Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$1/(x^6 + 1)$ partial fraction decomposition, with computer? This is a re-post from StackOverflow, I was advised to post it here.
https://stackoverflow.com/questions/64101194/partial-fraction-decomposition
How do I find the constants A,B,C,D,K,S such that
$$
\frac{1}{x^6+1} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{x^2-\sqrt ... | What you got is indeed correct since we have $$
\frac{1}{3}\frac{1}{x^2+1} + \frac{-\frac{x}{2\sqrt{3}}+\frac{1}{3}}{x^2-\sqrt 3 x+1} + \frac{\frac{x}{2\sqrt{3}}+\frac{1}{3}}{x^2+\sqrt 3 x+1}
$$
$$=\frac{1}{3}\frac{1}{x^2+1}+\frac{2-x^2}{3(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}+1)}$$
$$=\frac{1}{3}\frac{1}{x^2+1}+\frac{2-x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Finding $\lim_{x\to -\infty}\sqrt{x^2+9x+1}-x$ In this question,
$$\lim_{x\to -\infty}\sqrt{x^2+9x+1}-x=\lim_{x\to -\infty}\left(\sqrt{x^2+9x+1}-x\right)\cdot \frac{\sqrt{x^2+9x+1}+x}{\sqrt{x^2+9x+1}+x}=$$
$$=\lim_{x\to -\infty}\frac{9x+1}{\sqrt{x^2+9x+1}+x}=\lim_{x\to -\infty}\frac{\frac{9x+1}{\sqrt{x^2}}}{\frac{\sqrt... | Answer :
$\lim_{x \to - \infty } \sqrt{x^2+9x+1 }-x$ = $\lim_{x \to - \infty } - x(\sqrt{1+\frac{9}{x}+\frac{1}{x^2 }}+1)$=$-\infty$
Because
$\lim_{x \to - \infty } \frac{9}{x}$=$0$
And
$\lim_{x \to - \infty } \frac{1}{x^2 }=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $X_n \sim \text{Beta}(n, n)$, show that $[X_n - \text{E}(X_n)]/\sqrt{\text{Var}(X_n)} \stackrel{D}{\longrightarrow} N(0,1)$ Let $X_n \sim \mathbf{B}(n,n)$ (Beta distribution), with pdf
$$
f_n(x) = \frac{1}{\text{B}(n,n)}x^{n-1}(1 - x)^{n-1},~~ x \in (0,1).
$$
Knowing that $\text{E}(X_n) = 1/2$ and that $\text{Var}(X... | The answer is in portuguese because I'm a native portuguese speaker.
O último cálculo na demonstração acima é um problema computacional excessivamente difícil. Aqui a ideia é apresentarmos uma demonstração alternativa, que se dá pelos seguintes passos:\
1º: Mostramos que a densidade de $ Y_n $ converge para a densidade... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{... | For $x\geq1$ we obtain:
$$x^{12}-x^9+x^4-x+1=x^9(x^3-1)+x(x^3-1)+1>0.$$
For $0<x<1$ we have:
$$x^{12}-x^9+x^4-x+1=(1-x)+x^4(1-x^5)+x^{12}>0.$$
For $x\leq0$ it's obvious that $$x^{12}-x^9+x^4-x+1>0.$$
| {
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"url": "https://math.stackexchange.com/questions/3845812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Series of rational fractions I read in Gradshteyn I , Ryzhik I Table Of Integrals, Series And Products (7Ed , Elsevier, 2007)(Isbn 0123736374)(1220S)the following series
$$-\frac{x}{x-1}=\sum _{k=0}^{\infty } -\frac{x^{2^k}}{x^{2^{k+1}}-1}=\sum _{k=1}^{\infty } \frac{2^k x^{2^k-1}}{x^{2^k-1}+1}$$
the serie $$\frac{2^k ... | We show the following is valid for |x|<1:
\begin{align*}
\frac{x}{1-x}&=\sum_{k=0}^\infty\frac{2^kx^{2^{k}}}{1+x^{2^{k}}}\tag{1}\\
&=\sum_{k=0}^{\infty}\frac{x^{2^{k}}}{1-x^{2^{k+1}}}\tag{2}\\
\end{align*}
Since we have
\begin{align*}
\frac{2^{k+1}x^{2^{k+1}}}{1-x^{2^{k+1}}}=\frac{2^kx^{2^k}}{1-x^{2^k}}-\frac{2^kx^{2^k... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$? This question arose on the code golf StackExchange:
Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$?
In other terms: is there an integer that looks like $\color{blue}{1000...001000...}$ in both b... | For a prime $p$, let $\nu_p(n)$ denote the exponent of $p$ in the prime factorization of $n$. We will use the Lifting the Exponent lemma.
Firstly, note that
$$a=\nu_2(2^a+2^b)=\nu_2(10^c+10^d)=c,$$
so we have $2^a+2^b=10^a+10^d$. Let $b=a+x$ and $d=a+y$, and write
$$1+2^x=5^a(1+10^y).\tag{1}\label{eq1}$$
We observe tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find the zeros of $f(x)=x^3−4x^2+x−4$ I am to find the zeros and multiplicities of $f(x)=x^3−4x^2+x−4$.
The solution provided in the answers section of my book is 4 with multiplicity 1. I arrived at $2\pm\sqrt(8)$.
My working:
$$x^3-4x^2+x-4$$
$$x(x^2-4x+1)-4$$
Then, focusing on the quadratic in the middle, I used 3Bl... | I think the following at least is right: $$x^3-4x^2+x-4=x^2(x-4)+1(x-4)=(x-4)(x^2+1),$$ which gives $$\{4,\pm i\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluating $\frac{dg}{dθ}$ at $(r,θ)=(2\sqrt{2},\frac{π}{4})$ where $g(x,y)=\frac1{x+y^2}$ using chain rule?
Okay so my first step is to find the partial derivatives:
$$\frac{\partial \:}{\partial \:x}\left(\frac{1}{x+y^2}\right)=-\frac{1}{\left(x+y^2\right)^2}$$
$$\frac{\partial \:}{\partial \:y}\left(\frac{1}{x+y^2}... | It is easier if you substitute $x$ and $y$ directly $$g(r,\theta)=\frac{1}{x+y^2}=\frac{1}{32r\cos(\theta)+9r^2\sin^2(\theta)}$$
So $$\frac{\partial g}{\partial \theta}=-\frac{-32r\sin(\theta)+18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$
Or $$\frac{\partial g}{\partial \theta}=\frac{\parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A cyclic inequality of degree 10 Suppose that $x,y,z\geq 0$. I would like to prove that
$$(x^5+y^5+z^5)^2\geq (x+y+z)(x^3y^6+y^3z^6+z^3x^6).$$
I can prove this inequality using some standard methods. For example, I can let $x=1, y=1+p$, and $z=1+p+q$. Then Mathematica can simplify the difference of the two sides of the... | Another way:
After homogenization we need to prove that:
$$\sum_{cyc}(x^{10}+2x^5y^5-x^7z^3-x^6z^4-x^6z^3y)\geq0$$ or
$$\sum_{cyc}(x^7z^3-x^6z^3y)+\sum_{cyc}(x^{10}-2x^7z^3-x^6z^4+2x^5y^5)\geq0,$$ which is true by AM-GM and SOS:
$$\sum_{cyc}(x^7z^3-x^6z^3y)+\sum_{cyc}(x^{10}-2x^7z^3-x^6z^4+2x^5y^5)=$$
$$=\frac{1}{37}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answ... | I do not think it would take a long time to simplify these expression by hand. Rewrite it as
$$\frac {5}{2}\cdot\frac {10}{3}\cdot\frac {17}{4}\cdot\frac {26}{5}\text{ vs }\frac {11}{5}\cdot \frac {13}{4}\cdot\frac {13}{3}\cdot \frac {11}{2}$$
Denominators go away, and the factor $13$ in the numerator as well:
$$1700\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How does recurrence $a_n=\frac12(a_{n-1}+\frac\beta{a_{n-1}})$ become $b_n=\frac12\frac{b_{n-1}^2+\beta-\alpha^2}{b_{n-1}+\alpha}$ when $b_n=a_n-a$?
$a_n = \frac{1}{2}(a_{n-1} + \frac{\beta}{a_{n-1}})$ for $ n > 0$ with $a_0 = 1$. Changing variables in this recurrence, and letting $b_n = a_n - a$, we find by simple al... | $b_n = a_n - a$, $b_{n-1} = a_{n-1} -a$. Hence
$$\begin{align}b_n = a_n - a &= \frac12(a_{n-1} + \frac \beta{a_{n-1}})-a
\\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a})-a
\\&=\frac12(b_{n-1}+a + \frac \beta{b_{n-1}+a}-2a)
\\&=\frac12(b_{n-1}-a + \frac \beta{b_{n-1}+a})
\\&=\frac1{2(b_{n-1}+a)}((b_{n-1}+a)(b_{n-1}-a)+\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y ... | So the issue is you have not used $x<0$. By the way be careful when you use a variable like x. in one part you have used it for the domain and in one for the range set.
By using the quadratic formula you have obtained two solutions, so your question is which is the right one at $a=1$. When $a=1$
$$ 1= x-\frac{2}{x} $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Intersection of median and segment between two sides of a triangle
In triangle $ABC,$ $M$ is the midpoint of $\overline{BC},$ $AB=12,$ and $AC=16.$ Points $E$ and $F$ are taken on $\overline{AC}$ and $\overline{AB},$ respectively, and $\overline{EF}$ and $\overline{AM}$ intersect at $G.$ If $AE=2AF,$ then what is $EG/... | This is an application to the law of sines.
In $\triangle ABC$:
$$\frac{\sin\angle ABC}{AC}=\frac{\sin\angle ACB}{AB}$$
In $\triangle ABM$:
$$\frac{\sin\angle BAM}{BM}=\frac{\sin\angle ABC}{AM}$$
In $\triangle CAM$:
$$\frac{\sin\angle CAM}{CM}=\frac{\sin\angle ACB}{AM}$$
In $\triangle FAG$:
$$\frac{\sin\angle BAM}{FG}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
| $$\eqalign{1 - 2 \sin^2 x \cos^2 x &= 1 - (1-\cos^2 x) \cos^2 x - \sin^2 x (1 - \sin^2 x)\cr
&= 1 - \cos^2 x + \cos^4 x - \sin^2 x + \sin^4 x\cr
&= \cos^4 x + \sin^4 x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How do you find the min/max values of $Arg(z)$ for {$z:|z-(4+3i)|=2$}? How do you find the min/max values of $Arg(z)$ for {$z:|z-(4+3i)|=2$}?
I have drawn a diagram with a circle centre $(4+3i)$ and radius = $2$. I think to find the min/max values of $Arg(z)$ I need to find the points of tangency? I don't understand ho... | Let $y=mx$ be the equation of a line tangent to the circle $(x-4)^2 + (y-3)^2 =4$. Then substituting the first into the second, we get
$$
\begin{align}
(x-4)^2 + (mx-3)^2 &= 4 \\
(m^2 + 1) \, x^2 - (6m+8) \, x +21 &= 0
\end{align}
$$
Since the line is tangent to the circle, the discriminant of the above must be zero.
$... | {
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"url": "https://math.stackexchange.com/questions/3857410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equati... | Answer :
$\frac{x^3-4x^2-4x+16}{\sqrt{x^2 - 5x+4}}= \frac{x(x^2 - 4)-4(x^2 - 4)}{\sqrt{x^2 - 5x+4}}$=$\frac{(x^2 - 4)(x-4)}{\sqrt{x^2 - 5x+4}}$
$\sqrt{x^2 - 5x+4} = 0 $ if $ x =(1, 4) $
Suppose $x≠(1,4)$
$(x^2 - 4)(x-4)$ =0
$\Rightarrow$ $ (x - 2)(x+2)(x-4)=0$
$\Rightarrow $ the solution is ($x=2$ or $ x=- 2)$
Because... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Stuck on Mathematical Induction Proof I have the following question: Prove with mathematical induction that $3^n+4^n\le 5^n$ for all $n\ge 2$.
$$\text{Assume true: }3^n+4^n \le 5^n \text{. Prove that $3^{n+1}+4^{n+1} \le 5^{n+1}$} \\
= 3\cdot3^n+4\cdot4^n \\ =3\cdot3^n+4^n(3+1)\\
=3\cdot3^n+3\cdot4^n+4^n \\
=3(3^n+4^n... | How about this, with the same first step:
$$\begin{align*}
3^{n+1} + 4^{n+1} &\le 3 \cdot 3^n + 4 \cdot 4^n \\
&\le 5 \cdot 3^n + 5 \cdot 4^n \\
&\le 5(3^n + 4^n) \\
&\le 5 \cdot 5^n \\ &= 5^{n+1}
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Let $f(x)$ be a polynomial satisfying $\lim_{x\to \infty} \frac {x^4 f(x)}{x^8+1} =3$, $f(2)=5$, $f(3)=10$,$f(-1)=2$,$f(-6)=37$. Find $f(0)$ It’s clear that $f(x)$ is a 4th degree polynomial.
If $f(x)=ax^4+bx^3+cx^2+dx+e$, then $a=3$
From the rest of the given data, I can form four linear equations, which should give m... | It is clear that $f(x)$ is of deegree 4 and
$$\begin{align} f(2) &= 1 + 2^2
\\ f(-1) &= 1 + (-1)^2
\\ f(3) &= 1+ 3 ^ 2
\\ f(-6) &= 1+ (-6)^2 \end{align} $$
so $f(x) - (1+x^2)$ has roots $2,3,-1,-6$
$$ \Rightarrow f(x) - (1+ x^2) = A(x+1)(x-2)(x+6)(x-3) $$
now you can continue
| {
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"url": "https://math.stackexchange.com/questions/3860206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Prove by deduction $\displaystyle \frac{a^2 + b^2}{2} \geq ab$ I am currently working in the following question:
Prove by deduction $$\frac{a^2 + b^2}{2} \geq ab$$ for all $a,b \in \mathbb{R}$
My question is: can prove instead the following?
$$\frac{a^2 + b^2}{2} - ab \geq 0$$
My solution:
$$\frac{a^2 + b^2}{2} - ab ... | You are correct since $$\frac{a^2+b^2}{2}-ab\geq 0\iff\frac{a^2+b^2}{2}\geq ab.$$
Alternatively since $(a-b)^2\geq0$ for all $a,b\in\mathbb R$ we have $$a^2+b^2\geq2ab$$ or $$\frac{a^2+b^2}{2}\geq ab$$
as required.
| {
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"url": "https://math.stackexchange.com/questions/3860558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion.... | Rather than carry the notation we write $s=\sin(x)$ and $c=\cos(x)$. Subtracting the equation from $s^2+c^2=1$ yields $s^3-s^2 + c^3-c^2=0$ or $s(s^2-1)+c(c^2-1) = s(-c^2) + c(-s^2) = 0$ and so either $s=0$ or $c=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find... | First, we can omit the $\sqrt{n}$ part and only sum up to $n$. To see this,
$$
\sum_{k=n}^{n+\sqrt{n}} x_n \leq
\sum_{k=n}^{n+\sqrt{n}} \frac{4(n+\sqrt{n})^2}{4n^3} \leq \sum_{k=n}^{n+\sqrt{n}} \frac{(n+n)^2}{n^3} \leq
\sum_{k=n}^{n+\sqrt{n}} \frac{4}{n} \leq 4\frac{\sqrt{n}}{n} \mapsto 0
$$
The rest is the Riemannia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $n =3,$ write $\Delta^2$ as an element of $A = \mathbb{Q}[e_{1}, e_{2}, e_{3}.]$(manually) Here is the question I want to answer letter $(e)$ of it manually:
Let $B = \mathbb{Q}[x_{1}, ... , x_{n}] \cong \mathbb{Q}^{[n]}$ and $A = \mathbb{Q}[e_{1}, ... , e_{n}]$ where $e_{i} \in B$ is the elementary symmetric polyn... | What a strange hint. There's a much better way to do it, and I have no idea why the author didn't set $a_3 = 1$ from the start.
I learned this argument from David Speyer. Computing the discriminant as a polynomial in the $e_i$ is equivalent to computing the discriminant of the polynomial $f(x) = x^3 - e_1 x^2 + e_2 x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3862302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinatorics / Inclusion-Exclusion Principle I have this problem:
A bakery sells $4$ varieties of donuts but there are only $6$ chocolate and $7$ glazed.
How many ways can we buy $12$ donuts if the selection must include at least $3$ glazed donuts (order is irrelevant)?
In my attempt, I first subtracted the $3$ glazed... | The question implies that there are at least $12$ donuts of each of the other varieties available.
If we select three glazed donuts to begin with, then we must select nine more donuts from the six chocolate, four glazed, and unlimited (for our purposes) number of the other two varieties available. The number of ways w... | {
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"timestamp": "2023-03-29T00:00:00",
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If $1+n+n^2+n^3$ is a perfect square, then $n=1$ or $n=7$ I want to prove that if $1+n+n^2+n^3$ is a perfect square then $n=1$ or $n=7$.
I managed to prove that $1+n+n^2+n^3=(n^2+1)(n+1)$ and that $(n^2+1,n+1)$ is either $1$ or $2$.
I found out that it could not be $1$, and then $\frac{1}{2}(n^2+1,n+1)=1$.
From here I ... | I. If $1+n+n^2+n^3=z^2$, and odd $n=2y+1$, then$$1+n+n^2+n^3=8y^3+16y^2+12y+4=z^2$$
Since $4$ divides $z^2$, then if $z^2=4w^2$ $$2y^3+4y^2+3y+1=w^2$$ and factoring $$(y+1)(2y^2+2y+1)=w^2$$
Now $y+1$ and $2y^2+2y+1$ are relatively prime, since if integer $k$ divides $y+1$, then $k$ divides $2y(y+1)=2y^2+2y$, and hence ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove the equation $\int_{a}^{b} f(x) \mathrm{d} x = \frac{b-a}{2}[f(a)+f(b)]-\frac{(b-a)^{3}}{12} f^{\prime \prime}(\eta)$? In numerical analysis, we know that the following equation is true:
$$\int_{a}^{b} f(x) \mathrm{d} x = \frac{b-a}{2}[f(a)+f(b)]-\frac{(b-a)^{3}}{12} f^{\prime \prime}(\eta), \quad \eta \in... | The link Trapezoid rule shows the error term is bounded but is a little unsatisfactory since it does not prove equality in this particular case.
Simplifying to the interval [0,1], If we assume $f$ is twice continuously differentiable on $[0,1]$ then, using integration by parts we can proceed thus,
\begin{aligned}
\int_... | {
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Prove $0 < x < \pi /2 \implies \sin x > x/\sqrt{x^2+1}$ using Mean Value Theorem I'm solving the following problem:
Show that if $0 < x < \pi /2$ then $\sin x > \dfrac{x}{\sqrt{x^2+1}}$.
One of the hints given is to apply mean value theorem for $\sin (x)$ on the interval $[0,x]$
This is my attempt so far:
Let $f(x) =... | You are on the right path: square both sides of $\sin(x)>x (1-\sin^2(x))^{1/2}$ and get $\sin^2(x)>x^2(1-\sin^2(x))=x^2-x^2\sin^2(x)$, then add $x^2\sin^2(x)$ to both sides to obtain $\sin^2(x)(1+x^2)>x^2$. Now divide both sides by $(1+x^2)$ and take the square root.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5... | $$q = -27p - \color{red}{141} {\leftarrow (4)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq \frac{a^2+b^2+c^2}{ab+bc+ca}+10.$
Problem. (?) For $a,b,c$ be non-negative numbers such as $a \geq 2(b+c).$
Prove:$$5\Big(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\Big)\geq
\frac{a^2+b^2+c^2}{ab+bc+ca}+10.$$
My Solution.
We write the inequality as
$$\big... | Let $c=\min\{a,b,c\}$, $b=c+u$ and $a-2(b+c)=v$.
Thus, $u\geq0$, $v\geq0$, $a=4c+2u+v$ and
$$(ab+ac+bc)\prod_{cyc}(a+b)\left(5\sum_{cyc}\frac{a}{b+c}-\frac{a^2+b^2+c^2}{ab+ac+bc}-10\right)=$$
$$=745vc^4+2(2u^2+745uv+247v^2)c^3+(6u^3+1074u^2v+741uv^2+109v^3)c^2+$$
$$+(2u^4+329u^3v+355u^2v^2+109uv^3+8v^4)c+$$
$$+2uv(18u^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $e ^{-\frac{x^{2}}{2}}\leq \frac{2}{\sqrt{{e}}}\frac{1}{1+x^{2}}$ I have two inequalities today :
*
*$$\color{red}{a.~~~~~ e ^{-\frac{x^{2}}{2}}\leq \frac{2}{\sqrt{{e}}}\frac{1}{1+x^{2}}}$$
*$$\color{purple}{b.~~~~~ e ^{-\frac{x^{2}}{2}}\leq\sqrt{e}e ^{-|x|}}$$
I tried to relate these inequalities to this fa... | For $a$, you can instead consider $y = x^2$ for positive $y$ to get $$e^{-\frac{y}{2}} \le \frac{2}{\sqrt{e}} \frac{1}{1+y} \to (1+y)e^{-\frac{y}{2}} \le \frac{2}{\sqrt{e}}$$
You can then find the maximum of the left hand side by differentiating and finding the roots. It would end up being at $y = 1$, and $(1+1) e^{-\f... | {
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| Solve for $c$:
$$c=\frac{-b\pm\sqrt{12-3b^2}}{2}$$
To prove the inequality, we need to prove it for the larger of the two values of $c$ (the one with the sign "plus"):
$$b+c=\frac{b+\sqrt{12-3b^2}}{2}\le 2$$
i.e.
$$\sqrt{12-3b^2}\le 4-b$$
which is defined for $|b|\le 2$ - and in that interval the RHS is positive so we ... | {
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How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equatio... | Note that $$4^na_n-4^{n-1}a_{n-1}=\left(\dfrac{8}{3}\right)^{n-1}$$ now telescope.
Add: Let me compete the computation to get a closed form. After taking the summation $$4^na_n-a_0=\sum_{k=1}^n\left(\dfrac{8}{3}\right)^{k-1}=\dfrac{1-\left(\dfrac{8}{3}\right)^{n}}{1-\left(\dfrac{8}{3}\right)}$$ and hence $$4^na_n=a_0+\... | {
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Prove by Induction. For $n \in \mathbb{N}$, $10|(9^{n+1}+7^{2n})$. So far, this is what I have, but I'm confused as to how to 1. remove the 7 from inside the brackets to be able to substitute 10k and 2. make the whole thing divisible by 10 so that I can prove it.
Basic Step: Let $n = 1$. Therefore,
$$
9^{1+1} + 7^{2 \c... | Instead of trying to factor out numbers to make it divisible by 10, I was able to solve the equation by rearranging the formula that equals 10k to
$$
9^{+1} = 10 - 7^{2}
$$
If you substitue it into P(k + 1) you get:
$$
P(k+1) = 9^1 \cdot (10m - 7^{2k}) + (49 \cdot 7^{2k})
$$
Which reduces to
$$
P(k+1) = 90m + (40 \cdot... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $\int [e^{\tan x} (1-\tan x)^2\cos^2 x]dx$ Let $\tan x=t$
$$\int [e^t (1-\tan x)^2 \cos^2 x \frac{1}{\sec^2x} ]dt$$
$$\int [e^t (\cos x-\sin x)^2 \cos^2 x ]dt$$
$$\int [e^t (1-\sin 2x)\cos^2 x] dt$$
How do I solve further?
| $(1-\tan x)^2 = 1+\tan^2x-2\tan x $
So, if you let $\tan x = t \Rightarrow dx = \dfrac{dt}{\sec^2x} = \dfrac{dt}{1+t^2}$ and $\cos^2x = \dfrac{1}{1+t^2}$
$\begin{align}\Rightarrow I = \int e^{t}(1+t^2-2t)\dfrac{dt}{(1+t^2)^2} & = \int e^t\left[\dfrac{1}{1+t^2 }-\dfrac{2t}{(1+t^2)^2}\right]dt\end{align}$
Now, $\dfrac{d}... | {
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Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ wh... | Titu's lemma gives us
$$ \sum \frac{q}{ 3 + p } = \sum \frac{ q^2 } { 3q + pq} \geq \frac{ (p+q+r)^2 } { \sum 3q + pq } = \frac{ 9}{9 + pq+qr+rs}.$$
Since $(p+q+r) ^2 \geq 3 (pq+qr+rs)$, so $3 \geq pq+qr+rs $ and thus
$$ \sum \frac{q}{ 3 + p } \geq \frac{9}{9 + pq+qr+rs} \geq \frac{3}{4}.$$
The original problem c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simplifying the derivative of $\sin(\cos^2 x)\cos(\sin^2 x)$ Differentiating the term
$$\sin(\cos^2 x)\cos(\sin^2 x)$$
leads me through the chain and product rule to
$$-\sin(2x)\cos(\cos^2 x)\cos(\sin^2 x)+(-\sin(\sin^2 x)\sin(2x)\sin(\cos^2 x))$$
where the derivative of $\sin^2 x$ equals to
$$\frac{d}{dx} \sin^2 x = 2... | If we start with $$y=\sin \left(f(x)^2\right) \cos \left(g(x)^2\right)$$
$$y'=2 f(x) f'(x) \cos \left(f(x)^2\right) \cos \left(g(x)^2\right)-2 g(x) g'(x)\sin
\left(f(x)^2\right) \sin \left(g(x)^2\right)$$ If $f(x)=\cos(x)$ and $g(x)=\sin(x)$, then
$$2 f(x) f'(x)=-2 \sin (x) \cos (x)=-\sin(2x)$$
$$2 g(x) g'(x)=+2 \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881523",
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Probability that the cards are in AP
Box $1$ contains three cards bearing numbers $1, 2, 3$ ; box $2$ contains five cards bearing numbers $1, 2, 3, 4, 5$ ; and box $3$ contains seven cards bearing numbers $1, 2, 3, 4, 5, 6, 7$. A card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the ... | As others have commented, $(3,2,1)$ is missing. Instead of listing the choices, observe that $x_1,x_2,x_3$ are in AP iff $x_1+x_3=2x_2$. This implies either both $x_1,x_3$ are odd or both are even since $x_1+x_3$ is even. We have $2$ odd values of $x_1$ and $4$ odd values for $x_3$, giving $2\times 4$ pairs. We also ha... | {
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Unifying a real-valued function over two branches I have a function that I want to evaluate numerically. These three forms listed below are all identical algebraically for nonnegative real $u$ (aside from removable discontinuities at $u=0$ and $u=1$), but none avoids complex numbers for its entire domain.
$$\begin{alig... | Using the series expansion
$$\ln(x)=\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{x-1}{x+1}\right)^{2n+1}$$
gives, according to the third representation,
\begin{align}f(x)&=\frac1{\sqrt{x^2-1}}\ln\left(x+\sqrt{x^2-1}\right)\\&=\frac1{\sqrt{x^2-1}}\sum_{n=0}^\infty\frac2{2n+1}\left(\frac{x-1+\sqrt{x^2-1}}{x+1+\sqrt{x^2-1}}\r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Jordan normal form of $\left(\begin{smallmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{smallmatrix}\right)$ I want to find the Jordan normal form of $A=\begin{pmatrix} 4 & 1 & 1 \\ -2 & 1 & -2 \\ 1 & 1 & 4 \end{pmatrix}$, but somewhere I think that I make a mistake (I am quite new to the computation of the Jordan ... | Since you have two linearly independent eigenvectors, you know that there are two blocks for $\lambda=3$. That is already enough to tell you what the Jordan form looks like.
So it is not really necessary to look for generalized eigenvectors, if you only want Jordan form. But still, it might be useful to try this - espe... | {
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$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$ Proof Does anyone know hot to prove this inequality?
Having: $a, b, c \gt 0$
$$\frac{b^2-a^2}{c+a} + \frac{c^2-b^2}{a+b} + \frac{a^2-c^2}{b+c} \ge 0$$
I tried with the AM-GM inequality but I couldn't get any improvement. I'm on a still point and I ... | Because
$$\sum \frac{a^2}{c+a} = \sum \frac{c^2}{c+a}.$$
Therefore, we need to prove
$$2\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqslant \frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}.$$
Suppose $a \geqslant b \geqslant c,$ then
$$a^2 \geqslant b^2 \geqslant c^2 \quad \text{and} \quad... | {
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"timestamp": "2023-03-29T00:00:00",
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prove $\ e^{n^2} \ge n^n $ for every n by induction Prove for any $\ n \ge 1 $
$$\ e^{n^2} \ge n^n $$
my attempt:
true for $\ n = 1 $ then for $\ n+ 1 $ :
$\ e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} \cdot e^{2n} \cdot e $
$\ ( n+1)^{n+1} = (n+1)^n\cdot (n+1) \le (n+n)^n\cdot(n+1) = (2n)^n \cdot(n+1) = 2^n \cdot n^n \cd... | You can continue your work by taking $\ln$ of both sides and we have $2n+1 \geq n\ln2 + \ln(n+1) \geq n/2 + n+1=3n/2 + 1$. Then we get that $n/2 \geq 0$ with reversible steps.
| {
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Simplifying $\cos^{-1}x +\cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$ A question has this equation: $$f(x) = \cos^{-1}x + \cos^{-1}\left(\frac{x}{2} + \frac{\sqrt{3-3x^2}}{2}\right)$$ and you're supposed to simplify it and find $f\left(\frac{2}{3}\right)$ and $f\left(\frac{1}{3}\right)$.
By taking $\cos... | $$
f(\cos\alpha)=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg(\frac{1}2\cos\alpha+\frac{\sqrt3}2\sin\alpha\bigg)\\
= \cos^{-1}(\cos\alpha) +\cos^{-1}\bigg(\cos\frac \pi 3\cos\alpha+\sin\frac{\pi}3\sin\alpha\bigg)\\
=\cos^{-1}(\cos\alpha)+\cos^{-1}\bigg[\cos\bigg(\frac\pi3-\alpha\bigg)\bigg]\\
= \pm\alpha+2\pi k_1 \pm\ (\frac\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887117",
"timestamp": "2023-03-29T00:00:00",
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Trouble Understanding Proof of Lagrange's Identity Using Binomial Formula I've been reading a book on elementary mathematics and am having trouble understanding a proof that they give for Lagrange's identity using the binomial formula.
Lagrange's identity is stated as :
\begin{equation}
\sum_{j=0}^{n} \binom{n}{j}^{2} ... | The binomial coefficient $\binom{2n}{n}$ is the coefficient of $x^n$ in the left-hand side.
In the right hand side you get $x^n$ from terms $x^i$ and $x^j$ where $i+j=n$. The term $x^i$ has coefficient $\binom{n}{i}$ and similarly for $x^j$, so you have
$$
\binom{2n}{n}=\sum_{i+j=n}\binom{n}{i}\binom{n}{j}
$$
Now obser... | {
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"timestamp": "2023-03-29T00:00:00",
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how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded? I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$
| Claim. If $a_n$ is increasing, then $a_n<3$.
Otherwise, if $a_n\ge 3$, for some $n\in\mathbb N$, then
Case A. $a_n=3$, then $a_{n+1}=\sqrt{a_n+6}=\sqrt{3+6}=\sqrt{9}=3$. Hence $a_n$ is not increasing.
Case B. $a_n>3$. Then $a_{n+1}>a_n>3$ and
$$
a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{\sqrt{(a_n+6}-a_n)(\sqrt{a_n+6}-a_n)}{\... | {
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Solution of |x| + 3x = 2 + 6i. Where is an error in my thinking? So I solved the equation: $|x| + 3x = 2 + 6i$ but I don't know where my error is. And I know there is an error because Wolfram Alpha shows that the only solution is $ x = 2i $ . In my calculations I have 2 solutions.
$$|x| + 3x = 2 + 6i$$
$$\sqrt{a^2 + b^... | When you square both sides of $\sqrt{a^2+4} = 2-3a$, an extra solution is generated.
In particular, the solution $a = 3/2$ is the solution to $\sqrt {a^2+4} = 3a-2$, which is a perfectly valid solution for the squared equation $a^2+4 = (2-3a)^2$.
However for this $a$, $\sqrt{a^2+4} > 0$ but $2-3a < 0$, rendering it inv... | {
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)... | The set of solutions is the segment $\{\alpha i\mid-1\le\alpha\le1\}$.
This follows from the following geometrical argument. $|z-i|$ is the distance from $z$ to $i$; $|z+i|$ is the distance from $z$ to $-i$. Which point $z\in\mathbb C$ can have the sum of distances to $i$ and to $-i$ equal to $2$, knowing that the dist... | {
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Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$
Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
I got this question on my exam today ... | So you begin by writing
$$
\tan 6\theta \equiv {\sin 6\theta \over \cos 6\theta}
$$
From there you apply De Moivre's Theorem which states:
$$
(\cos \theta +\iota\sin\theta)^n \equiv (\cos n\theta + \iota\sin n\theta)
$$
In this case we have $n = 6$, so:
$$
(\cos \theta +\iota\sin\theta)^6 \equiv (\cos 6\theta + \iota\... | {
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Is there a simple, but tight lower bound for the error made when $\sum_{n=1}^{k}\frac{1}{n^2}$ is used to approximate $\frac{\pi^2}{6}$? As a math-for-fun exercise, I've recently been seeking bounds for the error $R_k$ made when using $\sum_{n=1}^{k}1/n^2$ to estimate its beautiful sum $\pi^2/6$. Applying the Compariso... | In the same spirit as @Robert Israel in his answer, we could use
$$\sum_{n=N+1}^\infty \frac{1}{n^2}=\psi ^{(1)}(N+1)$$ and use the series expansion of the rhs
$$\psi ^{(1)}(N+1)=\frac{1}{N}-\frac{1}{2 N^2}+\frac{1}{6 N^3}-\frac{1}{30
N^5}+\frac{1}{42
N^7}-\frac{1}{30 N^9}+O\left(\frac{1}{N^{11}}\right)$$ which, ... | {
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Finding positive integer(s) $n$ for which $3^{n-1} + 5^{n-1} | 3^{n} + 5^{n}$: Is my solution correct? The problem is the following
Find all positive integers $n$ such that $3^{n-1} + 5^{n-1}$ divides $3^{n} + 5^{n}$.
And I attempted as the following:
As $3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n}$,
$$3^{n-1} + 5^{n-1}\ |\... | What you did is mostly correct, but your argument that $p = \frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$ is only odd for $n = 1$ is not true since for all odd $n$, the result is either not an integer or it's an odd integer. Instead, $p$ is odd, as you stated, but with $2 - p \gt 0$ from $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} ... | {
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Number Theory : Find the group of A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ Find the group A such that $A=\{x \in \mathbb{Z}:\frac{x^3-3x+2}{2x+1}\in \mathbb{Z}\}$ ?
Polynomial Long Division we get
$\frac{x^{2}}{2}-\frac{x}{4}-\frac{11}{8}+\frac{27}{8\left(2x+1\right)}$
but how i can fro... | Hint:
If $2x+1=y$
$$x^3-3x+2=(y-1)^3/8-3(y-1)/2+2=\dfrac{y^3-3y^2-9y+27}8$$ which will be divisible by $y$
$\iff y$ divides $y^3-3y^2-9y+27$ as $y$ is odd
| {
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"question_score": "2",
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"answer_id": 1
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Is there an elegant way to find the equation of a plane given its line and angle of intersection with another plane? This is a problem I give sometimes to my calculus students:
Let $\mathcal{Q}$ be the plane defined by $x+2y+3z=6.$ Find a plane $\mathcal{P}$ that meets $\mathcal{Q}$ at an angle of $\theta = \frac{\pi}... | a direction within the plane $Q$ that is orthogonal to the line is
$(1,2,3)$ crossed with $(-9,3,1);$ this is $7(-1,-4,3).$ Let us name $u= (-1,-4,3).$
Length of $u$ is $\sqrt{26},$ so make a unit vector
$$ v = \frac{1}{\sqrt{26}}(-1,-4,3) $$
The unit normal to $Q$ is
$$ w = \frac{1}{\sqrt{14}} (1,2,3) $$
Next l... | {
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Generating functions and a closed form for the Fibonacci sequence - the big picture I have spent the last few hours trying to understand one way of deriving a closed form for the Fibonacci sequence. As part of improving my mathematical maturity, I am trying to learn to see the "big picture" of what I'm doing as opposed... | It's a good approach. One thing that can be simplified a little bit is:
$$
f(x)=\frac{1}{(1-\alpha x)(1-\beta x)} = \frac{1}{\alpha - \beta} \cdot \frac{\alpha (1-\beta x) - \beta(1-\alpha x)}{(1-\alpha x)(1-\beta x)} = \frac{\alpha/(\alpha - \beta)}{1-\alpha x} - \frac{\beta/(\alpha - \beta)}{1-\beta x}.
$$
(And this... | {
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Suppose $x$ and $y$ are unequal real numbers
Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$
For the above question, I got $x$ not equal to $-1,0,1... | Intuitively, one should let $a=x+y$ and $b=x-y$.
Then from the condition $\sqrt {xy}=1$ one obtains $a^2-b^2=4$.
The other condition becomes
$$a = \sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}$$
Cubing both sides, we have
\begin{align}a^3 &= \frac ab + 3\left(\sqrt[3]{\frac ab}\right)^2\sqrt[3]{\frac ba}+ 3\sqrt[3]{\frac ab}... | {
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Can fractional/decimal radicals/roots exist? For questions like "What is the 1/2th root of x would the answer be $x^2$?
My logic is that since $$
\sqrt[\cfrac{1}{2}]{x}=x^{1/{(\cfrac{1}{2}})}
$$
Which simplifies to $x^2$.
So as a general rule it could be $$
\sqrt[\cfrac{1}{a}]{x}=x^{1/{(\cfrac{1}{a}})}
=x^a
$$
And with... | The definition is:
$$\sqrt[a]{x} = x^{\frac{1}{a}}.$$
Hence, if $a = \frac{1}{b}$, then:
$$\sqrt[\frac{1}{b}]{x} = x^{b}.$$
As a consequence:
$$\sqrt[\frac{1}{2}]{x} = x^2.$$
Notice that $a = \frac{1}{b}$ implies that, if for example $b = \frac{22}{91}$, then $a = \frac{91}{22}$. If for example $b = \sqrt{3}$, then $a... | {
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1... | Let $$S=1^1+3^2+\cdots+(2n-1)^2=\sum_{k=1}^{n}(2k-1)^2=\sum_{k=1}^{n}[4 k^2-4k+1]$$
$$S=4\sum_{k=1}^n k^2- 4\sum_{k=1}^{n} k+\sum_{k=1}^{n} 1$$
$$S=4\frac{n(n+1)(2n+1)}{6}-4 \frac{n(n+1)}{2}+n=\frac{n(4n^2-1)}{3}$$
| {
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"answer_id": 1
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Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function I am trying to derive the Hardy-Ramanujan asymptotic formula
$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$
from Radmacher's formula for the partition function $p(n)$ given by
$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}... | Rough sketch: Take the first term, and you'll get:
$$\begin{align}p\left(n\right)&\sim\frac{1}{\pi\sqrt{2}}A_{1}\left(n\right)\cdot\frac{d}{dx}\left(\frac{\sinh\left(\pi\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right)\\&\sim\frac{1}{\pi\sqrt{2}}\frac{d}{dx}\left(\frac{\sinh\left(\pi\s... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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How does $(abc) = (ac)(ab)$? I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$
However, I'm lost on how $(abc) = (ac)(ab)$ occurs. ... | We can also conveniently calculate the product of transpositions in cycle notation by switching to the two-line notation of permutations.
Considering the first identity $(abc)=(ac)(ab)$ we have the representation
\begin{align*}
(abc)\equiv\begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}\qquad\qquad&
(ac)\equiv\begin{pmatrix}a&... | {
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Triangle related problem
In the figure below if $tan \alpha=\frac{m}{n}$, where $m$ and $n$ are relatively prime, find the value of $m$ and $n$.
My approach is as follow
$\frac{{SinA}}{{14}} = \frac{{SinB}}{{15}} = \frac{{SinC}}{{13}}$
$\Delta APB$
$\angle PBA = B - \alpha $
$\angle APB = 180 - \left( {\alpha + B -... | P is first Brocard point of the triangle.
We have the relation $$\cot \alpha = \cot A + \cot B + \cot C$$
To quickly find these, we partition the triangle as below
so $$ \cot B = \dfrac{5}{12}, \cot C = \dfrac{3}{4} $$
and in any $\triangle$, $$ \cot A\cot B + \cot B\cot C +\cot C\cot A =1$$
Update :
Following @JeanM... | {
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Prove by induction $n!<4n^2+3$ How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?
Here´s my try:
The equation is valid for every $n\leq4$.
n=5: n!=5!=120>103
n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$
Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.
Any ... | Assuming you want to prove that for each $n\geq 5$ one has $n! > 4n^2+3$ just note that it easily holds for $n=5$. Now assume that it holds up to $n \geq n-1 \geq \ldots 5.$ Then
$$(n+1)! = (n+1)n! > (n+1)(4n^2+3).$$
It is only left to prove that $(n+1)(4n^2+3) > 4(n+1)^2+3$ for each $n\geq 5.$ In fact,
$$(n+1)(4n^2+3)... | {
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Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
| By binomial theorem$$5^{6n+6}-2^{2n+2}$$
$$={(3+2)}^{6n+6}-2^{2n+2}$$
$$=3k+2^{6n+6}-2^{2n+2}$$
$$=3k+2^{2n+2}({(3-1)}^{4n+4}-1)=3k+2^{2n+2}(3m)$$
| {
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Three inequalities for number $e$: $\frac{e}{2n+2}I have to prove the following inequalities for number $e$:
$$\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n<\frac{e}{2n+1}<\frac3{2n}$$
Here's what I have tried:
$\textbf{First inequality:}$ $\frac{e}{2n+2}<e-\left(1+\frac1n\right)^n$
I use that $\log{\left(\frac{n+1}n\right... | Update: The book I mentioned below provides more than what you need. Here I give only the things you need to prove the two inequalities.
Step 1: We prove $a_n=\left(1+\frac 1n \right)^{n+\frac 12}$ is decreasing. You can check out this post Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? but the book's method is differ... | {
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"source": "stackexchange",
"question_score": "3",
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Elementary algebraic system with integer and positive rational unknowns The problem is to determine the solutions of the system $x+y+z=3$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=5$, given that $x$ is integer and $y,z$ are positive rationals.
From the hypotheses I have obtained that $y$ and $z$ are rational solutions o... | (There was an old problem which looks similar.)
Solution:
First note that x<3, so if $x>0$ we can easily find $x=1$ is not good; and $x=2$ yields $(y,z)=(\frac 13, \frac 23)$ or $(\frac 23, \frac 13)$.
Now suppose $x<0, w=-x, w \in \mathbb N$. Your discriminant becomes $$(5w+1)(w+3)(5w^2+12w+3)=(5w^2+16w+3)(5w^2+12w+3)... | {
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The 'ratio' of a 2x2 matrix Define the 'ratio' of a 2x2 matrix $$A= \begin{pmatrix} a & b\\ c & d \end{pmatrix} $$
to be $\frac{b}{c}$ when $c\neq 0$. Show that the ratio of $A^n$ is equal to the ratio of $A$, when the ratio of $A^n$ is well-defined.
My instinct is to go with a proof by induction, but I really can't se... | Theorem
There are $x_n$ such that
$$ A_n :=
\begin{pmatrix}
a_n & b x_n \\
c x_n & d_n
\end{pmatrix}
= A^n
$$
Proof by induction
For $n=1$ it is trivial: $x_1 := 1$.
Let the theorem be true for a $n$.
Then on the one hand
$$ A^{n+1} = A^n
\cdot
\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}
=
\begin{pmatrix}
a_n & b x_n ... | {
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I cannot solve these separable differential equations There is no explanation in the book on how to solve these and I can't find any help online. Step-wise calculators also don't make sense.
(1) $y'=(y^2-1)x, \; y(0)=0$,
(2) $xy'=y^2-2y, \; y(1)=1 \; x\geq0$.
I'll post the question and write my attempts underneath stra... | For the first one, note that you can re-write
\begin{align}
y' &= (y+1)(y-1)x \\
\implies \displaystyle \int \frac{dy}{(y+1)(y-1)} &= \int x dx
\end{align}
Whence
\begin{align}
\frac{1}{2}\displaystyle \int \frac{1}{y-1} - \frac{1}{y+1}&= \int x dx
\end{align}
Thus
$$\ln\frac{y-1}{y+1} = x^2+C$$
Then high school algebr... | {
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Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\si... | Using the spherical coordinates (with the convention you used) we have, $$yz= r^2\sin\theta\cos\theta\sin\varphi,$$ $$x^2+y^2+z^2 = r^2$$ $$\,dx\,dy\,dz = r^2\sin\theta\,dr\,d\varphi\,d\theta.$$ Also, the region $E$ is bounded by $x^2+y^2+z^2-x = 0$, which is a sphere centered at $(\tfrac{1}{2},0,0)$ with radius $\tfra... | {
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"source": "stackexchange",
"question_score": "2",
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Finding the determinant of a matrix with $0$s on the diagonal, $1$s in the first row and column, and $x$ elsewhere. How do you attempt to solve the deterimnant of
$$D_n=\begin{vmatrix}
0 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & x & x & \cdots & x \\
1 & x & 0 & x & \cdots & x \\
1 & x & x & 0 & \ddots & \vdots \\
\vdots & \... | We have $$x^2D_n = \begin{vmatrix}
0 & x & x & \cdots & x \\
x & 0 & x & \cdots & x \\
x & x & 0 & \cdots & x \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x & x & x & \cdots & 0
\end{vmatrix} = x^n\begin{vmatrix}
0 & 1 & 1 & \cdots & 1 \\
1 & 0 & 1 & \cdots & 1 \\
1 & 1 & 0 & \cdots & 1 \\
\vdots & \vdots & \vdots... | {
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"source": "stackexchange",
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Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\e... | Your start of the question Find $\sqrt{1+ \sqrt 2}$ is misleading.
You're supposing that $\alpha = \sqrt{1+ \sqrt 2}$ belongs to the field extension $\mathbb Q(\sqrt 2) / \mathbb Q$... And you proved that it's not the case.
In fact $\alpha$ is an element of degree $2$ over $\mathbb Q(\sqrt 2)$ because $\alpha$ is a roo... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Steps for Velocity given vector I am asked to find velocity for the space curve
$$r(t) = \langle \cos(3t), 1-\sin(3t), 3\sin(3t)+\cos(3t)\rangle$$
at $t=\pi / 4$
I have differentiated each component and substituted $t$ in said components. Are those the correct steps to find the answers?
I got this as my answer:
$$\lang... | The steps you stated you did are the correct ones. In particular, differentiating each component leads to
$$\frac{dr(t)}{dt} = \langle -3\sin(3t), -3\cos(3t), 9\cos(3t) - 3\sin(3t) \rangle \tag{1}\label{eq1A}$$
Next, $t = \frac{\pi}{4}$ means $3t = \frac{3\pi}{4}$, with $\sin(3t) = \frac{1}{\sqrt{2}}$ and $\cos(3t) = -... | {
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"source": "stackexchange",
"question_score": "2",
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Finding specific interval The question is to find $[a,b]$ such that $$\forall x,y : x\in [a,\frac {a+b}{2}],y\in[\frac{a+b}{2},b]\Rightarrow x+y,xy \in [a,b]$$
I tried like below
$$a \leq x \leq\frac {a+b}{2}\\
\frac {a+b}{2} \leq y \leq b\\
\to \frac {3a+b}{2} \leq x+y \leq \frac {a+3b}{2}\\\to
a\leq \frac {3a+b}{2}\... | $$a \le \frac{3a+b}2 \iff 0 \le a+b$$
$$\frac{a+3b}{2}\le b \iff a + b \le 0$$
Hence we have $a+b=0$.
$$a=-b$$
$|x| \le b$, $|y| \le b$, we want $|xy| \le b$
$$b^2 \le b$$
$$b^2-b \le 0$$
$$b(b-1) \le 0$$
Hence,
$$0 \le b \le 1, a = -b$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
$\underset{x\to 1}{\text{lim}}\int_0^x \frac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \, \mathrm dt=\frac{ \pi }{\sqrt{2}}$
Define $f(x)=\dfrac{x+1}{(x-1)^2}$. Prove $\lim\limits_{x \to
1}\displaystyle\int_0^x \dfrac{\sqrt{t} f(t)}{\sqrt{f(x)-f(t)}} \,
{\rm d}t=\dfrac{\pi }{\sqrt{2}}$.
We can obtain
$$\lim_{x \to 1}\int_0^... | I found a possible method here which is straightforward. I will just carry it here.
\begin{aligned}
\lim _{x \rightarrow 1} u(x) &=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sqrt{t} \cdot \frac{1+t}{(t-1)^{2}}}{\sqrt{\frac{1+x}{(x-1)^{2}}-\frac{1+t}{(t-1)^{2}}}} d t \\
&=\lim _{x \rightarrow 1} \int_{0}^{x} \frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2} \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+... | Note
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} dx
= \int_{0}^{1} \frac{1+\frac1{x^{2}}}{x^{2}+\frac1{x^2}} dx
=\int_0^1 \frac {d(x-\frac1x)}{(x-\frac1x)^2+2}
= \frac\pi{2\sqrt2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Can such a matrix be singular? Let $A$ be an $n\times n$ matrix that satisfies
i. All diagonal entries of $A$ are positive, even integers,
ii. All non-diagonal entries of $A$ are positive, odd integers,
iii. $A$ is symmetric: $A_{ij}=A_{ji}$.
iv. If $i\neq j$, then $A_{ij}<A_{ii}$ and $A_{ij}<A_{jj}$.
Question: can suc... | I found:
$$
\begin{pmatrix}
15 & 0 \\
0 & 15 \\
11 & 11
\end{pmatrix}
\cdot
\begin{pmatrix}
14 & 1 & 11 \\
1 & 14 & 11 \\
\end{pmatrix}
=
\begin{pmatrix}
210 & 15 & 165 \\
15 & 210 & 165 \\
165 & 165 & 242
\end{pmatrix}
$$
The general idea is to start with a product of rank-$(n-1)$ matrices and to get the confi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3949158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Why do diagonal entries of odd potencies of adjacency matrices of forests sum up to zero? I came across the statement that the trace (sum of diagonal entries) of an odd potency of an adjacency matrix of a forest is always zero. I have tried out a few options and it seems to be correct for forests or trees but not for o... | If $A$ is the adjacency matrix of a graph, then $A^n_{ij}$ counts the number of paths of length $n$ from vertex $i$ to vertex $j$. If you are in a forest, because there are no cycles, every step along a path is either one step closer or one step father away from any particular vertex. Because of this, if we start at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $h(x) = f(x) + g(x) = 0$, $f(x)$ and $g(x)$ having roots that are negatives of each other. Let $f(x) = x^2 +bx+ 9$ and let $g(x) = x^2 +ax+c, a, b, c ∈ R$. The roots of $f(x) = 0$ and
$g(x) = 0$ are negatives of each other. If $h(x) = f(x)+g(x)$, then solve the equation $h(x) = 0$.
I'm not sure how t... | One can certainly use Vieta's formulas, but I'll give a self-contained solution. (My argument essentially reproves them though.)
Suppose $r_1, r_2$ are the two roots of $f$. By the Fundamental Theorem of Algebra (and using the fact that $f$ is monic--i.e., the leading coefficient of $x^2$ is $1$), we may write
$$f(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$ I don’t know how to begin solving. Can I get a hint?
My failed attempt
Let $t =\tan (x/2)$
Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve
| Continuing from your method is messy but possible:
Let $\frac{(3-t^2)}{(3+t^2)^2 } = \frac{A}{(3+t^2)} + \frac{B}{(3+t^2)^2}$. Then $3-t^2 = (3+t^2)A + B$. When $t = 0, 3 = 3A + B$, and when $t = 1, 2=4A+B$, so $A = -1, B= 6$. Thus:
$$I = -2 \int_0^1 \frac{1}{3+t^2} \ dt + 12 \int_0^1 \frac{1}{(3+t^2)^2} \ dt $$
For th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3954751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find the Expectation and Variance of 4 Independent Dice We were given this seatwork:
With four independent dice:
a) the expected value of the sum of the rolls,
b) the expected value of the product of the rolls, and
c) the variance of the sum of the rolls .
I was able to answer a and b, but I don't know how to get the v... | You have found the expected value of a dice roll to be $E(X)=3.5$, thus the variance of one roll is equal to:
\begin{align}
\operatorname{Var}(X) &= \sum_{k=1}^6 \frac{1}{6}\left(k - 3.5\right)^2 \\
&= \frac{35}{12} \approx 2.92.
\end{align}
Since the rolls of the 4 dies are independent the variance of their sum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perf... | Let $K= \mathbb{Q}(p^{\frac{1}{3}}) \cong \mathbb{Q}[x]/(x^3-p).$ We have the following:
*
*$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{1}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{1}{3}}$ is $x^3-p.$
*$\text{Tr}_{K/\mathbb{Q}} (p^{\frac{2}{3}}) = 0 $, as the minimal polynomial of $p^{\frac{2}{3}}$ is $x^3-p^2.$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number,... | Step 1 (in the question): $\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
Step 2: $C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
In step 1, $-\frac{1}{2}, -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Integral of $1 / \sqrt {(a x + b) (p x + q)}$ I am working on this:
$\displaystyle \int \dfrac 1 {\sqrt {(a x + b) (p x + q)} } \ \mathrm d x$
valid where $(a x + b)(p x + q) > 0$.
I am specifically interested in the case where $\dfrac {b p - a q} p < 0$.
I do this:
Let $u = \sqrt {a x + b}$
$\leadsto x = \dfrac {u^2 -... | Amplified comment: "Convert an $\arcsin$ to an $\arctan$."
I learned this long ago because certain computer languages had only $\arctan$ as built-in function, and you had to use formulas if you wanted $\arcsin, \arccos$, etc. (BASIC had ATN arctangent, but none of the other "arc" trig functions)
So here it is:
$$
\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$ for positive values such that $ab+bc+ca=3$ This problem is from my teacher.
Know that $a,b,c>0$, $ab+bc+ca=3$.
Prove that: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+2b+c}\leq\frac{6}{abc}$$
I tried to change the number '$6$' into $2... | By holder: $$\sum_{cyc}\sqrt[3]{\frac{1}{a}+\frac{2}{bc}+a+3b+c}\le \sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1) \cdot (1+1+1)}$$ It hence suffices to prove $$\sqrt[3]{\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})(1+1+1)(1+1+1)}\le \frac{6}{abc}$$ $$\sum_{cyc}({\frac{1}{a}+\frac{2}{bc}+a+3b+c})\le \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.
Is there a shorter/simpler solution... | Yes, there is a shorter solution that relies on the identity $$(x+y)(x^n + y^n) = (x^{n+1} + y^{n+1}) + xy(x^{n-1} + y^{n-1}),$$ which is easily verified by multiplication. Then if we let $f_n = x^n + y^n$, this may be written $$f_{n+1} = (x+y)f_n - xy f_{n-1} = f_1 f_n - xy f_{n-1}.$$
Next, observe that $f_0 = 2$ for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Simplify $\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}$ The double square roots can be removed by $8+2\sqrt7 = (\sqrt7+1)^2$ and $$4-\sqrt7 = \frac{1}{4}(16-4\sqrt7) = \frac{1}{4}(14+2-2\sqrt2\cdot\sqrt14) = \frac{1}{4}(\sqrt14-\sqrt2)^2$$The expression simplifies to $\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sq... | Let $\sqrt{8+2\sqrt{7}}=\sqrt{x}+\sqrt{y} \implies x+y+2\sqrt{xy}, \implies x+y=8, xy=7 \implies x=1,y=7$ and let $\sqrt{4-\sqrt{7}}=\sqrt{p}-\sqrt{q} \implies p+q=4, pq=7/4 \implies p=7/2, q=1/2.$
Then
$$F=\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}= \frac{\sqrt{7/2}-\sqrt{1/2}}{1+\sqrt{7}-\sqrt{2}}$$
Someth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Complete factorisation of $x^8-2x^4\cos (4\theta)+1$ with roots of unity
Factorize completely $x^8-2x^4\cos (4\theta)+1$ using complex numbers and $n$-th root of unity.
The answer given is
$$\prod_{r=0}^3 \left(x^2-2x\cos\left(\theta+\frac{r\pi}{2}\right)+1\right)\,.$$
Can anyone please help . I just want to know how... | With the 4th root of unity $e^{\pm i\frac{a+2\pi r}4} $ for $x^4 - e^{\pm ia}=0$, factorize as follows
\begin{align}
x^8-2x^4\cos (4\theta)+1
=&\> (x^4-e^{i 4\theta})(x^4-e^{-i 4\theta}) \\
= &\prod_{r=0}^{3}(x-e^{i\frac{4\theta+2\pi r}4})
\prod_{r=0}^{3}(x-e^{-i\frac{4\theta+2\pi r}4}) \\
= &\prod_{r=0}^{3}(x-e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Complex and real matrix for eigenvalues This is a short one but: Consider the real matrix $\alpha = \begin{pmatrix}
7 &3 &-4 \\
-2&-1 &2 \\
6&2 &-3
\end{pmatrix}$ and let $\beta \in M_n (\mathbb{C})$ be the same matrix but considered as a complex matrix.
Problem
with use of a computer program calculate the eige... | The characteristic polynomial of your matrix is
$$-x^3+3x^2-x+3 = -(x-3)(x^2+1)$$
The real root of this polynomial is $3$ which is an eigenvalue for $\alpha$ and $\beta$:
$$\begin{pmatrix}
7 &3 &-4 \\
-2&-1 &2 \\
6&2 &-3
\end{pmatrix}
\begin{pmatrix}
1 \\ 0 \\ 1
\end{pmatrix} = 3\begin{pmatrix}
1 \\ 0 \\ 1
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many variables in the multinomial expansion have an exponent that's different from 2? Multinominal: $(a + b + c + d)^{10}$
Question: How many variables in the multinomial expansion have an exponent that's different from 2?
I want to solve it using generating functions.
So we can write: $t_1 + t_2 + t_3 + t_4 = 10$,... | $(1 + x + x^3 + x^4 + ...)$ can be simplified as $[(1 - x)^{-1} - x^2]$.
Now, we had to find coefficient of $x^{10}$ in ${[(1 - x)^{-1} - x^2]}^4$
${[(1 - x)^{-1} - x^2]}^4$ = $\binom{4}{0}$ $[({1 - x}) ^ {-4}]$ - $\binom{4}{1}$ $[({1 - x}) ^ {-3}]$ $x^2$ + $\binom{4}{2}$ $[({1 - x}) ^ {-2}]$ $x^4$ - $\binom{4}{3}$ $[(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given positive numbers $x_1,...,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$
Given positive numbers $x_1,\dots,x_n$ such that $x_1\cdots x_n=1$ prove that $\frac{1}{n-1+x_1}+\cdots+\frac{1}{n-1+x_n}\le 1$.
I tried solving this question through substitution, e.g. $a_1=\... | EDIT: As pointed out, this answer contains a flawed application of Cauchy-Schwarz. In particular, the L.H.S. of the inequality can be zero, while the R.H.S. is positive, an absurdity. I tried to remedy this flaw but ultimately failed. Hence, I have come up with a new approach that proceeds by a "smoothing principle" ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Evaluate:$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$ Evaluate:$$\sum_{n=2}^{\infty}\frac{\tan \theta_{n}}{3^n\left(3-\tan^2\theta_{n}\right)}$$
where $$\theta_{n}=\frac{\theta}{3^n}$$ and $0<\theta<\pi$
I did try to find relation between $\tan 3x$ and $\tan x$
$$\tan 3x-3\tan x=\frac... | Note that
$$ \frac{\tan x}{3-\tan^2 x} = \frac{\cot x - 3\cot(3x)}{8}. $$
So we have
$$ \frac{\tan \theta_n}{3^n(3-\tan^2 \theta_n)}
= \frac{1}{8}\left( \frac{\cot \theta_n}{3^n} - \frac{\cot \theta_{n-1}}{3^{n-1}} \right). $$
Summing this from $n = 2$ to $\infty$ yields a telescoping series with the value
$$ \sum_{n=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\fra... | $$\frac{n}{n+1}-\frac{n+1}{n}=\frac{n2-(n^2+2n+1)}{n(n+1)}=-\frac{n+(n+1)}{n(n+1)}=-\bigg(\frac{1}{n+1}+\frac{1}{n}\bigg)$$
$$\therefore lim_{n\rightarrow\infty} \frac{n}{n+1}-\frac{n+1}{n}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
Let $a = \frac{1 + \sqrt{2009}}{2}$ . Find the value of $(a^3 - 503a - 500)^5$ .
What I Tried: We have :-
$$ (a^3 - 503a - 500)^5 = [a(a^2 - 3) - 500(a - 1)]^5$$
$$= \Bigg(\Bigg[\frac{1 + \sqrt{2009}}{2}\Bigg]\Bigg[\frac{1009 + \sqrt{200... | $a$ is a root of a quadratic equation with roots $$\frac{1 \pm \sqrt{2009}}{2}$$
That is, $a$ satisfies the following equation: $$x^2 - x - 502 = 0 \tag 1$$ Using this, we observe
$$\begin{align}(a^3 - 503a - 500)^5 &= (a(\color{red}{a^2})-503a-500)^5 \\&\overset 1= (a(\color{red}{a+502})-503a-500)^5 \\&= (\color{blue}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\co... | The system
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
is a linear system of two (independent) equations in two variables, and it is readly checked that $x=\sin\theta, y=\cos\theta$ is a solution. Therefore it is the unique solution.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Express roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta, $
If roots of the equation $ax^2+bx+c=0$ are $\alpha, \beta, $ find roots of equation $acx^2-b(c+a)x+(c+a)^2=0$ in terms of $\alpha, \beta$
Here's what I have tried so far,
I know that $\alpha+ \beta=\frac{-b}{a} $ and $\alpha \beta=\fra... | Solve the two equations to find
$$ay^2+by+c=0\implies y=\dfrac{-b\pm\sqrt{b^2-4ca}}{2a}\ \ \ \ (1)$$
$$cax^2-b(c+a)x+(c+a)^2=0\implies x=(c+a)\cdot\left(\dfrac{b\pm\sqrt{b^2-4ca}}{2ca}\right)\
\ \ \ (2)$$
Considering the opposite signs of the roots,
$$\dfrac yx=-\dfrac c{c+a}$$
Considering the same signs of the root... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\fra... | Let $x=\sin{u}$. Substitute $\arcsin{x}=u$ to change variables, notice that $d(\cos^3{u})=-3\sin{u}\cos^2{u}\,du$ and get the integral by parts
$$
\int x\sqrt{1-x^2}\arcsin{x}\,dx =
\int{u}\,(\sin{u}\cos^2{u}\,du) =
-\frac{1}{3}\int{u}\,d(\cos^3{u}) =
\frac{1}{3}\Bigl(\int\cos^3{u}\,du - u\cos^3{u} \Bigr)
$$
After b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
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Difficulties solving this integral: $ \int_0^1 \frac{\ln(x+1)} {x^2 + 1} \, \mathrm{d}x $ by differentiation under the integral sign So in the book Advanced Calculus Explored, by Hamza E. Asamraee. The next integral appears as an exercise to solve by differentiating under the integral sign:
$$ \int_0^1 \frac{\ln(x+1)} ... | This would work:
$$I(a) = \int_0^1 \frac{\ln(ax+1)}{x^2+1} dx \\
I’(a) =\int_0^1 \frac{x}{(x^2+1)(ax+1)} dx \\ \overset{\text{partial fractions}}= \\ \frac{-2\ln |ax+1| +\ln(x^2+1)+2a\tan^{-1} x}{2(a^2+1)} \bigg |_0^1 \\ =-\frac{\ln(a+1)}{a^2+1}+\frac{\ln 2}{2a^2+2}+\frac{\pi}{4} \frac{a}{a^2+1}
$$
Integrating from $0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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The floored quadratic equation $a \lfloor x^2 \rfloor + b \lfloor x \rfloor + c = 0$ What are the roots of $f(x) = a\lfloor x^2 \rfloor +b\lfloor x \rfloor + c$?
(where $a \neq 0$ and $a,b,c \in \mathbb{R}, f: \mathbb{R} \rightarrow \mathbb{R}$ as usual)
My failed approach:
Let $\delta = x - \lfloor x \rfloor$ and get
... | Fix $a,b,c$ reals with $a\neq 0$, let us look at some necessary conditions that $$(\dagger)\quad\quad a \lfloor x^2\rfloor + b \lfloor x\rfloor +c =0$$ will have a root:
(1) It is necessary that $a,b,c$ satisfies an integral relation $$aN + b M +c =0$$ for some integers $N,M$. Otherwise we will never have a root. For i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The common difference is equal to the common ratio. Four numbers are in A.P. The first, the second and the fourth are in G.P. Find the numbers if the common difference is equal to the common ratio.
Let the terms of the A.P. be $a_1,a_1+d,a_1+2d,a_1+3d$ and the terms of the G.P. will be $a_1,a_1+d,a_1+3d.$ We know that ... | Let the four numbers be $a_1, a_2, a_3, a_4$. The common difference $d$ will make the terms equal to $a_1, a_1+d, a_1+2d, a_1+3d$, and the terms of the geometric sequence should be $a_1, a_1+d, a_1+3d$ as you correctly mentioned.
Since $d=r$, $a_1, a_1r, a_1r^2$ are also the terms of the geometric sequence.
So we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Is there any way of calculating this long formula? $$\left\{\frac{-x^4+y^4-y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\sqrt{x^2+y^2}\sqrt{x^2+y^2-1}}\right\}\left\{1+\left\{\frac{y^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)}\right\}^2\right\}+\left\{\frac{-y^4+x^4-x^2}{\left(x^2+y^2\right)\left(x^2+y^2-1\right)\... | Your question is equivalent to asking whether the last term (without the minus sign) and the sum of the first two terms are equal. Expressing these two in the format of one big fraction, the denominators on both sides of the equation are the same: $(x^2+y^2)^{7/2} (x^2+y^2-1)^{7/2}$, therefore comparing the numerators ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\int_0^n f(x)\, dx - [\frac{1}{2} f(0) + f(1) +\dots + f(n-1) + \frac{1}{2} f(n)]$ If $f(x)$ is increasing, strictly concave, and twice continuously differentiable for $x > 0$, does the difference
$$
d_n = \int_0^n f(x)\, dx - [\tfrac{1}{2} f(0) + f(1) + f(2) +\dots + f(n-1) + \tfrac{1}{2} f(n)]
$$
conv... | Consider
$$
f(x)=-\frac{1}{(x+1)(x+2)}.
$$
Clearly $f$ is increasing and concave in $[0,\infty)$.
Then
$$
\frac{1}{2}f(0)+f(1)+\cdots+f(n-1)+\frac{1}{2}f(n)=\cdots=-\frac{3}{4}+\frac{1}{n+1}-\frac{1}{2(n+1)(n+2)}\\=-\frac{3}{4}+\frac{1}{n}+{\mathcal O}(n^{-2})
$$
while
$$
\int_0^n f(x)\,dx=\int_0^n \frac{dx}{x+2}-\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrate $\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$.
Integrate the following integral:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$$
I did this question a while ago and the answer is correct. However, when I ... | I think you are asking to verify that
$$\frac{\tan \left(x\right)}{\sin ^2\left(x\right)}.$$ RIGHT?
if so, $$\frac{\sin(x)/\cos(x)}{\sin ^2\left(x\right)}=\frac{1}{\sin(x)\cos(x)}=\frac{2}{2 \sin(x)\cos(x)}=\frac{2}{\sin(2x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove or disprove $(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}$ The following question I read in a book, but the book does not give proof. I doubt the correctness of the result
let $p>3$ be prime number. prove or disprove
$$(x+1)^{2p^2}\equiv x^{2p^2}+\binom{2p^2}{p^2}x^{p^2}+1\pmod {p^2}\tag{1}$$... | Let $p = 5$. For any integer $k$, we have
$$\begin{equation}\begin{aligned}
& ((5k + 1)(5k + 2))((5k + 3)(5k + 4)) \\
& \equiv (25k^2 + 15k + 2)(25k^2 + 35k + 12) \\
& \equiv (15k + 2)(10k + 12) \\
& \equiv 150k^2 + 200k + 24 \\
& \equiv -1 \pmod{25}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Using this for each ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$ Use the one to one property of logs to solve $\ln(x^2-10)+\ln(9)=\ln(10)$.
I get $x=\sqrt{11}$ or $x=10$ whereas my textbook says it's $x=\pm\frac{10}{3}$.
My working - initial attempt:
$$x^2-10+9=10$$
$$x^2-1=10$$
$$x^2=11$$
$$x=\sqrt{11}$$
My w... | Note that\begin{align*}\ln(x^2-10)+\ln(9)=\ln(10)&\iff\ln\bigl(9(x^2-10)\bigr)=\ln(10)\\&\iff9x^2-90=10\\&\iff x^2=\frac{100}9\\&\iff x=\pm\frac{10}3.\end{align*}
The error from your first attempt lies in assuming that$$\ln(x^2-10)+\ln(9)=\ln(10)\iff x^2-10+9=10,$$whereas the error in your second attempt lies in assumi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain. $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$
I know I have to switch the $f(x)$ and the $y$:
$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\fra... | a)
The function $f$ is not injective, therefore not bijective and cannot have an inverse therefore.
But you can calculate the individual branches (i. e. partial inverses) of $f$.
b)
You didn't rearrange your equation correctly.
Let $y$ denote $f^{-1}(x)$.
$$\sin^2\left(\frac{2y+1}{3}\right)=x$$
$$\left(\sin\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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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.