Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Minimal value of the product $(x+y)(y+z)$
Find the min. value of the product $\left(x + y\right)\left(y + z\right)$ given that $xyz\left(x + y + z\right) = 1$ and $x,y,z > 0$.
I want to solve this problem using geometry of a triangle:
*
*Let $a,b,c$ be the sides of the triangle and we can substitute
$a = x + y\,,\,... | Let $a=x+y, \, b=y+z ,\, c = z+x, \, p = \frac{a+b+c}{2},$ then
$$(x+y)(y+z) = ab,$$
and
$$1 = \sqrt{xyz(x+y+z)} = \sqrt{p(p-a)(p-b)(p-c)} = S = \frac{1}{2} ab \sin C.$$
But $\sin C \leqslant 1,$ so
$$2 = ab \sin C \leqslant ab.$$
Therefore $ab \geqslant 2.$ Equality hold for $x = 1, \, y = \sqrt 2 -1, \, z = 1.$
P/s.... | {
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"url": "https://math.stackexchange.com/questions/3995870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that $x_{2017} + x_{2023} = 990$ ----- (1) and $x_{n+1} = 3x_n + \... | $$x_{n+1} = 3x_{n} + \sqrt{8x_{n}^2+2}$$
$$x_{n+1}-3x_{n} = \sqrt{8x_{n}^2+2}$$
By squaring both sides,
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=2$$
Since this is a general expression, we can write -
$$x_{n+1}^2+x_{n}^2-6x_{n}x_{n+1}=x_{n}^2+x_{n-1}^2-6x_{n-1}x_{n}$$
Simplifying this we get,
$$x_{n+1}+x_{n-1}=6x_{n}$$
Now our... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Alternative approaches to prove the following inequality For $a,b,c \in \mathbb{R^+},$ prove that
$$\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3.$$
I managed to prove this problem using the technique of isolated fudging. In ... | We can assume that $a+b+c = 1$ and so with $$ f(x) = \left(\frac{2x}{1-x}\right)^{2/3}$$ the inequality becomes $$ f(a) + f(b) + f(c) \geq 3.$$ Now observe that $f(x) \geq 3x$ for $x \in [0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Prove that $\frac{a^{2}+1}{b}+\frac{b^{2}+1}{a} \geq 4$ $a$ and $b$ are both positive Real Numbers .
I saw this in a math olympiad, and within two days i couldn't solve it.
There is a simple case where:
$$\frac{a^{2}+1}{a}+\frac{b^{2}+1}{b} \geq 4 $$
We can solve it like this:
$$(a-1)^{2}\geq 0$$
$$\Leftrightarrow a^{2... | I would start with noticing that $a^{2} + 1 \geq 2\sqrt{a^{2}} = 2a$.
Similarly, we have that $b^{2} + 1 \geq 2b$.
Then we have that
\begin{align*}
\frac{a^{2} + 1}{b} + \frac{b^{2} + 1}{a} \geq 2\left(\frac{a}{b} + \frac{b}{a}\right) \geq 4\sqrt{\frac{a}{b}\times\frac{b}{a}} = 4
\end{align*}
Hopefully this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$ without using L'Hospital or Taylor's series Find limit without using L'Hospital or Taylor's series:
$$\displaystyle \lim_{x \to 0} \frac{\ln(1+\sin(2x^5))}{\tan^3(3x) \cdot (5^{x^{2}} - 1)}$$
$$\displaystyle \lim_{x \to 0} \frac{... | You can do it using equivalents and some standard limits: rewrite the fraction as
$$\frac{\ln(1+\sin(2x^5))}{\sin(2x^5)}\frac{\sin(2x^5)}{\tan^3(3x)} \cdot \frac{x^2}{(5^{x^{2}} - 1)}\cdot\frac1{x^2}.$$
By substitution, the first factor tends to $1$. Also by substitution, $\:\dfrac{5^{x^2}-1}{x^2}$ tends to the deriva... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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check if a function is holomorphic Let $f:\mathbb{C} → \mathbb{C}$ be defined as
$$f(z)= \begin{cases}\frac{x^3y(y − ix)}{x^6 +y^2}&\text{ if }z = x + iy= 0\\ 0 &\text{ if }z=x+iy=0\end{cases}$$
And holomorph at $z = 0$
I tried to apply the definition of holomorphic application, which is as follows:
$\displaystyle \lim... | Set $f=u + iv$, so $u=\frac{xy^2}{x^4-1}$ and $v=\frac{-x^2y}{x^4-1}$.
calculating the partial derivatives we have:
$$\frac{\partial u}{\partial x} = \frac{y^2(x^4-1)-xy^2(4x^3)}{(x^4-1)^2}$$
And
$$\frac{\partial v}{\partial y}= \frac{-2xy(x^4-1)-(-x^2y)(4x^3)}{(x^4-1)^2}$$
as the first equation is already different, i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3998716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do we prove the proposed expression is nonnegative? So the expression which I am interested in is given by
\begin{align*}
f(b,\theta) = \frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}
\end{align*}
where $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ ... | Denote $A = (1-\theta)(2b-1)^2 + \theta$.
Since $(2b-1)^2 \le 1$, we have $A \ge (1-\theta)(2b-1)^2 + \theta (2b-1)^2 = (2b-1)^2$.
We have
\begin{align}
f(b, \theta) &= \frac{2b-1}{b\sqrt{A}} + \frac{\sqrt{A} - 1}{2b^2(\theta - 1)} \\
&= \frac{2b-1}{b\sqrt{A}} + \frac{A - 1}{2b^2(\theta - 1)(\sqrt{A} + 1)}\\
&= \frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$
How to evaluate:
$$\int_{0}^{1} \frac{\ln(1 + x + x^2 + \ldots + x^n)}{x}\mathrm d x$$
Attempt:
$$\int_{0}^{1}\frac{\ln(1 + x + x^2 + \ldots + x^n)}{x} \mathrm dx
= \int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\mathrm d x$$
Any hints ... | I derive under the integral
$$f(n)=\int_{0}^{1}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx$$
$$f'(n)=\int_{0}^{1}\frac{\partial}{\partial n}\frac{\ln(1 -x^{n+1}) - \ln(1 - x)}{x}\,dx=$$
$$=\int_0^1\frac{x^n \log x}{x^{n+1}-1}\,dx=\frac{\pi ^2}{6 (n+1)^2}$$
$$f(n)=\int\frac{\pi ^2}{6 (n+1)^2}\,dn=C-\frac{\pi ^2}{6 (n+1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
} |
Calculate integral $\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx$.
Calculate integral $$\int_0^{\pi/2} \frac{\cos^3x}{\sin^2x + \cos^3x}dx.$$
My direction: Since this integral can't calculate normally, I tried to use the property following:$$\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx.$$
Then, I have
$$I=\int_0... | Rewrite the integral with $t=\tan\frac x2$
$$I=\frac\pi2-\int_0^{\pi/2} \frac{\sin^2x}{\sin^2x + \cos^3x}dx
=\frac\pi2 +8 \int_0^1 \frac{t^2}{t^6-7t^4-t^2-1} dt$$
Note that $t^6-7t^4-t^2-1$ is cubic in $t^2$, with one real root $r=7.159$ (analytically solvable with the Cadano’s formula). Then, factorize
$$ t^6-7t^4-t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
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Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denomin... | HINT.-Use the identity $A^3-B^3=(A-B)(A^2+AB+B^2)$ with $A=\sqrt[3]{11-x}$ and $B=x-1$ and multiplying numerator and denominator by $(A^2+AB+B^2)$ you' ll get again the undeterminated form $\dfrac{(4-4)(4+4+4)}{8-8}$. But the new denominator being equal to $11-x-(x-1)^3$ you have its derivative is equal to $-13\ne0$ so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$. It is given that we want to write five digits numbers using $0,0,0,1,1,2,2,2,2,3,3,3,4,4$.
How many ways are there to write it $?$
My thought In the first place, it seems cumbersome question a little bit , because i think that we should write them separatel... | Here is a method which still requires a bit of work, but which is still more efficient then naive brute force.
Start with a simpler question:
For each $k=1,2,\dots,5$, how many $k$ digit numbers can you make using $3, 3, 4, 4, 4$?
With a little work, the answers are
k
1
2
3
4
5
# numberss with k digits using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Binomial distribution closed by linear operations? I understand that normal distributions are closed by linear operations, but is binomial distribution as well?
In other words, if $X$ is a binomial random variable, does $aX + b$ follow a binomial distribution for any real number $a$ and $b$?
| Is Y=aX+b binomial for all real a,b? No, if a=2 then the support of Y is 0,2,4,...,2n which is not binomial.
Do the values of Y follow "binomial probabilities"? Yes. If $X\sim Bin(n,p), Y=aX+b$, then the support of Y is $aX+b, X=0,1,...,n$ and the pmf is $f(Y)={n\choose \frac{y-b}{a}}p^{\frac{y-b}{a}}(1-p)^{1-\frac{y-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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show $(1+\frac{x}{n})^n \leq e^x$ I want to show that $(1+\frac{x}{n})^n \leq e^x$ for all $x \geq -n$, $n\in \mathbb{N}$.
I already showed that positive case ($x>0$) and zero case $(x=0)$.
Not sure how to approach the negative case, because in the case $x>0$ I used that $x^n$ is an increasing function on $(0, \infty)$... | Assuming you know that
\begin{align}
\lim_{n\rightarrow \infty}\left( 1+ \frac{x}{n}\right)^n = e^x,
\end{align}
then it suffices to prove that
\begin{align}
\left( 1+ \frac{x}{n}\right)^n \le \left( 1+ \frac{x}{n+1}\right)^{n+1}
\end{align}
for all $n \in \mathbb{N}$.
By Bernoulli's inequality, we have that
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$ How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... ... | General term is:
$$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$$
Now we give values 1, 3,... n :
$1\times2\times3\times4=1^4+6\times1^3+11\times 1^2+6\times 1$
$3\times4\times5\times6=3^4+6\times3^3+11\times 3^2+6\times 3$
.
.
.
$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$
Now we sum up:
$A=S_4+6S_3+11S_2+6S_1$
We use well known formu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4025455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the total length of two line segments of two overlapping right triangles
Find the length of $AE+EB$ ?
(A) $\frac{128}{7}$
(B) $\frac{112}{7}$
(C) $\frac{100}{7}$
(D) $\frac{96}{7}$
(E) $\frac{56}{7}$
My solution:
For $\Delta AEB$ :
$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$
$\angle ABD = sin^{-1}(\frac... | Yes there is, drop a perpendicular down from $E$ to $AB$ and denote the point as $X$. Let $EX = x$ and $AX = 12-y$ and $BX = y$. Since $\triangle EBX $ ~ $\triangle DBA$, we have that $\frac9x = \frac{12}y$. Also, since we have $\triangle CBA$ ~ $\triangle EXA$, we have that $\frac5x = \frac{12}{12-y}$. Now, solving f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$ $$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$
I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?
Thanks!
| Cubing both sides, we wish to prove:
$$\sqrt[3]{2}-1 = \bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3}$$
We have:
$$\bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3} = \frac{3}{1 + 3\sqrt[3]{2} + 3\sqrt[3]{2}^{2} + 2}=\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{1}{1+1\cdot\sqrt[3]{2} + \sqrt[3]{2}^{2}}\cdot\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2... | $$I=\int_{0}^{1} \frac{r^3 dr}{\sqrt{1-r^2}}$$
Let $r=\sin t \implies dr= \cos t dt$, then
$$I=\int_{0}^{\pi/2} \sin^3 t dt=\int_{0}^{\pi/2} \sin t(1-\cos^2 t) dt$$
$$I=\int_{0}^{\pi/2} (\sin t -\cos^2 t \sin t) dt$$
$$ I=-\cos t+\frac{\cos^3 t}{3}|_{0}^{\pi/2}=-(-1+1/3)=2/3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4030391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Step by step method for $\int\frac{x^2+1}{x^2-1}\,dx$ $$\int\frac{x^2+1}{x^2-1}\,dx$$
What is the step-by-step way of solving this integral problem?
I tried using substitution which was $x^2-1=t^2$, but end up with an even more complicated equation.
Substituting trigonometric functions for example :$x^2=\sec^2(a)$ so t... | This can be evaluated using partial fraction decomposition. Observe how
$$\begin{align*}\frac {x^2+1}{x^2-1} & =\frac {x^2-1+2}{x^2-1}\\ & =1+\frac 2{x^2-1}\end{align*}$$
The final fraction can be broken down by assuming the partial fractions follow the form
$$\frac 2{(x+1)(x-1)}=\frac A{x+1}+\frac B{x-1}$$
Multiplying... | {
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"url": "https://math.stackexchange.com/questions/4031935",
"timestamp": "2023-03-29T00:00:00",
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given $a+b+c=3$ prove that $abc(a^2+b^2+c^2) \leq 3$ I am preparing for inmo and I came accross this problem while solving a worksheet, but couldn't solve it, pl help me...
Problem-
Prove that if a,b,c are non negative real numbers such that a+b+c=3, then $abc(a^2+b^2+c^2)\leq3$ ... (0)
Developments-
Firstly applying... | Assume $a=\min\{a,b,c\} \rightarrow 0\le a \le 1.$
By AM-GM, we have $$abc\left(a^2+b^2+c^2\right) =\dfrac{1}{3} a\cdot
3bc\cdot\left(a^2+b^2+c^2\right)$$
$$\le \dfrac{1}{12} a \left[a^2+bc+\left(b+c\right)^2\right]^2\le \dfrac{1}{12} a \left[a^2+\dfrac{\left(3-a\right)^2}{4}+\left(3-a\right)^2\right]^2\le 3,$$
so it s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplification - Pre Algebra I have $$\frac{x^4+2x}{x^4}$$ which becomes $$\frac{x^3+2}{x^3}$$ after pulling out the common factor. I wanted to know why you you don't simplify $\frac{x^4+2x}{x^4}$ to $2x$ and the two $x^4$ cancel each other out, or is this because you must apply the denominator fraction to all numerat... | You can either do
$$\frac{x^4+2x}{x^4} = \frac{x(x^3+2)}{x(x^3)} = \frac{x^3+2}{x^3}$$
and then split the fraction as
$$\frac{x^3+2}{x^3} = \frac{x^3}{x^3} + \frac{2}{x^3} = 1 + \frac{2}{x^3}$$
Alternatively, split up first, and cancel common factors:
$$\frac{x^4+2x}{x^4} = \frac{x^4}{x^4} + \frac{2x}{x^4} = 1 + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036341",
"timestamp": "2023-03-29T00:00:00",
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How do I find a number $\frac{a^4}{4}$ to be added to an odd integer $N$ to make it a Perfect Square? Finding the least number to be added to an integer $N$ to make it a Perfect Square is simple:
https://www.geeksforgeeks.org/least-number-to-be-added-to-or-subtracted-from-n-to-make-it-a-perfect-square/
But how do we pr... | Given $n=pq$ with $p$ and $q$ prime and $p\leq q$, if $a$ is an integer such that
$$n+\frac{a^4}{4}=c^2,$$
for some positive integer $c$, then $\tfrac{a^4}{4}$ is an integer and so $a$ is even, say $a=2b$. Then also
$$pq=n=c^2-\tfrac{a^4}{4}=c^2-4b^4=(c-2b^2)(c+2b^2).$$
Then either $c-2b^2=1$ and $c+2b^2=pq$, or $c-2b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$.
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$
I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. Bu... | In respect of the stated comment, I add our second case (and so the whole solution) I left for OP to solve.
First, check $m=1$.
If $m≥2, m-2=k$ and $n=2n_1$, then
$$3×2^{k}+1=n_1^2, n_1≥2$$
Case $-1.$
$n_1=3a+1$, where $a≥1.$
$$3×2^k=9a^2+6a$$
$$2^k=3a^2+2a$$
$$a≥1 \Longrightarrow \begin{cases} a=2b \\ k≥3 \end{cases}$... | {
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"answer_id": 0
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For real $a$, $b$, $c$ all greater than $1$, show $\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^a} \;\ge\; \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ An inequality question from an olympiad book:
If $a,b,c$ are real numbers satisfying: $a>1$, $b>1$ and $c>1$, then prove that:
$$\frac{a^a}{b^b}+\frac{b^b}{c^c}+\frac{c^c}{a^... | Note that $$0=\ln\dfrac{a^a}{b^b}+\ln\dfrac{b^b}{c^c}+\ln\dfrac{c^c}{a^a}=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}$$
Let (WLOG) $\;\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$
We have $\;\dfrac{a}{b}\ge\dfrac{b}{c}\ge\dfrac{c}{a}$
Thus, $\;ac\ge b^2\;$, $a^2\ge bc\;$ and $\;ab\ge c^2\;$
Combining these $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
help solving a geometric sequence after manipulation Evaluate the sum. $\sum_{n=2}^\infty \frac{n(-3)^n}{4^{n+1}}$
I have manipulated it to be in a more common form I thought I could solve but I'm still having trouble getting the correct sum. I'm using wolfram to check against. It says the sum evaluates to $\frac{99}{7... | $1+x+x^2+x^3+...=\frac{1}{1-x} \iff \sum_{n\geq0}x^n=\frac{1}{1-x}\Rightarrow \sum_{n\geq1}nx^{n-1}=\frac{1}{(1-x)^2}$
$\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n}=-\frac{3}{4}\sum_{n\geq2}n\Big(-\frac{3}{4}\Big)^{n-1}=-\frac{3}{4} \Big( \frac{1}{(1+\frac{3}{4})^2}-1\Big)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction: $9\mid (4^{3n} + 8)$ for all integers $n \geq 0$. I cannot seem to figure this one out.
My though process so far is the following:
$\textit{Proof}$: For the base case n = 0, we have $9 \mid (4^0 + 8)$. This is true.
$\textit{Inductive step:}$ Suppose for $k \geq 0$ we have that $9\mid (4^{3k} + 8)$.... | Well, we need to use $4^{3k} +8$ is divisible by $9$.
SO we have somehow isolate $4^{3k} + 8$ from $4^{3k}4^3 + 8$.
Only way I see to do that is $4^{3k}4^3 + 8= (4^3 - 1)4^{3k} + \color{blue}{(4^{3k} + 8)}$
Now we know $9$ divides $\color{blue}{(4^{3k} + 8)}$ so it suffices to show $9$ divides $(4^3-1)4^{3k}$ as well.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$ Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
I attempted to solve this question as follows:
$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le... | $a=y+z,b=x+z,c=x+y$ then we have to prove $$\sum 2z{(y+z)}^2\le 3(x+y)(y+z)(z+x) \tag 1$$ Now note that $$\sum (2x{(z+y)}^2-8xyz)=\sum 2x{(y-z)}^2$$ and $$(x+y)(y+z)(x+z)-8zxy=\sum z{(x-y)}^2$$ therfore after subtracting $24xyz$ from both sides of $(1)$ it remains to rpove $$\sum z{(x-y)}^2\ge 0$$ which is true
Note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verif... | $$g(x)=\frac{(1+x)^2}{3+x^2}=1+2\frac{x-1}{3+x^2}$$
$$\text{$g$ is continuous for all real numbers and has a derivative} \\ \text{therefore we can find the max and min} \\ g'=2\frac{3-x^2+2x}{(3+x^2)^2}; \quad g'=0\Rightarrow (x-3)(x+1)=0 \\ \text{After that it's obvious that the range of $g$ is } [f(-1),f(3)]=[0,\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4051812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Factoring $x^4 + 12x^3 + 46x^2 + 59x + 18$
How do I factorize the following?
$$x^4 + 12x^3 + 46x^2 + 59x + 18$$
I've tried looking for a root by trial and error to no avail.
The answer is
$$(x^2 + 5x + 2)(x^2 + 7x + 9)$$
| The first step is check for linear factors. By the rational roots theorem, the only possible rational roots are $\pm 1, \pm 3, \pm 9, \pm 2, \pm 6, \pm 18$. Once you have eliminated linear factors, the only possible factorization is as two quadratics.
With that out of the way, a good general technique to look for fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4056738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proof for The Limit $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4},\space n\in\Bbb{N}$ I am revising my knowledge on the topic of real analysis by attempting some simple proofs. The question requires for the proof of the Limit for $\lim_{n\to\infty}\frac{3n^2-n}{4n^2+1}=\frac{3}{4}$ using the definition of converg... | Check:
With $n=\dfrac1{\epsilon}$,
$$\left|\frac{\dfrac3{\epsilon^2}-\dfrac1{\epsilon}}{\dfrac4{\epsilon^2}+1}-\frac{3}{4}\right|=\left|\frac{3-\epsilon}{4+\epsilon^2}-\frac{3}{4}\right|=\left|\frac{4+3\epsilon}{4(4+\epsilon^2)}\right|\epsilon<\left|\frac{4+3\epsilon}{16}\right|\epsilon<\epsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? Let $k>2$, is a natural number, does $\lim_{k \to \infty}[(\sum_{n=1}^{k}\frac{1}{n})-\ln k]$ exists? If $k \to \infty$, then I think no, because $\ln k$ will go to infinite. And $\sum \frac{1}{n}$ diverge too, and so it's $\infty - \infty$. But some... | Note that
$$
- \log k = - \log \left( {\frac{2}{1} \cdots \frac{{k - 1}}{{k - 2}}\frac{k}{{k - 1}}} \right) = - \sum\limits_{n = 2}^k {\log \left( {\frac{n}{{n - 1}}} \right)} = \sum\limits_{n = 2}^k {\log \left( {1 - \frac{1}{n}} \right)} .
$$
Thus,
$$
\sum\limits_{n = 1}^k {\frac{1}{n}} - \log k = 1+\sum\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\int_0^{\pi/2} \log(1 - x \cot x) \, dx\;$ (Is there a closed form?) I am interested in knowing if it is possible to find a closed-form solution to the following challenging log-cotangent integral
$$\int_0^{\frac{\pi}{2}} \log(1 - x \cot x) \, dx$$
I very much doubt a closed form in terms of known mathema... | Approximating the integral using rational approximants
Using integration by parts, we easily get
$$I = \int_0^{\pi/2} \ln(1 - x\cot x) \mathrm{d} x
= - \frac{\pi^3}{24} + \frac{\pi}{2}\ln 2 - \int_0^{\pi/2} \frac{x^3}{\tan x - x} \mathrm{d} x.$$
$I \approx -3.353337262889472...$
We focus on
$$I_1 = \int_0^{\pi/2} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Evaluate this limit $ \lim_{x\to\infty}\left (\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x}$ Please help to evaluate this limit
$$ \lim_{x\to\infty} \left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x},$$
where $0 \leq a$ and $a \not= 1$.
I tried to logarithm from both sides, and apply taylor series but so... | Case 1: If $a>1$ then from $\text{exp}^{\large{\log{u(x)}}}=u(x)$ we deduce
\begin{align}
\left(\frac{1}{x}\frac{a^x - 1}{a - 1} \right)^\frac{1}{x} &=
\exp\left(\frac{\log\left(\frac{1}{x}\frac{a^x - 1}{a - 1}\right)}x\right)\\&=
\exp\left(\frac{\log\left(\frac 1x\right)+\log\left(\frac{a^x - 1}{a - 1}\right)}x\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Algebraic manipulation of equations and determinant $$a_{11}^2 + a_{12}^2 + a_{13}^2 = 1$$
$$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$$
$$k(a_{11}(a_{33}a_{22}-a_{23}a_{32})+a_{12}(a_{21}a_{33}-a_{31}a_{23})+a_{13}(a_{22}a_{31}... | The trick is to express $a_{1i}^2$ as $a_{1i}.k(...)$
$\dfrac{a_{11}}{a_{33}a_{22}-a_{23}a_{32}} = \dfrac{a_{12}}{a_{21}a_{33}-a_{31}a_{23}} = \dfrac{a_{13}}{a_{22}a_{31}-a_{21}a_{32}} = k$
$a_{11}^2=a_{11}k({a_{33}a_{22}-a_{23}a_{32}})$
$a_{12}^2=a_{12}k({a_{21}a_{33}-a_{31}a_{23}})$
$a_{13^2}=a_{13}k({a_{22}a_{31}-a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the volume of tetrahedron $ABCD$ if $AB=AC=AD=5$, $BC=3$, $CD = 4$, and $BD = 5$ I am trying to solve this question with out using calculus:
Calculate the volume of a tetrahedron $ABCD$ where $AB = AC = AD = 5$ and $BC = 3, CD = 4$, and $BD = 5$.
I managed to find the area of the base triangle which came up t... | I have another analytical geometry explanation for the height of the tetrahedron:
Consider that $C(0,0,0), B(3,0,0), D(0,4,0).$
Let $A(x,y,z)$. The constraints are:
$$\begin{cases}AB^2&=&25\\AC^2&=&25\\AD^2&=&25\end{cases} \ \ \iff \ \ \begin{cases}(x-3)^2+y^2+z^2&=&25& \ \ (1) \\
x^2+(y-4)^2+z^2&=&25& \ \ (2) \\
x^2+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$ leading to contradictory solutions I have the following equation
$$\frac{(x-a)^2}{(x-2a)^2}=\left\lvert{ {\frac{x-a}{x-2a}} }\right\rvert$$
And to find the critical values for the absolute value function I carried out polynomial long division and ... | For the case you describe, you didn't mess anything up: what you found tells you there is no solution for $ \ x > 2a \ $ . (However, your characterization of the rational function after polynomial division is incomplete.)
In your original equation,
$$\frac{(x-a)^2}{(x-2a)^2} \ = \ \left\lvert{ {\frac{x-a}{x-2a}} }\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4082680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end wit... | To be divisible by $5$, the five-digit number must end in $0$. There are $360=6\cdot5\cdot4\cdot3$ integers that can be made by arranging four of the available non-zero digits before the zero. These $360$ five-digit numbers might have remainder $0$, $1$, or $2$ when divided by $3$, and since the six available digits ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4084536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
In which of the intervals is $\sqrt{12}$ In which of the intervals is $\sqrt{12}:$
a) $(2.5;3);$
b) $(3;3.5);$
c) $(3.5;4);$
d)
$(4;4.5)$?
We can use a calculator and find that $\sqrt{12}\approx3.46$ so the correct answer is actually b. How can we think about the problem without a calculator (if on exam for example)? I... | $$\sqrt{12} = 3 + (\sqrt{12} - 3) = 3 + \frac{(\sqrt{12})^2 - 3^2}{\sqrt{12} + 3} = 3+\frac{3}{\sqrt{12}+3}$$
and since $3 < \sqrt{12} < 4$, $\frac{3}{3 + 4} < \frac{3}{\sqrt{12} + 3} < \frac{3}{3 + 3}$. Thus $\sqrt{12} < 3 + \frac{1}{2}$ so $b$ is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
False proof that $\frac{13}{6}=0$
At start, the length of a line segment is $a_0=0$. When $3$ hours have elapsed since start, its length is $a_3$. When $1$ hour has elapsed since start, its length increased by $\frac{a_3}{2}$ with respect to $a_0$ (call the new length $a_1$). When $2$ hours have elapsed since start, i... | The red equation gives $a_3 = 0$. So dividing by $a_1 = \dfrac{a_3}{2}$ is the same as dividing by $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4088006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Simplifying $\frac{2\sqrt n + \frac{1}{\sqrt n}-3}{2\sqrt n -1}$ Consider the term $$\dfrac{2\sqrt n + \dfrac{1}{\sqrt n}-3}{2\sqrt n -1} \tag1$$
I simplified it to
$$\dfrac{4 n-4 \sqrt{n}+\dfrac{1}{\sqrt{n}}-1}{4 n-1} \tag2$$
or also
$$\dfrac{2 n+1-3 \sqrt n}{2 n-\sqrt{n}} \tag3$$
but apparently one can simplify this ... | As you have $\frac{1}{\sqrt n}$ in the numerator, first multiply numerator and denominator by $\sqrt n$ and check if it gets simplified.
$$\frac{2\sqrt n +\frac1{\sqrt n} -3}{2\sqrt n -1}=\frac{2n + 1 -3\sqrt n}{\sqrt n(2\sqrt n - 1)}$$
If you let $\sqrt n = x$, the numerator becomes, $2x^2 - 3x +1 = 2x^2-2x-x+1 = (2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Show $(y/2)/\sqrt{1-x^y}$ is bounded by one for $x\in(0,3/4)$ and $y\in(0,1)$ I would like to show that for all $x\in(0,3/4)$ and $y\in(0,1)$ that $f(x,y) = \frac{y}{2\sqrt{1-x^y}}$ is bounded above by one.
My thought is that for any $x$ I can try to show it an increasing function of $y$ on $(0,1)$. Then it would be bo... | I think I might have figured it out:
Consider the function
\begin{align*}
g(x,y) = \frac{y}{2\sqrt{1-x^y}}.
\end{align*}
This function is clearly increasing (thanks @Eric) in $x$ for all $y\in(0,1)$, so it suffices to set $x=3/4$.
Thus, define
\begin{align*}
f(y) = \log \left( \frac{y}{2\sqrt{1-(3/4)^y}} \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Interpreting Riemann sums as integrals Can you interpret any Riemann sum as an infinite number of definite integrals, all giving you the same value when you plug in the limits? For example, this limit of a sum:
$$\lim_{n \to \infty} \frac{3}{n} \left[ 1+ \sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{n+3(... | The integral that you have written does not equal the limit when you use Reimann Sums, I believe your book is incorrect?
The middle integral is:
$$
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x
$$
which means you can't factor out a $\sqrt{3}$. So I'll answer the question using the correct Reimann Sum.
Using the Left point me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4091343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find reduction formula for $I_n=\int\frac{1}{x^n \sqrt{x^2-1}}dx$ I have tried manipulating the integral but couldn't come up with an answer that matched textbook answer.
I just need a hint. Thanks in advance.
$\int\frac{dx}{x^n \sqrt{x^2-1}}$
This is what I tried:
Let $I_{n-2}=\int\frac{dx}{x^{n-2} \sqrt{x^2-1}... | Reduce the integral as follows
\begin{align}
I_n= &\int\frac{1}{x^n \sqrt{x^2-1}}dx \\
=& \int\frac{(1-x^2)+x^2}{x^n \sqrt{x^2-1}}dx
=-\int \frac{\sqrt{x^2-1}}{x^n }dx + I_{n-2} \\
=&\ \frac1{n-1}\int \sqrt{x^2-1} \>d\left(\frac{1}{x^{n-1} }\right)+ I_{n-2} \\
\overset{ibp}=&\ \frac1{n-1}\frac{\sqrt{x^2-1}}{x^{n-1} }+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Weird "hidden answer" in $2\tan(2x)=3\text{cot}(x)$ The question is
Find the solutions to the equation $$2\tan(2x)=3\cot(x) , \space 0<x<180$$
I started by applying the tan double angle formula and recipricoal identity for cot
$$2* \frac{2\tan(x)}{1-\tan^2(x)}=\frac{3}{\tan(x)}$$
$$\implies 7\tan^2(x)=3 \therefore x=... | Factorize the equation as follows
\begin{align}
2\tan(2x)-3\cot(x)
=& \frac{2\sin2x}{\cos 2x} - \frac{3\cos x}{\sin x}\\
=& \frac{2\sin2x\sin x-3 \cos x\cos2x }{ \sin x\cos 2x}\\
=& \frac{\cos x(10\sin^2x-3 )}{ \sin x\cos 2x}\\
\end{align}
where the factor $\cos x =0$ captures the solution $x=\frac\pi2$ and $10\sin^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
A biquadratic equation to have at least two real roots Which of the equations have at least two real roots?
\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}
I wasn't able to notice something clever, so I solved each of the equations. The first one has $2$ r... | There is a test that can be made for biquadratics without the need for square-roots that is akin to a test for the existence of real zeroes of a quadratic polynomial (there is an evident connection with the Rule of Signs that Gregory discusses). There, "completing the square" produces
$$ u^2 \ + \ Bu \ + \ C \ \ = \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4093406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Number of possible values of $4x-z$ if $x+y+z=20$ Given that $x, y,z$ are non negative integers such that $x+y+z=20$. If $S$ is the set of all possible values of $4x-z$.Find $n(S)$.
My try:
By stars and bars number of non negative integer solutions of $x+y+z=20$ is $\binom{22}{2}=210$
Among these $210$ ordered triplets... | For a given value of $x$, $z$ can range from $0$ up to $20-x$. This means that
$$D_a := \{4a-z \mid a + y + z = 20\} = \{5a-20, 5a-19, \ldots, 4a\}$$
(note that $5a-20 \leq 4a$ when $0 \leq x \leq 20$) where $a$ is fixed. Notice also that $4a \geq 5(a+1)-20$ for $a \leq 15$. This means that $D_a$ and $D_{a+1}$ overlap ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Derive reduction formula for $\int \ln^n(x+\sqrt{x^2+1}) \, dx $ I wrote something like $$I=\int(x)' \ln^n {\left(x+\sqrt{x^2+1} \right)} \, dx$$ and $$I=x\ln(x+\sqrt{x^2+1})-n \displaystyle \int x \frac{\ln^{n-1} {\left(x+\sqrt{x^2+1} \right)}}{\sqrt{x^2+1}}dx.$$
I noticed that ${\left[\ln{{\left(x+\sqrt{x^2+1} \right... | Let's use integration by parts.
$\int u_1u_2 \,dx=u_1 \int u_2\,dx-\int u_1' (\int u_2\,dx)\,dx$
Sometimes choosing $u_1,u_2$ in certain ways makes integration by parts simpler. In this case, I'll take logarithmic function as $u_1$ and $1$ as $u_2$ to start with.
Let
$\begin{align} I_n=&\int \ln^n(x+\sqrt{x^2+1}) \ti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
\infty} (a_n-3)6^n$.
First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the for... | Ths approach shows that the results in question emerge quite naturally from the standard fixed-points-stability analysis of the evolution equation. Also the dependence of the limit of $b_n = (a_n -3)6^n$ from the inital value $a_1$ is discussed.
Following the common procedure to analyse the evolution equation
$$a_{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 4
} |
Is it possible to write any permutation as the product of transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$ I understand any permutation $(a_1 \text{ } a_2 ... a_k)$ in $S_n$ where $k \leq n$ can be written as a product of $(a_1 \text{ } a_k)(a_1a_{k-1})...(a_2 \text{ } a_1)$. Similarly, $(a_1a_2...a_k... | Solution 1:
It is obvious that we can write any permutation as a product of transpositions.
So, we need only to show that any transposition is a product of special transpositions $(1 \text{ } 2),(2 \text{ } 3),...(n-1 \text{ } n)$.
Let $1\leq i<j\leq n$.
$(i\text{ }j)=(i\text{ }i+1)\cdots(j-2\text{ }j-1)(j-1\text{ }j)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4102522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Get $f(x)=u_x\frac{x}{u}$ from ODE for $u$ Consider the Cauchy-Euler ODE
\begin{align*}
\frac{1}{2}x^2u_{xx}+xu_x-u=0.
\end{align*}
Guessing $u(x)=Cx^n$ gives
\begin{align*}
\frac{1}{2}n(n-1)Cx^n+nCx^n-Cx^n=0,
\end{align*}
which we can solve to get
\begin{align*}
n_1&=-2,\\
n_2&=1.
\end{align*}
Given initial conditions... | The general solution to your ODE is
$$
u(x)=Ax+Bx^{-2}
$$
and so
$$
f(x)=1-\frac{3 B}{A x^3+B}
$$
which is not in general constant. That is why you do not find $f_x=0$ directly from the ODE.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4103216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine all ordered pair (x,y) of positive integers such that $y^{2}−(x+ 2)2^x=1$. This is what I have come up so far,
*
*y has to be odd. Since $(x+2) 2^{x}$ is always even, hence $(x+2)2^x+1$ is always odd. Thus, y must be an odd number.
*$$y^2 \equiv1(mod 4)$$Since all square numbers either congruent to 0 or ... | $y^2 - 1 = (x+2)2^x$
$(y-1)(y+1)= (x+2)2^x$.
Now $y- 1, y+2$ must both be even and $\gcd(y-1,y+1) = 2$ so
So one of $y-1$ or $y+1$ must be divisible by $2$ but not any higher power of $2$. And the other must be divisible by a power of $2$ that is at least $2^{x-1}$.
The very least the term divisible by $2^{x-1}$ can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4104628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $xy = ax + by$, prove the following: $x^ny^n = \sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^nb^{n-k}x^k + a^{n-k}b^ny^k),n>0$ If $xy = ax + by$, prove the following: $$x^ny^n = \sum_{k=1}^{n} {{2n-1-k} \choose {n-1}}(a^nb^{n-k}x^k + a^{n-k}b^ny^k) = S_n$$ for all $n>0$
We'll use induction on $n$ to prove this.
My app... | The approach by induction is fine. We just need one additional twist (one more induction).
We want to show for $n\geq 1$
\begin{align*}
\color{blue}{(xy)^n=\sum_{k=1}^n\binom{2n-1-k}{n-1}\left(a^nb^{n-k}x^k+a^{n-k}b^ny^k\right)}\tag{1}
\end{align*}
provided
\begin{align*}
\color{blue}{xy=ax+by}
\end{align*}
The base st... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4106937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $x^{x^{x+1}}=\sqrt{2}$, then evaluate $x^{x^{p}}$, where $p = 2x^{x+1}+x+1$ I can't figure out how to give a proper form to this expression to use the root of two.
If
$$x^{x^{x+1}}=\sqrt{2}$$
find the value of $W$ if
$$W=x^{x^{p}} \quad\text{where}\; p = 2x^{x+1}+x+1$$
EDIT: This is an algebraic manipulation probl... | $x^{x^x} = \sqrt{2} \implies x^x = log_x\sqrt2$
Now:
$W = x^{x^{2x^{x+1}+x+1}} = x^{\left(x^{2x^{x+1}}x^xx\right)} = x^A$
Let the exponent be A for the time being:
$A = x^{2x^{x+1}}x^xx = \left(x^{x^{x+1}}\right)^2x^xx = \left(x^{x^xx}\right)^2x^xx$
Since $x^x = log_x\sqrt2$, we have $A = \left(x^{xlog_x\sqrt{2}}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4107777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Find $S = a + b$ such that for $\forall m \in \left[a\sqrt{\frac{15}{7}} + b\sqrt{\frac{7}{15}}; 2\right)$ then $2x^2 + 2x - mf(x) + 5 = 0$ has root
Let $f(x)$ be continuos on $\mathbb R$ satisfy $f(0) = 2\sqrt{2}$ and $f(x) > 0, \forall x \in \mathbb R$ and $f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$.
For all $m \in \left[a... | COMMENT.-Once find out $f(x)=\sqrt{(x^2+x+3)^2-1}$ the function $g(x)=2x^2+2x-mf(x)+5$ has derivative $g'(x)=(2x+1)\left(2-\dfrac{m(x^2+x+3)}{f(x)}\right)$ so take an extremum at $x=-\dfrac12=-0.5$ and $g(-0.5)=4.5-m\dfrac{\sqrt{105}}{16}$
It follows that $g(x)$ has at least a (real) root in a little interval for $m$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
what is $a^2+9=b^2+16=1+(a+b)^2$ solve for $a,b$ This is for a geometry question, and through a construction arrived at this equation. I could not solve it and after plugging it into wolfram got the correct answer but can anyone show a method for finding a,b
they are $\frac{5}{\sqrt3}$ and $\frac{2}{\sqrt3}$ respective... | Maybe try this way:
$b^2+16=1+(a+b)^2\Rightarrow 15=a(a+2b)$ and $a^2+9=1+(a+b)^2\Rightarrow 8=b(2a+b)$
$\frac{8}{15}=\frac{b(2a+b)}{a(a+2b)}=\frac{x(2+x)}{1+2x}$, where $x=\frac{b}{a}$ (Assuming $a$ and $b$ are positive).
On solving, we get $x=\frac{2}{5}=\frac{b}{a}$.
Thus, $5b=2a$.
As $a^2+9=b^2+16=(\frac{2a}{5})^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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How am I supposed to expand $\sin^2 A + \sin^4 A = 1$ into $1 + \sin^2A = \tan^2A$? My question is how can i expand
$$\sin^2 A + \sin^4 A = 1$$
into:
$$1 + \sin^2A = \tan^2A$$
I tried quite a few ways I know but all of them kinda felt random. i am not sure how to share my trials here. I am quite beginner in trigonometr... | Hint
If $\sin^2A + \sin^4A = 1 \implies \sin^4A = \cos^2 A$
Can you proceed from here?
Additional Info
$$\sin^2A(1-\cos^2A) = \cos^2A$$
$$\sin^2A = \cos^2A + \sin^2A\cos^2A$$
Divide by $\cos^2 A$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.
Prove that $\frac{\sin A}{\sin B}+\frac{\sin B}{\sin A} \leq \frac{A}{B}+\frac{B}{A}$ for acute angles, $A$ and $B$.
I'm confused about how to do this since we can't say $\frac{\sin A}{\sin B}\leq \fra... | Here is another solution: let $f(t) = t + 1/t$, and observe that it is increasing on $[1,\infty)$, and decreasing on $(0,1]$.
Since $f(1/t) = f(t)$ for any $t > 0$, we can assume w.l.o.g. that $B \leq A$.
Now, observe that the function $g(t)=\frac{\sin(t)}{t}$ is decreasing on $(0,\frac{\pi}{2})$, and thus:
$$
g(A) \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4118112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding limit of term involving binomial coefficient I am required to find the limit of
$$
\frac{\sqrt{n}}{2^n}\cdot \binom{n}{\frac{n}{2} + \sqrt{n}}
$$
I am given the following hints:
*
*Stirling's approximation
$$
n! \approx \sqrt{2\pi n}\cdot\left(\frac{n}{e}\right)^n
$$
*For $k \rightarrow \infty$, with $x$ be... | Ignoring the $\sqrt{2\pi}$ factor, you can collect terms as follows: $$2 \cdot\left( \frac{n}{2\sqrt{n^2/4 -n}} \right)^{n+1} \cdot \left( \frac{n/2-\sqrt n}{n/2 +\sqrt n} \right)^{\sqrt n} $$ Now, the first factor is further simplified to $$\left( \frac{\sqrt n}{\sqrt {n-4}} \right)^{n+1} \\ = \frac{1}{ (1-\frac 4n)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120773",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Maximizing $x^2y$ given $x^2+y^2=100$, without using the AM-GM inequality and calculus tools Problem says:
Let $x^2+y^2=100$, where $x,y>0$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
I know these possible tools:
*
*AM-GM inequality
*Calculus tools
Here, I want to escape from all of the ... | An alternative approach without calculus. Considering that $x^2y = \lambda$ and $x^2+y^2=10^2$ are analytic, at it's maximum, $x^2y$ should be tangent to $x^2+y^2=10^2$ then
$$
\frac{\lambda}{y}+y^2-10^2=0
$$
so, at tangency $y^3-10^2y+\lambda = (y-r_1)^2(y-r_2)$
Equating coefficients we have
$$
\cases{
r_1^2r_2 +\lamb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involvin... | Partial Answer
\begin{align}
\sum_{k=3}^\infty\frac{\ln k}{k^2-4}&=\sum_{k=3}^\infty\frac{\ln k}{k^2\left(1-\frac4{k^2}\right)}\\
&=\sum_{k=3}^\infty\frac{\ln k}{k^2}\sum_{l=0}^\infty\frac{2^{2l}}{k^{2l}}\\
&=\sum_{l=0}^\infty2^{2l}\sum_{k=3}^\infty\frac{\ln k}{k^{2l+2}}&&(\text{By Fubini's theorem})\\
&=\sum_{l=0}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 6,
"answer_id": 3
} |
How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
| Another way to notice the factorization
$$x^4-2x^3-x^2+2x+1=0$$
Since $x=0$ is not the root of the equation, divide by $x^2$ to get
$$x^2 -2x-1 + \frac{2}{x} + \frac{1}{x^2} = 0$$
Rewrite it as
$$x^2+\frac{1}{x^2} - 2\left(x-\frac{1}{x}\right) - 1 = 0$$
or
$$\left(x-\frac1x\right)^2 + 2 - 2\left(x-\frac{1}{x}\right) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find set of vectors orthogonal to $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ $\mathbf{Question:}$ Find set of vectors orthogonal to $\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}$
$\mathbf{My\ attempt:}$
The vector is in $R^3$ so we can let vector $\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix} $ represent ortho... | The space of vectors orthogonal to $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ can be defined by $span(\begin{bmatrix}
-1 \\
1 \\
0 \\
\end{bmatrix} , \begin{bmatrix}
-1 \\
0 \\
1 \\
\end{bmatrix} )$ where the dimension of the space is 2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are the following systems of inequalities the same? Suppose $x, y \in \mathbb{R}$, and $\mathcal{S_1}$ is a system of inequalities:
\begin{align*}
\mathcal{S_1} &= \begin{Bmatrix}
x - y \geq 1\\
-x + 2y \geq 1\\
3x - 5y \geq 2
\end{Bmatrix}\\
&= \begin{Bmatrix}
x \geq 1 + y\\
x \leq 2y-1\\
x \geq \frac{2+5y}{3}... | Yes, $S_3=S_1$. We need only to show that
$$5≤y<\infty$$
Therefore, we have,
$$3(-x+2y)+3x-5y≥3\times 1+2 $$
$$\iff y≥5.$$
But, this notation is problematic.
$$5≤y≤\infty$$
The correct one can be written as follows:
$$5≤y<\infty$$
Or,
$$y\in[5,+\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Differentiate $x^{a^x}$ without logarithmic differentiation? Problem. Compute $\frac{d}{dx} x^{a^x}$.
Method 1: Logarithmic Differentiation (Correct):
\begin{align*}
y&= x^{a^x} \\
\ln(y)&=a^x \cdot \ln(x) \\
\frac{1}{y} \frac{dy}{dx} &=a^x \frac{1}{x} + \ln(x) \ln(a) a^x \\
\frac{dy}{dx} &= y\left(\frac{1}{x} a^x + \l... | It is not a misapplication of the chain rule, but of the power rule. The power rule does not work for non-constant exponents; by that logic the derivative of $x^x$ would be $x\cdot x^{x-1}=x^x$, which is flat-out wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove $\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8}$ As the title says, prove $$\sum_{k=1}^{n-1}\frac{(n-k)^2}{2k} \geq \frac{n^2\log(n)}{8},$$ for $n>1$. This inequality is from Erdős, "Problems and results on the theory of interpolation". I, Lemma 3.
My attempt: since $H_{n-1}=\sum_{k=1}^{n-1}\frac{1}... | Using the stronger inequality, for $n>1$ that:
$$H_{n-1}>\frac{1}{2}\left(1-\frac1n\right) +\log n$$
(see proof below) we get:
$$\frac{n^2H_{n-1}}{2}-\frac{3n(n-1)}4>\frac{n^2\log n}2 +\frac{n(n-1)}4-\frac{3n(n-1)}4\\
=\frac{n^2\log n}2-\frac{n(n-1)}2$$
So you need:
$$\frac{3n^2\log n}8>\frac{n(n-1)}2$$
or $$\frac{\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $\gcd(a, b) = 1$. Determine the value of $a + 10b.$
Given a circle $ω$ with center $O$ and length of radius $r$. There is
a point $A$, $B$ on the circle and point $C$ outside the circle so
that $BC$ tangent to the circle $ω$. Then $AC$ intersects the ci... | The ratio ${\large{\frac{r^2}{AP^2}}}$ can be computed by brute force as outlined below.
*
*Assume $y=1$.
Since the ratio ${\large{\frac{r^2}{AP^2}}}$ is scale invariant, we can assume $y=1$.
*Compute $x$.
By power of a point we get $(x+1)^2=(1-x)(2)$, so
$$x=\sqrt{5}-2$$
*Compute $\cos\bigl(\angle{BCP}\bigr)$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Modified exponential summation How do we prove that
$$1+\frac{2^3}{2!}+\frac{3^3}{3!}+\frac{4^3}{4!}+\cdots=5e$$
I'll post the answer to this question as a knowledge share.
| Let $S$ denote the sum in question.
$$S=\sum\limits_{r=1}^{\infty}\left(\frac{r^3}{r!}\right)$$
Here, the general term
$$\begin{equation}\begin{aligned}
t_r &= \frac{r^3}{r!}\\
&=\frac{r^2}{(r-1)!}\\
&=\frac{(r-1+1)^2}{(r-1)!}\\
&=\frac{(r-1)^2}{(r-1)!}+\frac{1}{(r-1)!}+\frac{2(r-1)}{(r-1)!}\\
&=\frac{r-1}{(r-2)!}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof of $\int_{0}^{1}\dfrac{\log(x)\log(1 + 2x)}{x(1+x)} = -\dfrac{7}{6}\zeta(3)$ From quite sometime I've been struggling on proving
$$\int_{0}^{1}\dfrac{\log(x)\log(1 + 2x)}{x(1+x)}\,dx = -\dfrac{7}{6}\zeta(3)$$
I have tried partial fraction decomposition which produces a simple integral and a difficult one. I'm qui... | Assuming $0 \leq x\leq 1$ a CAS gives for the antiderivative
$$\text{Li}_3\left(1+\frac{1}{2
x}\right)-\text{Li}_3\left(2+\frac{1}{x}\right)+\text{Li}_3(-2 x-1)+\text{Li}_3(-2
x)+\text{Li}_3(-x)+\left(\text{Li}_2\left(2+\frac{1}{x}\right)-\text{Li}_2\left(1+
\frac{1}{2 x}\right)\right) \log \left(\frac{1}{x}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can we use Vieta's formula in solving Trigonometric equations? **The value of $$\sec\frac{\pi}{11}-\sec\frac{2\pi}{11}+\sec\frac{3\pi}{11}-\sec\frac{4\pi}{11}+\sec\frac{5\pi}{11}$$
is ...
My Approach
I used the fact that $$\sec (\pi-x)=-\sec x$$ to simplify the equation to $$\sec\frac{\pi}{11}+\sec\frac{3\pi}{11}+\sec\... | Observe that $\cos(2n+1)\pi=-1$ for any integer $n$
If $11x=(2n+1)\pi,\cos6x=-\cos5x$
With $\cos x=c,$
$$\cos6x=2\cos^23x-1=2(4c^3-3c)^2-1=?$$
Using Prosthaphaeresis Formulas,
$$\cos5x+\cos x=2\cos3x\cos2x=2(4c^3-3c)(2c^2-1)$$ to find
$$0=\cos6x+\cos5x=32c^6+16c^5-48c^4-20c^3+18c^2+5c-1$$
So, the roots of $$32c^6+16c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the global minimum of the multivariable function using only algebraic tools Problem says:
Find the global minimum of
$$\begin{align}f(x,y): &= x^2 + y^2 + \alpha xy + x + 2y\end{align}$$
where, $\alpha\in\mathbb R$.
The things I have done:
Let,
$$f(x,y):=x^2+x(\alpha y+1)+2y+y^2$$
*
*Algebraic tool I will ... | Hint.
Another approach is to include $f(x,y)$ in the quadratic class as
$$
f(x,y) = (x-x_0,y-y_0)'M(x-x_0,y-y_0)+c
$$
Choosing $M = \left(
\begin{array}{cc}
1 & \frac{\alpha }{2} \\
\frac{\alpha }{2} & 1 \\
\end{array}
\right)$ and equating coefficients we arrive at
$$
\cases{
x_0 = -\frac{2 (\alpha -1)}{\alpha ^2-4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Filling in details for calculation of the limit $\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right)$ I want to evaluate the following limit using asymptotics
\begin{equation}
\lim _{x\rightarrow \infty } x^{2}\left(\sqrt[7]{\frac{x^{3} +x}{1+x^{3}}} -\cos\frac{1}{x}\right... | You can make life easier letting $x=\frac 1y$. This makes the expression
$$A=\frac 1{y^2} \left(\frac{\sqrt[7]{1+y^2} }{\sqrt[7]{1+y^3} } - \cos(y)\right)$$ Now, using the binomial theorem or Taylor series for the surds and Taylor series for the cosine, you have
$$A=\frac 1{y^2} \left(\frac{1+\frac{y^2}{7}-\frac{3 y^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4142617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Hermite interpolation of even degree Using Hermite interpolation determine the fourth degree polynomial $p(x)$ for which
\begin{equation}
p(0) = p'(0) = 0, \quad p(1) = p'(1) = 1, \quad p(2) = 1
\end{equation}
I can only seem to find content on how to apply Hermite interpolation to data points with odd degr... | Let $P_0(x) =x^2, P_1(x)=(x-1)^2, P_2(x)=x-2$.
Let $A_0(x) = 0, A_1(x) = 1+(x-1)=x, A_2(x)=1$.
We want to find $p(x)$ such that $$p(x) \equiv A_i(x) \pmod{P_i(x)}$$
We let $$Q(x)=x^2(x-1)^2(x-2)$$
Denote $Q_i(x) = \frac{Q(x)}{P_i(x)}.$ We have
$$Q_0(x) =(x-1)^2(x-2)$$
$$Q_1(x) = x^2(x-2)$$
$$Q_2(x)=x^2(x-1)^2$$
Let $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need help with this! $\lvert x-y \rvert\lt\epsilon^2 \Rightarrow \lvert\,\lvert\sqrt x \rvert - \lvert \sqrt y \rvert \,\rvert \lt \epsilon$ Given $\Bbb R$, $ \epsilon \gt 0$. Prove that any positive $\Bbb R$
$$\lvert x-y \rvert\lt\epsilon^2 \Rightarrow \bigl\lvert\lvert \sqrt x \rvert - \lvert \sqrt y \rvert \bigr\ver... | Whenever I see something like $a^2 - b^2$ in these kind of problems, I often try to see if factoring it into $(a-b)(a+b)$ is fruitful.
In this problem, we do not have $x^2 - y^2$, but we do have $|x - y|$ and also $|\sqrt{x} - \sqrt{y}|$, which suggested to me that using $|x - y| = |(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$f(a)=b,f(b)=c,f(c)=a$, find $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Let $f(x)=x^2-x-2$. It is given that $a, b,c \in \mathbb{R}$ such that
$$f(a)=b,f(b)=c,f(c)=a$$ Then find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
My attempt:
We have:
$$\begin{array}{l}
a^{2}-a-2=b \\
b^{2}-b-2=c \\
c^{2}-c-2=a
\end{arra... | From the condition, we need $f\circ f\circ f(a)=a$. Now
$$f(x) = x^2-x-2=(x-2)(x+1)$$
$$f\circ f(x) = (x^2-x-4)(x^2-x-1)=x^4-2x^3-4x^2+5x+4$$
$$\implies f\circ f\circ f(x)=(x^4-2x^3-4x^2+4x+2)(x^4-2x^3-4x^2+4x+5)$$
$$\implies f\circ f\circ f(a)-a=a^8-4a^7-4a^6+26a^5+3a^4-54a^3-3a^2+34a+10 \\=(a^2-2a-2)(a^3-3a-1)(a^3-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solve Integral -$ \int \sqrt{3x^2 + 2x}\ dx$ I am trying to work out the integral without just skipping straight to the formula:
$$\int \sqrt{u^2-a^2} du = \frac{u}{2}\sqrt{u^2-a^2} - \frac{a^2}{2} \ln\left|u+\sqrt{u^2-a^2}\right|+C $$
I did all the substitutions, etc.
$$u=x+1/3,\,a=1/3,\,a\sec\theta=u,\,du=\sec\thet... | Yes your answer is correct. Here $a=\frac 13$, so $\frac 1a=3$, as you have written. Note that integration by parts can also be used to solve the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Triple integral between $z^2 = {(x - 1)}^2 + y^2$ and the unit sphere. I have to compute $\int_D f$, where $D$ is the region in ${(0 , \infty)}^3$ between the cone $z^2 = {(x - 1)}^2 + y^2$ and the sphere $x^2 + y^2 + z^2 = 1$, and $f : D \to \mathbb{R}$ is given by $f(x , y , z) = z \sqrt{x^2 + y^2}$.
My attempt: If I... | Cylindrical coordinates is easier.
Equation of the surface of the cone, $z^2 = (x - 1)^2 + y^2$
Equation of the surface of the sphere, $x^2 + y^2 + z^2 = 1$
At intersection,
$z^2 = (x - 1)^2 + y^2 = 1 - x^2 - y^2 \implies x^2 - x + y^2 = 0$
This is the projection in xy plane, of the intersection of cone with the sphere... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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System of two quadratics equation, $P(x)$ and $Q(x)$ If $P(x) = ax^2 + bx + c$ and $Q(x) = – ax^2 + dx + c$, $ac \ne 0$, then the equation $P(x) . Q(x) = 0$ has
(A) Exactly two real roots
(B) At least two real roots
(C) Exactly four real roots
(D) No real roots
My approach is as follow Let $T(x)=P(x).Q(x)$
$T\left( x \... | Set $PQ=\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)$ to zero, to get $$\left( ax^2+bx+c\right) \left( -ax^2+dx+c\right)=0.$$
Check the discriminant for both $P$ and $Q$.
For $P$, you get $D_P=b^2-4ac$. For $Q$, you get $D_Q=d^2+4ac$.
Suppose $P$ has no real roots, such that $D_P=b^2-4ac < 0 \implies4ac>b^2$. In th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying $\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2 $ Hi can someone help me please simplify the following showing the working out step by step?
$$
\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2
$$
I can't get the... | Just do it:
$$
\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2
$$
$$=\left({ {x} + 2+\frac{1}{ {x}}}\right) - \left(x-2+\frac1x\right)$$
$$=4$$
using $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$ with $a=\sqrt x$ and $b=\dfrac1{\sqrt x}$.
ADDENDUM to answer question in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations a... | Written in a more compact form
$$P_n=2 \left(\frac{n}{2 n+1}\right)^{1-\frac{1}{2 (n+1)}}$$
Take logarithms and compose Taylor series
$$\log(P_n)=\frac{\log (2)-1}{2 n}+\frac{3-4 \log (2)}{8
n^2}+O\left(\frac{1}{n^3}\right)$$
Continue with Taylor
$$P_n=e^{\log(P_n)}=1-\frac{1-\log (2)}{2 n}+\frac{4+(\log (2)-6) \log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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Let $c$ be a positive real number for which the equation $x^4-x^3+x^2-(c+1)x-(c^2+c)=0$ has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$ Let $c$ be a positive real number for which the equation
$x^4-x^3+x^2-(c+1)x-(c^2+c)=0$
has a real root $\alpha$. Prove that $c=\alpha ^2 - \alpha$
I tried to to solve usin... | Direct substitution with $\alpha=x$ and $c=x^2-x$ leads to.
$x^4-x^3+x^2-x(x^2-x+1)-((x^2-x)^2+x^2-x)$
$=(1-1)x^4+(2-2)x^3+(2-2)x^2+(1-1)x=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4155518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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A First Order Homogenous Differential Equation The following problem is from Schaum's book on Differential Equations.
Problem:
Solve the following differential equation.
$$ y' = \frac{ x^4 + 3x^2y^2 + y^4}{x^3 y} $$
Answer:
We rewrite this equation as:
$$ \frac{dy}{dx} =
\frac{ 1 + 3 \left( \dfrac{y^2}{x^2}\right) + \... | Check the step after $$-\frac {1}{2((\frac yx)^2+1)}=\ln|C_2x|$$
You should get $$-\frac {1}{2(x^2+y^2)}=\frac {\ln|C_2x|}{x^2}$$
That is not what you wrote.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The solutions of the equation $|p^3-q^2|=2^n$ Let's consider the following equation
$$|p^3-q^2|=2^n$$
where $p$, $q$ are distinct primes and $n$ a non-negative integer. The equation admits some trivial solutions:
$$|2^3-3^2|=2^0$$
$$|3^3-5^2|=2^1$$
$$|5^3-11^2|=2^2$$
$$|17^3-71^2|=2^7$$
Clearly, not all values of $n$ a... | We consider the elliptic curves $y^2 = x^3\pm2^n$ with parameter $n$ where $x$ and $y$ are prime number .
Searching the integer points $(x,y)$ with $\boldsymbol{n \lt 100}$ and $\boldsymbol{x \lt 10^{10}}$, we found the integer points $(x,y,n)=(3, 5, 1)$ and $(5, 11, 2)$ for $y^2 = x^3-2^n.$
Similarly, we found the in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4159319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to solve $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ I want to solve the expression $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ to get a much simpler and neater result. I have tried to manipulate this expression such as using sum/difference formulas, but it didn't help (and made the expression even more messy).
Here is wha... | $$\frac{1}{2\sin{50^{\circ}}} + 2 \sin{10^{\circ}}= \frac{1}{2\sin{50^{\circ}}} + 2 \sin{(60^{\circ}-50^{\circ})}=\frac{1}{2\sin{50^{\circ}}} + \sqrt{3} \cos{50^{\circ}} -\sin{50^{\circ}} =$$
$$=\frac{1-2 \sin^2{50^{\circ}}}{2\sin{50^{\circ}}}+\frac{2\sqrt{3} \cos{50^{\circ}}\sin{50^{\circ}}}{2\sin{50^{\circ}}}=\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find continuous function $f$ such that $\int_0^{\infty} f(x) \left(\frac{x^2}{1+x^2} \right)^n dx =\infty$ for all integers $n \geq1$. Let $f(x)$ be a continuous function on $[0,\infty)$.
I want to find $f$ such that $\int_0^{\infty} f(x) \left(\frac{x^2}{1+x^2} \right)^n dx =\infty$ for all integers $n \geq1$.
Let $g_... | Note that
$$
\left( {\frac{{x^2 }}{{1 + x^2 }}} \right)^n \ge \left( {\frac{{n^2 }}{{1 + n^2 }}} \right)^n \ge \frac{1}{2}
$$
for $n\geq 1$ and $x\geq n$. Taking, for example $f(x)=\frac{1}{x+1}$ (which is continuous for $x\geq 0$ and is bounded)
\begin{align*}
\int_0^{ + \infty } {f(x)\left( {\frac{{x^2 }}{{1 + x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166799",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solve the differential equation: $(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$ Solve the differential equation: $(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$
My first step was to solve for $y^\prime$ to get:
$$y^\prime=\frac{y}{x}\pm\sqrt{(\frac{y}{x})^2-1}$$
Solving each individually yields (using substitution $u = \frac{y}{x... | $$(y^\prime)^2 - \frac{2y}{x}y^\prime+1=0$$
This is D'Alembert 's differential equation:
$$y=xf(y')$$
where $$f(y')=\dfrac {1+y'^2}{2y'}$$
From your attempt you should get
$$ u+\sqrt {u^2-1}=cx$$
$$u^2-1=(cx-u)^2$$
$$c^2x^2-2cxu+1=0$$
$$c^2x^2-2cy+1=0$$
$$ \implies y= \dfrac {c^2x^2+1}{2c}$$
There is also a singular s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4167202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the range of possible values of $\sqrt{a^2+a+1}-\sqrt{a^2-a+1}$
Let $x=\sqrt{a^2+a+1}-\sqrt{a^2-a+1},x\in \mathbb R$. Find range of possible values of $x$.
I tried drawing the graph and obtained this:
Through which the answer came out to be $(-1,1)$.
What should be the procedure through algebra?
| Multiply your expression by
$$\frac{ \sqrt{a^2+a+1}+\sqrt{a^2-a+1} } {\sqrt{a^2+a+1}+\sqrt{a^2-a+1}}$$
To get
$$\frac{2a}{ \sqrt{a^2+a+1}+\sqrt{a^2-a+1}}.$$
For $a$ positive, divide top and bottom by $a$ and push
the $1/a$ inside the square roots:
$$\frac{2}{ \sqrt{ 1+ \frac{1}{a} +\frac{1}{a^2} } +
\sqrt{ 1 - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solving ODE-systems with Matrix exponential is wrong? Originally I've learned that the solution of a systems of coupled ODE:
$$\underbrace{\left[\begin{array}{cc}{y_1}'(x)\\ \vdots \\{y_n}'(x)\end{array}\right]}_{y'(x)}=
\underbrace{\left[\begin{array}{cccc}&a_{1\,1} &\cdots &a_{1\,n}
\\ &\vdots \quad &&\vdots \\
&a_{... | We have
$$\left(\begin{array}{cc}{y_1}'(x) \\ {y_2}'(x)\end{array}\right) = \left(\begin{array}{cc} 4 & 10\\ 8 & 2 \end{array}\right)\,\left(\begin{array}{cc}{y_1}(x) \\ {y_2}(x)\end{array}\right)$$
The eigenvalues are
$$\lambda_1 = 12, ~~\lambda_2 = -6$$
The eigenvectors are
$$v_1 = \begin{pmatrix} 5 \\ 4 \end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Series sum of $\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$) What is the sum of following series $\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$)
*
*$\frac{3}{8}$
*$\frac{3}{10}$
*$\frac{3}{14}$
*$\frac{3}{16}$
My Attempt:
$\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac... | Your error is that the geometric series starts with $n=1$, not $n=0$, so it should be $\sum_{n=1}^\infty \frac{1}{6^n} = \frac{1}{6} \frac{1}{1-\frac{1}{6}} = \frac{1}{5}$, not $\frac{6}{5}$. Making this correction will give you one of the answers in the multiple choice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrating $\int{\left( x+2 \right)\sqrt{x-1}dx}$ I’m stuck on this relatively simple indefinite integral but, unfortunately, cannot figure out why my answer is wrong.
Here it is:
\begin{align}
\int{\left( x+2 \right)\sqrt{x-1}dx}&=\int{\left( \left( x-1 \right)+3 \right)\frac{3}{2}\frac{1}{{{\left( \sqrt{x-1} \righ... | If you differentiate $\sqrt{x-1}$, what you get is $\dfrac1{2\sqrt{x-1}}$ rather than $\displaystyle\frac23\sqrt{x-1}^3$. If you want to get $\displaystyle\frac23\sqrt{x-1}^3$, then you should differentiate $\dfrac4{15}\sqrt{x-1}^5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Induction Squares Proof Prove that for every positive integer $n$ there exist positive integers $$a_{11}, a_{21}, a_{22}, a_{31}, a_{32}, a_{33}, \dots ,a_{n1}, a_{n2},\dots,a_{nn}$$
such that
$$
a_{11}^2 = a_{21}^2 + a_{22}^2 = a_{31}^2 + a_{32}^2 + a_{33}^2 = a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2.
$$
We're doing a ... | We prove by induction.
The statement holds true for $n = 2$. Suppose it holds true for $n$, there exists the $\{a_{ij}\}_{1\le j \le i\le n}$ such that
$$
\begin{align}
a_{11}^2 & = a_{21}^2 + a_{22}^2\\
& = a_{31}^2 + a_{32}^2 + a_{33}^2\\
& = \cdots\\
& =a_{n1}^2 + a_{n2}^2 + \cdots + a_{nn}^2 \\
\end{align}
$$
For $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$. For a,b satisfying $a^2+b^2=2$. Prove that $3a+3b+ab\geq -5$.
My attempt: We have
$$2(a^2+b^2)\geq (a+b)^2$$
so $$-2\leq a+b \leq 2$$
In other hand $$ab=\frac{(a+b)^2-2}{2}=(a+b)^2-1$$
| Let $a = \sqrt{2} \cos u, b = \sqrt{2} \sin u$ which immediately satisfies the condition. Using Simon's Favorite Factoring Trick, $9 + 3a + 3b + ab = (a + 3)(b + 3)$, so by C-S:
$$(\sqrt{2} \cos u + 3)(\sqrt{2} \sin u + 3) ≥(\sqrt{\sqrt{2} \cos u} \sqrt{\sqrt{2} \sin u} + \sqrt{3} \sqrt{3})^2$$
$$= (\sqrt{2 \cos u \sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Finding $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$ Find $n,m\in N$ such that $|\sqrt{e} - \frac{n}{m}| < \frac{1}{100}$.
I wrote this proof:
Let $f(x)=e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$
$|\sqrt{e}-P_N(\frac{1}{2})|=|R_N(\frac{1}{2})| < \frac{1}{100}$
From the Taylor theorem we get that there exi... | You can just compute the continued fraction getting $[1;1,1,1,5,1,1,9,1,1,13,\ldots]$. If you stop after the $5$ this evaluates to $\frac {28}{17}$, which is off by less than $0.0017$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4171798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove $ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $ without expansion It's easy to prove that if $a,b,c \neq 0 $: $$ \frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff (a+b)(b+c)(c+a)=0 $$
as $\frac{1}{a} +\frac{1}{b} +\frac{1}{c} = \frac{1}{a+b+c} \iff \frac{ab+bc+ca}{ab... | not an answer :
If we assume : $$a+b+c=abc$$
Then we obtain :
$$ab+bc+ca=1$$
Now substitute :
$$\frac{1}{a}=u$$
$$\frac{1}{b}=v$$
$$\frac{1}{c}=w$$
We obtain:
$$\frac{1}{uv}+\frac{1}{vw}+\frac{1}{wu}=1$$
Or :
$$uvw=u+v+w$$
So I recall the tangent formula :
$$\tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)$$
So there is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4174900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Computation of $\int_{0}^{\infty}\frac{x}{\left(1+y^2x^2\right)\left(1+a^2x^2 \right)}dx$ The following integral is simple to compute
$$I=\int_{0}^{\infty}\frac{\arctan(x)}{1+x^2}dx \,\,\tag{1}$$
letting $$u=\arctan x \Rightarrow du=\frac{dx}{1+x^2}$$
$$\int\frac{\arctan(x)}{1+x^2}dx=\int u \, du = \frac{u^2}{2}$$
Ther... | I've left some ideas below. However I'm having difficulty confirming them numerically.
I'll consider the integral in your title, namely
$$F(x,y)=\int_0^\infty \frac{t}{(1+x^2t^2)(1+y^2t^2)}\mathrm{d}t$$
What we do first is use a substitution $s=t^2\implies \mathrm ds/2=t~\mathrm dt$. So,
$$F(x,y)=\frac{1}{2}\int_0^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4176556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My a... | Using the Extended Euclidean Algorithm as described in this answer and adapted to polynomials yields
$$
\begin{array}{r|r|r|r}
\bbox[5px,border:2px solid #C00]{x^2+x+1}&\bbox[5px,border:2px solid #C00]{1}&0\\
\bbox[5px,border:2px solid #090]{x^2-x+1}&0&\bbox[5px,border:2px solid #090]{1}\\
2x\phantom{{}+0}&1&-1&1\\
1&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Proof the equality $\prod_{r=1}^{mn}\left(x+mn -r\right)=\prod_{k=1}^{m}\prod_{l=1}^{n}\left(x+mn-(1+ml-k) \right)$
I came across a proof of Gauss multiplication formula for the Gamma function which relies on the following indentity (without a proof)
$$\frac{\Gamma \left(x+mn \right)}{\Gamma \left(x \right)}=m^{mn}\pr... | Just notice that any natural number $\le nm $ can be uniquely written in the form $ml - k’$, where $1\le l \le n$, $0 \le k’ < m$ are integers (this is just a division by $m$ with remainder). Changing $k’$ on $k - 1$, we can see, that each factor of the left product in (4) appears exactly one time in the right product.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving an equation in quaternions How to solve equation $x^4=1$ in quaternions?
I know how to solve that equation in complex numbers but I have no idea how to do it in quaternions.
I have also a question about general way to solve equations in quaternions. And is there similar theorem to fundamental algebra theorem w... | Let me consolidate my two comments into an answer that may be a bit clearer to the OP.
You want to find all the roots of the polynomial $x^4 - 1$ in the quaternions $\Bbb{H}$. Note that $x^4 - 1 = (x^2 - 1)(x^2 + 1)$, just as in $\Bbb{R}$ or $\Bbb{C}$ and note that $\Bbb{H}$ has no non-trivial zero divisors: if $X, Y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solution set to non-linear equation of $3$ variables I have the following trigonometric equation of $3$ variables:
$$f(\theta,\lambda,\phi)=3 \cos (\theta ) \cos (\lambda ) \cos (\phi
)-(\cos (\theta )+3) \sin (\lambda ) \sin
(\phi )$$
$$-\sin (\theta ) \cos (\lambda )+\sin
(\theta ) \cos (\phi )+3 \cos (\thet... | We prove $f(\theta,\lambda,\phi) \le 0$ for each $\theta,\lambda,\phi$ and that equality holds exactly when you conjecture it does.
Write $$f(\theta,\lambda,\phi) = \cos(\theta)a+\sin(\theta)b+\cos(\lambda)\cos(\phi)-3\sin(\lambda)\sin(\phi)-7$$ for $a = 3\cos(\lambda)\cos(\phi)-\sin(\lambda)\sin(\phi)+3$ and $b = \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Centers of circumcircle define an equilateral triangle Let $ABC$ be a triangle with side $a,b,c$ and angles $\alpha, \beta,\gamma$.
It holds that $\alpha, \beta, \gamma<\frac{2\pi}{3}$.
At the side $\overline{BC}$ there is an equilateral triangle $\triangle BCA''$ at the outter side, i.e. $A''$ is the point for which ... | Join $B'C$ and $A'C$
$\angle A'CB'=\angle A'CB+\angle BCA+\angle ACB'$
$\angle A'CB'=30^{\circ}+\gamma+30^{\circ}$
$\angle A'CB'=60^{\circ}+\gamma$
The fact used above is that the line joining circumcenter of an equilateral triangle to any of the vertices bisects the angle corresponding to that vertex.
Now apply cosin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4183541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\t... |
$\,\\\;\;\;\;\;\;\;\; 2\,S \;=\; x \cdot \sqrt{100-x^2} \;=\; h \cdot 10 \;\le\; 5 \cdot 10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
ARML $1994$ Polynomial Manipulation Question
(ARML $1994$)
If $x^5 + 5x^4 + 10x^3 + 10x^2 - 5x + 1 = 10$, and $x \neq -1$,
compute the numerical value of $(x + 1)^4$.
I came across this problem in a packet about polynomial manipulation and I couldn't find a way past expanding $(x+1)^4$ and subtracting it from the pol... | Remember your binomial coefficients, and notice that the left-hand side of the equation is equal to $(x+1)^5-10x$. So we can rewrite that equation as follows:
\begin{align*}
(x+1)^5-10x&=10\\
(x+1)^5-10(x+1)&=0\\
(x+1)\left[(x+1)^4-10\right]&=0 \, .
\end{align*}
Since $x \neq -1$, the first factor above is nonzero, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4186610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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.