Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
A periodic summation question
The periodic series can be solved by doing
I understand the solution but it is little lengthy process. Can't we apply any law to get the solution direct?
| Splitting the sum and performing an index shift gets you there pretty quick.
$$
\begin{align*}
\sum_{n=1}^{20} a_n
&= \sum_{n=1}^{20} \left(\frac{1}{n}-\frac{1}{n+2}\right)
= \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=1}^{20} \frac{1}{n+2} \\
&= \sum_{n=1}^{20} \frac{1}{n} - \sum_{n=3}^{22} \frac{1}{n}
= \sum_{n=1}^{2} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of largest two angles All the inner angles of a 7 sided polygon are obtuse, their sizes in degrees being distinct integers divisible by 9. What is the sum (in degree) of the largest two angles?
| Sum of all angles is $180^\circ \cdot (n-2) = 900^\circ$.
Denote angles as $a,b,c,d,e,f,g$ ($a<b<c<d<e<f<g$).
If all angles are obtuse and are integer numbers divisible by $9^\circ$, then they can be
$99^\circ, 108^\circ, 117^\circ, 126^\circ, 135^\circ, 144^\circ, 153^\circ, ...$
A).
If $a\ge 108^\circ$, then $b\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/474333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true? I found the following relational expression by using computer:
For any natural number $n$,
$$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor.$$
Note that $\lfloor... | I want to proof this only by using basic arithmetic and therefore avoiding Jensen's Inequality and Taylor's Theorem.
It's simpler if one replaces $n$ by $n-2$. On then gets
$$ \lfloor \sqrt{n-2}+\sqrt{n-1}+\sqrt n+\sqrt {n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt {25n-1}\rfloor$$
It is necessary to show that
$$ \sqrt {25n-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 4,
"answer_id": 3
} |
Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$ How to prove the following equality?
$$\sum_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}.$$
| Addendum. The partial fraction identity at the top can be shown by induction on $m$. The case $m=1$ is trivial. The induction step gives
$$\frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k} \frac{1}{n+k+1} \frac{1}{n+m+1}
\\ = \frac{1}{(m-1)!} \sum_{k=0}^{m-1} (-1)^k \binom{m-1}{k}
\frac{1}{m-k} \left( \frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/477174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Find the shortest distance between the point $(8,3,2)$ and the line through the points $(1,2,1)$ and $(0,4,0)$ "Find the shortest distance between the point $(8,3,2)$ and the line through the points $(1,2,1)$ and $(0,4,0)$"
$$P = (1,2,1), Q = (0,4,0), A = (8,3,2)$$
$OP$ = vector to $P$
$$PQ_ = (0,4,0) - (1,2,1)$$
I fou... | The equation of the plane passing through the points $P$, $Q$, and $A$ is given by
$\begin{vmatrix}
x & y & z & 1 \\
1 & 2 & 1 & 1 \\
0 & 4 & 0 & 1 \\
8 & 3 & 2 & 1
\end{vmatrix}=0$.
After you compute the determinant, you get $3x-6y-15z+24=0$. Then the area $S$ of the triangle defined by the points is given by $S=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/478765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Closed-form expression for this definite integral Does this definite integral have a closed-form expression?
\begin{align*}
I &= \int_0^\infty \sqrt{ \frac{1}{2} \frac{1}{x} \left( \frac{1}{(1+x)^2} + \frac{z}{(1+xz)^2} \right) } \, dx \\
&= \frac{1}{\sqrt{2}} \int_0^\infty \frac{1}{x} \sqrt{ \frac{x}{1+x} \left( 1-\fr... | $$
{\rm I}\left(z\right)
\equiv
\int_{0}^{\infty}\sqrt{\vphantom{\LARGE A^{A}}\frac{1}{2x}
\left\lbrack%
\frac{1}{\left(1 + x\right)^{2}} + \frac{z}{\left(1 + xz\right)^2}
\right\rbrack\,}\ {\rm d}x\,,
\qquad
\begin{array}{|rclcl}
\,\,{\rm I}\left(0\right)
& = &
{\rm I}\left(\infty\right)
& = &
{\sqrt{2\,} \over 2}\,\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/479993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Chinese remainder theorem :Algebraic solution I need help in a question:
It is required to find the smallest $4$-digit number that when divided by $12,15$, and $18$ leaves remainders $8,11$, and $14$ respectively. Here's how I've attempted:
Let the number be $a$, then $$a=12p+8 = 15q+11 = 18r+14$$
Hence, $p=(5q+1)/2$ a... | Given:
$$x \equiv 8 \pmod{12} \Rightarrow x = 12a + 8 \\
x \equiv 11 \pmod{15} \Rightarrow x = 15b + 11\\
x \equiv 14 \pmod{18} \Rightarrow x = 18c + 14$$
We solve the equations pairwise:
$$1) \ 12a+8=15b+11 \Rightarrow 4a-5b=1 \Rightarrow \begin{cases}a=5n-1 \\ b=4n-1\end{cases}$$
$$2) \ 12a+8=18c+14 \Rightarrow 2a-3c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Calculate $ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $ How evaluate the following limit?
$$ \lim_{x \to 4} \frac{3 -\sqrt{5 -x}}{1 -\sqrt{5 -x}} $$
I cannot apply L'Hopital because $ f(x) = 3 -\sqrt{5 -x} \neq 0 $ at $x = 5$
| Let $\sqrt{5+x}=3+h$
Then $5+x=9+6h+h^2$
Let $\sqrt{5-x}=1+k$
Then $5-x=1+2k+k^2$
Adding gives: $5+x+5-x=9+6h+h^2+1+2k+k^2$
$0=6h+h^2+2k+k^2$
When $x=4$, $\sqrt{5+x}=\sqrt 9 =3 \Rightarrow 3+h=3 \Rightarrow h=0$
Similarly $k=0$
Required expression is $\frac {3-\sqrt{5+x}}{1-\sqrt{5-x}}=\frac h k$
$6h=-2k-k^2-h^2$
$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/480816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
limit of a recursive sequence:2 Let $$x_k = \frac{A}{1-C} x_{k-1} + \frac{B}{1-C}x_{k-2},$$ where $A, B, C$ are positive reals such that $A + B + C =1$. Let $$x_1 = 1$$ and $$x_2 = 1 + y,$$ with $y$ is positive. Which conditions should I impose on the parameters $A$, $B$, $C$, $y$ such that the sequence $x_k$ goes to i... | Let $a=\frac{A}{1-C}$ and $b=\frac{B}{1-C}$; then $a+b=1$. As was noted in the comments, this makes $x_k$ a convex combination of $x_{k-1}$ and $x_{k-2}$, and since $a,b>0$, $x_k$ lies strictly between $x_{k-1}$ and $x_{k-2}$. Thus, $1\le x_k\le 1+y$ for all $k\ge 1$, and the sequence is bounded.
The characteristic pol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/483104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate this limit: $\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac12$? How do I evaluate the limit of $$\lim_{x\to 0}\frac{\sqrt{x+1}-1}{x} = \frac{1}{2}$$? As $x$ approaches $0$, I know the answer is $\frac{1}{2}$, but I got this question wrong. I think you have to multiply by the conjugate of the numerator? C... | $\require{cancel}$
Multiply numerator and denominator by the conjugate of the numerator: $$\sqrt{x+1} + 1$$ then evaluate the limit.
When we multiply by the conjugate, recall how we factor the difference of squares: $$(\sqrt a - b) \cdot (\sqrt a + b) = (\sqrt a)^2 - b^2 = a - b^2$$
$$\dfrac {\sqrt{x+1} - 1}{x} \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/489699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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What should be added to $x^4 + 2x^3 - 2x^2 + x - 1$ to make it exactly divisible by $x^2 + 2x - 3$? I'm a ninth grader so please try to explain the answer in simple terms .
I cant fully understand the explanation in my book .
It just assumes that the expression that should be added has a degree of 1.
I apologize if t... | Imagine that we add $ax+b$ to the given polynomial, obtaining a new polynomial
$$P(x)= x^4 +2x^3-2x^2+x-1+ax+b.$$
Note that $x^2+2x-3=(x+3)(x-1)$. So if $x^2+2x-3$ divides our new polynomial $P(x)$, then $P(1)=0$ and $P(-3)=0$.
We have
$$P(1)=a+b+1, \qquad\text{and}\qquad P(-3)=-3a+b+5.$$
Solve the system of linear ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/490744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Use Laplace transform to solve the following initial–value problems. Use Laplace transform to solve the following initial–value problem.
$y′′′′ + 2y′′ + y = 0, y(0) = 1, y′(0) = −1, y′′(0) = 0, y′′′(0) = 2$
Answer
$s^4 L(s) - s^3y(0) -s^2 y'(0) - s y''(0) - y'''(0) +2[s^2L(s)-sy(0)-y'(0)] +L(s) \\\\$
I get the partial ... | Great job getting to this point, now we just have to get the answer in forms we can work with or use the formal definitions for inverse Laplace transforms. I will use known forms.
We have (split up the numerators):
$$\dfrac{s-1}{s^2 +1}+\dfrac{s+1}{(s^2+1)^2} = \dfrac{s}{s^2+1} - \dfrac{1}{s^2+1} + \dfrac{s}{(s^2+1)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/492513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse I'm finding the area of an ellipse given by $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$. I know the answer should be $\pi ab$ (e.g. by Green's theorem). Since we can parameterize the ellipse as $\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})$... | Your question has been answered, so now we look at how to find the area, using your parametrization, which is a perfectly good one.
The area is the integral of $|y\,dx|$ (or alternately of $|x\,dy|$. over the appropriate interval.
We have $y=b\sin\theta$ and $dx=-a\sin\theta\,d\theta$. So the area is
$$\int_0^{2\pi} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/493104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
lcm in $\mathbb{Z}[\sqrt{-5}]$ does not exists
I need to show that lcm of $2$ and $1+\sqrt{-5}$ does not exists in $\mathbb{Z}[\sqrt{-5}]$
Getting no idea about how to start, I was thinking when does lcm cannot exists!
| I suppose you know the squared modulus function $N(a+bi)=a^2+b^2$. In $\mathbb Z[\sqrt{-5}]$ this becomes
$$
N(a+b\sqrt{-5})=a^2+5b^2
$$
The values of $a^2+5b^2$ are sums of $a^2\in\{1,4,9,16,...\}$ and $5b^2\in\{5,20,...\}$ so we have $a^2+5b^2\in\{1,4,5,6,9,13,...\}$. In particular no element $x\in\mathbb Z[\sqrt{-5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/497775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Show that the equation $x^2+xy-y^2=3$ does not have integer solutions. Show that the equation $$x^2+xy-y^2=3$$ does not have integer solutions.
I solved the equation for $x$:
$x=\displaystyle \frac{-y\pm\sqrt{y^2+4(y^2+3)}}{2}$
$\displaystyle =\frac{-y\pm\sqrt{5y^2+12}}{2}$
I was then trying to show that $\sqrt{5y^2+1... | Just note that $5y^2$ is a multiple of $5$ and hence it ends on $0$ or $5$. Thus, $5y^2+12$ ends in $2$ or $7$, but there are no perfect squares that have this endings.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\cos 36^{\circ} = (1+ \sqrt 5)/4$? Given $4 \cos^2 x -2\cos x -1 = 0$.
Use this to show that $\cos 36^{\circ} = (1+ \sqrt 5)/4$, $\cos 72^{\circ} = (-1+\sqrt 5)/4$
Your help is greatly appreciated! Thanks
| To derive this from fundamentals, note that
$$\sin{108^{\circ}} = \sin{72^{\circ}}$$
then use a double-angle and triple-angle forumla:
$$\sin{2 x} = 2 \sin{x} \cos{x}$$
$$\sin{3 x} = 3 \sin{x} - 4 \sin^3{x}$$
In this case, $x=36^{\circ}$. Setting the above two equations equal to each other results in the quadratic equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/498600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can I show that $\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix}$? Well, the original task was to figure out what the following expression evaluates to for any $n$.
$$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}^{\large n}$$
By trying out different values of $n$, I found t... | Use induction on $n$.
(1) Prove the base case (trivial), perhaps even establish the case for $n = 2$ (two base cases here are not necessary, but as you found, it helps reveal the pattern.)
(2) Then assume it holds for $n = k$.
(3) Finally, show that from this assumption, it holds for $n = k+1$.
You've established t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 7,
"answer_id": 5
} |
sum of irrational numbers - are there nontrivial examples? I know that the sum of irrational numbers does not have to be irrational. For example $\sqrt2+\left(-\sqrt2\right)$ is equal to $0$. But what I am wondering is there any example where the sum of two irrational numbers isn't obviously rational like an integer an... | The examples below are not quite what you're asking for (because the results, while not obviously rational, wind up being nice integers), but they may nonetheless be amusing:
$$\sqrt{3 \; + \; 2\sqrt{2}} \; - \; \sqrt{3 \; - \; 2\sqrt{2}} \; = \; 2$$
$$\sqrt[3]{3\sqrt{21} \; + \; 8} \; - \; \sqrt[3]{3\sqrt{21} \; -\; 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/499784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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What is the smallest value of $x^2+y^2$ when $x+y=6$? If $ x+y=6 $ then what is the smallest possible value for $x^2+y^2$?
Please show me the working to show where I am going wrong!
Cheers
| When Usually found yourself stuck at such questions:
One can Go for Hit and Trial Method:
Since the value of
x+y = 6
Start with lowest possible combination you can think of
For x=1, y=5 : $x^2+y^2=26$
For x=2, y=4 : $x^2+y^2=20$
For x=3, y=3 : $x^2+y^2=18$
For x=4, y=2 : $x^2+y^2=20$
For x=5, y=1 : $x^2+y^2=26$
So th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/502034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
} |
Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$ Show that $\gcd(a + b, a^2 + b^2) = 1$ or $2$ if $\gcd(a, b)=1$.
I have absolutely no clue where to start and what to do, please provide complete proof and answer.
| We use repeatedly the identity $\gcd(x,y)=\gcd(x,y+kx)$ for any integer $k$.
For $k=-(a+b)$, we have $$\gcd(a+b,a^2+b^2)=\gcd(a+b,-2ab)$$
But $\gcd(a,b)=1$ so $\gcd(a,a+b)=1$ and similarly $\gcd(b,a+b)=1$. Combining, we get $\gcd(ab,a+b)=1$. Hence $\gcd(a+b,a^2+b^2)=\gcd(a+b,-2)$, which is $1$ or $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to prove $(1+x)^n\geq 1+nx+\frac{n(n-1)}{2}x^2$ for all $x\geq 0$ and $n\geq 1$? I've got most of the inductive work done but I'm stuck near the very end. I'm not so great with using induction when inequalities are involved, so I have no idea how to get what I need...
\begin{align}
(1+x)^{k+1}&\geq (1+x)\left[1+kx+... | We need to show $(1+x)^{k+1}\ge1+(k+1)x+\frac{(k+1)k}2x^2$
You already have
$$
\begin{align}
(1+x)^{k+1}&\geq1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3
\end{align}$$
$$=1+(k+1)x+x^2\frac{k(k+1)}2+\frac{k(k-1)}{2}x^3$$
which is $$\ge1+(k+1)x+x^2\frac{k(k+1)}2$$ if $\displaystyle\frac{k(k-1)}{2}x^3\ge0$ whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/507537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital $$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$
I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.
But it doesn't work :/
| $$\frac{x - \sin(x)}{x - \tan(x)} = \frac{x - \sin(x)}{x^3} \cdot \frac{x^3}{x - \tan(x)}$$
Let $x = 3y$ and $x\to 0 \implies y\to 0$
$$\lim_{x\to0} \frac{x - \sin(x)}{x^3} = L $$
$$L = \lim_{y\to0}\frac{3y - \sin(3y)}{(3y)^3} = \lim_{y\to0} \frac 3 {27} \frac{y - \sin(y)}{y^3} + \lim_{y\to0} \frac{4}{27} \frac{\sin^3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Matrix Algebra Question (Linear Algebra) Find all values of $a$ such that $A^3 = 2A$, where
$$A = \begin{bmatrix} -2 & 2 \\ -1 & a \end{bmatrix}.$$
The matrix I got for $A^3$ at the end didn't match up, but I probably made a multiplication mistake somewhere.
| Here's a different approach that avoids matrix multiplication in favor of determinants:
We want to find $a$ such that $A^3=2A$. This is equivalent to $A^3-2A=0$, and to $(A^2-2I)A=0$.
Thus either $A=0$, which is impossible, or $A^2-2I$ is a zero divisor.
Thus \begin{align}0&=\det (A^2-2I) \\ &= \det(A+\sqrt2I)\det(A-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/508896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How would I reduce my matrix even further? How would I reduce my matrix even further?
| I usually do also pivot reduction:
\begin{align}
\large
\begin{bmatrix}
2 & 2 & 1 & 2\\
-1 & 2 & -1 & -5\\
1 & -3 & 2 & 8
\end{bmatrix}
\xrightarrow{E_1(1/2)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 & 1\\
-1 & 2 & -1 & -5\\
1 & -3 & 2 & 8
\end{bmatrix}
\\
\large
\xrightarrow{E_{21}(1)}
&\large
\begin{bmatrix}
1 & 1 & 1/2 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/509860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $\frac{\cos x}{\cos y}=\frac{a}{b}$ then $a\tan x +b\tan y$ equals If $$\frac{\cos x}{\cos y}=\frac{a}{b}$$ Then $$a \cdot\tan x +b \cdot\tan y$$ Equals to (options below):
(a) $(a+b) \cot\frac{x+y}{2}$
(b) $(a+b)\tan\frac{x+y}{2}$
(c) $(a+b)(\tan\frac{x}{2} +\tan\frac{y}{2})$
(d) $(a+b)(\cot\frac{x}{2}+\co... | I am hoping that my calculations haven't gone wrong. Here is a method to proceed.
From your method you have $$\frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{y-x}{2}} = -\cot\Bigl(\frac{x+y}{2}\Bigr)\cdot\Bigl(\frac{x-y}{2}\Bigr)=\frac{a+b}{a-b}$$
Now note that
\begin{align*}
\tan(x) &=\tan\left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $x^2 + xy + y^2 \ge 0$ by contradiction Using the fact that $x^2 + 2xy + y^2 = (x + y)^2 \ge 0$, show that the assumption $x^2 + xy + y^2 < 0$ leads to a contradiction...
So do I start off with...
"Assume that $x^2 + xy + y^2 <0$, then blah blah blah"?
It seems true...because then I go $(x^2 + 2xy + y^2) - (... | Suppose that $x^2+xy+y^2<0$; then $x^2+2xy+y^2<xy$, so $(x+y)^2<xy$. Subtracting $3xy$ from both sides of the original inequality, we see that $x^2-2xy+y^2<-3xy$, so $(x-y)^2<-3xy$. Squares are non-negative, so on the one hand $xy>0$, and on the other hand $-3xy>0$ and therefore $xy<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
How to find the maximum of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b}+\frac{1}{a+c}$ given certain constraints. Let $a,b,c\ge 0,$ and such $a+b+c=1$. Find the maximum of:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a+b}+\dfrac{1}{a+c}$$
My try:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a+b}+\df... | Hint:
*
*Kuhn-Tucker Multipliers For inequality constraints
*Lagrange Multipliers For equality constraints
Another Hint:
The constraint $a+b+c=1$ makes a variable redundant. For example, $a$ may be substituted with $a = 1-b-c$ (which you have correctly identified), so only the inequality constraints need to be co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/514497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Series Convergence What does this series converge to?
$$ \sqrt{3\sqrt {5\sqrt {3\sqrt {5\sqrt \cdots}}}} $$
and also this?
$$ \sqrt{6+\sqrt {6+\sqrt {6+\sqrt {6+\sqrt \cdots}}}} $$
And, generally speaking, how should one approach these kind of questions?
| For $$\sqrt{6 + \sqrt{6 + \sqrt{6 +\dots}}}:$$
Let \begin{align*}
x &= \text{the given equation}\\
&= \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}
\end{align*}
Since the series is infinite, we can write
$x = \sqrt{6 + x}$,
or $$x^2 - x - 6 = 0.$$
Therefore, $x = 3$ or $x = -2$
Since the answer cannot be negative, reject $x ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Find an angle of an isosceles triangle $\triangle ABC$ is an isosceles triangle such that $AB=AC$ and $\angle BAC$=$20^\circ$. And a point D is on $\overline{AC}$ so that AD=BC, , How to find $\angle{DBC}$?
I could not get how to use the condition $AD=BC$ , How do I use the condition to find $\angle{DBC}$?
EDIT 1: Wi... | One way to calculate this is to write sin laws for two triangles $ABD$ and $BDC$. Call the angle $\angle ABD=x$. Then we have:
$$
\frac{AD}{\sin x}=\frac{BD}{\sin 20},\frac{BC}{\sin (20+x)}=\frac{BD}{\sin 80}
$$
Using $AD=BC$ and $\sin 80=\cos 10$ we get the following:
$$
\frac{\sin x}{\sin 20}=\frac{\sin (20+x)}{\sin ... | {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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Derivation of the quadratic equation So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
| Maybe we don't know that we should create a term that can be squared. But we do know that it would be easy to solve if we eliminate the $x$ term. We can use substitution of variables to eliminate it.
$ax^2+bx+c=0$
$x=y+d$
$a(y^2+2dy+d^2)+b(y+d)+c=0$
$ay^2+ad^2+(2ad+b)y+bd+c=0$
We want: $2ad+b=0$
$d=-\frac{b}{2a}$
$x=y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/515767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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What the rest of the division $1^6+2^6+...+100^6$ by $7$?
What the rest of the division $1^6+2^6+...+100^6$ by $7$?
$1^6\equiv1\pmod7\\2^6\equiv64\equiv1\pmod7\\3^6\equiv729\equiv1\pmod7$
Apparently all the leftovers are $one$, I thought of using Fermat's Little Theorem, however the $(7,7 k) = 7$, so you can not gener... | Split the sum into $1^6$ to $7^6$, then $8^6$ to $14^6$, and so on up to $92^6$ to $98^6$, plus the little tail $99^6+100^6$.
Each of the full chunks of length $7$ has shape $(7k+1)^6+(7k+2)^6+\cdots +(7k+6)^6+(7k+7)^6$. So the sum of the remainders of such a chunk on division by $7$ is the same as the remainder of $1... | {
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"url": "https://math.stackexchange.com/questions/518018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this $p\equiv 1\pmod 8$ let $m,n$ are positive integer numbers,and such $m<n$,if
$$p=\dfrac{n^2+m^2}{\sqrt{n^2-m^2}}$$
is prime number.
show that
$$p\equiv 1 \pmod 8$$
My try:
let $$p=\dfrac{n^2+m^2}{\sqrt{n^2-m^2}}=\dfrac{n^2-m^2+2m^2}{\sqrt{n^2-m^2}}=\sqrt{n^2-m^2}+\dfrac{2m^2}{\sqrt{n^2-m^2}}$$
Then I ... | If $p$ divided either of $n$ and $m$, then it must also divide the other, so dividing by $p$ yields (with $n = p\nu,\, m = p\mu$)
$$1 = \frac{\nu^2 + \mu^2}{\sqrt{\nu^2-\mu^2}},$$
from which it follows that $\nu = 1,\, \mu = 0$ contradicting the positivity of $n$ and $m$. So $p$ divides neither $n$ nor $m$.
Now, for
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/519105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Find a good candidate for a closed-form solution of this recurrence relation: $P(n-1)+n^2$. I want to find a candidate for this recurrence relation:
$$
P(n) = \left\{\begin{aligned}
&1 &&: n = 0\\
&P(n-1)+n^2 &&: n>0
\end{aligned}
\right.$$
Starting from 0 the first 8 values are 1,2,6,15,31,56,92,141.
I can't figure ou... | Write $P(n) - P(n-1) = n^2 = 2{n \choose 2} + {n\choose 1}$. But ${n \choose 2} = {{n+1} \choose 3}-{n \choose 3}$, and ${n\choose 1} = {{n+1} \choose 2} - {n\choose 2}$, so we have $P(n) - P(n-1) = Q(n) - Q(n-1)$, where $Q(n) = 2{{n+1} \choose 3} + {{n+1} \choose 2} $. Since $P(0) = 1$ and $Q(0) = 0$, we can now con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/519929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to. Given that $3x\equiv7\pmod{11}, 5y\equiv9\pmod{11}$. Find the number which $x+y\pmod{11}$ is congruent to. I'm thinking $20\equiv9\pmod{11}$, But I am having trouble find a number $3x$ that is divisible by $3$? Is there a b... | From the first congruence, by multiplying by $4$, we conclude that $x\equiv 28\equiv 6\pmod{11}$.
From the second congruence, multiplying by $2$, conclude that $-y\equiv 18\equiv 7\pmod{11}$, so $y\equiv -7\equiv 4\pmod{11}$.
Add. We get $x+y\equiv 10\pmod{11}$.
Another way: Multiply the first congruence by $5$, the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Integral $\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}$ I need your help with this integral:
$$\int_0^\infty\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2\,x^2+2}.$$
I wasn't able to evaluate it in a closed form, although an approximate numerical evaluation suggested its value could be $\frac{\pi}{4}$.
| Let us introduce the notation
$$\mathcal{I}=\int_0^{\infty}\frac{dx}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}.$$
*
*Now observe that
\begin{align}\frac{1}{\frac{x^4-1}{x\cos(\pi\ln x)+1}+2x^2+2}&=
\frac{1}{x^2+1}\cdot\frac{1}{\frac{x^2-1}{x\cos(\pi\ln x)+1}+2}=\\
&=\frac{1}{x^2+1}\cdot\frac{\cos(\pi\ln x)+\frac1x}{x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/521685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Logarithm Problem : Find the number of real solutions of the equation $2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)=1$
Find the number of real solutions of the equation $2\log_2\log_2x+\log_{\frac{1}{2}}\log_2(2\sqrt{2}x)=1$
My approach :
Solution : Here right hand side is constant term so convert it into log... | I think you went off the rails in the second line. Here's what I get
$$2 \log_2{\log_2{x}} = \log_2{\log_2^2{x}}$$
so that
$$2 \log_2{\log_2{x}} + \log_{1/2}{\log_2{(2 \sqrt{2} x)}} = \log_2{\frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}}}$$
and the equation becomes
$$\log_2{\frac{\log_2^2{x}}{\log_2{(2 \sqrt{2} x)}}} = 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/522227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $ For any $x \in \mathbb{R}$ and $m \in \mathbb{N} $ evaluate $\lfloor \frac{x}{m} \rfloor + \lfloor \frac{x+1}{m} \rfloor + \dots + \lfloor \frac{x+m-1}{m} \rfloor $.
Well if $x=m$ then we obviously have... | This is known as Hermite's identity. The result will always be $\lfloor x \rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A math contest problem $\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\frac{\ln(1-x)}x \ \mathrm dx$ A friend of mine sent me a math contest problem that I am not able to solve (he does not know a solution either). So, I thought I might ask you for help.
Prove:
$$\int_0^1\ln\left(1+\frac{\ln^2x}{4\,\pi^2}\right)\... | Here is a solution: Let $I$ denote the integral. Then
\begin{align*}
I &= - \int_{0}^{1} \log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}'(x) \, dx \\
&= \left[ -\log \left( 1 + \frac{\log^{2}x}{4\pi^{2}}\right) \mathrm{Li}_{2}(x) \right]_{0}^{1} + 2 \int_{0}^{1} \frac{\log x}{4\pi^{2} + \log^{2} x} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "179",
"answer_count": 4,
"answer_id": 2
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integrate $ \frac {(x^3 + 36)} {(x^2 + 36)}$ I know I have to use long division first, but I don't really know how to do it in this case
$$\int \frac{x^3 + 36}{x^2 + 36}dx$$
| HINT:
Using Partial fraction decomposition $$\frac{x^3+36}{x^2+36}=x+A+B\frac{\frac{d(x^2+36)}{dx}}{x^2+36}+C\frac1{x^2+36}$$
$$\implies \frac{x^3+36}{x^2+36}=x+A+\frac{2Bx}{x^2+36}+\frac C{x^2+36}$$
$$\implies x^3+36=(x+A)(x^2+36)+2Bx+C$$
$$\implies x^3+36=x^3+Ax^2+x(36+2B)+36A+C$$
Compare the coefficients of the diff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/523209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
$a+b+c=0;a^2+b^2+c^2=1$ then $a^4+b^4+c^4$ is equal to what?
I tried to solve this problem, and I get $a^4+b^4+c^4 = 2(a^2b^2 + 1)$ but I'm not sure if it's correct
| Given $a+b+c = 0$ and $(a^2+b^2+c^2) = 1$, Now $(a^2+b^2+c^2)^2 = 1^2 = 1$
$(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2) = 1.................(1)$
and from $(a+b+c)^2 = 0$,
we get $\displaystyle 1+2(ab+bc+ca) = 0\Rightarrow (ab+bc+ca) = -\frac{1}{2}$
again squaring both side , we get $\displaystyle (ab+bc+ca)^2 = \frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/524335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Standard inductive problem Question: Prove that $2^n \geq (n+1)^2$ for all $n \geq 6$.
I have tried to prove this below and I'm interested if my method was correct and if there is a simpler answer since my answer seems unnecessarily long for such a simple claim.
Inductive hypothesis $$2^n \geq (n+1)^2$$
We need to sho... | You have $2^{n+1}=2\times 2^{n}\geq 2(n+1)^2$
Now, for $n\geq 6,$
$$2(n+1)^2-(n+2)^2$$
$$=2(n^2+2n+1)-(n^2+4+4n)$$
$$=2n^2+4n+2-n^2-4-4n$$
$$=n^2-2\geq0$$
$$\therefore 2(n+1)^2\geq(n+2)^2$$
$$\therefore 2^{n+1}\geq(n+2)^2$$
:)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rabin's test for polynomial irreducibility over $\mathbb{F}_2$ I know that $f(x) = x^{169}+x^2+1$ is a reducible polynomial over $\mathbb{F}_2$. I want to show this using Rabin's irreduciblity test. First, I then have to check if $f$ is a divisor of $x^{2^{169}}+x$ and this is where I get stuck. That exponent is way to... | This is probably not what you wanted to see, but it is easy to see that your polynomial is divisible by $x^2+x+1=(x^3-1)/(x-1)$. Here
$$
x^{169}+x^2+1=(x^{169}+x)+(x^2+x+1)=x(x^{3\cdot56}+1)+(x^2+x+1),
$$
and clearly $x^{3\cdot56}-1$ is divisible by $x^3-1$ and hence by $x^2+x+1$.
It is usually easy to check the divisi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/528552",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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showing algebraic inequality with arithmetic and harmonic means Let x, y, z be positive real numbers. Prove that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{3}{2}$$
This problem appears to be simple, but upon further work and lots of failed attempts, I am stuck. I have tried using arithmetic and harmonic mean... | $$A=\dfrac a{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}$$
$$B= \dfrac b{b+c}+ \dfrac{c}{c+a}+\dfrac{a}{a+b}$$
$$C=\dfrac c{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+a}$$
By A-G inequality
$$A+B=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+c}+\dfrac{c+a}{b+a}\ge 3$$
$$A+C=\dfrac{a+c}{b+c}+\dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}\ge3$$
so
+
$$(A+B)+(A+C)\ge6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/530153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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A $2 \times 2$ matrix $A$ such that $A^n$ is the identity matrix So basically determine a $2 \times 2$ matrix $A$ such that $A^n$ is an identity matrix, but none of $A^1, A^2,..., A^{n-1}$ are the identity matrix. (Hint: Think geometric mappings)
I don't understand this question at all, can someone help please?
| Here's a cute way of looking at it: let
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$
then
$J^2 = -I. \tag{2}$
From (1) and (2) it follows that
$((\cos \theta)I + (\sin \theta)J)^n = ((\cos n \theta)I + (\sin n \theta)J); \tag{3}$
the proof of (3) is virtually identical to that of de Moivre's formula (s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/534198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
} |
Proof of $10^{n+1} -9n -10 \equiv 0 \pmod {81}$ I am trying to prove that $10^{n+1} -9n -10 \equiv 0 \pmod {81}$. I think that decomposing into 9 and then 9 again is the way to go, but I just cannot get there. Any help is greatly appreciated.
\emph{edit} I originally posted this a $9^n$ not $9n$. Apologies.
| $$
\begin{align}
&10^{n+1}-9n-10\\[9pt]
&=9\left(10\frac{10^n-1}{10-1}-n\right)\\
&=9\left(10\left(10^{n-1}+10^{n-2}+\dots+10+1\right)+9n-10n\right)\\
&=9\left(10\cdot9\left(\frac{10^{n-1}-1}{10-1}+\frac{10^{n-2}-1}{10-1}+\dots+\frac{10-1}{10-1}+\frac{1-1}{10-1}\right)+9n\right)\\
&=81\left(10\left(\frac{10^{n-1}-1}{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$ Let $a,b,c \in (0, \infty)$, with $a+b+c=3$. How can I prove that:
$$\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3 ?$$.
I try to use Cauchy-Schwarz rewriting the inequality like :
$$\sum_{cyc}\frac{a\sqrt{b}}{b} \geq \frac{(\... | We need to prove that
$$\sum_{cyc}\frac{a^2}{b}\geq3,$$ where $a$, $b$ and $c$ are positives such that $a^2+b^2+c^2=3.$
Indeed,
$$\sum_{cyc}\frac{a^2}{b}-3=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)-\left(3-a-b-c\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{9-(a+b+c)^2}{3+a+b+c}=\sum_{cyc}\frac{(a-b)^2}{b}-\frac{3(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/537934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
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Did I solve this limit problem correctly? $\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$ Need to solve this limit
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}$$
When I put $x$ in this expression I have indeterminate form $1^{\infty }.$
So I choose the way which was described o... | Your solution is correct halfway:
$$\lim_{x \to 3}\left(\frac{6-x}{3}\right)^{\tan \frac{\pi x}{6}}=\exp\left\{\lim_{x \to 3 }{\tan \frac{\pi x}{6}\left(\frac{6-x}{3}-1\right)}\right\}$$ Above you used two facts:
*
*$\lim(\exp)=\exp(\lim)$, which is the continuity of exponent
*$\frac{1+t}{t}\to1$ if $t\to0$: $$\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/539207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the coefficient of a power of x in the product of polynomials - Link with Combinations? I came across a new set of problems while studying combinatorics which involves restrictions to several variables and use of multinomial theoram to evaluate the number of possible combinations of the variables subjected to thos... | Consider for a moment the product
$$\left(x^0+x^1+x^2+x^3\right)\left(y^0+y^1+y^2+y^3+y^4\right)\left(z^0+z^1+z^2+z^3+z^4+z^5\right)\;.$$
When you multiply it out, you get a total of $4\cdot5\cdot6=120$ terms, each of the form $x^ay^bz^c$, where $0\le a\le 3$, $0\le b\le 4$, and $0\le c\le 5$.
Now change the $y$’s and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/540082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
finding determinant as an function in given matrix Calculate the determinant of the following matrix as an explicit
function of $x$. (It is a polynomial in $x$. You are asked to find
all the coefficients.)
\begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\
x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\
0 & 0 & 0 & x^{10} & x^{11}... | Hint:
$$\det\begin{pmatrix}A & 0\\ C & D\end{pmatrix} = \det\begin{pmatrix}A & B\\ 0 & D\end{pmatrix} = \det(A) \det(D)$$
Where $A,B,C,D$ are block matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/542148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finite Sum $\sum\limits_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}$
Question : Is the following true for any $m\in\mathbb N$?
$$\begin{align}\sum_{k=1}^{m-1}\frac{1}{\sin^2\frac{k\pi}{m}}=\frac{m^2-1}{3}\qquad(\star)\end{align}$$
Motivation : I reached $(\star)$ by using computer. It seems true, but I can't prove it... | Setting $\tan mx=0$ in Sum of tangent functions where arguments are in specific arithmetic series,
we get $$\binom m1\tan x-\binom m3\tan^3x+\cdots=0\ \ \ \ (1)$$
Multiplying both sides by $\cot^mx,$
$$\binom m1\cot^{m-1}x-\binom m3\cot^{m-3}x+\cdots=0\ \ \ \ (2)$$
$\tan mx=0\implies mx=n\pi$ where $n$ is any inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/544228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 8,
"answer_id": 0
} |
Prove that $ \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$. Prove that $\displaystyle \int_{0}^{1}\frac{\sqrt{1-x^2}}{1-x^2\sin^2 \alpha}dx = \frac{\pi}{4\cos^2 \frac{\alpha}{2}}$.
$\bf{My\; Try}::$ Let $x = \sin \theta$, Then $dx = \cos \theta d\theta$
$\displaystyle = \... | Here is a much nicer way than my first attempt. Use $x=\sin(\theta)$ and $u=\tan(\theta)$
$$
\begin{align}
&\int_0^1\frac{\sqrt{1-x^2}}{1-x^2\sin^2(\alpha)}\,\mathrm{d}x\\
&=\int_0^{\pi/2}\frac{\cos^2(\theta)}{1-\sin^2(\theta)\sin^2(\alpha)}\,\mathrm{d}\theta\\
&=\int_0^{\pi/2}\frac{\mathrm{d}\theta}{\sec^2(\theta)-\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/550145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Limit of $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$ What is the limit of this sequence $\frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$?
Where $a$ is a constant and $n \to \infty$.
If answered with proofs, it will be best.
| With $S_n = \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \frac{n}{a^n}$ and using the closed form of geometric sums,
$$
\begin{align}
\lim_{n \to \infty} S_n &= \frac{1}{a} + \frac{2}{a^2} + \frac{3}{a^3} + \cdots \\
&= \frac{1}{a} + \frac{1}{a^2} + \frac{1}{a^3} + \cdots \\
& \phantom{=\frac{1}{a}} + \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/551234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
The number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ Find the number of integral solutions for the equation $x-y = x^2 + y^2 - xy$ and the equation of type $x+y = x^2 + y^2 - xy$
| Rearranging we get $$x^2-x(y+1)+y^2+y=0$$ which is a Quadratic Equation in $x$
As $x$ must be real, the discriminant must be $\ge0$ i.e.,
$(y+1)^2-4(y^2+y)=-3y^2-2y+1\ge0$
$\iff 3y^2+2y-1\le0$
$\iff \{y-(-1)\}(y-\frac13)\le0$
$\iff -1\le y\le \frac13$
Now, use the fact that $y$ is integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How prove this inequality $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\ge x^2+y^2+z^2$ let $x\ge y\ge z\ge 0$,show that
$$\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge x^2+y^2+z^2$$
my try:
$$\Longleftrightarrow x^3y^2+y^3z^2+z^3x^2\ge xyz(x^2+y^2+z^2)$$
| Another way.
$$\sum_{cyc}\left(\frac{x^2y}{z}-x^2\right)=(y-z)^2\left(\frac{x}{z}+\frac{x}{y}-1\right)+(x-y)(x-z)\left(\frac{y}{z}+\frac{y}{x}-1\right)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How to find $x^4+y^4+z^4$ from equation? Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
| We have in general if
$$
s_1:=x+y+z\textrm{, }s_2:=x^2+y^2+z^2\textrm{, }s_3:=x^3+y^3+z^3
$$
and
$$
s_4:=x^4+y^4+z^4
$$
and if
$$
\sigma_1=x+y+z\textrm{, }\sigma_2=xy+yz+zx\textrm{, }\sigma_3=xyz\tag 1
$$
Then
$$
s_1=\sigma_1\textrm{, }s_2=\sigma_1^2-2\sigma_2\textrm{, }s_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\textr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/556726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the equation for a line tangent to a circle from a point outside the circle? I need to know the equation for a line tangent to a circle and through a point outside the circle. I have found a number of solutions which involve specific numbers for the circles equation and the point outside but I need a specific s... | Given a circle $x^2 + y^2 = r^2$ and the point (a,b)
The line from the origin to (a,b) is $y = \frac{b}{a} * x$
The line and the circle intersect at $( \frac{a*r}{\sqrt{a^2+b^2}}, \frac{b*r}{\sqrt{a^2+b^2}} )$
The slope of the tangent is $\frac{-a}{b}$
The equation of the tangent is
$y - \frac{b*r}{\sqrt{a^2+b^2}} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if:
$$a^2 = b(b+c)$$
where $a, b, c$ are the sides opposite to $A, B, C$ respectively.
I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\... | Double angle formula says
$$
\begin{align}
\sin(A)
&=2\sin(B)\cos(B)\\
&\implies\frac{\sin(A)}{\sin(B)}=2\cos(B)\tag{1}
\end{align}
$$
The formula for the sine of a sum yields
$$
\begin{align}
\sin(C)
&=\sin(A+B)\\
&=2\sin(B)\cos(B)\cos(B)+(2\cos^2(B)-1)\sin(B)\\
&=\sin(B)(4\cos^2(B)-1)\\
&\implies\frac{\sin(C)}{\sin(B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/557704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sqrt{x}$ is continuous on its domain $[0, \infty).$ Prove that the function $\sqrt{x}$ is continuous on its domain $[0,\infty)$.
Proof.
Since $\sqrt{0} = 0, $ we consider the function $\sqrt{x}$ on $[a,\infty)$ where $a$ is real number and $s \neq 0.$ Let $\delta=2\sqrt{a}\epsilon.$ Then, $\forall x \in do... | Formal Proof:
Let $\epsilon > 0$, $a = 0$. Choose $\delta = \epsilon^2$. Then $\left |x \right| < \delta$ implies $\left | \sqrt{x} \right | < \epsilon$, as desired.
Let $\epsilon > 0$, $a \in (0,\infty)$. Choose $\delta = \sqrt{a}\epsilon$. Then $\left |x - a \right | < \delta$ implies $\left | \sqrt{x} - \sqrt{a} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Find the maximum and minimum values of $A \cos t + B \sin t$ Let $A$ and $B$ be constants. Find the maximum and minimum values of $A \cos t + B \sin t$.
I differentiated the function and found the solution to it as follows:
$f'(x)= B \cos t - A \sin t$
$B \cos t - A \sin t = 0 $
$t = \cot^{-1}(\frac{A}{B})+\pi n$
Howe... | If $A=B=0$, then
$$
\min_t(A\cos t+B\sin t)=\max_t(A\cos t+B\sin t)=0.
$$
If $(A,B) \ne (0,0)$, let $\theta \in [0,2\pi)$ such that
$$
\cos\theta=\frac{A}{\sqrt{A^2+B^2}},\quad \sin\theta=\frac{B}{\sqrt{A^2+B^2}}.
$$
Then
$$
A\cos t+B\sin t=\sqrt{A^2+B^2}(\cos t\cos\theta+\sin t\sin\theta)=\sqrt{A^2+B^2}\cos(t-\theta).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/560711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A positive integer is divisible by $3$ iff 3 divides the sum of its digits I am having trouble proving the two following questions:
*
*If $p|N$, $q|N$ and gcd(p,q)=1, then prove that $pq|N$
*If $x$ is non zero positive integer number, then prove that $3|x$ if and only if 3 divides the sum of all digits of $x$.
For... | (1) The hypotheses $p|N$ and $q|N$ give us two integers $m,k$ so that $N=mp$ and $N=kq$. This implies $mp=kq$ so that $q|mp$. Now, since $gcd(p,q)=1$ and $q|mp$ we know that $q|m$ (think about why this is true if it's not clear). Then $m=sq$ for some integer $s$. Putting this all together, we have $N=mp=sqp=s(pq)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/564662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
What are the elements of the quotient ring $\mathbb{Z_3}[x]/(x^3 + x^2)$? $$R = \mathbb{Z_3}[x]/(x^3 + x^2).$$
As $\mathbb{Z_3}$ is a field we have that every polynomial in $\mathbb{Z_3}[x]/(x^3 + x^2)$ of degree less than ${x^3 + x^2}$ is a distinct element in $R$.
So I conclude the following are the elements of $R$.
... | As pointed out in the comments, you are correct. The ring $\mathbb{Z}_3[x]/(x^3 + x^2)$ has $27$ elements, and they are precisely the ones you wrote down.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/565415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum involving binomial coefficient I want to analyse the following expression for $x \geq 0$:
$$
\sum_{k = 0}^n (-1)^{k+n} \binom{n+k}{2k} x^k
$$
I expect and want to prove that for $x \geq 4$, the expression tends to infinity as $n \to \infty$. I tried coming up with a closed expression, however found it difficult to ... | First we try to find a closed form of the sum, preferably one whose asymptotics can be classified. Let $f_n$ be your sum. Then Sister Celine / Zeilberger's algorithm says that
$$f_n = (x-2) f_{n-1} - f_{n-2}.$$
The roots of the characteristic equation of this recurrence are
$$\rho_{1,2} = -1 + \frac{1}{2} x \pm \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/565810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series
$$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$
From what I see, each term is the inverse of the sum of $n$ natural numbers.
... | $$
\sum_{n = 1}^{N}{2 \over n\left(n + 1\right)}
=
2\sum_{n = 1}^{N}\left({1 \over n} - {1 \over n + 1}\right)
=
2\left[1 - {1 \over 2} + {1 \over 2} - {1 \over 3} + \cdots + {1 \over N}
-
{1 \over N + 1}\right]
$$
$$
\sum_{n = 1}^{N}{2 \over n\left(n + 1\right)}
=
2\left(1 - {1 \over N + 1}\right)\quad\to\quad \color{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/567736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Recurrence sequence limit I would like to find the limit of
$$
\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1}, & n \ge 1
\end{cases}
$$
I tried to use this - $\lim \limits_{n\to\infty}a_n=\lim \limits_{n\to\infty}a_{n+1}=L$, so $L=\lim \limits_{n\to\infty}a_n\dfrac{n^2+2n}{n^2+2n+1}=L\cdot... | $\begin{cases}
a_1=\dfrac3{4} ,\, & \\
a_{n+1}=a_{n}\dfrac{n^2+2n}{n^2+2n+1} & n \ge 1
\end{cases}
$
$a_{n+1}
=a_{n}\left(\dfrac{n^2+2n}{n^2+2n+1}\right)
=a_{n}\left(\dfrac{n(n+2)}{(n+1)^2}\right)
$
so
$\begin{align}
a_{n}
&=a_1\prod \limits_{k=1}^{n-1} \dfrac{a_{k+1}}{a_k}\\
&=a_1\prod \limits_{k=1}^{n-1} \dfrac{k(k+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/568572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Determinant of a n x n Matrix - Main Diagonal = 2, Sub- & Super-Diagonal = 1 I'm stuck with this one - Any tips?
The Problem:
Let $n \in \mathbb{N}.$ The following $n \times n$ matrix:
$$A = \left( \begin{array}{ccc}
2 & 1 & & & & ...\\
1 & 2 & 1 & & & ...\\
& 1 & 2 & 1 & & ...\\
& & 1 & 2 & 1 & ...\\
& & ... | Why not develop directly wrt the first column? The subindex means the order of the square matrix:
$$\begin{vmatrix}2 & 1 & & & & ...\\
1 & 2 & 1 & & & ...\\
& 1 & 2 & 1 & & ...\\
& & 1 & 2 & 1 & ...\\
& & & 1 & ... & 1\\
... & ... & ... & ... & 1 &2\end{vmatrix}_n=2\begin{vmatrix}2 & 1 & & & & ...\\
1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/571664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit of a matrix and showing if it exists For the matrix
\[
A=\frac{1}{10}\cdot \begin{pmatrix}
1 & 7 \\
7 & 1
\end{pmatrix}\]
I try to find the limit of $A^n$ when $n$ goes to infinity.
| Assuming you are good with guessing eigenvalues there is a a similar way which in this case is kind of more easy.
Because of
\[ \frac{1}{10}\begin{pmatrix} 1 & 7 \\ 7 & 1 \end{pmatrix} \cdot
\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{4}{5} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix}\]
and
\[ \frac{1}{10}\begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/572832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that there is no natural number $n$ such that $3^7$ is the largest power of $3$ dividing $n!$
Show that there is no natural number $n$ such that $7$ is the largest power $a$ of $3$ for which $3^a$ divides $n!$
After doing some research, I could not understand how to start or what to do to demonstrate this.
We ha... | For $n=15,\left[\frac{15}{3} \right]+\left[\frac{15}{3^2} \right]+\left[\frac{15}{3^3} \right]+\;...=6$
for $n=18$ (the next multiple of $3$) $\left[\frac{18}{3} \right]+\left[\frac{18}{3^2} \right]+\left[\frac{18}{3^3} \right]+\;...=8$
If $n\geq 18$ then $\left[\frac{n}{3} \right]+\left[\frac{n}{3^2} \right]+\left[\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/574898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Limit $\lim_{n\to \infty} n(\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{(2n)^2})$ I need some help finding the limit of the following sequence:
$$\lim_{n\to \infty} a_n=n\left(\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\cdots+\frac{1}{(2n)^2}\right)$$
I can tell it is bounded by $\frac{1}{4}$ from below and decreasin... | Notice
$$\frac{1}{(n+k-1)(n+k)} \ge \frac{1}{(n+k)^2}\ge \frac{1}{(n+k)(n+k+1)}$$
We have
$$\begin{align}a_n = n \sum_{k=1}^{n}\frac{1}{(n+k)^2}\le & n\sum_{k=1}^n\frac{1}{(n+k-1)(n+k)}= n \sum_{k=1}^n\left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)\\ = & n \left(\frac{1}{n}- \frac{1}{2n}\right) = \frac12\\ \text{AND}\quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proof of Heron's Formula for the area of a triangle
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.
Could you please provide the proof of this formula?
Thank you in advance.
| Nobody can provide the proof but many can provide a proof or perhaps many proofs.
Notice that the area must be a 2nd-degree homogeneous function of $a$, $b$, and $c$, for example, if you multiply $a$, $b$, and $c$ by $9$ then you multiply the area by $9^2=81$, etc.
Next, notice that the area must be $0$ if $a+b=c$: if ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/576831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$ Given that $z= \cos \theta + i\sin \theta$, prove that $\Re\left\{\dfrac{z-1}{z+1}\right\}=0$
How would I do this?
| Using $\displaystyle\cos2A=2\cos^2A-1=1-2\sin^2A $ and $\displaystyle\sin2A=2\cos A\sin A,$
$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}=\frac{-2\sin^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}{2\cos^2\frac{\theta}2+i2\sin\frac{\theta}2\cos\frac{\theta}2}$$
$$=\frac{2i\sin\frac{\theta}2\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/579140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Distribution of function of two random variables Let $X$ be the number on a die roll, between 1 and 6. Let $Y$ be a random number which is uniformly distributed on $[0,1]$, independent of $X$. Let $Z = 10X + 10Y$.
What is the distribution of $Z$?
| Suppose $Y$ is another discrete random variables on $[0,1]$ with $\Pr(Y=0) = \Pr(Y=1) = \frac{1}{2}$. It helps to build the table of possible values of $Z = 10 X + 10 Y$:
$$
\begin{array}{c|cccccc} Z(X,Y) & X=1 & X=2 & X=3 & X=4 & X=5 & X+6 \cr \hline
Y=0 & Z=10 & Z=20 & Z=30 & Z=40 & ... | {
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"answer_id": 0
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$\varepsilon$-$\delta$ proof of $\sqrt{x+1}$ need to prove that $ \lim_{x\rightarrow 0 } \sqrt{1+x} = 1 $
proof of that is:
need to find a delta such that $ 0 < |x-1| < \delta \Rightarrow 1-\epsilon < \sqrt{x+1} < \epsilon + 1 $ if we choose $ \delta = (\epsilon + 1)^2 -2 $ and consider $ |x-1| < \delta = (\epsilon + 1... | You need to show that for all $\varepsilon > 0$ there exists a $\delta > 0$ such that $$0<|x|<\delta \implies |\sqrt{x+1}-1|<\varepsilon.$$
It's helpful in this case to multiply by the conjugate of $\sqrt{x+1}-1$, which is $\sqrt{x+1}+1$, so that the rightmost inequality becomes $$|\sqrt{x+1}-1|\cdot|\sqrt{x+1}+1|<\var... | {
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"answer_count": 3,
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De Moivre's formula to solve $\cos$ equation Use De Moivre’s formula to show that
$$
\cos\left(x\right) + \cos\left(3x\right) = 2\cos\left(x\right)\cos\left(2x\right)
$$
$$
\mbox{Show also that}\quad
\cos^{5}\left(x\right) = \frac{1}{16}\,\cos\left(5x\right) + \frac{5}{16}\,\cos\left(3x\right) + \frac{5}{8}\,\cos\left(... | Let me first show you another problem using DeMoivre's formula, to give you the idea of how to do the first two parts. Namely, I will show that $$\cos(4x)=8\cos^4x-8\cos^2x+1,$$ using only DeMoivre's formula and the Pythagorean identity $\sin^2x=1-\cos^2x$.
First off, we know that $$\cos(4x)+i\sin(4x)=(\cos x+i\sin x)^... | {
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How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola? How to prove $\frac{y^2-x^2}{x+y+1}=\pm1$ is a hyperbola, knowing the canonical form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=\pm1$ where $a$ and $b$ are constants? Thanks !
| Taking the '+' sign, $$y^2-x^2=x+y+1\implies\left(y-\frac12\right)^2-\left(x+\frac12\right)^2=1^2$$
$$\implies\frac{\left(y-\frac12\right)^2}{1^2}-\frac{\left(x+\frac12\right)^2}{1^2}=1^2$$
Similarly for the '-' sign,
$$y^2-x^2=-(x+y+1)\implies\left(x-\frac12\right)^2-\left(y+\frac12\right)^2=1^2$$
Can you recognize $a... | {
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"source": "stackexchange",
"question_score": "1",
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How find this $M(2)$ let matrix $$A(x)=\begin{bmatrix}
1&x\\
x&1
\end{bmatrix}$$
and consider the infinte matrix product
$$M(t)=\prod_{n=1}^{\infty}A(p^{-t}_{n})$$
where $p_{n}$ is the nth prime
Evaluate $M(2)$
My try: since
$$M(2)=\prod_{n=1}^{\infty}A(\dfrac{1}{p^2_{n}})=\begin{bmatrix}
1&\dfrac{1}{2^2}\\
\dfrac{1}{2... | Your matrices form a commuting family and therefore can be simultaneously diagonalized. Let $P$ be the orthogonal matrix which simultaneously diagonalizes the family. We have
$$P = \begin{pmatrix}-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{pmatrix}$$
Note that the eigenvalue... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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2nd order differential equation I'm working on the following 2nd order ODE:
$$
x^2 y''+2(2x-1/b^2) y'+2(1-(a/b)^2)y=0,
$$
where $b\neq 0$. It's very similar to the equation for the generalized Bessel polynomials (see http://en.wikipedia.org/wiki/Bessel_polynomials ); There's a slight difference in the factor in front o... | Mathematica tells me that after multiplying by a (horrible) power it's hypergeometric:
$
y(x)=c_1 \left(\frac{1}{x}\right)^{\frac{3 b^2-b \sqrt{8
a^2+b^2}}{2 b^2}} \, _1F_1\left(\frac{3}{2}-\frac{\sqrt{b^4+8 a^2 b^2}}{2
b^2};1-\frac{\sqrt{b^4+8 a^2 b^2}}{b^2};-\frac{2}{b^2 x}\right)$
$
\quad +\quad c_2
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/585616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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does this series converge? $\sum_{n=1}^\infty{\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha} $
show the the following series converge\diverge
$\sum_{n=1}^\infty{\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha} $
all the test i tried failed (root test, ratio test,direct comparison)
please dont use integrals ... | Hint rationalize:
$$\left( \sqrt[3]{n+1} - \sqrt[3]{n-1} \right)^\alpha = \left( \frac{2}{(\sqrt[3]{n+1})^2+ \sqrt[3]{n-1}\sqrt[3]{n+1}+ (\sqrt[3]{n-1})^2} \right)^\alpha$$
Compare with $$\sum\frac{1}{n^{\frac{2\alpha}{3}}}$$
| {
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"source": "stackexchange",
"question_score": "1",
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Checking on some convergent series I need some verification on the following 2 problems I attemped:
I have to show that the following series
is convergent: $$1-\frac{1}{3 \cdot 4}+\frac{1}{ 5 \cdot 4^2 }-\frac{1}{7 \cdot 4^3}+ \ldots$$ .
My Attempt: I notice that the general term is given by $$\,\,a_n=(-1)^{n}{1 ... | Another approach for the last series is the alternating series test which proves the convergence since
i) $ a_{n+1} < a_n, \quad \forall n\geq 2, $
ii) $ a_n\longrightarrow_{n\to \infty} 0. $
Note: You can prove the $a_n$ is decreasing by using the first derivative test for the function
$$ f(x)=\frac{\ln x}{x^2}. ... | {
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"source": "stackexchange",
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Find $\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$ I am to find the limit of
$$\lim_{x \to - \infty} \left(\frac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)$$
so I used:
$$\lim_{x \to -\infty} = \lim_{x \to \infty}f(-x)$$
but I just can't solve it to the end.
| $\lim_{x \to -\infty} \left(\dfrac{4^{x+2}- 2\cdot3^{-x}}{4^{-x}+2\cdot3^{x+1}}\right)
=\lim_{x \to -\infty} \left(\dfrac{3^{-x}(4^{x+2}\cdot3^x- 2)}{3^{-x}(4^{-x}\cdot3^x+2\cdot3^{2x+1})}\right)
=\lim_{x \to -\infty} \left(\dfrac{4^{x+2}\cdot3^x- 2}{(3/4)^x+2\cdot3^{2x+1}}\right)$
The numerator converges to $-2$ while... | {
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"question_score": "1",
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Proof by Induction for a $f_3 + f_6 + · · · + f_{3n} = \frac{1}{2} (f_{3n+2} - 1)$ The Fibonacci numbers are defined as follows: $f_0 = 0,\ f_1 = 1$, and for $n ≥ 2,\ f_n = f_{n−1} +f_{n−2}$. Prove that for every positive integer $n$,
$$f_3 + f_6 + \ldots + f3_n = \frac{1}{2} (f_{3n+2} - 1) $$
Here are my steps:
Basis ... | $$f_3 + f_6 + · · · + f_{3n} =\frac{1}{2}(f_{3n+2} - 1) $$
$$f_3+f_6 + ···+f_{3n}+f_{3(n+1)} =\frac{1}{2}(f_{3n+2} - 1)+f_{3n+3}= $$
$$=\frac{1}{2}((f_{3n+2}+f_{3n+3})+f_{3n+3}-1)=\frac{1}{2}((f_{3n+4}+f_{3n+3}) - 1)= $$
$$=\frac{1}{2}(f_{3n+5} - 1)=\frac{1}{2}(f_{3(n+1)+2} - 1) $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$
I know the answer is $2^n - 1$, but how to simplify it?
| Forget about all the smart generic formulas. Just rewrite the last summand $1$ as $2-1$. You get
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2 + 2 - 1.
$$
Now group the $2$'s together. $2 + 2 = 2^2$, so you get
$$
2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2^2 - 1
$$
Now group $2^2$ and $2^2$ together to get $2^2 + 2^2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/590733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this reccurence relation? Let a,b,c be real numbers. Find the explicit formula for $f_n=af_{n-1}+b$ for $n \ge 1$ and $f_0 = c$
So I rewrote it as $f_n-af_{n-1}-b=0$ which gives the characteristic equation as $x^2-ax-b=0$. The quadratic formula gives roots $x= \frac{a+\sqrt{a^2+4b}}{-2}, \frac{a-\sqrt{a^2+... | If $a=1$, then $f_n=f_0+nb$, otherwise since $f_n=af_{n-1}+b$ we can subtract $\frac{b}{1-a}$ from both sides to get
$$
f_n-\frac{b}{1-a}=a\left(f_{n-1}-\frac{b}{1-a}\right)
$$
therefore,
$$
\begin{align}
f_n
&=\frac{b}{1-a}+a^n\left(f_0-\frac{b}{1-a}\right)\\
&=a^nf_0+b\frac{1-a^n}{1-a}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/592215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Determine coordinates $x_c$ and $y_c$
Answer: $x_c=18.2$ and $y_c=9.5$.
Who can solve with explanation?
| Given the distributive property of the centroid, calling $A_i, i=1,2,3$ the areas of the triangle, rectangle and circle, respectively, and $P_i=(x_i,y_i)$ the respective centroids, we have
\begin{align}
x_c&=\frac{A_1x_1+A_2x_2-A_3x_3}{A_1+A_2-A_3}\\
&=\frac{a^2\cdot\frac{2}{3}a+4a^2\cdot2a-\pi r^2\cdot2a}{a^2+4a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/598094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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finding the limit of a series with 3rd root this is not a homework question.
how do i reach the limit of:
i tried taking out $n^3$ and then i got $n-n$ which is $0$, but the true answer is $2/3$.
i can't understand why the answer is $2/3$ and what method to use here.
thank you very much in advance,
yaron.
| Perhaps easier to simplify first. We want to get rid of the cube roots, so remember that $a^3-b^3 = (a-b) \cdot \left(a^2 + ab + b^2\right)$. Note that
$$
\begin{split}
\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}
&= \frac{\left(\sqrt[3]{n^3+2n^2+4} - \sqrt[3]{n^3+1}\right)
\left( {(n^3+2n^2+4)^{2/3} + (n^3+1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex Analysis - Cauchy Principal Value of Improper Integral I am having some difficulty with a P.V. Cauchy Integral problem, and after working on it for hours, I just cannot seem to find what I might be doing wrong.
Here is the problem:
Evaluate the Cauchy principal value of the given improper integral:
$$\int_{-\in... | You should get $2\pi i \ Res(f(z)|_{z=2i} = 2\pi i \frac{d}{dz} \frac{1}{(z+2i)^2}|_{z=2i} = 2\pi i \frac{-2}{(4i)^3} = 2\pi i \frac{-2}{-64i} = \frac{\pi}{16}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/600593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Spring Calculation - find mass
A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass tha... | A short, non insightful answer:
$$ \omega_0^2 = \frac{k}{m} $$
$$\alpha=\frac{c}{2m}$$
If the system is critically damped, $\omega_0^2 = \alpha^2$, so:
$$\frac{3}{m}=\frac{81}{4m^2} $$
$$m=6.75kg$$
If the system is overdamped damped, $\omega_0^2 \lt \alpha^2$
If the system is underdamped damped, $\omega_0^2 \gt \alpha... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What is $ \lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2} $? I have limit:
$$
\lim_{(x,y)\to(2,2)}\frac{x^4-y^4}{x^2 - y^2}
$$
Why is the result $8$ ?
| The limit you are looking up to has a 'hole' in its value as $(x,y)\rightarrow(2,2)$. In order to eradicate that, we have to factor out the hole which is quite easy in this case. It can be done as follows:
$$\lim_{(x,y)\rightarrow(2,2)}\dfrac{x^4-y^4}{x^2-y^2}=\lim_{(x,y)\rightarrow(2,2)}\dfrac{(x^2+y^2)(x^2-y^2)}{x^2-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Prove $\sin^2\theta + \cos^2\theta = 1$ How do you prove the following trigonometric identity: $$ \sin^2\theta+\cos^2\theta=1$$
I'm curious to know of the different ways of proving this depending on different characterizations of sine and cosine.
| Consider a right-angled triangle, $\Delta ABC$, where $\angle BAC = \theta$,
By the Pythagorean theorem,
$$ {AC}^2+{BC}^2 = {AB}^2 $$
Dividing by $AB^2$,
$$
\require{cancel}
\begin{align}
&\Rightarrow \frac{AC^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AB^2}{AB^2}\\
&\Rightarrow \Big(\frac{\text{opposite}}{\text{hypotenuse}... | {
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"url": "https://math.stackexchange.com/questions/607103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "37",
"answer_count": 16,
"answer_id": 9
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Infinite Series $\sum\limits_{n=1}^\infty\frac{H_{2n+1}}{n^2}$ How can I prove that
$$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$
I think this post can help me, but I'm not sure.
| Using the following nice rule: $$\sum_{n=1}^\infty a_{2n}=\sum_{n=1}^\infty a_{n}\left(\frac{1+(-1)^n}{2}\right)$$
We get
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n+1}}{{(2n)}^2}=4\left(\frac12\sum_{n=1}^\infty\frac{H_{n+1}}{n^2}+\frac12\sum_{n=1}^\infty(-1)^n\frac{H_{n+1}}{n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/609056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
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solving an expression based on sin $\theta$ If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$, then $x$ must be equal to what?
What does the following solution mean?
$0 \le \sin^2 \le 1$
This implies $0 \le \frac{x^2 + y^2 + 1 }{ 2x } \le 1$
This implies $\frac{(x - 1)^2 + y^2 }{2x} \le 0 $
This implies $x = 1$.
Can you ... | Certainly, for any value of $\theta$, we have $-1 \leq \sin \theta \leq 1$, and therefore, $0 \leq \sin^2 \theta \leq 1$, which bounds the left-hand side of your equation
$$
\sin^2 \theta = \frac{x^2+y^2+1}{2x}.
$$
Since the left-hand side is bounded, so must be the right-hand side, since they are equal. Thus,
$$
0 \le... | {
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"source": "stackexchange",
"question_score": "1",
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Applying Fermat's Little Theorem: $6^{1987}$ divided by $37$ Find the remainder when $6^{1987}$ is divided by $37$.
Because 37 is prime we have: $6^{36}$ mod $37 = 1$. I tried to get a nice combination like: $1987 = 36 * 55 + 7$, so we would have $(6^{36})^{55}6^{7}=6^{1987}$.
Then, I've taken mod $37$, which is: $6^{1... | $$6^2\equiv-1\pmod{37}$$
So $6^7\equiv (-1)^3\cdot 6\equiv 31\pmod{37}$
You could have utilized this from the start, since $6^2\equiv-1\implies 6^4\equiv1$. Thus, you could have written $1987=496\cdot 4+3$ and $$6^{1987}=(6^4)^{496}\cdot6^3\equiv 6^3\equiv 6^26\equiv-6\equiv31\pmod{37}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: Exactly a quarter of 3-part partitions of numbers >2 equal to 0, 2, 10 mod 12 will make a triangle. Consider perimeters $>2$ equal to $0$, $2$, or $10 \mod(12)$. The sequence starts $10, 12, 14, 22, 24, 26, 34, 36, 38, 46, 48, ...$ and we can look at the three part partitions that make triangles.
$\{\{2, 4, 4\... | Let $P_\ell(n)$ denote the number of partitions of $n$ with exactly $\ell$ parts.
The case $n=0\mod 12$:
According to wikipedia, the number of partitions with exactly three parts is equal to
$$P_3(n)=\left[\frac{(n+3)^2}{12}\right]-\left\lfloor \frac n2+1\right\rfloor$$
where $[]$ denotes the nearest integer and $\lflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/613202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding a vector close to vector $\vec{b}$ using $A^T$ and $A\vec{x}=\vec{b}$ I'm having a hard time understanding the rest of the steps after $A^TA\vec{x}=A^T\vec{b}$ to find $\vec{x}$
Problem:
Find the vector in $W= span\ \left(\right.\
\left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right]$ ,
$\left[ \begin{a... | First bullet point: This is called the normal equations.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ if........... Help please: If $\sin\alpha+\sin\beta= \sqrt{3} (\cos\beta-\cos\alpha)$ then show that $\cos^2\frac{1}{2}(\alpha-\beta)=\frac{3}{4}$ please tell me how can... | Use the following identities:
$$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}2\right)\cos\left(\frac{\alpha-\beta}2\right)\\
\cos\beta-\cos\alpha=-2\sin\left(\frac{\beta+\alpha}2\right)\sin\left(\frac{\beta-\alpha}2\right)$$
So the first condition is equivalent to
$$\cos\left(\frac{\alpha-\beta}2\right)=-\sqrt{3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$0Let $a_n$ be a sequence of positive real numbers such that
$$a_n<\frac{a_{n-1}+a_{n-2}}{2}$$
Show that $a_n$ converges.
| It is clear that $(a_n)_{n\in\mathbb{N}}$ is bounded between $\ \max \{a_1,a_2\} $ and $0.$ Therefore by the Bolzano-Weierstrass theorem, $\ (a_n)_{n\in\mathbb{N}}\ $ has at least one limit (accumulation) point in the interval$\ [\ 0,\ \max \{a_1,a_2\}\ ].$
Now suppose, by way of contradiction, that $\ (a_n)\ $ has at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/614882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
Doubts in Trigonometrical Inequalities I'm now studying Trigonometrical Inequalities, and I've just got struck when I have modified arguments to my trigonometrical functions, for example:
$\sqrt{2} - 2\sin\left(x - \dfrac{\pi}{3} \right) < 0$ when $-\pi < x < \pi$
With some work I've got: $\sin\left(x - \dfrac{\pi}{3} ... | First of all, you said that $\sin x = \frac{\sqrt{2}}{2}$ when $x = \frac{\pi}{4}, \frac{3\pi}{4}$ then concluded that $\sin x > \frac{\sqrt{2}}{2}$ when $\frac{\pi}{4} < x < \frac{3\pi}{4}$. While this is true, you should give some explanation here as it could be the case that $\sin x < \frac{\sqrt{2}}{2}$ for $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help finding the $\lim\limits_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$ I need help finding the $$\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \sqrt[5]{x}}$$
I did the following:
$$\begin{align*}
\lim_{x \to \infty} \frac{\sqrt[3]{x} - \sqrt[5]{x}}{\sqrt[3]{x} + \... | Your major mistake : $x^a + x^b \ne x^{a+b}$. This is very wrong and a common mistake.
Did you know that $\sqrt[n]{x} = x^{1/n}$? Therefore the intial expression can be written as : $$\frac{x^{1/3} - x^{1/5}}{x^{1/3} + x^{1/5}} = \frac{x^{1/3}(1 - x^{1/5-1/3})}{x^{1/3}(1+x^{1/5-1/3})} = \frac{1-x^{-2/15}}{1+x^{-2/15}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/615505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Probability Of Rolling A Strictly Increasing Sequence On A Six-Sided Die
By rolling a six-sided die 6 times, a strictly increasing sequence of
numbers was obtained, what is the probability of such an event?
I have no ideas on how to attack this. It says, an increasing sequence for the $6$ times were obtained. So th... | For each roll of the die, there are $6$ possible outcomes — ${1,2,3,4,5,6}$. You roll $6$ times, so there are $6$ possible outcomes for the first, for each of these there are $6$ possible outcomes for the second roll, and so on. This means there are $6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 \cdot 6 = 6^6$ sequences possible. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that at least two of these inequalities are true: $|a-b|\le2$, $|b-c|\le2$, $|c-a|\le2$. It's given that: $$\begin{cases}a,b,c>0\\a+b+c\le4\\ab+bc+ac\ge4\end{cases}$$
Prove without using calculus that it's true that at least two of these are correct inequalities:$$\begin{cases}|a-b|\le2\\|b-c|\le2\\|c-a|\le2\end{... | It is easy!
first, we know that
$a^2+b^2+c^2<=8$, therefore
$2a^2+2b^2+2c^2<=16$ ............. ine1
and we know $ab+ba+ca>=4$,so
$-2ab-2bc-2ca<=-8$ ......... ine2
so ine1+ine2, we get
$(a-b)^2+(b-c)^2+(a-c)^2<=8$
so ,it is easy to explain your inequalities:
$\begin{cases}(a-b)^2\le4\\(b-c)^2\le4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$, where $x$, $y$ and $z$ are positive integers Find all solutions of ${\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1$ , where $x,y,z$ are positive integers.
Found ten solutions $(x,y,z)$ as ${(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(... | They are the only possible solution. Proof is as follows:
Suppose that $d = gcd(x,y)$ and $x=d ~ x0$, $y=d ~y_0$ where $x_0$ and $y_0$ are co-prime.
Substituting in the original equation we get
$$ 1/x + 1/y + 1/z=1 \Rightarrow -d\,x_0\,y_0\,z+y_0\,z+x_0\,z+d\,x_0\,y_0 =0$$
Solving for $d$:
$$d=\frac{\left( y_0+x_0\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/616639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"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.