Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculus Question: $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$ Can anyone help me to find $\int_0^{\frac{\pi}{2}}\tan (x)\log(\sin x)dx$? Any help would be appreciated. Thanks in advance.
| Rewrite
$$
\begin{align}
\int_0^{\Large\frac{\pi}{2}}\tan x\ln(\sin x)\,dx=\int_0^{\Large\frac{\pi}{2}}\frac{\sin x}{\cos x}\ln(\sin x)\ dx.
\end{align}
$$
Let $\,u=\cos x$, then $\,du=-\sin x\,dx$. For $\,0 < x < \frac{\pi}{2}$, we have $\,0 < u < 1$. Now, the integral turns out to be
$$
\begin{align}
\int_0^{\Large\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 0
} |
Limit of term with roots How can I prove the following:
$$\lim_{k\rightarrow \infty} \frac{2}{3 (\sqrt{k+1} - \sqrt{k-1})(\sqrt{k+2} + \sqrt{k})} = \frac{1}{3}$$
| We multiply by the conjugate of $\sqrt{k+1} - \sqrt{k-1}$ and we use the asymptotic limit
$$\sqrt{k+a}\sim_\infty \sqrt k$$
we find
$$\lim_{k\rightarrow \infty} \frac{2}{3 (\sqrt{k+1} - \sqrt{k-1})(\sqrt{k+2} + \sqrt{k})} =\lim_{k\rightarrow \infty}\frac23\frac{2\sqrt k}{4\sqrt k}=\frac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/796071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating function - What is the coefficient of $x^{26}$? What is the coefficient of $x^{26}$ given the function:
$${\left[ {{{1 + {x^{10}}} \over {{{(1 - x)}^7}}}} \right]^2}$$
My work:
$${\left[ {{{1 + {x^{10}}} \over {{{(1 - x)}^7}}}} \right]^2} = \left( {1 + 2{x^{10}} + {x^{20}}} \right){1 \over {{{(1 - x)}^{14}}}... | You are perfectly right. Using the notation $[x^n] f(x)$ for "the coefficient of $x^n$ in (the series expansion for) $f(x)$" we can write:
\begin{align}
[x^{26}] \left( \frac{1 + x^{10}}{(1 - x)^7} \right)^2
&= [x^{26}] (1 + x^{10})^2 \cdot (1 - x)^{-14} \\
&= [x^{26}] (1 + 2 x^{10} + x^{20})
\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/797868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the maximum of $\|Ax\|$ for all unit vectors $x \in\mathbb R^2$? $A = \left[\begin{matrix} 1 & 0 \\ 3 & -1 \\ \end{matrix} \right]$
How to find the maximum of $\|Ax\|$ for all unit vectors $x \in\mathbb R^2$?
My professor told me the answer is: $\sqrt{\frac{11+ \sqrt{117} }{2}}$
I know this has something t... | Let $x = (a,b)$, then $||Ax|| = ||(a, 3a - b)|| = \sqrt{a^2 + (3a - b)^2}$.
Consider: $f(a,b) = a^2 + (3a - b)^2 = a^2 + 9a^2 - 6ab + b^2 = 10a^2 - 6ab + b^2$ subject to: $g(a,b) = a^2 + b^2 = 1$. Using Lagrange Multiplier method:
$\nabla f = \lambda\cdot \nabla g$:
$(20a - 6b, 2b - 6a) = (2a\lambda, 2b\lambda)$. So:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Matrix equality has a certain solution I am wondering about the following matrix equality
$$ \begin{pmatrix} 0 \\ 1 & \lambda_{1} \\ & 1 & \lambda_{2} \\ && \ddots & \ddots \\
&&& 1 & \lambda_{k} \\ &&&& 1 & 0 \\
&&&&& \ddots & \ddots \\
&&&&&& 1 & 0 \end{pmatrix} \begin{pmatrix} c_{1} \\ c_{2} \\ c_{3} \\ \vdots \\ ... | My own answer:
Let $c_{1} = 1$, then the equation can be rewritten to
$$
\left( \begin{array}{ccccc|cc} \lambda_{1} &&&&&& \\ 1&\lambda_{2}&&&&& \\ &1&\lambda_{2} &&&& \\ && \ddots& \ddots & & & & \\ &&&1 & \lambda_{k} \\ \hline &&&& 1 & 0 \\ &&&&& \ddots & \ddots \\ &&&&&&1 & 0\end{array} \right) \begin{pmatrix} c_{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/801329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I rewrite vectors in other basis' given change of coordinate matrices? $\displaystyle β= \begin{bmatrix}2\\2\\\end{bmatrix}$,$\displaystyle \begin{bmatrix}4\\-1\\\end{bmatrix}$
$\displaystyle C= \begin{bmatrix}1\\3\\\end{bmatrix}$,$\displaystyle \begin{bmatrix}-1\\-1\\\end{bmatrix}$
From B to C Change of coordin... | Let the basis vectors for $B$ be $\mathbf{b}_1$ and $\mathbf{b}_2$. Then one can write the $\beta$ vectors as
$$
\begin{align}
\boldsymbol{\beta}_1 &= 2\,\mathbf{b}_1 + 2\,\mathbf{b}_2 = \beta_{11}^b\mathbf{b}_1 + \beta_{12}^b\,\mathbf{b}_2 \\
\boldsymbol{\beta}_2 &= 4\,\mathbf{b}_1 - \,\mathbf{b}_2 = \beta_{21}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you evaluate this limit? $\ln({x^3+2x^2+x})+ \frac{2}{x}$ How do you evaluate the following?
$$\lim_{x \to 0^+} \left [ \ln({x^3+2x^2+x})+ \frac{2}{x} \right ]$$
If I plug in $x$, I get $\infty-\infty$, which is undetermined, and I haven't been able to get the limit at a more manageable form. Can you please help... | Lets put $x = 1/y$ so that $y \to \infty$ as $x \to 0^{+}$. Then we can see that
\begin{align}
f(x) &= \log(x^{3} + 2x^{2} + x) + \frac{2}{x}\notag\\
&= \log\left(e^{2y}\left(\frac{1}{y^{3}} + \frac{2}{y^{2}} + \frac{1}{y}\right)\right)\notag\\
&> \log\left(e^{2y}\left(\frac{1}{y^{3}} + \frac{2}{y^{3}} + \frac{1}{y^{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction integral Question:
$\int \dfrac{5 }{(x+1) (x^2 + 4) } dx $
Thought process:
I'm treating it as a partial fraction since it certainly looks like one.
I cannot seem to solve it besides looking at it in the "partial fraction" way.
My work:
1) Focus on the fraction part first ignoring the $\int $ and $d... | Split the fraction.
$$\color{blue}{\dfrac{5 }{(x+1)(x^2 + 4)} = \dfrac{P(x)}{(x+1)}+\dfrac{Q(x)}{(x^2+4)}}$$
The largest partial fraction's denominator is an order 2 polynomial, so the order of the partial fractions' numerators must become order 2 when cross multiplied with the other partial fractions' denominators. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the two points where the shortest distance occurs on two lines Find the point P on $\vec{AB}$ and point Q on $\vec{CD}$ such that $\vec{PQ}$ is the shortest distance
between the lines AB and CD, given $\vec{AB} = \begin{pmatrix}
1\\
0\\
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}
,\vec{C... | Once you know that the normal vector is $n$, the vector equation
$$
\begin{pmatrix}
1\\
0\\
2\\
\end{pmatrix}
+ u\begin{pmatrix}
-2\\
2\\
1\\
\end{pmatrix}
+w\begin{pmatrix}
-3\\
-2\\
-2\\
\end{pmatrix}
= \begin{pmatrix}
0\\
1\\
1\\
\end{pmatrix}
+ v\begin{pmatrix}
2\\
-1\\
-2\\
\end{pmatrix}
$$
is equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Zero of Polynomial If $(a-b)$, $a$, and $(a+b)$ are the zeroes of the polynomial $x^3-3x^2+x+1$ then what are the values of $a$ and $b$?
I have taken $f(x)=x^3-3x^2+x+1$ and equated $f(a-b)$, $f(a)$ and $f(a+b)$ to zero. But I could not reduce the equations.
| So you got the equations
*
*$a^3 - 3a^2 b + 3ab^2 - b^3 - 3a^2 + 6ab - 6b^2 + a - b + 1 = 0$
*$a^3 - 3a^2 + a + 1 = 0$
*$a^3 + 3a^2 b + 3a b^2 + b^3 - 3a^2 - 6ab - 6b^2 + a + b + 1 = 0$
The challenge is how to systematically simplify these equations. It is too much to try and solve for one variable to plug into ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a formula for $0 · 1 · 2 + 1 · 2 · 3 + 2 · 3 · 4 + \dots +n(n + 1)(n + 2)$, for $n \in \mathbb N$ $$\sum\limits_{i=1}^n i(i + 1)(i + 2)$$
$$\sum\limits_{i=1}^n i^3 + 3i^2 + 2i$$
$$\sum\limits_{i=1}^n i^3 + 3\sum\limits_{i=1}^ni^2 + 2\sum\limits_{i=1}^ni$$
$$= (\frac14)n^4 + (\frac12)n^3 + (\frac14)n^2 + n^3 + 3(\f... | Required formulas:
$$\sum_{i=1}^n i^3=\left(\frac{n(n+1)}2\right)^2$$
$$\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}$$
$$\sum_{i=1}^n i=\frac{n(n+1)}2$$
I will start off at the third step.
$$\sum_{i=1}^n i^3+3\sum_{i=1}^n i^2+2\sum_{i=1}^n i$$
$$=\left(\frac{n(n+1)}{2}\right)^2+3\cdot \frac{n(n+1)(2n+1)}{6}+2\cdot \frac{n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How would you find $\tan(3\cdot\arctan(x))$? How would you find $\tan(3\cdot\tan^{-1}(x))$ as quickly as possible?
I don't really understand how to use $\tan(2x)$ or any other identity for that matter to solve this.
| Begin with $$\tan(2.\arctan(x))=\frac{2\tan(\arctan(x))}{1-\tan^2(\arctan(x))}=\frac{2x}{1-x^2}.$$
Next,
$$\begin{align}\tan(3.\arctan(x))&=\tan(2\arctan(x)+\arctan(x))\\&=\frac{\tan(2\arctan(x))+\tan(\arctan(x))}{1-\tan(2\arctan(x))\tan(\arctan(x))}\\&=\frac{\frac{2x}{1-x^2}+x}{1-\frac{2x}{1-x^2}x}\\&=\frac{3x-x^3}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Series Question: $\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$ How to compute the following series:
$$\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^3}$$
I tried to use partial fraction
$$\begin{align}\frac{n^2}{(4n^2-1)^3}&=\frac{1}{64(2n+1)}-\frac{1}{64(2n-1)}+\frac{1}{64(2n+1)^2}+\frac{1}{64(2n-1)^2}\\&-\frac{1}{32(2n+1)^3}+\... | From where you left off...:
$\displaystyle \sum_{n=1}^\infty \dfrac{1}{32}\cdot \left(\dfrac{1}{(2n-1)^3} - \dfrac{1}{(2n+1)^3}\right) = \dfrac{1}{32}\cdot \left(1 - \dfrac{1}{3^3} + \dfrac{1}{3^3} - \dfrac{1}{5^3} + ... \right) = \dfrac{1}{32}$, and
$\displaystyle \sum_{n=1}^\infty \dfrac{1}{64}\cdot \left(\dfrac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Integral $\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\frac{17\pi^4}{360}$ Hi I am trying to integrate $$
\mathcal{I}:=\iint \limits_{{x,y \ \in \ [0,1]}} \frac{\log(1-x)\log(1-y)}{1-xy}dx\,dy=\int_0^1\int_0^1 \frac{\log(1-x)\log(1-y)}{1-xy}dx \,dy
$$
A closed form does exist. I tried to w... | Consider the double integral
\begin{align}
I = \int_{0}^{1} \int_{0}^{1} \frac{\ln(1-x) \ \ln(1-y)}{1-xy} \ dx dy
\end{align}
which, upon expansion into series form, becomes
\begin{align}
I &= \sum_{n=0}^{\infty} \ \int_{0}^{1} x^{n} \ln(1-x) dx \ \int_{0}^{1} y^{n} \ln(1-y) dy \\
&= \sum_{n=0}^{\infty} \left( \int_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
writing a number as a sum of odd integers How many ways are there of writing $n$ as a sum of odd integers, where the order doesn't matter?
For example, there are $2$ ways of writing $3$: $(1,1,1)$ and $(3)$.
| A list of the first 10 cases to study:
1- 1 (1)
2- 1 (1,1)
3- 2 (1,1,1);(3)
4- 2 (1,1,1,1);(3,1)
5- 3 (1,1,1,1,1);(3,1,1);(5)
6- 4 (1,1,1,1,1,1);(3,1,1,1);(3,3);(5,1)
7- 5 (1,1,1,1,1,1,1);(3,1,1,1,1);(3,3,1);(5,1,1);(7)
8- 6 (1,1,1,1,1,1,1,1);(3,1,1,1,1,1);(3,3,1,1);(5,1,1,1);(5,3);(7,1)
9- 8 (1,1,1,1,1,1,1,1,1);(3,1,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/811429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $3x^2+2x\sin(x) + x^2\cos(x) > 0$ for all $x\neq 0$ I got this question:
Show that for all $x\neq 0$, $3x^2+2x\sin(x) + x^2\cos(x) > 0$
I tried to show it but got stuck.
| If $x > 0$, then: $f(x) = x(3x + 2\sin x + x\cos x)$. We need to prove:
$3x + 2\sin x + x\cos x > 0$ when $x > 0$.
$3x + 2\sin x + x\cos x \geq 3x + 2\sin x - x = 2(x + \sin x) > 0$ because $x + \sin x > 0$ as $(x + \sin x)' = 1 + \cos x \geq 0$. So: $x + \sin x > 0 + \sin0 = 0$.
If $x < 0$ we again need to show: $3x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to find measurements of angle CBD?
I got $x + CBD = 110^{o}$. I think I should find measurements of angle CBD, so how to find it?
Do you have any ideas? Please help me!
|
*
*$\angle ACB = 180^\circ-(10^\circ+70^\circ)-(60^\circ+20^\circ) = 20^\circ$ and $\angle AEB = 180^\circ-60^\circ-(50^\circ+30^\circ) = 40^\circ$.
*Draw a line from point $E$ parallel to $AB$, labeling the intersection with $AC$ as a new point $F$ and conclude $\Delta CEF\sim\Delta ABC$,
\begin{align}
\angle CE... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Small part of loop invariant i'm not getting Show that $y = \frac{1}{2} \cdot z \cdot (z + 1)$ is a loop invariant for the following WHILE loop:
y := 0;
z := 0
WHILE $\neg$(z = x) DO
z := z + 1;
y := y + z
In other words, proof that
$\{y = \frac{1}{2} \cdot z \cdot (z + 1)\}$
z := z + 1;
y := y + z
$ \{y = \frac{1}... | I'm not sure to understand your question ...
If we have :
$y = \frac{1}{2} \cdot z \cdot (z + 1)$
then, adding $(z+1)$ to both sides, we get :
$y +(z + 1) = \frac{1}{2} \cdot z \cdot (z + 1) + (z+1) = \frac{1}{2}[ z \cdot (z + 1) + 2 \cdot (z+1)] = \frac{1}{2}[ (z + 2) \cdot (z+1)] $;
it's all ...
In your calculati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do I derive the formula for the sum of squares? I was going over the problem of finding the number of squares in a chessboard, and got the idea that it might be the sum of squares from $1$ to $n$. Then I searched on the internet on how to calculate the sum of squares easily and found the below equation:$$\sum_{i=0}... | Set $S = \sum_{i=1}^{n}i^2$, and noting that
\begin{align}
S = 1^2 + 2^2 + \ldots + (n-1)^2 + n^2 \\
= 1 + 2 + 3 + \ldots + (n-1) + n \\
+2 + 3 + \ldots + (n-1) + n \\
+3 + \ldots + (n-1) + n \\
\ldots\\
(n-1) + n \\
+ n
\end{align}
Where there are n lines in that second sum. From this we get
\begin{align}
S &= \sum_{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Skanavi 2.003, Difference in answers, which is right? In Skanavi book i have a exercise to simplify an equation
$$\begin{align}
((\sqrt[4]{p} - \sqrt[4]{q})^{-2} + (\sqrt[4]{p} + \sqrt[4]{q})^{-2}) : \frac{\sqrt p + \sqrt q}{p-q}
\end{align}$$
Solving:
Replacing:
$$ a =\sqrt[4]{p} $$
$$ b=\sqrt[4]{q} $$
Equation is ch... | Since you have fourth roots you can do the following, to eliminate roots from the denominator.
$$\frac {2(a^2+b^2)}{a^2-b^2}=\frac {2(a^2+b^2)}{a^2-b^2}\cdot\frac {a^2+b^2}{a^2+b^2}=\frac {2(a^2+b^2)^2}{a^4-b^4}$$
Substitute your values into this and you will see that the two answers are equal to each other.
Alternati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Adding combinations Show non-numerically that:
$${2\choose2} + {3\choose2} + {4\choose2} + {5\choose2} = {6\choose3}$$
The answer is as follows, but I have no idea how it was done:
$$
\begin{eqnarray}
&\phantom{=}& {2\choose2} + {3\choose2} + {4\choose2} + {5\choose2} \\
&=& {3\choose3} + {3\choose2} + {4\choose2} + {5... | If you want to choose $k$ items from $n$ items, you can do either way:
*
*choose $k$ items from the first $n-1$ items.
*choose the last item, and then choose $k-1$ items from the first $n-1$ items.
That means
${n\choose k} = {n-1\choose k} + {n-1\choose k-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/818794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Determinant of specially structured block matrix How do you compute the determinant of the following $nd \times nd$ block matrix?
$$M = \begin{bmatrix}A+B & A & A & \dots & A & A\\ A & A+B & A & \dots & A & A\\ A & A & A+B & \dots & A & A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ A & A & A & \dots & A+B &... | It has the same determinant as the matrix
$$
\begin{pmatrix}
B & 0 & \cdots & -B \\
0 & B & \cdots & -B \\
\vdots & \vdots & \ddots & \vdots\\
A & A & \cdots & A+B
\end{pmatrix},
$$
which has the same determinant as
$$
\begin{pmatrix}
B & 0 & \cdots & 0 \\
0 & B & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots\\
A & A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\lceil \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}\rceil = 2n^2+n+3$ First, the question: Prove that
$$\Bigg\lceil \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}\Bigg\rceil = 2n^2+n+3$$
The motive to this question is the following.... | I do not know how much this could help you : for large values of $n$, the expansion of $$F(n)= \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}$$ is given by $$2 n^2+n+3-\frac{1}{4 n^2}+\frac{1}{16 n^3}+\frac{9}{16 n^4}-\frac{45}{64
n^5}+O\left(\left(\frac{1}{n}\right)^6\right)$$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int_{0}^{2\pi}{\sin^8x}\ {dx}$ I have started doing integration by parts:
$$\int_{0}^{2\pi}{\sin^8(x)}{dx}
= \int_{0}^{2\pi}{\sin^7(x)}\cdot{\sin(x)dx}
= \int_{0}^{2\pi}{\sin^7(x)}\cdot{d(-\cos(x))}
= \left. -\cos(x) \cdot \sin^7(x) \right|_0^{2\pi}
+ \int_{0}^{2\pi}{\cos(x)}{d(\sin^7(x))}
= \left. -... | Consider Beta function
$$
\text{B}(x,y)=2\int_0^{\Large\frac\pi2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\ d\theta=\frac{\Gamma(x)\cdot\Gamma(y)}{\Gamma(x+y)}.
$$
Rewrite
$$
\int_0^{\large2\pi}\sin^8x\ dx=4\int_0^{\Large\frac\pi2}\sin^8 x\ dx,
$$
then
$$
\int_0^{\large2\pi}\sin^8x\ dx=2\cdot\frac{\Gamma\left(\dfrac92\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
inequality $\prod\limits_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$ $n$ is a positive integer, then
$$\prod_{k=1}^n\frac{2k-1}{2k}\lt\frac1{\sqrt{3n}}$$
with mathematical induction, we can prove this.
But I would love to find a wonderful method without mathematical induction
Thank you!
| Building on André Nicolas' idea, you may rewrite $\prod_{k=1}^n\frac{2k-1}{2k}$ as
$$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{7}{8}\cdots\frac{2n-1}{2n}=\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdots (2n-1)2n}{(2\cdot 4 \cdot 6\cdot 8\cdots 2n)^2}=\frac{(2n)!}{(2^n\cdot 1\cdot 2\cdot 3\cdots n)^2}=\frac{1}{4^n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Finding the explicit notation of $f(n)$, based on it's recursive description. I came across this problem on a HackerRank challenge.
The function $f(n)$ is
*
*$1$ if $n = 0$
*$2f(n - 1)$, if $n$ is odd
*$f(n -1) + 1$, if $n$ is even
I solved the problem using a recursive function and it worked just well. However,... | To get a closed form, we will first rewrite the recurrence as:
\begin{align*}
f_n &= 2\, f_{n-2}+1+ \left(n\mod 2\right)
\end{align*}
Next, we will use generating functions, suppose $G(x)= \sum_{n\ge 0}f_n\, x^n$
\begin{align*}
\sum_{n\ge 2} f_n x^n &= 2\, \sum_{n\ge 2}f_{n-2}x^n+ \sum_{n\ge 2}x^n+ \sum_{n\ge 2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$ Evaluate the sum $\sum^{\infty}_{n=1} \frac{n^2}{6^n}$
My approach :
$= \frac{1}{6}+\frac{2^2}{6^2}+\frac{3^2}{6^3} +\cdots \infty$
Now how to solve this I am not getting any clue on this please help thanks.
| Starting with
\begin{align}
\frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}
\end{align}
then differentiate and multiply by $t$ to obtain
\begin{align}
\frac{t}{(1-t)^{2}} = \sum_{n=0}^{\infty} n t^{n}.
\end{align}
Repeating leads to
\begin{align}
\sum_{n=0}^{\infty} n^{2} \ t^{n} = \frac{t(1+t)}{(1-t)^{3}}.
\end{align}
Now ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Which of the following functions on R are uniformly continous?
$a)\frac {1}{x^2+1} $
$b)\cos^3x$
$c)\frac {x^2}{x^2+2} $
$d)x\sin x$
$a)|\frac{1}{x^2+1}-\frac{1}{y^2+1}|\leq|\frac{|x-y|(|x|+|y|)}{(1+x^2)(1+y^2)}|\leq \frac{|x-y||x|}{(1+x^2)(1+y^2)} +\frac{|x-y||y|}{(1+x^2)(1+y^2)}<\frac{|x-y|}{1+y^2}+\frac{|x-y|}{1+x... | Hint If a function $f$ is lipschitzian then it's uniformly continuous and notice that if $f'$ is bounded then $f$ is lipschitzian. We can apply the above remark to a,b,c).
Now for d) let $x_n=2n\pi$ and $y_n=x_n+\frac\pi2$ then we have
$$y_n-x_n=\frac\pi2\xrightarrow{n\to\infty}\frac\pi2$$
and
$$f(y_n)-f(x_n)\xrighta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/830993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving combinatorical problem using characteristic polynomial
How many $6$ length strings above $\left\{1,2,3,4\right\}$ are there such that $24$ and $42$ aren't allowed.
The suitable recurrence relation for this problem is: $a_{n+2} = 2a_{n-1} + 4a{n-2}$. Hence, the characteristic polynomial is: $x^2 -2x -4 = 0$.... | We may solve such kind of problems using directed graph/finite automaton.
Consider the following graph:
\begin{align*}
A = \left[\begin{array}{rrrrrrr}
& \mathrm{I} & \mathrm{O} & \mathrm{2} & \mathrm{4} & \mathrm{24} & \mathrm{42}\\
\mathrm{I} & 0 & 2 & 1 & 1 & 0 & 0 \\
\mathrm{O} & 0 & 2 & 1 & 1 & 0 & 0 \\
\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Confused about a linear equation So I am working through some notes on Linear Algebra and I cant seem to follow this one part. The question asks to
Solve:
$x+y-z+2w=-20$
$2x-y+z+w=11$
$3x-2y+z-2w+27$
I don't have a problem with putting the equation into matrix form and even reducing it. The way the notes explains it i... | You have four unknowns, so a solution is a vector with four components.
The system, after the final reduction, can be written as
$$
\begin{cases}
x+w=-3\\
y+4w=-19\\
z+3w=-2
\end{cases}
$$
Since $w$ is a free variable, you can set it to any value you want; therefore the solutions are
$$
\begin{cases}
x=-3-w\\
y=-19-4w\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove $a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}$ I was asked to prove an inequality:
For any $n$ positive numbers $\{a_i\}$ with $a_{1}a_{2}\cdots a_{n} = 1$ and $m \geq n-1$ be a non-negative integer,
$a_1^m + a_2^m + \cdots + a_n^m \geq \frac{1}{a_1} + \frac{1}... | We will prove that
$$a_1^m + a_2^m + \cdots + a_n^m \ge a_1^{n-1} + a_2^{n-1} + \cdots + a_n^{n-1} \geq \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}.$$
Left-hand side inequality:
Obviously we have $(x^{n-1}-1)(x^{m-n+1}-1) \ge 0$ for any $x>0$ (note that $m\ge n-1$). Expanding we get
$$x^m\ge x^{n-1} + (x^{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is wrong in the following calculation for the inverse of a matrix? $\left[\begin{array}{ccc|ccc}
0 & 3 & 0 & 1 & 0 & 0\\
4 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 5 & 0 & 0 & 1
\end{array}\right]$
Exchange the first and the second columns:
$\left[\begin{array}{ccc|ccc}
3 & 0 & 0 & 0 & 1 & 0\\
0 & 4 ... | You shoul work on columns or on rows, never on both.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/832787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help Evaluating $\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$ Does anyone know how to evaluate the following limit?
$$
\lim_{x \to \infty} \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}
$$
The answer is $\frac{1}{2}$, but I want to see a step by step solution if possible.
| Multiply by the conjugate, then reduce:
\begin{align*}
\lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right)
&= \lim_{x\to\infty} \left(\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right) \cdot \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}} \\
&= \lim_{x\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/834471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Hyberbolic and Circular (Trig) Functions: Why no parabolic? There are circular (trig) functions which determine all the points on a unit circle:
and which relate to the area swept out by an angle subtended on the circle. -- These functions can of course be extended to relations to ellipses as well.
There are also hype... | Note that
(1) For $\cos(\phi)$ and $\sin(\phi)$ the argument is a 'surface'.
(2) For $\cosh(\zeta)$ and $\sinh(\zeta)$ the argument is a 'surface'.
So you should look for functions where the argument is a 'surface'.
Consider $y = x^2 - c$, where $c$ is positive.
For a given $x = \xi$ we have an arc formed by:
(3a) the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to prove inequality $\frac{a}{a+bc}+\frac{b}{b+cd}+\frac{c}{c+da}+\frac{d}{d+ab}\ge 2$ Question:
Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$
when I solved this problem, I have see following three variables inequality:
Assumming that $a,b,c>0,a+b+c... |
Let $$a,b,c,d>0,a+b+c+d=4$$
show that
$$P=\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\ge 2$$
Now using Cauchy Schwarz:
$$\left(\dfrac{a}{a+bc}+\dfrac{b}{b+cd}+\dfrac{c}{c+da}+\dfrac{d}{d+ab}\right)((a+bc)+(b+cd)+(c+da)+(d+ab))\ge (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$$
Now:
$$P\ge\frac{4+2(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Markov chains by hand If I have a starting point: $A_T=[0,1]$ at $T=1$ and a one step transition matrix of:
$B=\left[ \begin{align} &\frac34 & \frac14& \\& \frac1{20}& \frac {19}{20} &\end{align} \right]$
I can compute $A_nB=A_{n+1}$
And keep reiterating this, but it is slow by hand, I assume there is a faster way? Wou... | $B$ can be diagonalized as $B = PDP^{-1}$, where
$P = \begin{bmatrix}-5 & 1\\1 & 1\end{bmatrix}$, $P = \begin{bmatrix}7/10 & 0\\0 & 1\end{bmatrix}$, $P^{-1} = \begin{bmatrix}-1/6 & 1/6\\1/6 & 5/6\end{bmatrix}$.
Then $B^n = PD^nP^{-1} = \begin{bmatrix}-5 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}(7/10)^n & 0\\0 & 1\end{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using calculus of residues, prove that $\int z^2log[(z+1)/(z-1)]dz$ Using calculus of residues, how can it be proven that
$$
\int z^2\log\left[\frac{z+1}{z-1}\right]\;dz
$$ taken round the circle $\left\vert z\right\vert=2$ has the value $\frac{4\pi i}{3}$?
| Note: I'm assuming that $\log \frac{z+1}{z-1}$ is defined on $\mathbb{C}\setminus [-1,1]$, or at least that a branch-cut that lies entirely inside the disk $\{ z : \lvert z\rvert < 2\}$ is used. For arbitrary branch-cuts intersecting the contour, the integral depends on the exact choice of the branch-cut.
One way is to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/838837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Generating Function for the number of ways representing positive integer with odd numbers I had an exam and this question struck me out of nowhere, making me sad :)
Let $f(n)$ denote the number of possibilities representing $n$ using odd positive single-digit numbers $[1,3,...,9]$
For example, $f(6)=4$ because:
$6=1+1+... | Since you are not making a difference based on the order of the numbers, the only important thing here is how many of each numbers is present in a partition. That means we can model 1's with this generating function:
$$1+x+x^2+x^3+\cdots=\frac{1}{1-x}$$
$1=x^0$ is the case when we don't take any ones, $x$ is the case w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Which are the two last digits of $a_n$? We have the sequence $(a_n)$ with $a_1=3$ and the recursive formula:
$$a_{n+1}=3^{a_n} , \forall n$$
Which are the two last digits of $a_n$ ?
How can I find them? I have to find $a_n \pmod {100}$ ,right?So,do I have to find the formula of $a_n$ ?
Could I use the Eulers'Theorem?
W... | As for integer $\displaystyle n>0, a_n\equiv-1\pmod4\equiv3,a_n-1\equiv2\pmod4\implies\frac{a_n-1}2\equiv1\pmod2$
$$a_{n+1}=3^{a_n}=3\left(10-1\right)^{\frac{a_n-1}2}\equiv3\left(-1+\binom{\frac{a_n-1}2}110\right)\pmod{100}$$
$$\implies a_{n+1}\equiv15a_n-18\pmod{100}$$
Now let us find the period of $\displaystyle15\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Reducing a matrix of the form J-n*In, where J is the all-ones matrix. How can I reduce a matrix like this? The goal is to prove that the dimension of the generated space is 1. It's the matrix of the homogeneous system.
$\left( \begin{array}{cccccc}1-n&1&\dots &\dots&\dots&1 \\ 1 & 1-n & 1 & \dots & \dots & 1 \\ \vdots ... | The matrix diagonalizes to $$\begin{pmatrix}
0&0&0&0\\
0&-n&0&0\\
0&0&-n&0\\
0&0&0&-n\\
\end{pmatrix}$$
Let $A=\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
\end{pmatrix}$
And if $X=\begin{pmatrix}
x_1\\
\vdots \\
x_n
\end{pmatrix}$ let $\sum X=x_1+\cdots x_n$.
Then we have $$AX=\sum X \begin{pmatrix}
1\\
\vd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial multiplication modulo polynomial Suppose we are working on finite field $F_{16}$ and have pritimive polynomial $z^4+z+1$. I stuck at how to compute polynomial modulo. For example, we have $z^5+z+1$ mod $z^4+z+1$. I use the usual division, I obtain the remainder is $-z^2+1=z^2+1$ because each coefficient is ... | I am assuming that $z^4+z+1$ is the irreducible polynomial used to construct the field $\mathbb{F}_{16}$ (this is different from the concept of primitive element). In that case $z^4+z+1 \equiv 0$.
\begin{align*}
z^4+z+1 & \equiv 0 \pmod{z^4+z+1}\\
z^4 & \equiv z+1 \pmod{z^4+z+1}\\
z^5 & \equiv z^2+z \pmod{z^4+z+1}\\
z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Differentiation of a function $f:\mathbb{Q}\to \mathbb{Q}$(Rational Calculus) Assume that $f:\mathbb{Q}\to \mathbb{Q}$ is given such that $\forall a\in \mathbb{Q}$ the following limit, exists
\begin{equation}
\lim_{x\to a} \frac{f(x)-f(a)}{x-a}\in \mathbb{R}
\end{equation}
Is it true to say that the above limit ... | It is false: I would define something piecewise-constant and discontinuous at irrational $x$ such as $f : \mathbb{Q} \to \mathbb{Q}$ by:
$f(x) = 1$ for $x > \frac{1}{\sqrt{2}}$
$f(x) = \frac{1}{2}$ for $\frac{1}{3\sqrt{2}}<x<\frac{1}{2\sqrt{2}}$
$f(x) = \frac{1}{3}$ for $\frac{1}{4\sqrt{2}}<x<\frac{1}{3\sqrt{2}}$, and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/845203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Finding closed form for $1^3+3^3+5^3+...+n^3$ I'd like to find a closed form for $1^3+3^3+5^3+...+n^3$ where $n$ is an odd number.
How would I go about doing this?
I am aware that $1^3+2^3+3^3+4^3+...+n^3=\frac{n^2(n+1)^2}{4}$ but I'm not too sure how to proceed from here.
My gut feeling is telling me to multiply the ... | If
$\color{blue}{n = 1}$, $\sum$ = 1
$\color{blue}{n=2}$, $\sum = 1^3 + 3^3 = 28 = 2^2 \cdot 7 = \color{blue}{2}^2(2\cdot \color{blue}{2}^2-1) $
$\color{blue}{n = 3}, \sum = 1^3 + 3^3 + 5^3 = 153 = 3^2 \cdot 17 = \color{blue}{3}^2(2\cdot \color{blue}{3}^2 - 1) $
$\color{blue}{n = 4}, \sum = 1^3 + 3^3 + 5^3 + 7^3 = 49... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/848087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Express $\log_5 288$ in terms of decimal logarithms $\log 2$ and $\log 3$ Assuming $a=\log 2$ and $b=\log 3$ (log is the base 10 logarithm). I have to find $\log_5 288$. How can I do this?
I've tried transforming $\log2$ to $\frac{\log_5 2}{\log_5 10}$ and same for $b$. Then it's $$\frac{5\log_5 2+2\log_5 3}{\log_5 10... | $$\log 2=\frac{\log_5 2}{\log_5 10}=\frac{\log_5 2}{\log_5 2+\log_5 5}=\frac{\log_5 2}{\log_5 2+1}\qquad \log 3=\frac{\log_5 3}{\log_5 10}=\frac{\log_5 3}{\log_5 2+1} $$
Hence
$$\log_5 2=\frac{\log 2}{1-\log2}=\frac{a}{1-a}\qquad \log_5 3= \log 3 (\log_5 2+1)=b\left(\frac{a}{1-a}+1\right)=\frac{b}{1-a}\ .$$
Hence
$$\lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Producing a CDF from a given PDF So I have this PDF:
$$
f(x)=
\begin{cases}
x + 3 & \text{ for } -3 \leq x < -2\\
3 - x & \text{ for } 2 \leq x < 3\\
0 & \text{ otherwise}
\end{cases}
$$
To make this a CDF, I have integrated the PDF from $-\infty$ to some value, $x$.
$$
F(x)=
\begin{cases}
\frac{x^2}{2} + 3x + \fra... | The cdf $F(x)$ is given by
$$F(x)
=\begin{cases}
0 & \text{ for } x <-3\\
\int_{-3}^{x}(t+3) \, dt & \text{ for } -3 \leq x <-2\\
\frac{1}{2} & \text{ for } -2 \leq x <2\\
\frac{1}{2}+\int_{-2}^{x}(3-t) \, dt & \text{ for } 2 \leq x <3\\
1 & \text{ otherwsie}
\end{cases}
$$
My suggestion is to plot the pdf $f(x)$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Picture/intuitive proof of $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$? Is there a nice geometric, intuitive or picture proof as to why the easily algebraically provable identity $\cos(3 \theta) = 4 \cos^3(\theta)-3\cos(\theta)$ is true?
Note I'm not looking for a computational proof like the one linked to, more ... | Enhancing my diagram for the angle-sum formula (currently featured in Wikipedia) to use three angles will get you pretty close ...
Thus,
$$\begin{align}
\cos(\alpha+\beta+\gamma) &= \cos\alpha \cos\beta \cos\gamma - \sin\alpha \sin\beta \cos\gamma - \sin\alpha \cos\beta\sin\gamma - \cos\alpha \sin\beta\sin\gamma \\
\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Proof of the inequality $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}$ I am currently attempting to prove the following inequality
$\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b} \geq \dfrac{3}{2}$ for all $ a,b,c>0$
My instinctive plan of attack is to use the AM/GM inequality with $x_1=\dfrac{a}{b+c}$ et... | Use AM > HM on $\frac{a+b+c}{a+b}, \frac{a+b+c}{b+c}$ and $\frac{a+b+c}{c+a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 5
} |
How to determine the eigenvectors for this matrix I have the matrix $$\left( \begin{array}{ccc}
-\alpha & \beta \\
\beta/K & -\alpha/K
\end{array} \right)$$ for which the eigenvalues are $$\lambda_{1,2}=-\dfrac{\alpha}{2}-\dfrac{\beta}{2K}\pm\dfrac{\sqrt{(K\alpha-\beta)^2+4K\alpha\beta }}{2K}$$ I have a problem with f... | To find the eigenvalues of a matrix $A$, we solve $\mathrm{det}(A-\lambda I)=0$ for $\lambda$. In this case:
$\begin{align}
\left|\begin{array}{cc} -\alpha-\lambda & \beta \\ \frac{\beta}{K} & \frac{-\alpha}{K}-\lambda \end{array}\right| &= (-\alpha-\lambda)\left(\frac{-\alpha}{K}-\lambda\right)-\frac{\beta^2}{K} \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/856361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can you show that the LHS equals the RHS in this equation, by showing how I can get the expression on the RHS? $$
\frac{1^2+2^2+...+(n-1)^2}{n^3} = \frac{(n-1)n(2n-1)}{6n^3}
$$
Can someone show me step by step how I can transform the LHS to the RHS? If possible, using high school-level math.
I have now edited the title... | Consider the geometric progression
$$
\sum_{k=1}^{n-1} x^k=\frac{x^{n}-x}{x-1}.\tag1
$$
Differentiating $(1)$ with respect to $x$ yields
$$
\sum_{k=1}^{n-1} kx^{k-1}=\frac{(n-1)x^{n}-nx^{n-1}+1}{(x-1)^2}.\tag2
$$
Multiplying $(2)$ by $x$ yields
$$
\sum_{k=1}^{n-1} kx^{k}=\frac{(n-1)x^{n+1}-nx^{n}+x}{(x-1)^2}.\tag3
$$
D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How find this integral $I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$ Find this integral
$$I(a,b)=\iiint_{x^2+y^2+z^2\le 1}(ax+by)^2\,dx\,dy\,dz$$
since
$$(ax+by)^2=a^2x^2+b^2y^2+2abxy$$
so
$$I=I_{1}+I_{2}+I_{3}$$
where
$$I_{1}=\iiint_{x^2+y^2+z^2\le 1}ax^2\,dV=a\iiint_{x^2+y^2+z^2\le 1}x^2\,dV$$
since
$$\iiin... |
With (almost) no computations:
The domain $x^2+y^2+z^2\leqslant1$ is invariant by the rotations of the plane $(x,y)$ hence $$I(a,b)=(a^2+b^2)\,J,$$ where
$$
J=\int\!\!\!\iint_{x^2+y^2+z^2\leqslant1}x^2\mathrm dx\mathrm dy\mathrm dz.
$$
By symmetry with respect to the $(x,y,z)$ axes, $3J=K(1)$, where
$$
K(t)=\int\!\!\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve: $\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$ $n$ is an integer variable satisfying $$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$ How can I find $n$?
| the equation
$$\frac{2^{n+1}}{n+1}=\frac{4+2^n}{3}$$
is equivalent to
$$2^{n-2}=\frac{n+1}{5-n}$$
because $2^{n-2}>0$ follow that $\frac{n+1}{5-n}>0,0<n<5$
we need to check for $n=1,2,3,4$
for $n=1$ we get $$2^{-1}=\frac{2}{4}=\frac{1}{2}$$
for $n=2$ we get $$2^0=\frac{3}{3}=1$$
for $n=3$ we get $$2^1=\frac{4}{2}=2$$
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{alig... | using the RMS GM inequality
$$\sqrt{\frac{x^2+y^2+z^2}{3}}\geq\sqrt[3]{xyz}$$
with equality if and only if $x=y=z$, plugging in your values for $xyz$ and$x^2+y^2+z^2$ we get that
$$\sqrt{\frac{3}{3}}\geq\sqrt[3]{1}$$
$$1=1$$
thus the only possible solutons are those you already stated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/864430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Upper bound of $\sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}$ I am looking for an upper bound of the following sum
$$
S_p:= \sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}.
$$
The upper bound should be independent of $p$, of course. Numerical experiments indicate that
$$
S_p \le \frac53
$$
with the maximum attained for $p=3,... | First, note that that the change of summation index $j\leftarrow p+1-j$ shows that
$$
S_p=\frac{p+1}{2^{p+1}}\sum_{j=1}^p\frac{2^j}{j}
$$
Thus
$$
\frac{2^{p+1}}{p+1}S_p=\sum_{j=1}^p\frac{2^j}{j}=\frac{2^p}{p}+\frac{2^{p}}{p}S_{p-1}
$$
Similarly$$\eqalign{
\frac{2^{p+2}}{p+2}S_{p+1} &=\frac{2^{p+1}}{p+1}+\frac{2^{p+1}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/867678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve x in logarithm equation I am trying to solve $x$ for $2log_{10} (x-4) - log_{10}4(x-1) = 0$
I have the key with the answer 10 and have confirmed this is correct using Wolfram Alpha but which steps should I take to reach that answer?
| Start with original problem:
$$2 \log_{10} (x-4) - \log_{10}4(x-1) = 0$$
A rule of logarithm: $2 \log_{10} (x-4)=\log_{10}(x-4)^2$:
$$ \log_{10} (x-4)^2 - \log_{10} (4x-4)= 0$$
Another rule of logarithm: Express as the left side of the equation as a single product:
$$ \log_{10} \frac{(x-4)^2}{4x-4}= 0$$
Convert to exp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/870921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ Suppose $n$ is an integer which is not divisible by $5$. Prove that $n^4 \equiv 1 \pmod{5}$.
| $$n^4-1 = (n-1)(n+1)(n^2+1)$$ The factors $n-1$ and $n+1$ take care of $n \equiv \pm 1 \mod 5$, while if $n \equiv \pm 2 \mod 5$, $n^2 + 1 \equiv 2^2 + 1 \equiv 0 \mod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 15,
"answer_id": 13
} |
Is this graph connected Define the following graph on the vertex set ${\mathbb N}_{\geq1}\>$:
Two numbers $a$, $b\in {\mathbb N}_{\geq1}$ are connected by an edge (written $a \ \mathcal{R} \ b)$ if and only if $a+b \ | \ ab-1$.
Clearly $1$ is isolated. Can we connect all integers greater than $2$ to $2$?
For example: ... | A partial answer :
We have $a \sim b$ if and only if there exist a sequence of integers $a_1, \ldots, a_n$ such that $a \ \mathcal{R} \ a_1 \ \mathcal{R} \ \cdots \ \mathcal{R} \ a_n \ \mathcal{R} \ b$. The relation $ ab-1 = c (a + b) $ can be written as $(a-c)(b-c)=c^2+1$ and can be solves $a=c+d$ and $b=c+ \dfrac{c ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 4,
"answer_id": 0
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For which $a$ does this system have one solution, infinite solutions, or no solution? Consider the system of linear equations:
\begin{cases} x+y+az=1 \\ x+ay+z=1 \\ ax+y+z=1 \end{cases}
For which $a$ does the system have:
*
*no solution
*one solution
*an infinite number of solutions?
So first I reduced the m... | If $a = 1$, the system reduces to $x + y + z = 1$, and it has more unknowns than equation, hence infinite solutions. If $a \neq 1$, then subtract the second from the first equation:
$(a-1)(z-y) = 0$. So: $z-y = 0$, and $y = z$. Similarly subtract the third from the second equation: $(a-1)(y-x) = 0$. So $y - x = 0$, an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How find this integral $I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\frac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$
Find the value:
$$I=\int_{0}^{\frac{\pi}{2}}(\ln{(1+\tan^4{x})})^2\dfrac{2\cos^2{x}}{2-(\sin{(2x)})^2}dx$$
I use computer have this reslut
$$I=-\dfrac{4\pi}{\sqrt{2}}C+\dfrac{13\pi^3}{24\sqrt{2}}+\dfrac{9}{... | Here is an approach. I'll continue from the step you have already reached with the correction suggested in Fabien comment which is considering the integral
$$ I = \int_{0}^{\infty} \frac{\ln^2(1+x^4)}{1+x^4} dx. $$
To evaluate the above integral we consider the integral
$$ F = \int_{0}^{\infty} (1+x^4)^{\alpha} dx = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/873333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c... | We need to show: $\left(\dfrac{b^2}{a^2}\right)^{x} + \left(\dfrac{c^2}{a^2}\right)^{x} > 1$ for $0< x = \dfrac{n}{2}< 1$, and $a^2 = b^2 + c^2$. Let $u = \dfrac{b^2}{a^2}$, and $v = \dfrac{c^2}{a^2}$, then $0 < u,v < 1$, and $u+v = 1$, and we now prove: $u^x + v^x > 1$.
Consider $f(x) = u^x + v^x$ on $(0,1)$. We have:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Prove that $\frac{d}{dx}(z^kJ_k(z))=z^kJ_{k-1}(z)$ for every $ k \geq \frac{1}{2}$. Then write an explicit expression for (a) $J_\frac{1}{2}(x)-J_\frac{5}{2}(x)$, and (b) an expression for $J_1+2J_3$ in terms of $J_0$.
Prove that $\frac... | This is only a partial answer and would be far too long for a comment, but I must quote which is limited in comments.
How are x and z related? Are the $J_k$ the Bessel functions of the first kind?
For them it is correct that $$\frac{d}{dx}(x^kJ_k(x))=x^kJ_{k-1}(x)$$
see http://dlmf.nist.gov/10.6#ii or Abramowitz and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\tan x $ if $\sin x+\cos x=\frac12$ It is given that $0 < x < 180^\circ$ and $\sin x+\cos x=\frac12$, Find $\tan x $.
I tried all identities I know but I have no idea how to proceed. Any help would be appreciated.
| Divide through by $\cos x$ to get
\begin{align}
\tan x + 1 & = \frac 12 \sec x \tag{1}\\
\text{Square both sides: } \qquad\qquad\tan^2 x + 2 \tan x + 1 & = \frac 14 \sec^2 x = \frac{1}{4} (\tan^2 x + 1) \\
3 \tan^2 x + 8 \tan x + 3 & = 0 \\
\tan x & = \frac{-8 \pm \sqrt{28}}{6} = \frac{-4 \pm \sqrt{7}}{3}.
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
} |
How do you calculate this limit $\lim_{n\to\infty}\sum_{k=1}^{n} \frac{k}{n^2+k^2}$? How to find the value of $\lim_{n\to\infty}S(n)$, where $S(n)$ is given by $$S(n)=\displaystyle\sum_{k=1}^{n} \dfrac{k}{n^2+k^2}$$
Wolfram alpha is unable to calculate it.
This is a question from a questions booklet, and the options fo... | Clearly,
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^n\frac{k}{n^2+k^2} &=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\frac{k}{n}}{1+\frac{k^2}{n^2}}
\stackrel{\text{Riemann sum}}\longrightarrow \int_0^1 \frac{x\,dx}{1+x^2}=\left.\frac{1}{2}\log (1+x^2)\right|_0^1\\
&=\frac{1}{2}\log 2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/879611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of $ \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx$ Evaluation of $\displaystyle \int \frac{x^2+n(n-1)}{(x\cdot \sin x+n\cdot \cos x)^2}dx$
$\bf{My\; Solution:}$ Using $\displaystyle (x\cdot \sin x+n\cdot \cos x) = \sqrt{x^2+n^2}\left\{\frac{x}{\sqrt{x^2+n^2}}\cdot \sin x+\frac{n}{\sqrt{x^2+n^2}}... | This is a shorter, but perhaps uglier, solution:
Using the identities $n^2+x^2=(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2$ and
$n=\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x)$, we can rewrite the integral as
$\displaystyle \int\frac{(n\sin x-x\cos x)^2+(n\cos x+x\sin x)^2-[\sin x(n\sin x-x\cos x)+\cos x(n\cos x+x\sin x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Finding a limit to negative infinity with square roots: $\lim\limits_{x\to -\infty}(x+\sqrt{x^2+2x})$ Find the limit of the equation
$$\lim_{x\to-\infty} (x+\sqrt{x^2 + 2x})$$
I start by multiplying with the conjugate:
$$\lim_{x\to-\infty} \left[(x+\sqrt{x^2 + 2x})\left({x - \sqrt{x^2 + 2x}\over x - \sqrt{x^2+2x}}\righ... | For better clarity set $\dfrac1x=-y\implies y\to0^+$
$$\lim_{x\to-\infty}(x+\sqrt{x^2+2x})=\lim_{y\to0^+}\frac{\sqrt{1-2y}-1}y$$
$$=\lim_{y\to0^+}\frac{1-2y-1}{y(\sqrt{1-2y}+1)}$$
Hope you can take it home from here
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/881145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Upper bound on $ \binom{a}{m+1}\sum ^m_{j=0} \binom{a-m-1}{j}/\binom{b}{j+m+1}$ Given $a,b,m$ such that $0<2m<a<b$.
I would like to find out upper bound of $$S = \binom{a}{m+1}\sum ^m_{j=0} \frac{\binom{a-m-1}{j}}{\binom{b}{j+m+1}}$$
Anyone can help me please?
Thank you so much.
| I begin by using this identity: $\int_0^1 t^{\alpha} (1-t)^{\beta}=\frac{!}{\alpha+\beta+1}\frac{1}{\binom{\alpha+\beta}{\alpha}}$, for some positive $\alpha$ and $\beta$. This can be alternatively expressed as follows: for some positive $n$ and $k$, we have $\frac{1}{\binom{n}{k}}= (n+1) \int_0^1 t^{k}(1-t)^{n-k} dt$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A problem on Mean Value Theorem If $f''(x)$ exists on $[a,b]$ and $f'(a)=f'(b)$, then :
$$f(\frac{a+b}{2})=\frac 1 2[f(a)+f(b)]+\frac{(b-a)^2}{8}f''(c)$$
for some $c\in(a,b)$.
I tried but was unable to think of a function and was unable to use the given condition except for Rolle's Theorem which does not yield anything... | there exists $x_0$ and $y_0$ in $[a,b]$ such that
\begin{align}
f(a) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(a - \frac{a+b}{2}) + \frac{1}{2}f''(x_0)(a-\frac{a+b}{2})^2 \\
f(b) = f(\frac{a+b}{2}) + f'(\frac{a+b}{2})(b - \frac{a+b}{2}) + \frac{1}{2}f''(y_0)(b-\frac{a+b}{2})^2
\end{align}
Multiplying by $\frac{1}{2}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Decompose a fraction in a sum of two Let's say that I have this fraction:
$$ \frac{2x}{x^2+4x+3}$$
I would like to decompose in two fraction:
$$ \frac{A}{x+3} + \frac{B}{x+1}$$
Which is the procedure for that? :)
| Add both:
$$\frac{2x}{x^2+4x+3}=\frac{Ax+A+Bx+3B}{x^2+4x+3}$$
$$\frac{2x}{x^2+4x+3}=\frac{(A+B)x+(A+3B)}{x^2+4x+3}$$
Then how this must be valid for all $x$ then $A+B=2$ and $A+3B=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/882733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluate the limit of $\ln(\cos 2x)/\ln (\cos 3x)$ as $x\to 0$ Evaluate Limits
$$\lim_{x\to 0}\frac{\ln(\cos(2x))}{\ln(\cos(3x))}$$
Method 1 :Using L'Hopital's Rule to Evaluate Limits (indicated by $\stackrel{LHR}{=}$. LHR stands for L'Hôpital Rule)
\begin{align*}
\lim _{x\to \:0}\left(\frac{\ln \left(\cos \left(2x\... | As told in answers, if you are just concerned by $$\lim_{x\to 0}\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}$$ the shortest way is L'Hopital's rule using the hints given by RecklessReckoner and JimmyK4542. You will then arrive to $$\lim_{x\to 0}\frac{\ln\Big(\cos(ax)\Big)}{\ln\Big(\cos(bx)\Big)}=\frac{a^2}{b^2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Trigonometric problem: $2^{\sin{x}} + 2^{\cos{x}} \ge 2^{(1-1/{\sqrt2})}$ Show that:
$$\large2^{\sin{x}} + 2^{\cos{x}} \ge 2^\left({1-\frac{1}{\sqrt{2}}}\right)$$
This looks like an am gm problem to me where we should be using the fact that am is more that or equal to gm but I am having problem solving this equation af... | By the am-gm inequality,
\begin{align}
\frac{2^{\sin x}+2^{\cos x}}{2}
&\ge 2^{\frac{1}{2}(\sin x+\cos x)}\\
\end{align}
Multiplying both sides of the inequality by $2$ yields
$$2^{\sin x}+2^{\cos x}\ge2\cdot 2^{\frac{1}{2}(\sin x+\cos x)}$$
Since $\color\red{\sin{x}+\cos{x}=\sqrt{2}\sin(x+\pi/4)}$, we have
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\sum \limits_{cyc} \frac{1+b^2+c^4}{a^+b^2+c^3}\geq 3$
If $a,b,c$ are positive real numbers, prove $$\sum \limits_{cyc} \frac{1+b^2+c^4}{a+b^2+c^3}\geq 3$$
Additional info: We should only use Cauchy (preferred to used at least once and more than AM-GM) and AM-GM. We are not allowed to use induction.
Things I... | Using Cauchy-Schwarz, note
$$LHS = \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a+b^2+c^3)^2} \ge \sum_{cyc} \frac{(1+b^2+c^4)(a+b^2+c^3)}{(a^2+b^2+c^2)(1+b^2+c^4)} = \sum_{cyc} \frac{a+b^2+c^3}{a^2+b^2+c^2}$$
As the denominators of the cyclic sum are now the same, it only remains to show that
$$\sum_{cyc} (a + b^2+c^3) ... | {
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If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$. If $4x^2+y^2=40$ and $xy=-6$, find the value of $2x+y$.
I tried the following,
As we know, $(a+b)^2=a^2+b^2+2ab$
Therefore, $(4x^2+y^2)^2=(4x^2)^2+(y^2)+2(4x^2*y^2)$
$=16^4+y^4+8x^2y^2$
What should I do know? Please help. I am stuck.
| Hint:
$(2x+y)^2-4xy=4x^2+y^2$
You can take it from here?
| {
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If $\gcd(a,b)=1$, then every odd factor of $a^2 + 3b^2$ has this same form EDIT: Please see EDIT(2) below, thanks very much.
I want to prove by infinite descent that the positive divisors of integers of the form $a^2+3b^2$ have the same form. For example, $1^2+3\cdot 4^2=49=7^2$, and indeed $7,49$ can both be written i... | For your second question. Not unique. $$28=4^2+3\cdot 2^2=5^2+3\cdot 1^2=1^2+3\cdot 3^2.$$
$$84=6^2+3\cdot 4^2=3^2+3\cdot 5^2.$$
| {
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How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? I am able to solve equation of the type $ |5x+1|=|11-2x|$. I square both the side and my equation becomes $ (5x+1)^2=(11-2x)^2 $ further simplification gives me $ (5x+1)=\pm (11-2x)$.
I get have difficulties in solving... | In this case, $|x + 1| - |1 - x| = 2$, we have,
$$
|x + 1| = \begin{cases} x+1 \ \text{if} \ x \geq -1\\ -x-1 \ \text{if} \ x < -1 \end{cases}
\quad \text{and} \quad
|1 - x| = \begin{cases} 1 - x \ \text{if} \ x \leq 1\\ x-1 \ \text{if} \ x \geq 1 \end{cases}
$$
If $x < -1, \ |x + 1| =-x-1$ and $|1 - x| = 1 - x$, then
... | {
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Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
$$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$
I stumbled upon this question some days ago and been... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that:
$$\frac{3v^2}{w^3}\geq\frac{9u^3-9uv^2+w^3}{u^4}+\frac{7}{4u},$$
which is a linear inequality of $v^2$,
which says that it's enough to prove our inequality for an extremal value of $v^2$,
which happens for equality case of two variables.
Let... | {
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How to calculate limit of a function having factorial in denominator For $n$ tending to infinity find the following limit
$$\frac{2^n}{n!}.$$
I have a feeling that it is multiplication of many numbers with the last one turning to $0$ but the first one is finite so limit should be $0$. But I am not sure and neither am I... | $$0<\frac{2^n}{n!}=\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{4} \cdot \dots \cdot \frac{2}{ n} \leq \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \dots \cdot \frac{2}{3}=\frac{2}{1} \cdot \frac{2}{2} \cdot \left (\frac{2}{3} \right )^{n-2}=2 \left ( \frac{2}{3} \right )^{n... | {
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"source": "stackexchange",
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If the sum of cubes of $a,b,c,d$ is $1$, then $\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$
$a,b,c,d>0$ satisfying $a^3+b^3+c^3+d^3=1$. Prove
$$\frac{1}{1-bcd}+\frac{1}{1-cda}+\frac{1}{1-dab}+\frac{1}{1-abc}\le \frac{16}{3}$$
I tried to go the normal way, by Cauchy-Schwarz, but t... | The AM-GM inequality yields
$$ abcd \le \frac{1}{4 \sqrt[3]{4}} \iff abc \le \frac{1}{4 \sqrt[3]{4}d} \iff \frac{1}{1-abc} \le \frac{1}{1 - \frac{1}{4 \sqrt[3]{4} d}}. $$
It follows that
$$ \sum_{cyclic} \frac{1}{1-abc} \le \sum_{cyclic} \frac{4 \sqrt[3]{4} d}{4 \sqrt[3]{4} d -1}.$$
Therefore it is sufficient to show ... | {
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Apostol (6.25.40): Find $\int {x^{-2}}\sqrt{2 - x - x^2} dx$ I've been struggling with this exercise from Apostol for some time (Section 6.25, Question 40). The integral is
$$
\int\frac{\sqrt{2-x-x^2}}{x^2}\, dx
$$
with a Hint of "multiply numerator and denominator by $\sqrt{2-x-x^2}$".
NB: the answer supplied in the ... | For the integral $\int\dfrac{dx}{x\sqrt{2-x-x^2}}$, we can substitute $x=\dfrac{1}{t}$, so that $dx=-\dfrac{1}{t^2}dt$ and the integral becomes $\int\dfrac{-\frac{1}{t^2}dt}{\frac{1}{t}\sqrt{2-\frac{1}{t}-\frac{1}{t^2}}}=-\int\dfrac{dt}{\sqrt{2t^2-t-1}}$ which can be evaluated easily. I think this should work out.
| {
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Evaluation of $\int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
Evaluation of $\displaystyle \int\frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx$
My Try:: Let $\displaystyle I = \int \frac{\ln(\cos x+\sqrt{\cos 2x})}{\sin^2 x}dx = \int \ln(\cos x+\sqrt{\cos 2x})\cdot \csc^2 xdx$
So $\displaystyle I = -\ln\left(\cos x... | Hint: After integrating by parts, try substituting $u=\sin{x}$. Also, it will be helpful to recall the trigonometric identity, $\cos{(2x)}=\cos^2{x}-\sin^2{x}=2\cos^2{x}-1=1-2\sin^2{x}$.
Let $I$ be your integral, then:
$$\begin{align}
I
&=-\cot{(x)}\ln{\left(\cos{x}+\sqrt{\cos{(2x)}}\right)}-\int\frac{\cot{x}\left(\sin... | {
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How to solve $(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$
Solve $$(x-3)\left(\frac{\mathrm dy}{\mathrm dx}\right)+y=6e^x, x>0$$
I have a very similar problem like this on my homework, and I have no clue how to set it up or even start. How could I set this up?
| $$(x-3)(\frac{dy}{dx})+y=6e^x \Rightarrow \frac{dy}{dx}+\frac{1}{x-3}y=\frac{6 e^x}{x-3} \Rightarrow y'(x)+\frac{1}{x-3}y(x)=\frac{6 e^x}{x-3} $$
The solution of the homogeneous problem is the following:
$$y'_h(x)+\frac{1}{x-3}y_h(x)=0 \Rightarrow \frac{dy_h}{dx}=-\frac{1}{x-3}y_h \Rightarrow \frac{dy_h}{y_h}=-\frac{dx... | {
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"source": "stackexchange",
"question_score": "1",
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What is the probability that a person has both of these attributes? The probability that a blue-eyed person is left-handed is $\frac{1}{7}$.
The probability that a left-handed person is blue-eyed is $\frac{1}{3}$ .
The probability that a person has neither of the attributes is $\frac{4}{5}$.
What is the probability t... | Draw a Venn diagram!
Let probabilities be as follows:
$a$=Blue only
$b$=Blue and Left
$c$=Left only
$d$=None
From probabilities given,
$$\begin{align}
\frac b{a+b}=\frac 17 \quad \Rightarrow a&=6b\\
\frac b{b+c}=\frac 13 \quad \Rightarrow c&=2b\\
d&=\frac 45\\
\end{align}$$
As the probabilites sum to 1,
$$\begin{alig... | {
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"answer_id": 3
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Integral of $R(R^2+y^2)^{-3/2}$ with respect to $y$ $$\int_0^\infty \frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}dy$$
The indefinite integral seems to be $$\frac{-R}{\sqrt{R^2+y^2}}+C$$
$R$ is a constant
| Using $y=Rx$, then $x=\tan(\theta)$ we get
$$
\begin{align}
\int_0^\infty\frac{R}{\sqrt{R^2+y^2}\left(R^2+y^2\right)}\mathrm{d}y
&=\frac1R\int_0^\infty\frac{\mathrm{d}x}{\sqrt{1+x^2}\left(1+x^2\right)}\\
&=\frac1R\int_0^{\pi/2}\cos(\theta)\,\mathrm{d}\theta\\
&=\frac1R
\end{align}
$$
It seems that you have miscomputed ... | {
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"answer_count": 3,
"answer_id": 1
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Proving that a number is non-negative? The numbers $a$,$b$ and $c$ are real.
Prove that at least one of the three numbers
$$(a+b+c)^2 -9bc \hspace{1cm} (a+b+c)^2 -9ca \hspace{1cm} (a+b+c)^2-9ab$$
is non-negative.
Any hints would be appreciated too.
| $$\sum [(a+b+c)^2-9bc]=3\sum[a^2+b^2+c^2-ab-bc-ca]=\frac32\sum (a-b)^2\ge0$$
If each $(a+b+c)^2-3bc<0,$ $$\sum [(a+b+c)^2-3bc]<0$$
| {
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Finding the sum of $3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3$ I see this:
$$A=3+4\cdot 3+4^2\cdot 3+\dots +4^{\log n-1} \cdot 3=3\cdot ([4^{\log n}-1]/3)=n^2-1$$
The base of logarithm is $2$, and $n$ is $2,4,8,\dots$
Anyone could describe me how this sum was calculated? Some hints or some tutorial for this?
| Maybe we need to take a few steps back ( though I imagine this has all been covered in your course). If $a$ is any real number other than $1$ and $r$ is a positive integer, then $1 + a + a^{2} +\ldots +a^{r-1} = \frac{a^{r}-1}{a-1}.$ If you need to verify that, multiply the left side by $a-1$ and notice that there is a... | {
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Can one use logarithms to solve the equations $2=3^x + x$ and $2=3^x x$? Could someone explain how would you solve:
$$2=3^x + x$$
and
$$2=3^x \cdot x$$
I can only solve halfway through.
And why is
$$10^{\log (x)}= x$$
Thanks
| By definition, a log is a quantity representing the power to which a fixed number (the base) must be raised to produce a given number.
Below is a simple example,
\[ 10^{2}=100 \]
So
\[ \log_{10} 100=\log_{10} 10^{2}=2\log_{10} 10= 2 \]
And
\[
10^{\log_{10} 100}=10^{2} = 100
\]
Generally
\[
b^{\log_{b}(x)}=x
\]
Also, t... | {
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Denest $\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$ Is it possible to denest following radical to sum of terms with smaller root count inside?
$\sqrt{20+10 \sqrt{2}-4 \sqrt{5}-2 \sqrt{10}}=\sqrt{\left(2+\sqrt{2}\right) \left(10-2 \sqrt{5}\right)}$
I've found that it cannot be denested into $a+b\sqrt{2}+c... | Well if you're still looking for something:
$$\sqrt{2+\sqrt{2}}=\sqrt{\frac{2+\sqrt{2}}{2}}+\sqrt{\frac{2-\sqrt{2}}{2}}$$
and
$$\sqrt{10-2\sqrt{5}}=2 \cdot \sqrt{\frac{5-\sqrt{5}}{2}}$$
so
$$\sqrt{\left(2+\sqrt{2}\right) \cdot \left(10-2\sqrt{5}\right)}=2\cdot \sqrt{\frac{5-\sqrt{5}}{2}} \cdot \left(\sqrt{\frac{2+\sqrt... | {
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How to find the polynomial such that ... Let $P(x)$ be the polynomial of degree 4 and $\sin\dfrac{\pi}{24}$, $\sin\dfrac{7\pi}{24}$, $\sin\dfrac{13\pi}{24}$, $\sin\dfrac{19\pi}{24}$ are roots of $P(x)$ .
How to find $P(x)$?
Thank you very much.
Thank you every one.
But consider this problem.
Find the polynomial with d... | You may try below :
Say $r_1, r_2, r_3, r_4$ are roots of polynomial,
$P(x) = \large x^4-\left(\sum r_1\right)x^3 + \left(\sum r_1r_2\right)x^2 - \left(\sum r_1r_2r_3\right)x + r_1r_2r_3r_4 $
Note that the roots define the polynomial only upto a constant factor.
| {
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"question_score": "8",
"answer_count": 5,
"answer_id": 3
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How to find ${\large\int}_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx$? Please help me to find a closed form for this integral:
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx
$$
Routine textbook methods for this complicated integral fail.
| Let's give a (sketch of the) proof of the following closed form evaluation.
$$
I=\int_1^\infty\frac{1-x+\ln x}{x \left(1+x^2\right) \ln^2 x} \mathrm dx=\frac{\ln2}6+\frac{\ln\pi}2-6\ln A+\frac2\pi G,
$$
where $A$ is the Glaisher–Kinkelin constant and $G$ is the Catalan constant.
Observe that, by the change of variabl... | {
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"question_score": "40",
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How to find $\int \frac{x^4-4}{x^2\sqrt{4+x^2+x^4}} \,\mathrm dx$ Integrate $$\int \frac{(x^4-4)}{(x^2\sqrt{4+x^2+x^4})}\mathrm dx$$
My try:
$$\int \frac{(x^2-4/x^2)}{(\sqrt{4+x^2+x^4})}\mathrm dx\\
=\int \frac{ (x^2-4/x^2)}{(\sqrt{(x^2+1/2)^2+15/4})}\mathrm dx\\$$
Let $t=x^2$
$$=\int \frac{(t^2-4)}{2t\sqrt t\sqrt{(t+... | $$\int \frac{(x^4-4)dx}{(x^2\sqrt{4+x^2+x^4})}$$
$$ \int \frac{(x^2-4x^{-2})dx}{(\sqrt{4+x^2+x^4})} $$
$$ \int \frac{(x-4x^{-3})dx}{(\sqrt{4x^{-2}+1+x^2})} $$
Take, $$ u = 1+x^2+4x^{-2} \implies du = (2x-8x^{-3})dx \implies du/2 = (x-4x^{-3})dx$$
So integral =
$$ \int \frac{du/2}{\sqrt{u}} = \sqrt{u} + C \implies \... | {
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"source": "stackexchange",
"question_score": "9",
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How to evaluate $\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$ How to evaluate: $$\int \frac{\mathrm{dx}}{x^4[x(x^5-1)]^{1/3}}$$
I have done a substantial work on it:
Let $x^5z^3=x^5-1$. So
$$x^5(z^3-1)=1\implies 5x^4(z^3-1)\mathrm{d}x+x^5(-3z^2\mathrm{d}z)=0\implies \mathrm{d}x=\frac{3xz^2\mathrm{d}z}{5(z^3-1)}$$
So:... | $$
\int \dfrac{1}{x^4\left[x\left(x^5-1\right)\right]^{1/3}}dx\tag{1}
$$
use the sub $u = 1/x^3\implies du = -\dfrac{3}{x^4} dx$.
thus Eq. (1) becomes
$$
\int \dfrac{1}{\left(\dfrac{1}{u^2}-\dfrac{1}{u^{1/3}}\right)^{1/3}}\dfrac{du}{-3} = \dfrac{1}{-3}\int \dfrac{u^{2/3}}{\left(1-u^{5/3}\right)^{1/3}}du
$$
then $v = \l... | {
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"source": "stackexchange",
"question_score": "4",
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How prove this inequality $\sum\limits_{cyc}\frac{1}{a+3}-\sum\limits_{cyc}\frac{1}{a+b+c+1}\ge 0$ show that:
$$\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)\ge 0$$
where $abcd=1,a,b,c,d>0$
I have show three variable... | Let's use Lagrange Multipliers:
Let use take $f$ to be:
$$f(a,b,c,d)=\dfrac{1}{a+3}+\dfrac{1}{b+3}+\dfrac{1}{c+3}+\dfrac{1}{d+3}-\left(\dfrac{1}{a+b+c+1}+\dfrac{1}{b+c+d+1}+\dfrac{1}{c+d+a+1}+\dfrac{1}{d+a+b+1}\right)$$
subject to constraint $abcd=1$:
$$g(a,b,c,d)=abcd=1$$
Here are the equations we need to solve:
$$\fr... | {
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Finding substitution in the integral $\int{\frac{2+3x}{3-2x}}dx$ In a problem sheet I found the integral $$\int{\frac{2+3x}{3-2x}}dx.$$
In the solution the substitution $z=3-2x$ is given which yields $x=\frac{3-z}{2}$ and $dx=-\frac{1}{2}dz$. We have
$$\int{\frac{2+3x}{3-2x}}dx = \int{\frac{2+3\left(\frac{3-z}{2}\right... | Hint: note that
$$ \frac{2+3x}{3-2x} = \frac{13}{2 (3-2x)}-\frac{3}{2} \tag{1}$$
Can you take it from here?
Addendum:
\begin{array}{cccccc}
\color{blue}{2} & \color{blue}{+} & \color{blue}{3 x }& | & \color{red}{3} & \color{red}{-} & \color{red}{2x} \\ \hline
-9/2 & + & 3 x & | & \color{purple}{-\frac{3}{2} } &... | {
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Is it true that $\int_0^1 \lfloor x^{-1} \rfloor^{-1} x^n dx = \frac{1}{n+1}(\zeta(2)+\zeta(3) + \dots + \zeta(n+2) ) - 1$? This question is inspired by the formula $$\displaystyle\int_0^1{\left\lfloor{1\over x}\right\rfloor}^{-1}\!\!dx={1\over2^2}+{1\over3^2}+{1\over4^2}+\cdots = \zeta(2)-1,$$
see for instance this qu... | To answer the side question.
Assume that $ \Re s >1$. Letting $t=x^{-1}$ so $dx=-t^{-2}dt$ the initial integral can be rewritten as:
$$\begin{align}\int_{1}^\infty \lfloor t\rfloor^{-1} t^{-(s+2)}dt &= \sum_{k=1}^\infty \frac{1}k\int_{k}^{k+1}t^{-(s+2)}dt \\&= \frac{1}{s+1}\sum_{k=1}^\infty \frac{1}{k}\left(k^{-(s+1)}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How do I integrate $\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}$ How do I evaluate this indefinite integral, for $|k| < 1$:
$$
\int\frac{\sqrt{1-k^2\sin^2 x}}{\sin x}\mathrm{d}x
$$
I tried the change of variable $t=\sin x$, and obtained two integrals, but I can't integrate either.
| Let $u=k\sin x$, so $du=k\cos x dx$. Then
$\displaystyle\int \frac{\sqrt{1-k^2\sin^2 x}}{\sin x} dx=k\int\frac{\sqrt{1-u^2}}{u}\cdot\frac{1}{\sqrt{k^2-u^2}}du=k\int\sqrt{\frac{1-u^2}{k^2-u^2}}\cdot\frac{1}{u}du$.
Now let $z=\sqrt{\frac{1-u^2}{k^2-u^2}}$, so $u^2=\frac{1-k^2z^2}{1-z^2}$ and $2udu=\frac{2z(1-k^2)}{(1-z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/908576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How to evaluate the following two integral combined with anti-trigonometric function and trigonometric function？ \begin{align*}
&\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} \\
&\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx}.
\end{align*}
A few days ago,my e-friend ask... | In fact, they are similar to Coxeter’s integrals, we use Sangchul Lee's result can get
\begin{align*}
\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx} & = \frac{{{11\pi ^2}}}{72},\\
\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} & = \frac{{{\pi ^2}}}{6}.
\end{align*}
We c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If the equation $2x^2-7x+12=0$ has two roots alpha and beta ,then the value of alpha/beta+beta/alpha is If the equation $2x^2-7x+12 =0$ has two roots $\alpha$ and $\beta$ ,
then the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$ is
note $x=\frac{-7+\sqrt{47}}{4},\frac{-7-\sqrt{47}}{4}$
then
$$\frac{\frac{-7+\sq... | First, note that your quantity can be rewritten as
$\frac{\alpha^2 + \beta^2}{\alpha \beta}$,
and the numerator and denominator here are both symmetric polynomials in $\alpha$ and $\beta$.
On the other hand, for any polynomial, the coefficients themselves are in fact symmetric polynomials in the roots. By dividing by t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$? How to get $f(x)$, if we know that $f(f(x))=x^2+x$?
Is there an elementary function $f(x)$ that satisfies the equation?
| Here’s a technique for finding the first few terms of a formal power series representing the fractional iterate of a given function like $f(x)=x+x^2$. I repeat that this is a formal solution to the problem, and leaves unaddressed all considerations of convergence of the series answer.
I’m going to find the first six te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 5,
"answer_id": 3
} |
Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational How can I prove that the sum $\sqrt 2+\sqrt[3] 3$
is an irrational number ??
| If $x=\sqrt2+\sqrt[3]3$, then
$$
\begin{align}
3
&=(x-\sqrt2)^3\\
&=x^3-3\sqrt2x^2+6x-2\sqrt2\\
(x^3+6x-3)^2&=2(3x^2+2)^2\\
0&=x^6-6x^4-6x^3+12x^2-36x+1
\end{align}
$$
Thus, $x$ is an algebraic integer. Since $2\lt x\lt3$, $x\not\in\mathbb{Z}$, so $x\not\in\mathbb{Q}$. In this answer, it is shown that a rational algebr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/913934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
Sum of a series problem involving cubes For any odd integer $n$, evaluate $n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3$
How would you go about solving such a problem?
Any help would be appreciated.
| I would use the following ideas:
(1): Standard formula: $1^3+2^3+\cdots+a^3=\left(\frac{a(a+1)}{2}\right)^2$
(2): Use (1) to get an expression for $2^3+4^3+6^3+\cdots+(2a)^3=2^3\cdot(1^3+2^3+\cdots+a^3)$
(3): Use (1) and (2) to get an expression for $1^3+3^3+5^3+\cdots+(2a-1)^3$ by noting that
$$1^3+3^3+5^3+\cdots+(2a-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/914621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
prove that $ a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$ I have:
let $a$, $b$ and $c$ be non-negative real numbers with sum $2$. Prove that
$$a^2 + b^2 + c^2 \ge 2\left( {a^3 b^3 + a^3 c^3 + c^3 b^3 + 4a^2 b^2 c^2 } \right)$$
I should determine whether this is a converg... | This inequality is come from? This inequality can use this
$$\Longleftrightarrow \frac{(a^2+b^2+c^2)(a+b+c)^4}{16}-2(\sum{a^3b^3}+4a^2b^2c^2)=\frac{3abc(a+b+c)}{4}\sum{(a-b)^2}+\sum\frac{c(2a+2b+c)(a^2+b^2-c^2)^2}{16}+\frac{(a-b)^2(b-c)^2(c-a)^2}{2}+\frac{a^2b^2c^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do you find the minimum percentage who have studied all four subjects? At college, 70% students studied Maths, 75% students studied English, 85% studied french and 80% studied german. What percentage at least must have studied all 4?
| Using the Generalized Principle of Inclusion-Exclusion, we compute:
$N(0)=100$
$N(1)=70+75+80+85=310$
$N(2)=x_1+x_2+x_3+x_4+x_5+x_6=X$
$N(3)=y_1+y_2+y_3+y_4= Y$
$N(4)=Z$
$100\binom{0}{0}-310\binom{1}{0}+X\binom{2}{0}-Y\binom{3}{0}+Z\binom{4}{0}= 0 $ Study Nothing
$\hphantom{100\binom{0}{1}-}310\binom{1}{1}-X\binom{2}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
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.