^2+b^2 \leq 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Why does a conjugate of a root remain a root of a polynomial? About quadratic equations, I have the follow question,
What is the polynomial, with rational coefficients, knowing that one root is $ x_1 = 1 + \sqrt 3$ ?
So, to solve the problem, I must know that the other root is $x_2 = 1-\sqrt3$. I see the resolution, ... | You can see this answer as a supplement to Bill Dubuque's, but I am avoiding any technical language.
The key to understanding why if $1+\sqrt{5}$ is a root of a quadratic equation with rational coefficients, then $1-\sqrt{5}$ is also a root, is understanding the symmetry between $\sqrt{5}$ and $-\sqrt{5}$. Here is what... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
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Max. and Min. value of $f(x,y) = \frac{x-y}{x^4+y^4+6}$ [1] Calculation of Max. and Min. value of $\displaystyle f(x) = \sqrt{x^3-6x^2+21x+18}$, where $\displaystyle -\frac{1}{2}\leq x\leq 1$
[2] Calculation of Max. and Min. value of $\displaystyle f(x,y) = \frac{x-y}{x^4+y^4+6}\;,$ where $x,y\in \mathbb{R}$
$\bf{My\; ... | If you want Max, $x-y>0$, you want Min,$x-y<0$
for Max:
$x-y \le |x-y| \le|x|+|y|$
your process can go on: $x^2+1\ge 2|x|,y^2+1\ge2|y| \implies x^4+y^4+6\ge 4(|x|+|y|)\implies \dfrac{x-y}{x^4+y^4+6} \le \dfrac{1}{4}$ when $x=-y=1$
you can get Min in same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Optimization with a few Variables (AMC 12 question) In the 2013 AMC 12B, question 17 says:
Let $a$,$b$, and $c$ be real numbers such that
$a+b+c=2$, and $a^2+b^2+c^2=12$
What is the difference between the maximum and minimum possible values of $c$?
I was wondering if there is a quick and easy solution using multivariab... | Here is an geometric interpretation: the locus of points $(a,b,c)$ such that $a+b+c = 2$ and $a^2+b^2+c^2 = 12$ is the intersection of a plane and a sphere in $\mathbb R^3$; i.e., it is a circle. The plane is perpendicular to the line $a = b = c$, and a little visualization leads us to conclude that the extrema of $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/669257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Trigonometry identities sum and substraction of cosine Someone can help me showing this identity?
$\frac{cos(a-b)}{cos(a+b)}=\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$
| Hint:
\begin{align*}
\frac{\cos(a-b)}{\cos(a+b)}&=\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b- \sin a\sin b}\\
&=\frac{\displaystyle\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b}}{\displaystyle\frac{\cos a\cos b- \sin a\sin b}{\cos a\cos b}}\\
&=\; ?
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$
$=3(a^2+4a+7) $
where do I go from here.. the solution I ha... | $$a^2+(a+2)^2+(a+4)^2+1=3a^2+12a+21\equiv3(a^2-1)\pmod{12}$$
Now, if $a$ is odd $a\pm1$ are even $\implies a^2-1$ will be divisible by $4$
In fact, $\displaystyle(2b+1)^2=4b^2+4b+1=8\frac{b(b+1)}2+1\equiv1\pmod8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Ways of writing $n=2a+b$ with $a$ and $b$ are non-negative integers For a non-negative number $n$, let $r_n$ be the number of ways of writing
$n = a + 2b$, where $a$ and $b$ are non-negative integers.
For example,
$5 = 1 \cdot 5 + 2 \cdot 0 = 3 \cdot 1 + 1 \cdot 2 = 1 \cdot 1 + 2 \cdot 2$,
so $r_5 = 3$.
My quest... | Hint: Let $n=2k$. Then we can use $0$ or $1$ or $2$ or $\dots$ or $k$ two-dollar coins, a total of $k$ choices.
Let $n=2k+1$. Then we can use $0$ or $1$ or $2$ or $\dots$ or $k$ two-dollar coins, a total of $k$ choices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Real number multiplicative inverses expressed in another form I've been asked to express the multiplicative inverse of $3 + \sqrt{5}$ in the form $c + d\sqrt{5}$, where $c,d$ are rational numbers.
I understand that for some rational numbers $c,d$ we must have:
$$1 = (3 + \sqrt{5})(c + d\sqrt{5}).$$
I was able to answer... | \begin{align}
Answer &= \frac{1}{3+\sqrt{5}}\\
\mbox{Multiply and Divide by Conjugate}&\\
&= \frac{1}{3+\sqrt{5}}\cdot \frac{3-\sqrt{5}}{3-\sqrt{5}}\\
&= \frac{3-\sqrt{5}}{(3)^2-(\sqrt{5})^2}\\
\end{align}
Hidden Answer:
$$\frac{3}{4}-\frac{\sqrt{5}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two-variable limit question $\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}$
How to solve this two-variable limit? Thanks :D
| Let
\begin{equation}
f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}
\end{equation}
If $y=0$, $\lim_{(x,y)\rightarrow(0,0)}f(x,y)=0$. If $y\neq0$, say approach the $(0,0)$ do not along with x-axis, we have:
\begin{equation}
f(x,y)=\frac{x^2y^2}{(x^2+y^4)\sqrt{x^2+y^2}}\leq\frac{x^2y^2}{2|x|y^2\sqrt{x^2+y^2}}=\frac{|x|}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/679428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Factoring quartic equation $$x^4+6x^2+25=0$$
How could I factor it into $(x^2-2x+5)(x^2+2x+5)=0$?
I got the result looking into horrible formulas on wikipedia, but I suppose there's a much easier way. Could you help me?
| Notice that
$$
x^4+6x^2+25=(x^4+6x^2+9)+16=(x+3)^2+4^2=(x^2+3+4i)(x^2+3-4i) \quad \forall x\in \mathbb{C}.
$$
Let $u \in \mathbb{C}$ such that $u^2=3+4i$. It is easy to show that $u\in\{2+i,-2-i\}$. We have
\begin{eqnarray}
x^4+6x^2+25&=&(x^2+u^2)(x^2+\bar{u}^2)=(x+iu)(x-iu)(x+i\bar{u})(x-i\bar{u})\\
&=&(x+iu)\overline... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express $\;f(x)=\frac{x − 1}{x + 1}\;$ as the sum of an even and an odd function. In homework there is such problem:
Express $\;f(x)=\dfrac{x − 1}{x + 1}\;$ as the sum of an even and an odd function.
(Simplify as much as possible.)
This function is not even and neither odd. Also if we take it as division of 2 funct... | I know it's late, but you could also simply multiply both numerator and denominator by $(x-1)$.
$$ \begin{align}
\frac{x-1}{x+1} &= \frac{x-1}{x+1} \cdot \frac{x-1}{x-1} \\
&= \frac{x^2-2x+1}{x^2-1} \\
&= \frac{x^2+1}{x^2-1} + \frac{2x}{1-x^2} \\
\end{align} $$
The left-hand term is even since it is of the form ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integer solutions of $ x^3+y^3+z^3=(x+y+z)^3 $ Consider the equation
$$ x^3+y^3+z^3=(x+y+z)^3 $$
for triples of integers $(x, y, z) $.
I noticed that this has infinitely many solutions: $ x, y $ arbitrary and $ z=-y $.
Are there more solutions?
| Too long for a comment. Interestingly, for $k>3$, there are non-trivial solutions to,
$k=4;\quad x_i = -2634, 955, 1770, 5400$: (Jacobi-Madden equation)
$$x_1^4+x_2^4+x_3^4+x_4^4 = (x_1+x_2+x_3+x_4)^4$$
$k=5;\quad x_i = - 3, - 54,24,28,67 $:
$$x_1^5+x_2^5+\dots+x_5^5 = (x_1+x_2+\dots+x_5)^5$$
$k=6;\quad x_i = -4170,... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factoring a difference of 2 cubes I am trying to factorize the expression $(a - 2)^3 - (a + 1)^3$ and obviously I would want to put it in the form of $(a - b)(a^2 + ab + b^2)$
So I start off with the first $(a - b)$ and I get $(a - 2) - (a + 1)$ which I simplify from $(a^2 + a -2a -2)$ to $(a^2 -3a -2)$
Now I'm up to $... | you have $$(a - 2)^3 - (a + 1)^3=[(a - 2) - (a + 1)][(a - 2)^2+(a - 2)(a + 1) + (a + 1)^2]
\\=[-3][a^2-4a +4+a^2 - a -2 + a^2 +2a+ 1]
\\=-3(3a^2-3a+3)
\\=-9(a^2-a+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/689671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. We have to find all triples $(p; q; r)$ of primes such that $pq = r+ 1$ and $2(p ^ 2+q ^ 2) =r ^ 2 + 1$. This question was asked in the 2013 mumbai region RMO but i could not find a solution to it. Can you please help me out w... | Well, $r^2+1=2(p^2+q^2)$ is even, so $r^2$ is odd, and so $r$ is odd. But then $pq=r-1$ is even. Thus, $p$ and/or $q$ must be an even prime.
What happens next depends on whether you are considering negative numbers as potential primes. I will proceed as though you are. Let's assume that $p$ is an even prime, so that $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/689884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve simple trigonometry equation. So we are learning trigonometry in school and I would like to ask for a little help with these. I would really appreciate if somebody can explain me how I can solve such equations :)
*
*$\sin 3x \cdot \cos 3x = \sin 2x$
*$2( 1 + \sin^6 x + \cos^6 x ) - 3(\sin^4 x + \cos^4 ... | Are your formulae things like $\sin^2 x + \cos^2 x = 1$?
If so then you need to spot where you can apply them.
For example $$\sin^2 x - \sin^4 x + \cos^4 x = 1$$
$$\sin^2 x (1- \sin^2 x) + \cos^4 x = 1$$
$$\sin^2 x \cos^2 x + \cos^4 x = 1 $$
$$(\sin^2 x + \cos^2 x) \cos^2 x = 1 $$
$$ \cos^2 x = 1 $$
$$ \cos x = \p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$ How can you evaluate $$\int\limits_0^{\pi/2}\log\cos(x)\,\mathrm{d}x\;?$$
| Here is another solution: As we have $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}=\frac{e^{2ix}+1}{2e^{ix}}$ we get:
$$I=\int_0^{\frac{\pi}{2}} \ln(\cos(x))dx=\int_0^{\frac{\pi}{2}} \ln(e^{2ix}+1)dx-\int_0^{\frac{\pi}{2}} \ln(2)dx-\int_0^{\frac{\pi}{2}} ixdx$$
By using the series expansion of $\ln(x)$ and by calculating the two s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 9,
"answer_id": 1
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Show that the gcd of $(n^2, n^2 + n + 1) = 1$ We know $(a,\;b)$ = $(a-kb,\;b)$, with $(x,\;y)$ meaning the greatest common divisor of x and y.
$(n^2,\;n^2 + n + 1)$ = $(n^2,\;(n+1)^2 - n)$
Let $d\;|\;n^2$ and let $d\;|\;((n+1)^2 -\;n)$
Then $d\;|\;n+1$, since by linear combination theorem $d\;|\;n^2$ and $d\;|\;(n^2+n+... | Use the Euclidean Algorithm for Greatest Common Divisor:
$$n^2 + n + 1 = n^2 + (n+1)$$
$$n^2 = (n-1)(n+1) + 1$$
Hence the greatest common divisor is $1$.
Another way is to assume $d \mid n^2$ and $d \mid n^2 + n + 1$, where $d$ is prime number. Now $d$ divides their difference so we have:
$$d \mid n^2 + n + 1 - n^2 \im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/692116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How prove for any $k$ then have $a^2_{1}+a^2_{2}+a^2_{3}+\cdots+a^2_{k}=m^3$ for any positive integer $k$,there exsit $m\in N$ and $a_{i}\in N,i=1,2,\cdots,k$,such
(1): $a_{i}\neq a_{j},i\forall i\neq j$,
(2):
$$a^2_{1}+a^2_{2}+a^2_{3}+\cdots+a^2_{k}=m^3$$
My idea: if $k=1$,then we let $a_{1}=1$,then $$a^2_{1}=1^3=m^... | Let $1^2+2^2+\cdots+k^2=m$ and multiply through by $m^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Pre Calculus Expression The questions is:
$$\dfrac{3(x+2)^2(x-3)^2 - 2(x+2)^3(x-3)}{(x-3)^4}$$
My answer is: $$\dfrac{3(x+2)^2 + 6x^2-4}{(x-3)^2}$$
Am I right? If not, where have I failed?
| Trusty maxima tells me:
factor(ratsimp((3*(x + 2)^2 * (x - 3)^2 - 2 * (x + 2)^3 * (x - 3))/(x - 3)^4));
is:
$$
\frac{(x - 13) (x + 2)^2}{(x - 3)^3}
$$
A trick that helps catch silly errors while simplifying is to replace some simple values, like $x = 0$ and $x = \pm 1$ in the expresssions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove an equation about binomial coefficients Could we prove:
$ \sum_{k} \binom{2k}{k}\binom{n+k}{m+2k} \frac{(-1)^k}{k+1} = \binom{n-1}{m-1}$ when $m,n \in N$
| The generating function for the Central Binomial Coefficients is
$$
\sum_{k=0}^\infty\binom{2k}{k}x^k=(1-4x)^{-1/2}\tag{1}
$$
Integrating $(1)$ and dividing by $x$ yields
$$
\sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{k+1}=\frac{1-\sqrt{1-4x}}{2x}\tag{2}
$$
Sum the formula against $x^m$:
$$
\begin{align}
&\sum_{m=-\infty}... | {
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"timestamp": "2023-03-29T00:00:00",
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} |
$\gcd(x^2+1,x^2+4x+5)$ Is there anything I can tell about $\gcd(x^2+1,x^2+4x+5)$ for any given integer $x$? I believe I've seen similar questions in the past, though I don't remember any details or what to search for. I did write a computer program to check the first $50$ values which seem to alternate between $1$ an... | If $d$ divides both
$d$ will divide $x^2+4x+4-(x^2+1)=4(x+1)$
$d$ will divide $4(x^2+1)-4(x+1)(x-1)=8$
Now if $x$ is even $x^2+1$ is odd $\implies(x^2+1,x^2+4x+5)=1$
If $x$ odd, $x^2\equiv1\pmod8\iff x^2+1\equiv2\implies(x^2+1,x^2+4x+5)=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/697369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fourier Series for $|\cos(x)|$ I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.
I understand its an even function, so all the $b_n$s are $0$
$$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$
$$a_n = \frac 2 \pi \int _0^\pi |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2... | You must breakup the integral into three intervals:
$\left[-\pi \cdots -\frac{\pi}{2} \right]$,
$\left[-\frac{\pi}{2} \cdots \frac{\pi}{2} \right]$, and
$\left[\frac{\pi}{2} \cdots \pi \right]$
Which represent the regions where the sign of $\cos x$ changes.
\begin{equation*}
\left\vert \cos x\right\vert =
\begin{case... | {
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"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
A formula for bit strings with no $k$ consecutive $1$'s using generating functions The following problem is taken from the book generatingfunctionology (P.28) by Herbert S. Wilf.
Let $f(n,m,k)$ be the number of strings of $n$ $\,0$'s and $1$'s that contain exactly $m\,$ $1$'s, no $k$ of which are consecutive.
(a) F... | First of all, some unsolicited advice: your derivation for (b) need not be so messy. It greatly simplifies things to observe that
$$
f(n,m,k) = f(n-1,m,k) + f(n-1,m-1,k) - f(n-k-1,m-k,k) \tag{1}
$$
holds not just when $0 \le k \le m \le n$, $k < n$. In fact, it holds for any $n \ge 0, m \ge 0, k \ge 1$, except in the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/701923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find all $x$ and $y$ such that $10xy^2=-x^3+4x$ and $13x^2y=-2y^3+4y$
Find all $x$ and $y$ such that $10xy^2=-x^3+4x$ and $13x^2y=-2y^3+4y$.
Well $(0,0)$ is a solution. And we can simplify as:
$10y^2=-x^2+4$ and $13x^2=-2y^2+4$
this is equal to:
$10y^2+x^2-4=0$ and $13x^2+2y^2-4=0$
and therefore
$10y^2+x^2-4=13x^2+2y... | $\displaystyle10xy^2=-x^3+4x\iff x(10y^2+x^2-4)=0$
If $\displaystyle x=0,0=-2y^3+4y\iff2y(y^2-2)=0\implies y=0$ or $y^2=2$
Else $10y^2+x^2-4=0\ \ \ \ (1)$
Similarly start from $\displaystyle13x^2y=-2y^3+4y\iff y(13x^2+2y^2-4)=0$
Finally solve for $x^2,y^2$ from $(1)$ and $13x^2+2y^2-4=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\epsilon$-$\delta$ limit proof that $\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$ I need to prove that $$\lim_{n\to \infty}\frac{n^2-n+2}{3n^2+2n-4}=\frac{1}{3}$$ using the epsilon definition.
I'm having specific trouble understanding how to make it less than epsilon once I've simplified the equation.
| Note that
$$
\begin{align}
\left|\frac{n^2-n+2}{3n^2+2n-4} - \frac{1}{3}\right|&=
\frac{1}{3}\left|\frac{3n^2-3n+6}{3n^2+2n-4} - 1\right|\\&=
\left|\frac{-5n+10}{3(3n^2+2n-4)}\right|
\end{align}
$$
Given an $\epsilon>0$, we need to find an $n$ such that this expression is less than $\epsilon$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Primes $p$ such that $p^2$ divides $x^2 + y^2 + 1$ Call a prime $p$ awesome if there exist positive integers $x$ and $y$ such that $p^2$ divides $x^2+y^2+1$.
Observation: $2$ is not awesome, because $x^2+y^2+1\not\equiv 0$ (mod $4$). But $3$ is awesome, because $9$ divides $27=5^{2}+1^{2}+1$. So my question is:
Are t... | Well time factor here is probably only 3.
Written in the ideal equation: $X^2+Y^2+1=3Z^3$
Solutions can be written using the following equation Pell. $p^2-2(k^2+k+1)s^2=1$
Then the solutions are of the form:
$Y=2p^2+2(k+2)ps-1$
$X=2p^2+2(2k+1)ps-1$
$Z=2p^2+2(k+1)ps-1$
$k$ - what some integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/705611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$ Task is to find $$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}).$$
I tried to make it with double-angle formula:
$\prod_{k=0}^{\infty} \cos(x \cdot 2^{-k}) = \frac{\prod_{k=0}^\infty \sin(x2^{1-k})}{2^\infty \cdot \prod_{k=0}^\infty \sin(x \cdot 2^{-k})} $
I'm sad to admit, tha... | For the partial products we have
$$\begin{align}
\prod_{k = 0}^N \cos \frac{x}{2^k} &= \frac{\sin \frac{x}{2^N}\cos\frac{x}{2^N}}{\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-1}}}{2\sin \frac{x}{2^N}} \prod_{k=0}^{N-1} \cos \frac{x}{2^k}\\
&= \frac{\sin \frac{x}{2^{N-1}}\cos \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/706749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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What is the exact coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$? What is the coefficient of $x^{12}$ in $(2+x^3+x^6)^{10}$?
I figure you need to pick $x^3$ 4 times so $C(10,4)$...but what happens with the other numbers/variables???
Can someone explain to me how this is done properly?
Thanks.
EDIT:
$(x + y)^n = C(n,k) \... | In combinatorial terms, you're looking for the number of partitions of $12$ that use only $0, 3$, and $6$ and have $10$ parts. You then weight these partitions by multinomial coefficients. There are $3$ such partitions:
\begin{align*}
12 &= 6 + 6 + 0 + \cdots + 0\\
&= 6 + 3 + 3 + 0 + \cdots + 0\\
&= 3 + 3 + 3 + 3 + 0... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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An integral identity for $\frac{x^{a-1}}{x^b-1}$ via. partial fractions Can somebody please confirm or correct the following? If $a$ and $b$ are both positive integers such that $a<b$ and $b$ is even then we can write
$$\frac{x^{a-1}}{x^{b}-1}=\frac{1}{b\left(x-1\right)}+\frac{\left(-1\right)^a}{b\left(x+1\right)}+\fra... | Mistake found
The rest of the partial fractions should have been found like so:
$$\frac{e^{\frac{2ak\pi}{b}i}}{x-e^{\frac{2k\pi}{b}i}}+\frac{e^{-\frac{2ak\pi}{b}i}}{x-e^{\frac{-2k\pi}{b}i}}=\frac{2\cos\frac{2ak\pi}{b}\left(x-\cos\frac{2k\pi}{b}\right)-2\sin\frac{2ak\pi}{b}\sin\frac{2k\pi}{b}}{x^2-2x\cos\frac{2k\pi}{b}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/712078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find the inverse of a symmetric tridiagonal Toeplitz matrix? Let $a, b>0$, and the matrix $A_{n \times n}$ and such
$$A=\begin{bmatrix}
a&b&0&\cdots&0&0\\
b&a&b&\cdots&0&0\\
0&b&a&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b&a&b\\
\cdots&\cdots&\cdots&\cdots&b&a
\end{bmatrix}$$... | Notice that
$$A=\begin{bmatrix}
a&b&0&\cdots&0&0\\
b&a&b&\cdots&0&0\\
0&b&a&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\
\cdots&\cdots&\cdots&b&a&b\\
\cdots&\cdots&\cdots&\cdots&b&a
\end{bmatrix} = a \begin{bmatrix}
1&b/a&0&\cdots&0&0\\
b/a&1&b/a&\cdots&0&0\\
0&b/a&1&\cdots&0&0\\
\cdots&\cdots&\cdots&\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/715411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How can I solve this question? Compute the value of the following improper integral. If it is divergent, type "Diverges" or "D".
$$\int_0^2 \frac{dx}{\sqrt{4-x^2}}$$
Do I make $u= 4-x^2$ then $du= -2x \, dx$
Not exactly sure..
| The limit of integration for which our integrand is undefined is $x = 2$. So, what we really need to compute is:
\begin{align}
\lim \limits_{t \to 2} \int_0^t\dfrac{dx}{\sqrt{4-x^2}}
\end{align}
First, let's worry about the integral. Consider the substitution $x=2\sin(\theta)$ (do a bit of trigonometry to see why this ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to divide polynomials? The title says it all : How to divide polynomials?
I don't understand the method below taught in my school.
Would any of you mind explaining it or even better, suggest an alternative way to solve this?
| This is an easier way to do polynomial division, at least I think it is easier.
Basically when we multiply polynomials (or any numbers really) we can sometimes do it in a table form, say if we wanted to multiply $x + 2$ and $x^2 + 3x + 4$ we might write a table like this:
$$
\\ \begin{array} {c|c|c|c}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/717179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solve two equations for $a$ and $b$ \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2(a)\, (c_3)\, (c_4) \\
\end{cases}
I am stuck at this point. Not sure on how to move forward. ( A small change made)
| The system \begin{cases}
c_2=\dfrac{c_1}{a} \left( \left(\dfrac{c_3}{b}\right)^3 - 1 \right) \\[2ex]
b^2 = a^2 + c_3^2 + 2ac_3c_4 \\
\end{cases}
is equivalent to
\begin{cases}
a=c_{1}\dfrac{c_{3}^{3}-x^{3}}{c_{2}x^{3}}\\[2ex]
b=x ,
\end{cases}
where $x$ is a solution of the following septic equation I've obtained in S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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evaluation of $\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)dx$
Compute the indefinite integral
$$
\int \cos (2x)\cdot \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\,dx
$$
My Attempt:
First, convert
$$
\frac{\cos x+\sin x}{\cos x-\sin x} = \frac{1+\tan x}{1-\tan x} = \tan \left(\fr... | \begin{array}{l}
\displaystyle \quad \int \cos 2 x \ln \left(\frac{\cos x+\sin x}{\cos x-\sin x}\right) d x
\\=\displaystyle \int \cos 2 x \ln \left(\tan \left(x+\frac{\pi}{4}\right)\right) d x \\
\displaystyle =\frac{1}{2} \int \ln \left(\tan \left(x+\frac{\pi}{4}\right)\right) d(\sin 2 x) \\
\displaystyle =\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/718719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 1
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Proving inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ Prove that for all $x,y>0$ the following inequality $x^xy^y \geq (\frac{x+y}{2})^{x+y}$ is true.
It smells like Jensen inequality, but all I can get is that $\frac{x+y}{2}ln(x) + \frac{x+y}{2} ln(y) \geq xln(\frac{x+y}{2})+yln(\frac{x+y}{2})$
| This proof won't use Jensen. Multiply both sides by $x^y y^x$ and regroup:
$$
\begin{alignat*}{}
&\Leftrightarrow &\ x^{x+y} y^{x+y} &\geq{} \left(\frac{x+y}{2}\right)^{x+y} x^y y^x\\
&\Leftrightarrow &\ \left(\frac{2xy}{x+y}\right)^{x+y} &\geq{} x^y y^x \\
&\Leftrightarrow &\ \frac{2xy}{x+y} &\geq{} x^{y/(x+y)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Problem with divisibility Proof of the following without induction :
$13| 3^{n+2} + 4^{2n+1}$
for every $n \in \mathbb{Z}^+$
Any help is apprciated.
| First, rewrite as:
$$\begin{align}3^{n+2} + 4^{2n+1} &= 9\cdot3^n + 4\cdot4^{2n}\\
&= 9\cdot3^n + 4\cdot16^{n}\end{align}$$
Then, use modular arithmetic to prove divisibility by $13$:
$$\begin{align}9\cdot3^n + 4\cdot16^{n}&= 9\cdot3^n + 4\cdot(13+3)^n\\
&\equiv 9\cdot3^n + 4\cdot3^n\pmod{13}\\
&= 3^n\cdot(9 + 4)\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can I prove this combinatorial identity without using Wilf-Zeilberger? I am trying to prove the following identity without using W-Z algorithm:
\begin{equation}
\sum\limits_{j=0}^n (-1)^{(n-j)} \frac{(n+j+1)!}{(n-j)!(j)!(j+1)!} = 1
\end{equation}
| As observed in the comments, your sum is equal to
$$
\sum_{j=0}^n (-1)^{n-j} \binom{n}{n-j} \binom{n+j+1}{j+1} \, .
$$
Given a (formal) power series $f(x)$, let $[x^m] f(x)$ denote the coefficient of $x^m$. Note that
$$
(1 - x)^n = \sum_{j \geq 0} (-1)^j \binom{n}{j} x^j
$$
and
$$
\frac{1}{(1 - x)^{n+1}} = \sum_{j \g... | {
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"source": "stackexchange",
"question_score": "3",
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Find constants of function I have this equality :
$$f(x)=\frac{9}{(x-1)(x+2)^2}$$
I am required to find the constants A, B and C so that,
$$f(x) = \frac{A}{(x-1)} + \frac{B}{(x+2)} + \frac{C}{(x+2)^{2}} $$
How do we go about solving such a question?
I am not sure on how to solve such questions. Approach and Hints to ... | $$\dfrac{9}{(x-1)(x+2)^2}=\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$$
Multiplying by the GCD(which means multiplying all the terms by $(x-1)(x+2)^2$)
$$9=A(x+2)^2+B(x-1)(x+2)+C(x-1)$$
Let $x=-2$;
$$9=A(-2+2)^2+B(-2-1)(-2+2)+C(-2-1)$$
$$9=A(0)+B(-3)(0)+C(-3)$$
$$9=-3C$$
$$C=-3$$
Now that we have C,we let x=1;
$$9... | {
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"url": "https://math.stackexchange.com/questions/729350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
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Find the value of $k$ such that $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$. $p(x)= kx^3 + 4x^2 + 3x - 4$ and $q(x)= x^3 - 4x + k$ , leave the same remainder when divided by $(x – 3)$.
(a) -1 (b) 1 (c) 2 (d) -2
I am getting the value of k: $-17/29$ after eq... | Find the value of $k$ if $x^3-4x^2+5x-k$ is completely divisible by $x-4$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Cesàro sum of $\sum\limits_{n = 0}^\infty {\cos n}=\dfrac{1}{2}$. Please check my work Thanks to the formula
http://functions.wolfram.com/ElementaryFunctions/Cos/23/01/0001/
Partial sums
$$s_m=\sum\limits_{k = 0}^m {\cos k = } \frac{{\sin \left( {\frac{1}{2}\left( {m + 1} \right)} \right)\cos \frac{m}{2}}}{{\sin \frac... | I haven't check your work, but it coincides with other results. Particularly, it is what you get when you do Ramanujan's summation of the series:
Let $f(x)=\cos x$. Then the various expressions for Ramanujan's summation are:
$$\sum _{x\ge0}f(x)=\sum _{n=1}^{\infty }{\frac {f^{(n-1)}(0)}{n!}}B_{n}(1)$$
$$\sum _{x\ge0}f(... | {
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"url": "https://math.stackexchange.com/questions/733766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Given the length of two altitudes and one side , find the area of triangle. Segments $BE$ and $CF$ are the altitudes in $\triangle ABC$.
$E$ is on line $AC$ and $F$ is on line $AB$.
$BC = 65$, $BE = 60$ and $CF = 56$.
Find $A(\triangle ABC)/100$.
By the Pythagorean theorem , $CE=25$ , and $BF= 33$.
If the length o... | We have that $\cot C=25/60$ and $\cot B=33/56$.
Now use the area formula:
$$\text{Area}=\frac{a^2}{2(\cot B + \cot C)}$$
Thus:
$$\frac{(ABC)}{100}=\frac{65^2}{200(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{8(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{\frac{10}{3} + \frac{33}{7}}=\frac{13^2}{\frac{70+99}{21}}=21$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove the ring $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ has inverse and is a field How can I prove that $\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}$ is of the form $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ (i.e. that $a+b\sqrt[3]{2}+c\sqrt[3]{4}$ is a field) for all rational $a,b,c$ and $a+b\sqrt[3]{2}+c\sqrt[3]{4} \neq 0$?
| This is very standard. Let $\phi(a,b,c)=a+b\sqrt[3]{2}+c\sqrt[3]{4}$. Then
you have
$$
\phi(a,b,c)+\phi(a',b',c')=\phi(a+a',b+b',c+c')
$$
$$
\phi(a,b,c)\phi(a',b',c')=\phi(aa'+2bc'+2cb',ab'+ba'+2cc',ac'+bb'+ca')
$$
$$
\phi(a,b,c)^{-1}=\phi(\frac{a^2-2bc}{D},\frac{2c^2-ab}{D},\frac{a^2-2bc}{D}), \
D=a^3+2b^3+4c^3-6abc
$... | {
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"url": "https://math.stackexchange.com/questions/734636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $a^n - b^n + c^n - d^n \ge (a-b+c-d)^n$ Following on from an earlier question, and in search of a conceptual insight, I asked myself:
Given real numbers $a \ge b \ge c \ge d \ge 0 \tag{1}$
Prove that
$a^n - b^n + c^n - d^n \ge (a - b + c - d)^n \text{ for all } \underline{n \in \mathbb{R}} \tag{2}$
First, I ... | As you already noted, the problem is equivalent to proving
$$(a+b+c+d)^\alpha-(b+c+d)^\alpha+(c+d)^\alpha-d^\alpha\ge(a+c)^\alpha$$
for $a,b,c,d\ge0$.
Lemma: If $\alpha\ge1$, $x\ge y\ge0$ and $r\ge0$, then $(x+r)^\alpha-(y+r)^\alpha\ge x^\alpha-y^\alpha$.
Proof: It suffices to show that $(x+r)^\alpha-x^\alpha$ is an in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Inequality with trigonometric functions Find all values for $a$ such that the following inequality holds:
$$\sin^6x + \cos^6x + a\sin x \cos x \ge 0$$
To be fair, I didn't manage to get anything helpful wiht my calculations. I tried to cube the basic trigonometric identity $\sin^2x + \cos^2x = 1$ in order to get a rep... | It is enough to find the minimum value of
$$
f_a(x):=\cos^6x+\sin^6x+a\cos x\sin x
$$
We have
\begin{eqnarray}
f_a(x)&=&\cos^6x+\sin^6x+a\cos x\sin x\\
&=&(\cos^2x+\sin^2x)^3-3(\cos^4x\sin^2x+\cos^2x\sin^4x)+a\cos x\sin x\\
&=&1-3\cos^2x\sin^2x(\cos^2x+\sin^2x)+a\cos x\sin x\\
&=&1-3\cos^2\sin^2x+a\cos x\sin x\\
&=&1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\int_0^a \sin(\tan^{-1}(b\sin\theta))\ d\theta$ How to compute
$$\int_0^a \sin(\tan^{-1}(b\sin\theta))\ d\theta$$
I've tried to rewrite it as
$$\int_0^a\frac{b\sin\theta}{\sqrt{b^2\sin^2\theta+1}}d\theta$$
But I'm still stuck here.
| The change of variables
$$
\cos\theta=\frac{b+1}{b}x
$$
gives
$$
\int\frac{b\sin\theta}{\sqrt{b^2\sin^2\theta+1}}d\theta=-\sqrt{b^2+1}\int\frac{dx}{\sqrt{1-x^2}}.
$$
Let's check it:
$$
\sin\theta\,d\theta=-\frac{b+1}{b}dx,
$$
$$
b^2\sin^2\theta+1=b^2(1-\cos^2\theta)+1=b^2(1-\frac{(b+1)^2}{b^2}x^2)+1=(b^2+1)(1-x^2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the fuction $g$. If $f:x \mapsto x^2 + 3$, find function $g$ such that $gf:x \mapsto 2x^2 + 3$.
I don't know how to do it, there is no such example in my book.
Help?
| $$f(x)=x^2+3,\quad g(f(x))=2x^2+3\\
\implies g(x^2+3)=2x^2+3=2(x^2+3)-3\\
\implies g(x)\text{ could be }g(x):=2x-3$$
If $gf=g(x)\times f(x)$, then
$$g(x)\times f(x)=2x^2+3\\
\implies g(x)=\dfrac{2x^2+3}{x^2+3}=1+\dfrac{x^2}{x^2+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/747179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that if $p$ is a prime number $> 3$ then $24 \mid p^2-1$ Hi guys can someone help me with this ?(Without using Modular arithmetic)
Show that if $p$ is a prime number $>3$ then $24$ $\mid$ $p^2-1$
| If $p$ is a prime greater than $3$, than it is equal to $1, 3, 5, 7$ $\pmod 8$
THis implies $p^2 \equiv 1 \pmod 8$ (because $1^2 \equiv 1$, $3^2 \equiv 1, 5^2 \equiv 1, 7^2 \equiv 1 \pmod 8$)
Also, $p \equiv 1, 2 \pmod 3 \Rightarrow p^2 \equiv 1 \pmod 3$
THis implies that $p^2 - 1 $ is divisible by both $8$ and $3$, an... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How prove this $\cos{x}+\cos{y}+\cos{z}=1$ Question:
let $x,y,z\in R$ and such $x+y+z=\pi$,and such
$$\tan{\dfrac{y+z-x}{4}}+\tan{\dfrac{x+z-y}{4}}+\tan{\dfrac{x+y-z}{4}}=1$$
show that
$$\cos{x}+\cos{y}+\cos{z}=1$$
My idea: let $$x+y-z=a,x+z-y=b,y+z-x=c$$
then
$$a+b+c=\pi$$
and
$$\tan{\dfrac{a}{4}}+\tan{\dfrac{... |
An interesting view of the surface $ \cos x + \cos y + \cos z = 1 $ Three lines ( in red ) are added for clarity. We can see that even along the plane $ x + y + z = \pi $ we must add a network of lines to encompass the simple periodicity condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How find this system $a^2+b^2=3,a^2+c^2+ac=4,b^2+c^2+\sqrt{3}bc=7$ Find the this system real solution
$$\begin{cases}
a^2+b^2=3\\
a^2+c^2+ac=4\\
b^2+c^2+\sqrt{3}bc=7
\end{cases}$$
I think that one can use Geometry to solve this system. Maybe there exist an algebraic method.
$$a^2+b^2=\sqrt{3}^2$$
$$a^2+c^2-2ac\cos{(1... | Here is an algebraic solution that yields all the real answers.
Let us introduce the complex numbers:
$$x=\frac{-b+i a}{\sqrt{3}},\quad y=\frac{2a+c}{4}-i\frac{\sqrt{3}}{4}c.$$
The first two equations are equivalent to the statement: $\vert x\vert=\vert y\vert=1$, and the third equation tells us that
$$\vert \sqrt{3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
If all eigenvalues are 1 or -1, is then $A^{12}=I$? True or false: If all the eigenvalues of A are either $\lambda=1$ or $\lambda = -1$ then $A^{12}$= I
If we have a matrix $$\mathbf A = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ this has eigenvalues of 1 or -1 then $$\mathbf A = \begin{pmatrix}\lambda-1&0\\0&\lambda+1\en... | Compute powers of the matrix $A$, where
$$A=\begin{pmatrix} 1 & 1\\0&1\end{pmatrix}.$$
You will find that
$$A^n=\begin{pmatrix} 1 & n\\0&1\end{pmatrix}.$$
Remark: If a $2\times 2$ matrix $A$ has both $1$ and $-1$ as eigenvalues, then we will have $A^2=I$. But we can find a $3\times 3$ matrix that has both $1$ and $-1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/752613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Integral $\frac{1}{\pi}\int_0^{\pi/3}\log\big( \mu(\theta)+\sqrt{\mu^2(\theta)-1} \big)\ d\theta, \quad \mu(\theta)=\frac{1+2\cos\theta}{2}.$ Hi I am trying to calculate this integral:
$$
I=\frac{1}{\pi}\int_0^{\pi/3}\log\left( \frac{1+2\cos\theta}{2}+\sqrt{\bigg( \frac{1+2\cos\theta}{2} \bigg)^2-1} \right)\ ... | The integral can be expressed in terms of a series of Gauss hypergeometric functions. It is doubtfull that it would be possible to go further on this way.
$I = \frac{1}{\pi}\int_0^{\pi/3}\ln\left(\mu(\theta)+\sqrt{\mu^2(\theta)-1}\right)d\theta = \frac{1}{\pi}\int_0^{\pi/3}\cosh^{-1}(\mu(\theta))$, where $\mu(\theta) =... | {
"language": "en",
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"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Find the first two iteration of the Jacobi method for the following linear system, using $x^{(0)} = 0$ $$3x_{1} - x_{2} + x_{3} = 1,$$
$$3x_{1} + 6x_{2} + 2x_{3} = 0,$$
$$3x_{1}+3x_{2}+7x_{3} = 4$$
So, from this I got
T = \begin{bmatrix}
0 & \frac{-1}{3} & \frac{1}{3} \\[0.3em]
\frac{1}{2} & ... | First iteration:
$$0.333333, \text{ }0.,\text{ }0.571429$$
Second iteration:
$$0.142857,\text{ }-0.357143\text{ },0.428571$$
So, it looks like somehow the sign of your $x_3$ on the first iteration got corrupted.
We rewrite the system as:
$$x_1 = \dfrac{1}{3} x_2 -\dfrac{1}{3}x_3 + \dfrac{1}{3} \\ x_2 = -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/755589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find value range of $2^x+2^y$ Assume $x,y \in \Bbb{R}$ satisfy $$4^x+4^y = 2^{x+1} + 2^{y+1}$$, Find the value range of $$2^x+2^y$$
I know $x=y=1$ is a solution of $4^x+4^y = 2^{x+1} + 2^{y+1}$ , but I can't go further more. I can only find one solution pair of $4^x+4^y = 2^{x+1} + 2^{y+1}$. It seems very far from sol... | Cauchy-Schwarz inequality allows us to conclude $$((a-1)^{2}+(b-1)^{2})(1+1) \geq (a-1+b-1)^{2}$$ and hence $4\geq (a+b-2)^{2}$ and finally $a+b \leq 4$ with equality holding only when $a=b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Explanation of recursive function Given is a function $f(n)$ with:
$f(0) = 0$
$f(1) = 1$
$f(n) = 3f(n-1) + 2f(n-2)$ $\forall n≥2$
I was wondering if there's also a non-recursive way to describe the same function.
WolframAlpha tells me there is one:
$$g(n) = \frac{(\frac{1}{2}(3 + \sqrt{17}))^n - (\frac{1}{2}(3 - \sqrt{... | $$\begin{align} f(n) &= 3\,f(n-1) + 2\,f(n-2) \\
f(n-1) &= 1\,f(n-1) + 0\,f(n-2)\end{align}$$
$$\begin{bmatrix} f(n) \\ f(n-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} f(n - 1) \\ f(n - 2) \end{bmatrix}$$
$$\begin{bmatrix} f(n) \\ f(n-1) \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ 1 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Trigonometric substitution Been out of touch with trigonometry for some time now. Need help proving this expression.
$$\sin^{2}\left(\frac{x}{2}\right) = \frac{1}{2}(1-\cos\left(x\right))$$
Any help will be appreciated. Thanks.
| From the identity $\sin^2\theta+\cos^2\theta=1$ and $\cos 2\theta=\cos^2\theta-\sin^2\theta$, rewrite the LHS as
$$
\begin{align}
\frac{1}{2}(1-\cos x)&=\frac{1}{2}\left(\sin^2\left(\frac x2\right)+\cos^2\left(\frac x2\right)-\left(\cos^2\left(\frac x2\right)-\sin^2\left(\frac x2\right)\right)\right)\\
&=\frac{1}{2}\le... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to prove $x^3$ is strictly increasing I am trying to use $f(x)=x^3$ as a counterexample to the following statement.
If $f(x)$ is strictly increasing over $[a,b]$ then for any $x\in (a,b), f'(x)>0$.
But how can I show that $f(x)=x^3$ is strictly increasing?
| Suppose $a<b$. We want to prove that $a^3 < b^3$.
Well, $a<b, \ \therefore \exists \delta>0$ such that $b = a+\delta$. So:
$b^3 = (a+ \delta)^3 = a^3 + 3a^2\delta + 3\delta^2a + \delta^3 = a^3 + (3a\delta(a+\delta)+\delta^3)$.
Now we consider five different cases:
*
*$\ a<-\delta$
*$\ a = -\delta$
*$\ -\delta<a<0$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
Solve $z^2 - iz = |z - i|$ I have the equation:
$z^2 - iz = |z - i|$
The solutions are $i$, $-\sqrt3/2 + i/2$, $\sqrt3/2 + i/2$
Can someone please walk me through or give me a hint...
| Let $z=x+iy$, $z^2-iz=|z-i|$ implies $0=\Im (z^2-iz)=2xy-x$ then $x=0$ or $y=\frac{1}{2}$.
If $x=0$, then we have $-y^2+y=|y-1|$ i.e. $-y(y-1)=|y-1|$ then $y=1$. So $z=i$ satisfies the equation.
Note that $z^2-iz=|z-i|$ implies $|z||z-i|=|z^2-iz|=|z-i|$, then $|z|=1$. So, if $y=\frac{1}{2}$, then $x=\pm\sqrt{1-\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Definite integral (mixture of functions) I have problem with this integral and I generally don't know how to approach it:
$$\int_{-2}^2 (x^4+4x+\cos(x))\cdot \arctan\left(\frac{x}{2}\right)dx$$
I know that I probably have to make some substitution, but really don't know which.
Thanks for your help.
| Using lab bhattacharjee's result. The integral can be solven by using integration by parts.
Let
\begin{align}
u=\arctan\frac{x}{2}\qquad\rightarrow\qquad du=\frac{2\,dx}{x^2+4}\qquad\text{and}\qquad dv=x\,dx\qquad\rightarrow\qquad v=\frac{1}{2}x^2
\end{align}
Then
\begin{align}
8\int_0^2x\arctan\frac {x}{2}\, dx&=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/759316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate this expression: $∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$?
How to evaluate: $$∫\frac{\sin2x}{(1 - \cos2x)^4}\:\mathrm{d}x$$ If anyone knows the answer, please help.
| Use $1-\cos 2x=2\sin^2 x$ and $\sin 2x=2\sin x\cos x$, then u will get
$$\int \frac{\sin 2x}{(1-\cos 2x)^4}dx=\int \frac{\cos x}{2^3 \sin ^7x }dx$$
Now letting $u=\sin x\Rightarrow du=\cos x dx$
$$\int \frac{\cos x}{2^3 \sin ^7x }dx= \frac{1}{2^3}\int\frac{du}{u^7}=\frac{1}{2^3}.\frac{-1}{6u^6}+C=\frac{1}{2^3}.\frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/761695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$. If $\sin A = \cfrac{3}{5}$ with $A$ in QII, find $\sec2A$.
I'm getting $\sec2A=\cfrac{25}{7}$. Is that correct?
| \begin{align}
\sec2A&=\frac{1}{\cos2A}\\
&=\frac{1}{1-2\sin^2A}\\
&=\frac{1}{1-2\left(\frac{3}{5}\right)^2}\\
&=\frac{1}{1-2\left(\frac{9}{25}\right)}\\
&=\frac{1}{1-\frac{18}{25}}\\
&=\frac{1}{\frac{7}{25}}\\
&=\frac{25}{7}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
No. of real solutions of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $ How many real solutions are there of the equation $2 \cos (\frac{x^2 + x}{6}) = 2^x + 2^{-x} $?
Please illustrate.
| AM-GM inequality says: $2^x + 2^{-x} \geq 2 \geq 2cos\left(\dfrac{x^2 + x}{6}\right)$. so equation occurs when: $2^x = 2^{-x}$ or $x = 0$, and this is the solution of the equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
When is the sum of consecutive squares a prime? For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?
I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$
| The sum of $x$ consecutive squares starting from $n$ is $$n^2+(n+1)^2+...+(n+x-1)^2=xn^2+2n\sum_{k=1}^{x-1} k +\sum_{k=1}^{x-1} k^2 $$
$$\Rightarrow n^2+(n+1)^2+...+(n+x-1)^2=xn^2+nx(x-1)+\frac{(x-1)x(2x-1)}{6}$$
If $x\equiv 1,5\ (mod\ 6)\ $ then we have $(x-1)(2x-1)\equiv0\ (mod\ 6)$
which means that $\frac{(x-1)(2x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A Horrible looking limit I have the following limit question:
$$\lim_{x \rightarrow 1 }\frac {({\rm log} (1+x)-{\rm log}\space 2)(3\times4^{x-1}-3x)}{[(7+x)^{1/3}-(1+3x)^{1/2}]{\rm sin}\space \pi x}$$
This has the form $\frac {0}{0}$ and hence I can apply L'Hospital's Rule. But, this is too large and that way it can go... | Put $x=y+1$ and take the limit $y\to 0$, so that we can use some standard limits and simplify the expression. The given limit will be:
\begin{align*}
& \lim_{y\to 0} - \frac{\displaystyle \log\left(1+\frac{y}{2}\right)\Big(3\, \left(4^y-1-y\right)\Big)}{\left(\left(8+y\right)^{1/3}-\left(4+3\, y\right)^{1/2}\right)\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/764539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove the inequality $\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$
If $a,b,c\in\mathbb R^+$ prove that:
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{a+c}+\frac{c^2+1}{a+b}\ge 3$$
| By C-S and AM-GM we obtain
$$\sum_{cyc}\frac{a^2+1}{b+c}=\sum_{cyc}\frac{a^2}{b+c}+\sum_{cyc}\frac{1}{b+c}\geq$$
$$\geq\frac{(a+b+c)^2}{2(a+b+c)}+\frac{9}{2(a+b+c)}=\frac{1}{2}\left(a+b+c+\frac{9}{a+b+c}\right)\geq$$
$$\geq\frac{1}{2}\cdot2\sqrt{\frac{9(a+b+c)}{a+b+c}}=3.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating the integral $\int\frac{3x^{2}-x+2}{x-1}\;dx$ As the title suggests, the following integral has been given to me
$$\int\frac{3x^{2}-x+2}{x-1}\;dx$$
Yet I still get the wrong answer every time.
Can someone calculate it step-by-step so I can compare it to my own answer?
| Another hint :
\begin{align}
\frac{3x^{2}-x+2}{x-1}&=\frac{3x^{2}-3x+2x+2}{x-1}\\
&=\frac{3x^{2}-3x}{x-1}+\frac{2x}{x-1}+\frac{2}{x-1}\\
&=\frac{3x(x-1)}{x-1}+2\left(\frac{x-1+1}{x-1}\right)+\frac{2}{x-1}\\
&=3x+2\left(1+\frac{1}{x-1}\right)+\frac{2}{x-1}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/767569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $\cos 36° > \tan 36° $ How do we prove that $\cos 36° > \tan 36° $ ? Please help . Thank you.
| $\cos36 > \tan36 \iff \cos36 > \dfrac{\sin36}{\cos36} \iff \cos^2{36} > \sin36 \iff 1 - \sin^2{36} > \sin36 \iff \sin^2{36} + \sin36 - 1 < 0 \iff \sin36 < \dfrac{\sqrt{5} - 1}{2} (*)$. We prove $(*)$ true. let $x = \sin18$, then using $\cos36 = \sin54$ gives: $\cos(2\cdot 18) = \sin(3\cdot 18) \iff 1 - 2x^2 = 3x - 4x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Maximum value of $abc$ for $a, b, c > 0$ and $ab + bc + ca = 12$ $a,b,c$ are three positive real numbers such that $ab+bc+ca=12$.
Then find the maximum value of $abc$
| Or, the same answer also comes by intuition. since a,b,c are positive numbers, $(ab).(bc).(ca)$ is maximum only when $ab=bc=ca=\frac{12}{3}=4$. therefore, $ abc={((ab)(bc)(ca))}^\frac{1}{2}=4^\frac{3}{2}=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve the following recursive relation by using generating functions $a_n - 9a_{n-1} + 26a_{n-2} - 24a_{n-3} = 0, n \ge 3, a_0 = 0, a_1 = 1,a_2 = 10$
I have tried solving it by the normal way, but I have no idea how to solve it by generating functions. Please give me a detailed answer.
| Check Wilf's "generatingfunctionology". Write the recurrence without subtraction in indices:
$$
a_{n + 3} - 9 a_{n + 2} + 26 a_{n + 1} - 24 a_n = 0
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$, sum over $n \ge 0$ and recognize sums like:
$$
\sum_{n \ge 0} a_{n + m} z^n
= \frac{A(z) - a_0 - a_1 z - \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{2^{4n + 2} + 1}{5}$ is composed for all $n \ge 2$ The problem is to prove that $$\frac{2^{4n + 2} +1 }{5}$$ is composed for every $n \ge 2$.
I've tried this way: write $2^{4n + 2} + 1$ as $1 - (-4)^{2n + 1}$ and $5 = 1 - (-4)$, then $$\frac{2^{4n + 2} + 1}{5} = \frac{1 - (-4)^{2n + 1}}{1 - (-4)} = 1 -... | Check you can understand/prove the following
$$\frac{2^{4n+2}+1}5=\frac{3\cdot\left(2^{4n+1}-2^{4n}+\ldots-2^2+2\right)}5$$
But $\;5\nmid 6\;$ , so $\;5\;$ must divide the other factor as the whole thing is an integer since
$$4\cdot16^n+1=0\pmod 5$$
and thus the expression equals $\;6k\;,\;\;k\in\Bbb N\;$ ...composite.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Integrating a top heavy function How do you integrate this function?
$$\int\frac{x^3}{(x+5)^2}dx$$
I have tried it myself by substitution but I can't seem to get rid of the $x$s.
| Long division:
$$
\begin{array}{cccccc}
& & x & - & 10 \\ \\
x^2+10x+25 & ) & x^3 \\
& & x^3 & + & 10x^2 & + & 25x \\ \\
& & & & -10x^2 & - & 25x \\
& & & & -10x^2 & - & 100x \\ \\
& & & & & & 75 x
\end{array}
$$
So we have
$$
\frac{x^3}{x^2+10x+25} = x - 10 + \frac{75x}{(x+5)^2} = x - 10 + \frac{A}{x-5} + \frac{B}{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Why a linear numerator for fractions with irreducible denominators? For example: (2x^3+5x+1)/((x^2+4)(x^2+x+2)) breaks down to (ax+b/(x^2+4))+(cx+d/(x^2+x+2)). I have been told that since the denominators are irreducible, the numerators will be either linear or constant. Now my question is for something like (2x^3+5x+1... | The reason that we always make a partial fraction's numerator at least $1$ degree less than that of the denominator is that, if the numerator's degree is greater than or equal to that of the denominator, the highest degree can be factored out as a non-fraction term.
Taking your $\displaystyle \frac{2x^3+5x+1}{x^2-4}$ f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. Please help me get the answer to this question.
Prove $f(x)=\sqrt{2x-6}$ is continuous at $x=4$ by using precise definition. ($\epsilon-\delta$ definition of limits.)
| We have $|x-4|< \delta$.
$|f(x)-f(4)|=|\sqrt {2x-6}-\sqrt {2}|=|\sqrt{2}(\sqrt{x-3}-1)|$
Now, multiplying the numerator and denominator by $(\sqrt{x-3}+1)$,
$=\sqrt{2} |\frac{x-4}{\sqrt{x-3}+1}$|
Now, ${\sqrt{x-3}+1}$ is always greater than or equal to $1$. Thus, $\frac{x-4}{\sqrt{x-3}+1}\leqslant (x-4)$.
Therefore, $|... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$\int_0^{\pi\over2}{\cos x\sin x\over(x+1)}dx$ = ${1\over2}({1\over2}+{1\over\pi+2}$- $\int_0^\pi{\cos x\over(x+2)^2}dx)$ $$A=\int_0^\pi{\cos x\over(x+2)^2}dx$$
Then prove that, $\displaystyle\int_0^{\Large\pi\over2}{\cos x\sin x\over(x+1)}dx={1\over2}\left({1\over2}+{1\over\pi+2}-A\right)$
| Rewrite:
$$
\int_0^{\Large\pi\over2}{\cos x\sin x\over x+1}dx=\frac12\int_0^{\Large\pi\over2}{\sin 2x\over (x+1)}dx
$$
Now, use IBP by taking $u=\dfrac1{x+1}\;\Rightarrow\;du=-\dfrac{dx}{(x+1)^2}$ and
$$
dv=\sin2x\ dx\quad\Rightarrow\quad v=-\frac12\cos2x
$$
Hence
$$
\begin{align}
\int_0^{\Large\pi\over2}{\sin 2x\over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/776702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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.