Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common... | Alternatively:
$$\frac{d}{dx}\left(x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}\right)=\frac{d}{dx}\left((6x^2-x^3)^{\frac13}\right)=\frac{1}{3}(6x^2-x^3)^{-\frac{2}{3}}(12x-3x^2)=\\
\frac{4x-x^2}{(6x^2-x^3)^{\frac{2}{3}}}=\frac{4x-x^2}{x^{\frac{4}{3}}(6-x)^{\frac23}}=\frac{4-x}{x^{\frac{1}{3}}(6-x)^{\frac23}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that the given linear system has this specific solution if $\det \neq 0$
Show that the linear system $\begin{pmatrix} a_{11} & a_{12}\\
a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x_1\\ x_2
\end{pmatrix}= \begin{pmatrix} b_1\\ b_2 \end{pmatrix} \,\,\,\,\,\,
\text{ with }\,\,\,\, A= \begin{pmatrix} a_{11} &... | You are almost there; simply write out what the two determinants
$$\det \begin{pmatrix} b_1 & a_{12}\\ b_2 & a_{22}
\end{pmatrix}\qquad\text{ and }\qquad\det
\begin{pmatrix} a_{11} & b_1\\ a_{21} & b_2 \end{pmatrix}$$
are. You will find that these are precisely the coefficients of the vector
$$\begin{bmatrix}
a_{22}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of $\dfrac{a+b+c}{b-a}$
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
Find the minimum value of $\dfrac{a+b+c}{b-a}$
Attempt:
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the express... | Alt. hint: $\;f(-2) = 4a - 2 b + c \ge 0 \iff 2a + c \ge 2(b-a)\,$, then:
$$
\dfrac{a+b+c}{b-a} = 1 + \frac{2a+c}{b-a} \;\ge\; 1 + 2 = 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20... | Using the identity from Pascal's Triangle, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{k+a}{b}
&=\sum_{k=0}^n(-1)^k\left[\binom{k+a+1}{b}-\binom{k+a}{b-1}\right]\\
&=\sum_{k=1}^{n+1}(-1)^{k-1}\left[\binom{k+a}{b}-\binom{k+a-1}{b-1}\right]\\
&=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=1}^{n+1}(-1)^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
How to find all rational solutions of $\ x^2 + 3y^2 = 7 $? I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at ... | There are only six solutions which are also algebraic integers, and they can each be found from whichever one you discover first.
For instance, $$(2 + \sqrt{-3}) \left( \frac{-1 + \sqrt{-3}}{2} \right) = \frac{-5 + \sqrt{-3}}{2}$$ and $$\frac{-5 + \sqrt{-3}}{2} \left( \frac{-1 + \sqrt{-3}}{2} \right) = \frac{1 - 3 \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
} |
A unit square has its corner chopped off to form a regular polygon with eight sides. What is the area of the polygon?
A unit square has its corner chopped off to form a regular polygon with eight sides. What is the area of the polygon? Source: ISI BMATH UGA 2017 paper
A regular polygon with 8 sides can be divided int... | An image makes everything simpler to understand:
Because the square side length is 1,
$$b + a + b = 1$$
Solving for $b$ we get
$$b = \frac{1 - a}{2} \label{NA1}\tag{1}$$
Squaring this (noting that we are limited to positive $a$ and $b$, i.e. that $a \gt 0$ and $b \gt 0$) we get
$$b^2 = \frac{1 - 2 a + a^2}{4} \label{N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2775575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a_1 = 3$ and $a_{n+1}=a_n^2-2$, then $\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1 a_2 a_3}+\cdots = \frac{3-\sqrt{5}}{2}$ Let $a_1=3$，and $a_{n+1}=a_n^2-2$ (for $n=1, 2, \cdots$). Show that
$$\frac{1}{a_1}+\frac{1}{a_1 a_2}+\frac{1}{a_1a_2a_3}+\cdots+\frac{1}{a_1a_2 \cdots a_n}+\cdots=\frac{3-\sqrt{5}}{2}.$$
My tho... | Notice that $a_1=3>2$. Suppose that $a_k>2$, then $a_{k+1}=a_k^2-2>2^2-2=2$. By mathematical induction, we have $a_n>2,$ further, $a_n^2>a_n+2$. Hence, $a_{n+1}=a_n^2-2>(a_n+2)-2=a_n.$ This shows that $a_n$ is an increasing integer sequence. Therefore, $a_n \to +\infty$ as $n \to \infty$.
Thus, we have $$\lim_{n \to \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2776422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $ a^n + b^n + c^n = d^n + e^n + f^n $ by induction If $a,b,c,d,e,f$ are six real numbers such that:
$$ a + b + c = d + e + f $$
$$ a^2 + b^2 + c^2 = d^2 + e^2 + f^2 $$
$$ a^3 + b^3 + c^3 = d^3 + e^3 + f^3 $$
Prove by mathematical induction that:
$$ a^n + b^n + c^n = d^n + e^n + f^n $$
I tried solving this q... | Consider the polynomial $p(x)=x^3-sx^2+ux-v$ where $s=a+b+c=d+e+f$, $u=ab+bc+ca=\frac 12\left((a+b+c)^2-a^2+b^2+c^2\right)=de+ef+fd$ and $v=abc=\frac 13\left((a^3+b^3+c^3)-s(a^2+b^2+c^2)+u(a+b+c)\right)=def$
Then $p(a)=p(b)=p(c)=p(d)=p(e)=p(f)=0$ and you can use $$a^rp(a)+b^rp(b)+c^rp(c)=0$$ to obtain an expression for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)dx?$ How may one demonstrate this integral is equal to this result?
$$\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)\mathrm dx=\pi\zeta(2)-2\pi[1-\ln^2(2)]$$
Where do I start from?
| This answer will primarily show how Waiting most likely got his answer. To start off, let's first define $I$ to be the integral under question, and using the basic logarithmic rules we all learned in first grade, $I$ can be transformed as
$$\begin{align*}I & =\int\limits_0^1dx\,\arccos\sqrt{1-x}\log^2(1-x)-\int\limits_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the Limit of $\prod_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n+x})^{n+x}}$ Consider the following productions
$$\prod_{n=1}^{\infty}\frac{1+\frac{1}{n}}{1+\frac{1}{n+x}}$$
and
$$\prod_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n+x})^{n+x}}$$
I know that the above are ... | For the first one, rewrite the product:
\begin{equation*}
\begin{split}
\prod_{n=1}^{\infty} \frac{(n+1)(n+x)}{n(n+x+1)}& = \frac{(1+1)(1+x)}{1(1+x+1)} \times\frac{(2+1)(2+x)}{2(2+x+1)} \times \frac{(3+1)(3+x)}{3(3+x+1)}\times ...\\
& = \frac{2(1+x)}{1(2+x)} \times\frac{3(2+x)}{2(3+x)} \times \frac{4(3+x)}{3(4+x)}\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Intersection between $y=\frac{x^2+1}{x+1}$ and $y=x+2$ on $y$-axis I stumbled upon this problem: Considering the function $f:(-1;\infty)\rightarrow R$, $$f(x)=\frac{x^2+1}{x+1}$$ Find the sum of the values on positive $y$-axis where the tangent of the function is perpendicular to the line $y=x+2$
After letting $(0,a... | $f(x)=\frac{x^2+1}{x+1}$
$\implies f'(x) = \frac{2x(x+1)-x^2-1}{(x+1)^2}$
$\implies f'(x) = \frac{2x^2+2x-x^2-1}{(x+1)^2}$
$\implies f'(x) = \frac{x^2-1+2x}{(x+1)^2}$
set this to $-1$ as tangent of $f(x)$ is perpendicular to $y=x+2$
$\frac{x^2-1+2x}{(x+1)^2}=-1$
$x^2-1+2x=-(x^2+1+2x)$
$2x^2+4x =0$
$2x(x+2)=0$
$\therefo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Take $\epsilon>0$, I want to find $\delta>0$ such that:
$$\lVert (x-1,y-2)\rVert <\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert... | $\lim_{(x,y) \to (1,2)}\dfrac{3x-4y}{x+y}
=-\dfrac{5}{3}
$
I like to let
variables go to zero,
so let
$x = 1+u, y=2+v$.
Then
$\begin{array}\\
\dfrac{3x-4y}{x+y}+\dfrac{5}{3}
&=\dfrac{3(1+u)-4(2+v)}{1+u+2+v}+\dfrac{5}{3}\\
&=\dfrac{3u-4v-5}{3+u+v}+\dfrac{5}{3}
\qquad\text{You can see here that the limit will be -5/3}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown of... | No, you cannot conclude that $a^2$ is a multiple of $11$. You can instead rewrite
$$
a^2=10b^2-2ab=2(5b^2-ab)
$$
so $2\mid a$. Write $a=2c$, with $c$ integer. Then
$$
4c^2=2(5b^2-2bc)
$$
or $5b^2=2(c^2+bc)$. Since $2\nmid 5$, we conclude $2\mid b$.
This is a contradiction to $a$ and $b$ being coprime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=... | Integrate by parts instead as follows
\begin{align}\int\sqrt{x^2-x}\ dx
= &\int\frac{\sqrt{x^2-x}}{2(x-\frac12)}d[(x-\frac12)^2]\\
= &\ \frac12(x-\frac12) \sqrt{x^2-x}-\frac18\int \frac 1{ \sqrt{x^2-x}}dx \\
=& \ \frac12(x-\frac12) \sqrt{x^2-x}
-\frac18\tanh^{-1}\frac{\sqrt{x^2-x}}{x-\frac12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Matrix Exponentiation $A^{15}$ A is a 2x2 matrix. Let it satisfy $A^2 = A - I$, where $I = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$. I want to find $A^{15}$:
My Approach:
I isolated $A$:
$A^2-A=-I$
$A(A-I)=-I$
From that, I just tried to solve the system that would generate:
$\begin{bmatrix}a & b \\ c & d\end{bmatrix... | We can show immediately that $A^3=-I$, so $A^{15}=(A^3)^5=(-I)^5=-I$, for instance:
$$
\begin{aligned}
A^3+I
&=(A+I)\underbrace{(A^2-A+I)}_{=0}\\
&=0\ .
\\&\qquad\qquad\text{Alternatively:}
\\
A^{15}+I
&=(A^3+I)(A^{12}-A^9+A^6-A^3+I)\\
&=(A+I)\underbrace{(A^2-A+I)}_{=0}(A^{12}-A^9+A^6-A^3+I)\\
&=0\ .
\end{aligned}
$$
O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatri... | The matrix is diagonalizable, since it is a $4\times4$ matrix with $4$ distinct eigenvalues. Therefore, the correct option is a).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Generating Function for $(4,9,16,25,36,.....)$ I have a sequence $(4,9,16,25,36,...)$ it is being generated by $a_n=(n+1)^2)$
I have found the generating function for $n^2$ here:
Proving the generating function of $n^2$.
I know I can shift a sequence:
$(1,4,9,16,25,....)$ to the right $(0,1,4,9,25,...)$ via $x^1*A(x)$... | We consider a power series $A(x)$ with constant term $a_0=0$.
\begin{align*}
A(x)=\sum_{j=j_0}^\infty a_jx^j
\end{align*}
A power series $B(x)=\sum_{j=j_0}^\infty a_{j+1}x^j$ can be derived via
\begin{align*}
A(x)&=\sum_{j=j_0}^\infty a_j x^j=\sum_{j=j_0-1}^\infty a_{j+1} x^{j+1}\\
&=x\sum_{j=j_0}^\infty a_{j+1}x^j+a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove using calculus or otherwise the following inequality Prove that$$ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2$$where $$x\in\left(0,\frac{\pi}{2}\right]$$
My Attempt:$$ f'(x)=\frac{ x\sin x(x-\sin x)-(1-\cos x)(\sin x-x\cos x)}{(x-\sin x)^2}=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin... | I did by using brute force
$$f'(x)=\frac{\sin x\left(x-\frac{1-\cos x}{2\sin x}(1+\sqrt{5+4\cos x})\right)\left(x-\frac{1-\cos x}{2\sin x}(1-\sqrt{5+4\cos x})\right)}{(x-\sin x)^2}=\frac{\sin x\left(x-\tan \frac{x}{2}(1+\sqrt{5+4\cos x})\right)\left(x-\tan \frac{x}{2}(1-\sqrt{5+4\cos x})\right)}{(x-\sin x)^2}$$
Let $$g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to ... | Put $a = \sin x, b = \cos x \implies a^2+b^2=1\implies \dfrac{a^2}{b}+\dfrac{b^2}{a} > \dfrac{a^2}{1} + \dfrac{b^2}{1} = a^2+b^2 = 1$ . In fact this is a different inequality, but still looks nice though. To return to the question above, observe that $\dfrac{a}{b}+\dfrac{b}{a} \ge 2$ by AM-GM inequality. Thus $(a^2+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Proving there exists $k$ such that $p(n)p(n+1)=p(k)$ Exact Question
Let $p(x)$ be monic quadratic polynomial over $\mathbb{Z}$. Show that for any integer $n$, there exists an integer $k$ such that $p(n)p(n+1)=p(k)$
Expanding the polynomial just creates a jumbled mess. Is there any intuitive way for this? Any hints ap... | Let $p(x)=x^2+bx+c$ due to monicity (if this is even a word). Then $$\begin{align}p(n)p(n+1)&=(n^2+bn+c)((n+1)^2+b(n+1)+c)\\&=(n^2+bn+c)(n^2+(2+b)n+1+b+c)\\&=(n^2+bn+c)^2+(n^2+bn+c)(2n+1+b)\\&=(n^4+2bn^3+(b^2+2c)n^2+2bcn+c^2)+(n^3+(1+3b)n^2+(b^2+b+c)n+c+bc)\\&=n^4+(1+2b)n^3+(1+b^2+3b+2c)n^2+(b^2+b+2bc+c)n+(bc+c+c^2)\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^... | Let $f(x) = x^{\sqrt{2}}$. Let's accept for the moment(actually, Yves' answer justifies this statement) that
$$f'(x) \sim \frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{2\sqrt{2}}$$
, reminiscent of the definition of the derivative.
Now, it then follows that
$$\lim_{x \to \infty} \frac{(x+\sqrt{2})^{\sqrt{2}}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving recurrence relation of this form by iteration for closed form The recurrence relation is of the form:
$$ G(n) = 5G(n-1) +\frac{4^n}{4^2},\quad G(1)=3$$
My answer always differs from that on wolfram Alpha where is my mistake??
My steps are
$ G(n) = 5^2G(n-2) +\frac{5\cdot 4^n}{4^3}+\frac{4^n}{4^2} $
$G(n)= 5^3G(... | You have extra $4$ here: $+4^n(\frac{5^n}{5^2\cdot 4^n\cdot \color{red}{4}}+$.
Preserving the degrees of $4$:
$$\begin{align}G(n) = &5G(n-1) +\frac{4^n}{4^2}=5\left(5G(n-2) +\frac{4^{n-1}}{4^2}\right)+\frac{4^n}{4^2}=\\&5^2G(n-2) +\frac{5\cdot 4^{n-1}}{4^2}+\frac{4^n}{4^2}=5^2\left(5G(n-3) +\frac{4^{n-2}}{4^2}\right)+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.
I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?
| $$
3^{log_4x+\frac{1}{2}}+3^{log_4x-\frac{1}{2}}=\sqrt{x}\\
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 3^{log_4x}=\sqrt{x}\\
$$
Write $3 = 4^{\log_4 3}$:
$$
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 4^{\log_4 (3) \cdot log_4x}=\sqrt{x}\\
$$
Take log to base 4:
$$
\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) +\log_4 (3) \cdot \log_4x=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Triple Integrals in Spherical Coordinates, problem with boundaries Calculate
$$\iiint \frac{1}{{x^2+y^2+z^2}}dA$$
Where
$x^2+y^2+(z-2)^2\le1$
I've used spherical coordinates, like this:
$x=\rho\sin\phi\cos\theta$; $y=\rho\sin\phi\sin\theta$; $z=\rho\cos\phi$ and $J=\rho^2\sin\phi$
but then I am having a rough time wit... | By letting $z=2+w$ the problem boils down to computing
$$\iiint_{x^2+y^2+w^2\leq 1}\frac{d\mu}{x^2+y^2+w^2+4w+4} $$
or, by setting $w=\rho\cos\theta,y=\rho\sin\theta\sin\varphi,x=\rho\sin\theta\cos\varphi$,
$$ 2\pi\int_{0}^{1}\int_{0}^{\pi}\frac{\rho^2\sin\theta}{\rho^2+4\rho\cos\theta+4}\,d\theta\,d\rho.$$
Let us focu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why does the circle intersect the line?
Is there an easy geometric way to prove that the circle is tangent to the line $\overleftrightarrow {CD}$, where $C=(3,-6)$, $D=(6,-2)$, and $B=(6,0)$? I can do this by using calculus but I think there has to be a nicer/shorter solution.
Thanks in advance.
Edit: I found that $CE... | Yes, you set the equation for the circle equal to the equation of the line and solve for x. If there is only one solution, the line is tangent and if two solutions the line cuts through the circle.
$\sqrt{6^2 - x^2} = \frac{4}{3}x - 10$
$36 - x^2 = \frac{16}{9}x^2 - \frac{80}{3}x + 100$
$\frac{25}{9}x^2 - \frac{80}{3}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 11,
"answer_id": 5
} |
Mistake in solving this simple inequality? Suppose have the the inequality, where $x,y$ are variables,
$$
\frac{x+y}{2}\leq \frac{\alpha}{\beta+\gamma}(x+y)
$$
Clearly, this is true if $\frac12 <\frac{\alpha}{\beta+\gamma}$, but if I try to solve it by adding/subtracting I get
$$
x(1-\frac{2\alpha}{\beta+\gamma})\leq ... | Method 2:
$x(1-\frac{2\alpha}{\beta+\gamma})\leq y(\frac{2\alpha}{\beta+\gamma} -1) = -y(1-\frac{2\alpha}{\beta+\gamma})$
Case 1: $1 = \frac {2\alpha}{\beta + \gamma}$
Then $0 = 0$ and nothing can be determined.
Case 2: $1 > \frac {2\alpha}{\beta + \gamma}$
Then $x\le -y$
Case 3: $1 < \frac {2\alpha}{\beta + \gamma}$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
System of differential equations verification I have a two questions about these types of equations:
*
*Is on checking if I had solve one case correctly
*A simpler way to solve/represent the solutions of it by using Jordan forms I think
So I want to solve this system
$x' = -23x -8y$
$y' = 60x +21y$
I first find e... | $$y=c_1e^{\lambda _1t }v_1 +c_2 e^{\lambda _2t }v_2=$$
$$ c_1e^{t } \begin{pmatrix}
1 \\
-3
\end{pmatrix} +c_2 e^{2t } \begin{pmatrix}
2 \\
-5 \end{pmatrix} $$
You can find $c_1$ and $c_2$ using a given initial value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of $a+2b+3c$ where $a,b,c$ are roots of a cubic equation. Let $a$, $b$, and $c$ be positive real numbers with $a<b<c$ such that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=216$. Find $a+2b+3c$.
I solved it like this:
$$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$$$$ab+bc+ca=47$$
$$a^3+b^3+c^3-3abc=(a+b+c)(... | Make the change: $a=x+4, b=y+4,c=z+4$. Then the equations will be:
$$\begin{cases} x+y+z=0 \\
(x+4)^2+(y+4)^2+(z+4)^2=50 \\
(x+4)^3+(y+4)^3+(z+4)^3=216 \end{cases} \Rightarrow
\begin{cases} x+y+z=0 \\
x^2+y^2+z^2=2 \\
x^3+y^3+z^3=0 \end{cases}$$
Express $x+y=-z$. Square it and subtract the second. Cube it and subtract... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Given $f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$, show that it is differentiable at $(0,0)$. I was given the function:
$f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$
I need to show it is differentiable at $(0,0)$.
I started using the method of differentials and infinitesimal functions:
$\Delta f=f(0+\Delta x,0+\Delta y)-f(0... | Another approach using the definition of differentiability:
To use the horrible limit
$$\lim_{(x,y)\to(0,0)}{\dfrac{f(x,y)-[f(0,0)+f'_x(0,0)(x-0)+f'_y(0,0)(y-0)]}{\sqrt{{(x-0)}^2+{(y-0)}^2}}}\tag 1$$
we need first to calculate the partial derivatives at $(0,0)$ using the fact that $f$ is $C^1(\text{Dom}(f))$. Let's beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $\lambda_{n} = \int_{0}^{1} \frac{dt}{(1+t)^n}$ , then value of $\lim_{n \to \infty} \lambda_{n}^{1/n}$?
Let $$\lambda_n = \int_0^1\frac{dt}{(1+t)^n} \quad \forall n \in \Bbb{N}.$$
To evaluate the integral,
Let $1+t = u$ then $dt =du$, when $t =0$ we get $u=1$ and when $t = 1$ we get $u = 2$; so that $\lambda_{n}... | $$\lim_{n \to \infty} \frac{1}{n} \log \frac{1-2^{-(n-1)}}{n-1} = \lim_{n \to \infty} \frac{1}{n} \log(1-2^{-(n-1)}) - \lim_{n \to \infty} \frac{1}{n} \log(n-1) = 0,$$
so $\lambda_n^{1/n} \to 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding The Zeros Of $\frac{z^2\sin z}{\cos z -1}$ $$f(z)=\frac{z^2\sin z}{\cos z -1}$$
$$\frac{z^2\sin z}{\cos z -1}=0\iff z^2\sin z=0$$ so $z=\pi k$ and $z=0$ are zeros, to find the order we must derive $\frac{z^2\sin z}{\cos z -1}$?
| Note that the denominator $\cos z - 1$ has zeroes at $z = 2m\pi$, so not all zeroes of $\sin z$ are zeroes of $f(z)$. To show this, apply L'Hopital's rule
$$ \lim_{z\to 2m\pi \ne 0} \frac{z^2\sin z}{\cos z - 1} = \lim_{z\to 2m\pi \ne 0}\frac{2z\sin z + z^2\cos z}{-\sin z} = \text{DNE} $$
Hence, the zeroes of $\sin z$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof for $\sum_{k=2}^\infty \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$ How can this identity be derived?
I have been searching the internet but I have no clue where to find a proof for this identity. Any help is highly appreciated.
$$\sum_{k=2}^\infty \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$$
| Using the residue theorem, we have
$$\begin{align}
\oint_{|z|=N+1/2}\frac{\cot(\pi z)}{(z^2+1)}\,dz&=2\pi i \sum\text{Res}\left(\frac{\cot(\pi z)}{z^2+1}\right)\\\\
&=2\pi i \left(\frac{2\cot(\pi i)}{2i}+\sum_{n=-N}^N \frac{1}{\pi(n^2+1)}\right)\tag1
\end{align}$$
As $N\to \infty$, the integral on the left-hand side o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1... | $$\lim_{x\to 1}\frac{x^2+x-2}{1-\sqrt{x}}=\lim_{y\to 1}\frac{y^4+y^2-2}{1-y}=-\lim_{y\to 1}\frac{y^4+y^2-2}{y-1}$$
And that last limit is, by definition, the derivative $\frac{d}{dy}(y^4+y^2)=4y^3+2y$ at $y=1.$
So the result is $-6.$
Or you can just factor:
$$\frac{y^4+y^2-2}{y-1}=\frac{y^4-1}{y-1}+\frac{y^2-1}{y-1}=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How did they get the term $(n+1)^3$ in the step of inductive proof which says $\sum_{k=1}^{n+1} k^3=\sum_{k=1}^n k^3 + (n+1)^3$? I'm struggling to understand on what was done in this inductive step. How did they get the $(n+1)^3$ term?
Proof Solution
| $\sum\limits_{k=1}^{\color{red}{n+1}} k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + \color{blue}{(n+1)^3}=$
$[1^3+2^3 + 3^3 + ...... + n^3] + \color{blue}{(n+1)^3} =$
$\sum\limits_{k=1}^{\color{red}{n}} k^3 + \color{blue}{(n+1)^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$.
I am rather stuck though. Is there a general method for going about this? I always find myself having to g... | Brute-forcing the first factorization by expanding and collecting the powers of $\,x\,$:
$$
\begin{align}
(x+y+z)^3 xyz - (xz+xy+yz)^3 &= yz \,x^4 - (y^3+z^3) \,x^3 + yz(y^3+z^3)\,x - y^3 z^3 \\
&=yz(x^4-y^2z^2)-(y^3+z^3)x(x^2-yz) \\
&= yz(x^2-yz)(x^2+yz)-(y^3+z^3)x(x^2-yz) \\
&= (x^2-yz)(\ldots)
\end{align}
$$
Ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluating a complex integral with the goniometric function I was evaluating a this complex integral via gamma function:
$\int_0^\infty \sin (x^p) \,dx$ $\;$for $p \gt 1$,
so I expressed it as an imaginary part of $\int_0^\infty \exp(-ix^p) \, dx$ $\;$for $p \gt 1$
*
*The formula of the gamma function is $\Gamma (z)... | You can first substitute $u=x^p$:
\begin{align}
I=\int_0^\infty \sin (x^p)\,dx&=\frac{1}{p}\int_0^\infty u^{\frac{1}{p}-1}\sin u \,du \\ \\
&=\frac{1}{p} \Im\int_0^\infty u^{\frac{1}{p}-1} e^{iu}\,du \\ \\
&=\frac{1}{p}\Gamma\left(\frac{1}{p}\right) \Im i^{1/p} \\ \\
&=\frac{1}{p}\Gamma\left(\frac{1}{p}\right) \Im e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1... | You're both right.
Note that
$$
x^3+2x^2+x = x (x + 1)^2
$$
$$
3x^2+4x+1 = (3 x + 1) (x + 1)
$$
On the other hand, you might have started with
$$
(x^3+2x^2+x)^4 = x^4 (x + 1)^8
$$
whose derivative is
$$
4x^3 (x + 1)^8 + 8x^4 (x + 1)^7
= 4x^3 (x+1 +2x) (x + 1)^7
= 4x^3 (3x+1) (x + 1)^7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Solving $y'''-y=\frac{1}{3}\left(e^x+e^{wx}+e^{w^2x}\right)$ by power series I have the following differential equation
$$y'''-y=\frac{1}{3}\left(e^x+e^{wx}+e^{w^2x}\right)$$
where $w=e^{2i\pi/3}$
I am trying to solve by power series. So let $y(x)=\sum_{n=0}^\infty a_n\frac{x^n}{n!}$. Then $y'''(x)=\sum_{n=0}^\infty ... | Indeed, the term with $x^n$ in the right-hand side is
$$
\frac{x^n}{3n!}(1+w^n+w^{2n})
$$
since $w^3=1$, writing $n=3q+r$, we have
$$
1+w^n+w^{2n}=1+w^r+w^{2r}
$$
If $r=0$, this equals $3$; if $r=1$ or $r=2$ this equals $0$.
The left-hand side is
$$
\sum_{n=0}^{\infty}(a_{n+3}-a_n)\frac{x^n}{n!}
$$
so we must have $a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that all eigenvalues of this pentadiagonal matrix are double degenerate I am trying to show in general that the following pentadiagonal matrix $\mathbf{M}$ has double degenerate eigenvalues,
\begin{equation}
\mathbf{M} =
\left[ \begin{array}{cccccccc}
4 & 0 & a & 0 & 0 & 0 & 0 & 0 \\
0 & 3 & 0 & a & 0 & ... | I should have made this only a comment, sorry:
Mathematica easily grinds out the characteristic polynomial of $\bf M$, $$a^8-6 a^6 \lambda ^2+34 a^6 \lambda -46 a^6+11 a^4 \lambda ^4-122 a^4
\lambda ^3+497 a^4 \lambda ^2-882 a^4 \lambda +577 a^4-6 a^2 \lambda
^6+94 a^2 \lambda ^5-596 a^2 \lambda ^4+1950 a^2 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1... | Calling
$$
\cos x = \frac{e^{ix}+e^{-ix}}{2}
$$
we have
$$
e^{3ix}+e^{-3i x} = e^{2ix}+e^{-2i x}
$$
or calling $z = e^{ix}$
$$
z^6+1 = z^5+z\to z^6-z^5-z+1 = (z-1)^2(z^4+z^3+z^2+z+1) = (z^5-1)(z-1) = 0
$$
so the solutions are obvious.
$$
x = \frac{2\pi}{5}k,\;\; \mbox{for}\;\; k=0,1,2,\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
} |
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
My attempt:
Let I = $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx$
$u=7-x\implies I=\frac 72 \int_3^4 \frac 1{\sqrt{(7x-10-x^2)^3}}dx$ Now I have to approximate this...
I got to this poi... | $$ \int_{3}^{4}\frac{x}{\sqrt{(x-2)^3 (5-x)^3}}\stackrel{x\mapsto\frac{7}{2}+z}{=}\int_{-1/2}^{1/2}\frac{z+7/2}{\sqrt{\left(9/4-z^2\right)^3}}\,dz=\frac{7}{2}\int_{-1/2}^{1/2}\frac{dz}{\sqrt{(9/4-z^2)^3}} $$
since the integral of an odd integrable function over a symmetric interval (with respect to the origin) equals z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Floor value of sum whose general term satisfy recursive relation.
Consider the sequence $x_{n}$ given by $\displaystyle x_{1} = \frac{1}{3}$ and $x_{k+1}=x^2_{k}+x_{k}$
and Let $\displaystyle S = \frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots \cdots +\frac{1}{x_{2008}}$. Then $\lfloor S \rfloor $ is
Try: From $\displaystyle ... | Let $y_n=\frac1{x_n}$. Then
$$\tag1y_{n+1}=y_n-\frac1{\frac1{y_n}+1}=y_n-\frac{y_n}{y_n+1}=y_n\cdot\left(1-\frac1{y_n+1}\right).$$
Note that this means $0<y_{n+1}\le \frac a{a+b}y_n$ if $0<y_n\le \frac ab $.
Thus if $y_n\le \frac ab<1$ then $$\tag2\sum_{k=n}^\infty y_n\le y_n\sum_{k=0}^\infty\left(\frac a{a+b}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le... | Tangent line to $y=\frac{2x-7}{2x+5}$ at $x=5$ and $x=7$ are: $y=\frac{8}{75}x-\frac13$ and $y=\frac{24}{361}x-\frac{35}{361}$, respectively.
Refer to the graph:
$\hspace{0.5cm}$
Hence:
$$\int_5^8\frac{2x-7}{2x+5}dx<\int_5^6 \left(\frac{8}{75}x-\frac13\right)dx+\int_6^8 \left(\frac{24}{361}x-\frac{35}{361}\right)dx\app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y =... | A varied version of @Oscar Lanzi's answer:
Rewrite the equation
$$a^2+a-(4b^2+4b)=0$$
$$\Delta=1+4(4b^2+4b)=(4b+2)^2-3\overset{def}{\equiv} K^2-3$$
If $a$ is an integer, then there exists some integer $M$ such that $\Delta=M^2$.
$$K^2-3=M^2\iff K^2-M^2=3$$
So there will only be a set of $(K,M)$ or $(a,b)$.
| {
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"url": "https://math.stackexchange.com/questions/2831439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):
*
*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^... | Rewrite the equation of the circle subbing in for $y=mx+c$:
\begin{align*}
(x-a)^2+(y-b)^2&=x^2-2ax+a^2+y^2-2by+b^2\\
&=x^2-2ax+a^2+(mx+c)^2-2b(mx+c)+b^2\\
&=x^2-2ax+a^2+m^2x^2+2mcx+c^2-2bmx-2bc+b^2\\
&=(m^2+1)x^2+2(m(c-b)-a)x+(a^2+b^2+c^2-2bc)=r^2\\
\end{align*}
Therefore solve
$$(m^2+1)x^2+2(m(c-b)-a)x+(a^2+b^2+c^2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Finding critical points of $f(x,y)= \sin x+\sin y + \cos(x+y)$
Find the critical points of function$$
f(x,y)=\sin x + \sin y + \cos(x+y),$$
where $0<x<\dfrac{\pi}{2}$, $0<y<\dfrac{\pi}{2}$.
What I have done:
$$f_{x}=\cos(x)-\sin(x+y),\\
f_{y}=\cos(y)-\sin(x+y).$$
From $f_{x}=0$, $\cos(x)=\sin(x+y)$. From $f_{x}=0$,... | $$f_x = \cos x - \sin(x+y) = 0\\
f_y = \cos y - \sin(x+y) = 0$$
Subtracting one from the other we get $\cos x = \cos y$ and with the restrictions of $x,y$ to $(0,\frac {\pi}{2})$ we can say $x= y$ and $\cos x - \sin 2x = 0$
$$\cos x(1-\sin x) = 0\\x = \frac {\pi}{6}$$
$\frac {\pi}{2}$ would also solve the equation bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?
Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it
| Let $AB=x$ and $AD$ be bisector of $\Delta ABC$.
Thus, $$\Delta ABD\sim\Delta CBA,$$ which gives
$$\frac{BD}{x}=\frac{x}{x+2}$$ or
$$BD=\frac{x^2}{x+2},$$ which gives
$$DC=x+2-\frac{x^2}{x+2}=\frac{4x+4}{x+2}$$ and since
$$\frac{AB}{AC}=\frac{BD}{DC},$$ we obtain:
$$\frac{x}{5}=\frac{\frac{x^2}{x+2}}{\frac{4x+4}{x+2}},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find absolute maximum and minimum values by parametrizing the boundaries $f(x,y) = 2\cos x + 3\sin y$ $; R= {(x , y): 0 \leq x \leq 2\pi \\\mbox{and}\\ 0 \leq y \leq \pi} $
I need to find the absolute maximum value and absolute minimum value in the region $R$, and I do have to parametrize the boundary pieces of $R$ to... | Making the parameterizations
$$
x = \pi(1+\sin (u))\\
y = \frac{\pi}{2}(1+\sin (v))
$$
we have
$$
f(u,v) = 3 \cos \left(\frac{1}{2} \pi \sin (v)\right)-2 \cos (\pi \sin (u))
$$
so the stationary points are the solutions for
$$
\left\{
\begin{array}{rcl}
2 \pi \cos (u) \sin (\pi \sin (u))=0 \\
-\frac{3}{2} \pi \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Size of partial quotients in continued fraction expansion of $\sqrt{n}$ It is a well-known result that the partial fraction expansion of $\sqrt{n}$ for some non-square natural number $n$ is periodic of the form
$$[a_0, \overline{a_1, a_2, \ldots, a_{l-1}, 2a_0}]$$
where $a_0 = \lfloor\sqrt{n}\rfloor$ and $a_1, \ldots, ... | It is more convenient to look at this in terms of
$$\omega = \sqrt{n} + \lfloor \sqrt{n}\rfloor = \bigl[\overline{q_0,q_1,\dotsc,q_{l-1}}\bigr]\,.$$
It is clear that we have $q_0 = 2a_0$ and $q_k = a_k$ for $k \geqslant 1$, since $\omega$ differs from $\sqrt{n}$ exactly by the integer $a_0$.
Proposition: The partial qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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expected sum of the numbers that appear on two dice, each biased so that a 3 comes up twice as often as each other number? This question is already posted here,but i want to check my approach.
Question
What is the expected sum of the numbers that appear on
two dice, each biased so that a $3$ comes up twice as often
as... | This is the same as two seven sided dice with two 3s.
Expected value $E= 2\cdot \frac{1+2+3+3+4+5+6}{7} = 6.8571$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appre... | Solving $$y=\frac{x^2-3x+c}{x^2+3x+c}$$ and $$y=7$$
we put $$\frac{x^2-3x+c}{x^2+3x+c}=7$$
to get $$x^2-4x+c=0$$ which should have exactly one root.
So putting $Discriminant=0$
we get $c=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}... | Just another way, note for $x\in(0,1)$,
$$f(x)=\frac1{1-x}-\frac2{1+x}-\frac98(3x-1)=\frac{(3x-1)^2(3x+1)}{8(1-x^2)}\geqslant0$$
while the inequality is just $f(a)+f(b)+f(c)\geqslant 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Probabilities of selection with replacement. Say you have a bag with 4 balls, 2 of which are red and 2 are blue. You select 2 balls randomly from the bag and you want to know the probability of selecting exactly 1 red ball.
Without replacement, there are two simple methods:
*
*$\binom{2}{1}\frac{2}{4}\frac{2}{3} = \... | With replacement we just have independent trials with success probability $\frac{2}{4} = \frac{1}{2}$. We do two trials, one of which must be a success (red) and one a failure. We have $\binom{2}{1}$ ways of choosing the trial with a success and all sequences with one sucess and one failure have probability $\frac12 \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2844456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Complex Partial Fraction Decomposition The question I need help with is:
Prove that
$$\sum_{k=0}^{6}\frac{1-z^{2}}{1-2z\cos\left(\frac{2k\pi}{7}\right)+z^{2}}=\frac{7(z^{7}+1)}{1-z^{7}}$$
I have already tried brute forcing this by combining the LHS into a single fraction. While this worked, it is an extremely long proo... | (Edit: the following was posted before the OP added "I have already tried brute forcing this ...".)
I don't see the elegant solution offhand, but the problem can certainly be brute-forced as sketched below. Note that the first term of the sum (excluding the $\,\,1-z^2$ factor ) is $\,\dfrac{1}{(z-1)^2}\,$ then the rest... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Show that the system has a limit cycle. The full question reads:
Show that the system with $\dot{x} = x-y-(x^2 + \frac{3}{2} y^2)x$ and $\dot{y} = x+y-(x^2 + \frac{1}{2} y^2)y$ has a limit cycle.
I wanted to check the correctness of my method. My thought was to find the equilibrium points (If I am correct, the only one... | Another approach is to proceed using the Poincare-Bendixon theorem.
Our aim then is to show that the system contains only one unstable equilibrium point, and that every trajectory of the system remains in a closed bounded subset.
From your original problem, let
$$
f_1(x,y) = x - y - x^3 - \frac{3}{2}y^2{x} \nonumbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the residue Finding the fourth derivative in order to get residue seems me very complicated, is there another way? $$Res\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right) }^{ 5 } } \right) =\lim _{ z\rightarrow i }{ \frac { 1 }{ 4! } \frac { { d }^{ 4 } }{ { d }{ z }^{ 4 } } \left( { \left( z-... | Since the residuum is the coefficient of $(z-i)^{-1}$ in the Laurent-series expansion of $\frac{e^{iz}}{(z^2+1)^5}$ at $z=i$, we can expand the function and extract the coefficient.
We obtain
\begin{align*}
\color{blue}{\mathrm{Res}}&\color{blue}{\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorization Manipulation Let $x$ and $y$ be real numbers. Consider $t=x^2+10y^2-6xy-4y+13$. So what is the $t$ as smallest number? The solution is $9$.
My trying:
$$t=x^2+10y^2-6xy-4y+13$$
$$=x^2+9+10y^2-6xy-4y+4$$
$$=(x+3)^2-3x+2y(5y-2)-6xy+4$$.
So let $x=-3$ and $y=0$. From this I got $t=13$. My answer is false... | Use Gauß' method to write a quadratic form as a linear combination of squares of linear forms:
\begin{align}
t&=x^2+10y^2-6xy-4y+13 =(x-3y)^2 -9y^2+10y^2-4y+13 \\
&=(x-3y)^2+y^2-4y+13 =(x-3y)^2+(y-2)^2-4+13\\
&=(x-3y)^2+(y-2)^2+9.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got a... | A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=\dfrac{y+1}{x+y-2}$
(Note: $x+y-2 \not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $\le 1$ (radius).
Distance to $(0,0):$
$d =\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Can we conclude that two of the variables must be $0$?
Assuming $$a^2+b^2+c^2=1$$ and $$a^3+b^3+c^3=1$$ for real numbers $a,b,c$, can we conclude that two of the numbers $a,b,c$ must be $0$ ?
I wonder whether mathworld's result that only the triples $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ satisfy the given equation-system ... | Without loss of generality assume that $a\not=0$ and $b\not=0$ such that $a^2+b^2+c^2=1$ and $a^3+b^3+c^3=1$. Then $|a|<1$, $|b|<1$ and $|c|<1$ (otherwise $a^2+b^2+c^2>1$). Therefore
$$1=|a^3+b^3+c^3|\leq |a|^3+|b|^3+|c|^3<|a|^2+|b|^2+|c|^2=1$$
Contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
| Let $\dfrac{a^2+b^2+c^2}{ab+bc}=k>0$ as $a,b,c>0$
$\iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $\ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)\ge0\iff k^2\ge\dfrac{4(a^2+c^2)}{(a+c)^2}$$
the equality occurs if $a=\dfrac{k(a+c)}2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2\ge0\imp... | {
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"url": "https://math.stackexchange.com/questions/2853202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve ... | We can assume that $x,y,z$ are sides of a triangle because otherwise the LHS is negative.
By Heron's formula we have
$$A^2 = s(s-x)(s-y)(s-z) = \frac1{16}(x+y+z)(x+y-z)(y+z-x)(x+z-y)$$
where $A$ is the area and $s$ is the semiperimeter.
Now recall this inequality:
$$4\sqrt{3}A \le \frac{9xyz}{x+y+z}$$
or $$A^2 \le \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find Principal Part Of $\frac{ze^{iz}}{(z^2+9)^2}$ Find the principal part of $\frac{ze^{iz}}{(z^2+9)^2}$ at $z_0=3i$
Can I say that $$\frac{ze^{iz}}{(z^2+9)^2}=\frac{\frac{ze^{iz}}{(z+3i)^2}}{(z-3i)^2}$$
And look at $g(z)=\frac{ze^{iz}}{(z+3i)^2}$?
| Note that both parts $\frac{1}{(z-3i)^2}$ as well as $g(z)$ contribute to the principal part of the function.
We obtain
\begin{align*}
\frac{ze^{iz}}{(z^2+9)^2}&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(z+3i)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(6i+z-3i)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(6i)^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving the equality case in triangle inequality Background
When plotted on a real number line, it may be deduced that if
$$a,b,c \in \mathbb{R} $$
$$a < b < c$$
then
$$\left| {a - c} \right| = \left| {a - b} \right| + \left| {b - c} \right|$$
Problem
But the problem is with the proof. How can the above statement be ... | To prove it you do cases
$|a-c| = \pm (a-c)$ and $|a-b| + |b- c| = \pm(a-b) + \pm (b-c)$.
consider the ways that $a,b,c$ may be ordered:
(your case is a subcase of case 6: so you can skip to the very last case if you want.)
1: Assume $a \ge b \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) + (b-c) = a-c = |a-c|$.... | {
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"url": "https://math.stackexchange.com/questions/2855210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $a^4+b^4+1\ge a+b$.
Prove that
$$a^4+b^4+1\ge a+b$$
for all real numbers $a,b$.
What I've tried:
1.I checked how AM-GM may help but doesn't look like it's useful here.
*I've tried:
$$(a^2+b^2)^2 -2(ab)^2+1 \ge a+b$$
But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too ... | Proof
Since $$\left(a^4-a^2+\frac{1}{4}\right)+\left(b^4-b^2+\frac{1}{4}\right)=\left(a^2-\frac{1}{2}\right)^2+\left(b^2-\frac{1}{2}\right)^2 \geq0,$$
hence $$a^4+b^4+1 \geq a^2+b^2+\frac{1}{2}.\tag1$$
Since $$\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)=\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}... | {
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"url": "https://math.stackexchange.com/questions/2857994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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How to evaluate $\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$? How do I start with evaluating this-
$$\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$$
What should be my first attempt at this kind of a problem where-
*
*The denominator and numerator are of the same degree
*Denominator involves fractional exponent like $3/2$.
Note:I am... | Using @Fimpellizieri 's brilliant observation on @Dahaka 's ingenious post:
\begin{equation}
\frac{d\,[f(x)]^a}{dx}=\frac{d\,f^a}{df}\frac{d\,f}{dx}=af^{a-1}\cdot f'=a\frac{f'}{f^{1-a}}\tag{1}
\end{equation}
we find the integral in the question belongs to a family that can be found from using $(1)$.
First consider $I=... | {
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"url": "https://math.stackexchange.com/questions/2858747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factorize $a(y) := xy^{3}+xy^{2}+(x+1)y+x \in GF(2)[x]_{x^{2}+x+1}[y]$ Could someone please help me to find irreducible factors of $a(y) := xy^{3}+xy^{2}+(x+1)y+x \in GF(2)[x]_{x^{2}+x+1}[y]$?
In $GF(2)[x]_{x^{2}+x+1}[y]$, we have $0,1,x,x+1$. So we use these in $a(y)$ one after the other:
$0:x⋅0^{3}+x⋅0^{2}+(x+... | @hardmath, I'll try :-)
In $GF(2)[x]_{x^{2}+x+1}[y]$, we have $0,1,x,x+1$. So we use these in a(y) one after the other:
*
*$0:x⋅0^{3}+x⋅0^{2}+(x+1)⋅0+x=x$ => not a root.
*$1:x⋅1^{3}+x⋅1^{2}+(x+1)⋅1+x=x+x+x+1+x=4x+1=1$ => not a root.
*$x:x⋅x^{3}+x⋅x^{2}+(x+1)⋅x+x=x4+x3+x2+x+x=x4+x3+x2 =0$ => x is a root.
*$x+1:x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder of a Polynomial when divided by a polynomial with nonreal roots What will be the remainder when $x^{2015}+x^{2016}$ is divided by $x^2+x+1$ without using the fact that $x^2+x+1$ has roots as non real cube roots of unity.
| Hint: using that $a^3-1=(a-1)(a^2+a+1)$ and so $\,x^{3k}-1=(x-1)(x^2+x+1)(\ldots)\,$:
$$
\big(x^{2016}\color{red}{-1}\big)+\big(x^{2015}\color{red}{-x^2}\big)\color{red}{+1+x^2} \\
=\left(\left(x^3\right)^{672}-1\right)+x^2\left(\left(x^3\right)^{671}-1\right)+\big(x^2\color{blue}{+ x} + 1\big)\color{blue}{- x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Logarithm equation
How to solve this logarithm equation?
$$\frac12\cdot[\log(x) + \log(2)) + \log[\sqrt{2x} + 1] = \log(6).$$
The answer is $2$.
I've tried to solve it, but I don't know how to proceed:
$\frac12\log(2x) + \log[\sqrt{2x} + 1] = \log(6)$
$\log([(2x)^\frac12] + \log[\sqrt{2x} + 1] = \log(6)$
$\log[\sq... | Square both sides of the equation $\sqrt{2x}=6-2x$ to obtain $2x=36-24x+4x^2$ which will give you this quadratic equation $4x^2-26x+36=0$. If we simplify the equation further by dividing by the common factor $2$ we obtain $2x^2-13x+18=0$. This should be easy to solve if you use the quadratic formula or by factorizing t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $... | By standard limits
*
*$\frac{\sin x}x \to 1$
*$\frac{1-\cos x}{x^2} \to \frac12$
we have that
$$\frac{\cos x-\cos(3x)}{\sin(3x^2)-\sin(x^2)}=\frac{\frac{\cos x-1+1- \cos(3x)}{x^2}} {\frac{\sin(3x^2)-\sin(x^2)}{x^2}}=\frac{-\frac{1-\cos x}{x^2}+9\frac{1- \cos(3x)}{(3x)^2}} {3\frac{\sin(3x^2)}{3x^2}-\frac{\sin(x^2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 6
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Proving $\sum_{j=0}^n (-1)^j {n \choose j} F_{s+2n-2j} = F_{s+n} $, where $F_n$ is the $n$-th Fibonacci number
$$\sum_{j=0}^n (-1)^j {n \choose j} F_{s+2n-2j} = F_{s+n} $$
($F$ is Fibonacci number).
I have been trying to prove this by mathematical induction.
First I assume this is true for n.
If I could prove it wor... | Here is a proof by induction over $n$. To recap, we want to show
$$F_{s+n} = \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2n-2j} \tag{1}$$
The base case, $n=0$, is trivial. So suppose (1) holds for some value of $n$. We want show it holds for $n+1$. Since $s$ is arbitrary, we may substitute $s+1$ for $s$ in (1):
$$F_{s+n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the values of $x$ and $y$ that satisfies $\sin(x+y)=\sin x+\sin y$. I know that in general the following equality does not hold: $\sin(x+y)=\sin x + \sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$\sin(x+y)=\sin x\cos y +... | use that $$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ and $$\sin(x+y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
compute $\frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$ compute the summation
$\sum_ {n=1}^{\infty} \frac{n}{3.5........(2n+1)}= ?$
My attempts : i take $a_n =\frac{n}{3.5........(2n+1)}$
Now =$\frac{1}{2}$ .$\frac{(2n+ 1) - 1}{3.5........(2n+1)}= \frac{1}{2}(\frac{1}{3.5........(2n-1)} -\fra... | Your work so far is good. Just keep going.
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{2} \left(\frac{1}{1 \cdot 3 \cdot 5 \cdots (2n-1)} - \frac{1}{1 \cdot 3 \cdot 5 \cdots (2n+1)}\right)
= \frac{1}{2} \left(\frac{1}{1} - \frac{1}{1 \cdot 3} + \frac{1}{1\cdot 3} - \frac{1}{1 \cdot 3 \cdot 5} + \frac{1}{1 \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2871600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get i... | We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$\begin{cases}a=q-10 \\ b=3q+3 \\ c=-q\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the maximum of the value$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\frac{a^n_{1}+a^n_{2}+\cdots+a^n_{n}}{n}}$
Let $n$ be give a positive integer, $a_{i}
\ge 0$,such that $a_{1}+a_{2}+\cdots+a_{n}=n$. Find the maximum value of
$$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\dfrac{a^n_{1}+a^n_{... | dezdichado's answer is correct, I'll just write out all the computations.
Clearly the maximum value is $n^{n-1}$.
Let $X = \frac{1}{n}\sum a_i^n$ and $Y=\prod a_i$. We would like to show that $\left(n^{n-1}-1\right)\sqrt[n]{Y}+\sqrt[n]{X}\le n^{n-1}$
Expanding and using AM-GM: $$n^n=\left(\sum a_i\right)^n=\sum a_i^n+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$ via Contour Integration? In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $\text{Proposition (1)}$ is sound ?
$\text{Proposition (1)}$
$$\int_{0}^{\infty} \frac{\cos(x)}{x^... | Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$\text{Proof}$
To proceed one must consider that,
$$ \int_0^\infty\,\frac{\cos(x)}{x^2+3}\,\text{d}x=\frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\,\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2... | $\newcommand{\I}{\mathfrak{I}}$$\newcommand{\dx}{\mathrm dx\,}$Let’s take the time to generalize this. Denote the generalized integral as$$\I=\int\limits_0^{\infty}\dx\frac 1{x^4+ax^2+b^2}$$Now factor out an $x^2$ from the denominator and complete the square to get
$$\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac 1{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating... | Your solution is fine.
Another possible way to do it would be using that, for $t>0$,
$$\tan^{-1}(t)+\tan ^{-1}\left(\frac{1}{t}\right)=\frac \pi 2$$ making
$$A=(x+2)\tan^{-1}(x+2) -x \tan^{-1}(x)$$ $$A=(x+2)\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x+2}\right) \right)-x\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&... | You had the right idea.
Of yourse, in this case, there was only one parameter, and even that appeared in the form of simple expressions ($a$ and $a^2$). With more complicated problems, I suggest you start in a different way.
Compute the determinant of the matrix of coeeficients on the LHS:
$$\det\begin{pmatrix} 1 & ... | {
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"source": "stackexchange",
"question_score": "2",
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Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.
I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
| $\displaystyle z=\frac{3i}{\sqrt{2}-i+1}$
$\displaystyle =\frac{3i}{\sqrt{2}-i+1}\times\frac{\sqrt{2}+i+1}{\sqrt{2}+i+1}$
$\displaystyle =\frac{-3+(3+\sqrt{2})i}{(\sqrt{2}+1)^2+1}$
$\displaystyle =\frac{-3+(3+\sqrt{2})i}{4+2\sqrt{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find all value of integers n for which the expression is a perfect square.
Find all integers $n$ for which the expression $$n^4+6n^3+11n^2+3n+31$$ is a perfect square.
My approach :
First I simplified the expression and equated it to some square number $$(n^2+3n+1)^2 - 3(n-10) = k^2$$ Now if we try to remove the t... | Strategy: try to find two perfect squares such that the expression $f(n)=n^4+6n^3+11n^2+3n+31$ is strictly between them for all but finitely many $n$.
$(n^2+3n+1)^2$ is important to compute, as you also pointed out.
It is $n^4+6n^3+11n^2+6n+1$.
So our impression is that $f(n)$ is close to this, and that if $f(n)$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Taking square roots modulo $2^N$ I was trying to solve $y^2 - y \equiv 16 \pmod{512}$ by completing the square.
Here is my solution.
\begin{align}
y^2 - y &\equiv 16 \pmod{512} \\
4y^2 - 4y + 1 &\equiv 65 \pmod{512} \\
(2y-1)^2 &\equiv 65 \pmod{512} \\
2y - 1 &\equiv \pm 33 \pmod{512} &\text{Found by pointwi... | Can you solve it in $p$-adics?
For any $2$-adic integer $n$ we have the familiar Taylor/binomial-power series for $\sqrt{1+8n}$:
$\sqrt{1+8n}=1+(1/2)(8n)-...$
All omitted terms vanish $\bmod 512$ when $n$ is a multiple of $8$, thus with $n=8$
$\sqrt{65}\equiv 1+(1/2)(8×8) \bmod 512$
$\sqrt{65}\equiv 1+32\equiv 33 \bmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$?
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$
Obviously the standard approach would be to use the chain rule and simpl... | Implicit differentiation:
$$\sin(f(x)) = \frac{2x}{1 + x^2}$$
$$\cos(f(x)) f'(x) = \frac{2(1 - x^2)}{(1 + x^2)^2}$$
$$
f'(x) = \frac{2(1 - x^2)}{\cos(f(x))(1 + x^2)^2} =
\frac{2(1 - x^2)}{\sqrt{1 - \sin^2(f(x))}(1 + x^2)^2} =
\frac{2(1 - x^2)}{\sqrt{(1 - x^2)^2/(1 + x^2)^2}(1 + x^2)^2}\cdots
$$
and:
*
*$1 - x^2 < 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Complex number question - spurious solutions...
$z_1 = 2 + 3i$ and $z_2 = 3 - 4i$
The complex number $z = x + iy$ is such that $\frac{z + z_1}{2z - z_2} = 1$.
Find the value of $x$ and the value of $y$.
Method 1:
$$z + z_1 = 2z - z_2 $$$$\Rightarrow x + iy + 2 + 3i = 2x + 2iy - 3 + 4i$$
Equating real and imaginary co... | We have that
$$\frac{a}{b}=1\implies \frac{ab}{b^2}=1\implies ab=b^2\implies b=a\:\text{or}\: b=0.$$ But $b=0$ is not a solution of $\frac{a}{b}=1.$
For the same reason
$$\dfrac{1}{z}=2\implies \dfrac{\bar{z}}{z\bar{z}}=2\implies \bar{z}=2|z|^2.$$ $z=0$ is a solution of $\bar{z}=2|z|^2$ but not of $\dfrac{1}{z}=2.$
Con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x... | WLOG, let $b=ax,x\ge 0$. Then:
$$a+\sqrt{b} = b + \sqrt{a} \iff a+\sqrt{ax}=ax+\sqrt{a} \iff \\
\sqrt{a}(\sqrt{x}-1)=a(x-1) \stackrel{x\ne 1}{\iff} 1=\sqrt{a}(\sqrt{x}+1) \iff a=\frac{1}{(\sqrt{x}+1)^2}.$$
Hence:
$$a+b=a+ax=\frac{1}{(\sqrt{x}+1)^2}+\frac{x}{(\sqrt{x}+1)^2}=\frac{x+1}{(\sqrt{x}+1)^2}\le 1, x\ge 0.$$
Equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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How can I show that these two polynomials are coprime? Let $n , m \in \mathbb{N}$ and let $k = \gcd(n , m)$. Then there exist $p , q \in \mathbb{N}$ coprime such that $n = p k$ and $m = q k$. And it is easy to see that
$$
\frac{X^n - 1}{X^k - 1} = \sum_{i = 0}^{p - 1} X^{k i} = f(X) \qquad \mbox{ and } \qquad \frac{X^m... | The following is a step-by-step proof for $k=1$. The general case is proved in my other answer to the related question, and the details can be filled-in very similarly there.
It is given that $\,X^{p} = \left(X-1\right)f(X)+1\,$ and $X^{q} = \left(X-1\right)g(X)+1\,$.
Since $\,p,q\,$ are coprime, there exist integers $... | {
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"source": "stackexchange",
"question_score": "2",
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How can I factorize this: "$X^3 + X^2 + X - 3$" I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
\begin{align}X^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \\
&= (X - ... | Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$
First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2... | Necessarily $x,y,z\ge{\large{\frac{1}{4}}}$.
Without loss of generality, assume $x=\min(x,y,z)$.
Then $2x \le y+z$, hence
\begin{align*}
&(y+z)^2=4x-1\\[4pt]
\implies\;&(2x)^2\le 4x-1\\[4pt]
\implies\;&4x^2-4x+1\le 0\\[4pt]
\implies\;&(2x-1)^2\le 0\\[4pt]
\implies\;&(2x-1)^2=0\\[4pt]
\implies\;&x={\small{\frac{1}{2}}... | {
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"url": "https://math.stackexchange.com/questions/2894140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How does $\frac{x + \frac{1}{2}}{\frac{1}{x} + 2} = \frac{x}2$? How does simplifying $\dfrac{x+\frac12}{\frac1x+2}=\dfrac{x}2$
Plugging and chugging seems to prove this, but I don’t understand the algebra behind it.
How would you simplify $\dfrac{x+\frac12}{\frac1x+2}$ to get $\dfrac{x}2$?
I tried multiplying the expre... | I'd start by multiplying by $2/2$, to make the expression easier to read.
$$\frac{x + 1/2}{1/x + 2} = \frac{2x + 1}{2/x + 4} = \frac{2x^2 + x}{4x + 2}$$
Factoring the top and bottom:
$$\frac{2x^2 + x}{4x + 2} = \frac{x(2x + 1)}{2(2x + 1)}$$
And cancel:
$$\frac{x(2x + 1)}{2(2x + 1)} = \frac{x}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\c... | Notice that if $x_0$ is a root, then so is $-x_0$. Can you see what the sum is now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \c... | Assume $z\ne\pm1$. Then ${z-1\over z+1}\in i{\mathbb R}$ means that, viewed from $z$, the two points $\pm1$ are seen under a right angle. By Thales' theorem this is the case iff $z$ is lying on a circle with diameter $[{-1},1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Converting complex numbers into Cartesian Form 3 When calculating the real and imaginary parts of the complex number, do we take the angle as shown or the magnitude of it? I thought that we would just take the angle as shown, but apparently not, according to my textbook (unless its a typo):
$$z=2\sqrt{3}\operatorname{... | We take the angle as shown, not just its absolute value. Therefore, $4\operatorname{cis}\left(\frac\pi6\right)\neq4\operatorname{cis}\left(-\frac\pi6\right)$. It looks as if you are right and your textbook is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ ---------
1} \\ \mbox{Also, }... | so to sum up. the question is incorrect. there are 4 solutions for this because root of any number has 2 solutions. in this question there are 2 real solutions and 2 solutions are on the imaginary axis
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using $\epsilon$-$\delta$ approach, prove that $\lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3} $ How to prove that $$ \lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3}$$
By $\epsilon$-$\delta$ definition.
What i did is this:
Let $\epsilon$ be grater than zero; i need to find a $\delta(\epsilon)>0$ such that $|... | You may also use this: if $$|x-1|<\epsilon\\|y-3|<\epsilon$$ we may bound $|\dfrac{x}{y}-\dfrac{1}{3}|$ sufficiently close to zero. To show that we have $$1-\epsilon<x<1+\epsilon\\3-\epsilon<y<3+\epsilon\\$$choosing $\epsilon$ small enough we obtain$$\dfrac{1-\epsilon}{3+\epsilon}<\dfrac{x}{y}<\dfrac{1+\epsilon}{3-\ep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$
Here's how I tried it ... | (This is a followup on Stefan Lafon's answer, too long for a comment.)
Using that $\,|x|=1 \iff \bar x = \dfrac{1}{x}\,$ and $\,\cos(a-b)= \operatorname{Re}\left(\dfrac{x}{y}\right)=\operatorname{Re}(x \bar y) = \operatorname{Re}(\bar x y)\,$, it follows that:
$$
\begin{align}
\cos(a-b) + \cos(b-c) + \cos(c-a) &= \cos(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Is $s(x)$ a linear combination of $a \cdot p(x) $and $b \cdot q(x)$ Let $p(x)=x^2+5x-3$ , $q(x)=4x^2+18x+4,$ and $s(x)=x^2+8x+2$.
I want to find two constants $a$ and $b$ such that $$s(x)=a \cdot p(x) +b \cdot q(x)$$
My attempt: $$ap(x)+bq(x) = ax^2+5ax-3a+4bx^2+18bx+4b = (a+4b)x^2+(5a+18b)x+4b-3a$$
I put this in the ... | Yes your way is fine, indeed with respect to the standard basis $\{1,x,x^2\}$ we have
*
*$v_p=(-3,5,1)$
*$v_q=(4,18,4)$
*$v_s=(1,8,1)$
and we need to solve
$$av_p+bv_q=v_s \iff \begin{bmatrix}-3&4\\5&18\\1&4\end{bmatrix}
\begin{bmatrix}a\\b\end{bmatrix}
=\begin{bmatrix}2\\8\\1\end{bmatrix}$$
and by the augmented... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$
Calculate
$$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$
My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\l... | Yes it looks fine, as a similar alternative we can use
$$\log_{\sin{x}}{\cos{x}}=\frac12\log_{\sin{x}}{\cos^2{x}}=\frac12\log_{\sin{x}}{(1-\sin^2{x})}=\frac12\frac{\log{(1-\sin^2{x})}}{\log {\sin{x}}}$$
$$\log_{\sin{(x/2)}}{\cos{(x/2)}}=\frac12\frac{\log{(1-\sin^2{(x/2)})}}{\log {\sin{(x/2)}}}$$
then
$$\frac{\log_{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Closed form for $t_n = 6t_{n-1}-9t_{n-2}$ where $t_0 = 5$ and $t_1 = 9$
Consider the sequence defined by
$$
\begin{cases}
t_0=5\\
t_1=9\\
t_n=6t_{n-1}-9t_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $t_n$.
Your response should be a formula in terms of $n$, and should not contain terms such as $t... | Another widely used approach is using generating functions, i.e.
$$f(x)=\sum\limits_{n=0}\color{red}{t_n}x^n=
5+9x+\sum\limits_{n=2}t_nx^n=
5+9x+\sum\limits_{n=2}\left(6t_{n-1}-9t_{n-2}\right)x^n=\\
5+9x+6x\left(\sum\limits_{n=2}t_{n-1}x^{n-1}\right)-9x^2\left(\sum\limits_{n=2}t_{n-2}x^{n-2}\right)=\\
5+9x+6x\left(\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Marginal Pdf from joint pdf of discrete variable X,Y are random discrete variables with joint probability function of
$f(x,y) = \dfrac {1}{e^2\cdot y!\cdot (x-y)!} $ with $x \in\{ 0,1,2,...\}$ and $y\in\{0,1,2,...x\}$
Find marginals $f_X(x)$ , $f_Y(y)$.
I tried to compute firstly $f_Y(y)$ but i have a problem with the... | The marginal $f_X(x)$ is
$$f_X(x) =\frac{1}{e^2} \sum\limits_y \frac{1}{y!(x-y)!} = \frac{1}{x!e^2} \sum\limits_y \frac{x!}{y!(x-y)!} = \frac{1}{x!e^2} \sum\limits_y \binom{x}{y}(1)^y(1)^{x- y} = \frac{1}{x!e^2} (1+1)^{x} $$
So
$$f_X(x) = \frac{1}{x!e^2} 2^x$$
Similarly
$$f_Y(y) =\frac{1}{y!e^2} \sum\limits_{x=0}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x ... | With a Riemannian sum:
$$\sum_{k=1}^n\frac{n}{n^2+k}=n\frac 1{n^2}\sum_{k=1}^n\frac1{1+\dfrac k{n^2}}$$ can be seen as a Riemaniann sum truncated to the $n$ first terms among $n^2$. Then
$$\lim_{n\to\infty}n\int_0^{1/n}\frac{dx}{1+x}=\lim_{n\to\infty}n\log\left(1+\frac1n\right)=1.$$
No quite rigorous, though (because t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x... | With $$\displaystyle \sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$$ We get $$x^3-x-n=0$$
Thus $$n=x^3-x=x(x-1)(x+1)$$
Therefore for any natural number $x>1$ we have a natural number $n$ satisfying $$n=x^3-x=x(x-1)(x+1)$$
For example with $x=5$ we get $n=120$ and for $x=10$ we get $n=990.$
Another way of sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
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.