Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$
If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.
My attempt is as follows:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+... | Why are you taking the intersection of those three conditions? It doesn't make much sense to me.
For example, the second condition $\Delta\geq 0$ is not necessary. If $a>\sqrt{2}$ then we have $\Delta<0$ and the inequality is satisfied for all real $x$ (not only for $x>0$):
$$\underbrace{(a^2+a-3)(a-\sqrt{2})}_{>0}x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of cubes formula
$$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-4xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
I can't see where one could start to justify this with... | The most general homogeneous quadratic function of two variables is $ax^2+bxy+cy^2$. Cubing this gives $$a^3x^6+3a^2bx^5y+(3a^2c+3ab^2)x^4y^2+(b^3+6abc)x^3y^3+(3ac^2+3b^2c)x^2y^4+3bc^2xy^5+c^3y^6.$$(The fact that this sextic is homogeneous makes the above easy to deduce from combinatorics, without manually expanding ou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
What is the sum of $k^2(n - k)$ for $k = 1$ to $k = n$? What is the value of $\sum\limits_{k=1}^n k^2(n - k)$?
The problem I had was:
for a square grid of size $n \times n$ how many squares have their corners on the intersecting points of the grid. (There are $n \times n$ points, the square's length is $n - 1$).
For e... | Let's start by splitting the sum into two different sums:
$$
\sum_{k=1}^n k^2(n-k)=\sum_{k=1}^n k^2n-\sum_{k=1}^n k^3=n\sum_{k=1}^n k^2-\sum_{k=1}^n k^3
$$
Now we have $\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k^3=\left( \frac{n(n+1)}{2}\right) ^2=\frac{n^2(n+1)^2}{4}$ and together
$$
n\frac{n(n+1)(2n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dealing with degrees in decomposition into partial fractions I had to decompose $ \frac{2x^2}{x^4-1} $ into partial fractions in order to determine its antiderivative.
So, I said:
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}
$$
However, in the answer key, they said:
$$
\frac{2x^2}{x^4-1} =... | There are always alternatives, one does not expect an answer key to give every way that works. For finding an antiderivative,
$$
\frac{2x^2}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx}{x^2+1} + \frac{D}{x^2+1}
$$
is quite good in terms of recognizing things. $C$ leads to a substitution, while $D$ leads to an ar... | {
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"url": "https://math.stackexchange.com/questions/3400562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Integrate area enclosed by $x^4 + y^4 = x^3 + y^3 $ I want to calculate the area bounded by the region
$$x^4 + y^4 = x^3 + y^3 $$
with integral. And also it's perimeter.
Can somebody please help me with it?!
|
Express the curve $x^4 + y^4 = x^3 + y^3$ in its polar coordinates,
$$r(\theta) = \frac{\cos^3\theta+\sin^3\theta}{\cos^4\theta+\sin^4\theta}\tag{1}$$
Recognizing that a complete loop is formed starting at origin and varying $\theta$ from $-\frac{\pi}{4}$ to $\frac{3\pi}{4}$. the area integral is then,
$$A= \int_{-\fr... | {
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A Wallis-like formula for $\pi$: $(\frac21)^2(\frac23)^2(\frac43)^2(\frac45)(\frac65)^2(\frac67)^2(\frac87)(\frac89)^2\cdots$ I don't quite understand this and would appreciate a clear explanation, if possible.
If we begin with the Wallis product for $\frac{\pi}{2}$,
$$\prod_{n=1}^\infty \left( \frac{2n}{2n-1} \cdot \f... | We have to show that the product of those factors is (convergent and) equal to $2$.
The factors are precisely $\frac{p-\chi(p)}{p+\chi(p)}$, where $p$ runs through odd primes, and $$\chi(n)=\begin{cases}(-1)^{(n-1)/2},&n\text{ is odd}\\\hfill 0,\hfill&n\text{ is even}\end{cases}$$ is known as the non-trivial Dirichlet ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Then find the sum of all possible values of $abc$. Let $a, b, c$ be positive integers with $0 < a, b, c < 11$. If $a, b, $ and $c$ satisfy
\begin{align*}
3a+b+c&\equiv abc\pmod{11} \\
a+3b+c&\equiv 2abc\pmod{11} \\
a+b+3c&\equiv 4abc\pmod{11} \\
\end{align*}then find the sum of all possible values of $abc$.
What I trie... | $3a + b + c \equiv abc \pmod{11}$
$a + 3b + c \equiv 2abc \pmod{11}$
$a + b + 3c \equiv 4abc \pmod {11}$
So
$(3a+b+c)+ 3(a+3b+c) + (a+b+3c)\equiv 11abc$
$7(a+c)\equiv 0\pmod{11}$ As $11$ is prime you can divide by $7$. (This might not be the case if $\gcd(7,11)\ne 1$ but in this case you can.)
$a\equiv -c\pmod {11}$... | {
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How to find x from the equation $\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$ For m are real number,
Find x from the equation
$$\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x+\sqrt{x}}}$$
I tried to multiply $\sqrt{x + \sqrt{x}}$ to the both sides and I get
$$x + \sqrt{x} -... | From the simplified form of @gt6989b, we get $$\sqrt{x}- \sqrt{x-1}=m-1.$$
Squaring for rationalization one must demand that $$m \ge 1...(1),~~ x \ge 1....(2),~~~\sqrt{x} \ge m-1....(3)$$.The Eq. (3) gives a very interestiong condition which has been missed out in the answers by @Dinno Koluh and @trancelocation. When... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rewriting a binomial formula summation I am trying to obtain an explicit solution for $p$ from the following equation:
$p = 1 - \left[ \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} \frac{1}{k+1} \right]^{\frac{1}{{n}}}$.
Obviously, I know that summing a binomial pdf we have
$\sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} = ... | $$
\eqalign{
& \sum\limits_{\left( {0\, \le \,} \right)\,k\,\left( { \le \,n} \right)} {{1 \over {k + 1}}\left( \matrix{
n \cr
k \cr} \right)p^{\,k} q^{\,n - k} } = \sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)} {{1 \over {n + 1}}\left( \matrix{
n + 1 \cr
k + 1 \cr} \right)p^{\,k} q^{\,n - k} } = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3405671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Definite integration evaluation of $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$. $$\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$$
how to proceed please help
The answer given is $\dfrac{\pi}{4a^3b}$.
| Change variable to $t = \tan x$, we have
\begin{align}
&\quad\int_0^{\frac{\pi}{2}}\frac{\sin ^2 x}{\left(b^2\cos^2 x+a^2\sin^2 x\right)^2}\, dx\\&=\int_0^{\frac{\pi}{2}}\frac{\tan^2 x}{\left(b^2+a^2\tan^2 x\right)^2}\cdot\frac{1}{\cos^2 x}\,dx\\&=\int_0^\infty\frac{t^2}{\left(b^2+a^2t^2\right)^2}\,dt\\&=\left.-\frac{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!
*
*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because... | Convergence:
First, let $f(x)=\frac{x \ln x}{(x^2+1)^2}$ then $f$ is decrease on $x>t$ for some $t\in\mathbb{R}$
Use $$x>\ln x \Rightarrow f(x)\leq\frac{x^2}{(x^2+1)^2}$$
and we know
$$(x^2+1)^2=x^4+2x^2+1\geq x^4 \Rightarrow \frac{1}{(x^2+1)^2}\leq\frac{1}{x^4}$$
Apply this for our first inequality, then we get
$$0\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f \big(x ^ 2 + y ^ 2f (x)\big) = xf (y) ^ 2-f (x) ^ 2$
Find all functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left ( x ^ 2 + y ^ 2 f ( x ) \right ) = x f ( y ) ^ 2 - f ( x ) ^ 2 $$
for all $ x , y \in \mathbb R $.
Let $P(x,y)$ denote the functional equ... | It' easy to see that the $ f ( x ) = 0 $ and $ f ( x ) = - x $ are both solutions to the functional equation
$$ f \big( x ^ 2 + y ^ 2 f ( x ) \big) = x f ( y ) ^ 2 - f ( x ) ^ 2 \text . \tag 0 \label 0 $$
We show that those are the only solutions. Letting $ x = y = 0 $ in \eqref{0}, we find out that either $ f ( 0 ) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409897",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If A commutes with both of these matrices, then A must be a scalar multiple of the identity matrix I am working on the following problem:
Let $A$ be a $4 \times 4$ matrix with entries in a field of characteristic zero. Suppose that $A$ commutes with both $\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 3 & 0\\... | My answer can be treated as the supplement to the Loup Blanc's answer, I would like to express similar result in a more elementary way.
Denote both mentioned matrices as $D$ and $P$.
It is easy to check that if $A$ commutes with matrices $D$ and $P$ then it commutes also with any power of matrices $D$ and $P$, any ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Maximum value of $8v_1 - 6v_2 - v_1^2 - v_2^2$ subject to $v_1^2+v_2^2\leq 1$
Given that $g:\mathbb{R}^2 \to \mathbb{R}$ defined by
$$g(v_1,v_2) = 8v_1 - 6v_2 - v_1^2 - v_2^2$$
find the maximum value of $g$ subject to the constraint $v_1^2+v_2^2\leq 1.$
My attempt:
Note that
$$g(v_1,v_2) = 8v_1 - 6v_2 - v_1^2 ... | Your result is correct but it is not clear what you mean by $g$ is a decreasing function and "find the intersection between $(0,0)$ etc."
So, here is another way using Cauchy-Schwarz:
$$8v_1 - 6v_2 \leq \sqrt{8^2+6^2}\sqrt{v_1^2+v_2^2} = 10 \sqrt{v_1^2+v_2^2}$$
$$g(v_1,v_2)\leq 10\sqrt{v_1^2+v_2^2}- (v_1^2 + v_2^2)=t(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If equation rep $x^2+2y^2-5z^2+2kyz+2zx+4xy=0$ represent pair of plane. then $k$ is The values of $k$ for which the equation
$x^2+2y^2-5z^2+2kyz+2zx+4xy=0$
represents a pair of plane passing Through origin,is
what i try
$x^2+2y^2-5z^2+2kyz+2zx+4xy=(ax+by+cz)(px+qy+rz)$
and camparing coefficients
but it is very tediou... | We can assume that $a=p=1$, so we have $$x^2+2y^2-5z^2+2kyz+2zx+4xy=(x+by+cz)(x+qy+rz)$$
This should be true for all $y$, so it is true for $y=0$ also, and we get:
$$x^2-5z^2+2zx=(x+cz)(x+rz)$$ So $cr=-5$ and $c+r=-2$. Also it is true for all $z$ and specialy for $z=0$: $$x^2+2y^2+4xy=(x+by)(x+qy)$$ so $bq=2$ and $b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} ... | A variant, with lil' Fermat:
*
*As $20$ is not divisible by $11$, we have $20^{15}\equiv 1\mod11 $.
*$2^{30}\equiv 1\mod 31$ and $2^{15}\equiv-1\mod 31$, so $(-2)^{15}\equiv 5^{15}\equiv 1\mod 31$.
*$2^6\equiv 3\equiv 5^3$, so $\;2^{30}5^{15}\equiv 9^5\mod 61$. Now $9^2\equiv 20$, so $9^3\equiv 180\equiv-3$ and ul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Constrained minimum with inequalities Show that for all real positive $x$ and $y$ such that $x+y=1$, the following holds: $$\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\geq \frac{25}{2}.$$
I know this can probably be easily solved with Lagrange multipliers, but I am more interested in proving it using ineq... | This problem can be rewritten as:
$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq\frac{5}{2}$$
We can use inequality between square and arithmetic mean:
$$\sqrt{\frac{(x+\frac{1}{x})^2+(y+\frac{1}{y})^2}{2}} \geq \frac{(x+\frac{1}{x})+(y+\frac{1}{y})}{2}=\frac{1+\frac{1}{x}+\frac{1}{y}}{2}$$
We only have to... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\left |\frac{1}{\sqrt{n}} \sum_{i=1}^{[tn]}X_i-\frac{\sqrt{t}}{\sqrt{[tn]}} \sum_{i=1}^{[tn]}X_i\right |\overset{n}{\to}0$ in probability
As the title states, I would like to prove that
$$\left | \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor tn \rfloor }X_i - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \sum_{i=1}... | Here's an attempt to an answer using the points from stochasticboy321's comment above.
We will try to apply Chebyshev's inequality which in our case states, since the variance of $\sum_{i=1 } ^{\lfloor nt \rfloor } X_i$ is $\lfloor nt \rfloor$, that
$$P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i | \ge \epsilon \sqrt{\lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$ I'm trying to find the number of real roots of the equation $5+|2^x-1|=2^x(2^x-2)$.
Let $2^x=a$
$$|a-1|=a^2-2a-5$$
Then there are two cases
$$a-1=a^2-2a-5$$
And $$a-1=-a^2+2a+5$$
Solving both equations
$$a=1,-4,-2,3$$
Now -4 and -2 can be neglected so th... | There are two cases:
*
*$x\geq0$. The equation becomes $$2^{2x}-2^{x+1}-2^x-4=0$$$$(2^x-2)(2^x-1)= 6$$Clearly, $2^x-2>0$, which implies $x>1$. Let $y=2^x$. We have $y^2-3y-4=0$ which implies $$(y-4)(y+1)=0$$ So, $y=4$ and $x = 2$.
*$x<0$. The equation becomes $$6-2^x=2^{2x}-2^{x+1}$$$$2^{2x}-2^{x+1}+2^x-6=0$$$$(2^x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve this integral for free WiFi I saw this today, I checked in Mathematica and the integral comes out to $\pi$, but I have no idea how to solve it.
FREE Wi-Fi: The Wi-Fi password is the first $10$ digits of the answer.
$$\int_{-2}^2\left(x^3\cos\frac x2+\frac12\right)\sqrt{4-x^2}\ dx$$
| We have that
$$\int_{-2}^2 x^3 \cos\frac x2 \sqrt{4-x^2} dx =0$$
since the integrand is point symmetric in the origin.
Since $\sqrt{4-x^2}$ on $[-2;2]$ is the formula for the upper part of a circle we find that
$$\int_{-2}^2 \sqrt{4-x^2} dx=\frac 12 \pi r^2=\frac 12 \pi 2^2=2\pi$$
So the whole integral is:
$$\int_{-2}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 2,
"answer_id": 1
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Find the range of $f(x)=2|{\sin x}|-3|\cos x|$ Find the range of $f(x)=2|{\sin x}|-3|\cos x|$
My attempt is as follows:-
$$f(x)=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|\right)$$
Let's assume $z=\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|$
$$f(x)=\sqrt{13... | For $0\le x\le\dfrac\pi2,$
$$f(x)=\sqrt{2^2+3^2}\sin\left(x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\right)$$
Now $0\le x\le\dfrac\pi2\implies-\arcsin\dfrac3{\sqrt{2^2+3^2}}\le x-\arcsin\dfrac3{\sqrt{2^2+3^2}}\le\dfrac\pi2-\arcsin\dfrac3{\sqrt{2^2+3^2}}$
Again,
$\dfrac\pi2-\arcsin\dfrac3{\sqrt{2^2+3^2}}=\arccos\dfrac3{\sqrt{2^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428466",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the unit digit of $2124^{392}+3143^{394}*7177^{392}-8818^{394}$ What I do here is I find the remainder of each parcel in mod 10 and then do the maths and mod 10 again:
$2124^{392} \equiv 4^{392} \pmod{10}$
$$4^1 \equiv 4 \pmod{10} \\
4^2 \equiv 6 \\
4^3 \equiv 4 \\
4^4 \equiv 6 \\
(...)$$
Then I do $392 \pmod{2}$... | $4^1\equiv 4$
$4^2\equiv 6$
$4^3\equiv 4$
$\vdots$
When the exponent is odd, the remainder is $4$.
When the exponent is even the remainder is $6$.
$0$ is even and so is $392$, thus $4^{392}\equiv 6\pmod{10}$
As an alternative notice that $6^k\equiv 6\pmod{10}$,
so $4^{392} = 16^{196}\equiv 6\pmod{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428802",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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General formula for $e^x+\cos(x)$, $e^x+\sin(x)$, $e^x-\sin(x)$, $e^x-\sin(x)$ I have been able to derive the formal series for these four functions:
$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$
$e^x+\cos(x) = 2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4... | If the terms in the series for the two functions you're adding or subtracting had the same numbering scheme, you could easily add or subtract them term by term within the same numbering scheme.
The reason you cannot do this with the usual series for these functions is that the $n$th term in the series for $e^x$ is the ... | {
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"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Logarithm question - spurious solution So I have the equation that I want to solve
$$\log_2(8x) - \log_2(1+\sqrt{x}) = 3, \hspace{1mm} x>0$$
My solution is $$\log_2\left(\frac{8x}{1+\sqrt{x}}\right) = 3 \\ \Rightarrow x = 1 + \sqrt{x} \\ \Rightarrow x - \sqrt{x} - 1 = 0 \\ \Rightarrow \sqrt{x} = \frac{1}{2} \pm \frac... | $$ \sqrt{x} = \frac{1}{2} \pm \frac{\sqrt{5}}{2} $$
At this stage you must ditch the negative solution $\frac{1}{2} - \frac{\sqrt{5}}{2}$ beacause $\sqrt{x}$ is non negative.
| {
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"question_score": "3",
"answer_count": 2,
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Very tricky triple integral in spherical coordinates Evaluate $$\int_{-2}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{2-\sqrt{4-x^2-y^2}}^{2+\sqrt{4-x^2-y^2}}(x^2+y^2+z^2)^{3/2} \; dz \; dy \; dx$$ by converting to spherical coordinates.
We know that $(x^2+y^2+z^2)^{3/2} = (\rho^2)^{3/2} = \rho^3$. The range of $y$ tel... | The boundary of our sphere is $x^2 + y^2 + z^2 = 4z$
Plugging
$x = \rho\cos\theta \sin\phi\\
y = \rho\sin\theta\sin\phi\\
z = \rho \cos\phi$
We get:
$\rho^2 = 4\rho\cos\phi\\
\rho = 4\cos\phi$
And the sphere is above the $xy$ plane, or $\phi \le \frac {\pi}{2}$
$\int_0^{2\pi}\int_0^{\frac {\pi}{2}}\int_0^{4\cos\phi} (\... | {
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How to prove the identity $\sin2x + \sin2y = 2\sin(x + y)\cos(x - y)$ I've been trying to prove the identity $$\sin2x + \sin2y = 2\sin(x + y)\cos(x - y).$$
So far I've used the identities based off of the compound angle formulas. I'm not quite sure if those identities would work with proving the above identity.
Thank y... | \begin{align}
2\sin(x + y)\cos(x - y)
&= 2(\sin x \cos y + \cos x \sin y)\cdot
(\cos x \cos y + \sin x \sin y) \\
&= 2\sin x \cos x(\cos^2 y + \sin^2 y) +
2\sin y \cos y(\cos^2 x + \sin^2 x) \\
&= 2\sin x \cos x + 2\sin y \cos y \\
&= \sin 2x + \sin 2y
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3436711",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Find $\frac{\partial x}{\partial u}$, if $u=x+y^2$, $v=y+z^2$, $w=z+x^2$. I tried very hard but couldn't make a relation out them. I cross checked question too but in the book it's given as above. Hints will be appreciated too. Thanks.
| $\text{We have}$:
$u=x+y^2\tag 1$ $v=y+z^2\tag 2$ $w=z+x^2\tag 3$
$\text{Therefore,}$
$\begin{align}x &=u-y^2[\text{ from }(1)]\\&=u-(v-z^2)^2[\text{ from }(2)]\\&=u-[v-(w-x^2)^2]^2[\text{ from }(3)]\\&=u-[v-(w^2-2wx^2+x^4)]^2\\&=u-[v-w^2+2wx^2-x^4)]^2\end{align}$
$\begin{align}\implies\dfrac{\partial x}{\partial u}&=1... | {
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"answer_count": 1,
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Range of $a$ in $x^2-a=\sqrt{x+a}$
The equation $x^2-a=\sqrt{x+a}$ has real or
imaginary roots depending on the values of $a.$
Then range of $a$ for which the equation.
$(a)\;\; $ No real roots
$(b)\;\; $ One real root
$(c)\;\;$ Exactly two real roots
$(d)\;\;$ At least two real roots
what i try
$x^2-a=\sqrt{x+a}\Rig... | First of all, a quadratic equation has no real roots when its discriminant is negative. So you have $1+4a<0$ and similarly for the other inequality. However, these are not necessarily all the cases. The equation is equivalent to the system
$$\begin{align}
(x^2-a)^2 &= x+a \\
x^2-a &\geq0 \Leftrightarrow x^2\geq a
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440510",
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"answer_count": 1,
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prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^... | You could show by induction that modulo $30$
$5^{2k+1}\equiv5,$
$3^{2k+1}\equiv3$ if $k$ is even and $-3$ if $k$ is odd, and
$2^{2k+1}\equiv2$ if $k$ is even and $8$ if $k$ is odd.
| {
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"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Problematic inequality & hint I would like to ask for hint for proving following inequality:
$$x^3(1+x)+y^3(1+y)+z^3(1+z)\geq \frac{3}{4}(1+x)(1+y)(1+z)$$
for all $x>0$, $y>0$, $z>0$ such that $xyz=1$.
Generally, I tried to find some elementary solution, but even with some calculus I didn't solve it.
Edit. My attempt:
... | (Edited version). By AM-GM we have:
$$ \sum_{cyc}(\frac{x^3}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8})$$
$$ \geq \sum_{cyc}3 \sqrt[3]{\frac{x^3}{(1+y)(1+z)} \cdot \frac{1+y}{8} \cdot \frac{1+z}{8}}$$
$$=\frac{3}{4}(x+y+z)$$
we have:
$$
\sum_{cyc}(\frac{x^3}{(1+y)(1+z)}\geq \frac{3}{4}(x+y+z) - \sum_{cyc}(\frac{1+y}{8} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
\int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh... | Use your 2nd trial:
$$
\newcommand{\abs}[1]{\left\vert #1 \right\vert}
\newcommand\rme{\mathrm e}
\newcommand\imu{\mathrm i}
\newcommand\diff{\,\mathrm d}
\DeclareMathOperator\sgn{sgn}
\renewcommand \epsilon \varepsilon
\newcommand\trans{^{\mathsf T}}
\newcommand\F {\mathbb F}
\newcommand\Z{\mathbb Z}
\newcommand\R{\Bb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| Since: $$\sqrt{2 a^2+b+1}= \sqrt{2 a^2-a+3}\geq {3\over 4}a+{5\over 4}$$
and the same for the other root, we have $$...\geq ({3\over 4}a+{5\over 4})+
({3\over 4}b+{5\over 4})=4$$
Here is a detailed proof for $$\sqrt{2 a^2-a+3}\geq {3\over 4}a+{5\over 4}$$
It is equivalent to $$16(2a^2-a+3)\geq (3a+5)^2$$
and this is e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
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Limit with radicals, $\cos$, $\ln$ and powers $\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\f... | Your conclusion is wrong since we find an indeterminate form $\frac 0 0$ and we can't conclude that the limit is zero.
Indeed we have that by standard limits
$$\sqrt[6]{1-\cos{\frac{1}{x^3}}} = \frac1{x\sqrt[6]2}+O\left(\frac1{x^2}\right)$$
$$2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}} = \frac1x\log \frac32+O\left(\frac1{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I justify this without determining the determinant? I need to justify the following equation is true:
$$
\begin{vmatrix}
a_1+b_1x & a_1x+b_1 & c_1 \\
a_2+b_2x & a_2x+b_2 & c_2 \\
a_3+b_3x & a_3x+b_3 & c_3 \\
\end{vmatrix} = (1-x^2)\cdot\begin{vmatrix}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c... | \begin{align}
&\phantom {=}\,\ \begin{vmatrix}
a_1+b_1x & a_1x+b_1 & c_1 \\
a_2+b_2x & a_2x+b_2 & c_2 \\
a_3+b_3x & a_3x+b_3 & c_3
\end{vmatrix} \\
&=
\begin{vmatrix}
a_1 & a_1x+b_1 & c_1 \\
a_2 & a_2x+b_2 & c_2 \\
a_3 & a_3x+b_3 & c_3
\end{vmatrix}
+ \begin{vmatrix}
b_1x & a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 7,
"answer_id": 2
} |
Value of $L +\frac{153}{L}=$
If $\displaystyle L = \lim_{x\rightarrow 0}\bigg(\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}\bigg).$
Then value of $\displaystyle L +\frac{153}{L}=$
what i try
$$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}$$
from D L Hopital rule
$$L=\lim_{... | We have that
$$\ln(x+\sqrt{x^2+1})=\ln (\sqrt{x^2+1})+\ln\left(1+\frac x{\sqrt{1+x^2}}\right)=$$
$$=\frac12\ln (1+x^2)+\ln\left(1+x\left(1-\frac12x^2+O(x^4)\right)\right)=$$
$$=\frac12\ln (1+x^2)+\ln\left(1+x-\frac12x^3+O(x^5)\right)=$$
$$=\frac12x^2+x-\frac12x^2+O(x^3)=x+O(x^3)$$
then
$$\frac1{\ln(x+\sqrt{x^2+1})}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc}... | By C-S
$$\sum_{cyc}\sqrt{a^2+ab+b^2}=\sqrt{\sum_{cyc}(a^2+ab+b^2+2\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}}=$$
$$=\sqrt{\sum_{cyc}(2a^2+ab++2\sqrt{\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right)\left(\left(a+\frac{c}{2}\right)^2+\frac{3c^2}{4}\right)}}\geq$$
$$\geq\sqrt{\sum_{cyc}(2a^2+ab+2\left(\left(a+\frac{b}{2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450882",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?
$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$
I have ... | Using Taylor series you get $\log(1+1/x)=\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+\ldots$, thus $$(x^3+x+1)\log(1+1/x)=(x^3+x+1)(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+\ldots)=$$
$$x^2-\frac{x}{2}+\frac{1}{3}+1+\text{other terms that goes to 0 for }x\rightarrow\infty.$$
Hence
$$\lim x^2-\frac{x}{2}-(x^3+x+1)\log(1... | {
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"source": "stackexchange",
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Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$.
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.
Question from Jalil Hajimir
| Let $x=\frac{a}{\sqrt2},$ $y=\frac{b}{\sqrt2}$ and $z=\frac{c}{\sqrt2}.$
Thus, since we can assume that $(a^2-1)(b^2-1)\geq0,$ by C-S we obtain:
$$\frac{(x^2+1)(y^2+1)(z^2+1)}{(x+y+z)^2}=\frac{(a^2+2)(b^2+2)(c^2+2)}{4(a+b+c)^2}\geq$$
$$\geq\frac{3(a^2+b^2+1)(1+1+c^2)}{4(a+b+c)^2}\geq\frac{3(a+b+c)^2}{4(a+b+c)^2}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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What is the period of the $f(x)=\sin x +\sin3x$?
What is the period of the $f(x)=\sin x +\sin3x$?
$f(x)=\sin x+\sin 3x=2\frac{3x+x}{2}\cos\frac{x-3x}{2}=2\sin2x\cos x=4\sin x\cos^2x\\f(x+T)=4\sin(x+T)\cos^2(x+T)=4\sin x\cos^2 x$
how can I deduct this I have no idea
| $f(x)=f(x+2\pi)$ so it has a period which is a factor of $2\pi$.
$df/dx=4$ only when $x$ is a multiple of $2\pi$ so any period is a multiple of $2\pi$.
So the period is $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Prove by mathematical induction that $(3n+1)7^n -1$ is divisible by $9$ for integral $n>0$ $7^n(3n+1)-1=9m$
$S_k = 7^k(3k+1)-1=9P$
$\Rightarrow 7^k(3k+1) = 9P+1$
$S_{k+1} = 7\cdot7^k(3(k+1)+1)-1$
$= 7\cdot7^k(3k+1+3)-1$
$= 7\cdot7^k(3k+1) +21\cdot7^k -1$
$= 7(9P+1)+21\cdot7^k -1$
$= 63P+7+21\cdot7^k -1$
$= 63P+6+21\cdo... | To address your question directly:
You calculations are correct but - as you realized by yourself - the conclusion at the end is still not justified since there are fractions in "$= 9(7P +2/3+21\cdot7^k/9)$".
So, just go one step back and try to squeeze out factor $9$, for example, as follows:
\begin{eqnarray}63P+6+21\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Determinant of a 3*3 matrix of cosines I need help in evaluating the following determinant:
$$\begin{vmatrix}
\cos\frac{2\pi}{63} & \cos\frac{3\pi}{90} & \cos\frac{4\pi}{77} \\
\cos\frac{\pi}{72} & \cos\frac{\pi}{40} & \cos\frac{3\pi}{88} \\
1 & \cos\frac{\pi}{90} & \cos\frac{2\pi}{99}
\end{vmatrix}$$
I noticed that ... | We assume the second entry to be $\cos(3\pi/70)$.
Multiply the last row by $-\cos(\pi/72)$ and add it the the second row; multiply the last row by $-\cos(2\pi/63)$ and add it to the first row. Then use Laplace expansion to obtain
$$\begin{vmatrix}
-\cos(\frac{\pi}{90}) \cos(\frac{\pi}{72}) +
\cos(\frac{\pi}{40})& -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3459899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove this $\sum_{i=1}^{n}(x_{i})^{S-x_{i}}>1?$ Question:
Let $x_{i} \in (0,1),i=1,2,\cdots,n$. Show that
$$
x_{1}^{S-x_{1}}+x_{2}^{S-x_{2}}+\cdots+x_{n}^{S-x_{n}}>1
$$
where $S=x_{1}+x_{2}+\cdots+x_{n}$.
I have proved when $n=2$,because it use this Bernoulli's inequality
$$
(1+x)^a\le 1+ax,0<a\le 1,x>-1
$... | For $n=3$ we can use your work and the Canhang's idea.
Let $\{a,b,c\}\subset(0,1).$ Prove that:
$$a^{b+c}+b^{a+c}+c^{a+b}>1.$$
Proof.
Let $a+b+c\leq1.$
Thus, by Bernoulli
$$\sum_{cyc}a^{b+c}=\sum_{cyc}\frac{1}{\left(1+\frac{1}{a}-1\right)^{b+c}}\geq\sum_{cyc}\frac{1}{1+\left(\frac{1}{a}-1\right)(b+c)}>\sum_{cyc}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3461483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Stuck on a probability problem/Expectation of coin toss I'm stuck on the following problem that is due for tomorrow:
We're flipping 1 coin indefinitely. $X$ is a random variable that count the amount of coin tosses.
What is the expectation of the number of coin toss until we have the following sequence: THH (T: tail... | Generating Function Approach
Duration Until $\boldsymbol{THH}$
Any trial can uniquely be constructed from any number of $H$ atoms, then an arbitrary combination of $TH$ and $T$ atoms, then a $THH$ atom. Put this together in a generating function:
$$
\begin{align}
\overbrace{\ \ \frac1{1-x}\ \ }^\text{$H$ atoms}\overbra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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particular solution of $(D^2+4)y=4x^2\cos 2x$
Find a particular solution to the equation $$(D^2+4)y=4x^2\cos 2x$$
\begin{align}
y_p&=\frac{1}{D^2+4}(4x^2\cos 2x)\\
&=\frac{1}{D^2+4}\left[2x^2(e^{2ix}+e^{-2ix})\right]\\
&=2\left[e^{2ix}\frac{1}{(D+2i)^2+4}x^2+e^{-2ix}\frac{1}{(D-2i)^2+4}x^2\right]\\
&=2\left[e^{2ix}\f... | Hint: $$\dfrac{1}{1+D} = 1-D+D^2-\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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evaluation of Trigonometric limit
Evaluation of $$\lim_{n\rightarrow \infty}\frac{1}{n\bigg(\cos^2\frac{n\pi}{2}+n\sin^2\frac{n\pi}{2}\bigg)}$$
What i try
Put $\displaystyle \frac{n\pi}{2}=x,$ when $n\rightarrow\infty,$ Then $x\rightarrow \infty$
$$\frac{\pi^2}{2}\lim_{x\rightarrow \infty}\bigg(\frac{ x^{-1}}{\pi\cos... | \begin{align*}
\lim_{k\rightarrow\infty}\dfrac{1}{2k[\cos^{2}(2k\pi/2)+2k\sin^{2}(2k\pi/2)]}=\lim_{k\rightarrow\infty}\dfrac{1}{2k}=0,
\end{align*}
while
\begin{align*}
&\lim_{k\rightarrow\infty}\dfrac{1}{(2k+1)[\cos^{2}((2k+1)\pi/2)+(2k+1)\sin^{2}((2k+1)\pi/2)]}\\
&=\lim_{k\rightarrow\infty}\dfrac{1}{(2k+1)(2k+1)}\\
&... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find the recurrence relation solution. $a_n$ = $a_{n-1} + 3n - 5$ $a_n$ = $a_{n-1} + 3n - 5$ $,$ $a_0 = 7$
So far what I got is:
$a_0 = 7$
$a_1 = 7+3(1)-5$
$a_2 = 7+3(1)-5 + 3(2)-5$
$a_3 = 7+3(1)-5 + 3(2)-5 + 3(3)-5$
$a_4 = 7+3(1)-5 + 3(2)-5 + 3(3)-5 + 3(4)-5$
This is where I'm stuck I'm having trouble finding a patter... | Let $A(z) = \sum_{n=0}^\infty a_nz^n$. Multiply both sides of the recurrence by $z^n$ and sum over $n$ to obtain
$$
\sum_{n=1}^\infty a_nz^n = \sum_{n=1}^\infty a_{n-1}z^n + \sum_{n=1}^\infty 3nz^n - \sum_{n=1}^\infty 5z^n.
$$
By some algebra we obtain
$$
A(z) - a_0 = zA(z) +\sum_{n=0}^\infty 3nz^n - \sum_{n=1}^\infty ... | {
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Without calculating the square roots, determine which of the numbers:$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$ is greater.
Without calculating the square roots, determine which of the numbers:
$$a=\sqrt{7}+\sqrt{10}\;\;,\;\; b=\sqrt{3}+\sqrt{19}$$
is greater.
My work (I was wondering if there are other way... | Since $a$ and $b$ are both positive, it follows that $a<b \iff a^2<b^2$; namely $$a<b\iff17+2\sqrt{70}<22+2\sqrt{57};$$ that is, $a<b$ iff $2\left(\sqrt{70}-\sqrt{57}\right)<5.$ Continuing equivalent statements in this way, we get
$$
a < b
\iff 508-8\sqrt{3990}<25
\iff 60 \tfrac38 < \sqrt{3990}
\iff 3600+45+\tfrac9{... | {
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"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
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Find the minimal polynomial for $\cos(\frac{2\pi}{5})$ and $\sin(\frac{2\pi}{5})$ Let $\omega$ be the primitive 5th root of $1$, then $\cos(\frac{2\pi}{5}) = \frac{w+w^{-1}}{2}$ and $\sin(\frac{2\pi}{5}) = \frac{w-w^{-1}}{2i}$. How to find the minimal polynomial of $\frac{w+w^{-1}}{2}$ then? (without using the Chebysh... | Let $w=e^{2\pi i/5}$. Then $w^5=1$ and $w\ne1$ so $w^4+w^3+w^2+w+1=0$, $w^4=w^{-1},$ and $w^3=w^{-2}$.
Also, $\cos(2\pi i/5)=\dfrac{w+w^{-1}}2$,
so $\cos^2(2\pi i/5)=\dfrac{w^2+w^{-2}+2}4=\dfrac{w^2+w^3+2}4=\dfrac{-1-w-w^{-1}+2}4=\dfrac{1-2\cos(2\pi i/5)}{4}.$
Can you take it from here?
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Show that $2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$ The question states:
Show that: $$2(\sin y + 1)(\cos y + 1) = (\sin y + \cos y + 1)^2$$
This is what I have done
$2(\sin y + 1)(\cos y + 1) = 2(\sin y + \cos y + 1)^2$
L. H. S. = R. H. S.
From L. H. S.
$2(\sin y +1)(\cos y + 1) = 2(\sin y... | $$RHS - LHS=(\sin y + \cos y + 1)^2-2(\sin y + 1)(\cos y + 1) $$
$$=[(\sin y +1)+( \cos y + 1) -1]^2-2(\sin y + 1)(\cos y + 1) $$
$$=(\sin y +1)^2+( \cos y + 1)^2 + 1 -2(\sin y + 1)-2(\cos y + 1) $$
$$=[(\sin y +1)^2-2(\sin y + 1)+1]+[( \cos y + 1)^2 - 2(\cos y + 1)+1] -1 $$
$$=\sin^2 y+\cos^2 y -1 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Possible Jordan Canonical Forms Suppose I have a matrix $A \in M_{n \times n}(\mathbb{C})$ such that its minimal polynomial is either $x-1$ or $(x-1)^{2}$. What are its possible Jordan Canonical Forms? I was thinking that if its minimal polynomial is $x-1$, then its Jordan canonical form is $I_{n}$, the $n \times n$ id... | Consider the special case when $A$ is nilpotent:
If minimal polynomial has degree $m=1$, then largest Jordan block will be of size $1\times 1$, thus JCF is diagonal. This can be generalized to any matrix.
If minimal polynomial has degree $m=2$, then largest Jordan block will be of size $2\times 2$, and the rest of the ... | {
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"question_score": "3",
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Having problem to find all solutions using the Chinese Remainder Theorem Find all solutions using the Chinese Remainder Theorem.
x = 2 (mod 4)
x = 3 (mod 5)
x = 9 (mod 13)
My step is here:
a = 2 (mod 4) , a = 0 (mod 5) , a = 0 (mod 13)
b = 0 (mod 4) , b = 3 (mod 5) , b = 0 (mod 13)
c = 0 (mod 4) , c = 0 (mod 5) , c = ... | We have ,
$$x\equiv2 \mod 4 \implies \color{#d05}{x=4a+2}\\ $$
$$\begin{align}x&\equiv3 \mod 5 \\4a+2&\equiv3\mod5\\ 4a&\equiv 1\mod5 \\ a &\equiv 4 \mod 5\implies\color{#2cd}{a = 5b+4}\end{align} $$
$$\begin{align}x&\equiv9 \mod 13 \\ 4(5b+4)+2&\equiv 9\mod13 \\ 20b+18&\equiv9\mod13 \\ 20b&\equiv 4\mod 13 \implies... | {
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"answer_count": 3,
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Find a cubic polynomial satisfying the given conditions Find $p(x) = ax^3 + bx^2 + cx + d $, where a, b, c and d are real constants, satisfying the following conditions:
$1)\quad 4x^3 - 12x^2 + 12x - 3 \le p(x) \le 2019(x^3 - 3x^2 + 3x) - 2018,$ for all values of $x \ge 1$.
$2)\quad p(2) = 2011.$
My idea:
The conditio... | Continue with what you obtained,
$$4(x - 1)^3 + 1\le p(x) \le 2019(x - 1)^3 + 1 $$
and rearrange,
$$4(x - 1)^3 \le p(x) - 1 \le 2019(x - 1)^3 $$
Assume $p(x) - 1 = a(x-1)^3$ and substitute $p(2) = 2011$ to obtain $a = 2010$. Thus,
$$p(x) = 2010(x-1)^3+1$$
Edit:
Assume, instead, a general form $p(x)-1=\sum_{k=0}^{\infty... | {
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How to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$ I know how to prove $2x^2 + 2y^2 + 2z^2 -2xy -2yz -2xz \geq 0$ since I can split it up in the three terms $(x-y)^2, (y-z)^2, (z-x)^2$ which are all greater than or equal to zero, but how do I prove this without the last term: $2x^2 + 2y^2 + 2z^2 -2xy -2yz \geq 0$?
| It is $$x^2+y^2-2xy+y^2+z^2-2yz+z^2+x^2\geq 0$$
so
$$(x-y)^2+(y-z)^2+z^2+x^2\geq0$$
| {
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"answer_id": 0
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If $m$ and $n$ are integers, show that $\left|\sqrt{3}-\frac{m}{n}\right| \ge \frac{1}{5n^{2}}$ If $m$ and $n$ are integers, show that $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr| \ge \dfrac{1}{5n^{2}}$.
Since $\biggl|\sqrt{3}-\dfrac{m}{n}\biggr|$ is equivalent to $\biggl|\dfrac{ \sqrt{3}n-m}{n}\biggr|$
So I performed the fol... | One aims to give an optimal lower bound as follows:
Theorem. $|\sqrt{3}-\frac mn |\geq \frac 1{(2+\sqrt{3})n^2}$
Proof. Clearly the lower bound is achieved when $n=1,m=2.$ As in other proofs, one may assume without loss of generality that $m,n$ are positive and $n\geq 2$. Observe first that $$\frac 53<\sqrt{3}<\frac 74... | {
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If $\cos(A+B+C)=\cos A\cos B\cos C\neq 0$, then evaluate $\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$
If
$$\cos(A+B+C)=\cos A\cos B\cos C, \quad\text{with}\;A,B,C\neq \frac{k\pi}{2}$$
then
$$\left|\frac{8\sin(A+B)\sin(B+C)\sin(C+A)}{\sin 2A\sin 2B\sin 2C}\right|$$ is what integer?... | So, where you have left off,
$$pq+qr+rp=0$$ writing $\tan A=p,\tan B=q,\tan C=r$
we have $p,q,r$ are non-zero and finite
Now $$\dfrac{2\sin(A+B)}{\sin2C}=\cdots=\dfrac{p+q}{r}$$
$$\implies\dfrac{2\sin(A+B)\cdot2\sin(B+C)\cdot2\sin(C+A)}{\sin2A\sin2B\sin2C}$$
$$=\dfrac{(p+q)(q+r)(r+p)}{pqr}$$
$$=\dfrac{2pqr+p(pq+q... | {
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If $a^2>b^2$ prove that $\int\limits_0^{\pi} \frac{dx}{(a+b\cos x)^3}=\frac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$. Problem: If $a^2>b^2$ prove that $\int\limits_0^\pi \dfrac{dx}{(a+b\cos x)^3} = \dfrac{\pi (2a^2+b^2)}{2(a^2-b^2)^{5/2}}$.
My effort:
If we choose $$x=\tan\frac{\theta}{2}\Longrightarrow d\theta=\frac{2}{x^2+... | It is well-known that
$$\int_0^\pi\frac{dx}{t+\cos x}=\frac{\pi}{\sqrt{t^2-1}}$$
for $t>1$. Differentiating gives
$$-\int_0^\pi\frac{dx}{(t+\cos x)^2}=-\frac{\pi t}{(t^2-1)^{3/2}}.$$
Differentiating again gives
$$2\int_0^\pi\frac{dx}{(t+\cos x)^3}
%=-\frac{\pi(t^2-1)}{(t^2-1)^{5/2}}+\frac{3\pi t^2}{(t^2-1)^{5/2}}
=\fra... | {
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"source": "stackexchange",
"question_score": "8",
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If $PA - PB = \text{constant}$ then the locus of $P$ will be a hyperbola. The sound of a cannon firing is heard one
second later at a position B than at position A.
If the speed of sound is uniform, then
(A) The positions A and B are foci of a
hyperbola, with cannon's position on one
branch of the hyperbola
(B) the po... | EDIT This was my first answer to the original question. OP change his question afterward, so I'll update my answer.
The geometric definition of an hyperbola is :
the points such that the difference of the distance to two foci is
constant. This difference is equal to the distance between the apex of the hyperbola
I... | {
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"question_score": "2",
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Finding $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$ $\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x}$
I'm looking more into simplifying this than the solution itself (which I know involves using t=tan(x/2)).
I did:
$$\int \frac{\cos(2x) dx}{\cos^4x+\sin^4x} = \int\frac{\cos(2x)dx}{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x} = \int\fr... | Continues, we have $$ \int{ \frac{1}{2 - t^2} dt } $$.
assume $ t = \sqrt{2} \sin (y) $, we differensial and get $ dt = \sqrt{2} \cos (y) dy $
$$ \int{ \frac{1}{2 - t^2} dt } = \frac{1}{2} \int{ \frac{\sqrt{2} \cos (y) dy}{1 - \sin ^2 (y) } dt } = \frac{\sqrt{2}}{2} \int{ \sec(y) dy } = \frac{\sqrt{2}}{2} \sec y \tan y... | {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?
| We know that: $$(x^2+y^2)\ge 2xy$$
Multiplying both sides by $y^2$
$$(x^2+y^2)\cdot y^2\ge 2xy^3$$
And clearly $x^2+y^2 \ge y^2$. So
$$(x^2+y^2)\cdot(x^2+y^2)\ge(x^2+y^2)\cdot y^2\ge 2xy^3$$
implying that $$\boxed{(x^2+y^2)^2 \ge 2xy^3}$$
| {
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"url": "https://math.stackexchange.com/questions/3481757",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Expressing $\cos(x)$ and $\sin(x)$ in terms of $\tan(\frac{1}{2} x)$ May I know is there any quick way to express $\cos(x)$ and $\sin(x)$ in terms of $\tan(\frac{1}{2} x)$. I checked that $\cos(x)=\frac{1 - \tan^2(\frac{1}{2} x)}{1+\tan^2(\frac{1}{2} x)}$. I really want to know how could I figure this out quickly, as w... | For the sine, you have
$$\begin{align}
\sin x&=2\sin\frac x2\cos \frac x2 \\
&= \frac{2\frac{\sin \frac x2}{\cos \frac x2}}{\frac{1}{\cos^2 \frac x2}} \\
&= \frac{2\tan\frac x2}{1+\tan^2\frac x2}\\
\end{align}
$$
In the second line, we divide the numerator and denominator ($=1$) by $\cos^2\frac x2$, and we then use $1+... | {
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"question_score": "1",
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solutions of $a+b=c^2 , a^2+c^2=b^2$ ; $a,b,c$ are natural numbers So it all started with a fun observation, $12+13=5^2$ and these are Pythagorean triplets($5,12,13$), so I thought are there more such numbers?
with brute force I was able to get $(24,25,7)$ and $(40,41,9)$.
Then I was able to find 3 families of solutio... | $a^2 + c^2 = b^2$
$c^2 = b^2 - a^2 = (b-a)(a+b)$ but $a+b = c^2$ so if we assume $a+b \ne 0$, we have $b-a = 1$ and $b = a+1$ and we have
$a^2 + c^2 = (a+1)^2$ and $2a + 1 = c^2$
If we replace $c^2$ with $2a+1$ we have $a^2 + 2a + 1 = (a+1)^2$ which is always true. So $2a+1=c^2$ can be any number that is both; an odd ... | {
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"source": "stackexchange",
"question_score": "9",
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Close formula of Yukawa Potential in 3 dimension by integrating $\int_{0}^{\infty}{\frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}}dt$ In PDE, Yukawa potential can be calculate as
$$G^\mu(x) = \int_{0}^{\infty} \frac{1}{(4\pi t)^{\frac{d}{2}}}e^{-\frac{x^2}{4t} - \mu^2t}dt$$
When $d = 3$ we can get the cl... | Let
\begin{align}
I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{4t}-\mu^2 t\right).
\end{align}
Then we see that
\begin{align}
I(\mu, x) = \int^\infty_0 dt\ \frac{1}{(4\pi t)^{3/2}}\exp\left( -\frac{x^2}{2}\left(\frac{1}{\sqrt{2t}}-\frac{\mu}{|x|} \sqrt{2t}\right)^2\right)\exp\left(-\mu |... | {
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"source": "stackexchange",
"question_score": "1",
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$\log_{2} (3 \sin \theta) = 2.\log_{2}(-3\cos \theta) + 1$ what's the sum of possible $\theta$ $0 \leq \theta \leq 360$
$\log_{2} (3 \sin \theta) = 2.\log_{2}(-3\cos \theta) + 1$
$\log_{2} 3 + \log_{2} \sin \theta = \log_{2} 9 + \log_{2} \cos ^2 \theta + \log_{2} 2$
$\log_{2} \sin \theta - \log_{2} \cos ^2 \theta = \lo... | Clearly we need $\sin\theta>0,\cos\theta<0$
$$\implies90^\circ<\theta<180^\circ$$
Choose $180^\circ-\theta=t$
$$\log_2(3\sin t)=2\log_2(3\cos t)+1=\log_2(2(3\cos t)^2)$$
$$3\sin t=18(1-\sin ^2t)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\int \sqrt{\frac{5-x}{x-2}}\,dx$ with two different methods and getting two different results I tried Evaluating $\int \sqrt{\dfrac{5-x}{x-2}}dx$ using two different methods and got two different results.
Getting two different answers when tried using two different methods:-
M-$1$:
$$\int \dfrac{5-x}{\sqrt{... | Shown below is that the two results differ by a constant
$-\frac{3\pi}4$.
Define $f(x)$ as the difference of the two results
$$f(x)=\frac32\sin^{-1}\frac{2x-7}3-3\sin^{-1}\sqrt{\frac{x-2}3}$$
where $2<x\le5$. Then, evaluate
$$f’(x) = \frac3{2\sqrt{(5-x)(x-2)}}- \frac3{2\sqrt{(5-x)(x-2)}}=0$$
Thus, $f(x)$ is a constan... | {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Finding the volume of the tetrahedron with vertices $(0,0,0)$, $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. I get $8$; answer is $4/3$. The following problem is from the 7th edition of the book "Calculus and Analytic Geometry Part II". It can be found in section 13.7. It is
problem number 5.
Find the volume of the tetrahedron who... | Referring to this, now you must divide it by $3!$ to get the final answer.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Evaluate $\int \sqrt{\frac{\sin(x-a)}{\sin(x+a)}}dx$ $$\int \sqrt{\dfrac{\sin(x-a)}{\sin(x+a)}}dx$$
My attempt is as follows:-
$$\int \sqrt{\dfrac{\sin(x-a)\sin(x-a)}{\sin(x+a)\sin(x-a)}}dx$$
$$\int \sin(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}dx$$
Integrating by parts-
$$-\cos(x-a)\sqrt{\dfrac{1}{\sin^2x-\sin^2a}}+\int \... | $$\int\frac{\sin(x-a)}{\sqrt{\sin^2x-\sin^2a}}dx$$
$$=\cos a\int\frac{\sin x}{\sqrt{\cos^2a-\cos^2x}}dx-\sin a\int\frac{\cos x}{\sqrt{\sin^2x-\sin^2a}}dx$$
Now, substitute $\sin x$ in second integral and $\cos x$ in first integral to reduce to standard formulas.
| {
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Finding the determinant of a matrix given by three parameters.
Show that for $a,b,c\in\mathbb R$ $$\begin{vmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{vmatrix} = \begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = 4a^2b^2c^2. $$
There must be some trick, like using elementary row operations, to get the det... | Use the rule of Sarrus to show
$\begin{vmatrix}0&c&b\\c&0&a\\b&a&0\end{vmatrix}^2 = (2abc)^2$ and then show that $A^2 :=\begin{pmatrix}0&c&b\\c&0&a\\b&a&0\end{pmatrix}^2=\begin{pmatrix}b^2+c^2&ab&ac\\ba&c^2+a^2&bc\\ca&cb&a^2+b^2\end{pmatrix}$ and then use $\det(A*A)=\det(A)*\det(A)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Evaluating $\int \frac{1}{(1+x^4)\sqrt{\sqrt{1+x^4}-x^2}}dx$ I tried substitution $x=\dfrac{1}{t}$, then $z=t^2$, tried rationalizing the denominator but haven't been able to pull it off. I can't think of any trig substitution either. I prefer an intuitive approach rather than an "plucked-out-of-thin-air" counter-intui... | Rewrite the integral as
$$\int \frac{x\:dx}{x^2(1+x^4)\sqrt{\frac{\sqrt{1+x^2}-x^2}{x^2}}}$$
by pulling out an $x^2$ from the square root then multiplying the top and bottom by $x$.
Now let $x^2 = \sinh( t) \implies xdx = \frac{1}{2}\cosh(t) dt$:
$$\int \frac{\left(\frac{1}{2}\cosh(t)\right)\:dt}{\sinh(t)(1+\sinh^2(t))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Inequality $\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2} \geqslant \frac{x+y+z}{2}$ Help to prove this Inequality:
If x,y,z are postive real numbers then:
$\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2} \geqslant \dfrac{x+y+z}{2}$
I tied to use analytic method with convex function but no... | $$\sum_{cyc}\frac{x^3}{x^2+y^2}-\frac{x+y+z}{2}=\sum_{cyc}\left(\frac{x^3}{x^2+y^2}-\frac{x}{2}\right)=\sum_{cyc}\frac{x^3-xy^2}{2(x^2+y^2)}=$$
$$=\sum_{cyc}\left(\frac{x^3-xy^2}{2(x^2+y^2)}-\frac{x-y}{2}\right)=\sum_{cyc}\frac{y(x-y)^2}{2(x^2+y^2)}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Reducing an inequality to Schur's inequality Given this,
$a\left(a+b\right)\left(a+c\right)+b\left(a+b\right)\left(b+c\right)+c\left(a+c\right)\left(b+c\right)\ge 2ab\left(a+b\right)+2bc\left(b+c\right)+2ac\left(c+a\right)$
how do you obtain Schur's inequality?
It is suppose to reduce to the nice form of Schur's inequa... | RHS can be rearranged as $$2ab\left(a+b\right)+2bc\left(b+c\right)+2ac\left(c+a\right)$$
$$=2a^2(b+c)+2b^2(c+a)+2c^2(a+b)$$
Now, observe that $$a(a+b)(a+c)-2a^2(b+c)=a(a-b)(a-c)$$
and you are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3491229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamm... | Hint:
$$-(2x^3+1)^3=(x^2+x)^3$$
$$-8(x^3)^3-12(x^3)^2-6(x^3)-1=(x^3)^2+x^3+3x^3(-(2x^3+1))$$
Replace $x^3$ with $y$ to form a cubic equation in $y$ whose roots are $\alpha^3=a$ etc.
Let $z=\dfrac1a+\dfrac1b+\dfrac1c-\dfrac2c=\dfrac{ab+bc+ca}{abc}-\dfrac2c$
We can find the values of $abc, ab+bc+ca$ from the cubic equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Proving the convergence of an infinite product I’m trying to prove the Taylor series of $e$ using binomial expansion:
$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$
The steps I’ve tried so far are:
$$\left(1+\frac{1}{n}\right)^n \\
= 1 + \binom{n}{1}\frac{1}{n} + \binom{n}{2}\frac{1}{n^2} + \binom{n}{... | $$\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\prod_{k=1}^n\frac{1}{n}\left(n-k+1\right)=\frac{1}{n^n}\big[n(n-1)\cdots(1)\big]=\frac{n!}{n^n}$$
Hence, we have:
$$\lim_{n\to\infty}\frac{1}{n!}\prod_{k=1}^n\left(1-\frac{k-1}{n}\right)=\lim_{n\to\infty}\frac{n!}{n^n\cdot n!}=\lim_{n\to\infty}\frac{1}{n^n}=0$$
This should b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If both roots of the equation $ax^2-2bx+5=0$ are $\alpha$ and roots of the equation $x^2-2bx-10=0$ are $\alpha$ and $\beta$.
find $\alpha^2+\beta^2$
Both equations have a common root
$$(-10a-5)^2=(-2ab+2b)(20b+10b)$$
$$25+100a^2+100a=60b^2(1-a)$$
Also since the first equation has equal roots
$$4b^2-20a=
0$$
$$b^2=... | $b^2=5a$, then $\alpha=b/a$
Next we have $\alpha+ \beta=2b$ and $\alpha \beta =-10$
Then $$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha \beta=4b^2+20=20a+20 ~~~(*)$$
Next we have $$\beta=\frac{-10}{\alpha}=2b-\alpha \implies -\frac{10a}{b}=2b-\frac{b}{a}$$
$$\implies -10a^2=2ab^2-b^2 \implies -20a^2=-5a \implies a=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find the maximum number of draws in a tournament. Consider a round robin tournament between 8 teams. A win gives 3 points, tie gives 1 point (to both teams) and loss gives 0 points.
If the match between two teams, say A & B, is a tie, then A & B can't end up with the same number points after the tournament.
The task is... | Start by considering the possible outcomes for a player. All players play seven other players and they can tie, win or lose games.
This table shows the score achieved in different situations:
\begin{array}{|c|c|c|c|c|}
\hline
&Ties & Wins & Loses & Score \\ \hline
A&7& 0& 0&7\\ \hline
B&6& 0 &1 &6\\ \hline
C&6& 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3504877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Help in summing: $\sum_{k=m} ^{2m} ~k\cdot ~2^{-k} {k \choose m}$ While solving an interesting problem I require two sums
$$S_1=\sum_{k=m}^{2m} 2^{-k} {k \choose m} ~ \text{and} ~ S_2=\sum_{k=m}^{2m} k \cdot ~ 2^{-k} {k \choose m}$$
The former is know to be unity see inside the solutions of
How to prove that $\sum_{i... | Starting from
$$\sum_{k=m}^{2m} k 2^{-k} {k\choose m}$$
we have as in the answer that was first to appear
$$\sum_{k=m}^{2m} (k+1) 2^{-k} {k\choose m}
- \sum_{k=m}^{2m} 2^{-k} {k\choose m}
\\ = (m+1) \sum_{k=m}^{2m} 2^{-k} {k+1\choose m+1}
- \sum_{k=m}^{2m} 2^{-k} {k\choose m}.$$
For the first sum term we get
$$(m+1) [z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Given positives $a, b, c$, prove that $\frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$.
Given positives $a, b, c$, prove that $$\large \frac{a}{(b + c)^2} + \frac{b}{(c + a)^2} + \frac{c}{(a + b)^2} \ge \frac{9}{4(a + b + c)}$$
Let $x = \dfrac{b + c}{2}, y = \dfrac{c + a}{2... | Recall Nesbitt's inequality
\begin{eqnarray*}
\frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2}.
\end{eqnarray*}
Using Cauchy-Schwartz and Nesbitt gives
\begin{eqnarray*}
&\left( 2 \cdot \frac{a}{b+c}+1 \right)^2 +\left( 2 \cdot \frac{b}{c+a}+1 \right)^2 + \left(2 \cdot \frac{c}{a+b} +1 \right)^2\\ &\geq
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Difference in choosing some generator of principal ideal for solving $x^2+d=y^3$ So in solving $E : x^2 + 19 = y^3$ in the integers I determined that $(x + \sqrt{-19}) = I^3$ for some ideal in $$\mathscr{O}_K = \mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$$ and since the class number of $K$ is not $3$, that $I$ must... | To solve $$\left(a+b\frac{1+\sqrt{-19}}{2}\right)^3=x+\sqrt{-19}$$ for integers $a,b$ you should express the right-hand side in terms of the integral basis $1, \frac{1+\sqrt{-19}}{2}$ as well: $$x+\sqrt{-19} = (x-1)+2\frac{1+\sqrt{-19}}{2}.$$
(You could also equate coefficients w.r.t the $\mathbb{Q}$-basis $1,\sqrt{-19... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show: For every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
Show that for every $\alpha \in \mathbb{Q}$, the polynomial $X^2+\alpha$ has endless different roots in $\mathbb{Q}^{2\times2}$.
What I did:
Let $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix... | By definition,
$\alpha \in \Bbb Q \Longrightarrow \exists p, q \in \Bbb Z, \; \alpha = \dfrac{p}{q} = pq^{-1}; \tag 1$
set
$Y = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} ; \tag 2$
then
$Y^2 = \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix} \begin{bmatrix} 0 & -p \\ q^{-1} & 0 \end{bmatrix}$
$= \begin{bmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$a^3 + b^3 + c^3 = 3abc$ , can this be true only when $a+b+c = 0$ or $a=b=c$, or can it be true in any other case? Since
$$
a^3 + b^3 + c ^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc ),
$$ from this it is clear that if $(a+b+c) = 0$ , then $a^3 + b^3 + c ^3 - 3abc = 0$ and
$a^3 + b^3 + c ^3 = 3abc$ . Also if $a=... | By your work:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$
Thus, $$a^3+b^3+c^3-3abc=0$$ for $a+b+c=0$ or for
$$a^2+b^2+c^2-ab-ac-bc=0,$$ which is
$$(a-b)^2+(a-c)^2+(b-c)^2=0,$$ which gives $$a=b=c.$$
Id est, we have no another cases for equality occurring for real values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\sum_{i = 1}^na_1^2 = 1$. Calculate the maximum value of $\sum_{cyc}|a_1 - a_2|$.
Given reals $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $a_1^2 + a_2^2 + \cdots + a_{n - 1}^2 + a_n^2 = 1$ $(n \in \mathbb N, n \ge 3)$. Calculate the maximum value of $$\large |... | The maximal value is $2\sqrt{n-1}$ if $n$ is odd, and $2\sqrt{n}$ if $n$ is even. We can prove the following:
Let $a_1, \ldots, a_n$ be real numbers, $n \ge 2$. Then
$$ \tag{*}
|a_1 - a_2| + |a_2 - a_3| + \ldots + |a_{n - 1} - a_n| + |a_n - a_1|
\le c_n \sqrt{a_1^2 + \ldots + a_n^2}
$$
where $c_n = 2\sqrt{n-1}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\sum_{i=0}^{k}\binom{k-i}{b}\binom{i}{a-1}=\binom{k+1}{a+b}$ I am trying to prove this identity from an exercise:
Given $k$, $a$, $b$, prove that
$$
\sum_{i=0}^{k}\binom{k-i}{b}\binom{i}{a-1}=\binom{k+1}{a+b}
$$
However, I'm having trouble using existing identities to prove this. Any help would be much appr... | Start from
$$\sum_{q=a-1}^{k-b} {k-q\choose b} {q\choose a-1}
= \sum_{q=0}^{k+1-b-a} {k+1-a-q\choose b} {q+a-1\choose a-1}
\\ = \sum_{q=0}^{k+1-a-b}
{k+1-a-q\choose k+1-a-b-q} {q+a-1\choose a-1}
\\ = [z^{k+1-a-b}] (1+z)^{k+1-a} \sum_{q=0}^{k+1-a-b}
{q+a-1\choose a-1} \frac{z^q}{(1+z)^q}.$$
The coefficient extractor enf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the set of values of x for which $\lvert \frac {x-1}{x+1} \rvert <2 $ How to solve this inequality question involving modulus?
I can’t get the same answer as the book [answer below]
I know the properties of absolute values, that is
If $\lvert x \rvert <k$ , then $-k < x < k$.
So for this question, this is my worki... | If $x=a+ib$ where $a,b$ are real
We have $$\sqrt{(a-1)^2+b^2}<2\sqrt{(a+1)^2+b^2}$$
Squaring and simplifying we get
$$b^2+a^2+\dfrac{10a}3+1>0$$
$$b^2+\left(a+\dfrac53\right)^2>\dfrac{25}9-1=\left(\dfrac43\right)^2$$
So, $x$ needs to lie outside the circle $$y^2+\left(x+\dfrac53\right)^2=\left(\dfrac43\right)^2$$
If $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Eliminating parameters from polynomial relations Let $p =x^3+y^3+z^3$, $q= x^2y+y^2 z+z^2x$, $r=xy^2+yz^2+zx^2$, and $s=xyz$. I want to find some non-zero polynomial in $\phi$ in $p,q,r,s$ such that $\phi(p,q,r,s)=0$; that is, to eliminate $x,y,z$. I have $p^2-2qr-2ps+6s^2=x^6+y^6+z^6$ and $p^3-3q^3-3r^3-24s^3+18qrs=x^... | Multiply $r$ and $q$
\begin{eqnarray*}
rq &=& \sum x^3y^3+ 3(xyz)^2+xyz \sum x^3 \\
&=&\sum x^3y^3+ 3s^2 +sp. \\
\end{eqnarray*}
So
\begin{eqnarray*}
\sum x^3y^3= rq -3s^2 -sp \\
\end{eqnarray*}
Now multiply this by $p$
\begin{eqnarray*}
p(rq-sp-3s^2) = \sum_{perms} x^6y^3+ 3(xyz)^3 \\
\sum_{perms} x^6y^3= p(rq-sp-3s^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can any polynomials in the rational field be decomposed like this I' ve learned that the following examples can be used to decompose a
factor in this way:
x^5 - 5 x + 12 = (x - a) (a^4 - (5 a^3)/8 + (7 a^2)/8 + 1/8 (5 a^2 - 12 a) +
1/8 (5 a^3 - 12 a^2) + ((5 a^4)/16 - a^3/8 + (7 a^2)/16 -
3/16 (5 a^2 - 12 a... | Here:
In[5]:= qpoly = x^5 - 5 x + 12;
rt = First[x /. Solve[qpoly == 0, x]]
fax = Factor[qpoly, Extension -> rt] /. rt -> a
(% /. a -> Root[#1^5 - 5 #1 + 12 &, 1]) // Simplify
Out[5]=(1/16)*(x - a)*(-a^4 - a^3 - a^2 + (-a^4 - a^3 - a^2 + 3*a + 4)*x - 5*a + 4*x^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove $\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2$ with $\frac1a+\frac1b=1$. Let $ a, b> 0 $ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that$$\sqrt{a^2+4}+\sqrt{b^2+4}\leq\frac{\sqrt{2}}{4}(a+b)^2.$$
Obviously $a+b=ab\ge4$. Other than that, I do not know what trick to use to deal with the constraint. The Lagra... | Let $x=\frac{1}{a}$ and $y=\frac{1}{b}$. We want to show that $x+y=1$ and $x,y\geq 0$ imply
$$ y^2x\sqrt{4x^2+1}+x^2y\sqrt{4y^2+1}\leq \frac{\sqrt{2}}{4}. $$
Letting $x=\frac{1+t}{2},y=\frac{1-t}{2}$, this is equivalent to finding the maximum of
$$ f(t) =(1-t^2)\left[ (1-t)\sqrt{1+(1+t)^2}+(1+t)\sqrt{1+(1-t)^2}\right]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3521160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Minimum without computing the derivative. I am trying to find the minimum of the following function:
$$H(x)=\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+(x^3+\frac{1}{x^3})},\hspace{1cm}x>0$$
What I usually do to find extrema of a function is computing the derivative of the function and finding its ... | $$\displaystyle \bigg[\bigg(x+\frac{1}{x}\bigg)^3\bigg]^2-\bigg[x^3+\frac{1}{x^3}\bigg]^2$$
$$\bigg[\bigg(x+\frac{1}{x}\bigg)^3+x^3+\frac{1}{x^3}\bigg]\bigg[\bigg(x+\frac{1}{x}\bigg)^3-x^3-\frac{1}{x^3}\bigg]$$
$$H(x)=\bigg(x+\frac{1}{x}\bigg)^3-x^3-\frac{1}{x^3}=3\bigg(x+\frac{1}{x}\bigg)\geq 3\cdot 2=6$$
Because arit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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For real $a$, $b$, $c$ in $(0,1)$, if $a+b+c=2$, then $\frac{a}{1-a}\times\frac{b}{1-b}\times\frac{c}{1-c}\geq 8$
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$
Prove that
$$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$
I tried by substituting $x$ for$(1-a)$ and similarly for o... | For the edited question
Let $a$, $b$, $c$ be three real numbers such that $0<a,b,c<1$ and $$a + b + c = 2$$
Prove that
$$\frac a{1-a}×\frac b{1-b}×\frac c {1-c}≥8$$
The inequality is equivalent with
$$abc \geq 8(1-a)(1-b)(1-c)$$
or
$$ abc \geq 8\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do we integrate the expression $\frac{b^{3}}{1 - \theta(1-b)^{2}}$ where $0\leq b\leq 1$ and $\theta\in(0,1)$? MY ATTEMPT
In order to solve this integral, I have tried using the integration by parts method, as suggested by the expression
\begin{align}
\int_{0}^{1}\frac{b^{3}}{1-\theta(1-b)^{2}}\mathrm{d}b = \int_{0... | You can use partial fractions:
$$
\frac{1}{1-\theta(1-b)^2}=\frac12\,\left(\frac1{1-\sqrt{\theta}(1-b)}+\frac{1}{1+\sqrt{\theta}(1-b)}\right).
$$
Now, using the substitution $v=1-\sqrt{\theta}(1-b)$, we have $dv=\sqrt{\theta}\,db$, so the integral of the first half is
\begin{align}
\int_0^1\frac{b^3}{1-\sqrt{\theta}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Solve $\sin x + \cos x = \sin x \cos x.$ I have to solve the equation:
$$\sin x + \cos x = \sin x \cos x$$
This is what I tried:
$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$
$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$
$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$
$$1 + \sin(2x) ... | Note that when you square the equation
$$(\sin x + \cos x)^2 = (\sin x \cos x)^2$$
which can be factorized as
$$(\sin x + \cos x - \sin x \cos x)(\sin x + \cos x + \sin x \cos x)=0$$
you effectively introduced another equation $\sin x + \cos x =- \sin x \cos x$ in the process beside the original one $\sin x + \cos x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
} |
$\int_0^{\pi/2} \sec^a(t)\,dt= \frac{\sqrt{\pi}}{2\Gamma\left(1-\frac{a}{2}\right)}\Gamma\left(\dfrac{1-a}{2}\right)$ Inside the Wolfram Documentation page for the secant function, an identity is given which involves the gamma function, polygamma function, and Catalan's constant.
Notes on documentation page:
Some spec... | To derive the identity, it is a straight up application of the Beta function in the form
$$\operatorname{B}(m,n) = 2 \int_0^{\frac{\pi}{2}} \cos^{2m - 1} t \sin^{2n - 1} t \, dt.$$
For the secant integral, we have
\begin{align}
\int_0^{\frac{\pi}{2}} \sec^a t \, dt &= \int_0^{\frac{\pi}{2}} \cos^{-a} t \, dt\\
&= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3531201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Bernoulli First Order ODE I want to know if my answer is equivalent to the one in the back of the book. if so what was the algebra? if not then what happened?
$$x^2y'+ 2xy = 5y^3$$
$$y' = -\frac{2y}{x} + \frac{5y^3}{x^2}$$
$n = 3$
$v = y^{-2}$
$-\frac{1}{2}v'=y^{-3}$
$$\frac{-1}{2}v'-\frac{2}{x}v = \frac{5}{x^2}$$
$$v'... | Write:$$(x^2y)' = 5y^3$$ Now let $z=x^2y$ then we get $$z' = {5z^3\over x^6}\implies {z'\over z^3} = 5x^{-6}$$
So after integrating both sides we get $$ -{z^{-2}\over 2} = -x^{-5}+c'\implies {1\over 2x^4y^2} = {1-c'x^5\over x^5}$$
So $$ y^2 = {x\over 2+cx^5}$$
where $c=-2c'$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Hölder continuous implying the rate of convergence of the Cesàro mean Let $f$ be a periodic function which is Hölder continuous of order $0<\alpha<1$
And $\sigma_nf$ be its $n^{th}$ Cesàro mean. Let $ \|f \|=\|f\|_\infty+\sup_{x \neq y}\frac{|f(x)-f(y)|}{|x-y|^{\alpha}}$.
The I have to show that $\|\sigma_nf-f\|_{\inft... | The proof for the line can be applied to the circle using inequality $(4)$.
Preliminary Inequalities for $\boldsymbol{0\lt x\lt\frac\pi2}$
Inequality $\bf{1}$:
$$
\begin{align}
\frac{\sin^2(x)}{x^2}
&=\prod_{k=1}^\infty\cos^2\left(\frac{x}{2^k}\right)\tag{1a}\\
&=\prod_{k=1}^\infty\left(1-\sin^2\left(\frac{x}{2^k}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $a$ such that the minimum and maximum distance from a point to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ Context: I have to solve the following problem: find $a\in \mathbb{R},\, a>0\,\, /$ the minimum and maximum distance from $(4,2)$ to the curve $x^2+y^2=a$ are $\sqrt{5}$ and $3\sqrt{5}$ respectively.... | Perhaps a different approach altogether. If $(4,2)$ is inside the circle of radius $\sqrt{a}$, then the minimum and maximum distances combine into one diameter, hence
$$
2a = d = \sqrt{5} + 3\sqrt{5} = 4\sqrt{5}
$$
and the point of interest is $a = \sqrt{2^2+4^2} = \sqrt{20} = 2\sqrt{5}$ away, which means the point $(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
When does $2n-1$ divide $16(n^2-n-1)^2$? Find all integers $n$ such that
$\dfrac{16(n^2-n-1)^2}{2n-1}$
is an integer.
| Let $k = 2n-1$ then $$\dfrac{16(n^2-n-1)^2}{2n-1} = \dfrac{(4n^2-4n-4)^2}{2n-1}$$
$$ = \dfrac{((k+1)^2-2(k+1)-4)^2}{k} = {(k^2-5)^2\over k}$$
So $$k\mid (k^2-5)^2= k^4-10k^2+25$$ $$\implies k\mid 25 \implies k\in\{\pm 1, \pm 5,\pm 25\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Quadratic function with roots in $[0,1]$. Prove that $f(0) \geq \frac49$ or $f(1) \geq \frac49$ Let $a,b$ in $[0,1]$ be such that the polynomial $f(x) = (x-a)(x-b)$ satisfies $f(\tfrac12) \geq \frac1{36}$.
I have found a quite complicated proof of the following inequality using calculus:
$$f(0) \geq \frac49 \quad\text... | Note,
$$f(\frac12) =(\frac12-a)(\frac12-b)\ge \frac{1}{36}\implies 2ab-(a+b)+\frac{4}{9} \geq 0$$
Use $a+b\ge 2\sqrt{ab}$ to factorize above inequality as
$$(\sqrt{ab}-\frac{1}{3})(\sqrt{ab}-\frac{2}{3})\ge 0$$
which leads to either $\sqrt{ab}\le\frac{1}{3}$ or $\sqrt{ab}\ge\frac{2}{3}$. Examine the two cases below.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Rationalizing the denominator of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$
Simplify
$$\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$$
I think this should be expressed without square roots at the denominator. I tried to multiply by conjugate.
| So, you multiply by the denominator over the denominator to get rid of the first square root (while still just technically multiplying only by 1):
$$
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2+\sqrt{2+\sqrt{2}}}} =
\frac{2\sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}
$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$. Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$.
I'm thinking to show that $n$ is divisible by each of the prime factors of $930$, is that right? I'm stuck
| $$n= 5(1 +5+ 5^2+ ...5^{149}) = 5 {5^{150}-1\over 4}$$
so $5\mid 4n\implies 5\mid n$. Now:
$$5^{150}-1 = 125^{50}-1 = (125-1)(125^{49}+...+125^2+125+1) $$
so $31\mid 4n \implies 31\mid n$ and $$5^{150}-1 = 25^{75}-1 = (25-1)(25^{74}+...+25+1) $$
so $24\mid 4n \implies 6\mid n$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$
Show that $$\sum_{k=1}^{n}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4}$$
holds for all $n\in\mathbb{N^+}$.
Since this is a positive series, it suffices to show that $$\sum_{k=1}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}<\frac{1}{4},$$which is true by machine computing. WA g... | Someone gives a solution as follows, which is essentially equivalent to @Robert Z 's except some details.
\begin{align*} s(n):&=\sum_{k=1}^n\frac{\ln k}{(2k-1)(2k+1)}\\ &\leq\sum_{k=2}^{\infty}\frac{\ln k}{(2k-1)(2k+1)}\\ &=\frac{1}{2}\sum_{k=2}^{\infty}\left(\frac{\ln k}{2k-1}-\frac{\ln k}{2k+1}\right)\\ &=\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Problem regarding unique solution of differential equation
A unique solution to the differential equation $y = x \frac{dy}{dx} - (\frac{dy}{dx})^2$ passing through $(x_0,y_0)$ doesnot exist
then choose the correct option
$1.$ if $ x_0^2 > 4y_0$
$2.$ if $ x_0^2 = 4y_0$
$3.$ if $ x_0^2 < 4y_0$
$4.$ for any $(x_0 ... | Hint: First treat the differantial equation as a quadratic in $\frac{dy}{dx}$
\begin{eqnarray*}
\left( \frac{dy}{dx} \right)^2 -x \frac{dy}{dx} +y=0.
\end{eqnarray*}
Now plug this into the quadratic formula and ... ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the closed form of this quickly growing sequence? Is there a pattern in the following sequence:
$$3,19,451,22051,\dots$$
I tried setting some $f(n)$ to this sequence, where $n$ is the term number. Then I basically treated it like a matrix equation: multiplying, adding, subtracting and dividing to hopefully obta... | If I subtract $1$ from each term, I see $2,18,450,22050,...$. The ratio of successive terms is $9,25,49,...$
And now I think I see it : the ratio of successive terms, minus $1$, gives the list of odd squares, but starting from $3^2 = 9$.
In other words, if $f(n)$ denotes the $n$th number of the sequence, then $(f(n)-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.