Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Expressing $\int dx\, \cos(ax^2 + 2bx + c)$ in terms of Fresnel integrals I have difficulties with this integral, which is is related to Fresnel integrals ($a>0$):
$$ \int dx\, \cos(ax^2 + 2bx + c) = \\ =\sqrt{\frac{\pi}{2a}} \left[ \cos(\frac{ac-b^2}{a}) \mathscr{C} \left( \frac{\sqrt{2} (ax + b)}{\sqrt{a\pi}} \right)... | Start completing the square $$ax^2+2bx+c=a\left(x+\frac b{a}\right)^2-\left(\frac{b^2}{a}-c\right)$$ So, $\cos(ax^2+2bx+c)$ write $$\cos \left(a\left(x+\frac b{a}\right)^2\right)\cos\left(\frac{b^2}{a}-c\right)+\sin \left(a\left(x+\frac b{a}\right)^2\right)\sin\left(\frac{b^2}{a}-c\right)$$ Now, you have to change vari... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction on the equality $ \sum_{i=2}^{n+1} \frac{1}{i(i+1)}= \frac{n}{2(n+2)}$ Prove with induction principle: $$ \sum_{i=2}^{n+1} \frac{1}{i(i+1)}= \frac{n}{2(n+2)} \quad \forall n \geq 2\in \mathbb{N} $$
Starting prove $P(0)$:
$$ \sum_{i=2}^{3} \frac{1}{i(i+1)}= \frac{1}{2(2+1)}+\frac{1}{2(3+1)}= \frac{1}{4} $$
... | you have $$\frac{n}{2(n+2)}+\frac{1}{(n+2)(n+3)}=\frac{n^2+3n+2}{2(n+2)(n+3)}=\frac{(n+2)(n+1)}{2(n+2)(n+3)}=\frac{n+1}{2(n+3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a sq... | With AM-GM $$\frac { x }{ y } +\frac { y }{ x } \geq 2\sqrt { \frac { x }{ y } \frac { y }{ x } } = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
When does a prime $p=m^2+n^2$ divide the term $ m^3+n^3-4$? I came across a math olympiad type question that goes like this:
For what primes $p$ will $p=m^2+n^2$ and $p$ divide $m^3+n^3-4$?
I tried a few examples and think that $p=5$ is the only solution but am unable to prove it. The property that $p \equiv 1 \mod 4$ ... | Sorry for waking up old post...
Anyway, I think the following works.
Since $p\mid m(m^2+n^2)$ and $p\mid n(m^2+n^2)$ we have $p|3(m^3+m^2n+n^2m+n^3)$ Since \begin{align}3(m^3+m^2n+n^2m+n^3)&=(m+n)^3+2(m^3+n^3)\\&=(m+n)^3+8+ 2(m^3+n^3-4)\end{align}
we obtain $p|(m+n)^3+8$，while $$(m+n)^3+8=(m+n+2)(m^2+n^2+2mn-2m-2n+4)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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A proof of the identity $ \sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k} = \frac{n!}{(x + 0) (x + 1) \cdots (x + n)} $. I have to prove that
$$
\forall n \in \mathbb{N}_{0}, ~ \forall x \in \mathbb{R} \setminus \mathbb{N}_{0}:
\qquad
\sum_{k = 0}^{n} \frac{(-1)^{k} \binom{n}{k}}{x + k}
= \frac{n!}{(x + 0) (x + ... | Remark. What follows is an exact duplicate of an earlier post of
mine which I am unable to locate despite making a considerable
effort.
We seek to verify that
$$\sum_{k=0}^n \frac{1}{x+k} (-1)^k {n\choose k}
= n! \times \prod_{q=0}^n \frac{1}{x+q}.$$
Consider the function
$$f(z) = n! \frac{1}{x+z} \prod_{q=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything ... | For a fixed positive $k$ we should expect $\sqrt {y+k}$ to be very close to $\sqrt y$ when $y$ is very large, because when $d$ is not close to $0,$ we expect a big difference between $(d+\sqrt y)^2$ and $y,$ which means $d+\sqrt y$ is too big to be $\sqrt {y+k}$.
We can confirm this by observing that if $k,y$ are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Eigenvectors and matrix decomposition of a Quaternion Given the matrix representation of Quaternions
(re. e.g. to this other post)
$$
Q \ := \ \left(\begin{array}{rrrr}d&-c&b&a\\c&d&-a&b\\-b&a&d&c\\-a&-b&-c&d\end{array}\right) \ \
$$
what "meaning" or "role" can be given to the eigenvectors? and what to the decomposit... | $$\det(tI-A)= \begin{vmatrix}t-d&c&\!\!-b&-a\\\!\!-c&t-d&a&-b\\b&\!\!-a&t-d&-c\\a&b&c&t-d\end{vmatrix}=$$
$$(t-d) \begin{vmatrix}t-d&a&-b\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+c \begin{vmatrix}c&\!\!-b&-a\\\!\!-a&t-d&-c\\b&c&t-d\end{vmatrix}+b\begin{vmatrix}c&\!\!-b&-a\\t-d&a&-b\\b&c&t-d\end{vmatrix}-$$$${}$$
$$-a\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1917755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove inequality $ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$ for $a,b,c\ge 0$ Prove that:
$$ (a+b+c)(a^7+b^7+c^7)\ge(a^5+b^5+c^5)(a^3+b^3+c^3)$$
I already know that this can be proven using Cauchy Schwarz, but I don't really see how to apply it here. I'm looking for hints.
| If all of $a$, $b$ and $c$ are zero, then the inequality holds trivially, and so we can assume that at least one of $a$, $b$, and $c$ are positive.
The Cauchy-Schwarz inequality gives us that for any natural number $n$,
$$
\left(a^n + b^n + c^n\right)\left(a^{n+2} + b^{n+2} + c^{n+2} \right)
\geq \left(a^{n+1} + b^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1919468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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... | Hint: As mentioned in the comments, substitute: $cos(120+A)$ and $sin(120+A)$ :$$\color{blue}{cos(120+A)=\dfrac{-1}{2}cosA-\dfrac{\sqrt{3}}{2}sinA}$$
$$\color{red}{cos(240+A)=\dfrac{-1}{2}cosA+\dfrac{\sqrt{3}}{2}sinA}$$
The L.H.S becomes:
$${cos}^3A +(\color{blue}{\dfrac{-1}{2}cosA-\dfrac{\sqrt{3}}{2}sinA})^3+(\color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
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How to solve $\int \sqrt{a+x^2} dx $, $a>0$? By setting $a+x^2 = t^2$ I can get $2xdx=2dt$ so $dx=dt/\sqrt{t^2-a}$. But such substitutions only lead to iterate integrals of the form
$$
\int \frac{t^2}{\sqrt{t^2-a}}dt,
$$
substituting again $t^2-a=u^2$ we get
$$
\int udu +\int\frac{2}{\sqrt{u^2+a}}du.
$$
How can I sol... | You can set $\sqrt{a+x^2}=x+t$, so $a+x^2=x^2+2tx+t^2$ and
$$
x=\frac{a-t^2}{2t}=\frac{a}{2t}-\frac{t}{2},
\qquad
dx=\left(-\frac{a}{2t^2}-\frac{1}{2}\right)dt=
-\frac{a+t^2}{2t^2}\,dt
$$
Moreover
$$
\sqrt{a+x^2}=\frac{a-t^2}{2t}+t=\frac{a+t^2}{2t}
$$
Hence the integral becomes
$$
-\int\frac{(a+t^2)^2}{4t^3}\,dt=
-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Let A be an $n*n$ matrix. Prove that if $rank(A) = 1$, then $det(A + E) = 1 + trace(A)$ I feel like I've got the answer, but I've never been good at putting what I think into words.
$\begin{vmatrix}
n_{11} & n_{12} \\
n_{21} & n_{22}
\end{vmatrix} = 0 = n_{11}n_{22} - n_{12}n_{21}$
$\begin{vmatrix}
n_{11} + 1 & n_{12} ... | Suppose $\;A=(a_{ij})\;$ with rank$\,A=1\;$ and assume that the only row in $\;A\;$ which is linearly independent (= non all-zeros row) is the first one, so that the other ones are scalar multiples of this first one. Say the $\;i\,-$ th row is $\;k_i\;$ times the first one
$$|A+E|=\begin{vmatrix}1+a_{11}&a_{12}&\ldots&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Distributing 4 distinct balls between 3 people
In how many ways can you distribute 4 distinct balls between 3 people such that none of them gets exactly 2 balls?
This is what I did (by the inclusion–exclusion principle) and I'm not sure, would appreciate your feedback:
$$3^4-\binom{3}{1}\binom{4}{2}\binom{2}{1}^2+\bi... | Here is a variation based upon generating functions. Distributing $0$ up to $4$ distinguishable balls to three persons can be represented by
\begin{align*}
\left(1+t+\frac{1}{2!}t^2+\frac{1}{3!}t^3+\frac{1}{4!}t^4\right)^3\tag{1}
\end{align*}
The coefficient of $t^4$ multiplied with $4!=24$ in the expanded expressi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Find the value of $a$ and $b$ when value of $x\cdot y \cdot z$ is given for both conditions If $a,x,y,z,b$ are in $Arithmetic$ $Progression$ then the value of $x\cdot y \cdot z$ is $\frac{35}{3}$ and if they are in $Harmonic$ $Progression$ then the value of $x\cdot y \cdot z$ is $\frac{75}{7}$, where $a$ and $b$ are gi... | Hint: Due to symmetry reasons it is convenient to use the middle variable $y$ as center of the arithmetic and geometric progression.
We obtain
\begin{array}{c|c|c|c|cl}
a&x&y&z&b\qquad&\qquad x\cdot y\cdot z\\
\hline
y-2k&y-k&y&y+k&y+2k\qquad&\qquad(y-k)\cdot y\cdot (y+k)=\frac{35}{3}\\
yq^{-2}&yq^{-1}&y&yq&yq^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1926815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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My solution of this integral does not match. Where am I doing wrong? I've been trying to calculate the following integral. But I always get the wrong result....
$$
S(a, b) = \int_0^{2\pi}\frac{d\theta}{\left(a + b\cos\theta\right)^2},
\quad\quad\quad\mbox{for}\quad a > b > 0
$$
Assume substitution: $z = e^{i\theta}$. ... | Notice that $S(a,b)$ can be computed from a symmetry argument only. We have
$$\begin{eqnarray*} S(a,b)=2\int_{0}^{\pi}\frac{d\theta}{(a-b\cos\theta)^2}&=&2\int_{-\pi/2}^{\pi/2}\frac{d\theta}{(a+b\sin\theta)^2}\\&=&2\int_{0}^{\pi/2}\left[\frac{1}{(a-b\sin\theta)^2}+\frac{1}{(a+b\sin\theta)^2}\right]\,d\theta\\&=&4\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
limit of $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞ $a_n=\sqrt{n^2+2} - \sqrt{n^2+1}$ as $n$→∞
Both limits tend to infinity, but +∞ −(+∞) doesn't make sense. How would I get around to solving this?
| We have $$\lim_{n \to \infty }\sqrt{n^2+2} - \sqrt{n^2+1} = \lim_{n \to \infty} = \frac{n^2+2 - n^2-1}{\sqrt{n^2+2} + \sqrt{n^2+1}} = \lim_{n \to \infty} \frac{1}{\sqrt{n^2+2} + \sqrt{n^2+1}}\\ = \lim_{n \to \infty}\frac{1}{n} \lim_{n \to \infty} \frac{1}{\sqrt{1+\frac{2}{n^2}} + \sqrt{1+\frac{1}{n^2}}} = \frac{1}{2}\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Solve $\iint_D\sqrt{9-x^2-y^2}$ Where $D$ is the positive side of a circle of radius 3
Solve $\displaystyle\iint_D\sqrt{9-x^2-y^2}$ Where $D$ is the positive
side of a circle of radius 3 ($x^2+y^2=9,x\ge0,y\ge0$)
I tried to subsitute variables to $r$ & $\theta$:
$$x = r\cos\theta$$
$$y = r\sin\theta$$
$$E = \{0\le ... | Be careful that the transformed domain $E$ should be defined by $0\le r\le3$ and $0\le\theta\le\pi/2$.
Now the integrand function is
$$
r\sqrt{9-r^2(\cos^2\theta+\sin^2\theta)}=r\sqrt{9-r^2}
$$
and this independent of $\theta$. For the integral over $dr$ use the substitution $r=3\sin u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Expansions of meromorphic functions I found in google books the following expansion
$$\sum _{k=1}^{\infty } \left(\frac{1}{(x-\pi k)^3}-\frac{1}{\pi k+x}+\frac{1}{(\pi k+x)^3}+\frac{1}{\pi k-x}\right)+\frac{1}{x^3}-\frac{1}{x}=\cot ^3(x)$$
and
$$\sum _{k=1}^{\infty } \left(-\frac{2}{\pi ^4 k^4}+\frac{8}{3 \pi ^2 k... | You need to find the Laurent expansion of $\cot^3(z)$ at $0$ : $\cot^3(z) = \frac{1}{z^3} + \frac{a}{z^2}+\frac{b}{z}+h(z)$ where $h(z)$ is analytic around $z=0$. With the $\pi$-periodicity of $\cot(z)$, you get that $$g(z) = \cot^3(z) - \sum_{k=-\infty}^\infty \frac{1}{(z-k\pi)^3} + \frac{a}{(z-k\pi)^2}+\frac{b}{z-k\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simplifying inverse trigonometric expressions such as $\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$ Okay so I'm just looking for a short cut method or a method that is not so long for simplifying expressions like this
$$\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{... | $$
\frac{1-\tan^2(u)}{1+\tan^2(u)}=\cos(2u)\implies
\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)
=\left\{\begin{array}{}
\pi+2\tan^{-1}(x)&\text{if }x\lt0\\
\pi-2\tan^{-1}(x)&\text{if }x\gt0
\end{array}\right.
$$
$$
\frac{2\tan(u)}{1-\tan^2(u)}=\tan(2u)\implies
\tan^{-1}\left(\frac{2x}{x^2-1}\right)
=\left\{\begin{array}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $\lim_{(x,y)\to(1,1)} x^2+xy+y=3$
Prove that $$\lim_{(x,y)\to(1,1)} x^2 + xy + y = 3$$ using the epsilon-delta definition.
What I have tried:
Let $\epsilon > 0$ be arbitrary. We must show that for every $\epsilon$ we can find $\delta>0$ such that
$$0 < \|(x,y) - (1,1)\| < \delta \implies \|f(x,y) - 3\| < \epsi... | It's easier if you substitute $x=t+1$ and $y=u+1$, so
$$
x^2+xy+y-3=(t+1)^2+(t+1)(u+1)+(u+1)-3=
t^2+tu+3t+2u
$$
Now, if $t^2+u^2<\delta^2$, which is the same as $\sqrt{(x-1)^2+(y-1)^2}<\delta$, we surely have $|t|<\delta$ and $|u|<\delta$, so
$$
|t^2+tu+3t+2u|\le
|t^2+u^2|+|t|\,|u|+3|t|+2|u|\le
2\delta^2+5\delta
$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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... | You can simplify your proof by setting $x,y,z$ the non-unit elements of your set. Show that $x^2=y^2=z^2=1$, $xy=z$ and $yz=x$. From this you conclude that $xz=xxy=y$. Therefore an arbitrary product of $x,y,z$ (and their inverses) will be in $\{1,x,y,z\}$, which is therefore a subgroup.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let a parabola have a vertex of $T(5,\frac{1}{2})$ and with the x-axes it forms an angle of 45 degrees. Find the point which it traverses? I tried to do a lot of stuff to get my answer but always fail.
$$5 = -\frac{b}{2a} \Rightarrow b=-10a$$
$$\frac{1}{2}= -\frac{-b^2+4ac}{4a} \Rightarrow c=\frac{-49}{8}$$
$$\tan 45=... | In this answer, I will focus on the cases in which the axis of the parabola is horizontal or vertical. Firstly, I consider the case where the parabola has a vertical axis, i.e. parallel to the $y $-axis. Define the equation of the parabola as $y=ax^2+bx+c $.
Its derivative $2ax+b \ $ must be zero for $x=5 \ $, so we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Taking the integral by partial fractions of $\frac{4}{(x+a)(x+b)}$ Given this equation, how do I solve for the integral if there are so many variables that I can't find $A$ and $B$?
$$\int \frac4{(x+a)(x+b)}\mathrm{d}x$$
I got until the point:
$$ \frac4{(x+a)(x+b)} = \frac{A(x+b)+B(x+a)}{(x+a)(x+b)}$$
so taking the num... | Ok, you get to $$4 = {A(x+b)+B(x+a)}.$$
Let's transform it:
$$4 = {A(x+b)+B(x+a)} \Rightarrow \\
0x + 4 = x(A+B) + (Ab + Ba).$$
Then you need to solve:
$$\begin{cases}
A+B = 0\\
Ab + Ba = 4
\end{cases} \Rightarrow \begin{cases}
A = -B\\
A(b -a) = 4
\end{cases} \Rightarrow \begin{cases}
A = \frac{4}{b-a}\\
B = \frac{4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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!
| We can use the method of Lagrange Multipliers, which is generally useful when you want to minimize or maximize something given a constraint. We have that
$$\mathcal{L}(a,b,c,\lambda) = (a+b+c+abc)-\lambda(ab+ac+bc-1)$$
Taking partials, we get that
$$\frac{\partial \mathcal{L}}{\partial a} = 1+bc-\lambda(b+c)$$
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
dedekind ramanujan identities I get the following representation using laplace transform the remainder part it dificult to check numerically could you get some numerically result?
Sorry for my bad latex i have correct the formula before
$$\sum _{k=1}^{\infty } \frac{1}{k \left(e^{2 \pi k x}-1\right)}=\left(-\frac{\pi ... | Let $$q = e^{-\pi / x}$$ and consider the sum
\begin{align}
S &= \sum_{n = 1}^{\infty}\log\left(\frac{2\pi^{2}n^{2}\operatorname{cosech}(\pi n/x)}{x^{2}}\right) - \frac{\pi n}{x}\notag\\
&= \sum_{n = 1}^{\infty}\log\left(\frac{2\pi^{2}n^{2}e^{-\pi n/x}}{x^{2}\sinh(\pi n/x)}\right)\notag\\
&= \sum_{n = 1}^{\infty}\log\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the transition matrix for ordered basis of 2x2 diagonal matrices The problem:
For the vector space of lower triangular matrices with zero trace,
given ordered basis:
$B=${$$
\begin{bmatrix}
-5 & 0 \\
4 & 5 \\
\end{bmatrix},
$$
\begin{bmatrix}
-1 & 0 \\
1 & 1 \\
\e... | Thank you for your direction. I was able to use your ideas to find the correct solution to the problem. First I expressed B and C in terms of the basis
$\left[ \begin{matrix} -1 & 0 \\ 0 & -1 \end{matrix} \right]$,$\left[ \begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right]$
The basis B is thus equivalent to $\left[ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Calculate Gauss-Jordan of this matrix Consider the matrix:
$$A=\begin{bmatrix} 0 & 1 &1 & 0 &-1 &0 \\ 0 & 3 & 3 &-2 &-3 &2 \\ 0 & -1&-1&1&1&-1 \end{bmatrix}\in M_{3\times6}(\mathbb Q)$$
(A) Calculate the normal Gauss-Jordan, $A '$, the matrix and calculate $P $ also invertible and such that $A' = P A$.
(B) Expressing r... | Perform the operation $-3R_1+R_2$ and $R_1+R_3$, I obtain the following matrices
I obtain the matrix $$\begin{bmatrix} 0 & 1 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & -2 & 0 & 2 \\ 0 & 0 & 0 & 1 & 0 & -1 &\end{bmatrix}$$
Next, I perform the operation $2R_3+R_2,$
I obtain the matrix $$\begin{bmatrix} 0 & 1 & 1 & 0 & -1 & 0 \\ 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:
$$\sum_{n=0}^k \cos... | This is not an answer but it is too long for a comment.
Considering $$u_k=\sum_{n=0}^k \cos\left(x + \frac{n \,\pi}{k+1}\right)^{2k}$$ and computing the first terms (with some minor trigonometric simplifications)
$$\left(
\begin{array}{ccc}
k & u_k & \text{value}\\
0 & 1 & 1 \\
1 & \cos ^2(x)+\sin ^2(x) & 1 \\
2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1951708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solving equation in three variables please help me understand how the following equation with 3 variables and power of 2 is solved and what solution approach is the quickest.
$$3y^2 - 3 = 0$$
$$4x - 3z^2 = 0$$
$$-6xz+ 6z = 0 $$
| $3y^2-3=0.............(1)$
$4x-3z^2=0...........(2)$
$6z-6xz=0...........(3)$
First, solving the equation(1),
$3y^2-3=0\;\;\;\implies y^2=1\;\;\;\implies y=\pm1$
Now, solving the equation(3),
$6z-6xz=0\;\;\;\implies 6z(1-x)=0$
$z=0,\;1-x=0$
$z=0,\;x=1$
Solving the equation(2),
$4x-3z^2=0$
Put the value of $x$ in equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What are the intersection points for these two equations? For the sake of me, I can't get the solution! Not sure what I have done wrong. Anyway these are the two equations:
$$\sqrt{25-(x-13.1)^2}=\frac x2-4$$
and I went
$$-(x-13.1)^2=\frac{x^2}4+16-25$$
$$(x-13.1)^2=-\frac{x^2}4+9$$
$$x^2+26.2x-171.61=-\frac{x^2}4-9$$
... | You've made a (common) mistake on your first line. When expanding $(a+b)^2$ some students only write $a^2+b^2$ where in fact the correct expansion is $a^2+2ab+b^2$.
So you should have gotten $\frac{x^2}{4}-4x+16$ when you expanded $\left(\frac{x}{2}-4\right)^2$ instead of just $\frac{x^2}{4}+16$.
Also on your third lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of $2^{\sin^2x}+2^{\cos^2x}$ The question is what is the minimum value of
$$2^{\sin^2x}+2^{\cos^2x}$$
I think if I put $x=\frac\pi4$ then I get a minimum of $2\sqrt2$. But how do I prove this?
| You know that $\cos^2 x = 1 - \sin^2x$, so you can rewrite:
$$
2^{\sin^2x} + 2^{\cos^2x} = 2^{\sin^2x} + 2^{1-\sin^2x} = 2^{\sin^2x} + \frac{2}{2^{\sin^2x}}
$$
Now, let $2^{\sin^2 x} = y$, then we basically have to maximize $y + \frac{2}{y}$. But then, note that: $y + \frac{2}{y}$ has derivative $1 - \frac{2}{y^2}$, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Prove that $(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$
Let $a,b,c,d>0$. Prove that $$(a+b+c-d)(a+c+d-b)(a+b+d-c)(b+c+d-a)\le(a+b)(a+d)(c+b)(c+d)$$
I don't know how to begin to solve this problem
| The simplest method to solve this question should be the following:
Let
$$b+c+d-a=w\tag{1}$$
$$a+c+d-b=x\tag{2}$$
$$a+b+d-c=y\tag{3}$$
$$a+b+c-d=z\tag{4}$$
So, the given inequality is equivalent to
$$16wxyz \leq (w+x)(x+y)(y+z)(z+w)\tag{*}$$
It is not hard to see that at most one of $w,x,y,z$ is less than zero.
So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Calculating $\sum_{n=1}^∞ \frac{1}{(2 n-1)^2+(2 n+1)^2}$ using fourier series of $\sin x$ I have to calculate $\frac{1}{1^2+3^2}+\frac{1}{3^2+5^2}+\frac{1}{5^2+7^2}+...$ using half range Fourier series $f(x)=\sin x$ which is:
$f(x)=\frac{2}{\pi}-\frac{2}{\pi}\sum_{n=2}^\infty{\frac{1+\cos
n\pi}{n^2-1}\cos nx}$
I hav... | Since $f(0)=\sin 0 =0$ and $\cos nx=1$ at $x=0$:
$$0=\frac{2}{\pi}-\frac{2}{\pi}\sum_{n=2}^\infty\frac{1+\cos n\pi}{n^2-1}$$
also note that $\cos n\pi=(-1)^n$. Hence
$$1=\sum_{n=2}^\infty\frac{1+\cos n\pi}{n^2-1}=\sum_{n=2}^\infty\frac{1+(-1)^n}{n^2-1}$$
All you need to know is
$$1+(-1)^n=\begin{cases}2&n=2k\\0& n=2k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Verify the triple angle formula Verify the triple angle formula
$$\tan(3x) = \frac{3 \tan(x) − \tan^3(x)}{1 − 3 \tan^2(x)}$$
I have tried simplifying the right side by the following
$$\tan(3x) = \frac{\tan(x)(3 − \tan^2(x))}{1 − 3 \tan^2(x)}$$
but then I am getting stuck trying to verify the equation
| You can try.
$$\tan (3x) = \frac{\sin 3x}{\cos 3x} = \frac{\sin x \cos 2x + \cos x \sin 2x}{\cos x \cos 2x - \sin x\sin 2x} = \frac{\sin x(\cos^2 x- \sin^2x) + 2\sin x\cos^2 x}{\cos x(2\cos^2 x -1) - 2\sin^2 x\cos x} = \frac{\tan x(1-\tan^2 x) + 2\tan x}{2 - \frac{1}{\cos^2 x} - 2\tan^2 x} = \frac{3\tan x - \tan^3x}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I find the sum of a sequence whose common difference is in Arithmetic Progression? How do I find the sum of a sequence whose common difference is in Arithmetic Progression ?
Like in the following series :-
$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91$
And also how to find it's $n^{th}$ term ??
| $$
\begin{align}
& \text{Let } a = a_{0} \space\,,\space d = \frac{1}{n} (a_{n} - a_{n-1}) \space\Rightarrow\space a_{n} = a_{n-1} + d \cdot n \quad \colon n \ge 1 \\
& \small \color{blue}{ (a=0 \,, d=1) \Rightarrow a_{n} = a_{n-1} + n = \{ \text{1, 3, 6, 10, ...} \} } \\
& \\
& a_{n} = d (n) + a_{n-1} = d (n) + d (n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Show that $2^{n} \geq (n +2)^{2}$ for all $n \geq 6$ Edit: If it is hard to read what I have written the essence of my question is: How come that $2 \times 2^{k} - (k+3)^{2} \geq 2^{k}$ from the assumption that $2^{k} \geq (k+2)^{2}$?
Show that $2^{n} \geq (n +2)^{2}$ for all $n \geq 6$
I have excluded steps:
Assumpt... | If you have done the Base case and want to show that $\text{LHS}_{n+1} - \text{RHS}_{n+1} \geq 0$ you use the assumption that $2^{n} \geq (n+2)^{2}$ but also that $n \geq 6$.
In other words:
$2^{n+1} - (n+3)^{2} = 2^{n} \times 2 - (n+3)^{2}$.
Now due to the fact that $2^{n} \geq (n+2)^{2}$ it follows that if we subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $a^2 - 4a \equiv - 3 \pmod{p}$ then $a$ is congruent 1 or 3 modulo $p$
Assume $p$ is a prime and $a$ is a number such that
\begin{equation}
a^2 - 4a \equiv - 3 \pmod{p}
\end{equation}
Show that then we must either have $a \equiv 3 \pmod{p}$ or $a \equiv 1 \pmod{p}$.
If we assume $a \equiv 3 \pmod{p}$ we get ... | Hint:
$
a^2 - 4a \equiv - 3 \bmod{p} \iff a^2 - 4a + 3 \equiv 0 \bmod{p}
$.
Now $a^2 - 4a + 3=(a-1)(a-3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Diophantine equation $\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n$ Let $a,b,c$ and $n$ be natural numbers and $\gcd(a,b,c)=\gcd(\gcd(a,b),c)=1$.
Does it possible to find all tuples $(a,b,c,n)$ such that:
$$\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b} = n?$$
| Add +3 both sides
We get $(a+b+c)(\frac{ab+bc+ca}{abc})=n+3$
Since $gcd(ab+bc+ca,abc)=1$ beacuse $(a,b,c)=1$ we should have $(a+b+c)\vdots (abc)$ and then we can get solutions $(1,1,1),(1,1,2),(1,2,3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
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... | Consider using the alternating sum division rule. We need to have the sum of $5$ digits - the sum of $4$ digits to equal a number divisible by $11$. Denote the sum of $5$ digits by $O$ and the sum of the 4 digits as $E$.
Thus, we want $O - E = (45 - E) - E = 45 - 2E$ (sum of digits 1-9 is $45$) to be divisible by $11$... | {
"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": 3
} |
A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\t... | \begin{align}
\begin{vmatrix}
x^2 & y^2 & z^2 \\
(x+1)^2 & (y+1)^2 & (z+1)^2 \\
(x+2)^2 & (y+2)^2 & (z+2)^2 \\
\end{vmatrix} &=
\begin{vmatrix}
x^2 & y^2 & z^2 \\
2x+1 & 2y+1 & 2z+1 \\
4x+4 & 4y+4 & 4z+4 \\
\end{vmatrix}
\\ &=
\begin{vmatrix}
x^2 & y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
How to evaluate $\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ I have a stuck on the problem of L'Hospital's Rule,
$\lim\limits_{x\to 0} \frac{\arctan x - \arcsin x}{\tan x - \sin x}$ which is in I.F. $\frac{0}{0}$
If we use the rule, we will have
$\lim\limits_{x\to 0} \frac{\frac{1}{1+x^2}-\fra... | Let $\arcsin x = t$ so that $\sin t = x$ and $$\tan t = \dfrac{x}{\sqrt{1 - x^{2}}}$$ so that $$\arcsin x = t = \arctan\dfrac{x}{\sqrt{1 - x^{2}}}$$ and then we can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\tan x - \sin x}\notag\\
&= \lim_{x \to 0}\frac{\arctan x - \arcsin x}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How does $6^{\frac{5}{3}}$ simplify to $6\sqrt[3]{36}$? I was recently given a problem along the lines of the below:
Simplify $6^{\frac{5}{3}}$ to an expression in the format $a\sqrt[b]{c}$.
The answer, $6\sqrt[3]{36}$, was then given to me before I could figure out the problem myself. I'm wondering what the steps to... | \begin{eqnarray*}
6^{5/3} &=& \left(6^5\right)^{1/3} \\ \\
&=& \left(6 \times 6 \times 6 \times 6 \times 6\right)^{1/3} \\ \\
&=& \left(6 \times 6 \times 6\right)^{1/3} \times \left(6 \times 6\right)^{1/3} \\ \\ &=& 6 \times \left(36\right)^{1/3} \\ \\
&=& 6 \times \sqrt[3]{36}
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Supremum and infimum of asinx + bcosx I am trying to draw conclusions about the supremum and infimum of $F=\{a\sin x+b\cos x: x \in \mathbb{R}\}$ where $a,b \in \mathbb{R}$ are parameters. I am able to rewrite the function as
$$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}sinx+\frac{b}{\sqrt{a^2+b^2}}cosx\right).$$
Noti... | Your approach is fine, but you may also consider that if $f(x)=a\sin x+b\cos x$ and $a,b\neq 0$, the Cauchy-Schwarz inequality ensures
$$ |f(x)|\leq \sqrt{a^2+b^2}\sqrt{\sin^2 x+\cos^2 x} = \sqrt{a^2+b^2} $$
with equality attained at the points for which $|\tan x|=\left|\frac{a}{b}\right|$. Since $f(x)=-f(x+\pi)$, the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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combinatorics: sum of product of integer compositions I am trying to solve a problem from Stanley's book, it says:
Fix $k,n \in \mathbb{P}$. Show that:
\begin{align}
\sum a_1 a_2 \cdots a_k = \binom{n+k-1}{2k-1}
\end{align}
where the sum ranges over all compositions $(a_1 , a_2 , \ldots , a_k)$ of $n$ into $k$ parts.
I... | We have using generating functions the closed form
$$[z^n](z+2z^2+3z^3+\cdots)^k
= [z^n] \frac{z^k}{(1-z)^{2k}}
\\ = [z^{n-k}] \frac{1}{(1-z)^{2k}}.$$
Using the Newton binomial this becomes
$${n-k+2k-1\choose 2k-1} = {n+k-1\choose 2k-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Astonishing: the sum of two infinite products of nested radicals equal to $ \pi $. Recently, I found the following handwritten expression in an old math book of my family. Probably it belonged to my great grandfather Boris, who had a P.D. in mathematics.
$ \pi = \frac{4\sqrt{5}}{5}.\frac{2}{\sqrt{2 + \frac{4}{\sqrt{5}}... | Let $\theta=\arctan\frac{1}{2}$. Then $\cos\theta = \frac{2}{\sqrt{5}}$, hence $\sec\theta=\frac{\sqrt{5}}{2}$ and
$$\cos\frac{\theta}{2}=\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}},\qquad \cos\frac{\theta}{4}=\sqrt{\frac{1+\sqrt{\frac{1}{2}+\frac{1}{\sqrt{5}}}}{2}}$$
and so on. Perform a bit of maquillage, then recall that
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
"answer_id": 0
} |
need arithmetic problem proof if $a^2+b^2=a^3+b^3=a^4+b^4$
find a and b
| If $a=0$, it's easy to see that $b=0,1$. If $b=0$, by the same way, we have $a=0,1$.
Now assume $ab\neq0$, let $b=xa$ for some $x\neq0$, then we have
\begin{eqnarray*}
x^2(a^2+1)=x^3(a^3+1)=x^4(a^4+1).
\end{eqnarray*}
So we get $(a^2+1)(a^4+1)=(a^3+1)^2$, that is, $a^2(a-1)^2=0$, which implies that $a=0$ or $a=1$. By t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Need some hints for proving a logarithmic inequality. $$\frac{\log_ax}{\log_{ab}x} +
\frac{\log_b{x}}{\log_{bc}x} +
\frac{\log_cx}{\log_{ac}x} \ge 6$$
Did as you suggested and got this, im stuck again:
$$\log_ab + \log_bc + \log_ca \ge 3$$
| $$
\frac{\log_a x}{\log_{ab} x} = \frac{1/\log_x a}{1/\log_x(ab)} = \frac{\log_x (ab)}{\log_x a} = \log_a(ab) = \log_a a + \log_a b = 1 + \log_a b.
$$
Therefore
\begin{align}
& \frac{\log_ax}{\log_{ab}x} + \frac{\log_b{x}}{\log_{bc}x} + \frac{\log_cx}{\log_{ac}x} = (1+ \log_a b) + (1+ \log_b c) + (1+ \log_c a) \\[10pt]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Sum of a finite geometric series $$\frac{1}{4} V^\frac{1}{4} + \frac{1}{4} V^\frac{1}{2} + \cdots + \frac{1}{4} V^{n-\frac{1}{4}} + \frac{1}{4} V^n =$$
$$\frac{1}{4} V^\frac{1}{4} \bigg(1 + V^\frac{1}{4} + V^\frac{1}{2} + \cdots + V^{n-\frac{1}{4}}\bigg) =$$
$$\frac{1}{4} V^\frac{1}{4} \bigg(\frac{1-V^{\big(n-\frac{1}{... |
We obtain
\begin{align*}
\frac{1}{4}&v^{\frac{1}{4}}\left(1+v^{\frac{1}{4}}+v^{\frac{1}{2}}+\cdots+v^{n-\frac{1}{4}}\right)\\
&=\frac{1}{4}v^{\frac{1}{4}}\left(1+v^{\frac{1}{4}}+v^{\frac{2}{4}}+\cdots+v^{\frac{4n-1}{4}}\right)\\
&=\frac{1}{4}v^{\frac{1}{4}}\left(1+v^{\frac{1}{4}}+\left(v^{\frac{1}{4}}\right)^2+\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}=-\frac{1}{3}$ Evaluate the following limit
$$\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}$$
From wolfram alpha, answer is $-\frac{1}{3}$. From here, we obtain the answer by using either Taylor se... | Use substitution $y:=\frac{1}{x}.$ Then $x\to+\infty \implies y\to0.$ So, $$\lim_{n\to+\infty}\Biggl(n^2-\frac{1}{\sin^2\bigl(\frac{1}{n}\bigr)}\Biggr)=\lim_{y\to0}\Biggl(\frac{1}{y^2}-\frac{1}{\sin^2(y)}\Biggr) = \lim_{y\to0}\Biggl(\frac{\sin^2(y)-y^2}{y^2\sin^2(y)}\Biggr).$$ Applying L'Hopital rule 4 times yields the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How do i find the inverse of a 3 x 3 matrix? $$
A= \begin{bmatrix}
2 & 1 & -1 \\
-2 & -2 & 1 \\
0 & -2 & 1 \\
\end{bmatrix}
$$
Can someone show me the best way to approach this? Should I use pivoting? I tried using the formula, but I think that only works for 2 x 2 matrices.
| As an easy-to-understand process, you can note that $A.A^{-1} = I$ and then undertake parallel row operations on $A$ and $I$ to transform this into $I.A^{-1}=X$, where $X$ is the result of the same operations on $I$ that transformed $A$ into $I$. You can check, through considering the action of matrix multiplication, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the number of real roots of given equation.
*
*$x^9+x^7+x^5+x^3+x+1=0$
*$x^3+x-1=0$
I know that the complex roots exist in pairs and also that the number of distinct roots has to do something with change in sign of the equation, but how?
How should i proceed in such type of questions?
| for the equation $x^9+x^7+x^5+x^3+x+1=0$ we take the function f(x)=$x^9+x^7+x^5+x^3+x+1$...now see f ' (x)=$9x^8+7x^6+5x^4+3x^2+1$ >0 for all x$\in$R. So f(x) is stricktly increasing for all x$\in$R...so it wil obviously cut the X axis at exactly one point.so the equation will have only one real root as f(x) is contin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Trigonometric limit I couldn't find this limit , can someone help me?
$$\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}$$
| Use identities:
$$
\sin\theta -\sin\varphi=2\sin\left(\theta-\varphi\over2\right)\cos\left(\theta+\varphi\over2\right)\\
\arctan x-\arctan y=\arctan\left(x-y\over1+xy\right)
$$
Then,
$$
\lim_{x \to \infty} \frac{\arctan(x+1) - \arctan(x)}{\sin\left(\frac{1}{x+1}\right) - \sin\left( \frac 1x\right)}=
\lim_{x \to \infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate exponential limit involving trigonometric functions Calculate the following limit:
$$\lim_{x \rightarrow 0} \left( \frac{\tan x}{x} \right) ^ \frac{1}{\sin^2 x}$$
I know the result must be $\sqrt[3]{e}$ but I don't know how to get it. I've tried rewriting the limit as follows:
$$\lim_{x \rightarrow 0} e ^ {\l... | To evaluate
$$
\lim_{x\rightarrow 0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{\sin^2(x)}}
$$
we first observe that $\lim_{x\rightarrow 0}\frac{\tan(x)}{x}=\lim_{x\rightarrow 0}\frac{1}{\cos(x)}\frac{\sin(x)}{x}$. Both of these factors approach $1$ as $x$ approaches $0$. On the other hand, since $\sin(x)$ approaches $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1995940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proof that $\prod_{k=1}^{n-1}(1+\frac1k)^k = \frac{n^n}{n!}$ for all $n \in \Bbb N \ge 2$ I've tried to prove this for a while now, but I can't get it:
$\prod_{k=1}^{n-1}(1+\frac1k)^k = \frac{n^n}{n!}$ for all $n \in \Bbb N \ge 2$
Solution:
$\prod_{k=1}^{(n+1)-1}(1+\frac1k)^k=\frac{(n+1)^{(n+1)}}{(n+1)!}$
$\left(1+\fra... | First, we can write out terms of the product as
$$\prod_{k=1}^{n-1}\left(1+\frac1k\right)^k=\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)^3\cdots \left(\frac{n-2}{n-3}\right)^{}\left(\frac{n-1}{n-2}\right)^{n-2}\left(\frac{n}{n-1}\right)^{n-1}$$
Then, note that the cancellation of numera... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Let $m$ and $n$ be positive integers such that $\gcd(m,n)=1.$Compute $\gcd(5^m+7^m,5^n+7^n)$. Problem: Let $m$ and $n$ be positive integers such that $\gcd(m,n)=1.$Compute $\gcd(5^m+7^m,5^n+7^n)$.
My Attempt: I wrote a simple computer program and deduced that $\gcd(5^m+7^m,5^n+7^n)=2$ if $m+n$ is odd and $\gcd(5^m+7^... | Here's a simple case. See if you can generalize it:
$\gcd (5^3+7^3,5^7+7^7)=$
$\gcd (5^3+7^3,5^7+7^7-7^4 (5^3+7^3))=$
$\gcd (5^3+7^3,5^7-5^3*7^4)=$ note: $5\not \mid 5^3+7^3$ so
$\gcd (5^3+7^3,5^4-7^4)=$
$\gcd (5^3+7^3,5^4-7^4+7 (5^3+7^3))=$
$\gcd (5^3+7^3,5^4+5^3*7)=$
$\gcd (5^3+7^3,5+7)=$
$\gcd (5^3+7^3-7^2 (5+7),5+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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For what $a,b,c,d$ are $A$ and $B$ similar?
Consider the following two matrices with complex entries.
$$A = \begin{pmatrix} 1&a&0\\ 0&2&b \\ 0&0&2 \end{pmatrix}, B=\begin{pmatrix} 2&d&c \\ 0&c&d \\ 0&0&2 \end{pmatrix}$$
When are they similar?
My progress: First we see that $c$ has to be equal to $1$ if $A$ is similar... | You have made a mistake. It should be $1+d^2$ instead of $1+d$.
Matrices are similar iff they have the same Jordan form. The eigenvalues are $\{1,2,2\}$. The eigenvalue $1$ has, of course, a single scalar Jordan block, so the Jordan block for $2$ must have the same size for both. Since $\operatorname{rank}(B-2I)=2$ (as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The variable that is going to enter the basis with revised Simplex already is in it I am looking to solve the following program using the revised Simplex method :
$$
\begin{cases}
\begin{aligned}
\max \ & x_1&+ x_2\\
&x_1&-x_2&\le 1\\
&-2x_1&-x_2&\le -2\\
&-2x_1&+x_2&\le 2\\
\forall i, x_i
\end{aligned}
\end{cases}
$$
... | for $J_0=\{x_2;x_3;x_5\}$, the determinant is
$$\begin{vmatrix}
-1&1&0\\
-1&0&0\\
1&0&1
\end{vmatrix}=-1$$ this base is feasible.
Yet, for $J_1=\{x_2;x_4;x_5\}$ $$\begin{vmatrix}
-1&1&0\\
-1&0&1\\
1&0&0
\end{vmatrix}=0$$ It is therfore unfeasible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the largest possible integer n such that $\sqrt{n}+\sqrt{n+60}=m$ for some non-square integer m. The solution to this problem says that:
Squaring both sides gives us that $n(n+60)$ is a perfect square
I do not understand this, can someone explain me why this is true.
squareing gives:
$2n+60+2\sqrt{n}\sqrt{n+60}=m... | Obviously both $n$ and $n+60$ should be squares so its product $n(n+60)$ is a square. On the other hand, $n$ cannot be greater than some integer because if $n=x^2$ and $n+60=y^2$ then making $y=x+h$ we must have $2xh+h^2= 60$ which clearly put a bound for $x$.
This bound is easy to calculate and one has $x\lt 30$ beca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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What is the limit $\lim_{n\to\infty}\sqrt[n]{n^2+3n+1}$? Can you try to solve it? I tried to do something but I do not know how to continue it:
\begin{align}
& \lim_{n\to\infty}\sqrt[n]{n^2+3n+1}=\lim_{n\to\infty}(n^2+3n+1)^{1/n} = e^{\lim_{n\to\infty} \frac{1}{n}\ln(n^2+3n+1)} \\[10pt]
= {} & e^{\lim_{n\to\infty}\frac... | Hint: $$ \frac{1}{n} \ln \left( n^2+3n+1 \right) = \frac{1}{n} \ln \left( n \left(n+3+\frac{1}{n} \right) \right) = \frac{1}{n} \ln n + \frac{1}{n} \ln \left( n+3+\frac{1}{n} \right) $$
Write in the last expression:
$$ \frac{1}{n} = \frac{1}{n}-\frac{1}{n+3+\frac{1}{n}}+\frac{1}{n+3+\frac{1}{n}}= \frac{1}{n^2} \frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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CCRT = Constant case CRT: $\,x\equiv a\pmod{\! 2},\ x\equiv a\pmod{\! 5}\iff x\equiv a\pmod{\!10}$ Problem: Find the units digit of $3^{100}$ using Fermat's Little Theorem (FLT).
My Attempt: By FLT we have $$3^1\equiv 1\pmod2\Rightarrow 3^4\equiv1\pmod 2$$ and $$3^4\equiv 1\pmod 5.$$ Since $\gcd(2,5)=1$ we can multiply... | The phrase ‘Since gcd(2,5)=1 we can multiply the moduli’ is not clear at all. I would rather say something like ‘since $3^4\equiv 1\mod 2$ and $\bmod5$, we have $3^4\equiv 1\mod \operatorname{lcm}(2,5)=10$’ by the Chinese remainder theorem.
That said, why make things more complicated than they are?
$3^2\equiv -1\mod 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Solve for $x$ $2\log_4(x+1) \le 1+\log_4x$ $2\log_4(x+1) \le 1+\log_4x$
I did:
$$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$$
Using the quadrati... | $(x+1)^2 \ne x^2+1$, in general. If you fix that error the equation is now $x^2-2x+1 \le 0 \iff (x-1)^2 \le 0 \iff x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$
My attempt
\begin{align*}
&=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1... | Start by writing (all symbols $\sim$ are taken as $x \to +\infty$)
$$
\log(1+x) = \log x + \log \left( 1+ \frac{1}{x} \right) \sim \log x + \frac{1}{x} - \frac{1}{2x^2} + \frac{1}{3x^3}.
$$
Hence
$$
x \log (1+x) - x\log x + \arctan \frac{1}{2x} \sim 1 - \frac{1}{2x} + \frac{1}{3x^2} + \frac{1}{2x} - \frac{1}{24 x^3} \s... | {
"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": 0
} |
A polynomial of degree 3n has the value 2 at $0, 3, 6, \dots , 3n$, ... A polynomial of degree $3n$ has the value 2 at $0, 3, 6, \dots , 3n$, the value 1 at $1, 4, 7, \dots , 3n-2$ and the value 0 at $2, 5, 8, \dots , 3n-1$. Its value at $3n+1$ is $730$. What is $n$?
I've tried lots of approaches to this but I can't s... | By using Newton Forward Divided Difference Formula, the polynomial $P_n$ has fractional coefficients and it is equal to:
$$P_n(x)=\sum_{k=0}^{3n}\binom{x}{k}\Delta^kP_n(0).$$
Hence
\begin{align*}
&P_1(x)=2-x+3\binom{x}{3}\\
&P_2(x)=P_1(x)-9\binom{x}{4}+18\binom{x}{5}-27\binom{x}{6}\\
&P_3(x)=P_2(x)+27\binom{x}{7}-81\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How does this expression demonstrate mathematical telescoping? What exactly *is* telescoping? Here is the expression:
$x^n-y^n$
According to my professor, this is factorable, and an example for telescoping in mathematics. I've gleaned some sense of what Telescoping is from this answer: Mathematical Telescoping. Could s... | The way telescoping works is really simple. As an example I will show you how one can derive the formula to calculate the sum of the first $n$ terms of a geometric series.
We want to find $S_n = 1 + x + x^2 + x^3 + \cdots + x^n$
We first compute $xS_n = x + x^2 + x^3 + \cdots + x^n + x^{n+1}$
And now we calculate $xS_n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of Maclaurin series Find the sum of the infinite series
\begin{equation}
\sum_{n=2}^\infty\frac{7n(n-1)}{3^{n-2}}
\end{equation}
I think it probably has something to do with a known Maclaurin series, but cannot for the life of me see which one.. Any hints would be appreciated!
Edit: Using your hints, I was able to ... | You want
$\sum_{n=2}^{\infty} n(n-1) x^n
$
for a certain value of $x$.
Start with
$\sum_{n=2}^{\infty} x^n
=\dfrac{x^2}{1-x}
$
and differentiate it twice.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
What is the explanation for this visual proof of the sum of squares? Supposedly the following proves the sum of the first-$n$-squares formula given the sum of the first $n$ numbers formula, but I don't understand it.
| The first row of the first triangle is $1 = 1^2$, the second row sums to $2 + 2 = 2^2$, the third row sums to $3 + 3 + 3 = 3^2$ and so on. That means that the sum of the numbers in the triangle is $1^2 + 2^2 + 3^2 + \dots + n^2$. Now, the second and third triangles are the same, so the left-hand side is $3(1^2 + 2^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2014512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 4,
"answer_id": 0
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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... | The range of the function is $0 < f \le 3/4$.
You already figured out the maximum value of $f$ so I will not show the solution for that part.
As angle $A$ gets closer and closer to $0$ degrees, the value of $\sin(A)$ tends to $0$. $\sin(C)$ will tend towards a finite constant, since $C$ is going towards $120$ degrees. ... | {
"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": 2
} |
proving $ a^{8} - 1 = \left(a^{2} - 1\right)\left(a^{2} + 1\right) \left(a^{2}+\sqrt{\,2\,}a+1\right) \left(a^{2} - \,\sqrt{\,2\,}\,a +1\right) $
*
*For all real numbers it is true that
$$
a^{8} - 1 = \left(a^{2} - 1\right)\left(a^{2} + 1\right)
\left(a^{2} + \,\sqrt{\,2\,}\,a + 1\right)
\left(a^{2} - \,\sqrt{\,2\,}\... | \begin{align}
(a^8-1) &= (a^4-1)(a^4+1)
\\
&= (a^2-1)(a^2+1)(a^4+2a^2+1 - 2a^2)
\\
&= (a^2-1)(a^2+1)((a^2+1)^2 - 2a^2)
\\
&= (a^2-1)(a^2+1)((a^2+1) - \sqrt{2}a)((a^2+1) + \sqrt{2}a)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$. Find three distinct triples (a, b, c) consisting of rational numbers that satisfy $a^2+b^2+c^2 =1$ and $a+b+c= \pm 1$.
By distinct it means that $(1, 0, 0)$ is a solution, but $(0, \pm 1, 0)$ counts as... | $\frac{6}{7}, \frac{3}{7}, -\frac{2}{7}$ is the third solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do i find the roots of this polynomial equation? The polynomial equation is: $x^4-5x^3+5x^2+5x-6=0$.
How do i simplify this equation so that i can find its roots.
Please, can anyone teach me how to find roots of equations of degree 4 and degree 3.
| .The key is first substituting integral small values of $x$, which are factors of the constant term, which is $6$. In this case, we try to substitute $1$. We see that if $f(x) = x^4-5x^3+5x^2+5x-6$, then: $f(1)=0$ and $f(-1)=0$. Hence, $x-1$ and $x+1$ are factors of $x^4-5x^3+5x^2+5x-6$.
Now perform polynomial divisi... | {
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"url": "https://math.stackexchange.com/questions/2020951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
} |
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 ... | \begin{align*}
b+ai &= e^{i\alpha}+e^{i\beta} \\[5pt]
b-ai &= e^{-i\alpha}+e^{-i\beta} \\[5pt]
\frac{b+ai}{b-ai} &=
\frac{e^{i\alpha}+e^{i\beta}}{e^{-i\alpha}+e^{-i\beta}} \\[5pt]
\frac{(b+ai)(b+ai)}{(b-ai)(b+ai)} &=
\frac{e^{i(\alpha+\beta)}(e^{i\alpha}+e^{i\beta})}
{e^{i(\alpha+\beta)}(e^{-i\alpha}... | {
"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": 0
} |
How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1... | $$(c^2+s^2)^3=c^6+3c^4s^2+3c^2s^4+s^6=c^6+s^6+3c^2s^2(c^2+s^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 11,
"answer_id": 9
} |
Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$... | After your $" to be proven "$ step, follow the rules below:
1. Take $ (k+1)^{2} $ common so we have then $ (k+1)+\frac {k}{4} ^{2} $.
2. Then simplify the result to get $ \frac { k^{2}+4k+4}{4} $.
3.This is equivalent to $\frac {(k+2)}{2}^{2} $
4. Proof finished as we have our result equal to $ \frac {(k+1)(k+2)}{2} ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Prove: If $p$ is prime and $ (a,p)=1 $, then $\ 1 + a + a^2 + ... + a^{p-2}=0$(mod $p$) Prove: If $p$ is prime and $ (a,p)=1 $, then $\ 1 + a + a^2 + ... + a^{p-2}≡0 \pmod p$
As an example, for $a=2$ and $p=5$, then:
$$1+2+2^2+2^3=1+2+4+8=15≡0 \pmod 5$$
This can also be written as:
$$1 \pmod 5 + 2 \pmod 5 + 4 \pmod 5 +... | Notice that Fermat's Little Theorem tells us that if $p$ is prime and $\gcd(a,p)=1$, then
$$a^{p-1} \equiv 1 \bmod p$$
Clearly $a^{p-1}-1 \equiv 0 \bmod p$. Recall that $x^n-1 = (x-1)(1 + x + x^2 + \cdots + x^{n-1})$, and so
$$a^{p-1}-1 = (a-1)(1+a+a^2 + \cdots + a^{p-2})$$
We know that $a^{p-1}-1 \equiv 0 \bmod p$, i.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Finding the M.G.F of product of two random variables. We are given two independent standard normal random variables $X$ and $Y$. We need to find out the M.G.F of $XY$.
I tried as follows :
\begin{align}
M_{XY}(t)&=E\left(e^{(XY)t}\right)\\&=\int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{(XY)t}f_X(x)f_y(y)dxdy \\
&=... | Careless mistake at second last line:
\begin{align}
\dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-t^2y^2}{2}}e^{\frac{-y^2}{2}}dy &= \dfrac{1}{\sqrt{2\pi}}\int_{- \infty}^{\infty}e^{\dfrac{-(t^2+1)y^2}{2}}dy\\
&=\frac{1}{\sqrt{t^2+1}}
\end{align}
Edit:
There is actually a mistake earlier. Thanks, tmrlvi for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrating Logs Looking at: $$\int\frac{\sqrt{x}}{\sqrt{x}-3}$$
I use these steps to find the integral of the equation
$$\int\frac{\sqrt{x}-3+3}{\sqrt{x}-3}$$
$$\int\frac{\sqrt{x}-3}{\sqrt{x}-3}+3\int\frac{1}{\sqrt{x}-3}$$
Now substituting $u=\sqrt{x}-3$ and $du=\frac{1}{2\sqrt{x}}$
$$\int1+3(2)\int\frac{u+3}{u}$$
$$\... | \begin{align}
u & = \sqrt x - 3 \\[10pt]
u - 3 & = \sqrt x\\[10pt]
(u-3)^2 & = x \\[10pt]
2(u-3)\,du & = dx \\[10pt]
\int \frac{dx}{\sqrt x - 3} & = \int \frac{2(u-3)\,du} u = 2 \int \left( \frac u u - \frac 3 u \right) \, du = 2\int\left( 1 - \frac 3 u \right)\, du = \text{etc.}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2026602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
heterogeneous recurrence with f(n) as constant How to solve this $s_{n+1}=4s_{n-1}-3s_n+5$ where f(n)=5 conditions $s_0=-3$ $s_1=3$
I calculated the general solution $s_n=c_1*(-4)^n+c_2*1^n$ of this recurrence. The roots are $q_1=-4$ and $q_2=1$ but I have problem with particular solution with method of prediction .
I ... | $$s_{n + 2} = -3~s_{n + 1} + 4~s_{n} + 5$$
$$s_{n + 3} = -3~s_{n + 2} + 4~s_{n + 1} + 5$$
So
$$s_{n + 3} - s_{n + 2} = (-3~s_{n + 2} + 4~s_{n + 1} + 5) - (-3~s_{n + 1} + 4~s_{n} + 5)$$
$$s_{n + 3} = -2~s_{n + 2} + 7~s_{n + 1}$$
The roots are $-4$, $1$ and $1$, so the final form is:
$$s_n = A~(-4)^n + B~1^n + C~n~1^n$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a+b+c=3$, find the greatest value of $a^2b^3c^2$.
If $a+b+c=3$, and $a,b,c>0$ find the greatest value of $a^2b^3c^2$.
I have no idea as to how I can solve this question. I only require a small hint to start this question. It would be great if someone could help me with this.
| Questions like these are about tricks. Here's one you should remember.
Rewrite $a+b+c = 3$ as $2\frac{a}{2} + 3\frac b3 + 2\frac c2 =3$. Use the AM-GM inequality:
$$
\frac{2\frac{a}{2} + 3\frac b3 + 2\frac c2}{7} \geq \sqrt[7]{\frac{a^2b^3c^2}{4 \cdot 27 \cdot 4}}
$$
So we simplify:
$$
a^2b^3c^2 \leq 432 \left(\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
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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... | The system is not always solvable. The two first equations factor as
$$\tag{1} \begin{pmatrix} \cos\psi & \sin\psi \\ \sin\psi & -\cos\psi \end{pmatrix}
\begin{pmatrix} \sin\theta \\ \cos\theta \sin\phi \end{pmatrix}
= \begin{pmatrix} x \\ y \end{pmatrix} $$
where the first factor is just a reflection in a line through... | {
"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": 0
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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... | The general formula you gave is false. First notice the determinant you gave for $n=3$ is wrong, the first minor you computed has value $2\cdot 2- 1=3$, so it is in fact $6-2=4$. This shows the importance of checking the initial values correctly when using induction, you could be led to different answers.
A thing that ... | {
"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": 1
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Number of integral values of $x\in (\frac{\pi}{2},5\frac{\pi}{2})$ such that $\max(x^2+\cos x-1,1-x^2-\cos x)=\max(\cos x+x^2-1,x^2-1-\cos x)$ The number of integral values of $x\in(\frac{\pi}{2},5\frac{\pi}{2})$ such that
$$\max(x^2+\cos x-1,1-x^2-\cos x)=\max(\cos x+x^2-1,x^2-1-\cos x)$$ is satisfied is... ?
I unders... | For $x\in (\frac{\pi}{2},\frac{5}{2}\pi)$, we have that
$$\begin{align}(x^2+\cos x-1)-(1-x^2-\cos x)&=2(x^2+\cos x-1)\\\\&\ge 2(x^2-2)\\\\&\gt 2\left(\left(\frac{\pi}{2}\right)^2-2\right)\\\\&=\frac{1}{2}(\pi^2-8)\\\\&\gt 0\end{align}$$
and that
$$(\cos x+x^2-1)-(x^2-1-\cos x)=2\cos x\begin{cases}
\lt 0 & \text{if}\ x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help with $\int \cos^6{(x)} \,dx$ Problem:
\begin{eqnarray*}
\int \cos^6{(x)} dx \\
\end{eqnarray*}
Answer:
\begin{eqnarray*}
\int \cos^4{(x)} \,\, dx &=& \int { \cos^2{(x)}(\cos^2{(x)}) } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=& \int { \frac{(1+\cos(2x))^2}{4} } \,\, dx \\
\int \cos^4{(x)} \,\, dx &=&
\int { \fra... | If you're going to use lots of trig identities and repeatedly apply integration by parts, we could've more quickly undone the entire problem by using Chebyshev polynomials of the first kind, reducing it down to some basic integral. This method took me about 5 steps to do (minus simplifying):
$$\int32\cos^6(x)dx=\int\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1. Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1. I tried factoring this expression but couldn't get very far. It is simple to show for even n but odd n was more difficult, at least for me.
| The first step is to make explicit the algebraic dependencies among $\,3^{2n},4^n,6^n.\,$ Clearly all can be expressed in terms of the multiplicatively independent basis $\, x = 3^n\,$ and $\,y = 2^n\,$ as follows
$$\begin{align}
&3^{2n} = (3^n)^2 = x^2\\
&4^n =\ (2^n)^2 = y^2\\
&6^n = 3^n 2^n = xy\end{align}$$
Rewri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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... | I would go to polar coordinates, substituting $x=r\cos\theta$ and $y=r\sin\theta$. Then $\lVert(x,y)\rVert\rightarrow\infty$ becomes simply $r\rightarrow\infty$
$$f(r,\theta)=r^4(\cos^4\theta+\sin^4\theta)-3r^2\sin\theta\cos\theta$$
This is a polynomial in $r$, the coefficient of $r^4$ is positive (cannot have both $\s... | {
"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": 0
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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.
| Hint: let $x=y+1$, expand, regroup and collect, then the equation becomes:
$$y^4-y^3+y^2-y+1=0$$
Multiplying by $y+1 \ne 0$ gives $y^5 + 1 = 0$ having as solutions the $5^{th}$ roots of $-1$.
| {
"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": 0
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Find the sum of $\displaystyle\sum_{n=3}^\infty \frac{2^n-1}{3^n}$ My work:$$\sum_{n=3}^\infty \frac{2^n-1}{3^n}$$
$$a_1 = \frac{2^3-1}{3^3}$$
$$a_1 = \frac{7}{27}, r=\frac{2}{3}$$
$$S_N=\frac{\frac{7}{27}}{1-\frac{2}{3}} = \frac{7}{9}$$
The correct answer is $\frac{5}{6}$
| Strictly for fun, note that
$$2^n-1=(2-1)(2^{n-1}+2^{n-2}+\cdots+2^{n-n})=\sum_{k=1}^n{2^n\over2^k}$$
so that
$$\sum_{n=1}^\infty{2^n-1\over3^n}=\sum_{n=1}^\infty\sum_{k=1}^n\left(2\over3\right)^n{1\over2^k}=\sum_{k=1}^\infty\sum_{n=k}^\infty{1\over2^k}\left(2\over3\right)^n=\sum_{k=1}^\infty{1\over2^k}{(2/3)^k\over1-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the determinant using row operations. So I have to find the determinant of $\begin{bmatrix}3&2&2\\2&2&1\\1&1&1\end{bmatrix}$ using row operations. From what I've learned, the row operations that change the determinate are things like swaping rows makes the determinant negative and dividing a row by a value mean... | Why divide, when simple addition and subtraction will do? Subtract twice the bottom row from the middle and top rows and you are left with
$$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 0 & - 1 \\ 1 & 1 & 1\end{vmatrix} = 1-0+0=1$$
by evaluating by minors.
Even easier is if you take that matrix and add row 2 to row 3 and subtract ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How many solutions does $x^2 + 3x +1 \equiv 0\, \pmod{101}$ have? $x^2 + 3x +1 \equiv 0 \pmod{101}$. To solve this I found the determinant $D = 5 \pmod{101}$). Using the Legendre symbol,
$$\left(\frac{5}{101}\right) = \left(\frac{101}{5}\right) \equiv \left(\frac{1}{5}\right) \equiv 1,$$
$\therefore$ The equations have... | For prime $p$: If $A$ is a square modulo $p,$ take any B such that $A\equiv B^2\pmod p.$ Then $$x^2\equiv A \pmod p\iff (x-B)(x+B)\equiv x^2-B^2\equiv 0 \pmod p\iff$$ $$\iff (\;p|(x-B)\lor p|(x+B)\;)\iff x\equiv \pm B \pmod p.$$ If $A\not \equiv 0 \pmod p,$ then $B\not \equiv 0\pmod p.$ If $p\ne 2,$ and $B\not \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Calculating the convergence radius of a power series I've tried to calculate the convergence radius of the following power series:
$$\sum_{n=1}^{\infty}\frac{3^n+4^n}{5^n+6^n}x^n$$
The Cauchy–Hadamard theorem doesn't help in this situation (I think).
So what I did is I tried to apply the d'Alembert ratio test to it and... | I'd go about like this:
$$\lim_{n\to\infty}\frac{\frac{3^n+4^n}{5^n+6^n}}{\frac{3^{n+1}+4^{n+1}}{5^{n+1}+6^{n+1}}}=\lim_{n\to\infty}\frac{(3^n+4^n)(5^{n+1}+6^{n+1})}{(5^n+6^n)(3^{n+1}+4^{n+1})}$$
$$=\lim_{n\to\infty}\frac{((\frac{3}{4})^n+1)(5\cdot (\frac{5}{6})^n+6)}{((\frac{5}{6})^n+1)((\frac{3}{4})^n\cdot 3+4)}$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the last digit of $3^{1006}$ The way I usually do is to observe the last digit of $3^1$, $3^2$,... and find the loop. Then we divide $1006$ by the loop and see what's the remainder. Is it the best way to solve this question? What if the base number is large? Like $33^{1006}$? Though we can break $33$ into $3 \time... | You have
\begin{cases}3^1& =3\\ 3^2& =9 \\ 3^3&=27\\3^4&= 81\\3^5&= 243\end{cases} Thus the last digit repeats after $4$ steps. Since $1006=251\cdot 4+2$ it is
$$3^{1006}=(3^4)^{251}\cdot 3^2.$$ The last digit of $(3^4)^{251}$ is $1$ and the last digit of $3^2=9.$ So the answer you are looking for is $9.$
To compute ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
What is $\tan ^{-1} (5i/3)$
What is $\tan ^{-1} (5i/3)$
My progress: Let $\tan x= \dfrac{5i}{3}= \dfrac{\sin x}{\cos x}$
I tried using $\sin x= \dfrac{e^{ix}-e^{-ix}}{2}, \cos x= \dfrac{e^{ix}+e^{-ix}}{2}$ to show that $\dfrac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}= \dfrac{-5}{3}$ or $e^{2ix}= \dfrac{-1}{4}$, but I'm stuck ... | Let us write $z =\frac{5i}{3}$ and also $w =\arctan (\frac{5i}{3}) =\arctan z$. Hence $\tan w=z$ and we want to solve for $w$. We have $\sin w= \frac{e^{iw} -e^{-iw}}{2i}$ and $\cos w=\frac{e^{iw}+e^{-iw}}{2}$. So then we have $\tan w=\frac{e^{iw}- e^{-iw}}{i(e^{iw}+e^{-iw})}$.
Let $p =e^{iw}$ so that $\frac{1}{p} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Max. and Min. of $a\sin^2 x+b \cos^2 x$ $f(x)=a\sin^2 x+b \cos^2 x$
It is stated that, pls check link
If a > b, Maximum value = a and Minimum value = b
If a < b, Maximum value = b and Minimum value = a
How do we find this?
My attempt:
$$
f(x)=a\sin^2 x+b \cos^2 x=a\sin^2 x+b (1-\sin^2 x)=a\sin^2 x+b-b\sin^2 x\\=\sin^2 ... | $$f(x)=(a-b)\sin^2(x)+b$$
Once $0 \le\sin^2(x) \le 1$ the $f(x)$ has limits $a$ and $b$.
If $a<b$ then $f_{max}=b$ and $f_{min}=a$
If $b<a$ then $f_{max}=a$ and $f_{min}=b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Alternative way of calculating the multiplicative inverse Suppose I want to find the inverse of $7^{-1} \equiv 1 \mod 19 $, is there a quick way to do instead of extended euclidean algorithm? (Assume the numbers are not large).
Thank you
| I sometimes look at the two multiples that straddle the modular base.
$\color{blue}{2}\cdot 7 = 14 \equiv \color{green}{-5} \bmod 19 \\
\color{orange}{3} \cdot 7 = 21 \equiv \color{red}{2} \bmod 19 $
And then combine those to get the inverse:
$1\cdot \color{green}{-5} + 3\cdot \color{red}{2} = 1 \\
\implies (1\cdot \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On $119^2+120^2=13^4\,$ and $p=239$ We are familiar with, $$\frac\pi4=4\arctan\tfrac15-\arctan\tfrac1{239}$$Let $p=a+b=239$ and $(a,b,c,d)=(120,119,13,2).\,$ Some years back, I observed this rather long list of Diophantine relations and Pell equations that they satisfied,
$$120^2-119^2 = 239$$
$$120^2 + 119^2 = 13^4$$
... | Another relation is:
$$(2\cdot2)^4-2^4-239=1$$
This implies:
$$239\cdot13^4 + 13^4 + (2\cdot13)^4 = (2\cdot2\cdot13)^4$$
which, using the first two relations on your list, gives:
$$(120^2-119^2)(120^2+119^2)+13^4+(2\cdot13)^4=(2\cdot2\cdot13)^4$$
which is equivalent to the last relation on your list.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
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}}\... | You can consider system \begin{align} (\sqrt 3 - 1)\cos x+(\sqrt 3+ 1)\sin x &= 2\\ \cos^2 x + \sin^2 x &= 1\end{align} Geometrically, you are looking for intersection of a line and unit circle, so we are expecting at most two solutions (see this plot). Thus, we can safely square the first equation and eliminate extra ... | {
"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": 1
} |
Curious limits with tanh and sin These two limits can be easily solved by using De l'Hopital Rule multiple times (I think), but I suspect that there could be an easier way... Is there?
\begin{gather}
\lim_{x\to 0} \frac{\tanh^2 x - \sin^2 x}{x^4} \\
\lim_{x\to 0} \frac{\sinh^2 x - \tan^2 x}{x^4}
\end{gather}
Thanks... | Taylor expansions
$$\tanh^2x = x^2-\frac{2x^4}{3}+o\left(x^6\right)$$
$$\sin^2x = x^2-\frac{1}{3}x^4+ o (x^6)$$
$$\lim _{x\to \:0}\:\frac{\tanh ^2\:x\:-\:\sin ^2\:x}{x^4} = \lim _{x\to \:0}\:\frac{x^2-\frac{2x^4}{3}\:-\:\left(x^2-\frac{1}{3}x^4\right)+o(x^6)}{x^4} = \color{red}{-1/3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Graph $\text{Im}\left(\frac{1}{z}\right)=1$ I used the identity $z=x+iy$ which resulted in $\text{Im}\left(\frac{1}{x+iy}\right)$. Multiplying by the conjugate, I found that this was equal to $\text{Im}\left(\frac{x-iy}{x^2+y^2}\right)$, which by splitting this fraction into two terms is $\frac{-y}{x^2+y^2}=1$. Multipl... | Since $\operatorname{Im}z = \frac{1}{2i}(z-\bar z)$ the relation can be rewritten with equivalences at each step:
$$
\frac{1}{z}-\frac{1}{\bar z} = 2i \\
\bar z - z = 2i z \bar z \\
z \bar z-\frac{i}{2}z + \frac{i}{2}\bar z = 0\\
\left(z+\frac{i}{2}\right)\left(\bar z - \frac{i}{2}\right) = \frac{1}{4} \\
\left|z+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Some questions on the ring $(\mathbb{F}_3,+,\cdot)$
*
*Find an irreducible polynomial in $\mathbb{F}_3$ of degree 2.
The polynomials to analyse are $9$ in the form:
$X^2+aX+b$ where $a,b \in \mathbb{F}_3$
I find out that the irreducible ones are:
$$X^2+1$$
$$X^2+X+2$$
$$X^2+2X+2$$
*For the polynomial $p(X)$ found... | Your process for finding the multiplicative inverse is wrong. In particular, in the presence of the ideal $\langle X^2+1\rangle$, your implication that $a=0$ (for instance) is completely wrong - for instance, $X^3+X^2+1 + \langle X^2+1\rangle = 1+\langle X^2+1\rangle$.
You're better off first showing that every polynom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Manipulating a version of summations to produce the closed formula for the Catalan Numbers. Manipulate $g(x) = \frac{1}{2x}\left(1 - \sum_{k=0}^{\infty}{1/2 \choose k}(-4x)^k\right)$ into $g(x) = \sum_{n=0}^{\infty}{1/2 \choose n+1}(-1)^n2^{2n+1}x^n$
I know that I should substitute n + 1 for k to allow the summation to... | $$\begin{align*}
\frac1{2x}\left(1-\sum_{k\ge 0}\binom{1/2}k(-4x)^k\right)&=\frac1{2x}\left(1-\sum_{k\ge 0}\binom{1/2}k(-1)^k2^{2k}x^k\right)\\
&=\frac1{2x}\left(1-\sum_{k\ge -1}\binom{1/2}{k+1}(-1)^{k+1}2^{2k+2}x^{k+1}\right)\\
&=\frac1{2x}-\sum_{k\ge -1}\binom{1/2}{k+1}(-1)^{k+1}2^{2k+1}x^k\\
&=\frac1{2x}+\sum_{k\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How does this surface looks like? The formula is $({x^4 \over{a^4}} + {y^2 \over{b^2}} + {z^2 \over{c^2}})^2 = {x^2 \over{p^2}}$, where $a,b,c,p$ is positive. The left part is pretty similuar to ellipsoid, but there are few differences. I also have tried to use the difference of two squares and brackets expansion, but ... | As a first step, move the right-hand side to the left side, and use the conjugate rule. You will get
$$
\Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+\frac{x}{p}\Bigr)\cdot
\Bigl(\frac{x^4}{a^4}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x}{p}\Bigr)=0
$$
You will in general get two "blobs", symmetric with respect to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The largest value of $n$ such that $n+10| n^3+100$ If $n\in \mathbb{Z}$ and follow the property
$$n+10| n^3+100$$
then what is the largest value of $n$
Please help me to solve this!!!
| Note that
$$
n+10|n^3+10n^2\implies n+10|(n^3+10n^2)-(n^3+100)=10n^2-100.
$$
Similarly,
\begin{aligned}
&\quad n+10|(10n^2+100n)-(10n^2-100)=100n+100\\
&\implies n+10|(100n+1000)-(100n+100)=900.
\end{aligned}
So the largest possible $n$ is $900-10=\boxed{890}$. Verify that this works:
$$
\frac{890^3+100}{890+10}=783299... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.