Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
how to prove $\displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = .... $ How to prove $$ \displaystyle \frac{\sin (2n+1)\theta}{\sin \theta} = (2n+1) \prod_{k=1}^{n}\left(1 - \frac{\sin^2 \theta}{\sin^2 \left( \frac{k\pi }{2n+1} \right ) } \right ) $$
So far, I manage to prove $ \displaystyle \frac{\sin (2n+1)\theta... | First consider the following Lemma
Lemma :
$$ \prod_{k=1}^{n} \sin^2 \left(\dfrac{k\pi}{2n+1}\right) = \dfrac{2n+1}{2^{2n}} $$
Proof : Note that,
$ \displaystyle \prod_{k=1}^{2n} \sin \left(\dfrac{k\pi}{2n+1}\right) = \prod_{k=1}^{n} \sin \left(\dfrac{k\pi}{2n+1}\right) \cdot \prod_{k=n+1}^{2n} \sin \left(\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
epsilon-delta proof of $\lim_{x \to 4} \sqrt{x} = 2$ Prove $$\quad \lim_{x\to4}\sqrt{x}=2 $$
using the precise definition of limits. (Epsilon-Delta)
I think I proved this problem but when I look at the textbook to compare the proofs
they are quite different and I don't get how the book worked it all out.
So I am going ... | Answer to your where did $\geq2 > 1$ comes from:
It comes from the fact $\sqrt{x} \geq 0$, so $|\sqrt{x} + 2| \geq |0 + 2| = 2$ and of course $2 > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/310611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
solving $\sqrt{3-\sqrt{3+x}}=x$. Can we solve the following equation in $\mathbb{R}$ without expanding it into a fourth degree equation :
$$ \sqrt{3-\sqrt{3+x}} = x.$$
squaring both sides and squaring again is the only thing I could done, If you have any other idea just post hints.
| First, it must be $\,3+x\ge 0\Longleftrightarrow x\ge -3\,$ , for the inner-most square root to be defined on the real field.
Then it also must be
$$3-\sqrt{3+x}\ge 0\Longrightarrow9\ge 3+x\Longrightarrow x\le 6$$
for the outer square root to be defined, thus our definition domain is $\,-3\le x \le 6\,$ . Now directl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/311856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Quadratic equation $x^2-ax+2a = 0$ with Integers Roots Find Integer value of $a$ for which both roots of the equation $x^2-ax+2a = 0$ are also integers
My Try:: $\displaystyle x = \frac{a \pm \sqrt{a^2-8a}}{2}$. Now if the Roots are Integer . Then $a^2-8a=k^2$ where $k\in \mathbb{Z}$
Now How can I calculate after that.... | As you say, the question comes down to knowing when $a^2 - 8a - k^2 = 0$ for $k \in \mathbb{Z}$. Solving for $a$, $a = 4 \pm \sqrt{16 + k^2}$. As you can see, only for $k = 0$ or $k=3$ does this produce an integer solution for $a$ and so $a = 8$, $a = 0$, $a=9$ and $a=-1$ are possible.
The reason why those were the o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/315999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\sum\limits_{n=1}^{+\infty}\frac{1}{1^k+2^k+\cdots+n^k}$? There is post here asking for $$\sum_{n=1}^{+\infty}\frac{1}{1^2+2^2+\cdots+n^2}.$$
And the answer is $18-24\ln 2$.
It is easy to elvaluate that $\displaystyle\sum_{n=1}^{+\infty}\frac{1}{1+2+\cdots+n}=2$ and it is not difficult to justify the c... | I guess the way is pretty much the same, as there are always formulas to evaluate
$$\sum_{i=1}^n i^k, $$
For $k=3$ the result of the sum above is
$$\frac{4}{3} (-9+\pi^2)$$
For higher terms you will probably not always get a nice result, but for $k=5$ you have $$ 4\left(15-\pi^2 + 2\sqrt{3} \pi \tan\left(\frac{\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/317336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$ I need to show that
$$
\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}
$$
I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable... | You are likely expected to integrate by parts (twice)
$$ \begin{eqnarray}
\int \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} \mathrm{d} x
\\
&=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/318037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 6,
"answer_id": 3
} |
If $v$ is an eigenvalue of $A$ and $c$ is an eigenvalue of $B$, must $vc$ be an eigenvalue of $AB$? Let $v$ be an eigenvalue of $A$ and $c$ be an eigenvalue of $B$. Is the product of $v$ and $c$ equal to an eigenvalue of $AB$?
| $1$ is an eigenvalue of $
\begin{pmatrix}
1 & 0\\
0 & 2\\
\end{pmatrix},$
$3$ is an eigenvalue of $
\begin{pmatrix}
2 & 0\\
0 & 3\\
\end{pmatrix}$
but $3$ is not an eigenvalue of $
\begin{pmatrix}
1 & 0\\
0 & 2\\
\end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Exponential equation, $(3+2\sqrt2)^x+1=6(\sqrt2+1)^x$ http://i.stack.imgur.com/YP2Ha.png
$$(3+2\sqrt2)^x+1=6(\sqrt2+1)^x \qquad\qquad x\in\mathbb{R}$$
I managed to find one of the solutions (x=2), but I got stuck. I would really appreciate a step by step solution. Thanks in advance :)
| $(\sqrt2+1)^2=3+2\sqrt2$
If $\sqrt2+1=y, y^{2x}+1=6{y^x}\implies y^{2x}-6y^x+1=0, y^x=\frac{6\pm\sqrt{6^2-4}}2=3\pm2\sqrt2$
Now, $3+2\sqrt2=y^2\implies 3-2\sqrt2=\frac1{3+2\sqrt2}=y^{-2}$
If $y^x=3+2\sqrt2,y^x=y^2\implies x=2$ as $y=\sqrt2+1\ne0,\pm1$
Similarly for $y^x=3-2\sqrt2,y^x=y^{-2}\implies x=-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/324925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
An irregular 6 faced dice An irregular 6 faced dice is such that the probability that it gives 3 even numbers in 5 throws is twice the probability that it gives 2 even numbers in 5 throws. How many sets of exactly 5 trials can be expected to give no even number out of 2500 sets?
(1) Nearly 5
(2) Nearly 10
(3) Nearly 15... | Let the probability of an even number be $p$. Thus the probability of an odd number is $1-p$.
The probability of $3$ even in $5$ throws is $\binom{5}{3}p^3(1-p)^2$. The probability of $2$ even in $5$ throws is $\binom{5}{2}p^2(1-p)^3$.
We are told that $\binom{5}{3}p^3(1-p)^2=2\binom{5}{2}p^2(1-p)^3$. Solve for $p$. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Frequency response We consider the response $x(t)$ when the frequency of the input y(t) varies.
$$x''+bx'+kx=ky(t), y(t)=\cos \omega t$$
1) If $k=3.24$, what is the natural ($b=0$) frequency $\omega_n$?
I think we have $\omega_n=\frac{\sqrt{3.24}}{2\pi}$
2) If $k=3.24$ and $b=0.8$, what is the amplitude gain from the i... | Homogeneous equation is
$$
x''+bx+kx=0
$$
Characteristic equation of which
$$
\lambda^2+b\lambda+k=0
$$
which has solutions
$$
\lambda_{1,2}=-\frac b2 \pm \frac{\sqrt{b^2-4k}}2
$$
To have solution only in reals and keep order of ODE, you have two options.
First if $b^2-4k > 0$ you have two real solutions
$$
\lambda_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the appropriate base $$213_x=139_{10}$$
$$21_x=1021_4$$
How would I solve this
| $$213_x=139_{10}$$
$$2\cdot x^2+1\cdot x^1+3\cdot x^0=1\cdot 10^2+3\cdot 10^1+9\cdot10^0$$
$$2x^2+x+3=139$$
$$2x^2+x-136=0\Rightarrow x_{1,2}=\frac{-1\pm\sqrt{1+8\cdot 136}}{4}=\frac{-1\pm33}{4}$$
$$x_1=8$$ the base $$x_2=-17/2$$
is not a correct base see comments below.
similarly
$$21_x=1021_4$$
$$2x+1=4^3+2\cdot 4 +1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/327517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx$? How to find the follwing integral's value ?
$$\int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx$$
Actually, I don't know it can be represented as closed form.
| Here is another approach based on power series and beta function
$$ \int_{0}^{1}(\arcsin{x})(\sin{\frac{\pi}{2}x})dx = \frac{2}{\pi}\,\int_{0}^{1}\!{\frac{\cos \left( \frac{\pi x }{2} \right) }{\sqrt {1-
{x}^{2}}}}{dx}$$
$$ = \frac{2}{\pi} \sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac{\pi}{2}\right)^{2k}}{(2k)!}\int_{0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
|
Consider an equilateral triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Let $P'$, $P'_1$ and $P'_2$ be the reflection of the Point $P$ along $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$
Then we have
$${\triangle APB} \cong {\triangle AP'_2B}$$
$${\tri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/330333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Transforming matrix-equation to overdetermined minimum problem i have broken down my problem to plainmath and could really use some help.
Basis: I have an image. In this image I have several UV-XYZ pairs. So i know the 3d position of serveral Pixels.
Given the following equations from 3d to 2d space.
$ Z_{c} \begin{pma... | Let me do some rearranging and substituting to give you a more concise formula:
Your very first formula with the substitution
$$Z_{c} = \begin{pmatrix} r_{7} & r_{8} & r_{9} & t_{3} \end{pmatrix} \begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix}$$
gives
$$ \begin{pmatrix} U \\ V \\ 1 \end{pmatrix} \begin{pmatrix} r_{7} &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $\sum nx^n$ I am trying to find a closed form solution for $\sum_{n\ge0} nx^n\text{, where }\lvert x \rvert<1$.
This solution makes sense to me:
$\sum_{n\ge0} x^n=(1-x)^{-1} \\
\frac{d}{d x} \sum_{n\ge0} x^n = \frac{d}{d x} (1-x)^{-1} \\
\sum_{n\ge0} nx^{n-1} = (1-x)^{-2} \\
x \sum_{n\ge0} n x^{n-1} = x(1-x)^{-2}... | Here's a solution that use neither differentiation nor integraion. Since this question is marked as the duplicate target of How do I compute $\sum_{k=1}^{\infty} k \cdot p^k$, which no longer accepts new solution, I'm posting mine for fun.
Use summation by parts
$$S_N = \sum_{n=0}^N a_nb_n = a_N B_N - \sum_{n=0}^{N-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Find a general control and then show that this could have been achieved at x2 Determine the general form of $u_0, u_1 ~\text{and} ~ u_2$ if a system of difference equations of the form
$$x_{n+1} = Ax_n + Bu_n,$$
where:
$$A = \begin{pmatrix}
3 & 2 & 2 \\
-1 & 0 & -1 \\
0 & 0 & 1
\end{pmatrix}$$
and:
$$B = \be... | First of all we have forall $n\in N:$ $u_{n}\in R^{2}$.
So let
$$u_{0}=(u_{0,1},u_{0,2})^{t}$$ be given as well as $x_{0}=0$.
Then we get
$$
x_{1}=Ax_{0}+Bu_{0}=0+(0,u_{0,2},u_{0,1})^{t}$$
Again plugging this into the recursion we obtain
$$
x_{2}=Ax_{1}+Bu_{1}=A(0,u_{0,2},u_{0,1})^{t}+B(u_{1,1},u_{1,2}).$$
This has by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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finding the real values of $x$ such that : $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ How to find the real values of $x$ such that : $$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
| Elaborating some on what @Fredrik Meyer suggested, one can get:
$$\begin{align*}x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}\hspace{5pt}&\Rightarrow\hspace{5pt} x^2=2+\sqrt{2-\sqrt{2+x}}\hspace{5pt}\Rightarrow\hspace{5pt} x^4-4x^2+4=2-\sqrt{2+x}\\
&\Rightarrow\hspace{5pt}x^8+16x^4+4-8x^6+4x^4-16x^2=2+x\\
&\Rightarrow\hspace{5pt} x^8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$ Find all solutions of $x^8 \equiv 5 \pmod {3\cdot 11}$
Im not sure should I use primitive root or quadratic residue.
For primitive root, $U_{33} = \{1,2,4,5,7,8,10,13,14,16,17,19,20,23,25,26,28,29,31,32\}$
Thus $\phi_{(3\cdot11)}=20=2^2\times{5}$
But it takes way ... | Hint: Consider the equation modulo $3$. What are the solutions to $x^8 \equiv 5 \mod 3$?
Alternatively, you could reduce the above to $(x^4)^2 \equiv -1 \mod 3$, which means that for a solution, you need that $-1$ is a quadratic residue modulo $3$ and $x^4$ is its 'square root'. Is $-1$ a quadratic residue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/336378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that :
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ .
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $
What to do... | Hint:
From the identity : $\cos A= 1- 2 \sin^2(\frac{A}{2})$
$\sin^2(\frac{A}{2})=1-\cos A$
$1- \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2bc-b^2-c^2+a^2}{2bc}$
$= \dfrac{a^2-(b-c)^2}{2bc}=\dfrac{(a+b-c)(a-b+c)}{2bc}$
$a+b-c=\dfrac{(s-b)}{2}$, $a-b+c=\dfrac{(s-c)}{2}$
$2 \sin^2\dfrac{A}{2}=\dfrac{4(s-b)(s-c)}{2bc}$
$\sin^2(\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Distance: sphere related question Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere ?
How do we do here as I have four options as answer :
(a) $3+\dfrac{\sqrt{30}}{2}$
(b) $3+\dfrac{\sqrt{69... | Let the centers of your 3 spheres of radius 1 be $C_1, C_2, C_3$. Let the center of the sphere of radius 2 be $C_4$. These 4 points forms a tetrahedron with equilateral triangle base of length 2 ( $C_1C_2 = C_2C_3=C_3C_1=2$) , and a side length of 3 ($C_4C_1 = C_4C_2 = C_4C_3 = 3$). Let $O$ be the center of the equilat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
logarithm problem - four tuple How many distinct four tuple (a,b,c,d) of rational numbers are there with $a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7=2005$
Can we proceed like this :
Using $\log a +\log b = \log(ab)$ and $m\log a = \log a^m$
$\Rightarrow \log_{10}2^a \cdot 3^b \cdot 5^c \cdot 7^d = 2005$
Please gu... | You have
$$
2^a\cdot 3^b\cdot 5^c\cdot 7^d = 10^{2005} = 2^{2005}\cdot 5^{2005}.
$$
So this works if $a=c=2005$ and $b=d=0$.
To see if other solutions exist, observe that you'd get
$$
2^{a-2005}\cdot 5^{c-2005} = 3^{-b}\cdot 7^{-d}.
$$
All the exponents are rational, so we have
$$
2^{n_1/n_2}\cdot 5^{n_3/n_4} = 3^{n_5/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/343811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Infinite Factorization Power Series of $\sin(x)$ My teacher has given us a rather long problem and the last part is stumping me. How would one go about factoring the power series of sin(x)? Where:
p(x) = x - $\frac{x^{3}}{3!}$ + $\frac{x^{5}}{5!}$ - $\frac{x^{7}}{7!}$ ...
Better expressed as:
p(x) = $\sum_1^\infty (-1)... | I show here the method used by Euler.
*
*Initial Identities:
Euler showed that
\begin{eqnarray}
\frac{\sin z}{z} = \prod_{k=1}^{\infty} \left ( 1 - \frac{z}{k^2 \, \pi^2} \right )
\end{eqnarray}
He started by considering his formula
\begin{eqnarray}
e^z = \lim_{n \to \infty} \left ( 1 + \frac{z}{n} \right )^n,
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Exponential and logarithmic series: Find the sum of $2^2 + 3^2/ 2!+4^2/3!+...$ to infinity
Find the sum of the following series:
$ 2^2 + 3^2/2! + 4^2/3! + ...$ to infinity
The answer is given as $5e$ but I got it as $5e+1$
$T_n = 1/(n-2)! +3/(n-1)! + 1/(n)! $ for $n \ge 2$
and $T_1+T_2 + ... $ to infinity = $4 + e... | So, $$T_n=\frac{(n+1)^2}{n!}$$
Let $(n+1)^2=n(n-1)+Bn+C$
$\implies n^2+2n+1=n^2+n(B-1)+C$
$\implies B-1=2,B=3, C=1$
So, $$T_n=\frac{(n+1)^2}{n!}=\frac1{(n-2)!}+3\frac1{(n-1)!}+\frac1{n!}$$
Putting $n=0,1,2,3,\cdots$
$$T_0=\frac1{0!}$$
$$T_1=0+3\frac1{0!}+\frac1{1!}$$
$$T_2=\frac1{0!}+3\frac1{1!}+\frac1{2!}$$
$$T_3=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Dot Product Intuition I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ).
For example, suppose I have vector $\mathbf{a} = (a_1,a_2)$ and vector $\mathbf{b}=(b_1,b_2)$. What's the ... | By the scalar product (synonymously, the dot product) of two vectors $\mathbf{A}$ and $\mathbf{B}$, denoted by $\mathbf{A} \cdot\mathbf{B}$, we mean the quantity:
\begin{equation}
\mathbf{A} \cdot\mathbf{B} = |\mathbf{A}| |\mathbf{B}| cos(\mathbf{A},\mathbf{B})
\end{equation}
i.e., the product of the magnitudes of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/348717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "66",
"answer_count": 14,
"answer_id": 11
} |
Proving $x$ is divisible by $20$ I need to prove that $x$ divisible by $20$ if and only if $x=0\pmod4$ and $x=0 \pmod 5$
proving that if $x=0 \pmod 4$ and $x=0 \pmod 5$ than $x$ divisible by $20$ is by the Chinese theorem (am I right??)
But the other way - I dont understand why its not enough to be divisible only by on... | Hint $\rm\ \ 4,5\mid n\:\Rightarrow\: \dfrac{n}4,\dfrac{n}5\in\Bbb Z\:\Rightarrow\: \dfrac{n}4-\dfrac{n}5 = \dfrac{n}{20}\in\Bbb Z\:\Rightarrow\:20\mid n$
Edit $\ $ Regarding the new question in your edit, the above applies as follows
$\rm mod\ 4\!:\ f(n) = \color{#C00}{7^n}\!+4\cdot2^n\!+8^n\!-3^n\equiv \color{#C00}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/350670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
functional equation problem in competition: $ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $
Find all $ f : \mathbb Q \to \mathbb Q $ such that $ f ( 1 ) = 2 $ and
$$ f ( x y ) = f ( x ) f ( y ) - f ( x + y ) + 1 $$
for all $ x , y \in \mathbb Q $.
thank you very much!
| Put $y=1$. Than we have
$$f(x)=f(x)\cdot f(1)- f(x+1)+1$$
Hence
$$f(x)=2f(x)- f(x+1)+1$$
Hence
$$f(x+1)=f(x)+1$$.
So we have $f(0)=1$.
Now we say $x=-y$, we have
$$f(-x^2)=f(x)f(-x)$$
When we have
$$0=f(-1)=f(-2\cdot \frac{1}{2}) = f(-2) f(\frac{1}{2})- f(-\frac{3}{2}) +1$$
So we have
$$0= -1\cdot f(\frac{1}{2}) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the dimensions of a cuboid given the volume, surface area, and diagonal
Let the volume, surface area and length of the diagonal of a cuboid be as follows: $144$, $192$, $13$. Find the dimensions.
My trial: $$lbh = 144$$
$$2(lb+bh+lh) = 192 \implies lb+bh+lh = 96$$
$$l^2 + b^2 + h^2 = 13^2 = 169.$$
As $(l+b+h)... | $l,b$ and $h$ are the roots of cubic euqation equation :
$x^3+bx^2+cx+d=0$
$lbh= \dfrac{-d}{1}$
$lb+bh+hl=\dfrac{c}{1}$
$l+b+h= \dfrac{-b}{1}$
Plug in the values and use Vieta's formula.
Aliter:
$lbh=144$
$l^2+b^2+h^2=169$
$169 \equiv 1 \mod 4$
$l^2 \equiv 0 \mod 4, b^2 \equiv 0 \mod 4 $ and $h^2 \equiv 1 \mod 4$
$ \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/352022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $\frac{1}{2}1-\left(a_1+\frac{a_2}{2}+\cdots+\frac{a_n}{2^{n-1}}\right)$
Let $n>1$ be a positive integer and $\frac{1}{2}<a_{j}<1$ for $j=1,2,\ldots,n$. Show that $$(1-a_{1})(1-a_{2})\cdots (1-a_{n})>1-\left(a_{1}+\frac{a_{2}}{2}+\cdots+\frac{a_{n}}{2^{n-1}}\right).$$
I have no ideas.
| Let $b_i=1-a_i$, so that $0<b_i<\frac{1}{2}$. The inequality becomes:
$$b_1b_2 \ldots b_n>1-((1-b_1)+\frac{1-b_2}{2}+\ldots+\frac{1-b_n}{2^{n-1}})$$
$$1-\frac{1}{2^{n-1}}+b_1b_2 \ldots b_n>b_1+\frac{b_2}{2}+\ldots+\frac{b_n}{2^{n-1}}$$
We first prove that $\frac{1}{2}+b_1b_2>b_1+\frac{b_2}{2}$. This is equivalent to $(... | {
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"source": "stackexchange",
"question_score": "1",
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Solving for three unknowns given three linear equations Have to admit, I forgot how to do simple 9th grade algebra problems. I need to know how to do this stuff for an exam tomorrow. I am doing recurrences and the only thing I need to do is find $a, b$, and $c$ from the following
$$\begin{align*}
a + b + c &= 3\\
4a +... | Just use Gaussian elimination.
$a+b+c=3$
$4a+2b+c=6$
$9a+3b+c=13$
Subtracting equation 1 from equation 2 and equation 3:
$a+b+c=3$
$3a+b+0c=3$
$8a+2b+0c=10$
Subtract 2 times equation 2 from equation 3.
$a+b+c=3$
$3a+b+0c=3$
$2a+0b+0c=4$
We can see that $a=2$
Substitute this into the other equations
$2+b+c=3$
$3(2)+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2 } \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $
Evaluate
$$\int _{ 0 }^{ 1 }{ \left( { x }^{ 5 }+{ x }^{ 4 }+{ x }^{ 2
} \right) \sqrt { 4{ x }^{ 3 }+5{ x }^{ 2 }+10 } \; dx } $$
The question look's like there is a nice method to do ... | $\displaystyle{\large%
\int_{0}^{1}
\left(x^{5} + x^{4} + x^{2}\right)\,
\sqrt {4x^{3} + 5x^2 + 10\;}\;{\rm d}x\quad:{\Huge ?}}$
Following @jdh8 ( $\color{#0000ff}{\mbox{There is a missing}\; \color{#ff0000}{\large 3/2}\
\mbox{factor in his formula}}$ ).
$P \equiv a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}$
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358452",
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"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$ Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$
I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).
| Splitting the series into 2 yields
$$
\begin{aligned}
\sum_{n=0}^{\infty} \frac{(-1)^{2}}{(2 n+1)^{3}} &=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}-\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^{3}} \\
&=\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^{3}}+\sum_{n=-1}^{\infty} \frac{1}{(-4 n-3)^{3}} \\
&=\frac{1}{64}\left[\sum_{n=0}^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/359667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 5
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integral with $\log\left(\frac{x+1}{x-1}\right)$ I encountered a tough integral and I am wondering if anyone has any ideas on how to evaluate it.
$$\displaystyle \int_{0}^{\infty}\frac{x}{x^{2}+a^{2}}\log\left(\left|\frac{x+1}{x-1}\right|\right)dx=\pi\tan^{-1}(1/a), \;\ a>1$$
I tried breaking up the log and differentia... | This is a challenging integral with unexpected twists and turns in its evaluation. Ultimately, though, it all works out.
We begin by integrating by parts, something we normally have no business doing, but the infinities cancel:
$$\int_0^{\infty} dx \frac{x}{x^2+a^2}\log{\left(\frac{x+1}{x-1}\right)} = \\ \lim_{N \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Finding a solution to the recurrence relation $a_n = 5a_{n−2} − 4a_{n−4}$ Find the solution to $$a_n = 5a_{n−2} − 4a_{n−4}$$ with $$a_0 = 3$$
$$a_1 = 2$$ $$a_2 = 6$$ $$a_3 = 8$$
My answer: Observe that the degree of recurrence is 4. Hence, the characteristic equation is: $x^4 - 5x^2 +4 = 0$. Solving for $y=x^2$ we ge... | Just for kicks, use generating functions on this one. Define $A(z) = \sum_{n \ge 0} a_n z^n$. Your recurrence is:
$$
a_{n + 4} = 5 a_{n + 2} - 4 a_n \quad a_0 = 3, a_1 = 2, a_2 = 6, a_3 = 8
$$
Using properties of ordinary generating afunctions:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2 - a_3 z^3}{z^4}
= 5 \frac{A(z) - a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Need help solving an Integral Equation Need help solving:
$$
f(x) = x + \lambda \int_{0}^{1}y(x+y)f(y)dy
$$
keeping terms through $\lambda^{2}$,
(a) by using the Fredholm method
(b) by using the Neumann method
| Ok, I guess I'll try to help with the Neumann method. (Hopefully it's not too late for you.) First, note that if $\lambda$ was $0$, then $f(x) = x$. If both $\lambda$ and $x$ were small, then we'd hope that $f(x) \approx x$. So let's plug $f(x)\approx x$ into the left-hand side of your equation, to get a new approximat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Writing a series using Sigma notation How do I write $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}?$$
I have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.
| Notice that $2=\frac {2x^0}1$, so you want $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\frac {2x^0}1+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\sum_{i=1}^5\frac {(i+1)x^{i-1}}{2^{i-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Graph the polynomial $f(x)=x^4-3x^3-4x^2+3x-2$ using a graphing calculator Graph the polynomial $f(x)=x^4-3x^3-4x^2+3x-2$ using a graphing calculator?
a) The Range
b) The Real Zeros
c) The y-intercept
d) relative minimum and Relative Maximum
e) The interval where $f(x)\le0$
| With the help of the calculator is easy to establish:
$$
\begin{array}{c|ccccc} a & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline \\ f(a) & 115 & 16 & -5 & -2 &-5 & -20 & -29 & 10\end{array}
$$
This suggests the minimum is close to $-29$. In fact it is $-29.4136$. For very large $|x|$, the fourth power will dominate ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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prove a sequence is bounded above. I want to prove that $x_n=\left(1-\frac{1}{n}\right)^n$ is bounded above.
I know that $b_n=\left(1+\frac{1}{n}\right)^n$ is bounded above, because $b_n\rightarrow \mathcal e$. Indeed, it can be proved that $b_n<3$. Since $b_{n-1}=\frac{n^{n-1}}{(n-1)^{n-1}}$, $x_n=\frac{(n-1)^n}{n^n}=... | If $n\ge 1$, you have $0 \le 1-\frac{1}{n} \le 1+\frac{1}{n}$. Hence $(1-\frac{1}{n})^n \le (1+\frac{1}{n})^n$.
Indeed, $(1-\frac{1}{n}) (1+\frac{1}{n}) = 1-\frac{1}{n^2} \le 1$. So you have $(1-\frac{1}{n})^n (1+\frac{1}{n})^n \le 1$, or $(1-\frac{1}{n})^n \le \frac{1}{(1+\frac{1}{n})^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/366484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Integral of $\arccos$ within $\sin$ How to calculate the following integral:
$\int^{R}_{0}[2 \cos^{-1}(\frac{r}{2R}) -\sin(2 \cos^{-1}(\frac{r}{2R}) ) ] dr$.
This is a part of a complex formula.
| You can use substitution to solve this problem. Clearly the substitution is $u=\cos^{-1}\left(\frac{r}{2R}\right)$ and hence $\cos u=\frac{r}{2R}$. Then $r=0\to u=\frac{\pi}{2},r=R\to u=\frac{\pi}{3}$ and $dr=-2R\sin u$. Thus
\begin{eqnarray*}
I&=&-2R\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}(2u-\sin(2u))\sin udu\\
&=&2R\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the last digit of $3^{729}$ I'm practicing for my algebra exam but I stumbled on a question I don't know how to solve.
Let $N = 3^{729}$. What is the last digit of $N$?
The example answer says
Since $\gcd(3, 10) = 1$, check that
$3^4 = 81 = 1 \pmod {10}$:
Now, $729 = 182 \times 4 + 1,$ so we get we get... | $3^0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 1\equiv3^0\mod 10$
$3^1\equiv 3\times 1\equiv 3\equiv3^1\mod 10$
$3^2\equiv 3\times3\equiv 9\equiv3^2\mod 10$
$3^3\equiv 3\times9\equiv 7\equiv3^3\mod 10$
$3^4\equiv 3\times7\equiv 1\equiv3^0\mod 10$
$3^5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \equiv 3^1\mod 10$
$3^6\ \ \ \ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 6,
"answer_id": 3
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Prove that $3\sum\limits_{i=0}^k\binom{n}{3i}\leq2^n+2$ If $n\in \mathbb{Z^+}$ and $k$ is the largest integer for which $3k\leq n$, then is it true that $\sum_{i=0}^k\binom{n}{3i}\leq \frac{1}{3}(2^n+2)$?
My work: We can break this into two cases: $n=3k+1$ and $n=3k+2$.If $n=3k+1$, then we need to prove that $$\sum_{i=... | [This is essentially the same as Did's, but explicitly avoids the discrete Fourier Transform.]
We use the binomial theorem, which states that
$$ (x + 1 ) ^ n = \sum_{i=0}^n { n \choose i} x^ i $$
Since we want just those values where $i$ is a multiple of 3, we will use the fact that for $\omega \neq 1$ a cube root of 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/371551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Finding generating function for the recurrence $a_0 = 1$, $a_n = {n \choose 2} + 3a_{n - 1}$ I am trying to find generating function for the recurrence:
*
*$a_0 = 1$,
*$a_n = {n \choose 2} + 3a_{n - 1}$ for every $n \ge 1$.
It looks like this:
*
*$a_0 = 1$
*$a_1 = {1 \choose 2} + 3$
*$a_2 = {2 \choose 2} + 3... | As I said in my comment, you have:
$$\displaystyle a_n=3^n+\sum_{k=0}^{n-1}3^k{n-k \choose 2}$$
You can rewrite this as:
$$\displaystyle a_n=3^n+\sum_{k=1}^{n}3^{n-k}{k \choose 2}=3^n+3^n\sum_{k=1}^{n}3^{-k}{k \choose 2}=3^n\left(1+\sum_{k=1}^{n}\left(\frac{1}{3}\right)^{k}{k \choose 2}\right)$$
Also you have:
$$\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Stuck on Epsilon proof.. Using the $\epsilon-M $ definition of the limit, calculate
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$
Working so far:
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$
Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$
$$\left|\frac{3x^2+7}{x^2+x+8}... | Let $x>\max\{\sqrt{17},\frac{3}{\epsilon-1}\}$. We have $x>0$. This gives us
$$\left|\frac{-3x-17}{x^2+x+8} \right|= \left|\frac{3x+17}{x^2+x+8} \right|=\frac{3x+17}{x^2+x+8}<\frac{3x+17}{x^2}<\frac{3}{x}+1<\epsilon$$
This is incomplete as it assumes $\epsilon\neq 1$. When $\epsilon=1$ you can solve a quadratic and fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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(4x^2+2kx-5)/(x+2) remainder is 3 find value of k? 2 methods - first is long division by $(x+2)$, 2nd is to use remainder theorem
let $f(x) = 4x^2+2kx-5$ and $g(x) = x+2$
to find the remainder of $\frac{f(x)}{g(x)}$ where $g(x) = (x+c)$ we need to evaluate $f(-c)$
$f(-2) = 4(-2)^2+2k(-2) -5$
Because the remainder is 3,... | $4x^2+2kx-5=(x+2)2x+x(2k-4)-5$
$=(x+2)4x+(x+2)(2k-8)-2(2k-8)-5$
$4x^2+2kx-5=(x+2)(2x+2k-8)+11-4k$
Alternatively, putting $x=-2,4(-2)^2+2k(-2)-5=11-4k$ which we have obtained by the long division
So, $11-4k=3\implies k=2$
So, both of the methods are correct. In fact, they are not independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/375469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$ To split in to partial fractions, the expression $\frac{1}{x^2(x+a)^2}$
$\frac{1}{x^2(x+a)^2}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{(x+a)}+\frac{D}{(x+a)^2}$
One method of finding the values of the constants A,B,C & D is as follows.
$Ax(x+a)^2+B(x+a)^... | In this equation:
$$Ax(x+a)^2+B(x+a)^2+C(x^2)(x+a)+Dx^2=1
$$
one little thing you can do is to evaluate at $x = -a$ to get $0 +0 +0 +D(-a)^2 = 1$ to get $D$. After that can evaluate at $x=0$ to get $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Expand the trinomial $(x+y+z)^4$ using the Multinomial Theorem
Use the multinomial theorem to expand $(x+y+z)^4$.
To calculate the number of terms, you apply the following formula: $\binom{n+r-1}{n}$. Here $n=4$ and $r=3$. So $\binom{6}{4}=15$. I don't understand how they are getting $15$ terms.
The Multinomia... | Perhaps the other three terms that you are missing correspond to (1,1,2)?
| {
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"url": "https://math.stackexchange.com/questions/384831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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It is an easy question about integral,but I need your help. How to compute this integral?
$$ \int^{\pi}_{0} \frac{\sin(nx)\cos\left ( \frac{x}{2} \right )}{\sin \left ( \frac{x}{2} \right ) } \, dx$$
I need your help.
| Hint:
$\int_0^{\pi}\dfrac{\sin nx . \cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}=\int_0^{\pi}\dfrac{\sin n(\pi -x) . \sin \dfrac{x}{2}}{\cos\dfrac{x}{2}}=1$
$\sin (n \pi-nx)=\sin n{\pi} \cos nx-\cos{n \pi}\sin{nx}= (-1)^{n+1}\sin nx$,
$2I=\int_0^{\pi}\dfrac{\sin nx}{\cos \dfrac{x}{2} \cdot \sin \dfrac{x}{2}}$, when $n \in$ Ev... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}=2$
Find $\displaystyle\sqrt[3]{2+\frac {10} 9\sqrt 3}+\sqrt[3]{2-\frac {10} 9\sqrt 3}$.
I found that, by calculator, it is actually $\bf{2}$.
Methods to denest something like $\sqrt{a+b\sqrt c}$ seems to be useless here, what should I do?
| Let $x=\sqrt[3]{2+\frac {10}{9}\sqrt 3}, y=\sqrt[3]{2-\frac {10}{9}\sqrt 3}$. Then
$$
x^3+y^3=4,\quad xy=\frac{2}{3}.
$$
The rest is yours.
| {
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"url": "https://math.stackexchange.com/questions/386488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
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} |
Closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$. What is the closed form for $\sum_{n=-\infty}^{\infty}\frac{1}{(n-a)^2+b^2}$? We can use Fourier series of $e^{-bx}$ ($|x|<\pi$) to evaluate $\sum_{n=-\infty}^{\infty}\frac{1}{n^2+b^2}$. But this one, it seems to me it is tough to get the closed form.
| This one can also be done with the standard technique of using the $\pi \cot(\pi z)$ multiplier, integrating $$f(z) = \frac{\pi \cot(\pi z)}{(z-a)^2+b^2}$$ along a circle using the same technique as here.
We integrate $f(z)$ along a circle of radius $R$ with $R$ going to infinity and the integral disappears in the limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 4,
"answer_id": 2
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Determine the $n$-th power of matrix Determine the $n$-th power of the matrix. $$\pmatrix{2 & 2 & 0 \\ 1 & 2 & 1 \\1 & 2 & 1 \\ }$$ May you help me with the answer since the book states : $${1\over 6}\pmatrix{4+2\cdot4^n & 3\cdot4^n & -4+4^n \\ -2+2\cdot4^n& 3\cdot4^n & 2+4^n\\-2+2\cdot4^n & 3\cdot4^n & 2+4^n \\ }$... | I get for the symbolic form depending on n and the symbolic eigenvalues $\alpha,\beta,\gamma$ (where we insert later the known values $\small \alpha=0,\beta=1,\gamma=4$)
$$ A^n = \frac16 \small \begin{array} {|r|r|r|r}
4 \cdot \beta^n+2\cdot\gamma^n & -3\cdot\alpha^n+3\cdot\gamma^n & -4\cdot\beta^n+3\cdot\alpha^n+\gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Weird Identities with Scalar Product & Transpose: $\vec{a}\cdot\vec{b} = \vec{b}^T \cdot {a}^T$, $\vec{a}^T \cdot \vec{b} = \vec{b}^T \cdot \vec{a} $? Let's say I have a row vector $\vec{a}$ and a column vector $\vec{b}$:
\begin{align}
\vec{a}= \begin{pmatrix}4 & 5 & 6\end{pmatrix} \qquad \vec{b} = \begin{pmatrix}1\\2... | Vectors have a notation, they can be written only as a column matrix. In case of multiplying two such vectors, matrix multiplication is not defined as both would be 3x1 matrices.
\begin{gather}
\vec{u} = \left[ \begin{array}{c}
u_1 \\
u_2 \\
u_3 \\
\end{array} \right] & \vec{v} = \left[ \begin{array}{c}
v_1 \\
v_2 \\
v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Integration of a rational function from +/- infinity I am trying to calculate the integral
$$\int_{-\infty}^{\infty}{\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}}dx$$
where $\{a, b, c\}\in \mathbb{R}$. I have looked in a table of integrals for rational functions, but with no luck. Is there a smart trick I can utilize?... | Why don't you want to simplify it by putting the term under the integral in the following manner:
$$\frac{a+x}{b^2 + (a+x)^2}\frac{1}{1+c(a-x)^2}=\frac{\alpha +x}{\beta (c (x-a)^2+1)}-\frac{\gamma }{\delta +x}$$
with $$\alpha =\frac{4 a^2 c+a b^2 c+1}{b^2 c}$$
$$\beta =\frac{4 a^2 c+4 a b^2 c+b^4 c+1}{b^2 c}$$
$$\gamm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Why is this derivative incorrect? We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$
I would like to know why my approach is incorrect:
$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)} - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$
$$ = \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin... | Third line is: $ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)\cdot\cos^2(2x)}{\cos^2(2x) \cdot \sin^2(x)}$
instead of $\\\\ \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/392853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$
$$\lim_{R \to \infty} \int_0^R \frac{dx}{(x^2+x+2)^3}$$
Please help me in this integral, I've tried some substitutions, but nothing work.
Thanks in advance!
| First complete the square:
$$
x^2 + x + 2 = \left( x^2 + x + \tfrac{1}{4} \right) + \tfrac{7}{4} = \left( x + \tfrac{1}{2} \right)^2 + \left( \tfrac{\sqrt{7}}{2} \right)^2.
$$
Now, make the (inverse) trigonometric substitution:
$$
\tan t = \frac{x + \tfrac{1}{2}}{\tfrac{\sqrt{7}}{2}}.
$$
This choice of ratio is motivat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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$x^2+x+1$ is the cube of a prime. Please help me find all natural numbers $x$ so that $x^2+x+1$ is the cube of a prime number.(Used in here)
| This is never a complete solution, but larger than the size that a comment can afford
If $x^2+x+1=p^3$ where $p$ is prime.
Clearly, $p\ne2$
As $p$ divides $x^2+x+1,p$ will divide $(x-1)(x^2+x+1)=x^3-1$
$\implies x^3\equiv1\pmod p\implies ord_px=3\implies 3$ divides $p-1$
So, prime $p=3q+1$ for some natural number $q$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\cot^2 x$? How do you find $\int \cot^2 x \, dx$? Please keep this at a calc AB level. Thanks!
| Since the integrand $$\cot^2 x=\frac{\cos^2 x}{\sin^2 x}$$ is a rational fraction of $\sin x,\, \cos x$, you could use a universal standard substitution called the Weirstrasse substitution
$$
\begin{equation*}
\tan \frac{x }{2}=t,\qquad x =2\arctan t,\qquad dx =\frac{2}{1+t^{2}}dt
\end{equation*},
$$
which converts th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
What would be the value of $a$ and $b$ in following rational expression? If $(5 + 2\sqrt{3})/(7 + \sqrt{3}) = (a - \sqrt{3b})$,
How do I find the value of $a$ and $b$ where $a$ and $b$ are rational numbers?
| $$\frac{5+2\sqrt{3}}{7+\sqrt{3}} = a - \sqrt{3b}$$
$$5+2\sqrt{3}=(a-\sqrt{3b})(7+\sqrt{3})$$
$(a-\sqrt{3b})(7+\sqrt{3})=7a+a\sqrt{3}-7\sqrt{3b}-3\sqrt{b}$
Now we have two equotation for two unknown $a$ and $b$:
$5 = 7a-3\sqrt{b}$
and$2\sqrt{3}=(a-\sqrt{b})\sqrt{3} \implies 2=a-\sqrt{b}$.
Now
$\sqrt{b}=\frac{7a-5}{3}=a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Proving one function is greater than another How can I prove $f(x)$ $>$ $g(x)$ for all $x > 0$ given $f(x) = (x+1)^{2}$ and $g(x) = 4qx$ where $q$ is a constant in $(0, 1)$?
My approach was to show that $(x+1)^2 > 4qx$ for the interval endpoints, e.g. $q=0$ and $q=1$. E.g. $(x+1)^2 \geq 4x$ for all $x$ and $(x+1)^2 > 0... | $\begin{align}
f(x)-g(x)
&=(x+1)^2-4qx\\
&=x^2+2x+1-4qx\\
&=x^2+2(1-2q)x+1\\
&=x^2+2(1-2q)+(1-2q)^2-(1-2q)^2+1\\
&=(x+1-2q)^2+1-(1-2q)^2\\
\end{align}
$.
Since $0 <q < 1$,
$-1 < 1-2q < 1$
so $0 < (1-2q)^2 < 1$
so $1 > 1-(1-2q)^2 > 0$
so (finally)
$f(x) > g(x)$.
Another way to get this final inequality is
$1-(1-2q)^2
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Matrix Equation, Solving for Variables. I'm going through my exercises, and came across a problem that wasn't covered in our lectures. Here's the question:
$
\begin{align}
\begin{bmatrix}
a-b & b+c\\
3d+c & 2a-4d
\end{bmatrix}
\end{align}
=
$
$
\begin{align}
\begin{bmatrix}
8 & 1\\
7 & 6
\end{bmatrix}
\end{align}
$
Wha... | How familiar are you with basic linear algebra? As @response has posted, you have numerous algebra mistakes. However, as for the method used to solve the problem, I think it's worth noting that you can easily transform this system to one more familiar looking and then use standard technique of Gaussian elimination. Tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Is there any easy way to simplify the following term? $[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ]
+6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$
After the simplification , the result will be:
$3pq \cdot (p^2q + pq^2+2p^2-2pq+2q^2)$
I have tried to simplify it by using ... | Well... This bizzare thing $$[p (p+1) (p+2) (p+3) (p+q)^3] - 4[ p^2 (p+1) (p+2) (p+q)^2 (p+q+3) ]
+6[ p^3 (p+1) (p+q) (p+q+2) (p+q+3)] -3[ p^4 (p+q+1) (p+q+2) (p+q+3)]$$
Can be split like : $T_1-T_2-3T_2+3T_3+3T_3-3T_4$ , Where $T_i$ denotes the $i^{\text{th}}$ term.
$[p (p+1) (p+2) (p+3) (p+q)^3] - [ p^2 (p+1) (p+2) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/399871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve equation $\sqrt{s+13} - \sqrt{7-s} = 2$ Solve the equation
$$\sqrt{s+13}-\sqrt{7-s} = 2$$
I moved the $-\sqrt{7-s}$ to the right side
Thus, I had
$$\sqrt{s+ 13} = 2 +\sqrt{7-s}$$
I then squared both sides
$$\sqrt{s+ 13}^2 = \left(2 +\sqrt{7-s}\right)^2$$
Using the product formula $(x + y)^2 = x^2 + 2xy + y^2$
... | Clearly $7\ge s\ge-13\iff7-(-3)\ge s+3\ge-13-(-3)$
WLOG $s+3=10\cos 2y$ where $0\le2y\le\pi$
$\implies\sqrt{s+13}=\sqrt{10(1+\cos2y)}=2\sqrt5\cos y,\sqrt{7-s}=2\sqrt5\sin y$
$\implies2\sqrt5\cos y-2\sqrt5\sin y=2$
Squaring we get $$20(1-\sin2y)=4\iff\sin2y=\dfrac45\implies\cos2y=\pm\sqrt{1-\sin^22y}=\cdots$$
Check whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Solving Simple Mixed Fraction problem? How do you wrap your head around mixed fraction, does anyone knows how to figure out, can someone give me an example how it can be solved?
| Here is an example.
$$2 \frac{3}{11}=\frac{2 \cdot 11}{11}+\frac3{11}=\frac{22}{11}+\frac3{11}=\frac{25}{11}$$
Or conversely,
$$\frac{11}3=\frac{11-3 \cdot 3}{3}+\frac{3 \cdot 3}{3}=\frac{11-9}{3}+\frac{9}{3}=\frac{2}{3}+3=3 \frac{2}3$$.
In general,
$$x+\frac{y}{z}=\frac{x \cdot z}{z}+\frac{y}{z}=\frac{x \cdot z +y}z$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/400864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the given integral I have got 2 questions I can't seem to get the same answer to as the book.
31) $\displaystyle \int \sin(\sqrt x)\,\mathrm{d}x = -2\sqrt x \cos(\sqrt x)+2\sin(\sqrt x )+c$.
I just get the integral to be $\frac{1}{2}\cos(\frac{1}{x})+c$
and
36) $\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)... | Hints:
For the first one, take: $ t = \sqrt{x} $ and you get $$ 2\int{t \sin(t) dx} $$
For the second:
$$
\begin{align}
\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} &= \dfrac{1}{4} \displaystyle\int \frac{(x^2 + 4 - x^2) \ \mathrm{d}x}{x^2(4+x^2)} \\
&= \frac{1}{4} \left[ \displaystyle\int \frac{\mathrm{d}x}{x^2} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Can all points in the plane be represented like this? Solving a task regaring affine geometry, I've come across a problem:
Is it true that, for every point $(x,y)\in \mathbb{R}^2$, there exist $t\in \mathbb{R}, \alpha\in[0,2\pi]$, such that
$$x = t\cos\alpha + \sin \alpha \\ y = t\sin\alpha + \cos \alpha?$$
I don't ... | Let's play around and see what happens.
I write $a$ for $\alpha$,
$c$ for $\cos \alpha$,
and $s$ for $\sin \alpha$
cause I'm lazy.
We have
$x = t c + s$
and
$y = t s + c$.
$x^2+y^2 = t^2 c^2+2tcs + s^2 + t^2 s^2 + 2tcs + c^2
= t^2+1+4tsc$.
$xy = t^2 cs + tc^2+t s^2 + sc
= cs(t^2+1)+t$.
We can solve for $cs$ in each of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
addition with a variable (mod) Given $2+x \equiv 7 \pmod 3$.
$2 + 0 = 2$
$2 + 1 = 3$
$2 + 2 = 4$
.
.
.
$2 + 5 = 7$
so, the answer will be $x = 5, 8, 11, 14, 17,\dots$
Is this correct? Because somebody told me the answer should be $x = 2, 5, 8, 11, 14, 17,\dots$
| We are given that:
$(2+x) = 7 (\mod 3)$
Thus, $2+x-7 = 3k$ where $k$ is an integer.
Simplifying, yields:
$x-5 = 3k$
Therefore, $x\in (5,8,..)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Whether the given series are convergent or divergent: Whether the given series is absolutely convergent, conditionally convergent or divergent:
$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.
$b)\displaystyle\sum_{n=1}^{\infty}(-1... | For (b), $\frac{n}{n+2} \to 1$,
so the series is absolutely
fairly violently divergent.
The sum of consecutive even-odd terms is
$\frac{2n}{2n+2} - \frac{2n+1}{2n+3}
= \frac{2n(2n+3)-(2n+2)(2n+1)}{(2n+2)(2n+3)}
= \frac{4n^2+6n-(4n^2+3n+2)}{(2n+2)(2n+3)}
= \frac{3n-2}{(2n+2)(2n+3)}
$,
and the sum of consecutive odd-even... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Indefinite integral $\int \left(\frac{\arctan x}{\arctan x - x}\right)^3 \mathrm{dx}$ My imagination doesn't help me with
$$\int \left(\frac{\arctan x}{\arctan x - x}\right)^3 \mathrm{dx}$$
What tools should I use? W|A doesn't help either.
| Hint:
$\because0<\dfrac{\tan^{-1}x}{x}\leq1$ $\forall x\in\mathbb{R}$
$\therefore\int\biggl(\dfrac{\tan^{-1}x}{\tan^{-1}x-x}\biggr)^3~dx$
$=\int\biggl(\dfrac{\dfrac{\tan^{-1}x}{x}}{\dfrac{\tan^{-1}x}{x}-1}\biggr)^3~dx$
$=-\int\dfrac{\dfrac{(\tan^{-1}x)^3}{x^3}}{\left(1-\dfrac{\tan^{-1}x}{x}\right)^3}dx$
$=-\int\dfrac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/407142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A challenging logarithmic integral $\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx$ While playing around with Mathematica, I found that
$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$
Please help me prove this result.
| We can take $\ 2ab= a^2+b^2-(a-b)^2$ to get:
$$I= \frac12 \int_0^1 \frac{\ln^2 (1-x)}{1+x}dx+\frac12\int_0^1\frac{\ln^2(1+x)}{1+x}dx-\frac12 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx$$
By letting $\frac{1-x}{1+x}=t$ and expanding into power series the last one we get:
$$I=\frac12 J +\frac{\ln^3(1+x)}{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 9,
"answer_id": 1
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Prove inequality $(x+y+z-2xyz)^2 \le 2$
Problem: Prove inequality $(x+y+z-2xyz)^2 \le 2\ (1)$ with
$x^2+y^2+z^2 = 1 \land x,y,z \in \mathbb R$
I tried expand $LHS$ and have:
$$(1)\iff 1 - 2 (xy+yz+xz) + 4 xyz(x+y+z)-(2xyz)^2 \ge 0$$
Denote: $xy = a, yz = b,xz=c \implies (1) \iff1-2\sum a+ 4 \sum ab - 2abc \ge0$
Bu... | Expanding the $LHS$ we have
\begin{align}1+ 2(xy+ xz+ yz) - 4xyz(x+y+z) + 4x^2y^2z^2 \leq 3- 4xyz(x+y+z) + 4x^2y^2z^2
\end{align}
since $(xy+ xz+ yz) \leq (x^2+y^2+z^2)=1$ by Cauchy-Schwarz inequality. Then
\begin{align} 3- 4xyz(x+y+z) + 4x^2y^2z^2 &= 3- 4x^2y^2z^2(1/xy+1/yz+1/xz) + 4x^2y^2z^2\\& \leq 3- 32 x^2y^2z^2
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How do I find the exact value of $\cos\frac{\pi}{12}\cos\frac{5\pi}{12}\cos\frac{7\pi}{12}\cos\frac{11\pi}{12}$? I know that $\cos(6\phi)\equiv32c^6-48c^4+18c^2-1$ where $c=\cos\phi$.
I also know that when $\cos(6\phi)=0$, then $\phi=\frac{k\pi}{12}$ ($k = 1,3,5,7,9,11$).
How do I find the exact value of:
$$\cos\left(\... | HINT:
SO, the equation whose roots are $\cos\frac{r\pi}{12}$ where $r=3,9$ is
$$\left(c-\cos \frac{3\pi}{12}\right)\left(c-\cos \frac{9\pi}{12}\right)=0$$
$$\implies \left(c-\frac1{\sqrt2}\right)\left(c+\frac1{\sqrt2}\right)=0\text{ as }\cos \frac{9\pi}{12}=\cos\left(\pi- \frac{3\pi}{12}\right)=-\cos\frac{3\pi}{12}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/409396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\le\frac{2}{\sqrt{1+\sqrt{xy}}}$ Let $x,y>0$ and $xy\le 1$. Show that
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}\le\dfrac{2}{\sqrt{1+\sqrt{xy}}}.$$
This inequality have same follow methods?
I saw this.
Let $x,y>0, xy\le 1$.
$$\dfrac{1}{1+x}+\dfrac{1}{1+y}\le... | As you show
$$\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}},$$
it follows that
$$(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}})^2\le 2 (\frac{1}{1+x}+\frac{1}{1+y})\le \frac{4}{1+\sqrt{xy}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/409936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Finding Maximum Under Constraint Suppose $a$,$b$,$c$ satisfy $a+b+c=1$ and $a$,$b$,$c\in [0,1]$
Find the maximum value of $(a-b)(b-c)(c-a)$
| Another way is not to use calculus. WLOG, let $a=$MAX{$a,b,c$},then if we want max, it must >0, $→ a>c>b$, rewrite it to $\left(a-b\right)\left(c-b\right)\left(a-c\right)$, since $b\ge0$, so $0<a-b\le a,0<c-b\le c → \left(a-b\right)\left(c-b\right)\left(a-c\right)\le ac\left(a-c\right)$, then $b=0$ hold max,since $b=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
Given $x,y,z\geq0$ and $xy+yz+zx=1$. Show that $\displaystyle\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.
I've tried many things but all failed. The only thing I know is that the... | $z=\dfrac{1-xy}{x+y} \ge 0 \to xy\le 1$
LHS$=\sqrt{\dfrac{1}{x+y}}+\sqrt{\dfrac{x+y}{1+x^2}}+\sqrt{\dfrac{x+y}{1+y^2}}=f(x,y)$
$f'_{x}=\dfrac{1-x^2-2xy}{(x^2+1) \sqrt{x^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(y^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<1>
$f'_{y}=\dfrac{1-y^2-2xy}{(y^2+1) \sqrt{y^2+1} \sqrt{x+y}}+\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$ using contour integration The question is to solve the integral using concepts of contour integrals:
$$\frac{1}{2 \pi} \int_0^{2\pi} \frac{1 - r^2}{1 + r^2 - 2r \cos{t}} dt = 1$$
| Remark: the integrand $P_r(t)$ is known as the Poisson kernel and the result follows immediately from a term-by-term integration of the normally converging (when $0\leq r<1$) series $P_r(t)=\sum_{n\in\mathbb{Z}}r^{|n|}e^{in t}$ over $[0,2\pi]$. But since you want contour integration...
Hint: when $0<r<1$ use Cauchy's i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluting $\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} \operatorname dx$ $$\int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx$$
My try:: $\displaystyle \int_{1}^{2} \frac{\tan^{-1} x}{\tan^{-1} \frac {1}{x^2-3x+3}} dx = \int_{1}^{2}\frac{\tan^{-1}x}{\tan^{-1}(x-1)-\tan^{-1}(x-2)}dx$
N... | Observe that the quadratic $ x^2 -3x +3 $ has discriminant $$b^2-4ac=(-3)^2-4(1)(3)=9-12=-3$$ Hence no real solutions, which means it never crosses the $x$-axis, and since it opens up, this means it is always positive. Using this, and the following formula for $\arctan \frac{1}{u}$ $$\arctan \frac{1}{u}=\frac{1}{2}\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 2,
"answer_id": 1
} |
How find $a$ such that $x^2-\sqrt{a-x}=a$ has exactly two real solutions Consider the equation
$$ x^2-\sqrt{a-x}=a.$$ I wish to determine the values of $a$ for which the above equation has exactly two real solutions (for $x$).
My idea:
$$a-x=(x^2-a)^2=x^4-2ax^2+a^2\Longrightarrow f(x)=x^4-2ax^2+x+a^2-a=0$$ and we must ... | Hint:$x^2-a=\sqrt{(a-x)}\ge 0$.So we have $x\le a\le x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
Let $x+y+z=3,x,y,z\ge 0$,show that
$$\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$$
Additional information
I have seen the following problem:
$x,y,z>0,x+y+z=3$, prove that
$$\sqrt{x^2+y+2}+\sqrt{y^2+z+2}+\sqrt{z^... | Here is my sketch, mostly to convince that alternative approach (i.e. inequalities instead of search for minimum) really may be possible. At the other hand, useful inequalities have "differential" nature (that is, I think they may be most easily proven by differentiation, but there may be other ways to proof).
Let us ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
} |
Determinant of symmetric matrix Given the following matrix, is there a way to compute the determinant other than using laplace till there're $3\times3$ determinants?
\begin{pmatrix}
2 & 1 &1 &1&1 \\
1 & 2 & 1& 1 &1\\
1& 1 & 2 & 1 &1\\
1&1 &1 &2&1\\
1&1&1&1&-2
\end{pmatrix}
| Look at the matrix $$A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & -3 \end{pmatrix}.$$ It has rank two and its nonzero eigenvalues have sum $1$ (the trace) and product $$\det \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix} + \det \begin{pmatrix} 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
Thanks!
| This limit problem seems to cooperate rather nicely with the "conjugate factor" method. Multiply the numerator and denominator by the "conjugate" of the numerator to obtain
$$\lim_{x \rightarrow 1} \ \frac{\sqrt{x^2+3}-2}{x^2 - 1} \ \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2} $$
$$ = \ \lim_{x \rightarrow 1} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to solve this system of non-linear equations My knowledge of algebra is still insufficient to solve this problem. Any help in solving the system of equations would be greatly appreciated.
$$
xy(x+y)=30\\
x^3+y^3=35
$$
| Note that
$$
(x+y)^3
= x^3 + 3x^2y + 3xy^2 + y^3 = x^3+y^3 + 3xy(x+y)
= 35 + 3 \cdot 30 = 125
$$
and therefore $x+y = \sqrt[3]{125} = 5$ and $xy = \frac{30}{x+y} = 6$.
Thus $y = 5-x$ and $6 = xy = x(5-x)$, and so $x^2-5x+6=0$, which implies $x \in \{2,3\}$ and so $y \in \{3,2\}$. Thus, either $(x,y)=(2,3)$ or $(x,y) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/422129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x... | We have
$$\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}=\frac{\sin x^2\cos\frac{1}{x}+\sin\frac{1}{x}\cos x^2-\sin\frac{1}{x}}{x}$$
and since $\sin x^2\sim_0 x^2$ and $\left|\cos \frac{1}{x}\right|\leq 1$ and since $\cos x^2\sim_0 1-\frac{x^4}{2}$ we can see easily that the given limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/423938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Prove that $\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b} \ge a+b+c$ If $a,b,c$ are positive , show that $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$$
Trial: Here I proceed in this way $$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge \dfrac{2bc}{b+c}+\d... | We have by Titu's Lemma, for any $x,y$ reals and $a,b >0$
$ \dfrac{x^2}{a}+\dfrac{y^2}{b} \ge \dfrac{(x+y)^2}{a+b}$
$\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b} \ge a+b+c$ is quite direct. :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/424150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Linear Independence by row space.
Let $(1,1,-1), (2,1,0)$ and $(-1,0,1)$ be vectors, show if they are independent.
I wrote each vector on the rows of the matrix $A$.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -1 & 0 & 1 \end{pmatrix}$
Then I put $A$ on an echelon form.
$A=\begin{pmatrix} 1 & 1 & -1 \\ 2 & 1 & 0\\ -... | Edit: The row operations you performed and the row echelon matrix are indeed correct.
You're correct that since $r = 3 = n$ and hence, the vectors are linearly independent.
Note, since the echelon form of the matrix is now in upper triangular form, we can "read off" the determinant of the matrix in echelon form, which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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What is the general equation of the ellipse that is not in the origin and rotated by an angle? I have the equation not in the center, i.e.
$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$$
But what will be the equation once it is rotated?
| If you came here looking for how
$$c_0x^2 + c_1y^2 + c_2xy + c_3x + c_4y + c_5 = 0$$
relates to $h, k, a, b, A$ its the following:
\begin{eqnarray}
c_0&=&\frac{\cos^2(A)}{a^2} + \frac{\sin^2(A)}{b^2}\\
c_1&=&\frac{\sin^2(A)}{a^2} + \frac{\cos^2(A)}{b^2}\\
c_2&=&\frac{\sin(2A)}{a^2} - \frac{\sin(2A)}{b^2}\\
c_3&=&-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 6,
"answer_id": 1
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A constrained extremum problem Find the maximum possible value of $$A = a^{333} + b^{333}+c^{333}$$ subject to the constraints $$a+b+c=0$$ and $$a^2+b^2+c^2=1,$$
where $a,b,c\in \mathbb{R}$
Thank you for helping me.
| $a+b+c=0$, so at least two of $a,b,c$ have same sign, WLOG, let $bc \ge 0 $
$A=a^{333}+b^{333}+c^{333} = (-b-c)^{333}+b^{333}+c^{333}=- \sum_{i=1} ^{332} C_i b^{333-i}c^i$,
We want $A$ is max, so b,c must be negative, ,let $x=-b,y=-c $, $ \to x \ge 0,y \ge 0, xy \le \dfrac{1}{6}, x+y=\sqrt{\dfrac{1}{2}+xy} $
$ A=(x+y)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/428804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Prove by induction that $d_n=2^n+3^n$, where $d_n = 5d_{n-1}-6d_{n-2}$ I have one more induction question.
$d_0 =2 $ $d_1=5$
let $d_n=5d_{n-1} - 6d_{n-2}$
Prove that $d_n=2^n+3^n$
| Assuming $d_0=2$ and $d_1=5$ (which also starts the induction, incidentally):
let $n \geq 2$, and suppose the property holds for all $0\leq k< n$. Then
$$
\begin{align*}
d_{n} &= 5d_{n-1} - 6d_{n-2} \\
&= 5\left(2^{n-1}+3^{n-1}\right) - 6\left(2^{n-2}+3^{n-2}\right) \qquad\qquad\text{(induction hypothesis)} \\
&= 5\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $w^2 + x^2 + y^2 = z^2$, then $z$ is even if and only if $w$, $x$, and $y$ are even I'm trying to go through the MIT opencourseware Mathematics for Computer Science (6.042J). I've been stumped for half a day trying to figure it out. Something isn't clicking, and I could use some help.
Here is the problem:
Suppose t... | First of all, the sum of a set of integers will be even iff the number of odd summands is even(may be $0$).
So, here either none or two of $w,x,y$ are odd.
If none of them is odd, we are done.
If two of them are odd, say $w=2W,x=2X+1,y=2Y+1$ where $W,X,Y$ are integers.
$w^2+x^2+y^2=(2W)^2+(2X+1)^2+(2Y+1)^2=4(W^2+X^2+X+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/430554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Evaluation of $\lim\limits_{n\to\infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2}) $ Could you, please, check if I solved it right.
\begin{align*}
\lim_{n \rightarrow \infty} (\sqrt{n^2 + n} - \sqrt[3]{n^3 + n^2})
&= \lim_{n \rightarrow \infty} \sqrt{n^2(1 + \frac1n)}
- \sqrt[3]{n^3(1 + \frac1n)})\\
&= \lim_{n \rightarrow ... | Forget your last term and Just set $x = 1/n$ in the second
The first term of the series is $x/2 - x/3=x/6.$
Since $x=1/n,$ the result is just $1/6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Calculate residue at essential singularity I know you can calculate a residue at an essential singularity by just writing down the Laurent series and look at the coefficient of the $z^{-1}$ term, but what can you do if this isn't so easy?
For instance (a friend came up with this function): what is the residue at $z = 0... | We can look at the power series $$\frac{1}{\frac{1}{z} - 3} = \frac{z}{1 - 3z} = z + 3z^2 + ...$$ and $$\sin(z) = z - \frac{1}{6}z^3 + ...$$ so $$\frac{\sin(z)}{\frac{1}{z} - 3} = \Big(z + 3z^2 + ...\Big)\Big(z - \frac{1}{6}z^3 +...\Big) = z^2 + 3z^3 + ...$$ and $$\frac{\sin(\frac{1}{z})}{z - 3} = \frac{1}{z^2} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int \frac{\mathrm dz}{z^3 \sqrt{z^2 - 4}}$ $$\int \frac{dz}{z^3 \sqrt{z^2 - 4}}$$
$z = 4\sec\theta$
$dz = 4\sec\theta \tan d\theta$
$$\int \frac{\sec\theta \tan\theta}{4^3 \sec^3 \theta \tan \theta}$$
$$ \frac{1}{4^3} \int \frac{d \theta}{\sec^2 \theta}$$
I am again stuck, I have no idea how to proceed.
| Putting $z^2-4=u^2, zdz=udu$
$$\int \frac{dz}{z^3\sqrt{z^2-4}}=\int \frac{udu}{(u^2+4)^2u}=\frac{du}{(u^2+2^2)^2}$$
This can be addressed using Trigonometric identities $u=2\tan\theta$ like my other answer
Alternatively, $$\text{Let }I_n=\int \frac {du}{(u^2+a^2)^n}$$
$$=\int du\cdot\frac1{(u^2+a^2)^n}-\int\left(\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why such function does not exist? I could not prove the following:
A function $f \in \mathscr{C}^2([0, \pi])$, such that $$f(0) = f(\pi) = 0,\\
\int_0^{\pi} (f'(x))^2dx = 1,\\
\text{and }\int_0^{\pi} (f(x))^2dx = 2$$
Then such function does not exist.
I think that I have to use the Rayleigh quotient and have a co... | This was meant as a comment to O.L.'s answer, but it was too long:
Here is an array of the four values of $(a,b)$ for $f(x)=(a+bx)x(\pi-x)$ that satisfy the given conditions:
$$
\left(
\begin{array}{cc}
\frac{\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}-\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi
^5} , -\frac{\sqrt{\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Given $B \in M_{n\times n}(\mathbb R)$ is invertible and $B^2+B^4+B^7 = I$, find an expression for $B^{-1}$ in terms of only $B$. Given $B \in M_{n\times n}(\mathbb R)$ is invertible and $B^2+B^4+B^7=I$, find an expression for $B^{-1}$ in terms of only $B$.
I don't know where to start. Thanks in advance.
| Since $B^2 + B^4 + B^7 = I$, we have $B(B + B^3 + B^6) = I$; thus $B^{-1} = B + B^3 + B^6$. Note also that since $B^7 + B^4 + B^2 - I = 0$, any eigenvalue $\lambda$ of $B$ must satisfy $\lambda^7 + \lambda^4 + \lambda^2 - 1 =0$, hence cannot be $0$; thus the hypothesis that $B$ is invertable is extraneous, given that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/436977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to find the limits $\lim\limits_{h\rightarrow 0} \frac{e^{-h}}{-h}$ and $\lim\limits_{h\rightarrow 0} \frac{|\cos h-1|}{h}$? How to work around to find the limit for these functions :
*
*$$\lim_{h\rightarrow 0} \frac{e^{-h}}{-h}$$
*$$\lim_{h\rightarrow 0} \frac{|\cos h-1|}{h}$$
For the second one i think that ... | We know that $e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ Note that for $|h|<1$ and sufficiently small,
$$
\frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1
\leq
e^{-h}
\leq
\frac{1}{0!}(-h)^0+\frac{1}{1!}(-h)^1+\frac{1}{2!}(-h)^2
$$
which is the same as
$$
1-h
\leq
e^{-h}
\leq
1-h+\frac{1}{2}h^2
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Wolfram double solution to $\int{x \cdot \sin^2(x) dx}$ I calculated this integral :
$$\int{x \cdot \sin^2(x) dx}$$
By parts, knowing that $\int{\sin^2(x) dx} = \frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x) +c$. So I can consider $\sin^2(x)$ a derivative of $\frac{1}{2} \cdot x - \frac{1}{4} \cdot \sin(2x)$, and I... | By the double angle identity for cosine, they differ by a constant. Add a constant of integration $d$ to your answer, and the two answers will be equivalent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/437681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.
My (strange) proof:
$$
\begin{align*}
a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\
\sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\
\sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\
a^2+b... | (@HaiDangel told me.
https://diendantoanhoc.net/topic/182934-a3-b3-c3geqq-a2b-b2c-c2a/?p=731023)
A stronger version: Let $a, b, c$ be real numbers with $a + b \ge 0, b + c \ge 0$ and $c+a\ge 0$. Prove that
$$a^3 + b^3 + c^3 \ge a^2b + b^2c + c^2a.$$
I have an SOS expression:
\begin{align}
&a^3 + b^3 + c^3 - a^2b - b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Trigonometric problem : Eliminate $\theta$ and $\phi$ from the relation and find relation between p and q Question :
Eliminate $\theta$ and $\phi$ from the relation
$$\begin{align}
p \cot^2\theta + q \cot^2\phi &= 1 &(1)\\
p \cos^2\theta + q \cos^2\phi &= 1 &(2)\\
p \sin\theta &= q\sin\phi &(3)
\end{align}$$
Also f... | HINT:
$$p\sin\theta=q\sin\phi\implies p^2\sin^2\theta=q^2\sin^2\phi$$
$$\implies p^2\cos^2\theta-q^2\cos^2\phi=p^2-q^2\ \ \ \ (4)$$
Use $(2),(4)$ to solve for $\cos^2\phi,\cos^2\theta$
If $\cos^2\phi=y, \cot^2\phi=\frac{\cos^2\phi}{1-\cos^2\phi}=\frac y{1-y}$
Put the values of $\cot^2\phi,\cot^2\theta$ in $(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/438788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do I write a sum of cosines as a product of sines? I am trying to prove that
$$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle.
I tried up to the stage of
$$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$
but how do I proceed from here?
| Recall the identities
$$
\begin{align}
2\sin(x)\sin(y)&=\cos(x-y)-\cos(x+y)\tag{1}\\
2\sin(x)\cos(y)&=\sin(x-y)+\sin(x+y)\tag{2}
\end{align}
$$
First use $(1)$ on $2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)$, then $(2)$ on the results. Finally, use $A+B+C=\pi$.
$$
\begin{align}
&4\sin\left(\frac{A}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/439523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum\limits_{k=0}^{n-1}\dfrac{1}{\cos^2\frac{\pi k}{n}}=n^2$ for odd $n$ In old popular science magazine for school students I've seen problem
Prove that $\quad $
$\dfrac{1}{\cos^2 20^\circ} +
\dfrac{1}{\cos^2 40^\circ} +
\dfrac{1}{\cos^2 60^\circ} +
\dfrac{1}{\cos^2 80^\circ} = 40. $
How to prove mo... | Whenever I see a problem like this, I think Chebyshev polynomials.
The Chebyshev polynomials of the first kind, $T_m(x)$, are defined so that:
$$T_m(\cos \theta) = \cos m\theta$$
$T_m$ is an $m$th degree polynomial, and the roots of $T_m(x)-1$ are exactly $\cos 2\pi k/m$ for $k=0,1,\dots,m-1$.
When $m=2n$ is even, we c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 5,
"answer_id": 4
} |
Partial fractions for $\frac{t+1}{2\sqrt{t}(t-1)}$ How do I use partial fractions for the expression $\dfrac{t+1}{2\sqrt{t}(t-1)}$? Because I have to find the integral of it... Thank you
| Your given integrand isn't a rational function (yet) to use partial fractions, we must first obtain a ratio of polynomials. We can do this by substituting $u = \sqrt t$.
$$u = \sqrt t \implies t = u^2 \implies dt = 2u\,du$$
Now, substituting the above into our original integral gives us:
$$\int \frac{t+1}{2\sqrt{t}(t-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Maximize $(a-1)(b-1)(c-1)$ knowing that : $a+b+c=abc$. If : $a,b,c>0$, and : $a+b+c=abc$, then find the maximum of $(a-1)(b-1)(c-1)$.
I noted that : $a+b+c\geq 3\sqrt{3}$, I believe that the maximum is at : $a=b=c=\sqrt{3}$. (Can you give hints).
| $a=\dfrac{b+c}{bc-1}>0 \to bc>1 \implies $ two of ${a,b,c} >1 $, otherwise ,one will be negative.
WOLG, let $a>1,b>1$. if $c\le 1, \implies (a-1)(b-1)(c-1) \le 0$ so the possible max value will be got when $c>1$.
$x=a-1>0,y=b-1>0,z=c-1>0, x+y+z+3=xyz+x+y+z+xy+yz+xz+1 \to xyz+xy+yz+xz=2$
$xy+yz+xz \ge 3(xyz)^{\frac{2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/445293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to solve $x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$ solve the equalition
$$x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$$
I have seen some methods, for this problem.
my idea:
let $\sqrt[3]{x^3+x+1}=y,$ then
$\sqrt[3]{y^3+y+1}+y=-x-3$
and following have nice methods? t... | Let $y=\sqrt[3]{x^3+x+1}$ and $z=\sqrt[3]{y^3+y+1}$, then we translate the original equation into$$x+y+z+3=0$$ by following the procedure in the original post. We also have, by direct calculation:
$$z^3=y^3+y+1=x^3+x+1+y+1$$ so that $$z^3+z+1=x^3+(x+y+z+3)=x^3$$ So we have the suggestive equations$$y^3=x^3+x+1$$$$z^3=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/447922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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.