Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a... | Rewrite the equation $7X^4-14X^3-7X+2 = 0$ this way: $7X^3-14X^2-7+\frac{2}{X} = 0$. Replacing $X$ by $X_1$ then $X_2$ etc. and getting the sum will have:
$$
7\sum{X_i^3} -14\sum{X_i^2}-7\sum{X_i}+2\sum{\frac{1}{X_i}}=0 \tag1
$$
From $(1)$ should be easy to get the cube root's sum.
Note that $\sum{\frac{1}{X_i}}$ can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$
The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \... | If you saw something variable shows on the exponent, first take logarithm usually helps. Write
\begin{align}
& \left(\frac{\sin x}{x}\right)^{1/x^2} \\
= & \exp\left[\frac{1}{x^2}\log\left(\frac{\sin x}{x}\right)\right] \\
= & \exp\left[\frac{1}{x^2}\log\left(\frac{x - \frac{x^3}{6} + o(x^3)}{x}\right)\right] \\
= & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1535827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit with trigonometric function
Find $$\lim_{x \to \pi/4}\frac{1-\tan(x)}{\cos(2x)}$$
without L'Hôpital.
$$\lim_{x \to \pi/4}\frac{1-\tan(x)}{\cos(2x)}=\lim_{x \to \pi/4}\frac{\cos^2(x)+\sin^2(x)-\frac{\sin(x)}{\cos(x)}}{\cos^2(x)-\sin^2(x)}$$
How can I continue?
| Note that
$$
1-\tan x = \frac{\cos x - \sin x}{\cos x}=\frac{(\cos x - \sin x)(\cos x+ \sin x)}{\cos x(\cos x+ \sin x)}=\frac{\cos^2 x - \sin^2 x}{\cos x (\cos x+ \sin x)}=
\frac{\cos 2x}{\cos x (\cos x+ \sin x)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Possible values of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ given $x+y+z=1$ Given $x,y,z$ are real numbers and $x+y+z=1$ and $x$ is not equal to $z$, if $ {1\over x} + {1\over y} + {1\over z} = m $, which of the following values of $m$ are possible?
(A) 1
(B) 2
(C) 3
(D) All of these.
I don't know where to start. I thou... | (1) Consider the function $f(x)=x^3-x^2-2x+2=(x-1)(x^2-2)$ we have three real roots and $x+y+z=-{1\over -1}=1,{1\over x}+{1\over y}+{1\over z}=-{2\over-2}=1$.
(2) Consider the function $f(x)=x^3-x^2-40x+20$. Set the derivative equal $0$ we get $x=4$ and $x=-{10\over3}$ are two local extrema. When $x=4, f(x)<0$, when $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Square of a sum equals sum of cubes Consider a sequence $(a_n)$ of positive numbers such that
$$(a_1+\cdots+a_n)^2=a_1^3+\cdots +a_n^3,\quad n\ge 1.$$
Prove that $a_n=n$ for all $n\ge 1$.
| The question should specify that the sequence is increasing since $\{1,2,4,3,5\}$ is a counterexample.
Let's prove this by induction. The base case of $a_1 = 1$ holds since $a_1^2 = a_1^3$. Now assume that $(a_1,\ldots,a_k) = (1,\ldots,k)$ for some $k$. Then we have that $(1+2+\cdots+k+a_{k+1})^2 = \Bigg(\dfrac{(1+k)k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Converting Polar Equation to Cartesian Equation problem So I have
1. $$\frac{r}{3\tan \theta} = \sin \theta$$
2. $$r=3\cos \theta$$
What would be the Cartesian equation???
| the first
$$\frac{r}{3\tan \theta} = \sin \theta$$
$$\frac{r\cos \theta}{3\sin \theta} = \sin \theta$$
$$\frac{x}{3\sin \theta} =\sin \theta$$
$$x =3\sin^2 \theta$$
$$r^2x =3r^2\sin^2 \theta$$
$$x(x^2+y^2)=3y^2$$
The second
$$r=3\cos \theta$$
$$r^2=3r\cos \theta$$
$$r^2=3x$$
$$x^2+y^2=3x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\fr... | Using the recursion for Pascal's Triangle, $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, we get
$$
\begin{align}
a_n
&=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\dots\\
&=\binom{n-1}{0}+\binom{n-2}{1}+\binom{n-3}{2}+\dots\\
&+\binom{n-1}{-1}+\binom{n-2}{0}+\binom{n-3}{1}+\dots\\[4pt]
&=a_{n-1}+a_{n-2}
\end{align}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$ $$\lim\limits_{n\rightarrow\infty}\left(\frac{n^2 + 1}{n^2 - 2}\right)^{n^2}$$
Trying to solve this. At first thought it was $1$, but in wolfram it's $e^3$. Thanks
| Since $$\lim _{ n\rightarrow \infty }{ { \left( 1+\frac { 1 }{ n } \right) }^{ n } } =e$$
so
$$\lim _{ n\rightarrow \infty }{ { \left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-2 } \right) }^{ { n }^{ 2 } } } =\lim _{ n\rightarrow \infty }{ { \left( \frac { { n }^{ 2 }-2+3 }{ { n }^{ 2 }-2 } \right) }^{ { n }^{ 2 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $s$ for which the matrix is positive definite I am given the following Matrix
$$
A= \begin{pmatrix}
s & -4 & -4 \\
-4 & s & -4 \\
-4 & -4 & s \\
\end{pmatrix}
$$
and asked to find $s$ such that $A$ is positive definite. In finding it, I simply set the determinan... | The matrix $A$ will be positive definite if and only if all its eigenvalues are positive. You have
$$ A = \left( \begin{matrix} s & -4 & -4 \\ -4 & s & -4 \\ -4 & -4 & s \end{matrix} \right) = \left( \begin{matrix} -4 & -4 & -4 \\ -4 & -4 & -4 \\ -4 & -4 & -4 \end{matrix} \right) + \left( \begin{matrix} s + 4 & 0 & 0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$ Find the value of the expression $\tan\frac{\pi}{16}+\tan\frac{5\pi}{16}+\tan\frac{9\pi}{16}+\tan\frac{13\pi}{16}$
I identified that $\frac{\pi}{16}+\frac{13\pi}{16}=\frac{5\pi}{16}+\frac{9\pi}{16}=\frac{14\pi}{16}$
$\tan(\frac{\pi}{16}+... | let $\frac{\pi}{16}=A$ then $tan13A=-tan3A$ and $tan9A=-tan7A$
so
$$S=tanA-tan3A+tan5A-tan7A=\frac{-sin2A}{cosAcos3A}+\frac{-sin2A}{cos5Acos7A}$$
$$S=-2sin2A\left(\frac{1}{2cosAcos3A}+\frac{1}{2cos5Acos7A}\right)$$
$$S=-2sin2A\left(\frac{1}{cos4A+cos2A}+\frac{1}{cos12A+cos2A}\right)$$
$$S=-2sin2A\left(\frac{1}{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
| Integrate both sides. Let L.H.S be $f(x)$ and R.H.S be $g(x)$.
(1). L.H.S
$$\int \frac{ix^{i-1}}{1+x^{i}} = \ln (1+x^i)$$
This gives $$\int f(x) dx= \sum_{i=1}^\infty \ln (1+x^i) = \ln (\prod_{i=1}^\infty (1+x^i))$$
(2). R.H.S
$$\int \frac{(2i+1)x^{2i}}{1-x^{2i+1}} = -\int \frac{-(2i+1)x^{2i}}{1-x^{2i+1}} = -\ln (1-x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Finding the (complex) solutions $z$ to $z^4 = w$
What are possible solutions to $$z^4 = w$$ where $z \in \mathbb{C} $ and $w \in \mathbb{R}$?
Here is my attempt:
Write both numbers in polar form: $r^4(\cos 4 \theta + i\sin 4 \theta) = w(\cos 0 + i \sin 0) $.
Working with the conventions of $r \geq 0$ and $0 \leq \th... | Assume $z\in\mathbb{C}$ and $w\in\mathbb{R}$:
$$z^4=w\Longleftrightarrow$$
$$z^4=|w|e^{\arg(w)i}\Longleftrightarrow$$
$$z^4=we^{0i}\Longleftrightarrow$$
$$z=\left(we^{2\pi ki}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$z=\sqrt[4]{w}e^{\frac{2\pi ki}{4}}\Longleftrightarrow$$
$$z=\sqrt[4]{w}e^{\frac{\pi ki}{2}}$$
With $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| $\int \frac{x-1}{(x+3)(x^2+1)} dx= \int\frac{A}{x+3} + \frac{B(2x) + C}{x^2+1}\,dx $
and
$ x-1= A(x^{2}+1)+ (B(2x)+C)(x+3)$
which is simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Distribution of $Y$ given $X$ Let $X_1$ and $X_2$ be the numbers on two independent fair-die rolls. Let $X$ be the minimum and $Y$ the maximum of $X_1$ and $X_2$.
$(a)$ Find the distribution of $Y\mid X$
Here is my work:
Here is what I know:
$$P(Y = y\mid X=x) = \frac{P(Y=y, X=x)}{P(X=x)}$$
and $P(X=x)$ can be found by... |
So I said, $$P(Y=y\mid X=x) =
\begin{cases}
\frac{1}{36} \cdot P(X=x) & \text{for } x = y \\[4pt]
\frac{2}{36} \cdot P(X=x) & \text{for } x = 1,2,3,4,5,6, \ y > x
\end{cases}$$
Am I correct?
Almost but, no. You know that the conditional probability is the joint divided by the marginal, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
My question is, is this a double angle formula? And if so, how would I go about to solve it?
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
$$=2\sin{\pi \over 4}+\le... | Yes, you can use double angle identity to simplify as follows $$\left(\sin\frac{\pi}{8}+\cos\frac{\pi}{8}\right)^2=\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
$$=\left(\sin^2\frac{\pi}{8}+\cos^2\frac{\pi}{8}\right)+2\sin\frac{\pi}{8}\cos\frac{\pi}{8}$$
$$=1+2\sin\frac{\pi}{8}\cos\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
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Sum of a Series With Denominators of the form $(2^i) (3^j)(5^k)$ Can anyone solve this?
Find the sum of the series $1 + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \cdots,$ where the denominators are of the form $(2^i) (3^j)(5^k)$?
The test came with the next answer ... | Consider $$(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n} + \ldots)(1 + \frac{1}{3} + \frac{1}{9} + \ldots + \frac{1}{3^n} + \ldots)(1 + \frac{1}{5} + \frac{1}{25} + \ldots + \frac{1}{5^n} + \ldots)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why is my $\epsilon$-$\delta$ proof incorrect? Prove that $f(x)= \sqrt{4+x^2}$ is continuous at $x_0$ using the $\epsilon -\delta$ definition of continuity.
Textbook proof:
$|\sqrt{4+x^2}-\sqrt{4+x^2}|=\frac{|4+x^2-(4+{x_0}^2)|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\frac{|x^2-{x_0}^2|}{\sqrt{4+x^2}+\sqrt{4+{x_0}^2}}=\underbr... | Regarding your proof, there is nothing wrong with it. Only that dealing with $\delta=1$ is not needed, as you can always request $\delta<1$.
As for why the "results" are different? There is no result. No one says that for a given $f$ and a given $x_0$ and a given $\varepsilon$, there is a single $\delta$. Any $\delta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must... | You have an algebra error. When you put $x = -2$, the constraint equation collapses to $4C = 12$, so $C = 3$.
Your method for determining $B$ is correct – the constraining identity must hold for any $x$. Putting $x = 1$ is a fine choice as the arithmetic is easier. However the incorrect value for $C$ will affect the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Finding all possible integer values of $a,b,c$, given $2$ equations. If $a^2=bc+1$ and $b^2=ac+1$, how do I find all possible values of $a,b,c$. Assuming $a,b,c$ all are integers?
| Subtract one equation from the other. Then
$$a^2 - b^2 = bc - ac$$
$$(a + b)(a - b) = -(a - b)c$$
$$(a + b)(a - b) + (a - b)c = 0$$
$$(a - b)(a + b + c) = 0$$
Considering the case where $a = b$: We end up with $a^2 = ac + 1$ so that $a(a - c) = 1$. Since $a, c$ are integers, we only have to consider two cases: $a, a-c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd ... | From the initial data and recursion, compute $a_0=2,a_1=3,a_2=7,a_3=18$.
If we have an $S$ and $T$ so that $a_{n+1}=Sa_n+Ta_{n-1}$, then
$$
\begin{bmatrix}3&2\\7&3\end{bmatrix}\begin{bmatrix}S\\T\end{bmatrix}=\begin{bmatrix}7\\18\end{bmatrix}\implies\begin{bmatrix}S\\T\end{bmatrix}=\begin{bmatrix}3&2\\7&3\end{bmatrix}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
$\lim\limits_{n\to\infty}$ $(1+ {2\over n})^n$ = $e^2$ If I set N=$2\over n$, the equation becomes $(1+ N)^{2\over N}$, which I can take the natural log of and work down until I reach $e^2$. However, my teacher wants me to use subsequences, starting with $x_n$ = $(1+ {1\over n})^n = e$, to prove this and I am stuck.
| Let $a_n$ =$(1+2/n)^n$ and $a_{n+1}$=$1+$($\frac{2}{n+1})^{n+1} $
Proceed by their binomial expansion.
$a_n$=$1+n$($\frac{2}{n})$+n($\frac{n-1}{2!})$ $(\frac{2}{n})^{2}$+⋯+$\frac{n(n-1)(n-2)…}{n!}$ ($\frac{2}{n})^{n}$
$a_n=$1+2+($\frac{1}{2!})$ $2^2$(1-$\frac{1}{n})$+($\frac{1}{3!})$ $2^3$(1-$\frac{1}{n})$(1-$\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluating Limits - Finding Multiple Results Only one of which is Correct When I calculate the limit
$$\lim_{x\to\infty}\sqrt{x^2+x} - \sqrt {x^2-x}$$
I get $2$ answers for this question: $1$ and $0$ but $1$ is the right answer. I don't know why this is the case, however. If you multiply by the conjugate divided by th... | $
x\bigl(
\sqrt{1+\tfrac1x}
-
\sqrt{1-\tfrac1x}
\bigr)
$
as $x\to\infty$ is of the form $\infty\cdot0$, which is indeterminate hence your wrong conclusion.
A correct way to get rid of the indeterminate form is as you suggested earlier: indeed,
\begin{align}
\sqrt{x^2+x} - \sqrt{x^2-x}
={}&
\frac{
\left(\sqrt{x^2+x} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Induction Proof with Fibonacci How do I prove this?
For the Fibonacci numbers defined by $f_1=1$, $f_2=1$, and $f_n = f_{n-1} + f_{n-2}$ for $n ≥ 3$, prove that $f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n$ for all $n≥ 1$.
| the inductive step:
$$
f^2_{n+1} - f_{n+1}f_n - f^2_n = (-1)^n \Leftrightarrow \\
f_{n+1}f_{n-1} - f_n^2 =(-1)^n \Leftrightarrow \\
(f_{n+2}-f_n)(f_{n+1}-f_n) -f_n^2 = (-1)^n \Leftrightarrow \\
f_{n+2}f_{n+1}-f_nf_{n+3} = (-1)^n \Leftrightarrow \\
f_{n+1}^2+f_nf_{n+1}-f_nf_{n+1}-f_nf_{n+2}=(-1)^n \Leftrightarrow \\
f_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 5
} |
Solve $x^2\equiv -3\pmod {\!91}$ by CRT lifting roots $\!\bmod 13\ \&\ 7$ Question 1) Solve $$x^2\equiv -3\pmod {13}$$
I see that $x^2+3=13n$. I don't really know what to do? Any hints?
The solution should be $$x\equiv \pm 6 \pmod {13}$$
Question 2) $\ $ [note $\bmod 7\!:\ x^2\equiv -3\equiv 4\iff x\equiv \pm 2.\,$ Her... | For Question 2), you can use the Chinese Remainder Theorem on the one equation
$$
13x+7y=1\tag{1}
$$
Using the Extended Euclidean Algorithm, an implementation of which is in this answer, compute
$$
\begin{array}{r}
&&1&1&6\\\hline
1&0&1&-1&7\\
0&1&-1&2&-13\\
13&7&6&1&0
\end{array}\tag{2}
$$
that is,
$$
13(-1)+7(2)=1\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding Lyapunov function for a given system of differential equations I am being introduced to the Lyapunov functions in order to determine the stability of a given system. I know that finding a Lyapunov function is not easy, so I would like to ask for any trick or hint in order to find a Lyapunov function for
$$ \lef... | It is not quite clear why you put a bounty on this question, since @Evgeny answered it in the best possible way. However, if you are looking for a Lyapunov function, here it is (up to an additive constant):
$$
L(x,y)=\frac{2^{2/3} \left(1-\frac{x (x+4) \left(x^2-12 x+8 y+48\right)}{\sqrt[3]{x^3 (x+4)^3} \left(x^2-4 y\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Green's Theorem on Line Integral I am asked to find the line integral for the following field:
$$F = (e^{y^2}-2y)i + (2xye^{y^2}+\sin(y^2))j$$
On the line segment with points $(0,0),(1,2)$ and $(3,0)$. I have to do it with Greens theorem. This is the setup I have so far.
$$\int_C F \cdot dr = \iint_D \frac{\partial Q}{... | The partial derivative of $Q$ regarding $x$ lacks a factor $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate:
$$x^2=\frac{x+y}{x-y}$$
Preferring to avoid the quotient rule, I take away the fraction:
$$x^2=(x+y)(x-y)^{-1}$$
Then:
$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$
If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if ... | You should first find $y$ in terms of $x$ as follows $$x^2=\frac{x+y}{x-y}$$ $$x^3-x^2y=x+y$$$$ (1+x^2)y=x^3-x$$
$$y=\frac{x^3-x}{1+x^2}$$
now, differentiate w.r.t. $x$, $$\frac{dy}{dx}=\frac{(1+x^2)(3x^2-1)-(x^3-x)(2x)}{(1+x^2)^2}$$
$$=\frac{x^4+4x^2-1}{(1+x^2)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $x+y+z=6$ and $xyz=2$, then find $\cfrac{1}{xy} +\cfrac{1}{yz}+\cfrac{1}{zx}$
If $x+y+z=6$ and $xyz=2$, then find the value of $$\cfrac{1}{xy}
+\cfrac{1}{yz}+\cfrac{1}{zx}$$
I've started by simply looking for a form which involves the given known quantities ,so:
$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfra... | $$x+y+z=6$$
Divide both sides with $xyz$
$$\frac{x+y+z}{xyz}=\frac{6}{xyz}$$
$$\frac{x}{xyz}+\frac{y}{xyz}+\frac{z}{xyz}=\frac{6}{2}$$
$$\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion Prove that $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{1}{60}$ without using L Hospital rule or series expansion.
I tried $\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\lim_{... | As per the method suggested by Lab bhattacharjee, I got the answer to my problem.Thanks to him.
Let $L=\lim_{x\to 0}\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}......(1)$
So $L=\lim_{x\to 0}\frac{2+\cos 2x}{8x^3\sin 2x}-\frac{3}{16x^4}$
$L=\lim_{x\to 0}\frac{1+2\cos^2x}{16x^3\sin x\cos x}-\frac{3}{16x^4}$
$16L=\lim_{x\to 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Line tangent to circle
A circle with a radius of $2$ units has its center at $(0,0)$. A circle with a radius of $7$ units has its center at $(15,0)$. A line tangent to both circles intersects the $x$-axis at $(x,0)$. What is the value of $x$? Express your answer as a common fraction.
My problem with this question is ... | So there is solution using the $Ax+By+C=0$ notation for lines. Given a circle with center $(x_c,y_c)$ and radius $r$ the all the possible tangents are given by the line:
$$ x (\cos\psi) + y (\sin \psi) -(x_c \cos\psi + y_c \sin \psi + r) = 0 $$
In fact, every scalar multiple of the $(A,B,C)$ coefficients also describe ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Double sum of products of integers up to $n$ Suppose that $S$ is defined by
$$
S(n) = \sum_{i=0}^{n} \sum_{j=0}^{i} ij.
$$
I'm confused as to how $S(3) = 25$ from this summation. Can anyone expand on it as to how to get the answer?
| \begin{align*}
S(3) &= \sum_{i=0}^3\sum_{j=0}^i ij \\
&= 0\sum_{j=0}^0 j + 1\sum_{j=0}^1 j + 2\sum_{j=0}^2 j + 3\sum_{j=0}^3 j \\
&= 0 + (0+1) + 2(0+1+2) + 3(0+1+2+3) \\
&= 25.
\end{align*}
More generally,
\begin{align*}
S(n) &= \sum_{i=0}^n\sum_{j=0}^i ij \\
&= \sum_{i=0}^n i\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How find gcd polynomials? How to find gcd of polynomials $gcd(x^3+x^2-x-1,3x^2+2x-1)$ ??
I divide of polynomials. It worked like this $\frac 13 x - \frac19$,$ R\left( x\right) =-\frac89x -\frac89$
| HINT:
If $f(x)$ divides both,
$f(x)$ must divide $$3(x^3+x^2-x-1)-x(3x^2+2x-1)=x^2-2x-3=(x-3)(x+1)$$
Now $3x^2+2x-1=(3x-1)(x+1)$
OR
$x^3+x^2-x-1=(x+1)(x^2-1)=(x+1)^2(x-1)$
and
$3x^2+2x-1=(3x-1)(x+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Conditions under which $a+b+c$ divides $1-abc$ What are the conditions such that $a+b+c$ divides $1-abc$, where $(a, b, c)$ are nonzero integers ?
| The triples $(a,b,c)$ with the property that $a+b+c$ divides $1-abc$ can be characterized as triples $(a,b,d-a-b)$, where $d$ is a divisor of $1+ab(a+b)$. This is because $a+b+c=d$ for such triples, while
$$1-abc=1-ab(d-a-b)=1+ab(a+b)-abd$$
For example, if $a=b=4$, we have $1+ab(a+b)=129$, which has $8$ divisors: $d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of ways of forming 4 letter words using the letters of the word RAMANA
Question: Find the number of ways of forming 4 letter words using the letters of the word "RAMANA"
This can be solved easily by taking different cases.
*
*All 3 'A's taken: remaining one letter can be chosen in $^3C_1$ ways. Total possi... | Suppose you have a $4$ letter string composed of, say, $1$ distinct and $3$ identical letters
There would be $\frac{4!}{1!3!}$ permutations, also expressible as a multinomial coefficient, $\binom{4}{1,3}$
Similarly, for $2$ distinct, $2$ identical, and $3$ distinct, $1$ identical,
it would be $\binom{4}{2,2}\;$ and $\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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If $x-y = 5y^2 - 4x^2$, prove that $x-y$ is perfect square Firstly, merry christmas!
I've got stuck at a problem.
If x, y are nonzero natural
numbers with $x>y$ such that
$$x-y = 5y^2 - 4x^2,$$
prove that $x - y$ is perfect square.
What I've thought so far:
$$x - y = 4y^2 - 4x^2 + y^2$$
$$x - y = 4(y-x)(y+x) ... | Generalization. Let $a,b$ be integers. If there exists consecutive integers $c,d$ such that $a-b=a^2c-b^2d$, then $|a-b|$ is a perfect square.
Proof. If $c=d+1$, we have $a-b=a^2(d+1)-b^2d=(a-b)(a+b)d+a^2$, so $$a^2=(a-b)(1-d(a+b))$$ Now let $g=\text{gcd}(a-b,1-d(a+b))$. We have $g^2|a^2$, so $g|a$. Now we have $g|b$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
$3^{n+1}$ divides $2^{3^n}+1$ Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$.
I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt ... | This actually looks like an induction proof. Let $P(n)$ be the statement : $$3^{n+1} \ \mid \ 2^{3^n} + 1 $$
Then we can verify that it is indeed true for $n = 1$ since $9$ divides itself. So suppoose $P(n)$ is true. Now,
$$2^{3^{n+1}} + 1 = \left({2^{3^n}}\right)^3 + 1 $$
Notice that $$\left({2^{3^n} + 1}\right)^3 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
prove this inequality with $a+b+c=1$ Let $a,b,c>0,a+b+c=1$,show that
$$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\ge \dfrac{16}{3(a+b)(b+c)(c+a)}$$
| This inequality I think some time,Now I solve it.Following is my solution.
Use Holder inequality we have
$$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\cdot\sum_{cyc}c(a+b)^2\ge (a+b+b+c+c+a)^3$$
so
$$\left(\sqrt{\dfrac{a+b}{c}}+\sqrt{\dfrac{b+c}{a}}+\sqrt{\dfrac{c+a}{b}}\right)^2\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^n+y^n=2015$ Determine the natural numbers $x,y,n$ matching equality
$$x^n+y^n=2015.$$
I noticed for $n = 1$ the equation has solutions $(x, 2015-x), x$ integer.
For $n = 2$, given that $x$ and $y$ are different parities taking $x = 2k$ and $y=2m + 1$ we come to contradiction.
What must be done to $n\geq3$?
| Since $2015=5\cdot 13\cdot 31$, and the multiplicities of the primes congruent $3$ modulo $4$ are odd (only $p=31$ to consider here), it follows that $2015$ cannot be represented as the sum $x^2+y^2$, by Fermat's theorem. This does not only work for $2015$. Also, characterisation of two cubes is known, see here. Of cou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| $$\int{\frac{x^2 + 4}{x^2 + 6x +10}}dx=x-6\int{\frac{x+1}{x^2 + 6x +10}}dx$$
$$\int{\frac{x+1}{x^2 + 6x +10}}dx=\int{\frac{x+1}{(x+3)^2+1}}dx$$ now we can use $u$ substitution where $u=x+3$ and $du = dx$ so
$$\int{\frac{u-2}{u^2+1}}du=\int{\frac{u}{u^2+1}}du-2\int{\frac{1}{u^2+1}}du=\frac{\ln|u^2+1|}{2}-2\arctan(u)+C$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.
| This answer uses the hint using the polygamma function
$$
\psi^{(2)}(z) = -\int_0^1 \frac{t^{z-1}}{1-t}\ln^2t dt.
$$
First, using the expansion of $(1-t)^{-1}$, the polygamma function $\psi^{(2)}(x)$ can be written as
\begin{align}
\psi^{(2)}(z) &= -\sum_{n=0}^\infty \int_0^1 t^{n+z-1}\ln^2 tdt \cr
&= -\sum_{n=0}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric Ratios Of Multiple Angles If $2\tan A=3\tan B$ then prove that $$\tan(A-B)=\frac{\sin2B}{5-\cos 2B}.$$
I found that $\tan A=\frac{3}{2} \tan B$ and after that used the formula of $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$ but could not reach to the required answer.
| We have
$$\tan\left(A-B\right) = \dfrac{\tan(A)-\tan(B)}{1+\tan(A)\tan(B)} = \dfrac{\tan(B)/2}{1+\dfrac32 \tan^2(B)} = \dfrac{\tan(B)}{2+3\tan^2(B)} = \dfrac{\dfrac{\sin(B)}{\cos(B)}}{2+3 \cdot \dfrac{\sin^2(B)}{\cos^2(B)}}$$
Hence, we have
$$\tan\left(A-B\right) = \dfrac{\sin(B)\cos(B)}{2\cos^2(B)+3\sin^2(B)} = \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
solution of nested radical $\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$ This question is from my friend. he think that there is trigonometry involved to this equation.
$\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$
that is from $\ \ \ x = \sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt...} } }} } }$
I try to compare... | $$\begin{align}
\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } &=x\\
7+2\sqrt{7-2\sqrt{7-2x} } &=x^2\\
2\sqrt{7-2\sqrt{7-2x} } &=x^2-7\\
4\left(7-2\sqrt{7-2x} \right) &=x^4-14x^2+49\\
-8\sqrt{7-2x} &=x^4-14x^2+21\\
64(7-2x) &=x^8-28x^6+238x^4-588x^2+441\\
0 &=x^8-28x^6+238x^4-588x^2+128x-7\\
0&=\left(x^2+2x-7\right)\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series represen... | Your sum is $$\sum_{n = 1}^{\infty} \sum_{k = 1}^{n} \frac {k} {n!} = \sum_{n = 1}^{\infty} \frac {n + 1} {2 (n - 1)!} = \sum_{n = 0}^{\infty} \frac {n + 2} {2 n!} = \frac {3e} {2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 3
} |
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.
If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
| Expanding the whole thing, you get
$$
(2k+1)^2(2\ell+1)^2 -1 = 16 k^2 \ell^2+16 k^2 \ell+4 k^2+16 k \ell^2+16 k \ell+4 k+4 \ell^2+4 \ell
$$
Modulo 8, the terms multiplied by 16 disappear, and this becomes
$$
4 k^2+4 k+4 \ell^2+4 \ell = 4k(k+1) + 4\ell(\ell+1)
$$
and it only remains to see that both $\ell(\ell+1)$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 4
} |
I have a problem with solving this integral by partial fraction I have an example like this :
$\int \frac{1}{x^{4}}dx$ = $\int x^{-4}dx$ = $\frac{1}{-4 + 1}x^{-4 + 1}+C$ = $\frac{1}{-3}x^{-3}+C$
What if this?
$\int \frac{1}{2x-10}dx$ = ?
| $\displaystyle \int\! x^r \, dx=\frac{x^{r+1}}{r+1}+C$ for $\color{red}{r \ne -1}$.
Let $u = 2x-10$ then $dx = \frac{1}{2}du$ and
$\displaystyle \int \frac{1}{2x-10}dx=\frac{1}{2}\int \frac{1}{u}\,du =\frac{1}{2}\ln|u|+C = \frac{1}{2}\ln|2x-10|+C. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For all $n >0$, $ \left(1+\frac{1}{n} \right)^n = 1+ \sum_{k=1}^n\bigl[ \frac{1}{k!} \prod_{r=0}^{k-1}(1-\frac{r}{n}) \bigr]$ I am working through some problems on induction and I have been stuck on this one for a while. If anyone has any hints. I can show it is true for the $n=1$ and $n=2$ case but I am having difficu... | s(n+1)
Hints:
1.(1 + 1/n+1)^n+1 = (1 + 1/n)^n(1+ 1/n)...
2.Split up the sums
3.Use the binomial Theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find Complex Roots
Question:
Find the complex roots of $$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
What I have attempted:
$$ {(z^{12} -1)\over (z^4-1)(z^3-1)} = 0 $$
$$ {(z^{6} -1)(z^{6} +1)\over (z^2-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^{3} -1)(z^3 +1)(z^6+1)\over (z+1)(z-1)(z^2+1)(z^3-1)} = 0 $$
$$ {(z^3 +1)(z^6+... | A fraction is zero if and only if its numerator is zero, and well-defined if and only if its denumerator is non-zero.
Therefore
$$
\frac{z^{12} - 1}{(z^4 - 1)(z^3 - 1)} = 0
\iff
z^{12} = 1, z^4 \neq 1, z^3 \neq 1.
$$
The solutions of the first condition are given by the $12$-th root of unity, namely
$$
z = e^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer Square Root Algorithm I'm currently working my way through David M. Bressoud's "Factorization and Primality Testing", and I'm struggling with an exercise (exercise 5.7) that asks the reader to prove that the following algorithm produces the greatest integer less than or equal to the square root of $n$:
\begin{a... | Using the notation of the $a$ and $b$ is only necessary when programming.
You have correctly identified that the algorithm gives a sequence of values, which I will call (as you did) the $x_m$.
We thus have the relationship $x_{m+1} = \lfloor \frac{x_m^2 + n}{2 x_m} \rfloor$
Starting with $x_0=n$ gives $x_{1} = \lfloor ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I integrate this? (for calculate value of L-function ) I want to calculate the definite integral:
$$
\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx.
$$
Indeed, I already know that $\int_{0}^{1} \frac{x+x^{3}+x^{7}+x^{9}-x^{11}-x^{13}-x^{17}-x^{19}}{x(1-x^{20})}dx=L(\chi,1)=\f... | You may notice that, through the substitution $x=\frac{1}{z}$, we have:
$$ I=\int_{0}^{1}\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,dx = \int_{1}^{+\infty}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz $$
hence:
$$ I = \frac{1}{2}\int_{0}^{+\infty}\frac{1+z^6}{1-z^2+z^4-z^6+z^8}\,dz = \frac{1}{4}\int_{\mathbb{R}}\frac{1+z^6}{1-z^2+z^4-z^6+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Finding the sum of the infinite series whose general term is not easy to visualize: $\frac16+\frac5{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\cdots$ I am to find out the sum of infinite series:-
$$\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+................ | This is $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{1}{2}) - 1\right)$ where $_2F_1$ is the hypergeometric function. For these parameters this evaluates to $\frac{1}{2} \left(_2F_1(2/3,1;1;\frac{z}{2}) - 1\right) = \frac{1}{2} \left( (1 - \frac{z}{2})^{-2/3} - 1\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| As $a^2 + b^2 = 1$, there exists a $\theta$, such that $a = \cos\theta$ and $b=\sin\theta$
$$ a= \cos\theta = \frac{1-\tan^2\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$
$$ b= \sin\theta = \frac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$$
so exists $t = \tan\frac{\theta}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Pove that the angle between planes in which origin lies is acute if $a_1a_2+b_1b_2+c_1c_2<0$ Suppose we have two planes $$a_1x+b_1y+c_1z+d_1=0$$ and $$a_2x+b_2y+c_2z+d_2=0$$ where $d_1,d_2 >0 \ or \ <0$ then prove that the angle between planes in which origin lies is acute if $$a_1a_2+b_1b_2+c_1c_2<0$$
The angle betwe... |
The angle between the planes is given by $$cos\theta=\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}.\sqrt{a_2^2+b_2^2+c_2^2}} \tag1$$
More precisely, one of the angles between the planes is given by $(1)$.
So, if you mean that $\theta=\text{(the angle between the planes in which origin lies)}$, then we have
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the Fourier's coefficient $a_n$ of the Fourier's series of $\sin(x)$ on $(0,\pi]$, $0$on $(-\pi,0]$ Considering $g(x)$, periodical with a period of $2\pi$ defined by
\begin{equation*}
g(x)=
\begin{cases}
0 & \text{for $x \in (-\pi;0]$} \\
\sin(x) & \text{for $x \in [0;\pi)$}
\end... | For
\begin{equation*}
g(x)=
\begin{cases}
0 & \text{$x \in [-\pi,0)$} \\
\sin(x) & \text{$x \in [0,\pi)$}
\end{cases}
\end{equation*}
Write
$$g(x) = \frac12 a_0 + \sum_{n=1}^{\infty} \left (a_n \cos{n x} + b_n \sin{n x} \right ) $$
$$a_0 = \frac1{\pi} \int_0^{\pi} dx \, \sin{x} = \frac{2}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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integration $\int \frac{-4x+5}{\sqrt{12x-4x^2-8}}$
$$\int \frac{-4x+5}{\sqrt{12x-4x^2-8}}$$
My attempt:
$-4x^2+12x-8=-(2x-3)^2+1$
$$\begin{align}\int \frac{-4x+5}{\sqrt{1-(2x-3)^2}}\,dx&=\int \frac{-4x}{\sqrt{1-(2x-3)^2}}\,dx+\int \frac{5}{\sqrt{1-(2x-3)^2}}\,dx\\
&=
-4\int \frac{x}{\sqrt{1-(2x-3)^2}}\,dx+5\int \frac... | Noting that $-4x+5=-2(2x-3)-1$, you get:
$$\int \frac {-2(2x-3)}{\sqrt{1-(2x-3)^2}}\,dx -\int\frac{1}{\sqrt{1-(2x-3)^2}}\,dx$$
In the first, let $u=1-(2x-3)^2$, then $du=-4(2x-3)\,dx$ and in the second, let $v=2x-3$ and you get the integrals:
$$\frac{1}{2}\int \frac{du}{\sqrt{u}} - \frac{1}{2}\int \frac{dv}{\sqrt{1-v^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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limit as n goes to infinity of $\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$
How do you go about solving $$\lim_{n\to\infty}\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$$
I know that I have to fix the top so that it is not $(\infty - \infty$),
but if I multiple it by
$$\frac{\sqrt{n^3-3}+\sqrt{n^3+2n^2+... | When you are at the stage
$$
\lim_{n\to\infty}\frac{-2n^2-6}{
\sqrt{n+2}\,(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})
}
$$
pull $n^2$ from the numerator, $n$ from $n+2$ and $n^3$ from the other two radicals, so you have
$$
\lim_{n\to\infty}\frac{n^2\left(-2-\dfrac{6}{n^2}\right)}{
\sqrt{n}\,\sqrt{1+\dfrac{2}{n}}\,
\sqrt{n^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$
Let $x=\cos\theta$ and $y=\sin\... | HINT:
Let $\dfrac{4-\sin \theta}{7-\cos \theta}=u$
Use Weierstrass Substitution $t=\tan\dfrac\theta2$ to form a Quadratic Equation in $t$ on rearrangement.
As $t$ is real, the discriminant must be non-negative
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the ratio of empty to filled volume of the glass? The base diameter of a glass is $20$% smaller than the diameter at the rim. The glass is filled to half of the height. Then what is the ratio of empty to filled volume of the glass ?
| Let $r_1, r_2$ and $r_3$ be the top, bottom and middle radius respectively, and h be the hight of the cone.
Volume of glass= $\frac{1}{3}\pi h (r_1+r_2+r_1 r_2)$.
Let $r_1=10$, then $r_2=8$ and $r_3=9$.
So the ratio of volume of empty portion of glass to volume of filled portion of glass,
$=\frac{\frac{1}{3} \pi\frac{h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$ Find $\int_{1}^{2}\frac{x-1}{x^2\sqrt{x^2+(x-1)^2}}$
I tried to solve it by using the property $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$
Let $I=\int_{1}^{2}\frac{(x-1)dx}{x^2\sqrt{x^2+(x-1)^2}}$
$I=\int_{1}^{2}\frac{(2-x)dx}{(3-x)^2\sqrt{(3-x)^2+(2-x)^2}}$
But thi... | First step is put $x = u+\dfrac{1}{2}\Rightarrow \sqrt{x^2+(x-1)^2}=\sqrt{\left(u+\dfrac{1}{2}\right)^2+\left(u-\dfrac{1}{2}\right)^2}=\sqrt{2u^2+\dfrac{1}{2}}=\sqrt{2}\cdot \sqrt{u^2+\dfrac{1}{4}}$. Second step is to put $u =\dfrac{1}{2}\tan \theta$. I think we can go further at this point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Now am I doing induction correctly? Recursion: $L_n = L_{n-1} + n$ where $L_0 = 1$.
We guess that solution is $L_n = \frac{n(n+1)}{2} + 1$.
Base case: $L_0 = \frac{0(0+1)}{2} + 1 = 1$ is true.
Inductive step: Assume $L_n = \frac{n(n+1)}{2} + 1$ is true for some $n$. We will show that $L_{n+1} = \frac{(n+1)(n+2)}{2} + 1... |
Base case: $L_0 = \frac{0(0+1)}{2} + 1 = 1$ is true.
Inductive step: Assume $L_n = \frac{n(n+1)}{2} + 1$ is true for some $n$. We will show that $L_{n+1} = \frac{(n+1)(n+2)}{2} + 1$ given that $L_n = L_{n-1} + n$ is true.
Fine.
$L_{n+1} = \frac{(n+1)(n+2)}{2} + 1 = L_n + (n+1)$
Don't start with $L_{n+1}=\frac{(n+1)... | {
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"timestamp": "2023-03-29T00:00:00",
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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function satisfying $f(x+y^3)=f(x)+f(y)^3$
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function satisfying $f(x+y^3)=f(x)+f(y)^3$ for all $x,y\in\mathbb{R}$. If $f'(0)\ge0$, find $f(10)$.
If $x=y=0$, $f(0+0)=f(0)+f(0)$. So $f(0)=0$.
If $x=0$, $f(0+y^3)=f(0)+f(y)^3$. So $f... | Since $f'(0)$ exists, we know that $f'(0) = \displaystyle\lim_{y \to 0}\dfrac{f(y)-f(0)}{y-0} = \lim_{y \to 0}\dfrac{f(y)}{y}$ exists.
Then, for any $x \in \mathbb{R}$, we have $\dfrac{f(x+y^3)-f(x)}{y^3} = \dfrac{f(y)^3}{y^3} = \left(\dfrac{f(y)}{y}\right)^3 \to f'(0)^3$ as $y \to 0$.
Therefore, $f'(x)$ exists for a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If in a triangle $\alpha$ is pi/3 then $\frac 1c+\frac 1b\ge\frac2a$. If in a triangle $\alpha$ is pi/3 then $\frac 1c+\frac 1b\ge\frac2a$.
Since $\alpha$ is pi/3 that means $ b\le a \le c $.
Then i used AM and HM ,so $\frac 1c+\frac 1b\ge\frac {4}{c+b} $ .
and since $\alpha$ is pi/3 and $ b\le a \le c $ then $ \frac ... | From the Law of Sines, we know $\frac{\sin\alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin\gamma}{c}$, so
$$ \frac{1}{b}+\frac{1}{c} = \sin\alpha\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\frac{1}{a}.$$
It suffices to prove
$$\sin\alpha\left(\frac{1}{\sin\beta}+\frac{1}{\sin\gamma}\right)\ge 2.$$
Since $\alpha =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the maximum value of $\sqrt6 xy + 4yz$ given $x^2 + y^2 + z^2 = 1?$ Problem:
Let $x$,$y$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 1.$ Find the maximum possible value of $\sqrt6 xy + 4yz$
I don't know how to proceed with the question. Applying AM-GM inequality doesn't work because the second equatio... | Let $a,b\in\mathbb{R}$. We shall maximize $axy+byz$ subject to $x^2+y^2+z^2=1$ ($x,y,z\in\mathbb{R}$).
By Cauchy-Schwarz, $axy+byz=(ax+bz)y\leq \left(\sqrt{a^2+b^2}\sqrt{x^2+z^2}\right)|y|$. By AM-GM, $\sqrt{x^2+z^2}\,|y|\leq \frac{\left(x^2+z^2\right)+y^2}{2}=\frac{1}{2}$. That is, $axy+byz\leq \frac{\sqrt{a^2+b^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$ I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$
I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$
but it ... | Remember the identity: $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$.
$$\left(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1}\right)\sqrt{3n^3+1}$$
$$=\frac{\left(\left(n^3+\sqrt{n}\right)-\left(n^3-1\right)\right)\sqrt{3n^3+1}}{\sqrt[3]{\left(n^3+\sqrt{n}\right)^2}+\sqrt[3]{\left(n^3+\sqrt{n}\right)\left(n^3-1\right)}+\sqrt[3]{\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{a^3+2b^3}+\frac{b^3}{b^3+2c^3}+\frac{c^3}{c^3+2a^3} \geq 1.$
Prove that if $a,b,$ and $c$ are positive real numbers, then $$\dfrac{a^3}{a^3+2b^3}+\dfrac{b^3}{b^3+2c^3}+\dfrac{c^3}{c^3+2a^3} \geq 1.$$
This question seems hard since we aren't give... | Let $x=a^3$, $y=b^3$, $z=c^3$. By Cauchy-Schwarz inequality:
$$\left(\sum_{\text{cyc}}\frac{x}{x+2y}\right)\ge \frac{(x+y+z)^2}{\left(\sum_{\text{cyc}} x(x+2y)\right)}=1$$
Equality holds iff $\frac{\frac{x}{x+2y}}{x(x+2y)}=\frac{\frac{y}{y+2z}}{y(y+2z)}=\frac{\frac{z}{z+2x}}{z(z+2x)}$, i.e. iff $x+2y=y+2z=z+2x$,
i.e. i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Volume between hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and line $x = 2a$ around $y$ axis I'm trying to calculate the volume between the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the line $x = 2a$ around the $y$ axis using two methods but I'm getting different answers:
*
*Using volume of a solid... | The second integral is wrong.
\begin{equation*}
\int \left(x\left(\frac{2b}{a}\sqrt{x^2-a^2}\right)\right)\,dx
= \frac{2b}{3a}\int 3x\sqrt{x^2-a^2}\,dx
= \frac{2b}{3a}(x^2-a^2)^{3/2},
\end{equation*}
so the definite integral is
\begin{equation*}
V = 2\pi\int_a^{2a} \left(x\left(\frac{2b}{a}\sqrt{x^2-a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607599",
"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\tan{\frac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\frac{\pi\cos{x}}{4\cos{\alpha}}} > 1$
Prove the inequality $$\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} > 1$$
for any $x, \alpha$ with $0 \leq x \leq \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$... | We could also do this without casework using Jensen inequality, and using the fact that the first derivative of $\tan(x)$ is always positive and the second derivative is positive when $0 < tan(x) < \frac{\pi}{2}$.
So, by Jensen we have,
$\tan{\left(\dfrac{\pi\sin{x}}{4\sin{\alpha}}\right)}+\tan{\left( \dfrac{\pi\cos{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have... | Assume without loss of generality that $\displaystyle A , B, C $ belongs to the first quadrant.By the identity $\displaystyle sen^2x+cos^2x=1$, we can see that:
\begin{equation*}
tanx+cotx=\frac{1}{senxcosx}
\end{equation*}
So:
\begin{equation}
\frac{sen\frac{\beta}{2}sen\frac{\gamma}{2}}{sen\frac{\alpha}{2}co... | {
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"url": "https://math.stackexchange.com/questions/1609327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 2
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Is there an easy way to compute the determinant of matrix with 1's on diagonal and a's on anti-diagonal? \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
Thanks
| Partition the matrix into blocks $\begin{bmatrix}A&B\\C&D\end{bmatrix}$, where
$A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, $B = \begin{bmatrix}0&0&a\\0&a&0\\a&0&a\end{bmatrix}$, $C = \begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}$, $D = \begin{bmatrix}1&0&a\\0&1&0\\0&0&1\end{bmatrix}$.
The block matrix de... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Probability that $2^a+3^b+5^c$ is divisible by 4
If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4.
For a number to be divisible by $4$, the last two digits have to be divisible by $4$
$5^c= \_~\_25$ if $c>1$
$3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$
$2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=... | $2^{\color\red{a}}+3^\color\green{b}+5^\color\magenta{c}\equiv0\pmod4\iff$
*
*$\Big(\big(\color\red{a}=1\big)\wedge\big((\color\green{b}=2)\vee(\color\green{b}=4)\big)\Big)\vee$
*$\Big(\big(\color\red{a}\neq1\big)\wedge\big((\color\green{b}\neq2)\wedge(\color\green{b}\neq4)\big)\Big)$
Therefore, the probability ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Prove that $\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)=\left(\frac{p}{3}\right)$ Why does $\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)$ equal $\left(\frac{p}{3}\right)$?
| Use quadratic reciprocity :
$$ \left (\frac{p}{3} \right )=\left (\frac{3}{p} \right ) \cdot (-1)^{\frac{3-1}{2} \cdot \frac{p-1}{2}}$$
But we also know that : $$(-1)^{\frac{p-1}{2}}=\left (\frac{-1}{p} \right )$$ and thus the conclusion follows :
$$\left (\frac{p}{3} \right )=\left (\frac{3}{p} \right ) \cdot \left (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Writing sigma notation $\sum^n_{i=1} \frac {i}{2^i}$ in closed form What would be a way to find the closed form of $\frac {1}{2} + \frac {2}{4}+\frac {3}{8}+\cdots+\frac {n}{2^n}=\sum^n_{i=1} \frac {i}{2^i}=s$
I've looked at $\frac {s}{2}=\frac {1}{4} + \frac {2}{8}+\frac {3}{16}+\cdots+\frac {n}{2^{n+1}}$
And then $s-... | $$
\sum_{i=1}^n i x^i = \sum_{i=1}^n x \frac d {dx} x^i = x \frac d {dx} \sum_{i=1}^n x^i = x \frac d {dx}\ \frac{x - x^{n+1}}{1-x} = \cdots
$$
(Now evaluate the derivative and then substitute $\dfrac 1 2$ for $x$.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}... | Your mistake was not considering the restrictions on $x$ when you solved the individual cases.
Case 1: $x \geq 0$
If $x > 0$, then $x + 4 > 0$, so the direction of the inequality is preserved if we multiply by $x + 4$. Hence,
\begin{align*}
\frac{x}{x + 4} & < 4\\
x & < 4(x + 4)\\
x & < 4x + 16\\
-16 & < 3x\\
-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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How to solve this definite integral $\int_0^\pi \frac{\cos^9(x)}{\sin^3(x)+\cos^3(x)}dx$ I'm having trouble evaluating the following integral:
$$
\int^\pi_0 \frac{\cos^9(x)}{\sin^3(x)+\cos^3(x)}dx
$$
I tried to convert it into an algebraic function by multiplying the numerator and denominator by $\sec^{11}(x)$ and subs... | Assuming the usage of the Cauchy principle value, let's define
$$ I_1 := \int_0^\frac\pi2 \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
\\ I_2 := \text{p. v.} \int_\frac\pi2^\pi \frac{\cos^9(x)}{\cos^3(x)+\sin^3(x)} {\rm d}x
$$
For $I_1$, by making a substitution and adding it to itself,
$$
I_1 = \int_0^\frac\pi2 \fra... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Which Has a Larger Volume a Cylinder or a Truncated Cone? The following link describes a programming problem. However, I am unable to work out the maths for the problem.
We are given $r$ – radius of lower base and $s$ – slant height. The figure can be cylinder or truncated cone. You have to find as largest volume as p... | If the constant slant height and base radius are $s, r$ respectively and the variable angle which the slanting face makes with the vertical is $\theta$ then the height and top radius of the frustum are
$$h=s\cos\theta, R=r+s\sin\theta$$
The volume of the frustum is $$V=\frac13 \pi h (R^2+Rr+r^2)$$
Now $$\frac{dh}{d\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the domain of $\frac{1}{f(x)}$ in this case? Suppose $f(x) = \frac{x-1}{x-2}$ ,
Then $\frac{1}{f(x)} = \frac{1}{\frac{x-1}{x-2}}$
So would the domain be all real numbers excluding 1 and 2 or would the domain include 2? Since
$\frac{1}{\frac{x-1}{x-2}} = \frac{x-2}{x-1}$ ,
I guess I should also ask whether the d... | One typically understands "$\frac{1}{f(x)}$" as the composition of the two functions $x \mapsto f(x)$ and $x \mapsto \frac{1}{x}$. So we have to apply $f(x)$ first, hence 2 is not in the domain.
| {
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To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 ... | hint
Consider the power series
$$
f(x) := \sum\limits_{n=1}^\infty \dfrac{x^n}{2n-1}
$$
Investigate what differential equation it satisfies. Looking at it suggests that this is
maybe easier for
$$
g(x) := \frac{f(x^2)}{x} = \sum\limits_{n=1}^\infty \dfrac{x^{2n-1}}{2n-1}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\epsilon - \delta$ proof of limit of $\arctan \left(\frac {x+z}{y}\right)$ as $(x, y, z) \to (1, 2, -3)$ Find the limit and prove the limit for
$$\lim_{(x,y,z)\rightarrow(1,2,-3)}\arctan \left(\frac{x+z}{y}\right).$$
This is a homework problem. While I am comfortable with general epsilon-delta proof techniques, I am ... | We have
$$\begin{align}
\left|\arctan\left(\frac{x+z}{y}\right)-\arctan\left(-1\right) \right|&=\left|\arctan\left(\frac{x+y+z}{x-y+z}\right)\right|\\\\
&=\left|\arctan\left(\frac{(x-1)+(y-2)+(z+3)}{x-y+z}\right)\right|\\\\
&\le\left|\frac{(x-1)+(y-2)+(z+3)}{x-y+z}\right|\\\\
&\le \frac{3\sqrt{(x-1)^2+(y-2)^2+(z+3)^2}}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| If $k$ is odd, then $2^k$ is $2*2^{k-1}$. If $n$ is not a multiple of 3 then $n^2-1$ is since $n^2-1 = (n-1)(n+1)$ and at least one of them is a multiple of 3. This means that $2*2^{k-1} - 2$ is a multiple of $3$ as is $2*2^{k-1} + 1$.
Hope I helped.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the excircle and right triangle inequality
Let $\triangle{ABC}$ be a right triangle with $\angle{A}= 90^{\circ}$. Let $D$ be the intersection of the internal angle bisector of $\angle{A}$ with side $BC$ and $I_a$ be the center of the excircle of the triangle ABC opposite to the vertex $A$. Prove that $$\dfrac{DA... | Not the best solution but still:
Let E be point where excircle touches triangle $ABC$.
From right triangle $ EDI_a $ we have $ DI_a \geq r_a $
Using $ r_a=\frac{A}{s-a}$ and $A= \frac{1}{2}DA \cdot b \sin 45 + \frac{1}{2}DA \cdot c \sin 45 = \frac{\sqrt2}{4} DA \cdot (b+c) $ where A is area of triangle $ABC$ we get
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What does the homogeneous system of equations represent under certain conditions?
Consider the following linear equations
$ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$
1) $a+b+c \neq o$ and $a^2+b^2+c^2=ab+bc+ca$
2) $a+b+c \neq o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
3) $a+b+c = o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
4) $a+b+c = o$ and $... | The determinant is :
$$\begin{align}\begin{vmatrix}
a&b&c\\\ b&c&a \\\ c&a&b
\end{vmatrix}&=3abc-a^3-b^3-c^3
\\
&=-\frac 12(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right)
\\
&=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
\end{align}
$$
1)This one implies $a=b=c\ne0$ so they are identical planes.
2)Here the determinant is non-zero so th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$
Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$ given that:
$a_n=1/{{n}\choose{0}}+1/{{n}\choose{1}}+...+1/{{n}\choose{n}}$
The hint says to consider when $n$ is even and odd. When $n=2k$ I get:
$$a_{n}=1/{{2k}\choose{0}}+1/{{2k}\choose{1}}+...+1/{{2k}\choose{2k}}$$
$$=1+1/{{2k}\choose... | First, observe that the given recursion can be written as $$2n(a_n - 1) - (n+1)a_{n-1} = 0.$$ Second, observe that $$a_n - 1 = \sum_{k=0}^{n-1} \binom{n}{k}^{\!-1},$$ since the final term of $a_n$ is always $1$. Therefore, the relationship to be proven is equivalent to showing $$0 = S_{n-1} = \sum_{k=0}^{n-1} \frac{2... | {
"language": "en",
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Erasing numbers from the front of the row Numbers $1,2,\ldots,k$ are written in this order in a row. For $i=1,\ldots,k$, in the $i$th step, a random variable $V_i$ is drawn uniformly from the interval $[0,2i]$. If $V_i$ is greater than the first remaining number, that number is erased. What is the expected number of nu... | Let $E_{b,k}(a)$ be the expected number of erased numbers when you have already erased $a$ numbers and you are at the $b$th shot, out of $k$ shots.
Those can be defined for $1 \le a,b \le k+1$ by induction :
When you don't have any shot left, $E_{k+1,k}(a) = a$.
When you do you have $E_{b,k}(a) = \frac 1 {2b}((a+1) E_{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Equation with radicals and reciprocals Find all $x\in\mathbb{R}$ satisfying $$x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}.$$
Multiply both sides by $x^{1/2}$ to get $$x^{3/2} = \sqrt{x^2-1} + \sqrt{x-1}.$$Making the substitution $a = \sqrt{x^2 - 1}$, $b=\sqrt{x-1}$, we have $a+b = x^{3/2}$ and $a^2 - b^2 = x^2 - x$, s... | Square both sides first: $x^2 = x-\dfrac{2}{x} + 1+ 2\sqrt{x-1-\dfrac{1}{x}+\dfrac{1}{x^2}}\Rightarrow x\left(x-1-\dfrac{1}{x}+\dfrac{1}{x^2}\right)+\dfrac{1}{x} =2\sqrt{x-1-\dfrac{1}{x}+\dfrac{1}{x^2}}$. but by AM-GM inequality, the $LHS \geq RHS$. Thus equality occurs or the equation holds if $x\left(x-1-\dfrac{1}{x}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$ As stated in the tag, I'm trying to prove by induction the claim $F_{2n} = F^2_{n+1} - F^2_{n-1}$, where $F_{n}$ is the $n^{th}$ Fibonacci number. I've spent hours on the inductive step without substantial progress, and am hoping someone can provide a path to the des... | Method 1. By induction on $n$ we have $F_n=(a^n-b^n)/\sqrt 5$ where $a=(1+\sqrt 5)/2$ and $b=(1-\sqrt 5)/2=-1/a.$ Plug this into your formula.
Method 2. Let $M$ be the $2\times2$ matrix with top row $(1,1)$ and bottom row $(1,0).$ By induction on $n,$ the top row of $M^n$ is $(F_{n+1},F_n)$ and the bottom row of $M^n... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $N$ so that the sequence is the product of three consecutive numbers
Find the smallest natural number $N$ such that $13 \cdot 17 \cdot N$ is the product of three consecutive natural numbers.
$x(x+1)(x+2) = 13 \cdot 17 \cdot N$.
So let $x=N$, then, $N+1 = 13$ and $N+2 = 17$.
I am unsure how to proceed, can I get ... | By finding the gcd of 13 and 17 you can find a solution to $13x-17y=1$. In this case $x=4,y=3$.
Consider $N=3\cdot 4 \cdot 50$
$$13 \cdot 4 =52,17\cdot 3=51$$
$$x=50 \rightarrow x(x+1)(x+2)=50\cdot 51\cdot 52 = 13\cdot 17 \cdot (3\cdot 4\cdot 50)=13\cdot 17 \cdot N$$
Hence smallest value of $N=3\times 4\times 50 = 600$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$? I have problems to solve this limit
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$
I tried with Taylor:
$$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac... | Full solution:
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)= \lim _{x\to \infty }x^2\left(\sqrt{1-\frac{1}{x}}-\cos\left(\frac{1}{\sqrt{x}}\right)\right)=$$$$=\lim _{t\to 0}\frac{\sqrt{1-t^2}-\cos{t}}{t^4} =\lim _{t\to 0}\frac{1-t^2-\cos^2{t}}{t^4(\sqrt{1-t^2}+\cos{t})}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_{0}^{1}\frac{\sqrt[4]{x (1-x)^{3}}}{(1+x)^{3}}\mathrm{d}x$ How to evaluate the two integrals below:
$$\int_{0}^{1}\frac{\sqrt[4]{x\left ( 1-x \right )^{3}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
and
$$\int_{0}^{1}\frac{\sqrt[3]{x\left ( 1-x \right )^{2}}}{\left ( 1+x \right )^{3}}\mathrm{d}x$$
After a l... | Using the definition of Beta function
$$B\left ( p,q \right )=\frac{\Gamma \left ( p \right )\Gamma \left ( q \right )}{\Gamma \left ( p+q \right )}=\int_{0}^{1}x^{p-1}\left ( 1-x \right )^{q-1}\mathrm{d}x~ ~ ~ ~ \left ( \Re q>0,\Re p>0 \right )$$
In general,
$$I\left ( m,n \right )=\int_{0}^{1}\frac{\sqrt[n]{x^{m}\lef... | {
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"url": "https://math.stackexchange.com/questions/1629945",
"timestamp": "2023-03-29T00:00:00",
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Evaluate the triple integral $\iiint (x^2+y^2+z^2)\,dx\,dy\,dz$ I have to evaluate this integral. It is enough for me to know the correct limits to integration.
$$ \iiint_W (x^2 + y^2 + z^2) \,\mathrm dx\,\mathrm dy\,\mathrm dz$$
Conditions:
$$x\ge 0,\quad y \ge 0 ,\quad z \ge 0,\quad 0 \le x + y + z \le a,\quad (a>0... | *
*Your manipulation of the inequality is wrong: $0\leq x+y+z$ implies $x\geq-y-z$, not $\leq$.
*As @Hagen pointed out, you should NOT have bounds that depend on the variable of integration. Perhaps you meant to order the $d$s differently?
*Again as @Hagen pointed out in his answer, the bounds are much easier.
Her... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the number 14641 is the fourth power of an integer in any base greater than 6? Prove that the number $14641$ is the fourth power of an integer in any base greater than $6$?
I understand how to work it out, because I think you do
$$14641\ (\text{base }a > 6) = a^4+4a^3+6a^2+4a+1= (a+1)^4$$
But I can't unders... | As a matter of personal preference, I like $b$ to represent the base rather than $a$.
So then $(b + 1)^2 = b^2 + 2b + 1$. And $(b^2 + 2b + 1)^2 = b^4 + 4b^3 + 6b^2 + 4b + 1$.
These facts are true whether $b$ is an ordinary positive integer greater than 1 or a more "exotic" number, like $\sqrt{-2}$. But whether $(b + 1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
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Solve $3x(1-x^2)y^2\frac{dy}{dx}+(2x^2-1)y^3=ax^3$ I am solving this linear Differential equation which can be easily solve by using the formulas for the Bernoulli's Equations
I have solved till
$$\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)y^2}=\frac{ax^3}{3x(1-x^2)y^2}$$
$$y^2\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)... | After substitution $t(x) = y^3(x)$
$$x(1-x^2) t'(x) + (2x^2-1) t(x) = ax^3.$$
Let's find a particular solution in form $t_0(x) = bx:$
$$bx - bx^3 + 2bx^3 - bx = ax^3 \Rightarrow b = a$$
So the full solution has form $t(x) = ax + z(x),$
$$x(1-x^2) z'(x) + (2x^2-1) z(x) = 0 \Rightarrow$$
$$\log z = - \int \frac{2x^2 -1}{... | {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$ by algebraic methods I was trying to solve this equation without using calculus.
Is it possible to be solved by elementary algebraic methods?
$$\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$$
| Let $u=\sqrt[3]{q+\sqrt{p^{3}+q^{2}}}$, $v=\sqrt[3]{q-\sqrt{p^{3}+q^{2}}}$,
we have $y=u+v$ satisfying $y^{3}+3py=2q \ldots \ldots (*)$.
Take $y=\sqrt[3]{2}$, $q=7x$ and $p^{3}+q^{2}=19^{2}$ or $p^{3}=19^{2}-(7x)^{2}$.
Substitute into $(*)$,
\begin{align*}
2+3\sqrt[3]{19^{2}-(7x)^{2}} \times \sqrt[3]{2} &= 2(7x) \\
... | {
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"timestamp": "2023-03-29T00:00:00",
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Rational Points on Fibonacci-like Sequence of Polynomials Let $\{a_n\}$ be a sequence of polynomials in $\mathbb{Q}[x,y]$ with $a_0=0,a_1=1$, and
$$a_n=xa_{n-1}+ya_{n-2}$$
The first few look like
$$a_3:y+x^2$$
$$a_4:2xy+x^3=x(2y+x^2)$$
$$a_5:y^2+3x^2y+x^4$$
$$a_6:3xy^2+4x^3y+x^5=x(y+x^2)(3y+x^2)$$
Note that $a_6=0$ has... | First of all, you can do the same sort of trick to get a closed form for this as we use for any other linear recurrence.
The roots of $U^2-xU -y=0$ are $u_1,u_2=\frac{x\pm\sqrt{x^2+4y}}{2}$.
And $a_n = bu_1^n + cu_2^n$ for some $b,c\in R=\mathbb Q(x,y)[\sqrt{x^2+4y}]$.
Solving for $b$ and $c$, we get $b=-c$ and $b(u_1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$\sin2(x) - \tan(x) = 0$ , solve for $-180\le x\le 180$ I have been unable to solve the following question,
If $$\sin(2x) - \tan(x) = 0$$
Find $x$ , $-\pi\le x\le \pi$
So far my workings have been
Use following identity:
$$\sin(2x) = 2\sin(x)\cos(x)\\2\sin(x)\cos(x) - \tan(x) = 0\\2\sin(x)\cos(x) - \frac{\sin(x)}{\co... | Notice, $$\sin 2x-\tan x=0$$
$$\frac{2\tan x}{1+\tan^2x}-\tan x=0$$
$$\tan x\left(\frac{1-\tan^2 x}{1+\tan^2x}\right)=0$$
$$\color{blue}{\tan x\cos 2x=0}$$
Now, solving for $x$,
$$\tan x=0\iff x=n\cdot 180^\circ$$
where $n$ is any integer
For given interval $[-180^\circ, 180^\circ]$, setting $n=-1, 0, 1 $, one should... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Combinatorics: How do you find the coefficient in the given expression? The question asks me to find the coefficient of the term $x^6y^4$ in the expression $(xy^2+x^2+3y)^7$. This was pretty simple. This is how I did it:
$$(xy^2+x^2+3y)^7 = \sum_{a+b+c = n} (xy^2)^a + (x^2)^b + (3y)^c $$
$$ x^{(a+2b)}, y^{(2a+c)}, 3^c ... | You were almost there, mate.
Examine: $a+2b = 6,\, 2a+c = 4,\, a+b+c +d = 7$ for $0\leq a,b,c,d\leq 7$
We have $b=3-a/2$, $c=4-2a, d=7-a-b-c$ so we list the possibilities: $$\begin{cases}a=0, b=3, c=4, d=0 \\[1ex] a=2, b=2, c=0, d=3\end{cases}$$
So the term we require is: $x^{(0+6)}\, y^{(0+4)}\, 3^4\, 4^0 + x^{(2+4)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve a linear system of equation involving some recursion $$
\begin{align*}
x_{1} &= 1 + x_{2}\\
x_{2} &= 1 + \frac{1}{2} x_{3} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{i} &= 1 + \frac{1}{2} x_{i+1} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{n-2} &= 1 + \frac{1}{2} x_{n-1} + \frac{1}{2} x_{1}\\
x_{n-1} &= 1 + \frac{1}{2} x_{n} + \f... | Disclaimer
I'm gonna write couple of terms and then general equation for both forward and backward substitution, so you'll need to use mathematical induction to actually prove them
First, substitute $x_1$ to the second equation
$$
2 x_2 = 2 + x_3 + x_1 = 2 + x_3 + 1 + x_2 = 3 + x_3 + x_2 \implies x_2 = 3\cdot 2^0 + x_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solving $(4y+2x-5)dx+(6y+4x-1)dy=0$ using 2 methods produced 2 different answers! $$(4y+2x-5)dx+(6y+4x-1)dy=0,y(-1)=2$$
First method:
$$\frac{dy}{dx}=-\frac{4y+2x-5}{6y+4x-1}$$
let $Y=y-\frac{9}{2}$, $dY=dy$ and $X=x+\frac{13}{2}$, $dX=dx$;
$$\frac{dY}{dX}=-\frac{4Y+2X}{6Y+4X}$$
let $u=\frac{Y}{X}$ , $Y'=u'X+u$;
$$u'... | The indefinite integration formula
$$
\int \frac{dU}{U} = \ln |U| + c
\tag{1}
$$
(correctly) summarizes two formulas that hold in disjoint intervals,
$$
\left.
\begin{aligned}
\int \frac{dU}{U} &= \ln U + c &&(U > 0), \\
\int \frac{dU}{U} &= \ln (-U) + c && (U < 0).
\end{aligned}\right\}
\tag{2}
$$
(I'm deliberately... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is the integral value of $\frac{\tan 20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ}{\sin40^\circ}$? I have tried possibly all approaches.
I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ a... | Using $\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ=\tan60^\circ$ (Proof)
$$\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ$$
$$=\tan20^\circ+\tan40^\circ+\tan80^\circ-\tan20^\circ\cdot\tan40^\circ\cdot\tan80^\circ$$
$$=\tan20^\circ(1-\tan40^\circ\cdot\tan80^\circ)+\tan40^\circ+\tan80^\circ$$
$$=(1-\tan40^\circ\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + ... | Let $$P = \prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) =\prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \left(1+\frac{1}{r}\right) $$
So $$\prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) = \prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \prod^{n}_{r=2}\left(1+\frac{1}{r}\right)$$
Now Open These two products.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
A formula for length of representation of a number in a "base" without zeros If you had 2 items the sequence would go like this:
$$1,1,2,2,2,2,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5, \ldots$$
This is $\lfloor\log_2(n+2)\rfloor$.
What if I wanted to do for 3 items which goes like this: $$1... | Hint The number of elements of length $k$ is $3^k$, so the number of elements of length $\leq k$ is $$3 + 3^2 + \cdots 3^k = \frac{3}{2} (3^k - 1) .$$ Thus, the $n$th element in the sequence $A, B, C, AA, AB, \ldots$ has length $k$ iff $$\tfrac{3}{2} (3^{k - 1} - 1) < n \leq \tfrac{3}{2}(3^k - 1) .$$
Rearranging gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequa... | Let
$$ a_n = \frac{1}{n+1}+\frac{1}{n+3}+\ldots+\frac{1}{3n-1}. $$
We have $a_1=\frac{1}{2}$ and:
$$\begin{eqnarray*} a_{n+2}-a_n &=& \frac{1}{3n+5}+\frac{1}{3n+3}+\frac{1}{3n+1}-\frac{1}{n+1}\\&>&\frac{3}{3n+3}-\frac{1}{n+1} = 0\end{eqnarray*} $$
so the claim is trivial, since $a_2>a_1$ and $a_{n+2}>a_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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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.