Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Summation of infinite series, $\sum((3n+1)^{-1}-(3n+2)^{-1})$ Find the sum of:
$$\sum_{n=0}^\infty \left(\frac{1}{3n+1}- \frac{1}{3n+2}\right) $$
Anwser given was $\dfrac{\pi}{3\sqrt{3}}$.
Thanks in advance.
| \begin{align}
\sum_{n=0}^\infty \left(\frac{1}{3n+1}- \frac{1}{3n+2}\right)
&=\sum_{m=0}^\infty\left(\int^1_0x^{3n}-x^{3n+1}dx\right)\\
&=\int^1_0\frac{1-x}{1-x^3}dx\\
&=\int^1_0\frac{1}{(x+1/2)^2+3/4}dx\\
&=\frac{2}{\sqrt{3}}\left[\arctan\left(\frac{2x+1}{\sqrt{3}}\right)\right]^1_0\\
&=\frac{2}{\sqrt{3}}\left(\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/886112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\frac 1 {\sqrt{x^2 - a^2}}$ I'm trying to understand how to integrate:
$$\frac 1 {\sqrt{x^2 - a^2}}$$
I tried substituting $t$ as $\sqrt{x^2 - a^2}$, but I can't get an answer that way, and I don't know any other way to integrate other than by parts, which I don't think can be used here. How should I integ... | $$\begin{align}
\text{Integrating} &\int \frac {1\cdot dx}{\sqrt {(x^2-a^2)}}\\\\
\text{Put x} &= a \cdot \cosh (\theta)\\
\Rightarrow \theta &= \cosh^{-1} \left(\frac {x}{a} \right)\\
\text{Therefore}\\ dx &= a \cdot \sinh(\theta) \cdot d\theta\\\\
\text{Now}\\
&\int \frac {a \cdot \sinh (\theta) \cdot d \theta}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
How many different positive integer factors does have? How many different positive integer factors does $(2^7)(3^4)(7^3)(23^5)$ have?
Do we have to do any combinations between the powers here?
| What you are looking for is the number of the positive divisors of $2^7\cdot 3^4\cdot 7^3\cdot 23^5$. Then, the answer is
$$(7+1)(4+1)(3+1)(5+1)=8\cdot 5\cdot 4\cdot 6=960.$$
In general, if
$$N={p_1}^{\color{red}{q_1}}\cdot {p_2}^{\color{red}{q_2}}\cdots {p_k}^{\color{red}{q_k}}$$
where $q_i,k\in\mathbb N$ and $p_1\lt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/891005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a limit. What makes the equality right? I'm reading a proof of a limit calculation. The limit is:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x}$$
where $a,b>0$.
The aother claims that:
$$\lim\limits_{x\to 0}\left(\frac{a^x+b^x}{2}\right)^\frac{1}{x} =
\exp\left( \lim\limits_{x\to 0}\frac{\... | First use simple fact, that $\displaystyle\lim_{y \to 0}\frac{\ln(1+y)}{y}=1$, so:
$$\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{y}=\lim_{x \to 0}\frac{\ln(\frac{a^x+b^x}{2}-1+1)}{\frac{a^x+b^x}{2}-1} \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{y}=\\=1 \cdot \lim_{x \to 0}\frac{\frac{a^x+b^x}{2}-1}{x}$$
Now $\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Arithmetic Progression. Q. The ratio between the sum of $n$ terms of two A.P's is $3n+8:7n+15$. Find the ratio between their $12$th term.
My method:
Given:
$\frac{S_n}{s_n}=\frac{3n+8}{7n+15}$
$\frac{S_n}{3n+8}=\frac{s_n}{7n+15}=k$
$\frac{T_n}{t_n}=\frac{S_n-S_{n-1}}{s_n-s_{n-1}}=\frac{k\left(\left(3n+8\right)-\left(3\... | Hint.
We have that
$$
S_n=A_1+\cdots+A_n,
$$
where $A_n=Kn+L$, and hence
$$
S_n=K\frac{n(n+1)}{2}+Ln.
$$
Similarly, $s_n=a_1+\cdots+a_n$, where $a_n=kn+\ell$ and $s_n=k\frac{n(n+1)}{2}+\ell n$,
and
$$
\frac{S_n}{s_n}=\frac{K\frac{n(n+1)}{2}+Ln}{k\frac{n(n+1)}{2}+\ell n}=\frac{K(n+1)+2L}{k(n+1)+2\ell}=\frac{Kn+2L+K}{kn+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/894079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
| There is no simple factorization of $x^2+2xy-y^2$ nor $x^2-2xy-y^2$, although you can write:
$$\begin{align}x^2+2xy-y^2 &= (x+y)^2-2y^2 \\&= \left(x+y(1+\sqrt{2})\right)\left(x+y(1-\sqrt{2})\right)
\end{align}$$
and similarly:
$$x^2-2xy-y^2 = \left(x-y(1+\sqrt{2})\right)\left(x-y(1-\sqrt{2})\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
On the decomposition of stochastic matrices as convex combinations of zero-one matrices Let "stochastic" matrix be the matrix whose rows sum to one and deterministic matrix be a stochastic matrix whose all rows consist of a one and zero.
For example $\left [ \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 ... | Constructive proof:
Consider some nonzero substochastic matrix $P$ with equal-row-sums. Call $P_{is(i)}$ one of the minimal nonzero entries of row $i$, and consider the deterministic matrix $D$ such that $D_{is(i)}=1$ for every $i$ and $D_{ij}=0$ otherwise. Finally, define $m(P)=\min\{P_{ij}\mid P_{ij}\ne0\}$.
Then $P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to prove that a diffrensiation of a formula equals to another formula. QUESTION 1) if $y =\dfrac{ \sin x-x\cos x}{x\sin x+\cos x}$ show that $\dfrac{dy}{dx}= \dfrac{x^2}{(x\sin x+\cos x)^2}$
QUESTION 2) if $y = \dfrac{\tan x+1}{\tan x-1}$ show that $\dfrac{dy}{dx}= \dfrac{-2}{1-\sin 2x}$
| For the first, we get
\begin{align}
\frac{d}{dx}\left(\frac{ \sin x-x\cos x}{x\sin x+\cos x}\right)
&= \frac{(x\sin x + \cos x)(\cos x + x\sin x-\cos x)-(\sin x - x\cos x)(\sin x + x\cos x - \sin x)}{(x\sin x + \cos x)^2} \\
&= \frac{x^2\sin^2 x + x\sin x\cos x - (x\sin x\cos x - x^2\cos^2 x)}{(x\sin x + \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can I perform the quadratic formula on polynomial with complex coefficient? 2 weeks ago, we had a Math test on complex number. One of the question was:
Let $z=x+iy$ be a non-zero complex number, where $x,y \in \mathbb{R}$.
Given that $z+\frac{1}{z} = k$, where $k$ is a real number, show that if $y \neq 0$, then $| k |... | $$x^2-y^2+1+2xy(i)=xk+i(yk)$$
Equating the imaginary parts, $$yk=2xy\implies x=\frac k2\text{ as }y\ne0$$
Equating the real parts, $$x^2-y^2+1=xk\iff\frac{k^2}4-y^2+1=\frac{k^2}2\iff k^2=4(1-y^2)\le4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/897996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ordinary Differential Equation with trigs! I need help and alternative ways of solving this problem:
$$ y'(x^2 + 1) - 2xy = (x^2 + 1)\arctan x $$
Thank you for all kind of help and hints :)
| $$
\left(x^2+1\right)y' -2xy = \left(x^2+1\right)\arctan x
$$
$$
y' -\frac{2x}{x^2+1}y = \arctan x
$$
$$
\frac{g'}{g} = \frac{2x}{x^2+1} = \frac{d}{dx}\ln g(x)
$$
thus
$$
y\mathrm{e}^{-\ln g} = \frac{y}{g} = \int \frac{1}{g}\arctan x dx = \int \frac{1}{x^2+1}\arctan x dx
$$
$$
\frac{d}{dx}\arctan x = \frac{1}{x^2+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Summation of general series One of the problems in Donald Knuth's Art of Programming is phrased as follows:
Find and prove a simple formula for the sum $$\sum\limits_{n=0}^k\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}.$$
I have very little experience with summations. My method was to try to find some function such that
$$\frac{... | By computing the residues of $\frac{x^3}{x^4+4}$ in $x=\pm 1\pm i$ we have:
$$\frac{x^3}{x^4+4}=\frac{1}{2}\left(\frac{x-1}{(x-1)^2+1}+\frac{x+1}{(x+1)^2+1}\right)\tag{1}$$
hence:
$$\begin{eqnarray*}\sum_{n=0}^{k}\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}&=&\frac{1}{2}\sum_{n=0}^{k}(-1)^n\left(\frac{2n}{4n^2+1}+\frac{2(n+1)}{4(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Find arc length of curve on the given interval I was asked to find the arc length of the curve of the following curve:
$24xy = x^4 + 48$ from $x = 2$ to $x = 4$
This has turned out to be a very difficult problem, I get stuck using the arc length formula with the derivative I have calculated.
| We can put
$$\begin{cases} x&=t \\y&=\frac{1}{24}t^3+2t^{-1}\end{cases}$$
Arc length formula $$s=\int_{2}^{4}{\sqrt{(x')^2+(y')^2}\text{d}t}$$ give us
\begin{align*}
s&=\int_{2}^{4}{\sqrt{1+\left(\frac{1}{8}t^2-2t^{-2}\right)^2}\text{d}t}\\
&=\int_{2}^{4}{\sqrt{\left(\frac{1}{8}t^2+2t^{-2}\right)^2}\text{d}t}\\
&=\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I bring $\sin(x)$ to the following form? What steps do we take for the following?
$$\sin x = \frac{{2\tan\frac{x}{2}}}{1+\tan^2\frac{x}{2}}$$
| $$\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} = 2\sin\frac{x}{2}\cos\frac{x}{2}\frac{\cos\frac{x}{2}}{\cos\frac{x}{2}} = 2\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\cos^2\frac{x}{2} = \frac{2\tan\frac{x}{2}}{\frac{1}{\cos^2\frac{x}{2}}} = \frac{2\tan\frac{x}{2}}{\frac{\cos^2\frac{x}{2}+\sin^2\frac{x}{2}}{\cos^2\frac{x}{2}}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Should I throw the dice again if I have rolled 4? My math skills are very basic so it might be a stupid question, I had a discussion with my brother in law and now we have a 'math problem'.
We were playing a game with dices and he threw 4. The challenge was to throw the highest number, you can stop or throw again once,... | If you and your brother-in-law both have the same strategy, then the game is even. So the only question is, what happens if you re-throw on a 4 and your brother-in-law doesn't? Here is a table of the probability of each outcome:
1 2 3 4 5 6
You 1/9 1/9 1/9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/905438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 0
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Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
| Alternate form:
$$\frac{\frac{\sqrt{n}}{ \sqrt{n}}}{\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}}{\sqrt{n}}}=\frac{1}{\sqrt{\frac{n+\sqrt{n+\sqrt{n}}}{n}}}=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{n}}}=\frac{1}{\sqrt{1+ \frac{\sqrt{n+\sqrt{n}}}{\sqrt{n} \sqrt{n}}}}=\frac{1}{\sqrt{1+ \frac{1}{\sqrt{n}} \frac{\sqrt{n+\sqrt{n}}}{ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is If $\frac{5x}{2x^2+5x+1}=\frac13$ then the value of $x+\frac{1}{2x}$ is
note $5x=1$
$2x^2+5x+1=3$
or
$15x=2x^2+5x+1$
so $2x^2-10x+1$
so $\frac{5+\sqrt{23}}2$ or $\frac{5-\sqrt{23}}2$
| Here are the steps
$$
\frac{5x}{2x^{2}+5x+1}=\frac{1}{3}
$$
$$
15x=2x^{2}+5x+1
$$
$$
10x=2x^{2}+1
$$
What would we get if we divide both sides of the last equation by $2x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Catagorising a Differential Equation I have
$$
\frac{d^{2}}{d\epsilon^{2}}g^{\star}+\frac{\left(R^{2}+3\epsilon^{2}\right)}{\epsilon\left(R^{2}-\epsilon^{2}\right)}\frac{d}{d\epsilon}g^{\star}+\frac{\left(5R^{2}+3\epsilon^{2}\right)}{\left(R^{2}-\epsilon^{2}\right)^{2}}g^{\star}=0
$$
and I wonder if anyone could tell m... | Assume $R\neq0$ for the key case:
Let $r=\epsilon^2$ ,
Then $\dfrac{dg^\star}{d\epsilon}=\dfrac{dg^\star}{dr}\dfrac{dr}{d\epsilon}=2\epsilon\dfrac{dg^\star}{dr}$
$\dfrac{d^2g^\star}{d\epsilon^2}=\dfrac{d}{d\epsilon}\left(2\epsilon\dfrac{dg^\star}{dr}\right)=2\epsilon\dfrac{d}{d\epsilon}\left(\dfrac{dg^\star}{dr}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/912304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Verification of Laurent Series calculation I tried to calculate the Laurent series of these functions but I have no way to verify my answers.
i)
$$
\begin{align}
f(z)=\frac{e^{z^2 }-1}{z^4}, \mathbb{D}=\mathbb{C} \backslash \{0\}
\end{align}
$$
We know that $ e^z=\sum_{n=0}^\infty(\frac{1}{n!})z^n $ , hence
$$
f(... | You first computation is correct. The second is mostly correct, only at the end you dropped a minus sign. (It might have been better to absorb it into the sign in the sum and write $(-1)^{n-1}$ there rather than $(-1)^n$.)
In the third,
We also know that when we a function to the power of 2, its correspondent series w... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
= (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calcula... | The solution-attempt in the question is excellent (after your edit)!
But unfortunately you made a small mistake. After taking out the factor $(a-b)(c-a)$, the middle element should be $c-a-b$, not $a+b+c$. If you fix that, the remaining determinant can be expanded and easily factorised.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/913798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Taking Calculus in a few days and I still don't know how to factorize quadratics Taking Calculus in a few days and I still don't know how to factorize quadratics with a coefficient in front of the 'x' term. I just don't understand any explanation. My teacher gave up and said just use the formula to find the roots or so... | The term $4x^2+16x$ is almost square of $2x+4$, more precisely
$$
4x^2+16x=(2x+4)^2-4^2.
$$
Therefore, we have
\begin{align}
4x^2+16x-19&=(2x+4)^2-4^2-19\\
&=(2x+4)^2-35\\
&=(2x+4)^2-\left(\sqrt{35}\right)^2.
\end{align}
Knowing that
$$
p^2-q^2=(p-q)(p+q),\tag1
$$
then we have
\begin{align}
4x^2+16x-19
&=\left(2x+4-\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 10,
"answer_id": 2
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Rationalized limit denominator, still undefined (divide by zero), how to solve? I am trying to solve:
$$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \sqrt{5x-1}}$$
My first step is to multiply by the conjugate to rationalize the denominator.
$$\lim_{x \to 2}\frac{\sqrt{x+2} - \sqrt{3x-2}}{\sqrt{4x+1} - \... | Since we have
$$\begin{align}\sqrt{x+2}-\sqrt{3x-2}&=\frac{(\sqrt{x+2}-\sqrt{3x-2})(\sqrt{x+2}+\sqrt{3x-2})}{\sqrt{x+2}+\sqrt{3x-2}}\\&=\frac{-2(x-2)}{\sqrt{x+2}+\sqrt{3x-2}}\end{align}$$
$$\begin{align}\sqrt{4x+1}-\sqrt{5x-1}&=\frac{(\sqrt{4x+1}-\sqrt{5x-1})(\sqrt{4x+1}+\sqrt{5x-1})}{\sqrt{4x+1}+\sqrt{5x-1}}\\&=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Determine the irrational numbers $x$ such that both $x^2+2x$ and $x^3-6x$ are rational numbers I did not make any progress. The problem is from RMC 2008.
The only idea that I have is:
Try to find sets of irrational numbers such that every number in the set multiplied by another number in the set yields a rational numbe... | I add yet another solution hopefully a simple one.
Since $x^2+2x$ is rational so too is $x^2+2x+1=(x+1)^2$.
Let $y=x+1$, so $y^2$ is rational, then $x^3-6x=y(y^2-3)-3y^2+5$ is rational, which means that $y(y^2-3)$ is rational.
Now since $y$ is irrational and $y^2-3$ rational we must have $y^2-3=0$, thus $y=\pm \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
specific magma examples Give an example of a magma $S$ such that $S$ has a zero and $S$ has a left zero divisor that is not a right zero divisor
an example of a magma with an identity such that there is an element with exactly $2$ left inverses but only one right inverse
For the first, I was thinking $\{0,a,b\}$ where ... | Your answer to the first question may be right (thanks @celtschk for commenting):
$$
\begin{array}{c|ccc}
\cdot & 0 & a & b\\
\hline
0 & 0 & 0 & 0\\
a & 0 & a & 0\\
b & 0 & a & b
\end{array}
$$
Here $0$ denotes an absorbing element, $a$ is only a left zero divisor, while $b$ is only a right zero divisor.
Second questi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/919208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find a vector that bisects the smaller of the two angles formed by vectors <3,4> and <5,-12>. The solution is <8,-1>. I tried finding the angle between the two vectors, but wasn't sure what to do next.
| The cosines of the angles have to be the same and positive (for the smaller angle). Therefore
$$\frac{3 a + 4 b}{\sqrt{3^2 + 4^2} \cdot \sqrt {a^2 + b^2} } = \frac{5 a + (-12) b}{\sqrt{5^2 + (-12)^2} \cdot \sqrt {a^2 + b^2} }>0$$ or
$$\frac{3 a + 4 b}{5 } = \frac{5 a + (-12) b}{13 }>0$$
Check that $(a,b) = (8,-1)$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Proof by induction that $a_0 = 1, a_1 = 1, a_n=2a_{n-1} + 3a_{n-2}$ satisfies $a_n = \frac12 (3^n) + \frac12 (-1)^n$ The question:
The terms of a sequence are given recursively as
\begin{cases}
a_0 = 1,\\ a_1 = 1 \\a_n=2a_{n-1} + 3a_{n-2} \quad\text{ for } n \geq 2
\end{cases}
prove by mathematical induction
$a_n = \... | We have, by induction hypotheses, that :
$a_k=\frac 12(3^k)+\frac 12(-1)^k$
and :
$a_{k-1}=\frac 12(3^{k-1})+\frac 12(-1)^{k-1}$.
If we "plug them" into :
$a_{k+1}=2a_k+3a_{k-1}$
we get :
$$a_{k+1}=2\left[\frac 12(3^k)+ \frac 12(-1)^k\right]+3\left[\frac 12(3^{k-1})+ \frac 12(-1)^{k-1}\right] =$$
$$= 3^k+(-1)^k... | {
"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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Integral $\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}dx$ I have a problem with the following integral:
$$
\int_{0}^{1}\ln\left(\,3 + x \over 3 - x\,\right)\,
{{\rm d}x \over \,\sqrt{\,x\left(\,1 - x\,\right)\,}\,}
$$
The first idea was to use the integration by parts because
$$
\int{{\rm d}x \over \,\sqrt{x\... | Split the integral into two forms by expanding the logarithm function
$$
\int_{0}^1\frac{\ln\frac{3+x}{3-x}}{\sqrt{x(1-x)}}\ dx=\int_{0}^1\frac{\ln(3+x)}{\sqrt{x(1-x)}}\ dx-\int_{0}^1\frac{\ln(3-x)}{\sqrt{x(1-x)}}\ dx
$$
Let $t=\sqrt{x}\ \rightarrow\ dt=\dfrac{dx}{2\sqrt{x}}$, we have
$$
2\int_{0}^1\frac{\ln(3+t^2)}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 0
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Find solution of equation $(z+1)^5=z^5$ I attempt to solve the equation
$(z+1)^5=z^5$.
My first approach is to expand the left hand side but ı get more complicated equation. So I couldn't go further. Secondly, I write equation as, since $z\neq0$,
$(\frac{z+1}{z})^5=1$, put $\xi=\frac{z+1}{z}$
and attempt to solve e... | To exploit symmetry, put $z=w-\frac 12$, which gives $z+1=w+\frac 12$.
Solving:
$$\begin{align}(z+1)^5&=z^5\\
\left(w+\frac 12\right)^5&=\left(w-\frac 12\right)^5\\
2\left[5w^4\left(\frac12\right)+10w^2\left(\frac 12\right)^3+\left(\frac 12\right)^5\right]&=0\\
80w^4+40w^2+1&=0\\
w^2&=\frac{-40\pm\sqrt{1600-320}}{160}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find the integral $\int_{0}^{\frac{\pi}{4}}\frac{\sin{x}\cos{x}}{\sin{x}+\cos{x}}dx$ find the integeral
$$\int_{0}^{\frac{\pi}{4}}\dfrac{\sin{x}\cos{x}}{\sin{x}+\cos{x}}dx$$
I know
$$\dfrac{2\sin{x}\cos{x}}{\sin{x}+\cos{x}}=\dfrac{(\sin{x}+\cos{x})^2-1}{\sin{x}+\cos{x}}=\sin{x}+\cos{x}-\dfrac{1}{\sin{x}+\cos{x}}$$
an... | Here is a very slightly (and I mean very slightly) different approach compared to those already given.
Recognising
$$\sin x + \cos x = \sqrt{2} \sin \left (x + \frac{\pi}{4} \right ).$$
Now
\begin{align*}
(\sin x + \cos x)^2 &= \sin^2 x + 2 \sin x \cos x + cos^2 x\\
\Rightarrow 2 \sin^2 \left (x + \frac{\pi}{4} \right ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find two real numbers x and y such that x, y, x + y, x^2, y^2, x^2 - y^2, x^3, y^3, x^4, y^4 are all irrational but x^2 + y^2 is rational. Find two real numbers $x$ and $y$ such that $x, y, x + y, x^2 , y^2 , x^2 - y^2, x^3, y^3, x^4, y^4$ are all irrational but $x^2 + y^2$ is rational.
What is the approach to solve t... | If we want to avoid transcendentals, use $x=\sqrt{3}+\sqrt{2}$ and $y=\sqrt{3}-\sqrt{2}$.
The price is that we need to work a little harder to prove the irrationality of $x$, $y$, $x+y$, $x^2$, $y^2$, $x^2-y^2$, $x^3$, $y^3$, $x^4$ and $y^4$. But not much harder, there is a lot of symmetry.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align... | Using:
$$
1 -\sqrt{1-x} = \frac{x}{1+\sqrt{1-x}}\qquad \frac{1}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{x}
$$
we have:
$$
\frac{1 -\sqrt{1-x}}{\sqrt{1+x}-1} = \frac{\sqrt{1+x}+1}{1+\sqrt{1-x}} = 1 + \frac{\sqrt{1+x}-\sqrt{1-x}}{1+\sqrt{1-x}} = 1 + \underbrace{\frac{2x}{\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(1+\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Existence of solution in $x,y \in (a,b)$ of $(\frac { a+b}2)^{x+y}=a^xb^y$ Let $a<b$ be positive real numbers , then is it true that there exist $x,y \in (a,b)$ such that
$ \bigg(\dfrac { a+b}2\bigg)^{x+y}=a^xb^y$ ?
| We can solve the equation by
$$(x(t),y(t)) = \left(\ln\left(\frac{2b}{a+b}\right)t,\ln\left(\frac{a+b}{2a}\right)t\right)$$
Now $\frac{a+b}{2}\geq a$ and therefore $\ln\left(\frac{a+b}{2a}\right)\geq 0$
and by $\frac{a+b}{2}\leq b$ we conclude $\ln\left(\frac{2b}{a+b}\right)\geq 0$.
Furthermore $\ln\left(\frac{a+b}{2a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$. If $p > 5$ is a prime number, then the last digit of $p^4-1$ is $0$ (ex.: $7^4-1=2400$).
How do I prove this?
| By Fermat theorem
$$p^4\equiv1\pmod{5}\\p^4-1\equiv0\pmod{5}\\p^4\equiv1\pmod{2}\\p^4-1\equiv0\pmod{2}\\p^4-1\equiv0\pmod{10}$$
Since $p$ is odd so is $p^4$ clearly it gives $1$ as remainder by $2$.Also since $2$ and $5$ are co-prime we have that $0$ is remainder of $2\cdot 5$ which gives that the last digit is $0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $m\cos^2{\theta} + n\sin^2{\theta} < l \implies \sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta} < \sqrt{l} $ Prove that $m\cos^2{\theta} + n\sin^2{\theta} < l \implies \sqrt{m}\cos^2{\theta} + \sqrt{n}\sin^2{\theta} < \sqrt{l} $ for every $m, n, l >0$.
| This problem can be found in
LARSON, LOREN C., 1983: Problem-Solving Through Problems.
Springer-Verlag, p. 255, 267.
The author proposes two solutions: the first one is based on the Cauchy-Schwarz inequality; the second one on the arithmetic-mean−geometric-mean inequality. A third solution, which uses the concavit... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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When $x^2+6xy+y^2$ a square number? Find all natural numbers $x$ and $y$ such that $x^2+6xy+y^2$ is a square number.
For example, $(x,y)=(2,3)$ or $(x,y)=(3,10)$.
Obviously, we can consider $gcd(x,y)=1$.
| For non-trivial cases, $xy\ne0$
Let $x^2+6xy+y^2=(x+ky)^2$ where $k$ is any integer
$\iff y(6x+y)=y(2kx+k^2y)\implies6x+y=2kx+k^2y\iff x(6-2k)=y(k^2-1)$
So, $\dfrac x{k^2-1}=\dfrac y{6-2k}=m$(say an integer)
As $x,y>0$ if $m<0,$ we need $6-2k<0\iff k>3\ \ \ \ (1)$ and $k^2-1<0\iff -1<k<1\ \ \ \ (2)$
There can be no ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Another parametric integral relating to hyperbolic function if $0<a\leq1$, then canwe get a closed form of
$$I(a)=\int_0^\infty\frac{x}{\tanh x}\frac{1}{\cosh^2(ax)}dx.$$
In fact,if $a=1$,$I(a=1)=\pi^2/8$.
| If you are interested in particular cases:
$$\begin{align}
I\left(\frac14\right) & = \pi^2 + 2 \\
I\left(\frac13\right) & = -12\operatorname{Li}_2\left( \frac{2}{i\sqrt{3}-1} \right) -12\operatorname{Li}_2\left( -\frac{2}{i\sqrt{3}+1} \right) - \frac{5\pi^2}{8} \\
I\left(\frac12\right) & = \frac{\pi^2}{4} + 1 \\
I\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Simplifying a proof by contradiction: if $a\equiv 1\bmod 5$, then $a^2\equiv 1\bmod5$ Prove the following either by Direct Proof or by Contraposition:
Suppose $a\in\mathbb{Z}$, if $a\equiv 1\pmod 5$, then $a^2\equiv 1\pmod5$
Suppose $a\equiv 1\pmod 5$
Then $5|\left(a-1\right)$, therefore $a-1=5k$
$a^2-1=\left(a-1\ri... | If
$a\equiv 1\pmod 5$,
$a = 5n+1$
for some integer $n$.
Therefore
$a^2
= 25n^2+10n+1
= 5(5n+2)+1
$
so
$a^2\equiv 1\pmod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| Your answer and Geogebra's answers are equivalent.
$x^3(5+x)(5-x)=-x^3(5+x)(x-5)=-x^3(x+5)(x-5)$.
They just factored out a negative out of $(5-x)$ in order to make it $-(x-5)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Exercise in algebra - Express different terms
We have $ a = \dfrac{1}{\sqrt{1-b^2}}, c = \sqrt{\dfrac{1+b}{1-b}}, 0 \leq b < 1 $
Express $b$ in terms of $a$, $b$ in terms of $c$, $c$ in terms of $a$ and $a$ in terms of $c$.
So I want to do this in the quickest and algebraically "cleanest" way. Here's my own attempt:... | Now your 1 and 2 look fine.
For 3 : $$c=\frac{\sqrt{1+b}}{\sqrt{1-b}}\cdot\frac{\sqrt{1+b}}{\sqrt{1+b}}=\frac{1+b}{\sqrt{1-b^2}}=a(1+b)=a\left(1+\sqrt{1-\frac{1}{a^2}}\right)=a+\sqrt{a^2-1}.$$
For 4 : $$(c-a)^2=a^2-1\Rightarrow c^2-2ac=-1\Rightarrow a=\frac{c^2+1}{2c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:
$ \forall ... | I write an answer to reorganize the Slade's second answer for oneself.
What I would like to prove is the following proposition.
Proposition. Let $p$ be an odd prime. For every $t, k \in \mathbb{N}$, if $p \mid t - 1$ then
$$\nu_p\left(\frac{t^k - 1}{t - 1}\right) = \nu_p(k).$$
Once this has proved, the substitutio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Proving the Volume of an Ellipsoid This is the question:
The solid generated by rotating the region inside the ellipse with equation
$$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1 $$
around the $x$-axis is called an ellipsoid.
(a) Show that the ellipsoid has volume $\displaystyle \frac{4}{3} \p... | We know that the volume of the solid generated by rotating the curve $y=f(x)\space $ about the x-axis is given as follows
$$V=\int_{x_1}^{x_2}\pi y^2dx$$
Hence, the volume of the solid generated by rotating the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \space$ about the x-axis is obtained by setting $y^2=\frac{b^2}{a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the sequence $x_n = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} +\dots + \frac{\sin n}{2^n}$ Cauchy? $$x_{n} = \frac{\sin 1}{2} + \frac{\sin 2}{2^2} + \frac{\sin 3}{2^3} + ... + \frac{\sin n}{2^n}$$
I came across this sequence while studying, and while it is convergent, I'm curious as to whether or no... | For $n > m > 0$, we have
$\vert x_n - x_m \vert = \vert \sum_{m + 1}^n \dfrac{\sin i}{2^i} \vert \le \sum_{m + 1}^n \dfrac{\vert \sin i \vert}{2^i} \le \sum_{m + 1}^n \dfrac{1}{2^i}, \tag{1}$
since $\vert \sin i \vert \le 1$. But
$\sum_{m + 1}^n \dfrac{1}{2^i} = \dfrac{1}{2^{m + 1}} \sum_0^{n - m -1}\dfrac{1}{2^i} \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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number of matrices of rank 3? Let M be the space of all $4\times 3$ matrices with entries in the finite field of three elements.The number of matrices of rank 3 in M are?
A. $(3^4-3)(3^4-3^2)(3^4-3^3)$
B.$(3^4-1)(3^4-2)(3^4-3)$
C.$(3^4-1)(3^4-3)(3^4-3^2)$
D.$(3^4)(3^4-1)(3^4-2)$
which of the following is the answer??
| The first column can be any non-zero vector in $\mathbb{F}^4$. There are $3^4-1$ of those. Having chosen one of them, call it $c_1$, and then the second column needs to be any vector not in $span(c_1)$. There are $3^4-3$ of those. The third column needs to be outside $span(c_1,c_2)$, so it is one of $3^4-3^2$ options.
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Finding the integer solutions of the equation $3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14$ $
3\sqrt {x + y} + 2\sqrt {8 - x} + \sqrt {6 - y} = 14
$ .
I already solved this using the Cauchy–Schwarz inequality and got $x=4$ and $y=5$. But I'm sure there is a prettier, simpler solution to this and I was won... | The Cauchy-Schwarz inequality goes like this:
$LHS \leq \sqrt{3^2+2^2+1^2}\cdot \sqrt{\left(\sqrt{x+y}\right)^2+\left(\sqrt{8-x}\right)^2+\left(\sqrt{6-y}\right)^2}=14=RHS$.
Thus you have equality when $\dfrac{3}{\sqrt{x+y}}=\dfrac{2}{\sqrt{8-x}}=\dfrac{1}{\sqrt{6-y}} \to x=4,y=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How do I make this simple proof better (and more correct?) Let $x$ and $y$ be real numbers. If $x\cdot{y}>\frac{1}{2}$, then $x^2+y^2>1$.
Proof: We will prove with the direct method. Let $x$ and $y$ be real numbers. Since
$$ x\cdot{y}>\frac{1}{2} $$
it follows that
$$ 2xy>1,$$
which means
$$x^2+y^2 \geq 2xy.$$
Therefor... | There is a much simpler proof.
Since $xy > \frac{1}{2}$, $x$ and $y$ must be non zero numbers, which are both positive or negative.
Suppose $x>0$ and $y>0$.
If $x>1$, then also $x^2>1$, and $x^2 +y^2>x^2>1$ is obvious.
Take $0<x\leq 1$ and $y>0$.
Since $0<x\leq 1$, the function $f$ with $f(x)=\sqrt{1-x^2}$ is well de... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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integer $n$ for which $n^6+3n^5-5n^4-15n^3+4n^2+12n+3$ is a perfect Square Prove that the no integer $n\;,$ for which $n^6+3n^5-5n^4-15n^3+4n^2+12n+3$ is a perfect
Square.
My Try:: We can write $(n^6+3n^5-5n^4-15n^3+4n^2+12n+3) = (n^3+an^2+bn+c)^2$
Now Here we have to find values of $a,b,c$.
But
this become very comp... | Modulo $4$ we have
$$\eqalign{n^6+3n^5-5n^4-15n^3+4n^2+12n+3
&\equiv n^6-n^5-n^4+n^3+3\cr
&\equiv n^3(n+1)(n-1)^2+3\cr
&\equiv3\ ;\cr}$$
but a square can only be congruent to $0$ or $1$ modulo $4$.
Reason for the last step: if $n$ is even then $n^3$ is a multiple of $4$, while if $n$ is odd then $(n-1)^2$ is a mu... | {
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"url": "https://math.stackexchange.com/questions/950523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Evaluate this square root $\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$
I have no clue where to begin. I would appreciate a hint, the answer should be
$2\sqrt{5}$
In general, how do you evaluate
$\sqrt{a + b} + \sqrt{a - b}$?
Thanks!
| Notice that $(\sqrt{5}-1)^2 = 6-2\sqrt{5}$ and $(\sqrt{5}+1)^2 = 6+2\sqrt{5}$, hence:
$$ \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=(\sqrt{5}+1)+(\sqrt{5}-1)=2\sqrt{5}. $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $1(0!)+3.(1!)+7(2!)+13(3!) +21(4!) + \cdots $ n terms.... Question( from sequences) :
If $1(0!)+3.(1!)+7(2!)+13(3!) +21(4!) + \cdots $ n terms = $(4000)(4000!)$ Then what is the value of n.
How to proceed in this please suggest , will be of great help to me thanks...
| The summation is
\begin{align}
S_{n} = \sum_{r=0}^{n} \left( r(r+1) + 1\right) \, r!
\end{align}
and by selecting $n$ values it is seen that
\begin{align}
S_{0} &= 1 \\
S_{1} &= 0! + 3 \cdot 1! = 2 \cdot 2! \\
S_{2} &= 0! + 3 \cdot 1! + 7 \cdot 2! = 3 \cdot 3!
\end{align}
which leads to
\begin{align}
S_{n} = \sum_{r=0}... | {
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"answer_id": 1
} |
Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\p... | The solutions to $z^2 + z + 1 = 0$ are the complex $z = 1 \angle (\pm 1/3)$.
So
$$\begin{align}
z^{10} + z^{-10} & = \left(1\angle (\pm 1/3)\right)^{10} + \left(1\angle (\pm 1/3)\right)^{-10}
\\& = \left(1\angle (\pm 10/3)\right) + \left(1\angle (\mp 10/3)\right) \
\\ & = (1\angle 10/3) + (1\angle -10/3)
\\ & = (1\an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Leibniz test for convergence of non alternating series I am aware that one can use the comparison test and the integral test to show that the series $$\sum_{n=1}^{\infty}\frac{1}{n(n+3)}$$ converges. Is it possible to use the Leibniz test to show that the series converges?
| Your sequence $$\sum\frac1{n(n+3)}=\frac14+\frac1{10}+\frac{1}{18}+\frac1{28}+\frac1{40}+\cdots$$
$$=\sum\left(\frac1n-\frac{n+2}{n(n+3)}\right)=1-\frac{3}{4}+\frac12-\frac4{10}+\frac13-\frac{5}{18}+\frac14-\frac{6}{28}+\frac15-\frac{7}{40}\pm\cdots$$
$$=\sum\left(\frac{n+4}{n(n+3)}-\frac1n\right)=\frac{5}{4}-1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/957922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt... | Setting $\dfrac1x=h,$ the limit reduces to $$\lim_{h\to0^+}\frac{\sqrt{1+2h}-\sqrt{1-7h}}h$$
$$=\lim_{h\to0^+}\frac{1+2h-(1-7h)}{h(\sqrt{1+2h}+\sqrt{1-7h})}$$
$$=\lim_{h\to0^+}\frac9{\sqrt{1+2h}+\sqrt{1-7h}}$$
$$=\frac9{\sqrt{1+2\cdot0}+\sqrt{1-7\cdot0}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Find the number of common tangents to $y^2=2012x$.... Problem :
Find the number of common tangents to $y^2=2012x$ and $xy =(2013)^2$
Solution :
Common tangent will have slope equal to both curves.
therefore, differentiation both the curves we get the slopes .
$\therefore 2y\frac{dy}{dx}=2012 \Rightarrow \frac{dy}{... | Let $(s,t)$ be the point on $y^2=2012x$ and let $(u,v)$ be the point on $xy=(2013)^2$.
Then, we have
$$t^2=2012s,\tag1$$
$$uv=(2013)^2.\tag2$$
Since
$$y^2=2012x\Rightarrow 2y\cdot\frac{dy}{dx}=2012,$$
we know that the $y$-axis is tangent to this curve at the origin. However, since $y$-axis is not tangent to the curve ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $abc\ne0$ and $ab+bc+ac=0$ then $a+b+c\ne0$ I tried to do proof by contradiction, but problem is how to get from $ab+bc+ac$ to $a+b+c$
Assuming $a+b+c=0$
my approachs:
*
*Adding $ab+ac+bc=0$ and $a+b+c=0$ and try to factor
*Deriving $$a^2+ab+ac=0\\ac+bc+c^2=0\\ab+b^2+bc=0$$ and trying to derive somet... | If $ab+bc+ac=a+b+c=0$, then $a,b,c$ are the three roots of a cubic equation $X^3 - abc=0$. But any nonzero number has two non-real cube roots, so $abc=0$—assuming that at least two of $a,b,c$ are real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\int _0^1\int _0^{\left(1-x^n\right)^{1/n}}\left(-x^n-y^n+1\right)^{1/n}dydx$ Let $n>0$. How does one integrate
$$\int _0^1\int _0^{\left(1-x^n\right)^{1/n}}\left(-x^n-y^n+1\right)^{1/n}dydx$$
?
This integral represents the volume enclosed by $$x>0,y>0,z>0,x^n+y^n+z^n<1$$.
By substitution of $y=t\left(1-x^n\right)^{1... | Final steps thanks to Lucian starting from
$$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-x^n\right)^{2/n}dx$$
Let $x=s^{1/n}$ and we get
$$\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{n\Gamma \left(\frac{n+2}{n}\right)}\int _0^1\left(1-s\right)^{2/n}s^{1/n-1}ds$$
$$=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the value of $\sum_{m=1}^{19} \frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$ with $\zeta=e^{2\pi i/19}$? Given that $\zeta=e^{2\pi i/19}$, how to find the value of $$S=\sum_{m=1}^{19} \dfrac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$$?
All I could think of was to somehow factorize the denominator and apply some sort ... | Let $X$ be the set $\{\; \zeta, \zeta^2, \zeta^3, \ldots, \zeta^{18}\; \}$.
Since $19$ is a prime number, for any integer $m$ relative prime to $19$, the map
$$X \in x\quad \mapsto \quad x^m \in X$$
is a permutation of $X$. Together with the obvious identity $\sum\limits_{x\in X} x = -1$, we have
$$\begin{align}
&\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Code is not cyclic for any q I have code $C$ over $F_p$ with generator matrix which looks like
$G =
\begin{pmatrix}
0 &0& 0& 1& 0& 1& 1 &1\\
1& 0 &0& 0 &1 &0 &1& 1\\
1& 1& 0& 0& 0& 1& 0& 1\\
1 &1& 1& 0 &0 &0& 1 &0\end{pmatrix}$
I need to show that this code is not cyclic for any $p$.
I constructed vector which looks... | $C$ must contain every cyclic shift of its codewords. Consider the cyclic shift of the last row of $G$, $x = \left( {01110001} \right)$. If $C$ is a cyclic code, $x$ must be a linear combination of the rows of $G$, i.e.
${x^T} = a\left( {\begin{array}{*{20}{c}}
0\\
0\\
0\\
1\\
0\\
1\\
1\\
1
\end{array}} \right) + b\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
how to solve $\int\frac{1}{1+x^4}dx$ i want find the answer and method of solve of $\int\frac{1}{1+x^4}dx$.
I know $$\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$$,
How I can use this to solve of that integration.
| Hint:
Use the identity
$$1+x^4=(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)$$
and Partial fractions decomposition.
Edit:
Then
$$\dfrac{1}{1+x^4}=\dfrac{1}{(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)}\\=
\dfrac{Ax+B}{1+\sqrt{2}x+x^2}+\dfrac{Cx+D}{1+\sqrt{2}x+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Eigenvectors of $\left( \begin{array}{ccc} a & 0 \\ 0 & -b \end{array} \right)$ I calculated the eigenvalues of the following matrix to be $a$ and $-b$.
$J = \left( \begin{array}{ccc}
a & 0 \\
0 & -b \end{array} \right)$
But when I use the formula $(J - \lambda I)v = 0$ with either $a$ or $-b$ as eigenvalue to calculat... | if $a=b=0$
Then you have
$$\begin{pmatrix}a&0\\0&-b\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$
This means that every basis $(v_1,v_2)$ of $\Bbb{R}^2$ is a basis of eigenvectors. One can verify this by seeing that the dimension of the kernel of $A$ is $2$ and the fact that multiplying the zero matrix by $v_1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/973914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$ I know that it is true but not sure how to write the proof for: $1 + 4 + 9 ... + n^2 = (n/6)(n+1)(2n+1)$. I need help to guide me in the right direction.
Thanks in advance.
edit:
Okay at n=k I have $ 1+4+9 ... + k^2 = (k/6)(k+1)(2k+1)$
and at $n=k+1$ I have $((k+1)/6... | When $n=1$, it is Okay.
Suppose that $n=k$, $$1 + 4 + 9 \cdots + k^2 = (k/6)(k+1)(2k+1).$$
Then $n=k+1$, $$1 + 4 + 9 \cdots + k^2+(k+1)^2 = (k/6)(k+1)(2k+1)+(k+1)^2=((k+1)/6)(k+2)(2k+3).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/974355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integration by parts with Legendre Functions I need help deriving
$\int_{-l}^l [P_l^m(x)]^2 = \frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$
for the associated Legendre functions
I am supposed to use
$P_l^m(x) = (-1)^{-m}\int_{-l}^l \frac{(1-x^2)^{\frac{m}{2}}}{2^ll!}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l$
and
$P_l^m(x) = \frac{1}... | The Rodrigues' formulae you supplied have gone somewhat astray; they should be
$P_n^m(x)=\frac{1}{2^n n!} (x^2-1)^{m⁄2} [(x^2-1)^n ]^{(n+m)}$
and
$P_n^{-m}(x)=\frac{1}{2^n n!} (x^2-1)^{-m⁄2} [(x^2-1)^n ]^{(n-m)}$
Let us derive the value of $‖P_n^m‖^2$ for positive m:
$\newcommand{\partial}[1]{\left[#1\right]}$
$\newcom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\tan \{\frac{1}{2} \sin^{–1} (2x/ 1 + x^2) + \frac{1}{2} \cos^{–1} (1 – y^2/1 + y^2) \}$ is equal to. $$
\tan \left\{
\frac{1}{2} \arcsin \frac{2x}{1 + x^2} +
\frac{1}{2} \arccos \frac{1 – y^2}{1 + y^2}
\right\}
$$
is equal to.
Note:
i think $\sin a=2x/1+x^2$, $\cos b=(1 – y^2/1 + y^2)$
| Set $\arctan x=A\implies x=\tan A\ \ \ \ (1)$
Using definition of Principal values, $\displaystyle-\frac\pi2\le A\le\frac\pi2\ \ \ \ (2)$
and using Weierstrass substitution formula, $\displaystyle\frac{2x}{1+x^2}=\sin2A$
Now $\displaystyle\arcsin(\sin2A)=\begin{cases} 2A &\mbox{if } -\dfrac\pi2\le2A\le\dfrac\pi2\iff-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $n!^{n+1}$ divides $(n^2)!$ My attempt so far is by induction. Let $f(n) = \frac{(n^2)!}{n!^{n+1}}$, I will try showing that $f(n)$ is a positive integer for all $n$.
We have $f(0) = \frac{0!}{0!^{n+1}} = 1$.
Now assume for induction, that $f(n) = k$ for some positive integer k.
Next, I'll examine $f(n+1) = \... | Here is a combinatorial proof:
Arrange $n^2$ person in $n^2$ chairs in a row, we have $n^2!$ ways.
Alternatively, we can pick $n$ persons for the first $n$ chairs, then $n$ persons for chairs $n+1$ to $2n$, then $n$ persons for $2n + 1$ to $3n$, etc, then arrange each group of $n$ persons, we have $\left(\prod_{k=0}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Last 3 digits of $7^{12341}$ I know that I need to reduce $7^{12341} \pmod {1000}$
By Euler I have $7^{\phi(1000)}\equiv 7^{400}\equiv1\pmod{1000}$
That leaves me with the monster $7^{341}\pmod{1000}$
Is there a way to reduce this smoothly without working $7^2, 7^4, 7^8$ etc manually ?
| $$7^2=50-1$$
$$7^{20n+1}=7(7^2)^{10n}$$
Now $(7^2)^{10n}=(50-1)^{10n}=(1-50)^{10n}$
Using Binomial Theorem,
$(7^2)^{10n}=1-\binom{10n}150+\binom{10n}2(50)^2\cdots+(50)^{2n}$
$\implies(7^2)^{10n}\equiv1-500n+\dfrac{10n(10n-1)}250^2\pmod{1000}\ \ \ \ (1)$ for $n\ge2$
Again, $\dfrac{10n(10n-1)}250^2=50n^2(50^2)-5n(2500)\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/983926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Limit by L'hospital's rule I have to prove that:
$$\lim \limits_{x \to \infty} \frac{{\int_x^{\infty} \exp(-t^2/2)dt}}{\exp(-x^2/2)
(1/x)}=1$$
Should I use L'hospital rule, if yes what are the derivatives?
| First note that
$$
\int e^{-cx^2}dx=\sqrt{\frac{\pi}{4c}}\mathrm{erf}(\sqrt{c}x)
$$
So now we have
$$
\lim_{x \to \infty} \left[\frac{\int_x^{\infty} e^{-\frac{t^2}{2}}dt}{\frac{1}{x}e^{-\frac{x^2}{2}}}\right]=\lim_{x \to \infty} \left[\frac{\lim \limits_{b\to\infty} \int_x^b e^{-\frac{t^2}{2}}dt}{\frac{1}{x}e^{-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that the locus of a point P is a circle I'm struggling with this geometry question:
The fixed points A and B have coordinates $(-3a,0)$ and $(a,0)$ respectively. Find the equation of the locus of a point P which moves in the coordinate plane so that $AP = 3PB$. Show that the locus is a circle, S, which touches t... | I think you overworked here. The condition is
$$|AP|^2=9|BP|^2\iff (x+3a)^2+y^2=9\left[(x-a)^2+y^2\right]\iff$$
$$\iff 8x^2-24ax+8y^2=0\iff \left(x-\frac{3a}2\right)^2+y^2=\frac{9a^2}4$$
and we get that the points$\;P=(x,y)\;$ are the locus of circle with center $\;\left(\frac{3a}2\,,\,0\right)\;$ and radius $\;\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/987791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$
Find the set of integer solutions $(x,y)$ to
$$\frac{x^2+2xy+y^2}{x^2-y^2} >x+y$$
I can't seem to multiply both sides by the expression in the denominator.
Nor can I simplify and cancel any terms. How should I it?
| $$\frac{x^2+2xy+y^2}{x^2-y^2}=\frac{(x+y)^2}{(x-y)(x+y)}=\frac{x+y}{x-y} \ge (x+y) \implies \frac{1}{x-y}\ge 1$$
Without loss of generality, assume that $x \ge y$, and fiddle around with this expression.
Edit: We do the standard LHS-RHS trick:
$$\frac{x^2+2xy+y^2}{x^2-y^2}-(x+y)>0 \\ LHS= \frac{(x+y)^2}{(x-y)(x+y))}-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/988198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}$ for $0 < x < 1$ I stumbled upon this question while doing practice inequalities questions, and I do not know how to start...
Problem: Prove that
\begin{align*}
\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}
\end{align*}
for $0 < ... | Hints:
Notice that $\sqrt{2x^2-2x+1}=\sqrt{x^2+(x-1)^2} \ge \sqrt{(\frac{x+1-x}{2})^2}=\frac12$ and
$\frac{1}{x+\frac1x}\le \frac{1}{2x\frac1x}=\frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Remainder of $3^7/8$ I read here that the remainder of $\frac{ab}{c}$ is equal to the remainder of $\frac{a}{c}\frac{b}{c}$ implying that the remainder of $\frac{a^b}{c}$ is equal to the remainder of $[\frac{a}{c}]^b$. However, when I apply this here, I would get remainder of $[\frac{3}{8}]^7=3^7$ (while the correct an... | Remainder of $ \frac {ab}{c}$ = remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} $ is FALSE in most cases.
When remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} \lt c$, only then it is valid.
However when remainder of $ \frac {a}{c} $ $\times$ remainder of $ \frac {b}{c} \ge c$, then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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To calculate side of the Equilateral triangle The figure is an equilateral triangle. 3 line segments , which meet at a(any) point in the triangle , are of the length 5cm, 4cm, and 3 cm as shown in the figure. Find the side of the equilateral triangle.
| Theorem: Let $ABC$ be an equilateral triangle, with a point $P$ inside it such that $PC^2=PA^2+PB^2$, then $\angle APB=150^{\circ}$.
So to do that, we cleverly construct a point $D$ such that $APD=60^{\circ}$ and $AP=PD$. Hence, $APD$ is equilateral and $AP=AD$.
Now, as $60^{\circ}=\angle CAB=\angle DAP$, so, $\angle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice spec... | \begin{align}
\int_0^\infty \frac{\ln x}{x^n-1}\,dx
=& \>\frac1n\int_0^\infty \int_0^\infty \frac{1}{(1+y)(x^n+y)} dy \ dx \\
=&\>\frac{\pi }{n^2}\csc\frac\pi n \int_0^\infty\frac{y^{\frac1n-1}}{1+y}dy
=\frac{\pi^2}{n^2}\csc^2\frac\pi n
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Closed form for $I=\int_{0}^{\infty}\frac{x^n}{x^2+u^2}\tanh(x) \, dx$ solve
$$I=\int_{0}^{\infty}\frac{x^n}{x^2+u^2}\tanh(x) dx:0<n<2$$
I tried for $n=1$ :
$$I(v)=\int_{0}^{\infty}\frac{x}{x^2+u^2}\tanh(vx) dx$$
$$I'(v)=\int_{0}^{\infty}(\frac{1}{\cosh^2(vx)}-\frac{u^2}{(x^2+u^2)\cosh^2(vx)}) dx$$
$$I'(v)=-\sum_{k=1}... | As Lucian stated in the comments above, the integral converges only if $-2 < n <1$.
I'll do the case $n=-1$ using contour integration.
For $u>0$ and not equal to $\pi \left(k + \frac{1}{2} \right)$ (where $k$ is a nonnegative integer), consider the complex function
$$ f(z) = \frac{\tanh z}{u^{2}+z^{2}} \frac{1}{z}.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
How do I solve this differential equation? $y'=\frac{y}{x+y^3}$ So, hey.
The equation is $y'=\frac{y}{x+y^3}$
So, I get something like this: $y'\left(x+y^3 \right)-y=0$, which I can't actually solve. I must admit I am slightly confused how to attack this one.
What should I do?
| $y'=\frac{y}{x+y^3}\Leftrightarrow(x+y^3)dy=ydx$.
Case 1: $y=0$.
Case 2: $y\neq0$, then $$\frac{1}{y^2}[(x+y^3)dy-ydx]=ydy-xd\frac{1}{y}-\frac{1}{y}dx=d(\frac{1}{2}y^2-\frac{x}{y})$$
$\frac{x}{y}-\frac{1}{2}y^2=C$.
The solution to the ODE is $x=\frac{1}{2}y^3+Cy$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$; is $p$ divisible by $1997$? if $p,q\in \mathbb{N}$ and
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\dots+\frac{1}{1331}=\frac{p}{q}$$
why is $p$ divisible by $1997$?
| Let $S$ be the LHS. Using
$$
1-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{1}{2n}
$$
(which can be proved by induction), we can deduce that
$$
S-\frac{1}{1332}=\frac{1}{667}+\cdots+\frac{1}{1331}+\frac{1}{1332}\\
\implies S=\frac{2}{1332}+\frac{1}{667}+\cdots+\frac{1}{1331}=\frac{1}{666}+\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/993852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Closed form of $\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$ Find the closed form of $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^4$$
I know the closed form for smaller powers like $2, 3$ exists, but I'm not sure if there is a
closed form for this variant. Is it possible to tackle the question in an elementar... | The result is
\begin{align*}
S_{1^4,4}=\sum\limits_{n = 1}^\infty {\frac{{H_n^4}}{{{n^4}}}} = \frac{{13559}}{{144}}\zeta \left( 8 \right) - 92\zeta \left( 3 \right)\zeta \left( 5 \right) - 2\zeta \left( 2 \right){\zeta ^2}\left( 3 \right) + 26{S_{2,6}},
\end{align*}
Detailed process see the paper ``Multiple zeta valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
How do I write out this proof? The following works with any number greater than the number 9.
How do I write the following examples out in a proof?
Take the two digits in your age for example & add then add those two numbers.
So let's say you're 28. 2+8=10 Now subtract that total from your age:
28-10=18 now add t... | the main point here is that the remainder when a number is divided by $9$ is the same as the remainder when the sum of its digits is divided by $9$. in particular this means that if $\sigma(N)$ is the sum of digits (decimal) of $N$ then $N-\sigma(N)$ is divisible by $9$.
the number may be written as:
$$
N = \sum_{k=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
proving an invloved combinatorial identity How to prove following Identity?
$$\sum_{k=0}^n (-1)^k {n-k \choose k} m^k (m+1)^{n-2k} = \frac {m^{n+1}-1}{m-1}, m \ge 2$$
This seems very hard to me. Any idea about how to prove it combinatorialy?
[P.S: for $k > \lfloor {\frac n 2}\rfloor$ L.H.S evaluates to $0$. ]
| This is not a combinatorial proof but I still find it rather nice. Let us replace $m$ by $x$ to get a more general polynomial identity.
We can notice that
\begin{equation*}
(-1)^{k} \binom{n-k}{k} x^{k}(x+1)^{n-2k}
\end{equation*}
is the coefficient of $y^{k}$ in $(1+x-xy)^{n-k}$. Consequently the left hand of the iden... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Proving limits with existing results $\lim_{x\to a} \frac{\sin^2 x - \sin^2 a}{x-a} = \sin 2a$ So in my lecture yesterday I learnt how to prove that $\lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e$, but I'm lost as to how to apply the results to prove limits.
Any help would be greatly appreciated.
Use the re... | i) Let $y=x-a$. Then, when $x\to a$, $y\to 0$ and we have
$$
\lim_{y\to 0}\frac{\sin^2(y+a)-\sin^2{a}}{y} = \lim_{y\to 0}\frac{\sin^2 y\cos^2a+2\sin a\cos a\sin y\cos y +\sin^2 a\cos^2 y-\sin^2{a}}{y} = \lim_{y\to 0}\frac{\sin^2 y\cos^2a+2\sin a\cos a\sin y\cos y -\sin^2 a\sin^2 y}{y}=\lim_{y\to 0}\frac{\sin y\cos a\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to find all irreducible polynomials in Z2 with degree 5? I am totally lost on how to do this one. I am supposed to accomplish the following:
Find all irreducible polynomials in $\mathbb{Z}_2[x]$ with degree $5$.
I may use the fact that x, $x+1$ and $x^2+x+1$ are the irreducibles of degree less than or equal to 2
I... | The following argument is specific to the question asked. Although I am
by no means a trained professional, please do not attempt this at home with
polynomials of larger degree.
The irreducible polynomial must be of the form
$x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + 1$, that is, the degree-$5$ term
as well as the consta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/998563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Solving 2nd order ODE with Frobenius method - problems with summation symbol I'm trying to solve the ODE:
$$ y''(x) + \frac{2x}{(x-1)(2x-1)} y'(x) - \frac{2}{(x-1)(2x-1)} y(x) = 0 $$
I'm trying to find a solution by the Frobenius method, expanding a power series of the solution around $x = \frac 12$, that is in a serie... | Consider the differential equation
\begin{align}
(x-1)(2x-1) y''(x) + 2x y'(x) - 2 y(x) = 0
\end{align}
with solutions expanded around $1/2$. For this it is seen that
\begin{align}
y(x) = \sum_{n=0}^{\infty} a_{n} \left(x - \frac{1}{2} \right)^{n+\sigma}
\end{align}
which leads to
\begin{align}
0 &= 2(x-1)\left(x-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$ I am trying to calculate the following series:
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$
and I managed to reduce it to this term
$$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$
And here I am stuck. I tried writing down a fe... | $$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4}=\\\frac{1}{2n}-\frac{2}{2n+2}+\frac{1}{2n+4}=\\(\frac{1}{2n}-\frac{1}{2n+2})+(-\frac{1}{2n+2}+\frac{1}{2n+4})=\\(\frac{1}{2n}-\frac{1}{2n+2})-(\frac{1}{2n+2}-\frac{1}{2n+4})=\\\frac{1}{2}((\frac{1}{n}-\frac{1}{n+1})-(\frac{1}{n+1}-\frac{1}{n+2}))=\\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
How to prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$? Prove the inequality for a triangle with sides $a,b,c$ we have $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$$
Trial: Since $a,b,c$ are sides of a triangle I know $a+b>c,b+c>a,a+c>b$
| \begin{align*}\frac a{b+c}+\frac b{c+a}+\frac c{a+b}<2&\iff 3-2<1-\frac a{b+c}+1-\frac b{c+a}+1-\frac c{a+b}\\&\iff 1<\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}\end{align*}
Now
$$\frac {b+c-a}{b+c}+\frac {c+a-b}{c+a}+\frac {a+b-c}{a+b}>\frac {b+c-a}{a+b+c}+\frac {c+a-b}{a+b+c}+\frac {a+b-c}{a+b+c}=1$$
You... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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A question on cosine integral So I've read a book and found myself stumped in this integral:
$$\int_{0}^{\pi} \frac{\cos(n\theta)}{b^2-a^2\cos(2\theta)}\, d\theta=\begin{cases}
\,\,0 &,\quad\mbox{if}\,\, n\,\,\mbox{is odd}\\[20pt]
\,\,\dfrac{\pi}{\sqrt{b^4-a^4}}\left(\dfrac{\sqrt{b^2-\sqrt{b^4-a^4}}}{a}\right)^n&,\qua... | If $n=2k+1$, then
$$\int_{\pi/2}^\pi\frac{\cos((2k+1)\theta)}{b^2-a^2\cos(2\theta)}d\theta=\int_0^{\pi/2}\frac{\cos((2k+1)(\pi-\theta))}{b^2-a^2\cos(2(\pi-\theta))}d\theta=\int_0^{\pi/2}\frac{-\cos((2k+1)\theta)}{b^2-a^2\cos(2\theta)}d\theta$$
and the integral is $0$ because the convergence is trivial. So assume $n=2k$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we... | Following mookid's hint, we can also avoid the use of Lagrange multiplicators. Normalize so that $a+b+c=1$, and then use the inequality $\sqrt{\dfrac{a}{1-a}}\geq 2a$. This is equivalent to $a(2a-1)^2\geq 0$.
Hence $f(a,b,c)\geq 2(a+b+c)=2$. Equality cannot hold, since $a=b=c=\dfrac{1}{2}$ doesn't satisfy $a+b+c=1$. Bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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A combination integral and series resulting the inverse tangent integral $\def\Ti{{\rm{Ti}}_2}$I have been able to solve an integral problem, now I tried to use the other method to crack the integral and I have to prove the following expression
\begin{equation}
I=\sum_{n=1}^\infty \frac{z^{2n-1}}{2n-1}\int_0^{\Large\f... | I stress that this is not a solution, but such a comment was too long to fit above.
I got a different formula for
\begin{equation}
\int_0^{\Large\frac{\pi}{2}}\sin(2n-1)x\cot x\,dx=\frac{(-1)^{n-1}}{2n-1}+2\sum_{k=1}^{n-1}\frac{(-1)^{k-1}}{2k-1}
\end{equation}
than what you posted. But after close inspection, they can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 1,
"answer_id": 0
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Finding the locus of a $z=x+iy$ and its radius if $|z-1|=2|z+2-3i|$
If the point $P$ in the complex plane corresponds to the complex number $z=x+iy$ show that if $|z-1|=2|z+2-3i|$ then the locus of $P$ is a circle centre at $-3+4i$, and find the radius of the circle.
Putting them into cartesian equations, we have:
$$... | With thanks to Anurag A:
$$
(x-1)^2+y^2=4[(x+2)^2+(y-3)^2]\\
0=3x^2+18x+3y^2-24y+51\\
-(x+3)^2-(y-4)^2=-8\\
$$
The centre is therefore $(-3,4)$ or $-3+4i$ and the radius is $\sqrt8$ or $2\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^... | Let $z_+ = -\gamma + \sqrt{\gamma^2 - 1}$ and $z_- = -\gamma - \sqrt{\gamma^2 - 1}$. Then
\begin{align}
z_+ &= \frac{r^2 + 1}{2r} + \sqrt{\frac{r^4 + 2r + 1 - 4r^2}{4r^2}}\\
&= \frac{r^2 + 1}{2r} + \frac{r^2 - 1}{2r}\\
&= \frac{2r^2}{2r}\\
&= r\\
z_- &= \frac{r^2 + 1}{2r} - \sqrt{\frac{r^4 + 2r + 1 - 4r^2}{4r^2}}\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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For all positive real numbers, is $f(x)=\sqrt{x}+x+2$ one to one? I understand that in order to prove this to be one to one, I need to prove $2$ numbers, $a$ and $b$, in the same set are equal.
This is what I did:
$$\sqrt{a} + a + 2 = \sqrt{b} + b + 2$$
$$\sqrt{a} + a = \sqrt{b} + b$$
$$a + a^2 = b + b^2$$
How would I... | $f(x)=\sqrt{x}+x+2$ is strictly increasing on $(0,\infty)$, so it's one-one (suppose not, use the strict monotonicity to draw a contradiction).
Or, from your second step,
$$\sqrt{a} + a = \sqrt{b} + b\iff \sqrt{a} - \sqrt{b}+ a -b=0 \iff(\sqrt{a} - \sqrt{b})(1+\sqrt{a}+\sqrt{b})=0 $$
Since $1+\sqrt{a}+\sqrt{b}>0$, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
poisson gamma mixture Let $N$ be Poisson distributed with parameter $10B$, where $B \sim \Gamma(3,1)$ (i.e. $f(b)=\frac{b^2 e^{-b}}{2}$). Find the p.m.f of $N$. How should I manipulate $10B$ in the integration? What is its pdf?
| $\newcommand{\E}{\operatorname{E}}$We are given this conditional distribution:
$$
\Pr(N=n\mid B) = \frac{(10B)^ne^{-10B}}{n!}.
$$
We need to find a marginal distribution:
\begin{align}
\Pr(N=n) & = \E(\Pr(N=n\mid B)) = \E\left( \frac{(10B)^ne^{-10B}}{n!} \right) \\[8pt]
& = \int_0^\infty \frac{(10b)^n e^{-10b}}{n!} \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$. I tried it this way:
$$ 40\cosθ+9\sinθ=41 $$
$$ 9\sinθ=41-40\cos\theta $$
Squaring both the sides:
$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$
$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$
$$168... | Using Brahmagupta-Fibonacci Identity,
$$(9\sin\theta+40\cos\theta)^2+(40\sin\theta-9\cos\theta)^2=(40^2+9^2)(\sin^2\theta+\cos^2\theta)=41^2$$ as $41^2=40^2+9^2$
So, $40\sin\theta-9\cos\theta=0$ and $9\sin\theta+40\cos\theta=41$
Can you solve for $\cos\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
factorise the expression $x^3 - 3x^2 -4x + 12$
Factorize the expression $$x^3-3x^2-4x+12$$ Hence calculate the ranges of values of $x$ for which $x^3-3x^2>4x-12$.
I factorised it to obtain $(x-2)(x-3)(x+2)$ but I don't how how to get to the next step.
| $$
{x}^{3}-3\,{x}^{2}-4\,x+12={x}^{3}-4\,x-3\,{x}^{2}+12=x(x^2-4)-(3x^2-12)=
$$
$$
=x(x^2-4)-3(x^2-4)=(x-3)(x^2-4)=(x-3)(x-2)(x+2).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Inequality problem involving QM-AM-GM-HM or Cauchy Schwarz inequality Question: Prove that if $x$, $y$, $z$ are positive real numbers then the following inequality holds: $$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$I tried thinking of applying QM-AM, but I didn... | We want to show that:
$$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$
Note that, by the QM-AM Inequality:
$${\left(\frac{x+y}{2}\right)}^2 \leq \frac{x^2 + y^2}{2}.$$
Consequently:
$$\frac{x + y}{x^2 + y^2} \leq \frac{2}{x + y}.$$
Therefore, it remains to show tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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directional derivatives in (0,0) I've this function : $f(x,y)= \dfrac{(1+x^2)x^2y^4}{x^4+2x^2y^4+y^8}$ for $(x,y)\ne (0,0)$ and $0$ for $(x,y)=(0,0)$
It's admits directional derivatives at the origin?
| A function admits directional derivative at a point if its gradient $\nabla{f}$ exists at that point. The gradient of your function is given by,
$$\nabla{f}=\left(\begin{array}{cc} -\frac{2\, x\, y^4\, \left(2\, x^6 + 3\, x^4 - 2\, y^4 + 1\right)}{{\left(x^6 + x^2 + 2\, y^4\right)}^2} & \frac{4\, x^2\, y^3\, \left(x^6 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$?
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience.
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )
| Raising the given number to the third power and using the identity $$(a-b)^3=a^3-3a^2b+3ab^2-b^3=a^3-b^3–3ab(a–b)$$ you obtain $$\left(\sqrt[3]{2}-\sqrt[3]{4}\right)^3=2-4-3\sqrt[3]8\left(\sqrt[3]{2}-\sqrt[3]{4}\right)=-2-6\left(\sqrt[3]{2}-\sqrt[3]{4}\right)$$ or equivalently $$\left(\color{blue}{\sqrt[3]{2}-\sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Cylindrical coordinates on elliptic paraboloids. I want to compute the volume bounded by:
*
*the cylinder $x^2+4y^2=4$.
*the $z=0$ plane.
*the elliptic paraboloid $z = x^2 + 6y^2$.
I would like to use cylindrical coordinates. However I have never dealt with a problem in which $r$ bounds depend on $\theta$.
Here ... | Consider the change of variables $x=2u\Rightarrow dx = 2\,du$. Now we are working on a circular cylinder: $u^2+y^2=1$. Now let's take cylindric coordinates. Note that the Jacobian is $2r$. And the volume:
\begin{align}
V=\int_0^1\int_0^{2\pi}\int_0^{4r^2\cos^2\theta+6r^2\sin^2\theta} 2r\,dz\,d\theta\,dr = \int_0^1\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find all integers n such that $\;\frac{n^2-9}{n^2-5n+4}$ is an integer.
Find all integers such that $\;\dfrac{n^2-9}{n^2-5n+4}\;$ is an integer.
I am really struggling to figure this out. I can tell that -3,3, and 5 are solutions but I don't know how to show that these are the only solutions or if they even are the o... | Longer, but using only divisibility considerations:
This factors as $\frac{(n-3)(n+3)}{(n-4)(n-1)}$. Note $n$ cannot be $1$ or $4$.
So $|n-1|\geq1$, and if $p$ is a prime dividing $n-1$, then $p$ has to divide $n-3$ or $n+3$. So mod $p$, either $n\equiv1\equiv3$ or $n\equiv1\equiv-3$. Either way, $p$ must be $2$, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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} |
How can I calculate $\sum_{j=0}^{\infty}(j+1)(\frac{1}{1.05})^{j+1}$ I have to calculate the sum:
$$\sum_{j=0}^{\infty}(j+1)\cdot\left(\frac{1}{1.05}\right)^{j+1}$$
I know it is convergent from the ratio test.
| Let
$$f(x)=\sum_{k=0}^{\infty}(k+1)(x)^{k+1}=\sum_{k=0}^{\infty}(k+2-1)(x)^{k+1}=\sum_{k=0}^{\infty}(k+2)x^{k+1}-\sum_{k=0}^{\infty}x^{k+1}$$
for $|x|\lt1$
$$g'(x)=\sum_{k=0}^{\infty}(k+2)x^{k+1}=2x+3x^2+4x^3+\cdots$$
$$g(x)=x^2+x^3+x^4+\cdots=\frac{x^2}{1-x}$$
$$g'(x)=\frac{x(2-x)}{(1-x)^2}$$
$$p(x)=\sum_{k=0}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability of two out of three events happening? All events are independent.
$$\Pr(A) = \frac{9}{10}$$
$$\Pr(B) = \frac{9}{10}$$
$$\Pr(C) = \frac{6}{10}$$
What is the probability of at least two events happening?
I'd like to use negation, to negate the possibility that event no event happen plus the probab... | Your approach works, and your answer is correct.
It can also be calculated straight forwards as:
$\begin{align}
\mathsf P(AB\cup AC\cup BC)
& = \mathsf P(AB)+\mathsf P(A^cBC)+\mathsf P(AB^cC)
\\[1ex]
& = \mathsf P(A)\mathsf P(B) + \bigg(\mathsf P(A^c)\mathsf P(B)+\mathsf P(A)\mathsf P(B^c)\bigg)\mathsf P(C)
\\[0ex]
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)} $ Let $x,y,z$ be real numbers in the range of $(0,1)$.
Prove that
$$\frac{1}{x(1-y)} +\frac{1}{y(1-z)} +\frac{1}{z(1-x)} \ge \frac{3}{xyz+(1-x)(1-y)(1-z)}.$$
| It's obviously true after following substitution and full expanding.
$x=\frac{a}{a+1}$, $y=\frac{b}{b+1}$ and $z=\frac{c}{c+1}$, where $a$, $b$ and $c$ are positives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n ... | Ratio test:
$$
\frac{\left(\dfrac{2^{n+1}+3}{3^{n+1}-1}\right)}{\left(\dfrac{2^n+3}{3^n-1}\right)} \to \frac 2 3 \text{ as }n\to\infty,
$$
so the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove or disprove: $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ Is it true that $\sqrt{2\sqrt{3\sqrt{4\sqrt{\ldots\sqrt{n}}}}}<3$ for any positive integer $n$?
We cannot prove the statement using induction as it is, because the left-hand side is increasing while the right-hand side stays constant. So we need to mo... | You can think of the left hand side as the geometric mean of the multi-set with $2^{n-k}$ values $k$ for $k=2,\dots, n$, plus an additional one, for a total of $2^{n-1}$ elements. The AM/GM rule then gives:
$$\frac{1}{2^{n-1}}\left(1+\sum_{k=1}^n k2^{n-k}\right) = \frac{1}{2^{n-1}} + \sum_{k=2}^n\frac{k}{2^{k-1}}$$
as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How do you find the imaginary roots of a fourth degree polynomial that cannot be simplified? I started out with $f(x)=16x^6-1$, and I got it down to $64x^4+16x^2+4$ by synthetically dividing by roots $0.5$ and $-0.5$ How should I continue in order to find the other roots?
| $16x^6=1$
$(\sqrt[3]{4}x)^6=1$
$u=\sqrt[3]{4}x$
$u^6=1$
These roots are spaced at u=$\pm 1$ and $\pm \frac{1}{2} \pm \frac{i\sqrt{3}}{2}$ ($n^{th}$ roots of unity)
Or $+1,-1, +\frac{1}{2} -\frac{i\sqrt{3}}{2}, +\frac{1}{2} -\frac{i\sqrt{3}}{2} , - \frac{1}{2} + \frac{i\sqrt{3}}{2}, - \frac{1}{2} - \frac{i\sqrt{3}}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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