Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the equation of a circle tangent to two lines Find the equation of the circle tangent to the $x$-axis at the origin and tangent to the line $4x-3y+24=0$.
My Approach:
Let the equation of the required circle be:
$$x^2+y^2+2gx+2fy+c=0$$
Let the equation to the tangent at origin $(0,0)$ to the above circle be $gx+fy+... | As the circle touches the x-axis at (0,0) then the centre of the circle must be on the y-axis and the y coordinate of the centre must be equal to the radius. So the centre is $(0,r)$. So the equation is:
$$x^2+(y-r)^2=r^2$$
Then you want to find the point of intersection of the circle and the straight line. Lets rearra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4... | You're correct so far. What you need to finish is this
Hint: $4u^2+4u = 8v$
Solution:
$ 4u^2+4u=4(u+1)u=8\binom{u+1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Factorization of $x^5-1$ over $\mathbb F_{19}$ I've come across this question Factorization of $x^5-1$ over $F_{11}$ and $F_{19}$. The answer was good but I don't understand how to actually solve it.
It says that i can split $x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)$. If I expand this I get $x^4+(a+c)x^3+(ac+b+d)x^2+(ad+b... | Paying some time on this problem in order to give an elementary proof for the O.P. one has the cyclotomic $x^4+x^3+x^2+x+1$ is irreducible over $\Bbb F_{19}$.
We have
$$a+c=1\Rightarrow (a,c)\in S_1\subset \Bbb F_{19}\text { x }\Bbb F_{19}\\bd=1\Rightarrow (b,d)\in S_2\subset \Bbb F_{19}\text { x }\Bbb F_{19}$$
$$S_2=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for power series $y'' - 9y = 0$ I really need help with this, the solution from this equation is $y(x) = c_1 e^{3x} + c_2 e^{-3x}$. But I can't get to it, I obtain the next:
$$y(x) = \sum_{n=0}^{\infty}a_nx^n$$
$$y''(x) = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$$
Then the coefficients must be $a_{2m} = \frac{9^m a_0}... | Note that $a_0=y(0)=c_1+c_2$ and $a_1=y'(0)=3c_1-3c_2$. Then
\begin{align}
y(x) &= a_0 \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + a_1\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\\ \ \\
&= (c_1+c_2) \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + (3c_1-3c_2)\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\\ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find minimal value $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without derivatives. Find minimal value of $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without using the derivatives and without the formula for the distance between two points.
By using the derivatives I have found ... | If we complete the squares we get $$\sqrt{\left(x-{5\over 2}\right)^2 +{75\over 4}} + \sqrt{ (x-6\sqrt{3})^2 +36}.$$ This is the distance from the point $P(x,0)$ to the point $Q(5/2, -5\sqrt{3}/2)$ plus the distance from $P(x,0)$ to the point $R(6\sqrt{3},6).$ Choosing $x$ amounts to sliding $P$ along the $x$-axis. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $d$ must be a perfect square
Prove that if $a,b,c,d$ are integers such that $$(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2 = d,$$ then $d$ is a perfect square.
In order for $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2$ to be an integer, either $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ has to be the square root of a pos... | We expand and get
$$
d=(a+2^{1/3}b+2^{2/3}c)^2\\
=(a^2+4bc)+(2c^2+2ab)2^{1/3}+(b^2+2ac)2^{2/3}
$$
If $d$ is to be an integer, then from the expression above we must have $2c^2+2ab=b^2+2ac=0$. Assuming $a\neq0$, this means that
$$
b=-\frac{c^2}{a}\\
c=-\frac{b^2}{2a}
$$
Inserting one of these into the other, we get
$$
b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\t... | The direct formula for $3$ by $3$ determinants isn't so bad
$$\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{vmatrix}=aei+bfg+cdh-ceg-bdi-afh$$
so
$$\begin{vmatrix}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25 \\
\end{vmatrix}=225+576+576-729-400-256=-8.$$
Row operations and other similar tricks tend to speed ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Compute $\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$ Compute $$\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$$
Tried use substitution
$$\mu=5-3\cos x$$
But made the problem even complecated...Any help?
Thank you~
| $$I=\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$$
$\cos(-x)=\cos(x)$, hence we can write:
$$
I=\frac{1}{2}\int^{\pi}_{-\pi}\frac{1}{(5-3\cos x)^3}dx=\frac{1}{2}\int^{2\pi}_{0}\frac{1}{(5+3\cos x)^3}dx
$$
Last expression can be calculated through methods of complex analysis:
$$
I=\frac{1}{2}\int_{|z|=1}\frac{1}{\left(5+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution Review
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$; $x = 0$ when $y = 1$.
I solved this DE using the integration factor method. However, online calculators are giving me a different answer, where the... | We separate the variables
$\frac{dy}{y}=-3xdx$
which we integrate as follows
$ln(\frac{y}{C})=-\frac{3}{2}x^2$
and
$y=Ce^{-\frac{3}{2}x^2}$
if $x=0$, $y=1$ so $C=1$
and finally
$$y=e^{-\frac{3}{2}x^3}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplifying fraction of infinite series There is $\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{%
\left( jx\right) ^{k/5}}{k!}}{1+\left( jx\right) ^{1/5}%
+\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}$
(where $x\epsilon \mathbb{R}$, $x>10$ and $j=\sqrt{-1}$).
It is $\lim\limits_{k\... | We can use the power series representation of the exponential function
\begin{align*}
\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\qquad\qquad\qquad x\in\mathbb{C}
\end{align*}
in order to simplify the expression. We get
\begin{align*}
\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}&=\sum_{k=2}^\infty \frac{\left((jx)^\frac{1}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
| Clearly $1\le x\le5$
As $\dfrac{1+5}2=3, 1-3\le x-3\le5-3$
WLOG $x-3=2\cos2y,$ where $0\le2y\le\pi$
$\sqrt{x-1}=2\cos y$ and $\sqrt{5-x}=2\sin y$
Now $\sin y+\cos y=\sqrt2\sin\left(y+\dfrac\pi4\right)$ and $0\le y\le\dfrac\pi2$
$\implies\dfrac1{\sqrt2}\le\sin\left(y+\dfrac\pi4\right)\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
cube roots of unity Let $\omega \ne 1$ be a complex cube root of unity.
If $$(4 + 5\omega + 6\omega ^2)^{n^2 + 2} + (6 + 5\omega ^2 + 4\omega)^{n^2 + 2} + (5 + 6\omega + 4\omega ^2)^{n^2 + 2} = 0$$ then $n$ can be...
| The LHS of the given expression can be written thus:
$$\left( 4 + 5\omega + 6\omega ^2 \right)^{n^2 + 2}+ \left( (4 + 5\omega + 6\omega ^2)w \right)^{n^2 + 2}+\left( (4 + 5\omega + 6\omega ^2)w^2 \right)^{n^2 + 2}$$
$$=\left( 4 + 5\omega + 6\omega ^2 \right)^{n^2 + 2}.F \ \ \text{with} \ \ F=1+w^{n^2 + 2}+(w^2)^{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove the result of this integration? How to prove that?
$$\int_0^1 \frac{1 - e^{-t} - e^{-1/t}}{t}\ \text{d}t = \gamma$$
where $\gamma = 0.5772156649015328606065\ldots$ is the Euler-Mascheroni constant.
Additional question: is there a way to evaluate it via Residues Theorem too?
| Preliminary results.
Let us begin with a definition of the Euler-Mascheroni Constant
\begin{equation}
\gamma = \lim_{n \to \infty} H_{n} - \mathrm{ln}(n)
\label{eq:1}
\tag{1}
\end{equation}
where $H_{n}$ are the harmonic numbers defined as
\begin{equation}
H_{n} = \displaystyle\sum_{k=1}^{n} \frac{1}{k}
\label{eq:2}
\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{3}$ represented as continued fraction Am I incorrect in the assumption that I deal with $\sqrt{3}$ in the same way I would approach $\sqrt{2}$, by adding and subtracting $1$ such that:
$\sqrt{3}=1+\sqrt{3}-1=1+\cfrac{2}{1+\sqrt{3}}$
The table representation is $[1; 1,2]$.
This is what I have so far:
$\sqrt... | If you obtain a rational approximation of $\sqrt{3}$, for example $\frac{362}{209}$ you'll find its continued fraction is $$1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{3}}}}}}}$$ That 3 looks extraneous, so we might form the conjecture that $$\sqrt{3}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Finding the roots of any cubic with trigonometric roots.
Question: How would you find the roots of a cubic polynomial whose roots are only expressive in trigonometric forms?
I'm really confused on how you would solve it. Some examples:$$x^3+x^2-2x-1=0\\x_1=2\cos\frac {2\pi}7,x_2=2\cos\frac {4\pi}7,x_3=2\cos\frac {8\... | The strategy to solve a cubic equation is first to change it to the form $x^3+px+q=0$, i.e. to eliminate the squared term.
Cardano's method works well if there is only $1$ real root ($4px^3+27q^2>0$). If there are $3$ real roots ($4px^3+27q^2<0$), you can set $x=A\cos\theta$ ($A>0$). The equation becomes
$$A^3\cos^3\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why is $\lim_{n \rightarrow \infty} \bigg( \bigg| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} \bigg| \bigg)=e$? I know that the following is the correct limit, but I have difficulties in seeing just why this is.
$$\lim_{n\to\infty}\left| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}}\right|=e$$
| Starting with $2 n^2 + 2n + 1 = n^2 + (n+1)^2$ then
\begin{align}
\frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} &= \left(\frac{n}{n+1}\right)^{n^{2}} \, \left( \frac{n+2}{n+1} \right)^{(n+1)^{2}} \\
&= \frac{ \left(1 + \frac{1}{n+1} \right)^{(n+1)^{2}} }{ \left( 1 + \frac{1}{n} \right)^{n^{2}}} \\
&= \frac{e^{(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Performing LU factorization on Ax = b. Can someone show me step by step how this works? Consider matrix
$$
A= \begin{bmatrix}
0 & -2 & 1 \\
2 & 1 & -1 \\
-2 & -2 & 1 \\
\end{bmatrix}
$$
and resulting vector :
b = (3
0
1)
Perform LU factorization with row swapping, indic... | We are given:
$$A = \begin{bmatrix} 0 & -2 & 1 \\ 2 & 1 & -1 \\ -2 & -2 & 1 \end{bmatrix},b = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}$$
We are asked to solve $Ax = b$ using the LU factorization with pivoting. Our approach will be:
*
*$(1.)$ Compute $PA = LU$, if $PA = LU$, then $LU x = Pb$, so find $L, U, P$
*$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving Logarithm Equation Solve for x:$$x^2(\log_{10} x)^5 = 100$$
Here's what I've tried:
$(\log_{10} x)^5 = A \\ \log_{ \log x} A= 5 = \frac{\log_{10} A}{\log_{10} \log x}$
Not sure how to continue
| $$x^2(\log_{10}x)^{5}=10^2$$
$$x^{\frac{2}{5}}\log_{10}x=10^{\frac{2}{5}}$$
$$\log_{10}x^{x^\frac{2}{5}}=\log_{10}10^{10^\frac{2}{5}}$$
$$x^{x^\frac{2}{5}}=10^{10^\frac{2}{5}}$$
so the $$x=10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| Less bashy than Parcly Taxel's answer:
Since $P$ is divisible by $x^3$ and $P-1$ is divisible by $(x-1)^3$, we know that $P'$ is divisible by $x^2(x-1)^2$. Since $P'$ is degree $4$, it must be a constant multiple of this. Say $P'(x)=Ax^2(x-1)^2=Ax^4-2Ax^3+Ax^2$.
Then $P(x)$ is an antiderivative of this, namely $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Sequence of fractions that converges to $\sqrt{n}$ Begin with any two positive integers $a,b$. Let
$$s_1 = \frac{a}{b}$$
and recursively we define $s_{i+1}$ from $s_i$ as follows: If ${s_i}$ is the fraction $\frac{a'}{b'}$, then set
$$s_{i+1} = \frac{a'+2b'}{a'+b'}$$
(Notice that the actual value of the $s_{i+1}$ does ... | In the general case dividing by $b'$ in the nominator and the denominator we have that:
$$s_{i+1} = \frac{s_i + n}{s_i + 1}$$
Now consider two cases, namely $s_i \le \sqrt{n}$ and $s_i > \sqrt{n}$. Now let WLOG $s_i \le \sqrt{n}$. Then expressing $s_{i+2}$ by $s_i$ we have that:
$$s_{i+2} = \frac{s_{i+1} + n}{s_{i+1} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$ where $0\leq x\leq 360$ Solve
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot \sin 2x \cdot \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x \cdot \cos x ). (4 \sin x \cdot \cos x \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I pr... | Since
$$ \sin a \cdot \sin b = \frac{1}{2} \left[ \cos (a - b) - \cos (a+b) \right] $$
$$ \sin a \cdot \cos b = \frac{1}{2} \left[ \sin (a + b) + \sin (a - b) \right] $$
and
$$ \sin a - \sin b = 2 \sin\left[ \frac{1}{2} (a-b) \right] \cos \left[ \frac{1}{2} (a+b) \right] $$
we can use these identities to solve
$$ 4 \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The idea behind the sum of powers of 2 I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.
For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, ... | Here's a geometric intuition for why $2^0 + 2^1 + 2^2 + \dots + 2^n = 2^{n+1} - 1$:
Here's the idea. Notice that the boxes for $1 + 2 + 4$ are right next to a single box of size $8$. They perfectly fill that box once you add the gold square in, so $1 + (1 + 2 + 4) = 8$, meaning $1 + 2 + 4 = 8 - 1$. Similarly, the boxe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "56",
"answer_count": 11,
"answer_id": 3
} |
The least positive real number $k$ for which $k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$
The least positive real number $k$ for which $$k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$$
Where $x,y,z>0$
$\bf{My\; Try::}$ Here $$k\geq \frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{... | Let $x=y=z$.
Hence, $k\geq\frac{3}{2\sqrt2}$.
We'll prove that $k=\frac{3}{2\sqrt2}$ is valid.
Indeed, we need to prove that
$9(x+y)(y+z)(z+x)\geq8(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2$.
By C-S $(x+y+z)(xy+yz+zx)\geq(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2$.
Thus, it remains to prove that
$$9(x+y)(y+z)(z+x)\geq8(x+y+z)(xy+yz+zx)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that, for large N, the proportion of positive integers n≤N...
Show that, for large $N$, the proportion of positive integers $n ≤ N$
(a) not divisible by $2^7$is $1 − \frac{1}{2^7}$
(b) not divisible by any of $2^7, 3^4, 5^3, 7^2, 11^2, 13^2, 17^2, 19^2, 23^2$ is $(1-\frac{1}{3^4})(1-\frac{1}{2^7})\cdots(1-\frac{1... | It is pure probability at that point. not divisible by an of a,b,c,d, or e is the same as (not divisible by a and not divisible by b and ...) which are independent for large N so you have P(not divisible by a)*P(not divisible by b)P(not divisble by c)...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Coefficient of a term using binomial theorem I was just wondering how would I find the coefficient of any term let's say $x^3$ in the expansion of $(x^2+2x+2)^{10}$ using binomial expansion or any other technique. Please let me know if this can be found directly using a shortcut if any.
| We can also apply the binomial theorem twice in order to determine the coefficient of $x^3$ in $(x^2+2x+10)^{10}$.
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rig... | Note that if for some $n \geq 1$ the proposition is true then $5\cdot 10^{n} + 10^{n-1} + 3 = 9k$ for some $k$; hence
$$
5\cdot 10^{n+1} + 10^{n} + 3 = 90k - 27.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Inequality based on AM/GM Inequality Find the greatest value of $x^3y^4$
If $2x+3y=7 $ and $x≥0, y≥0$.
(Probably based on weighted arithmetic and geometric mean)
| $y$ is a function of $x$. That is, $y=(7-2x)/3.$ And $\frac {dy}{dx}=y'=-2/3.$
Let $f(x)=x^3y^4.$ Then $f'(x)=3x^2 y^4+4x^3y^3y'=3x^2y^4-(8/3)x^3y^3.$
For $xy\ne 0$ we have $f(x)>0$ and $f'(x)=f(x)(\frac {3}{x}-\frac {8}{3y}).$
So for $xy\ne 0$ we have $f'(x)>0\iff \frac {3}{x}-\frac {8}{3y}>0\iff 9y>8x\iff21-6x>8x \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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binomial congruence $\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0~or (-2)\pmod p$
Let $p\ge 5$ be a prime number. Show that $$\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0 \text{ or } -2\pmod p .$$
Examples:
If $p=5$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}=8\equiv -2\pmod 5 .$... | My solution:\begin{align*}&\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}=\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}\left(\dfrac{1}{2}+\frac{1}{2}\right)^{i}
=\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}\sum_{j=0}^{i}\binom{i}{j}\dfrac{1}{2^i}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(2i)!}{2^i\cdot i!\cdot j!(i-j)!}\\
&=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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$\lim _{x \to a} \frac{x^n-a^n}{x-a}$ when $n$ is irrational. Question on the theorem :$$\lim _{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}$$
Is it true when $n$ is irrational ? is there a proof ?
Ex:
$$\lim _{x \to a} \frac{x^{\sqrt{2}}-a^{\sqrt{2}}}{x-a}=\sqrt{2}a^{\sqrt{2}-1}$$
| Just another way to do it.
Define $$x=a(1+y)\implies A=\frac{x^n-a^n}{x-a}=\frac{a^n(1+y)^n-a^n}{ay}=a^{n-1}\frac{(1+y)^n-1}{y}$$ Apply the generalized binomial theorem or Taylor series around $y=0$ $$(1+y)^n=1+n y+\frac{n(n-1) }{2} y^2+O\left(y^3\right)$$ which makes $$\frac{(1+y)^n-1}{y}=n+\frac{n(n-1)}{2} y+O\lef... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding range of this expression If a line makes angles $\alpha,\beta,\gamma$ with positive axes,then the range of $\sin\alpha\sin\beta + \sin\beta\sin\gamma +\sin\gamma\sin\alpha$ is? I am a noob in finding range , so please help me from beginning.
| Take a unit vector $\mathbf u$ parallel to the line: its components will be $(cos\alpha , cos\beta , cos\gamma)$,
with $0 \leqslant \alpha , \; \beta ,\; \gamma \leqslant \pi$, and $cos^2\alpha + cos^2\beta + cos^2\gamma=1$.
We shall note that the $3$ angles are not independent (there are $2$ degrees of freedom on the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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The number of solutions of $x^2+2016y^2=2017^n$
The number of solutions of $x^2+2016y^2=2017^n$ is $k$. Write $k$ with $n$.
For $n = 1$, the only solution is $(1,1)$. For $n = 2$, it gets more complicated. Taking the equation modulo $2016$ we find that $x^2 \equiv 1 \pmod{2016}$. How do we continue?
Solving for $y$ w... | If $x^2+2016y^2$ is divisible by $2017$, then taking $\mod 2017$ shows that $x\equiv\pm y\mod2017$. Then
$\left(\dfrac{x\pm2016y}{2017}\right)^2+2016\left(\dfrac{y\mp x}{2017}\right)^2=\dfrac{x^2+2016y^2}{2017}$,
shown just by expanding out. We can choose the sign of each of those so that the numbers are actual integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\int\limits_{0}^{2\pi}\frac{a\cos x -1}{(a^2+1-2a\cos x)^{3/2}}dx.$ $$\int\limits_{0}^{2\pi}\frac{a \cos x -1}{(a^2+1-2a \cos x)^{3/2}}dx
= 2\int\limits_{0}^{\pi}\frac{a \cos x -1}{(a^2+1-2a\cos x)^{3/2}}dx.$$
If a<1, this integral doesn't converge. How to evaluate it for any other a? I think it can be e... | Note that $$I\left(a\right)=2\int_{0}^{\pi}\frac{a\cos\left(x\right)-1}{\left(a^{2}+1-2a\cos\left(x\right)\right)^{3/2}}dx
$$ $$=\frac{d}{da}\left(-2a\int_{0}^{\pi}\frac{1}{\sqrt{a^{2}+1-2a\cos\left(x\right)}}dx\right)
$$ and, since $a>1$, $$\int_{0}^{\pi}\frac{1}{\sqrt{\left(a+1\right)^{2}-2a-2a\cos\left(x\right)}}d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluation of $\int x^{26}(x-1)^{17}(5x-3)dx$
Evaluation of $$\int x^{26}(x-1)^{17}(5x-3) \, dx$$
I did not understand what substution i have used so that it can simplify,
I have seems it is a derivative of some function.
Help me, Thanks
| Sorry for the informal writing am rushing out.
Hope it helps!
$$
(x - 1)^{17} = \sum \left(
\begin{array}{c}
17 \\
r
\end{array}
\right) x^{17-r}(-1)^r
\\
x^{26}(x - 1)^{17} = \sum \left(
\begin{array}{c}
17 \\
r
\end{array}
\right) x^{43-r}(-1)^r
\\
x^{26}(x - 1)^{17}(5x−3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Multivariable limit - perhaps a trickier problem I am stuck on. I am trying to solve the following limit:
$\lim_{(x,y) \to (0,0)} \frac{x^4y^4}{(x^2 + y^4)^3}$
(This is a more challenging problem from Folland Calculus, it seems).
I am pretty sure this limit does not exist (however, this is just a guess, and I am not 10... | Use polar coordinates :
$$\frac{x^4y^4}{(x^2+y^4)^3} = \frac{r^8\cos^4\theta\sin^4\theta}{r^6(\cos^2\theta+r^2\sin^4\theta)^3} = r^2\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$
Now the problem remaining is : is the fraction bounded or not ? You have, for $\theta\ne\frac\pi2\mod\pi$ :
$$\frac{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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How to prove that $f(x)$ is equal to $ \frac{1}{2}(\exp(x) -\exp(-x)) $? $$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$
Since we know that $ \frac{1}{2}(\exp(x) - \exp(-x)) $ is same as
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{n}}{n!} - \sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}\right)$$
we can place our $f... | They split the sum in an even and an odd part. In general
$$
\sum_{n = 0}^{+\infty} a_n = \sum_{n=0,\;n\; {\rm even}}^{+\infty}a_n + \sum_{n=0,\;n\; {\rm odd}}^{+\infty}a_n = \sum_{n=0}^{+\infty}a_{2n} + \sum_{n=0}^{+\infty}a_{2n+1}
$$
This is particularly useful here because $1-(-1)^{2n} = 1-1 = 0$ and $1 -(-1)^{2n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to find minimal polynomial for finite field How can we find minimal polynomials for $\alpha $ in $GF(2^{n})$.
What is the general approach to find minimal polynomials. I know about minimal polynomials they are monic etc. In particular i want to know about primitive polynomials of $GF(32)$.
| There are $\frac{\phi(2^n-1)}{n}$ primitive polynomials of degree $n$ over $GF(2)$. For $n=5$ we have $30/5=6$ primitive polynomials, namely
$$
x^5+x^2+1,\; x^5+x^3+1,\; x^5+x^3+x^2+x+1,\;x^5+x^4+x^2+x+1,\;
$$
$$
x^5+x^4+x^3+x+1,\;x^5+x^4+x^3+x^2+1.\;
$$
Indeed, the product of all of them, together with the factor $(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Derivative with quotient rule Trying to get $\dfrac{d}{dx}\left[\frac{20e^x}{(e^x+4)^2}\right]$.
After quotient rule: $$f'(x)=20\dfrac{e^x(e^{2x}+8e^x+16)-e^x(2e^x+8e^x)}{(e^x+4)^4}\\\\\\= 20\dfrac{e^{3x}+8e^{2x}+16e^x-2e^{2x}-8e^x}{(e^x+4)^4}\\\\=20\dfrac{e^{3x}+6e^{2x}+8e^x}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+4)(e^x+2)}... | Using:
$$
\frac{d}{dx}(e^x+4)^2=2e^x(e^x+4)
$$
we have
$$
f'(x)=\frac{20e^x(e^x+4)^2-20\cdot 2e^x(e^x+4)}{(e^x+4)^4}=\frac{20 e^x(e^x+4)(e^x+4-2e^x)}{(e^x+4)^4}=
$$
$$
=\frac{20e^x(-e^x+4)}{(e^x+4)^3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ with $p>1$, and $x,y,z$ positive
By Jensen I got taking $p_1=p_2=p_3=\frac13$, (say $x=x_1,y=x_2,z=x_3$)
$\left(\sum p_kx_k\right)^p\le\sum... | Indeed, Jensen works.
Let $x+y+z=3$.
Hence, we need to prove that $\sum\limits_{cyc}f(x)\geq0$, where $f(x)=\frac{x^p}{3-x}-\frac{1}{2}$.
$f''(x)=\frac{x^{p-2}\left((p-2)(p-1)x^2-6p(p-2)x+9p(p-1)\right)}{(3-x)^3}$.
If $p=2$ so $f''(x)>0$.
If $p>2$ so since $f''(0)>0$ and $\lim\limits_{x\rightarrow3^-}f''(x)>0$ and $\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is the sequence $a_{n}=\prod\limits_{i=1}^{n}\left(1+\frac{i}{n^2}\right)$ decreasing?
let $$a_{n}=\left(1+\dfrac{1}{n^2}\right)\left(1+\dfrac{2}{n^2}\right)\cdots \left(1+\dfrac{n}{n^2}\right)$$
since $$a_{1}=2,a_{2}=\dfrac{15}{8},a_{3}=\dfrac{1320}{729}\cdots $$
I found
$$a_{1}>a_{2}>a_{3}>\cdots$$
I conjecture
$\... | Taking logarithms is often helpful when dealing with products, so let's try that here too. Grouping terms in the difference of the logarithms, we obtain
$$\log a_n - \log a_{n+1}
= \sum_{k = 1}^n \Biggl(\log\biggl(1 + \frac{k}{n^2}\biggr) - \log \biggl(1 + \frac{k}{(n+1)^2}\biggr)\Biggr) - \log \biggl(1 + \frac{1}{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | If $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$
Now write Identity as $$\sin^2 x+\cos^2 x+(-1) = 0\;,\text{ Then }\sin^6 x+\cos^6 +(-1)^3 = -3\sin^2 x\cos^2 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 4
} |
Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$ While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer,
$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$
it seems the general form is just
$$p \int_0^\infty \frac{1}{1+x^{p}}dx=\color{blue}{\frac{\pi... | The following close-form
holds for any integer $p\ge 2$
\begin{align}
\int_0^1 \frac1{1+x^p}dx
= \frac2p \sum_{k=1}^{[\frac p2]} ( \theta_k \sin2\theta_k + \cos2\theta_k \ln \cos\theta_k)
\end{align}
where $\theta_k= \frac{p-2k+1}{2p}\pi $. In particular
\begin{align}
& \int_0^1 \frac1{1+x^5} dx= \frac{\pi\sqrt{\phi}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$... | The second you've got down pat.
$$\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} = \lim_{x \to 0} \frac{\arctan 2x}{x}\frac{x}{\sin(2\pi x)} = \frac{1}{\pi}$$
The first is a little trickier. Use the binomial theorem to get...
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} = \lim_{x \to 0} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Find intersection of 2 linear equations I have two tables with data, both describing linear equations
Table 1:
x = 60, y = 0
x = 61, y = 1
x = 62, y = 2
x = 63, y = 3
x = 64, y = 4
x = 65, y = 5
x = 66, y = 6
x = 67, y = 7
x = 68, y = 8
x = 69, y = 9
x = 70, y = 10
Table 2:
x = 64, y = 0
x = 65, y = 2
x = 66, y = 4
x... |
Part 1:
For Table 1, we use the fact that the slope is the change in $y$ divided by the change in $x$. Or,$$m=\frac {y_2-y_1}{x_2-x_1}\tag1$$
Randomly choosing two points, say $(70,10)$ and $(69,9)$ and plugging them in, we get the slope as$$m=\frac {10-9}{70-69}=1$$
Thus, we now have $y=x+b$ for $b$ is the $y$-inter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Algebraic proof that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $
Prove that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.
Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.
I am not sure how to p... | We need to proof $1+a^9\le\frac{1}{a}+a^{10}$. You already proof that is sufficient and necessary that $(a-1)(a^{10}-1)\ge 0$.
This last inequality is true (when $a>0$) because:
$(a-1)(a^{10}-1)=(a-1)^2(a^9+a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1)$
And is trivial that $a^9+a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1> 0$, because $a>0$, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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How to prove that $\int_{-1}^{+1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$ I'am trying to prove that
$$\int_{-1}^{+1} (1-x^2)^n\:\mathrm{d}x = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$$
Here what I have done so far. We know that:
\begin{equation}
\sin^{2}x + \cos^{2}x = 1
\end{equation}
Let $x = \sin\alpha$ so $\mathrm{d}... | If I try to prove it for $n=1$ I get:
$$\int_{-1}^1(1-x^2)dx=(x-x^3/3)|_{-1}^{1}=2-2/3=4/3\ne\frac{3!1!^2}{2^3}=\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Proof: $\sum^n_{i = 0} {n\choose i} F_{i+m}$ is Fibonacci number I'm trying to solve the following problem:
Show
$$\sum^n_{i = 0} {n\choose i} F_{i+m}$$
is Fibonacci number.
I know many properties of binomial symbol and Fibonacci numbers but I have no idea how to start proving given formula.
| The task is to express
$$S_{n,m} = \sum_{q=0}^n {n\choose q} F_{q+m}$$
as a Fibonacci number. Using the generating function of the Fibonacci
numbers we find that
$$F_{q+m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+m+1}} \frac{z}{1-z-z^2} \; dz.$$
We thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| As a function of $a$, the minimum of the quadratic $4a^2+b^2+1-(2ab+2a+b)$ is $\frac34 (b - 1)^2 \ge 0$.
More precisely,
$$4a^2+b^2+1-(2ab+2a+b)=\left(2a-\frac{b+1}{2}\right)^2+\frac34 (b - 1)^2 \ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Binomial theorem coefficient question Use the binomial theorem to find the coefficient for $x^3$ on both sides of the expansion of:
$(1+x)^3$$(1+x)^3$ $=$ $(1+x)^6$
i. Hence show $(_3C_0)^2+(_3C_1)^2+(_3C_2)^2+(_3C_3)^2$ $=$ $_6C_3$
ii. Use the same argument with $(1+x)^n(1+x)^n = (1+x)^{2n}$ to prove
$\sum_{k=0}^n(_nC... | It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. We also use the notation $\binom{n}{k}$ instead of $_nC_k$.
We obtain
\begin{align*}
[x^3](1+x)^3(1+x)^3&=[x^3]\left(\sum_{j=0}^3\binom{3}{j}x^j\right)\left(\sum_{k=0}^3\binom{3}{k}x^k\right)\tag{1}\\
&=\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $x_{n+2} = \sqrt{x_{n+1}} + \sqrt{x_{n}}$
Let $x_0,x_1> 0$. Prove that the sequence defined by $x_{n+2} = \sqrt{x_{n+1}} + \sqrt{x_{n}}$ converges.
Here's my solution: it's easy to prove by induction that $$\forall n, 0<\min(4,u_0,u_1)\leq u_n\leq \max(4,u_0,u_1)$$
The sequence is therefore bounded wit... | If there is a limit, then $L=2\sqrt{L}$ so that $L=4$. Now
$$
x_{n+2}-4=\sqrt{x_{n+1}}-2+\sqrt{x_{n}}-2
=\frac{x_{n+1}-4}{\sqrt{x_{n+1}}+2} + \frac{x_{n}-4}{\sqrt{x_{n}}+2}
$$
If one can additional prove $x_k\ge 1$ for all $k$ beforehand, then
$$
|x_{n+2}-4|\le\frac13(|x_{n+1}-4|+|x_{n}-4|)
$$
To get a recursive bou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Question about Euler proof of divergence of the sum of the reciprocals of the primes. Consider the Euler proof :
$$\ln\sum_{n = 1}^{\infty}\frac{1}{n} = -\ln\sum_{p}\frac{1}{1-p^{-1}} = \sum_{p}\left(\frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \dots\right) = A + \frac{1}{2}B + \frac{1}{3}C + \dots = A + K,$$ where... | Yes, you are right. For all $d\geq 2$
$$\frac{1}{d}\sum_p\frac{1}{p^d}<\frac{1}{2^{d-1}}.$$
Infact, for $d\geq 2$,
$$\sum_p\frac{1}{p^d}\leq \frac{1}{2^d}+\sum_{n\geq 3}\frac{1}{n^d}< \frac{1}{2^d}+\int_{x\geq 2}\frac{dx}{x^d}= \frac{1}{2^d}+\frac{2}{(d-1)2^d}\leq \frac{3}{2^{d}}$$
which implies that
$$\frac{1}{d}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2040644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan... | You have
$$
\sin\theta+1=\frac{3}{2}\cos\theta
$$
and so
$$
3\cos\theta=2(1+\sin\theta)
$$
Note that $\cos\theta\ge0$, so $-\pi/2\le\theta\le\pi/2$. We can square to get
$$
9-9\sin^2\theta=4+8\sin\theta+4\sin^2\theta
$$
so
$$
13\sin^2\theta+8\sin\theta-5=0
$$
and therefore
$$
\sin\theta=\frac{5}{13}
\qquad\text{or}\qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2042019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove the following trignometric inequality for all real $x$ Prove the following trignometric inequality for all $x \in \Bbb R$
$$x^2 \sin(x) + x \cos(x) + x^2 + {\frac 12} >0$$
take $x$ in the form of radians.
This particular question is the seventh question of the 1995 Indian RMO.
| There may be more efficient ways but this is what I have:
If it is positive (i.e. nonnegative) then certainly when we multiply it by a positive function it will stay positive... to this end, multiply it by $\sin^2(x)\cos^2(x)$ this gives
$$x^2\sin^3\cos^2(x)+x\sin^2(x)\cos^3(x)+x^2\sin^2(x)\cos^2(x)+\frac{1}{2}\sin^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Probability that 2 numbers will occur exactly 4 times. A die is rolled 10 times, what is the probability that 2 numbers will occur exactly 4 times? So I know that the probability of rolling any one number exactly 4 times would be
$$\frac{10!}{4!6!}\cdot \left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^6,$$
so ... | No, you have ignored two effects. The first is that you seem to be trying to calculate the probability that two specific numbers occur four times each, and the problem asks for any two numbers. The second is that once one number is given to have occured exactly four times, the odds of the other number also doing so a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving basic system of equations $x, y$ are positive reals such that $y\sqrt{x^2 - y^2} = 48$ and $x + y + \sqrt{x^2 - y^2} = 24$. How to find $x$ and $y$? Squaring equations leads to complicated equations with polynomials of high degrees. Is there any way to omit it and solve it smarter?
| Call $x^2-y^2=z^2 \quad (1)$ then:
$yz=48 \quad(2)$ and $x+y+z=24 \quad(3)$
From $(1)$ we have:
$$x^2=y^2+z^2=(y+z)^2-2yz$$
and using $(2)$ and $(3)$ we get:
$$x^2=(24-x)^2-2.48 \Rightarrow x=10$$
and then $y=8$ or $y=6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to express $ lcm \big( 1, 2, \dots, n \big)$ in terms of factorials $1!, 2!, 3!, \dots, n!$? I observed that if you arrange the factorials up to 18 in a certain way it is possible to get the LCM of all numbers of to 18. With some effort I could show:
\begin{eqnarray*} \frac{18! \times 3! }{9! \times 6! }
&=& \frac... | $lcm \big( 1,2,3, \dots, n\big) = \prod_{p \leq n} p^i$ where p is prime and $i$ such that $p^i \le n$ and $p^{i+1}>n$
Using Arthurs comment:
$lcm \big( 1,2,3, \dots, n\big) = \prod_{p \leq n}\frac{(p^i)!}{(p^i-1)!}$
For example
$lcm \big( 1,2,3, \dots, 18\big) = \frac{17! \times 13! \times 11! \times 7! \times 5! \tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How does one get that $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$? While watching interesting mathematics videos, I found one of the papers of Srinivasa Ramanujan to G.H.Hardy in which he had written $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$.
The problem is that every term on the left is more than $\frac{1}{120}$ yet the sum ... | I think I have rediscovered it (after watching the video you linked and read wiki biography of Ramanujan).
Start with $$
\frac{1}{x+1} =1 -x +x^2 -x^3 +-\cdots, \quad |x| <1.
$$ and differentiate to get $$
-\frac{1}{(x+1)^2} =-1 +2x -3x^2 +4x^3 -+\cdots, \quad |x| <1. \\
\frac{2}{(x+1)^3} =2 \cdot 1 -3 \cdot 2x +4 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
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Prove that $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are in arithmetic progression under given condition. Suppose $(b-c)^2,(c-a)^2,(a-b)^2$ are in arithmetic progression.
Then show that $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are also in an arithmetic progression.
Please help.
| Since it is give that $(b-c)^2,(c-a)^2,(a-b)^2$ are in AP. So, you get that :
$$2(c-a)^2=(b-c)^2+(a-b)^2$$
$$\implies2c^2+2a^2-4ac=2b^2+c^2+a^2-2bc-2ab+2ac$$
$$\implies c^2+a^2-2ac=2b^2-2bc-2ab+2ac $$
$$\implies (c-a)^2=2(b^2-bc-ab+ac)$$
$$\implies (c-a)^2=-2(ab+bc-b^2-ac)$$
$$\implies (a-c)(c-a)=2(ab+bc-b^2-ac)$$
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if I so decide the limit? $\lim_{n\to +\infty}{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})}$ $$ \lim_{n\to +\infty}{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})} = \lim_{n\to +\infty}{\frac{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})*(\sqrt{n+1} + \sqrt{n-1})}{(\sqrt{n+1} + \sqrt{n-1})}} = \lim_{n\to +\infty}{\frac{\sqrt n * (n + 1 - n + 1)}{... | All your moves are fine. To gain more intuition for what is happening you might enjoy noting that to compute
$\lim_{n \to \infty} \left(\sqrt{ n^2 + n} - \sqrt{n^2 - n}\right),\;$ we see
$$\sqrt{n^2 + n} = \sqrt{\left(n+\frac{1}{2}\right)^2 - \frac{1}{4}}$$
$$\sqrt{n^2 + n} = \sqrt{\left(n-\frac{1}{2}\right)^2 - \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you find the real solutions to these simultaneous equations? I am looking for all real $(a,b,c)$ that satisfy the following
\begin{equation}
\left\{
\begin{array}{l}2a + a^2b = b\\
2b + b^2c = c\\
2c + c^2a = a\\
\end{array}
\right.
\end{equation}
I know that $a=b=c = 0$ is the only real solution to the problem... | Rearrange
\begin{equation}
\left\{
\begin{array}{cccc}
\tan y &=& b &=& \dfrac{2a}{1-a^2} \\
\tan z &=& c &=& \dfrac{2b}{1-b^2} \\
\tan x &=& a &=& \dfrac{2c}{1-c^2} \\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{cccc}
\tan y &=& \tan 2x \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Can an indefinite integral evaluate to zero? Problem : Evaluate $\displaystyle\int \frac{\sin x+\cos x}{\cos^2 x+\sin^4 x} \, dx $
I evaluated in the following way, and somehow got zero :
$$I=\int \frac{\sin x+\cos x}{\cos^2 x+\sin^2 x(1-\cos^2 x)} \, dx$$
$$I=\int \frac{\sin x}{\cos^2 x+\sin^2 x(1-\cos^2 x)}dx +\int ... | Let $$I =\int\frac{\sin x+\cos x}{\cos^2 x+\sin^4 x}dx$$
Now we can write $\cos^2 x+\sin^4 x = \sin^4 x-\sin^2 x+1 = \cos^4 x-\cos^2 x+1$
So $$I = \int\frac{\sin x}{\cos^4 x-\cos^2 x+1}dx+\int\frac{\cos x}{\sin^4 x-\sin^2 x+1}dx$$
Now put $\cos x=t\;,$ Then $\sin xdx = -dt$ and put $\sin x=u\;,$ Then $\cos dx = du$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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$x^3 = x$ for all $x \in R$, where $R$ is a ring. Prove that $6x = 0$ for all $x \in R$. I am very confused about this problem. Because, clearly 6x = 0 only has one solution, 0. But I can make
$x^3 = x$
$x^3 -x=0$ (Adding additive inverse of x to both sides)
$x(x^2-1)=0$ (Distributive property of the ring)
$... | $(x+1)^3=x^3+3x^2+3x+1=x+1=x+3x^2+3x+1$implies that $3x^2+3x=0$. $3(x^2+x)=0$.
$3((x+1)^2+x+1)=3(x^2+2x+1+x+1)=3(x^2+x)+6x+6=0$.
$(1+1)^3=1+1$, $8=2$ implies $6=0$, so $6x+6=6x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Is there a property similar to $ x^2 = (x-1)(x+1)+1$ for $ x^3$? I am looking for a way to decompose $x^3$ in a similar way.
| $$\small \begin{vmatrix} x+n & 1 & 0 & 0 & \ddots & 0 & 0 & 0 & 0\\ -n & x+n-2 & 2 & 0 & \ddots & 0 & 0 & 0 & 0\\ 0 & -n+1 & x+n-4 & 3 & \ddots & 0 & 0 & 0 & 0\\ 0 & 0 & -n+2 & x+n-6 & \ddots & 0 & 0 & 0 & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots\\ 0 & 0 & 0 & 0 & \ddots & x-n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.
| The pertinent part of the polynomial expansion can be done by hand without too much fuss:
$$\begin{align}
(1+x^2-x^3)^8&=(1+x^2)^8-8x^3(1+x^2)^7+{8\choose2}x^6(1+x^2)^6-{8\choose3}x^9(1+x^2)^5+\cdots\\
&=(1+x^2)^5((1+x^2)^3-8x^3(1+x^2)^2+28x^6(1+x^2)-56x^9)+\cdots\\
&=(1+x^2)^5((1+3x^2+3x^4+x^6)-8x^3(1+2x^2+x^4)+28x^6(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&... | With a hopefully obvious notation,
$$s_3=s_2c+c_2s=2sc^2+(1-2s^2)s=2s(1-s^2)+(1-2s^2)s=3s-4s^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2... | Since that is a quotient you can rewrite it as $$2^{(4n-n^2-1)-2n}= 2^{-(n-1)^2}$$
Notice indeed that $$(a-b)^2=a^2-2ab+b^2$$ so that you have $$(n-1)^2 = n^2-2n+1$$ so clearly also $$-(n-1)^2 = -n^2+2n-1$$ and by rewriting the middle term you get $$-(n-1)^2 = -n^2+4n-2n-1$$ because clearly $4n-2n =2n$ and as you can p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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find range of $a^2+b^2$ without trigonometric substution given $7a^2-9ab+7b^2=9$ and $a,b$ are real no. then find range of $a^2+b^2$ without trigonometric substution
from $7a^2-9ab+7b^2=9$
$\displaystyle ab = \frac{7(a^2+b^2)-9}{9}$
put into inequality $\displaystyle a^2+b^2 \geq 2ab$
$\displaystyle a^2+b^2 \geq \frac{... | If $a = 0$ or $b = 0$, then $a^2+b^2 = \dfrac{9}{7}$, and this is one of the many values in the range of $a^2+b^2$. So assume $ab \neq 0$, then put $x = \dfrac{a}{b} > 0$ ( we can assume $a, b > 0$ ), and rewrite the equation by dividing both sides by $ab$ to have: $\dfrac{9}{ab} = 7\left(x+\dfrac{1}{x}\right)-9 = 7u-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and ... | Here is my attempt at a helpful rule:
The expression $x\frac{y}{z}$ always means $x\times\frac{y}{z}$
except when $x,y,$ and $z$ are all integers written in decimal notation; then it means $x+\frac{y}{z}$.
So $n\frac{n^2+5}{4}$ means $n\times\frac{n^2+5}{4}$, but $3\frac14$ means $3+\frac14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 4
} |
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx $ Problem :
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx$.
My approach :
\begin{align}
&\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx \\
=& \ln(1^x +2^x +3^x +6^x) \frac{x^2}{2} - \int^1_{-1} \frac{1}{1^x+2^x+3^x+6^x}(2^x\log2+3^x \log3+6^x \log... | $$1+2^x+3^x+6^x=(1+2^x)(1+3^x)$$
Now $\ln(AB)=\ln A+\ln B$
Apply $\displaystyle\int_p^qf(x)\ dx=\int_p^qf(p+q-x)\ dx$
in $I=\displaystyle\int_{-1}^1x\ln(1+a^x)\ dx=\int_{-1}^1(-x)\ln(1+a^{-x})\ dx=\int_{-1}^1(-x)\{\ln(1+a^x)-x\ln a\}dx$
$\implies\displaystyle I=-I+\ln a\int_{-1}^1x^2\ dx$
| {
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"url": "https://math.stackexchange.com/questions/2074951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\sum\limits_{cyc}\frac{a}{a+b}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$$
I tried C-S, uvw, BW and more, but without success.
| @Michael Rosenberger Thanks, here is my $0.02:
Write the whole inequality in terms of $ab, ac, bc$:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{ac}{ac+bc}+\frac{ab}{ab+ac}+\frac{bc}{bc+ab}$$
Then use the following inequality proved here - as a Lemma inside Andreas's answer:
$$ \frac{a}{a+b} + \frac{b}{b+c} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x\to\pi/4} \frac{1-\tan(x)^2}{\sqrt{2}*\cos(x)-1}$ without using L'Hôpital's rule. Find $$\lim_{x\to\pi/4} \frac{1-\tan(x)^2}{\sqrt{2}\times \cos(x)-1}$$ without using L'Hôpital's rule.
I can solve it using L'Hôpital's rule, but is it possible to solve it without using L'Hôpital's rule?
| You can try the following:
\begin{align}
\frac{1-\tan^2({x})}{\sqrt{2} \cos{x}-1}\frac{\sqrt{2} \cos{x}+1}{ \sqrt{2} \cos{x}+1 }&=\frac{1-\frac{\sin^2{x}}{\cos^2{x}}}{2 \cos^2{x}-1} (\sqrt{2} \cos{x}+1)\\ &=\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x} (2\cos^2{x}-\sin^2{x}-\cos^2{x}) }(\sqrt{2} \cos{x}+1)\\
&=\frac{\sqrt{2} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
prove that: $\sqrt{2}=e^{1-{2K\over \pi}}\prod\limits_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$ show that
$$\sqrt{2}=e^{1-{2K\over \pi}}\prod_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$$
where K is the catalan's constant; $K=0.9156 ...$
My try:
take the ln
$${1\over2}\ln{2}=\left(1-{2K\over \pi}\rig... | Consider
\begin{align}
\sum_{n=1}^{\infty}4n \log\left({4n-1\over 4n+1}\right)+2 &=-
\sum_{n=1}^{\infty}4n \sum_{k=0}^\infty \frac{-2}{(2k+1)(4n)^{2k+1}}+2\\
&=-\sum_{k=1}^\infty \frac{2}{(2k+1)4^{2k}}\sum_{n=1}^{\infty}\ \frac{1}{n^{2k}}\\
&=-\sum_{k=1}^\infty \frac{2\zeta(2k)}{(2k+1)4^{2k}}\\
&= \frac{2K}{\pi}-1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Does $\sum \frac{a_n}{1+n a_n}$ converge if $a_n = \frac{1}{\sqrt{n}}$ if $n$ is a perfect square and $a_n = \frac{1}{n^2}$ otherwise? Given the sequence $\left\{ a_n \right\}$, where
$$a_n = \begin{cases} \frac{1}{\sqrt{n}} \ \mbox{ if } n \mbox{ is a perfect square} \\ \frac{1}{n^2} \ \mbox{ otherwise}, \end{cases}$... | Hint.
$$\sum_{n=1}^\infty\frac{a_n}{1+na_n}=\sum_{k=1}^\infty\frac{a_{k^2}}{1+k^2 a_{k^2}}+\sum_{n\text{ is not a square}} \frac{a_n}{1+na_n}$$
From what you've done we know
$$\sum_{k=1}^\infty\frac{a_{k^2}}{1+k^2 a_{k^2}}=\sum_{k=1}^\infty\frac 1{k(k+1)}<\infty$$
and
$$\sum_{n\text{ is not a square}} \frac{a_n}{1+na_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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On the convergence of the sequence $\,a_n=\displaystyle\sum_{k=1}^{n-1}\left(\frac{n}{k(n-k)}\right)^{2}$ Let the sequence
$$
a_n=\sum_{k=1}^{n-1}\left(\frac{n}{k(n-k)}\right)^{2}, \quad n\ge 2.
$$
Show that:
A. $\,a_n\le 4$.
B. $\,\{a_n\}_{n\in\mathbb N}$ is decreasing, and hence converging.
C. Find $\lim_{n\to\infty}... | $$\begin{align}
a_n= & \sum_{k=1}^{n-1}\left[\frac{n}{k(n-k)}\right]^{2} \\
= & \sum_{k=1}^{n-1}\left[\frac{(n-k)+k}{k(n-k)}\right]^{2} \\
= & \sum_{k=1}^{n-1}\left[\frac{1}{k}+\frac{1}{n-k}\right]^{2} \\
= & \sum_{k=1}^{n-1}\left[\frac{1}{k^{2}}+\frac{1}{(n-k)^{2}}+\frac{2}{k(n-k)}\right] \\
= & \sum_{k=1}^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit question related to integration Find the limit,
$$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
My try:
$$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx< \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{... | Hint/Intuition:
For $x,y\ge0, \lim_{n\to\infty} \left(x^n+y^n\right)^{1/n}=\max\{x,y\}$.
Alternative Approach:
$$I=\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx=2\int_{0}^{1/2}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
$$=2\int_0^{1/2}(1-x)\left(1+\left(\frac{x}{1-x}\right)^n\right)^{1/n}$$
Let $u=\frac{x}{1-x}\implies dx=\frac{du}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Can you prove that $\frac{a+b}{ab+1}$ is real if $|a|=1$, $|b|=1$, and $ab\ne-1$? Can you prove that $\dfrac{a+b}{ab+1}$ is real if $|a|=1$, $|b|=1$, and that $ab$ isn't equal to $-1$?
| $$\dfrac{a+b}{ab+1}= \dfrac{(a+b)(1+\overline{a}\overline{b})}{(ab+1)(1+\overline{a}\overline{b})}=\dfrac{(a+b)(1+\overline{a}\overline{b})}{|ab+1|^2}=\dfrac{a+b+|a|^2\overline{b}+\overline{a}|b|^2}{|ab+1|^2}=\dfrac{a+b+\overline{b}+\overline{a}}{|ab+1|^2}=\dfrac{2Re(a+b)}{|ab+1|^2} \in \Bbb R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal.
\begin{equation*}
A =
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
\end{equation*}
I know that a $n \times n$ matrix is called orthogonal ... | The system of equations has no solution at all, since $1+a^2=1$ implies $a=0$ and hence $2+ab=2\neq 0$. hence $A^TA\neq I_2$ for all $a,b$. This is, given the computation in the question, much more immediate than to use that the norm of a column vector must be $1$ for an orthogonal matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| If $x = a+b$, then $x^3 = (3ab)x + (a^3+b^3)$.
(Conversely, if $x^3 - px - q = 0 $ then we may find $a,b$, such that $3ab = p, a^3+b^3=q$.
using Cardano's method for solving the cubic equation.)
Letting $w = e^{\frac {2i \pi}{3}} $ denote a cube root of unity, notice that the numbers $a+b, wa + w^2b$, and $w^2a + w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 3
} |
Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}... | Let $f(x)=\frac{1}{x^2-a^2}$.
$f$ is continuous at $(-\infty,-|a|),(-|a|,|a|)$ and $(|a|,+\infty)$.
in each interval
$$2af(x)=\frac{1}{x-a}-\frac{1}{x+a}$$
and $$\int f(x)dx=\frac{1}{2a}\ln(\frac{|x-a|}{|x+a|}).$$
the final expression depends on which interval $J$ we want the antiderivative.
for example, if $a>0$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Induction with floor, ceiling $n\le2^k\implies a_n\le3\cdot k2^k+4\cdot2^k-1$ for $a_n=a_{\lfloor\frac{n}2\rfloor}+a_{\lceil\frac{n}2\rceil}+3n+1$ Via induction I need to prove an expression is true. the expression is:
$n \leq 2^k \longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot
2^k-1$
for all $n,k \in \mathbb{Z^+}$
$a... | Hint: Your induction should be for $k$, not for $n$
$P(k): n \leq 2^k \Longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot
2^k-1$
For $k=1$ it is trivial
Suppose $P(k)$ is true and prove that $P(k+1)$ is true
$P(k+1): n \leq 2^{k+1} \Longrightarrow a_n \leq 3 \cdot (k+1)2^{k+1} + 4 \cdot
2^{k+1}-1$
But if $n\le 2^{k+1}$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Absolute value rational inequality I just stumbled upon this particular question and cannot answer
$$\left\lvert \frac{2x+1}{x-2} \right\rvert<1$$
I know that there are some rules in order to answer this Problem
$\frac{2x+1}{x-2} $ if only the $\frac{2x+1}{x-2}≥ 0 $
and
$\frac{(-)2x+1}{x-2}$ if only the $\frac{2x+1}... | $|\frac {2x+1}{x-2}|<1$
To get rid of the absolute value:
$-1< \frac {2x+1}{x-2}<1$
When you mulitiply through by $x-2$ it is going to flip the direction of the inequalities if $x-2 < 0$
Suppose (x-2) > 0
$2-x < 2x+1 <x-2$ and $(x-2) > 0$
$2-x < 2x+1$ and $2x+1 <x-2$ and $(x-2) > 0$
$\frac 13 < x$ and $x <-3$ and $x >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n... | HINT Find the difference (an+1 - an) and study the sign of this difference. If it is positive, the sequence is increasing, otherwise it is decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Proving inequality for $n\in \mathbb{N}$ How can we prove the following: For all $x>0 $ and $n\in \mathbb{N}$,
$$\frac{x^n}{1+x+x^2+...+x^{2n}}\leq\frac{1}{2n+1}.$$
I was wondering if someone can help me. Thanks.
| As suggested by @fonfonx in the comments, you can write
\begin{align}
\frac{x^n}{1+x+x^2+...+x^{2n}} &= \frac{1}{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\ldots+\frac{1}{x}+1+x+\ldots+x^{n}}\\
&= \frac{1}{1+(\frac{1}{x}+x)+(\frac{1}{x^2}+x^2)+\ldots+(\frac{1}{x^n}+x^n)}\\
&\leq\frac{1}{1+2+2+\ldots+2}\\
&=\frac{1}{1+2n}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$ for $a$, $b$, $c$ positive and $1+a+b+c=2abc$
Given $1+a+b+c = 2abc$ and positivity of real numbers $a,b,c$, we are asked to prove that $$\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$$
If $d=a+b+c$ I got as far as to simp... | We need to prove next equivalent inequality $$\sum\limits_{cyc}\left(\frac{ab}{1+a+b}+1\right)\geq\frac{9}{2}$$ or
$$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9}{2}$$
Then by Cauchy-Schwarz inequality $$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9\prod\limits_{cyc}(a+1)}{\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
I don't know how to start...
| We can still use the classic "complete square" trick to do this without having to use the "double" variables as shown by the first answer. To this end, we have: $\left((x^2+4x+8) + \dfrac{3x}{2}\right)^2 - \dfrac{9x^2}{4}+2x^2=0\implies \left((x^2+4x+8)+\dfrac{3x}{2}\right)^2 = \left(\dfrac{x}{2}\right)^2$. Now using t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Sum of binomial coefficients involving $n,p,q,r$
Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$
Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.
Could some help me with this, thanks
| The proposed identity can be derived (putting $r=s$) from this more general one
$$
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{
m - \left( {r - s} \right) \cr
k \cr} \right)\left( \matrix{
n + \left( {r - s} \right) \cr
n - k \cr} \right)\left( \matrix{
r + k \cr ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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$a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a} \le 4$ I've got stuck at this problem which I found some days ago in a book about inequalities.
If $a, b, c, d ∈ [0, +\infty)$ and $a+b+c+d=4$, then prove
$$a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a} \le 4$$
I thought about Cauchy-Buniakowsky-Schwartz inequality bu... | Because by AM-GM $\sum\limits_{cyc}a\sqrt{b}\leq\sum\limits_{cyc}\frac{a+ab}{2}=2+\frac{(a+c)(b+d)}{2}\leq4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$?
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$.
I think this is a simple exercise, but I get this:
$(n+1)^2+1=n^2+2n+2$.
$n^2+2n+2 = (n^2+1)+(2n+1)$
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$
and $\displaystyle n^2+1 = \frac{n(2n+1)}{... | Let $d=\gcd(n^2+1, (n+1)^2+1)$
So $d$ divides their difference
$d|2n+1$ so by multiplication with $n$ one gets $d|2n^2+n$
But $d|2n^2+2$ so $d|n-2$ so $d|2n-4$ so $d|5$
By considering all the residues modulo $5$ of $n^2+1$ and $(n+1)^2+1$ we can see that $\gcd$ is $5$ if $n\equiv2$ mod $5$, and $1$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Principal ideal by considering a product How can I show by considering the product of $\zeta_7^3 + \zeta_7^2 +1$ with $\zeta_7^3 + \zeta_7 +1$ that $(2, \zeta_7^3 + \zeta_7^2 +1 )$ is a principal ideal of $O_{Q(\zeta_7)}$?
| Let $\zeta=\zeta_7$. Multiply the numbers together. Doing so gives you the following:
\begin{align*}
(\zeta^3+\zeta^2+1)(\zeta^3+\zeta+1)&=\zeta^6+\zeta^5+\zeta^3+\zeta^4+\zeta^3+\zeta+\zeta^3+\zeta^2+1\\&=(\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1)+2\zeta^3\\
&=2\zeta^3
\end{align*}
$\zeta$ is a unit in this rin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
| Note that
Since
$$
3x^2+3y^2y'=0
$$
we have
$$
y'=-\frac{x^2}{y^2}
$$
Therefore,
$$
\begin{align}
0
&=(x+y)'\\[6pt]
&=1+y'\\
&=1-\frac{x^2}{y^2}\\
\end{align}
$$
At $x=y=1$, we get a maximum of $2$.
$x=-y$ doesn't happen, but $\frac xy\to-1$ as $x\to\pm\infty$. Since $xy\le\frac{x^2+y^2}2$, we have $x^2-xy+y^2\ge\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Number of positive integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$ for distinct primes $p$ and $q$
Let $p$ and $q$ be distinct primes. Then find the number of positive integer solutions of the equation $$\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$$
We get $pq=\frac{xy}{x+y}$
Now $x+y$ must divide $xy$ as L.H.S.... | Since $x$ and $y$ are positive, we have $\frac1x<\frac1{pq}$ and $\frac1y<\frac1{pq}$, which implies $x>pq$ and $y>pq$. This suggests the substitution $x:=pq+a$ and $y:=pq+b$ with positive integers $a$ and $b$.
Under this substitution, the given equation can be rewritten into the equivalent equation $ab=p^2q^2$. As $p^... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.
Can anyone provide a hint about what should be done?
| Rewrite your equation so that there is only one square root on each side. We get
$$\sqrt{4x^2+5x+1}=2\sqrt{x^2-x+1}+9x-3.$$
Squaring both sides we get
$$
\begin{align}
4x^2+5x+1&=4(x^2-x+1)+4(9x-3)\sqrt{x^2-x+1}+(81x^2-54x+9)\\
&=85x^2-58x+13+4(9x-3)\sqrt{x^2-x+1}\end{align}.$$
We get
$$81x^2-63x+12=-4(9x-3)\sqrt{x^2-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Inequality related to sum of reciprocals: $\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$?
For every integer $n>1$, prove that :
$\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$
I don't seem to find any clue on how to relate the left side of the inequality to the right side.
I tried a little bit of AM-GM o... | for $n=2$ we have we have $$1+\frac{1}{4}>\frac{6}{5}$$ this is true since we have $$25>24$$
now we assume that is true:
$$\sum_{k=1}^n\frac{1}{k^2}>\frac{3n}{2n+1}$$
and we have to prove that
$$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3(n+1)}{2(n+1)+1}$$
we show that
$$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3n}{2n+1}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$ Prove that :
$$\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$$
| The trick is induction together with integration by parts. Denote our integral by $I_n$. Then
$$
I_{n+1}=\int_{-1}^1(1-x^2)^n(1-x^2)\text{d}x=I_n-\int_{-1}^1 x^2(1-x^2)^n\text{d}x.
$$
Integrating by parts by $u=x$ and $v'=x(1-x^2)^n$ we arrive at the recurrence $$I_{n+1}=\frac{2n+2}{2n+3}I_n.$$
Therefore $$I_{n+1}=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to express $\sqrt{2-\sqrt{2}}$ in terms of the basis for $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ I've already shown that $\sqrt{2-\sqrt{2}}\in\mathbb{Q}(\sqrt{2+\sqrt{2}})$ but I want to write $\sqrt{2-\sqrt{2}}=a+b\alpha+c\alpha^2+d\alpha^3$ where $\alpha=\sqrt{2+\sqrt{2}}$. By squaring $\sqrt{2-\sqrt{2}}$ I get a system ... | There's a trigonometry trick for this particular case.
Let $\alpha=\sqrt{2+\sqrt{2}}$; then
$$
\alpha=2\sqrt{\frac{1+\sqrt{2}/2}{2}}=2\cos\frac{\pi}{8}
$$
whereas
$$
\beta=\sqrt{2-\sqrt{2}}=2\sqrt{\frac{1-\sqrt{2}/2}{2}}=2\sin\frac{\pi}{8}=
2\cos\left(\frac{\pi}{2}-\frac{\pi}{8}\right)=
2\cos\frac{3\pi}{8}
$$
Now
$$
\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit
$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$
can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty... | I unusually use this general method:
Let $\displaystyle \lim_{x \to x_o}u(x)=1$ and $\displaystyle \lim_{x\to x_o}V(x)=\infty$, then
$\displaystyle \lim_{x \to x_o}u^V=\lim_{x \to x_o}\Big[ \Big(1+(u-1) \Big)^{\frac{1}{u-1}}\space \Big]^{(u-1)V}=\Big[ \lim_{x\to x_o}\Big( 1+(u-1)\Big)^{\frac{1}{u-1} \space} \Big]^{\dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate volume between two geometric figures I have a figure C that is defined as the intersection between the sphere $x^2+y^2+z^2 \le 1 $ and the cyllinder $x^2+y^2 \le \frac{1}{4}$.
How should i calculate the volume of this figure?
| The given cylinder and the given sphere share the sections given by $z=\pm\frac{\sqrt{3}}{2}$. If $|z|\leq\frac{\sqrt{3}}{2}$, the area of the section is $\frac{\pi}{4}$. If $\frac{\sqrt{3}}{2}\leq|z|\leq 1$, the area of the section is $\pi(1-z^2)$.
Integrating $1$ on sections, we get that the volume is given by:
$$ \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the limit $\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$ I have been trying to find the limit,
$$\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$$
and sort of succeeded. But my $0$ answer doesn't converge with what Wolfram says... | Without Taylor nor the generalized Binomial theorem:
Use the identity
$$a^{12}-b^{12}=\\
(a-b)(a^{11}+a^{10}b^{1}+a^{9}b^{2}+a^{8}b^{3}+a^{7}b^{4}+a^{6}b^{5}+a^{5}b^{6}+a^{4}b^{7}+a^{3}b^{8}+a^{2}b^{9}+ab^{10}+b^{11}).$$
By multiplying/dividing by the conjugate "dodecanomial", the numerator becomes the polynomial
$$(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Finding the 5th order taylor polynomial of a function without brute force I am currently stuck on this 5th order taylor polynomial question for almost an hour now, it's quite difficult if you brute force it but there was a hint given today by my tutor that there is an easier way to solve for the 5th taylor polynomial o... | Building on the hint $u=x^2$.
$\sin(u)=u-\frac{u^3}{3!}+\frac{u^5}{5!}$
$\frac{\sin(u)}{u}=1-\frac{u^2}{3!}+\frac{u^4}{5!}$
Now back substituting in $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{x^4}{3!}+\frac{x^8}{5!}+...$
Edit
Using $x=x^2$ as the substitution as op had.
$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
$\frac{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.
Find the minimum value of $a+b+c$.
Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
| I am assuming here $a=b=c$ since $1000$ is the cube of $10$.
$$(a+1)(b+1)(c+1) = 1000$$
now, $(a+1)^3 = 1000$;
$$a+1 = 10\iff
a=9$$
so, $a+b+c = 3*9 = 27$.
Just that we can get the integer value but I cannot say $27$ is the minimum value for $a+b+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Binomial theorem expansion. Why do the powers flip in certain circumstances? Okay given an expression like $(x+y)^5$
we have terms
$x^ky^{n-k}$
So we have ${5 \choose 0} x^0y^5 + {5 \choose 1} x^1y^4+ {5 \choose 2} x^2y^3+ {5 \choose 3} x^3y^2+ {5 \choose 4} x^4y^1+ {5 \choose 5} x^5$
However when we have $(3x^2+y)^5$ ... | Expansion of the expression $\left(x+y\right)^n$ gives :
$$\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}x^ky^{n-k}$$
but you can also write it as :
$$\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$
simply because addition and multiplication are commutative.
So you can put the exponent $n-k$ to whatever term you w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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.