Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that $ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, [\arcsin{\frac{x}{a}} / \arccos{\frac{x}{a}}]= \frac{\pi^2}{8} $? I found this very interesting result:
$$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}} = \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arccos{\frac{x}{a}} = \frac{\pi^2}{8}, $$
for $a>0.$ I wond... | We have that
$$\int_0^a \frac{dx}{\sqrt{a^2-x^2}} \,
\arcsin{\frac{x}{a}}=\int_0^1 \frac{dt}{\sqrt{1-t^2}} \, \arcsin{t}\\
=\int_0^1 \arcsin(t)\ d(\arcsin{t})=\frac{1}{2}\left[\arcsin^2{t}\right]_0^1=\frac{\pi^2}{8}.$$
For the other one you can integrate in a similar way or use the identity $\arcsin(t)+\arccos(t)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone els... | https://ernstchan.com/b/src/1457375466-129.pdf
The question is actually "what is the value of $z /\overline{z}$.
Which is easy. By theorem 2:
$|z /\overline{z}| = |z|/|\overline{z}| = \sqrt{x^2 + y^2}/\sqrt{x^2 + (-y)^2} = 1$
;I don't see if Lang ever stated this in the text but it should be obvious $|z| = |\overline ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Examining convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$ with mean value theorem $$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$$
I would like to examine covergence of this series using mean value theorem.
I would like to check my solution and optionally ... | It is OK. Alternatively, you may write, by the Taylor series expansion, as $n \to \infty$,
$$
\begin{align}
\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}&=\frac{1}{n}\sqrt{\left(1+O\left(\frac{1}{n}\right)\right)-\left(1+O\left(\frac{1}{n+1}\right)\right)}
\\\\&=\frac{1}{n}\sqrt{O\left(\frac{1}{n}\right)}
\\\\&=O\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might... | This sum can be done with some form of zeta function regularization. For $\Re s >1$, define:
$$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s) $$
Then, by analytic continuation, we can calculate:
$$\sum_{n=1}^\infty (-1)^{n-1} \sqrt{n} \to \eta\left(-\frac{1}{2}\right) = (1-2\sqrt{2})\zeta\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Smallest positive integer that gives remainder 5 when divided by 6, remainder 2 when divided by 11, and remainder 31 when divided by 35? What is the smallest positive integer that gives the remainder 5 when divided by 6, gives the remainder 2 when divided by 11 and 31 when divided by 35?
Also, are there any standard m... | The system of equations
$$ \left\{\begin{array}{ll} n \equiv 5 &\pmod{6} \\ n \equiv 2&\pmod{11} \\ n\equiv 31 &\pmod{35}\end{array}\right. $$
is equivalent to $n\equiv m\pmod{2\cdot 3\cdot 5\cdot 7\cdot 11}$ by the Chinese remainder theorem.
If $m_6, m_{11}, m_{35}$ are the smallest integer solutions of the systems
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 0
} |
Find large power of a non-diagonalisable matrix
If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$.
The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresp... | Since the given matrix is rather simple, you could also compute a few powers:
\begin{align*}
A^1&=
\begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}\\
A^2&=
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{pmatrix}\\
A^3&=
\begin{pmatrix}
1 & 0 & 0 \\
2 & 0 & 1 \\
1 & 1 & 0
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
Parabolic Representation The points $(-1/2, 0)$, $\left(0, \frac{\sqrt3}{2}\right)$, $(1/2, 0)$ are the vertices of an equilateral triangle. A parabolic equation that contains the three points is
$$y=
(-2 \sqrt3)x^2+\frac{\sqrt{3}}{2} \,.$$ How can the other two parabolas, that are congruent to the first parabola and ... | Since you are given the three points, there is a straight vectorial approach that you can follow.
Premised that
the parabola $y=x^2$, the parabola with vertex in $O=(0,0)$ and passing through
$U_{-1}=(-1,1)$ and $U_{1}=(1,1)$, can be written as:
$$
\mathop {OP}\limits^ \to \cdot \mathbf{u}_y = \left( {\mathop {O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$? How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$?
| Referring to the standard result, the number of factors of $p$ in $\frac{(3n)!}{n!n!n!}$ is given by
$$
\sum_{i \ge 1} \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \sum_{i \ge 1} \left\lfloor \frac{n}{p^i} \right\rfloor
= \sum_{i \ge 1} \left( \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \left\lfloor \frac{n}{p^i} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| If you have trouble approaching this, try some examples: Note that (for instance)
$$
1 \times \color{red}{2 \times 3} \times 4 = 24 = 5^2-1
$$
$$
2 \times \color{red}{3 \times 4} \times 5 = 120 = 11^2-1
$$
$$
3 \times \color{red}{4 \times 5} \times 6 = 360 = 19^2-1
$$
and observe that
$$
5 = \color{red}{2 \times 3} - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 6
} |
Multi-index sum property Exercise 1.2.3.29 in Donald Knuth's The Art of Computer Programming (3e) states the following property of a multi-indexed sum:
$$
\sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k = \frac{1}{3}S_3 + \frac{1}{2}S_1S_2 + \frac{1}{6}S_1^3,
$$
where $S_r = \sum_{i=0}^n a_i^r$.
I tried to prove it an... |
Here is a rather elementary approach based upon two aspects:
*
*Iterative calculation
We look at first at the simpler double sum
\begin{align*}
A_2&:=\sum_{i=0}^n\sum_{j=0}^ia_ia_j=\sum_{0\leq j\leq i\leq n}a_ia_j\\
&=\sum_{0\leq i\leq j\leq n}a_ia_j
\end{align*}
and obtain an expression in $S_1=\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to e... | As an addendum to the previous answer, it can be shown (through the FL-expansion of $\frac{\arcsin\sqrt{x}}{\sqrt{x}}$) that
$$\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)=4\text{ Im } \text{Li}_3\left(\frac{1+i}{2}\right)-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2).$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
Craft Inequality Originally the inequality looks like :
$$
\frac{1}{4}(\sqrt{\sum_{cyc}\sin(u)\sin(v)} - \sum_{cyc}\sin(u)\sin(v))
\geq\sum_{cyc}\cos(u)-(\sum_{cyc}\sin(\frac{3u}{2}))
$$
with $$u+v+w=\pi$$
after many transformations the inequality looks like :
$$
\sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2C+\sum_{\math... | I begin with a transformation :
$$x^2=1-C$$
$$y^2=1-B$$
$$z^2=1-A$$
We obtain :
$$ \sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2(1-x^2)+\sum_{\mathrm{cyc}}2(1-x^2)(1-y^2)\right)}-\frac{4}{2}\sum_{\mathrm{cyc}}((1-x^2))-\frac{1}{2}\sum_{\mathrm{cyc}}2((1-x^2)(1-y^2))\geq-\frac{1}{2}\sum_{\mathrm{cyc}}\sqrt{\frac{x^2}{2}}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equiv... | As @user236182 pointed out, we can write $a=p+q$, $b=q+r$, and $c=r+p$, with $p,q,r>0$. Letting $s = p+q+r$, we have
$$\frac{a}{b+c} = \frac{s-r}{s+r} = 1 - \frac{2r}{s+r} $$
and hence
$$\sum\limits_{\text{cyc}}{\frac{a}{b+c}}< 2 \iff \sum\limits_{\text{cyc}}{\left(1-\frac{2r}{s+r}\right)}< 2 \iff \sum\limits_{\text{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:
Let $a,b,c$ be positive numbers with ... | $$\frac{a + \frac 1 3 + \frac 1 3 + \frac 1 3}{4} \ge \sqrt[4]{a\cdot \frac 1 3 \cdot \frac 1 3 \cdot \frac 1 3}
$$
and similarly for b and c. So $(a+1)(b+1)(c+1) \ge 64\sqrt[4]{\frac{abc}{27^3}}$. You must show this is $\ge 64 abc$, which follows from your inequality (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full ... | Define
\begin{align*}
x&=\frac{1}{1+a}\\
y&=\frac{1}{1+b}\\
z&=\frac{1}{1+c}
\end{align*}
Then your problem transforms into: given $x+y+z=1$ prove $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1) \geq 8$. I believe you know the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a, b, c, d> 0$ then I try to prove the following inequality.. If $a, b, c, d> 0$ then prove that $$a^2 b^2 c^2 + b^2 c^2 d^2 + a^2 b^2 d^2 + a^2 c^2 d^2 \geq ab^2 c^2 d + ab^2 cd^2 + a^2 bcd^2 + a^2bc^2d$$
I'm using AM GM relation but I'm not getting the answer, probably due to a conceptual mistake?
| We may write $a^2b^2c^2 + b^2c^2d^2 + a^2b^2d^2 + a^2c^2d^2$ as
$$\frac{a^2b^2c^2 + b^2c^2d^2}{2} + \frac{b^2c^2d^2 + a^2b^2d^2}{2} + \frac{a^2b^2d^2 + a^2c^2d^2}{2} + \frac{a^2c^2d^2 + a^2b^2c^2}{2}$$
Apply the AM-GM inequality to each term to obtain the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a sq... | Let $a = x/y$, then this is equivalent to proving $a + 1/a \geq 2$. $a$ is positive so multiply both sides by $a$ we get
$$a^2 - 2a + 1 = (a - 1)^2 \geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Infinitude of solutions to Bézout's equation If $c=\gcd(a,b)$ for $a,b\in\Bbb Z$ it is possible to write $ax + by = c$ with $x,y\in\Bbb Z$. Is it possible to prove that there are infinitely many integer solutions for $x$ and $y$ without using Diophantine equations, and if so how?
(The particular problem I have is $1947... | Writing $ax + by = c$, you get infinitely many solutions as follows: Let $k\in \mathbb Z$ be arbitrary, then
\begin{align*}
c &= ax + by = ax + \frac{ab}{c}\cdot k + by - \frac{ab}{c}\cdot k\\
&= a\cdot \left(x+\frac{b}{c}\cdot k\right) + b\cdot \left(y-\frac{a}{c}\cdot k\right) \tag{1}
\end{align*}
is another solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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limit of product of $(a_1a_2.\dots a_n)^{\frac{1}{n}}$ How to calculate the following limit
$$ \lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right) \right]^\frac{1}{n} .$$
I was trying this by taking the $\log $ of the product and then limit but I am ... | This answer uses Sterling's approximation to get the limit:
\begin{align}
l=\;&\left[ \left( 1+\frac { 1 }{ n } \right) \left( 1+\frac { 2 }{ n } \right) \cdots \left( 1+\frac { n }{ n } \right) \right]^{ \frac { 1 }{ n } } \\
=\; & \left[ \frac{1}{n^n}(n+1)(n+2)\dots(n+n) \right]^{ \frac { 1 }{ n } } \\
= \;& \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 7
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Why isn't there an $x$ such that $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$? If $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$, then $x$ is equal to
$$\frac{-\pi}{2}\le \sin^{-1}x \le \frac{\pi}{2}$$
$$\frac{-\pi}{10}\le \frac{\sin^{-1}x}{5} \le \frac{\pi}{10}$$
This show that no value of such $x$ exist.
I want to know wh... | $\sin^{-1} x = \frac{5 \pi}{4}$ gives $x=\sin \frac{5 \pi}{4} = \sin (\pi + \frac{\pi}{4})$, then $x = -\sin \frac{\pi}{4} = -\frac{1}{\sqrt 2}$. Your error is restricting $\sin^{-1} x$ to the values $-\frac{\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2}$ (the principal value is within this range).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A faster way than tedious multiplying
Show that $(a+b+c)(a+b\epsilon +c\epsilon^2)(a+b\epsilon^2 + c\epsilon) = a^3 + b^3 + c^3 - 3abc$ If $$\epsilon^2 + \epsilon + 1 =0$$
The solution in the back of the book is given as
Proved by a direct check, taking into consideration that $\epsilon^2 = -\epsilon -1 \: \:$ and ... | One way would be by factorizing $a^3+b^3+c^3-3abc$.
$$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
where $\omega$ is the cube root of unity.
More on the factorization can be found here.
Then, clearly, $\epsilon=\omega$, and thus, $$\epsilon^3=1$$
Factorizing this gives,
$$(\epsilon-1)(\epsilon^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, for all integers $n\ge 1$. Let $x_1,\ldots, x_n$ be positive integers. Use mathematical induction to prove that
$$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$$ for all integers $n \ge 1$.
(Give... | Suppose the given inequality holds for an integer $n(\geq 1)$.
Then for $n+1$,
$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)=\left(\sum_{i=1}^{n} x_i+x_{n+1}\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\right)$$
Expanding this gives us
$$=\left(\sum_{i=1}^{n} x_i\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right... | The left hand side is the sum of the odd numbers from $2n+1$ up to $4n-1$. Thus when going $n\to n+1$, the first summand $2n+1$ is dropped, and two summands $4n+1$ and $4n+3$ are appended a the end.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help m... | First note that $\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta$. Observing that $$\cos(3A+3\cdot\tfrac23\pi)=\cos(3A+3\cdot\tfrac43\pi)=\cos3A,$$and adding the results of $\cos^3\theta$ for $\theta=A$, $\theta=A+\frac23\pi$, and $\theta=A+\frac43\pi$, using the fact that $\cos A+\cos(A+\frac43\pi)=2\cos(A+\frac23\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 6
} |
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet... | \begin{align*}
\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} &= \tan^2\theta + \cot^2\theta \\
&= \sec^2\theta - 1+ \csc^2\theta - 1\\
&= \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} - 2\\
&= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} -2\\
&= \sec^2\theta \csc^2\theta - 2
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Let $p,q$ be irrational numbers, such that, $p^2$ and $q^2$ are relatively prime. Show that $\sqrt{pq}$ is also irrational. Progress:
Since, $p,q$ are irrationals and $p^2$ and $q^2$ are relatively prime, thus, $p^2\cdot{q^2}$ cannot be a proper square, so, $pq$ is also irrational. Suppose, $pq=k$, then: $$\sqrt{k}\cd... | Suppose $\sqrt{pq}$ is rational, say $\sqrt{pq} = \dfrac mn$ where $m$ and $m$ are positive, relatively prime integers. Then $p^2 q^2 = \dfrac{m^4}{n^4}$ where $p^2q^2$ is an integer. Since $m$ and $n$ are relatively prime integers, then $m^4$ and $n^4$ are also relatively prime integers. It follows that $n^4=1$ and he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show by induction that $n! > n \cdot F_n$ for all $n > 3$ Show by induction that $n! > n \cdot F_n$ for all $n> 3$, where $F_n$ is the $n$th fibonacci number.
For the basis step, I put $n=4$ and got:
\begin{align*}
&4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\
&4 \cdot F_4 = 4 \cdot 3 = 12
\end{align*}
So the statement hold... | How about the following way?
It is sufficient to prove that $(n-1)!\gt F_n$ for $n\gt 3$.
Inductive step : we assume that $(k-1)!\gt F_k$.
Then, multiplying the both sides by $k$ gives
$$k!\gt kF_k\gt F_k+F_k\gt F_k+F_{k-1}=F_{k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Partition of a prime $p$ into primes raised to a power to be dvisible by $p$ Find all the partitions of prime $p$ in primes less than $p$. Raise each term in each partition by some power $k > 1$ to see if the sum of these terms will be divisible by $p$. Of course, $k$ can differ for each partition. For example, for $5$... | As McFry has already answered, the first question is answered by Fermat's little theorem: Since $a^p+b^p+\cdots+z^p\equiv a+b+\cdots+z$, there will always be a least power satisfying the OP's condition, and that least power will be no greater than $p$.
As for the second question, $p=7=2+2+3$ is an example where the le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Simplifying inverse trigonometric expressions such as $\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$ Okay so I'm just looking for a short cut method or a method that is not so long for simplifying expressions like this
$$\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{... | Revised for the last part
In your specific case
$$
\cos ^{ - 1} \left( {\frac{{x^2 - 1}}
{{x^2 + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2 - 1}}} \right)
$$
you may note that:
$$
\left\{ \begin{gathered}
x^2 - 1 = \operatorname{Re} \left( {\left( {x + i} \right)^2 } \right) \hfill \\
2x = \operator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Zeros of $a(1-x)x^b-(1-a)x(1-x)^b = 0$ What are the zeros of this equation?
$$a(1-x)x^b-(1-a)x(1-x)^b = 0$$
I was told that there were 3 zeros, however I've only been able to find 2 of them. Namely, $x=0$ and $x=1$. I can't think of any way to solve it algebraically and I'm not seeing any other obvious solutions. I a... | The equation has at either $3$ or $4$ roots, assuming that $0 < a < 1$.
In particular, the third root they were probably looking for is
$$
x = \frac{1}{1 + \sqrt[(b-1)]{r}},
$$
where $r = \frac{a}{1-a}$.
This root is between $0$ and $1$.
If $b$ is odd and $a \ne \frac12$, there is a second root,
$$
x = \frac{1}{1 - \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that $\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$ is a subgroup of $S_4$ Show that {$1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)$} is a subgroup of $S_4$
My attempt:
Let $H = \{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, where $1$ is the identity permutation. Then,
$1\circ(1, 2)(3,4) = (1, 2)(3,4) $
$1\circ(1, 3... | Let $S=\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, and relabel the elements $\sigma_1=1$, $\sigma_2=(1, 2)(3,4)$, $\sigma_3=(1, 3)(2,4)$, $\sigma_4=(1,4)(2, 3)$, where we note $\sigma_1=1=(1)(2)(3)(4)$ is the identity permutation.
Consider the product $(1,3)(2,4){\circ}(1,3)(2,4)$ in $S_4$. Remember when computing p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to find all solutions of $x^2-y^2=x+y$? For this equation $$x^2-y^2=x+y$$ I know there are 2 functional solutions, $y=x-1$, and $y=-x$.
The second solution we can see immediately because both sides will always be $0$.
The first solution is also easy to get:$$(x-y)(x+y)=x+y\ \ ///\div(x+y)\ \ \ [y\neq-x]\\x-y=1\\y=x... | $x^2 - y^2 = (x-y)(x+y) = x+y$
If $x+y=0$ then all $\{(x, -x)|x \in \mathbb R\}$ are solutions.
If $x+y\ne 0$ then $(x-y) = 1$ and $y = x -1$ so all $\{(x, x-1)|x \in \mathbb R\}$ are solutions.
So solutions are $\{(x, -x)|x \in \mathbb R\}\cup\{(x, x-1)|x \in \mathbb R\}$.
....
or $x^2 - y^2 = x+y$
$x^2 - y^2 -(x+y) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Arc Circle Intersection in Euclidean 3D While programming on a GPS based game I came upon the following problem for 3D euclidean space:
Assume we have an arc (given by a sphere center coordinate C and two euclidean coordinates $A$ and $B$ on that sphere, with $AC = AB$ and $A !=B$) and another sphere (given by an arbit... | I thought about it, and the second answer I proposed earlier is not all that bad. I'm going to treat the points $A$, $B$, $C$, $Q$, as coordinate triples, so that I can add and subtract them as vectors. I'm also going to write $r_2$ for $R$, the radius of the second sphere, and $r_1$ for the radius of the first. Here g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
| The way you would derive this identity is as follows:
Note that $\sin A + \sin B = 2\sin\big(\frac{A+B}{2}\big)\cos\big(\frac{A-B}{2}\big)$
Putting $A=50$ and $B=10$,
$$
\sin 50 + \sin 10 = 2\sin\big(30\big)\cos\big(20\big) = \cos 20
$$
Now, we know that $\cos 20 = 1- 2\sin^2 10$. So we substitute:
$$
\sin 50 + \sin 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$. Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$.
How can I use the first two equations to help solve the third? I'm stuck. Any solutions are greatly appreciated!
| When $a = b = c = \frac{1}{\sqrt3}$,
we have $ab + bc + ca = 1$ and $a + b + c + abc = \frac{10\sqrt3}{9}$.
Indeed, the minimum of $a + b + c + abc$ is $\frac{10\sqrt3}{9}$.
It suffices to prove that, for all $a, b, c\in (0, 1)$ with $ab + bc + ca = 1$,
$$a + b + c + abc\ge \frac{10\sqrt3}{9}.$$
Let $p = a + b + c, q =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Find limit of sequence sum Let us have sum of sequence (I'm not sure how this properly called in English): $$X(n) = \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+...+\frac{2n-1}{2^n}$$
We need
$$\lim_{n \to\infty }X(n)$$
I have a solution, but was unable to find right answer or solution on the internet.
My idea:
This can be re... | $$\frac12+\frac34+\ldots+\frac{2n-1}{2^n}=\left(1-\frac12\right)+\left(1-\frac14\right)+\left(\frac34-\frac18\right)+\ldots\left(\frac n{2^{n-1}}-\frac1{2^n}\right)=$$
$$1+1+\frac34+\ldots+\frac n{2^{n-1}}-\frac12-\frac14-\ldots-\frac1{2^n}\xrightarrow[n\to\infty]{}$$
$$\to\sum_{k=1}^\infty\frac k{2^{k-1}}-\sum_{k=1}^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Weighted sum of two dice such that the result is a random integer between $0$ and $35$
The numbers shown by 2 dice are labelled $d$ and $e$. $A, B$ and $C$ are constants, giving a score $S=Ad + Be + C$. Find $A, B$ and $C$ such that the range of possible values for $S$ covers all integers from $0$ to $35$, with an equ... | The maximum value of $S$ (assuming $A,B\geqslant0$) is $6(A+B)+C$, and the minimum value $A+B+C$. This gives us the system of equations
\begin{align}
6A+6B+C &= 35\\
A+B+C &= 0.
\end{align}
Gaussian elimination yields $C=-7$, and therefore $A+B=7$. So $(A,B,C)=(6,1,-7)$ and $(1,6,-7)$. Taking $(A,B,C)=(6,1,-7)$ yields
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:
$$\sum_{n=0}^k \cos... | First note that
$$
\left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} = \frac{1}{2^{2k}} \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right)
$$
Then,
\begin{align}
\sum_{n=0}^k \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} &= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ Prove that the solutions $y_1$ and $y_2$ for the initial value problem
$y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$
are:
$y_1=1-x$
$y_2={{-x^2}\over 4}$
And explain why the two solutions not contradiction with the theory of existence and uniqueness.
( sorry I don't spe... | Let $u=\sqrt{x^2+4y}$ ,
Then $y=\dfrac{u^2-x^2}{4}$
$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}$
$\therefore\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$
$\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$
$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{u}{2}$ with $u(2)=0$
$\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Greatest open ball in the unit $n$-cell
Let $n\geq2$ be an integer, let $C=[-1,1]^n$ and let $A$ be the set of all real numbers $r$ such that $r$ is the radius of some open ball $V$ such that $V$ is contained in $C$ and $V$ is disjoint to the open ball whose center is $\bf 0$ and radius is $1.$ Find $\sup A.$
I think... | Consider a point $p ∈ [-1, 1]^n \setminus B(\mathbf{0}, 1)$. Observe that $B(p, r) ⊆ [-1, 1]^n \setminus B(\mathbf{0}, 1)$ if and only if $r$ is not greater than the distance from $p$ to any of the faces of the cube and also not greater than the distance from $p$ to $B(\mathbf{0}, 1)$. But we can calculate those!
For $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to find the determinant of a 5x5 matrix How do I find the determinant of this?
$$\begin{bmatrix}
0& 6& −2& −1& 5\\
0& 0& 0& −9& −7\\
0& 15& 35& 0& 0\\
0 &−1 &−11& −2& 1\\
−2 &−2& 3& 0& −2\end{bmatrix}$$
I tried doing row reductions but every time I get a $0$ and I gain a number. I'm not really sure how to do this b... | By using a Laplace expansion along the first column the problem immediately boils down to computing $R=-2\cdot\det(M)$ with
$$ \det M=\det\begin{pmatrix}6&-2&-1& 5 \\ 0 & 0 & -9 & -7 \\ 15 & 35 & 0 & 0 \\ -1&-11&-2&1\end{pmatrix}=-5\cdot\det\begin{pmatrix}6&-2&1& 5 \\ 0 & 0 & 9 & -7 \\ 3 & 7 & 0 & 0 \\ -1&-11&2&1\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $\Pr(Y\le 2X)$ for two iid random variables of a given pdf Consider two independent random variables $X$ and $Y$.
Let both PDFs are defined by $$f_X(x)=\begin{cases}\displaystyle\frac{4-2x}{3}&\text{if }x\in(0,1),\\0&\text{otherwise.}\end{cases}$$
Find $\Pr(Y \le 2X)$.
I solved this the following process.
*
*I... | Yes, indeed, that's the right integral. But not the right value.
$$\begin{align}\mathsf P(Y\leq 2X) ~=~& \int_0^{1/2}\frac{4-2x}3\int_{0}^{2x}\frac{4-2y}3\,\mathsf d y\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\int_0^1 \frac{4-2y}3\,\mathsf d y\,\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{4-2x}3\frac{4(2x-x^2)}3\,\mathsf d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $x^3 \equiv 0,1,8 \mod{9}$ (Critique my solution) My solution is based on the idea that it is sufficient to consider the cubes less than 9. Ie $0^3 \equiv 0, 1^3\equiv 1, 2^3 \equiv 8$. Since these are the only cubes in the residue class$\mod{9}$ these are the only possibilities for $x^3 \mod{9}$
Is this a su... | According to that reasoning, since $1^3\equiv1,2^3\equiv8,3^3\equiv0\pmod{27}$, we have that $x^3\equiv0,1,8\pmod{27}$ for any $x$. But $4^3\equiv10\pmod{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove statement about sum We have such sum: $$\sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^p}, 0<p<1$$
How to prove that sum is between 1/2 and 1?
| Let $ \quad f(x) = 1/x^{p} \quad \colon \{ x \ge 1 \,, \space 0 \lt p \lt 1 \} \space\Rightarrow\space 1/x \lt 1/x^{p} \lt 1 $
$ \Rightarrow \quad f^{\prime}(x) = -p/x^{p+1} \lt 0 \space\Rightarrow\space f^{\prime\prime}(x) = p(p+1)/x^{p+2} \gt 0 \space\Rightarrow\space f(x) \space $ is convex
$$
\begin{align}
& \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$, then what is $\int_{-1}^{1} p(x)dx$? Q.If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$,Then $\int_{-1}^{1} p(x)dx$=__?
My attempt:
Let $p(x)$ be $ax^3+bx^2+cx+d$
$p(0)=d=2$
$p(1)=a+b+c+d=3$
$p(-1)=-a+b-c+d=4$
From them,we get $b=3.5$,$d=2$... | It could be helpful to do the integration step as well:
\begin{align}
\int_{-1}^1p(x)dx&=\int_{-1}^1\left(ax^3+bx^2+cx+d\right)\,\text{d}x\\
&=\color{blue}{\frac{a}{4}x^4\big|_{x=-1}^{x=1}}+\color{red}{\frac b3x^3 \big|_{x=-1}^{x=1}}+\color{green}{\frac c2 x^2\big|_{x=-1}^{x=1}}+\color{black}{dx\big|_{x=-1}^{x=1}}\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$
I know the condition for divisibility by $11$ but I couldn't guess how to apply it here.
Please help me in this regard... | A number is divisible by 11 if and only if the sum of the odd-position digits and the even-position digits differ by a multiple of 11. Now, the sum of the digits from 1 to 9 is odd, so the difference in any such number must be either 11 or 33. The only subsets of $\{1,2,3,4,5,6,7,8,9\}$ where the difference between the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 4
} |
How to show $\tanh^{-1}(3i)=i\tan^{-1}(3)$? I want to find all the complex values associated with $\tanh^{-1}(3i)$. On WolframAlpha, it computes $\tanh^{-1}(3i)=i\tan^{-1}(3)$. I know that $\tanh^{-1}(z)=\frac{1}{2} \log{\frac{1+z}{1-z}}$ and so
$$\begin{align}
\tanh^{-1}(3i)&=\frac{1}{2} \log{\frac{1+3i}{1-3i}}\\
&=\f... | put $\arg(-4+3i)=2\theta$
$\displaystyle\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=-\frac{3}{4} $
$\displaystyle\tan\theta=3$
$\displaystyle\theta=i\tan^{-1}3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Equation which has to be solved with logarithms I need some help how to solve these equations for $x$. I think I have to use logarithms but still not sure how to do it and would be really grateful if someone could explain me.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
= x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^... | 1) Making $2^{x-1}=a$ and $2^{|x-3|+2}=b$ you have $$4ax^2+b=4bx^2+a\iff(4x^2-1)(a-b)=0$$ This gives $x=\frac 12$ and $x\ge3$
2) You have two independent possibilities $$x^2-7x+5=1\\x^2-2x-15=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Supremum and infimum of asinx + bcosx I am trying to draw conclusions about the supremum and infimum of $F=\{a\sin x+b\cos x: x \in \mathbb{R}\}$ where $a,b \in \mathbb{R}$ are parameters. I am able to rewrite the function as
$$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}sinx+\frac{b}{\sqrt{a^2+b^2}}cosx\right).$$
Noti... | Checking with calculus, we have the extreme point of
\begin{align}
f(x) = a \cos x + b\sin x \ \ \Rightarrow \ \ f'(x) = -a\sin x + b\cos x.
\end{align}
Solving for the critical points
\begin{align}
b\cos x - a\sin x = 0
\end{align}
leads to the cases
Case 1: $a\neq 0$
Then it follows
\begin{align}
\tan x = \frac{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
combinatorics: sum of product of integer compositions I am trying to solve a problem from Stanley's book, it says:
Fix $k,n \in \mathbb{P}$. Show that:
\begin{align}
\sum a_1 a_2 \cdots a_k = \binom{n+k-1}{2k-1}
\end{align}
where the sum ranges over all compositions $(a_1 , a_2 , \ldots , a_k)$ of $n$ into $k$ parts.
I... | This is a supplement to @MarkoRiedel's answer which should clarify OPs question. At first we look at the compositions of $n=4$ which consists of two terms ($k=2$) and look at the corresponding terms of the generating function.
\begin{array}{crl}
1+3\qquad&\qquad x^1\cdot3x^3=&3x^4\\
2+2\qquad&\qquad2 x^2\cdot 2 x^2=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Difficult limit problem involving sine and tangent I encountered the following problem:
$$\lim_{x\to 0} \left(\frac 1{\sin^2 x} + \frac 1{\tan^2x} -\frac 2{x^2} \right)$$
I have tried to separate it into two limits (one with sine and the other with tangent) and applied L'Hôpital's rule, but even third derivative doesn'... | Since $\sin x$ and $\tan x$ are “almost equal” to $x$, separating into a sum of two limits seems like an interesting approach. The first limit is
$$
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=
\lim_{x\to0}\frac{x^2-\sin^2x}{x^2\sin^2x}
$$
Finding a suitable Taylor expansion of the numerator is easy:
$$
x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solution to System of Linear Differential Equations with Variable Coefficients I'm stuck trying to solve the system $\frac{dx(t)}{dt} = A(t)x(t)$ with $A(t) =\frac{1}{2}\begin{pmatrix}
ln(t+1) & ln(t-1)\\
ln(t-1) & ln(t+1)
\end{pmatrix}$.
I think I could try computing $e^{\int A(z) dz}$ and that should give me a fundam... | We shall consider our initial time at $t_0 = 1$ for this post.
Since
\begin{align}
A(t) = \frac{1}{2}
\begin{pmatrix}
\ln (t+1) & \ln (t-1)\\
\ln (t-1) & \ln (t+1)
\end{pmatrix}
\end{align}
then it follows
\begin{align}
\int^t_{1} A(s)\ ds = \frac{1}{2}
\begin{pmatrix}
(t+1)\ln (t+1) +(t+1) & (t-1)\ln(t-1) + (t-1)\\
(t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The value of following integral $$\frac {x^2\cos (x)}{1+e^x} dx$$ from $[-\pi,\pi] $ ? Now I converted to $$R\frac{e^{ln (x^2)+ix}}{1+e^x}dx $$ wherw R is real part now it isnt an even or odd so those manipulations are useless. $1+e^x=u $ thus $e^xdx=du $ but from here I can go nowhere.
| Following @Sangchul Lee's comment (although the idea is pretty straight forward to someone who is aware of basic calculus technics) we have:
\begin{align*}
\int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^x+1} \, {\rm d}x &\overset{u=-x}{=\! =\! =\! =\!} \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^{-x} +1} \, {\rm d}x \\
&= \int_{-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the coefficient of $x^{80}$ in the power series expansion $\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}$ Find the coefficient of $x^{80}$ in the power series expansion $$\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}.$$
I don't know how to find coefficients in power series, solutions are greatly appreciated!
| $$\frac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}=x^2(1+x^2+x^5)\frac{1}{4}\left(\frac{2}{1-x^2}+\frac{1}{(1-x)^2}\right)$$
then use the following
$${\frac {1}{1-x^2}}=\sum _{n=0}^{\infty }x^{2n}$$
$${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$. Maybe a very stupid question but I am stuck. Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$.
Is it enough to simply multiply, i.e. $z\bar{z} = \frac{1\times z}{z} = 1$? Showhow I feel this is not correct. I know that if $z = \pm 1$ or $z \pm i$ the... | $z = a + bi$
for $\frac{1}{z} = \bar{z}$
$\frac{1} {a + bi} = a - bi$
$\frac{1} {(a + bi)} \frac{a - bi}{a - bi} = a - bi$
$\frac{a - bi}{a^2 + b^2} = a - bi$
cancel and rearrange
$(a^2 + b^2) = 1$
recalling the formula for |z|
$|z| = \sqrt{a^2 + b^2} = \sqrt 1 = 1$
the other way
if |z| = 1 then $a^2 + b^2 = 1$
$z = a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
What is the curved asymptote of $\frac{x+1}{\sqrt{3x-2}}$? Please excuse any non-technical language I use...
I'm confused by the process for finding the curved asymptote (as $x$ gets really large) of the graph $y=\frac{x+1}{\sqrt{3x-2}}$.
I first looked at the (naive?) approach of separating the fraction up into two p... | Using equivalents, when $x$ is large $$x+1\sim x \qquad , \qquad \sqrt{3x-2}\sim \sqrt{3x}\qquad \implies y \sim \frac{x}{\sqrt{3x}}=\frac{\sqrt x}{\sqrt{3}}$$
Letting $x=\frac 1t$, you also could write $$y=\frac{1+\frac 1t}{\sqrt{\frac 3t -2}}=\frac 1{\sqrt t}\frac{1+t}{\sqrt{ 3 -2t}}$$ Now, use Taylor expansion ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inverse of function $f : \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ Matthew Szudzik's mapping function is another approach on mapping $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$, however, I have trouble finding the inverse of it. The function is
$$f(a, b)=\left\{\begin{array}{ll}a^2+a+b&\mbox{, if }a\geq b\\a+b^2&\mbox{... | $f(a,b) = a^2 + a + b=c; a\ge b$
$f(a,b) = a+b^2=c; a< b$.
So make wild guesses. If $a' = -b'$ we have $f(a',b) = a'^2 = c$ so $g(c) = (\sqrt{c}, -\sqrt{c})$ will work if $c$ is a perfect square. But obviously $c$ needn't be a perfect square. But we can be close.
Suppose $n^2 \le c < (n+1)^2$ or in other words $n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$ Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$
My Try:
if $y=2$ then $$x^2=2^{x-2}$$ and Taking log on both sides we get
$$2 \log x=(x-2) \log 2$$ i.e.,
$$2 \log_2 x=x-2$$ if $x=2^k$ then we get
$$2k=2^k-2$$ or
$... | For
$x^y=y^{x-y}
$
I find that
$x=8, y=2$
and
$x=9, y=3$
are the only solutions.
Must have $x > y$.
$x^y
=y^{x-y}
$
so
$(xy)^y = y^x
$.
Let
$x = ry$ where
$r > 1$.
$(ry^2)^y = y^{ry}$
so
$ry^2 = y^r$
or
$r = y^{r-2}$
or
$y = r^{1/(r-2)}$.
If $x = 2y$
then
$(2y)^y = y^y$
which has no solution.
If
$1 < r < 2$,
then
$r-2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How To Write A Formula For The First N Terms of a Sequence? So I have the sequence $$\{5, 15, 45,\cdots\}$$ and I figured out that the formula to find a particular term is $$S_n = 5 \times 3^{n-1}$$ but how do I use this to find the sum of the first $n$ terms?
|
Geometric Summation: Given the sequence$$a,ar,ar^2,ar^3,\ldots,ar^{n-1}\tag1$$
We have the sum as$$S=a\left(\frac {1-r^n}{1-r}\right)\tag2$$
Where $S$ is the total sum and $a$ is the first term of the sequence.
Proof: Multiplying $S$ by $r$, we obtain another equation$$Sr=ar+ar^2+ar^3+\ldots+ar^n\tag3$$
And subtr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value We have
$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$, then find the value of $\sum_{k=1}^{3} k f(k)$.
Could someone give me slight hint as how to find $f(x)$ here.
| $$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$
$$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{n^2}{n^2+x^2r^2}$$
$$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+x^2\cdot \frac{r^2}{n^2}}$$
$$= \lim_ {h \to 0} \text{h} \sum_{r=1}^{n} \frac{1}{1+x^2(rh)^2}$$
Then using Riemann ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$ Prove that, for all non-negative real numbers $x, y, z$ that satisfy $x + y + z = 1$,
$$x^2 y + y^2 z + z^2 x \leq \frac {4}{27}
$$
I'm having trouble with this question. I suspect it may have a fairly simple proof using t... | Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=b(a^2+ac+c^2)\leq b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An Identity for Pell-numbers The Pell-numbers are defined recursively by:
$P_0 = 0,
P_1 = 1$ and $
P_{n+2} = 2P_{n+1} + P_n$
I am stuck trying to prove the identity:
$P_{2n+1}^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
A proof would be great, otherwise a list of known Pell-identies would also help
| You know $P_{n+2} = 2 P_{n+1} + P_{n}$ $\Rightarrow$ $P_{n+2}-(1+\sqrt{2})P_{n+1}= (1-\sqrt{2})(P_{n+1}-(1+\sqrt{2})P_{n})$. In other word, we have
\begin{align*}
\frac{P_{n+2}-(1+\sqrt{2})P_{n+1}}{P_{n+1}-(1+\sqrt{2})P_{n}} = (1-\sqrt{2}).
\end{align*}
Since $P_0 = 0, P_1 = 2, P_2 = 2$, we deduce that
\begin{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate a determinant with pattern A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:
$$
\begin{vmatrix}
0 & 1 & \cdots &n-2&n-1 \\
n-1 & 0 & \cdots &n-3&n-2 \\
... | I take it this is a Toeplitz matrix (diagonals are constant).
Subtract row $n-1$ from row $n$, then row $n-2$ from row $n-1$, ... row $1$ from row $2$. Then subtract column $1$ from each other column.
Then add $1/n$ times the second, third, ..., last columns to the first column. The result should be upper triangula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $(1-\beta)(p+\beta)=1$
If $$(7+4\sqrt{3})^n = p+\beta,$$ where $n$ and $p$ are positive integers and $\beta$ is a proper fraction, then show that $$(1-\beta)(p+\beta)=1.$$
I cant even understand how to express the term in a positive number and a proper fraction. I would appreciate any hint.
| Fun question! The key realization is that
$$
(7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n \in \mathbb{Z}
$$
(do you see why?) and moreover, that
$$
0 < (7 - 4\sqrt{3}) < 1,
$$
so that
$$
0 < (7 - 4\sqrt{3})^n < 1,
$$
for all natural numbers $n$.
It follows from here that
\begin{align*}
p &= (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ $2\log_4(x+1) \le 1+\log_4x$ $2\log_4(x+1) \le 1+\log_4x$
I did:
$$2\log_4(x+1) \le 1+\log_4x \Leftrightarrow \log_4(x^2+1) \le 1+\log_4(x) \Leftrightarrow \log_4(\frac{x^2+1}{x}) \le 1 \Leftrightarrow 4 \ge \frac{x^2+1}{x} \Leftrightarrow 4x \ge n^2 +1 \Leftrightarrow 0\ge x^2 -4x +1$$
Using the quadrati... | write your inequation like that:
$$\log_4(x+1)^2\le \log_4 4+\log_4 x$$ thus $$\log_4 (x+1)^2\le \log_4 4x$$ this is equivalent to $$(x+1)^2\le 4x$$ can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$
My attempt
\begin{align*}
&=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1... | Let $f(x)$ be your function.
we have
$x\ln(1+\frac{1}{x})+\arctan(\frac{1}{2x})=$
$=x(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3})+(\frac{1}{2x}-\frac{1}{24x^3})+\frac{1}{x^3}\epsilon(x)$
$=1+\frac{1}{3x^2}(1+\epsilon(x))$
thus
$\ln(f(x))\sim \frac{1}{3}\arctan(x)\;\; (x\to+\infty)$
and
$$\lim_{x\to+\infty}f(x)=e^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
interesting scalene triangle trigonometry problem the problem is to find the minimum value of cos(2α)+cos(2β)+cos(2γ) in a scalene triangle. So what should the value of α,β, and γ be?
I already transformed the expression using γ=180∘−(α+β) to cos(2α)+cos(2β)+cos(2α+2β)
so how can I finish the problem if this is a good ... | Given $A+B+C = \pi$ and $$\cos 2A+\cos 2 B+\cos 2C = 2\cos(A+B)\cos(A-B)+\cos 2C$$
So $$\cos 2A+\cos 2B+\cos 2C = -2\cos C\cdot \cos(A-B)+2\cos^2 C-1$$
$$ = -2\cos C\left[\cos(A-B)-\cos(C)\right]-1=-2\cos C\left[\cos(A-B)+\cos(A-B)\right]$$
$$ = -4\cos A \cos B\cos C-1$$
So $$\cos 2A+\cos 2B+\cos 2C = -1-4\cos A\cos B\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Manipulation of conditions of roots of a quadratic equation. How do I write $a^5+b^5$ in terms of $a+b$ and $ab$. Also is there any general way of writing $a^n+b^n$ in terms of $a+b$ and $ab$?
| This is what I found when I first came across this question. Now let me see if I can find the steps to get Lozenges' expression. First of all, Dr. Sonhard's factorization is correct:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4).$$
Were Zlatan's sign doubt correct, we would get, I suppose, either $ab$ or $a-b$ as the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How does one solve a bivariate normal density function? If the exponent of $e$ of a bivariate normal density is
$$\frac{-1}{54} *(x^2+4y^2+2xy+2x+8y+4) \\\text{find } \sigma_{1},\sigma_{2} \text{ and } p \text{ given that } \mu_{1} =0 \text{ and } \mu_{2}=-1. $$
One must use this definition to solve.
A pair of rando... | If we substitute $\mu_1=0$ and $\mu_2=-1$ in the exponent of $e $ of the joint probability density formula, we get
$$\displaystyle -\frac 1{2(1-\rho^2)}\left[\left(\frac{x}{\sigma_1}\right)^2
-2\rho \left(\frac{x}{\sigma_1}\right) \left(\frac{y+1}{\sigma_2}\right) +\left(\frac{y+1}{\sigma_2}\right)^2\right]$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is h... | I would do it without brute force:
*
*Since $A^3=0$ we have
$$
e^A=I+A+\frac{A^2}{2!}.
$$
Similar for $B$.
*The LHS of your expression is then $e^Ae^B$. For commuting matrices it is true that
$$
e^Ae^B=e^{A+B}.
$$
*Finally notice that if $A^3=B^3=0$ and $AB=BA$ then $(A+B)^5=0$ (e.g. do the binomial expansion and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Disproving a ring Homomorphism Problem My question is:
$\varphi=\left\{
\begin{array}{c l}
T &\mbox{$\longrightarrow R$} \\
\begin{pmatrix}
a & 0\\
b & c
\end{pmatrix} & \mbox{$\longrightarrow$ $\begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}$}
\end{array}\right.$
Show that this function... | I will call $f$ to your $\varphi$. Look that if you have:
$$
A= \begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix},
B= \begin{pmatrix}
0 & 0\\
1 & 1
\end{pmatrix}
$$
Then,
$$\begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix}*\begin{pmatrix}
0 & 0\\
1 & 1
\end{pmatrix}= \begin{pmatrix}
0 & 0\\
0 & 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does one prove that this function is injective? As the title suggests, I am trying to prove that the following function is injective but at the moment I am stuck.
$$f(x)=-x^5-16x -1 $$
$$-x_1^5 - 16x_1 -1 = -x_2^5 - 16x_2 -1$$
$$x_1^5 + 16x_1= x_2^5 + 16x_2$$
What should be the next step?
Could you also tell me if ... | I suppose that your function is defined on $\mathbb{R}$. Assume that $f(x)=f(y)$. This means that $-x^{5}-16x-1=-y^{5}-16y-1$, so $x^5+16x=y^5+16y$. From this we get $$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4+16)=0.$$
Then, $x=y$ or $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$. If $x=y$ we're done. Otherwise, $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \fr... | It is sufficient to consider the function
$$f(x)=\sin (x)\sin(\frac{2\pi}{3}-x)$$ restreint to the domain $D=\{x|\space 0\lt x\lt\dfrac{2\pi}{3}\}$.
It is easy to find $f$ is positive and has a maximun at the point $x=\dfrac{\pi}{3}\in D$; furthermore the infimum of $f(x)$ is equal to $0$ taken to the neighborhood of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that the set of numbers whose Euler function is less than a certain value has an upper bound Let $S = \{n \in \mathbb{Z} \mid \varphi(n) \le 12 \}$ be the set of number whose Euler function is less or equal 12. Prove that $S$ has an upper bound.
In the example $S= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 1... | The Fundamental theorem of arithmetic allows to write $k = p_1^{\alpha_1} \cdot \ldots \cdot p_n^{\alpha_n}$.
As $\varphi(p) = p-1$, $p$ must be less than 13. So the only primes allowed are $\{2,3,5,7,11,13\}$. Consider every possibile combination:
*
*$p_1 = 13$ gives $13$ and $13 \cdot 2 = 26$
*$p_1 = 11$ gives $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$. Prove that $a,b$ are both divisible by $p$. [Solution verification] Problem:Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some $a,b\in \mathbb{Z}$. Prove that $a,b$ are both divisible by $p$.
My Attempt: $a^2+ab+b^2\equiv 0 \pmod p\R... | Here's a correct different solution. If $p=3k+2$ is an odd prime, $p\mid a^2+ab+b^2$, where $a,b\in\mathbb Z$, then $$a^2+ab+b^2\equiv 0\pmod{p}$$
$$\stackrel{\cdot 4}\iff (a+2b)^2\equiv -3a^2\pmod{p}$$
$$\iff (2a+b)^2\equiv -3b^2\pmod{p}$$
For contradiction, let either $p\nmid a$ or $p\nmid b$. Then either
$$\left((a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
| Mod $5$, we have that the only squares are $0,1,4$. The squares of these are $0,1$. So, we have each of $a^4,b^4,c^4,d^4\in\lbrace 0,1\rbrace $, and their sum is $0$ (becaue $5$ divides them). Because of this, each of $a^4,b^4,c^4,d^4\equiv 0\pmod{5}$. Because of this, we have that $a,b,c,d\equiv 0\pmod{5}$, so the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to p... | As shown in this post,
$$ \sum_{k=1}^n x^k = x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1}$$
Differentiate five times:
$$ \sum_{k=1}^n k (k-1)(k-2)(k-3) x^{k-4} = \frac{d^5}{dx^5} \left[ x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$
For left side,
$$ \sum_{k=1}^n k (k-1)(k-2)(k-3)(k-4) x^{k-4}$$
We can sta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 9
} |
If $ \sin \alpha + \sin \beta = a $ and $ \cos \alpha + \cos \beta = b $ , then show that $\sin(\alpha + \beta) = \frac {2ab } { a^2 + b^2} $ I've been able to do this, but I had to calculate $ \cos (\alpha + \beta) $ first. Is there a way to do this WITHOUT calculating $\cos(\alpha+\beta)$ first ?
Here's how I did it ... | How about:
$$\begin{array}{}
a&=\sin\alpha+\sin\beta&=2\sin\frac 12(\alpha+\beta)\cos\frac 12(\alpha -\beta) \\
b&=\cos\alpha+\cos\beta&=2\cos\frac 12(\alpha+\beta)\cos\frac 12(\alpha -\beta) \\
ab&=4\sin\frac 12(\alpha+\beta)\cos\frac 12(\alpha+\beta)\cos^2\frac 12(\alpha -\beta)&=2\sin(\alpha+\beta)\cos^2\frac 12(\al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots
Problem Statement:-
Find all the values of $a$ for which the equation
$$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$
possess at least two negative roots.
I know that there has been a post regarding this same problem here, but I have a dif... | Just posting an alternative solution, as the comments already helped you to debug your solution:
For quartic polynomials, there is a closed-form analytical solution for the zeroes. However horrible that formula may seem, you don't need it here. You only need the discriminant.
A polynomial $ax^4+bx^3+cx^2+dx+e=0$ has tw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$... | First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^3=(1(1+1)/2)^2$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^3=(n(n+1)/2)^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^3=$
$\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$
$\color\red{(n(n+1)/2)^2}+(n+1)^3=$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:
*
*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.
*Solve $... | What you have found is a simplification for $F(x)$, provided $x\neq 2, x\neq -2$. The denominator $(x^2 - 4)$ of the original function makes it undefined at $x = 2, \;x=-2:$ $(2^2-4) = (-2)^2 - 4 = 0$
So your simplification
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$ is valid, $\forall x \in \mathbb R \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How can simultaneous sinusoidal equations be solved? I have come across a set of simultaneous equations which I can't figure out how to solve. I have 3 equations and only two unknowns, but they are angular quantities and feature in the equations as sinusoidal functions of the angular quantities.
The system of equations... | Note that
$$
\begin{array}{l}
\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{\cos \psi \sin \theta + \cos \theta \sin \phi \sin \psi } \\
{\sin \psi \sin \theta - \cos \psi \cos \theta \sin \phi } \\
{\cos \phi \cos \theta } \\
\end{array}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction.
Theorem:
Let $A_n$ be a $n\times n$ matrix of the form:
$\begin{pmatrix}
2 & 1 & 0 & 0 && & \cdots & 0\\
1 & 2 & 1 & 0 && & \cdo... | Let $d_n = \det A_n$. Note that $d_1 = 2, d_2 = 3$.
For the induction step, look at the (clunky) picture below:
Note that $d_n = 2 d_{n-1} -d_{n-2}$.
Now show that $n \mapsto 3(n-1)$ satisfies the equation with the same
initial conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving inequality involving floor function We have this inequality (over real numbers) :
$$x^2-2x\le \frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x \rfloor + \lfloor -x \rfloor}$$
How we can solve it using both of algebraic and geometric methods ?
| $$\begin{cases}x= \lfloor x\rfloor+\{x\}\\-x= \lfloor -x\rfloor+1-\{x\}\end{cases}\Rightarrow\lfloor x\rfloor+\lfloor -x\rfloor=-1$$
$$\sqrt{1-\lfloor x\rfloor^2}=\begin{cases}0 \text{ if } -1\le x\lt0\\1\text{ if } 0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$ Hence
$$\frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Differential equation with integral that is hard to compute Task and my attempt to solve it
$$xy'=\sqrt{2x^2+y^2}-y$$
I have tried substitution like $y=z·x$ but finally got very difficult integral
$$ \int \frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{dx}{x}$$
at the left side.
Maybe i have to do something different to avoid di... | *
*Multiply by $\frac{\sqrt{2+z^2}+2z}{\sqrt{2+z^2}+2z}$:
$$
\int\frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{\sqrt{2+z^2}+2z}{(2+z^2)-4z^2}\,dz=\int\frac{\sqrt{2+z^2}}{2-3z^2}\,dz+\int\frac{2z}{2-3z^2}\,dz.
$$
The second integral is simple.
*In the first integral do the change of variables $t=z+\sqrt{z^2+2}$:
$$
\int\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve system of equation using matrices (4 variables) The Question:
Solve using matrices.
$$2w-2x-2y+2z=10\\w+x+y+z=-5\\3w+x-y+4z=-2\\w+3x-2y+2z=-6$$
My work:
$$
\begin{bmatrix}
2&-2&-2&2&10\\
1&1&1&1&-5\\
3&1&-1&4&-2\\
1&3&-2&2&-6\\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&4&-1&1&-1... | In the second matrix of your work, entry $a_{3,3}$ should be $2$, not $-1$, and entry $a_{3,5}$ should be $-17$, not $-11$. Also, the whole fourth row seems wrong. It appears that the four row operations performed in that step should have been:
\begin{align}
&1.\text{ Replace R1 with $\frac12\times$ R1.} \\
&2.\text{ R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $f(x,y)=x^4+y^4-3xy$ is coercive? How to show that $f(x,y)=x^4+y^4-3xy$ is coercive ?
This is my attempt :
$$f(x,y)=x^4+y^4-3xy$$
$$f(x,y)=x^4+y^4\left(1-\frac{3xy}{x^4+y^4}\right)$$
As $||(x,y)|| \to \infty $ , $\frac{3xy}{x^4+y^4} \to 0$
So $||(x,y)|| \to \infty $ , $f(x,y)=x^4+y^4-3xy \to \infty$.
I... | Your answer looks right.
Here's another way. Observe
\begin{align}
xy \le \frac{x^2+y^2}{2} \ \ \text{ and }\ \ x^4+y^4\geq \frac{1}{2}(x^2+y^2)^2
\end{align}
which means
\begin{align}
x^4+y^4-3xy \geq&\ x^4+y^4-\frac{3}{2}(x^2+y^2) \\
\geq&\ \frac{1}{2}(x^2+y^2)^2-\frac{3}{2}(x^2+y^2)\\
=&\ \frac{1}{2}\left(x^2+y^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find $g$ such that $g\circ f=h$ Let
$$f(x)=\dfrac{2x+3}{x-1} \quad \mbox{and}\quad h(x)=\dfrac{6x^2+8x+11}{(x-1)^2} $$
find polynom $g$ such that $g\circ f=h$
Indeed,
$$g\circ f=h \iff \forall x\in \mathbb{R}\quad g\circ f(x)=h(x) $$
\begin{align}
g\circ f(x)&=h(x)\\
g\left(f(x)\right)&=h(x)\\
g\left(\dfrac{2x+3}{x-1}... | It is wrong. After you have written the expression of $g(y)$ we should have
$$ g(y) = \frac{6(3+y)^{2} + 8(3+y)(y-2) + 11(y-2)^{2}}{25} $$
instead of your expression. On simplifying we get $g(y) = y^{2}+2$, i.e,
$$ g(x) = x^{2} + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$
Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$
I am having difficulty factoring the whole equation so that I can equate each factor to $0$ one by one and get the roots accordingly.
| Divide both sides by $x^4$ and change variable to $y = \frac{x-1}{x^2}$, we have
$$\begin{align}1 = 5y(1-y)
\iff & 4y(y-1) = -\frac45 \iff
(2y-1)^2 = \frac15\\
\implies & y = \frac12\left(1 \pm \frac{1}{\sqrt{5}}\right) = \frac{\sqrt{5}\pm 1}{2\sqrt{5}} = \frac{2}{5 \mp \sqrt{5}}
\end{align}
$$
Substitute $y$ back by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Computation of eigenvalues/vectors of a $9\times 9$ matrix I have a symmetric matrix (with real coefficients) and I need to compute its eigenvalues and eigenvectors. My matrix depends on 3 parameters $(\nu_1,\nu_2,\nu_3)$ that are not independent (in fact we have $\nu_1^2+\nu_2^2+\nu_3^2=1$). If I am using Maple to com... | One can do this with most of the CASs, by taking the condition $a^2+b^2+c^2-1:=0$ as another polynomial equation for computing a Gröbner basis. It depends on your specific matrix, whether or not the complexity of the system of polynomial equations is still manageable.
Edit: the characteristic polynomial of $B$ is given... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$
How to evaluate this integral?
$$\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$
M... | $$I=\frac{1}{8} \int_{-\infty}^{\infty}
\frac{\cosh(7\pi z)}{(1+z^2)\cosh^3(3\pi z)} \text{d}z$$
Now we're considering the function
$$f(z)=\frac{\cosh(7\pi z)\psi^{(0)}\left ( 1-iz\right ) }{\cosh^3(3\pi z)}$$
From the residue theorem and calculate the residues at $$z=\frac{i}{6},\frac{i}{2},\frac{5i}{6}$$
We finally ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Determine the vectors of components For the polynomial vector space $\mathbb{R}[x]$ of degree $\leq 3$ we have the following three bases:
$$B_1 = \{1 - X^2 + X^3, X - X^2, 1 - X + X^2, 1 - X\} , \\
B_2 = \{1 - X^3, 1 - X^2, 1 - X, 1 + X^2 - X^3\}, \\
B_3 = \{1, X, X^2, X^3\}$$
How can we determine the following vec... | Hint:
$$
B_1=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&0&1&1\\
0&1&-1&-1\\
-1&-1&1&0\\
1&0&0&0
\end{bmatrix}
$$
$$
B_2=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&1&1&1\\
0&0&-1&0\\
0&-1&0&1\\
-1&0&0&-1
\end{bmatrix}
$$
$$
B_3=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The Most general solution satisfying equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$ The most general value of $x$ satisfying the equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$, is found to be $x=2n\pi+\frac{7\pi}{4}$.
My approach:
$$
\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}=\si... | Here is what you got wrong:
$$\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}$$
So you got $\sin x<0$ and $\cos x>0$. Then you are in the $4$th quadrant.
$$\sin x=\frac{-1}{\sqrt{2}}=\sin \left(\frac{7\pi}{4}\right) \Rightarrow x=\frac{7\pi}{4}+2n\pi$$
P.S.: You got wrong because $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\... | converting the given equation in $$\tan$$ we get
$${\frac { \left( 1+\sqrt {3} \right) \left( 1- \left( \tan \left( x/2
\right) \right) ^{2} \right) }{1+ \left( \tan \left( x/2 \right)
\right) ^{2}}}+2\,{\frac { \left( \sqrt {3}-1 \right) \tan \left( x/2
\right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
| You have that $3y-2$ divides $y^2$.
Notice that $(3y-2,y^2)| (3y-2,y)^2$.
On the other hand, if $d$ divides $3y-2$ and $y$ we have $d|2$. So $(3y-2,y^2)| 4$.
Therefore $3y-2|4$
Therefore $y=0,1$ or $2$.
If $y=0$ we have $x=0$.
If $y=1$ we have $x=\pm 1$
If $y=2$ we have $x=\pm 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to factorize $a^2-b^2-a+b+(a+b-1)^2$? The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.
| $$(a+b-1)^2=a^2+b^2+1-2a-2b+2ab$$
$$\implies a^2-b^2-a+b+(a+b-1)^2=2a(a+b)-\{2a+(a+b)\}+1$$
$$ =2a\{a+b-1\}-\{a+b-1\}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$
$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$
Instead of expanding the integrand, or doing integration by part, is there any faster way to ... | You can do this with substitution and Cavalieri's formula:
$$
\int_0^1 u^n \,du = \frac{1}{n+1}
$$
For the first one, let $u= \frac{x-a}{b-a}$. Then $x = (b-a)u +a$, which means $x-a = (b-a)u$ and $x-b = (b-a)(u-1)$. Also $dx=(b-a)\,du$. So
\begin{align*}
\int_a^b (x-a)(x-b)\,dx
&= (b-a)^3 \int_0^1 u (u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
How to see there exists const $C$ such that $\frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C(1-x^2)^m\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n$ This comes up in relation to Legendre functions. The claim is made that for $n =0,1,2,3,\cdots$ and $m=0,1,2,3,\cdots,n$, there is a constant $C_{n,m}$ such that
$$
\frac{d^{n-m}}{dx^{n-m}}(1... | Suppose we seek to determine the constant $Q$ in the equality
$$ Q_{n,m} \left(\frac{d}{dz}\right)^{n-m} (1-z^2)^n
= (1-z^2)^m \left(\frac{d}{dz}\right)^{n+m}
(1-z^2)^n$$
where $n\ge m.$ We will compute the coefficients on $[z^q]$ on the LHS
and the RHS. Writing $1-z^2 = (1+z)(1-z)$ we get for the LHS
$$\sum_{p=0}^{n-m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help verify the proof that $\min \left({\lfloor{N/2}\rfloor, \lfloor{(N + 2)/p}\rfloor}\right) = \lfloor{(N + 2)/p}\rfloor$ for $N \ge p$ Well, I think that I have this correct, however my proof seams cumbersome, so can someone check these results or suggest a simplified proof.
Prove that ($N \ge p$)
\begin{equation*}
... | Show that
\begin{equation*}
\left\lfloor{\frac{N}{2}}\right\rfloor < \left\lfloor{\frac{N + 2}{p}}\right\rfloor
\end{equation*}
is not valid. Let $N = k\, p + m$ where $k \in \mathbb{Z}^{+}$ and $m \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{p - 1}\right\}$ by the Quotient Remainder Theorem. Then show this holds f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it... | Expanding $(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})$ gives $1+x+x^2+\cdots +x^{2^{n+1}-1}$ because every integer $k$ with $0 \le k \le 2^{n+1}-1$ has a unique binary representation (as a sum of powers of $2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
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Fractions in Questions and Answers
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