Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate an expression with assumption Let $a,b,c$ be non-zero real numbers and satisfy
\begin{equation}
\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}-\frac{a^3+b^3+c^3}{abc}=2.
\end{equation}
Evaluate
\begin{equation}
M=[(a+b)^{2018}-c^{2018}][(b+c)^{2018}-a^{2018}][(a+c)^{2018}-b^{2018}].
\end{equation}
My impression is ... | For a way to factor the original condition without using hints from the second part of the question, eliminate denominators and collect the powers of $\,a\,$ to get:
$$
\begin{align}
0 &= 2abc + (a^3+b^3+c^3) - ab(a+b) - bc(b+c) - ca(c+a) \\
&= a^3 - a^2 (b + c) - a (b^2 - 2 b c + c^2) + b^3 - b c (b + c) + c^3 \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $a_n = \frac{n}{2n+1}+\frac{1}{n^3}$ is a Cauchy Sequence I want to show that
$$a_n = \frac{n}{2n+1} + \frac{1}{n^3}$$
is a Cauchy sequence.
My attempt:
$$|a_m-a_n|=|(\frac{m}{2m+1}-\frac{n}{2n+1}+\frac{1}{m^3}-\frac{1}{n^3})|$$
$$\leq\frac{m}{2m+1}+\frac{n}{2n+1}+\frac{1}{m^3}+\frac{1}{n^3}$$
$$<\frac{1}{2}... | You split too much, instead:
\begin{align}\left|\frac{m}{2m+1}-\frac{n}{2n+1} + \frac{1}{m^3}-\frac{1}{n^3}\right|
&\leq \left|\frac{m}{2m+1}-\frac{n}{2n+1}\right| + \left|\frac{1}{m^3}-\frac{1}{n^3}\right|\\
&\leq \left|\frac{m(2n+1) - n(2m+1)}{(2m+1)(2n+1)}\right| + \left|\frac{n^3-m^3}{m^3n^3}\right|\\
&= \frac{|m-n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$ given $4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$
$$4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47$$ Solve for $x$.
I tried to put every single monomial under $x^2$ denominator, but that did not get me to anything I could solve for.
I appreciate any help.
| \begin{equation}
4x^2 + 12x + \frac{12}{x} + \frac{4}{x^2} = 47
\end{equation}
Multiply by $x^2$, i.e.
\begin{equation}
4x^4 + 12x^3 -47x^2 + 12x + 4= 0
\end{equation}
Notice that for $x = 2$ and $x = \frac{1}{2}$, we get zero. So
\begin{equation}
4(x-2)(x-\frac{1}{2})(x - a)(x-b)
=
4x^4 + 12x^3 -47x^2 + 12x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Differential equation: $(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $
Attempt:
After expanding, everything is neat except: $3x^2y dy - 3y^2 x dx$
I can't convert it to exact differential.
Also, there's weird symmetry in the equation wrt the coefficients of $x^2$ and... | Hint:
$$(2x^2 -4 + 3y^2 -3)x dx = (3x^2 -6+ 2y^2 -2 )y dy$$
and use substitutions $x^2-2=u$ and $y^2-1=v$. You will find a homogeneous differential equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all matrices that commute with $A$
Given $$A = \begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.
Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to fi... | With
$N = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \tag 1$
we have
$A = 3I + N; \tag 2$
then
$AB = BA \tag 3$
implies
$(3I + N)B = B(3I + N), \tag 4$
or
$3B + NB = 3B + BN, \tag 5$
whence
$NB = NB; \tag 6$
with
$B = \begin{bmatrix} b_1 & b_2 & b_3 \\ b_4 & b_5 & b_6 \\ b_7 & b_8 & b_9 \end{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Roots of unity and large expression
Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}.$$
I have tried combining the first and third terms & first and last ter... | Expand 2nd and 4th fraction with $\omega $ and $\omega ^2$ respectively: $$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}=\frac{\omega}{1 + \omega^2} + \frac{\omega^3}{\omega+ 1} + \frac{\omega^3}{1 + \omega} + \frac{\omega}{\omega^2+1}$$
$$=2\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determine if a system of equations is independent, dependent or inconsistent Is there a way to determine the nature of a system of equations without solving it? For example, given the system
\begin{cases}
2x + y - 4z = 6 \\[4px]
y - 2z = 2 \\[4px]
4x + 3y - 10z = -3
\end{cases}
Can I tell that this system ... | Not really. You can see whether the three equations are independent by computing the determinant
$$
\det\begin{bmatrix}
2 & 1 & -4 \\
0 & 1 & -2 \\
4 & 3 & -10
\end{bmatrix}
$$
but, unfortunately, it is $0$. Besides, computing a determinant with Laplace expansion is a very expensive computation.
Much simpler is going w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2887926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 ... | We need to solve
$$8\sin^2x\cos{x}=\sqrt3\sin{x}+\cos{x}$$ or
$$2\sin2x\sin{x}=\cos(x-60^{\circ})$$ or
$$\cos{x}-\cos3x=\cos(x-60^{\circ})$$ or
$$2\sin30^{\circ}\sin(30^{\circ}-x)=\cos3x$$ or
$$\cos(60^{\circ}+x)=\cos3x.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Difficulty proving integral reduction formula So I tried proving this integral reduction formula but to no avail:
If $$I_n=\int \frac{x^n}{\sqrt{ax+b}}dx$$, then
$$\int \frac{x^n}{\sqrt{ax+b}}dx=\frac{2x^n\sqrt{ax+b}}{a(2n+1)}-\frac{2nb}{a(2n+1)}I_{n-1}$$
I tried integrating by parts and my attempt went as followed:
$... | You are almost there. Just move the $-2nI_n$ term to the other side, then divide everything by $2n+1$
$$I_n=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}\\
I_n+2nI_n=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2nb}{a}I_{n-1}\\
I_n=\frac{2x^n\sqrt{ax+b}}{a(2n+1)}-\frac{2nb}{a(2n+1)}I_{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expanding product of binomials $(z^k + z^{-k})$ Suppose $z$ is a complex number, and consider the product
$$f_m(z)=\prod_{k=1}^m \left(z^k + \frac 1 {z^k} \right),$$
for $m = 1,2,\dots$ . Of course, one should be able to expand this into a sum of terms that are either powers of $z$ or of $1/z$. It is easy to see that t... | This is not an answer, but maybe help you to find an answer.
Since the function is same as
$$
f_{m}(z) = \frac{1}{z^{m(m+1)/2}}\prod_{k=1}^{m}(1+z^{2k}),
$$
it is enough to find an explicit formula for the coefficients of the polynomial
$$
g_{m}(z) = \prod_{k=1}^{m}(1+z^{k}).
$$
Note that $f_{m}(z) = g_{m}(z^{2})/z^{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Convergence and limit of recursive sequence Consider the sequence $(a_n)_{n \in \mathbb{N}}$ which has startvalue $a_0 > -1$ and recursive relation:
$$a_{n+1} = \frac{a_n}{2} + \frac{1}{1+ a_n}$$
How to prove the convergence and find the limit?
_
I think you need to show the convergence with a Cauchy sequence. It is al... | Solution $\blacktriangleleft$. Let
$$
f(x) = \frac x 2 + \frac 1 {1+x}, \quad [x >-1],
$$
then
$$
f'(x) = \frac 12 - \frac 1{(1+x)^2} = \frac {(x+1+\sqrt 2)(x+1-\sqrt 2)} {2(1+x)^2},
$$
hence $f(x) \searrow$ on $(-1, \sqrt 2 -1)$, $\nearrow$ on $(\sqrt 2 - 1, +\infty)$, therefore $f(x) \geqslant f(\sqrt 2 -1) = \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$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}$$
If $f(x)= x... | We have
$$a+\sqrt b=b+\sqrt a\iff a-b=(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=\sqrt a-\sqrt b.$$
Then $a=b$ (which is not allowed) or $\sqrt a+\sqrt b=1$.
Now,
$$a+b=a+(1-\sqrt a)^2$$ has an extremum found by taking the derivative on $a$,
$$1-\frac{1-\sqrt a}{\sqrt a}=0$$
which gives $a=\dfrac14$. But it turns out that this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Expected number of parallel tosses of $N$ unfair coins until all coins landed head at least once I am trying to understand this answer, but it doesn't work when I plug in the numbers. Given the recurrence relation
\begin{align*}
E_n &= \dfrac{\displaystyle 1+\sum_{k=1}^{n-1}q^kp^{n-k}E_{k}}{1-q^n} \\
E_1 &= \frac{1... | The recurrence relation given by @saulspatz in the comments is correct. Solved for $E_n$, it is:
$$ E_n = \dfrac{1+\sum\limits_{k=1}^{n-1}\binom{n}{k}p^{n-k}q^{k}E_k}{1 - q^n} $$
with $ E_1 = \frac{1}{p} $. This form correctly predicts $ E_2 = \frac{8}{3} $ and $ E_3 = \frac{22}{7} $, as well as $ E_5 = \frac{2470}{651... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.
If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?
I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it... | I'll suppose $a$, $b$, $c>0$ as that always seems to be assumed in these
sorts of problems. Then your inequality is equivalent to
$$(a^2+b^2+c^2)^2\ge abc(a+b+c)=a^2bc+b^2ac+c^2ab.$$
As $bc\le\frac12(b^2+c^2)$ etc. we get
$$a^2bc+b^2ac+c^2ab\le a^2b^2+a^2c^2+b^2c^2$$
and that is clearly less than $(a^2+b^2+c^2)^2$.
We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How does recurrence relation works with system of equations? There are $$ a+b+c+d = 2\\2a+2^2b+2^3c+2^4d = 5\\ 3a+3^2b+3^3c+3^4d = 6\\4a+4^2b+4^3c+4^4d = 1$$
then I'm given
$$C_{n}= a+bn+cn^2+dn^3$$
from linear recurrence relation with repeated roots said that
$$(x-1)^4 = \sum_{k=0}^{4}\binom{4}{k}(-1)^{4-k}x^... |
How do I use recurrence relations to find out the correct $C_6$
You did find the correct $\,C_6 = -8\,$.
$\;C_{n}== 4C_{n-1}-6C_{n-2}+4C_{n-3}-C_{n-4} \quad\quad\forall n \geq 5$
Correct, thus far.
but wolfram alpha told me that $C_{6} =-48$
What did you ask WA?
WA's answer to $\,c_{n} = 4 c_{n-1} - 6 c_{n-2} + 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Understanding a particular method of solving generalized version of Pell's equation So, I have understood how to solve Diophantine equations of the form
$$x^2-Dy^2=1.$$
However, when I was reading the solution of the generalized Pell's equation
$$x^2-Dy^2=c,$$
I got stuck. I knew how to solve it using continued fractio... | The point is to first find a single solution to $x^2-15y^2=61$. Then every other solution is obtained by multiplying by units. I won't prove this here.
First, solving $l^2\equiv15\pmod{61}$ yields the identity
$$25^2=15+61\cdot10\qquad\text{ so }\qquad \frac{(25+\sqrt{15})(25-\sqrt{15})}{10}=61.$$
Now suppose we have i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do I eliminate $x$ and $y$ from the system $x^2 y= a$, $x(x+y)= b$, $2x+y=c$ to get a single equation in $a$, $b$, $c$? Alright, a homework problem.
I'm stuck at this question,
Eliminate $x$ and $y$ from the given equations to get a single
equation in terms of $a$ , $b$ and $c$
$$\begin{align}
x^2 y &= a \\
x(x... | I will work with 2nd and 3rd equations to solve for $x$ and $y$.
From (3) we get,
$$y=c-2x$$
Substituting the value in (2) we get,
$$x(x+(c-2x))=b$$
$$\implies x^2-cx+b=0$$
$$x=\dfrac{1}{2}(c+\sqrt{c^2-4b}) \text{ or, }\dfrac{1}{2}(c+\sqrt{c^2-4b})$$
If $x=\dfrac{1}{2}(c+\sqrt{c^2-4b})$,then,
$$y=-\sqrt{c^2-4b}$$
Now,w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$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... | Hint: This hint also handles complex roots. Let $a,b,c\in\mathbb{C}$. Prove that, for a complex number $z$, $x:=z$ is a solution to $$x^4-a\,x^2\,\sin^2(bx)+c=0$$ if and only if $x:=-z$ is a solution of the above equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Regarding bijectivity of $f(x)=\frac{x}{1-x^2}$ Consider the map $f:(-1,1)\to \mathbb R,$ $$f(x)=\frac{x}{1-x^2}$$
Munkres claims that it is an order preserving bijection.
To see it's order preserving, assume $x< y$. Then $x^2<y^2, 1-x^2 >1-y^2$, so $\frac{1}{1-x^2}<\frac{1}{1-y^2}$.
To show it is a bijection, I guess ... | Suppose that $x \neq y$ but $f(x)=f(y)$ for some $x$ and $y$.
It follows that $\frac{x}{1-x^2}=\frac{y}{1-y^2}$, that is, $y(1+xy)=x(1+xy)$ and since $1+xy\neq 0$ it follows that $x=y$, but we assumed $x \neq y$ so when $x\neq y$ we have $f(x)\neq f(y)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Power series solutions
Construct two linearly independent, power series solutions to the ODE
$$u''+zu'+u=0.$$
Hence find the solution which satisfies $u(0)=1$ and $u'(0)=1.$
I have come up with the solution for the coeffecient, however I am not sure why we multiply them with $z^2$ instead of $z.$
| Let us consider your differential equation:
$$\frac{d^2u(z)}{dz^2}+z\cdot\frac{du(z)}{dz}+u(z)=0$$
Apply the reverse product rule:
$$\frac{d}{dz}\frac{du(z)}{dz}+\frac{d}{dz}(z\cdot u(z))=0$$
Integrate with respect to $z$:
$$\int\frac{d}{dz}\left(\frac{du(z)}{dz}+z\cdot u(z)\right) dz=0$$
$$\frac{du(z)}{dz}+z\cdot u(z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\left|\operatorname{Re}(3+i+2\bar z^2 -iz)\right| \le 6$, when $|z| \le 1$ So I started off with $z = x+iy$ where $x$ and $y$ are reals.
$\left|\operatorname{Re}\left((3+i+2(x-iy)(x-iy)-i(x+iy)\right)\right|$
$\left|\operatorname{Re}(3+i+2(x^2-2iyx-y^2)-ix+y)\right|$
$\left|\operatorname{Re}(3+i+2x^2-4iyx-2y... | Note that $3+2x^2-2y^2+y\leq5+y-4y^2$ if $x^2+y^2\leq 1$. Therefore,
$$3+2x^2-2y^2+y\leq 5+y-4y^2=5+\frac{4y(1-4y)}{4}\leq 5+\frac{1}{4}\,\left(\frac{4y+(1-4y)}{2}\right)^2=\frac{81}{16}$$
for $y\in[-1,+1]$, where we have applied the AM-GM Inequality on the right-hand inequality. We also have $3+2x^2-2y^2+y\geq 3+y-2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } ... | Divide both the numerator and denominator on the left side by $n^2$.
As n approaches $\infty$, the 1/n and 2/n tend to zero, giving $\frac{(1)(1)}{(2)(2)}$ = $\frac{1}{4}$.
The limit $\lim_{n\to\infty}(\frac{n+1}{n})^n$ = $\lim_{n\to\infty}(1+\frac{1}{n})^n$ = e.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
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How to prove that $f(x) = \frac{-2x+1}{(2x-1)^2-1}$ is one-to-one on $(0,1)$? How do I prove that the function
$$f(x) = \frac{-2x+1}{(2x-1)^2-1}$$
is one-to-one on the interval $(0,1)$? I have simplified the function $f(x)=f(y)$ and have a multivariable quadratic equation equal to zero. How do I conclude that the funct... | $f(x) = \frac{-2x+1}{(2x-1)^2-1}$
$f'(x) = \frac {(-2x+1)'[(2x-1)^2 -1]- (-2x+1)[(2x-1)^2 - 1]'}{[(2x-1)^2 - 1]^2}$
$=\frac {-2[(2x-1)^2 - 1]-(-2x + 1)2(2x-1)2}{[(2x-1)^2 - 1]^2}=\frac {2[1-(2x-1)^2] +4(2x -1)^2}{{[(2x-1)^2 - 1]^2}}$
Now if $0 < x < 1$ then $-1 < 2x -1 < 1$ and $0\le (2x -1)^2 < 1$ and $0< 1-(2x-1)^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Try to find eigenvalues, orthogonal vectors when you have big matrix Let
$$A =\begin{bmatrix}
1& 1& 1& 1& 1& 1& 1& 1\\
1& 1& -1& -1& 1& -1& -1& 1\\
1& -1& 1& -1& -1& 1& -1& 1\\
1& -1& -1& 1& -1& -1& 1& 1\\
1& 1& -1& -1& -1& 1& 1& -1\\
1& -1& 1& -1& 1& -1& 1& -1\\
1& -1& -1& 1& 1& 1& -1& -1... | Since $A^TA = 8I$, clearly the canonical vectors $\{e_1, \ldots, e_8\}$ are eigenvectors for $A^TA$. Your previous question then implies that $\{Ae_1, \ldots, Ae_8\}$ are orthogonal vectors and these are precisely the columns of $A$.
Since $\{Ae_1, \ldots, Ae_8\}$ is orthogonal, in particular it is linearly independent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The limit of $\lim_{n \to \infty}\left(1 + \frac{1}{\sqrt{n}}\right)^n$ I am approaching the question from an inequality perspective. In other words, I just want to see if the equation has an upper bound or lower bound.
After expanding the equation using binomial expansion, I get the term that $$\left(1 + \frac{1}{\sqr... | You can use the binomial expansion of the equation:
$$\left(1 + \frac{1}{\sqrt{n}}\right)^n = \sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac{1}{\sqrt{n}}\right)^{n-k} $$
Since all terms are positives, you can fix a lower bound:
$$\sum_{k=0}^n \begin{pmatrix} n\\k \end{pmatrix} 1^k\times\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
complex numbers equation deg 4 Find the sum of squares of elements of set $A$ if: $$A=\Big{\{}\big|z^n+\frac{1}{z^n}\big|;\:n\in\mathbb{N},\:z\in\mathbb{C},\: z^4+z^3+z^2+z+1=0\Big{\}}.$$
| Since $z^5=1$ (so $|z|=1$) it is not difficult to find $$A=\{ 2,|z+1|,|z^2+1|\}$$
If we devide $z^4+z^3+z^2+z+1=0$ with $z^2$ and put $t=z+{1\over z}$( so $t$ is a solution of $t^2+t-1=0$), then
$$|z+1|^2 = |z^2+2z+1| = |z||z+2+{1\over z}| = |t+2|$$
and $$|z^2+1|^2 = |t|^2$$
so we have $$S= 4+|z+1|^2+|z^2+1|^2= 4+|t+2|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} ... | You obtained the equivalent form $$\frac{4x^2-4x-8}{(x^2+2)(x+4)}\geq 0$$
Here, $x^2+2$ in the denominator is positive for all $x\in \mathbb{R}$, so you can cancel it.
However, you need to make a case distinction in terms of the sign of $x+4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x... | $n = 0$ would fit. Now let's assume $n > 0$.
I think you are on the right track. If the given expression, call it $y$, is a natural number, then $x = y^2 - n$ must also be an integer - and it is strictly positive, so it must be a natural number.
Then from $x^3-x-n = 0$ and $x$ being an integer, we conclude that $x$ mus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Solving $2^x=x^3$ Algebraically
How can I solve $2^x=x^3$ algebraically?
I could take $\log_2(\cdot)$ on both sides, but I'd still be stuck.
| This equation follows the general equation
$$a^x = x^{b}$$
and can be solved, for $a \neq 1$, as follows:
\begin{align}
a^{x} &= x^{b} \\
e^{x \, \ln a} &= x^{b} \\
x^{b} \, e^{- x \, \ln a} &= 1 \\
x \, e^{- x \, \ln a/b} &= 1 \\
- \frac{x \, \ln a}{b} \, e^{- x \, \ln a/b} &= - \frac{\ln a}{b} \\
u \, e^{u} &= - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all natural solutions $(a,b,c) $ such that $a^2−b, b^2−c, $ and $c^2−a $ are all perfect squares Find all natural solutions $(a,b,c)$ such that
$a^2-b$,
$b^2-c$,
$c^2-a$
are all perfect squares.
| Say $a>b$
Since $b<a\leq 2a-1$ we have: $$(a-1)^2<\underbrace{a^2-b}_{=x^2}<a^2\implies a-1 <x<a$$
which is impossible.
So $a\leq b$. With the same procedure we see that $b\leq c$ and $c\leq a$, so $a=b=c$
So we have $$(a-1)^2\leq a^2-a =x^2<a^2$$ and thus $x=a-1$ so $a^2-2a+1= a^2-a \implies a=1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find solutions of $2x^3\ge 1-6x^2$ The inital function I want to study is
$$\frac{x^3+1}{x+2}$$
Its derivative is
$$\frac{2x^3+6x^2-1}{(x+2)^2}$$
I want to solve the following inequality graphically:
$$2x^3\ge 1-6x^2$$
I tried drawing the graphs of both the functions. But couldn’t find a way to find the minimum $(0,1... | Exact solutions with CAS Maxima:
$$2 \cos{\left( \frac{\operatorname{atan}\left( \frac{\sqrt{7}}{3}\right) -3 {\pi} }{3}\right) }-1\le x \le \cos{\left( \frac{\operatorname{atan}\left( \frac{\sqrt{7}}{3}\right) +{\pi} }{3}\right) }-1,$$
$$x \ge \cos{\left( \frac{\operatorname{atan}\left( \frac{\sqrt{7}}{3}\right) -{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using Gram-Schmidt to find the QR decomposition I'm having problems doing a QR decomposition of a matrix...
Let $A=\begin{bmatrix}
{1} & {1} & {0} \\
{0} &{1} &{1} \\
{1} & {0} &{1}
\end{bmatrix}$
Find the QR decomposition for A
Here's what I've been doing:
I choose this basis, $B=\left \{(1,0,1), (1,1,0), ... | Suppose that we have $A$ as the following matrix
$$ A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \tag{1} $$
modified Gram Schmidt is
$$ v_{1} =a_{1} \tag{2}$$
$$ r_{11} = \| v_{1} \| = \sqrt{ 2}\tag{3}$$
$$ q_{1} = \frac{v_{1}}{r_{11}} = \langle \frac{1}{\sqrt{2}} , 0 , \frac{1}{\sqrt{2}} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the generating function for sequence $1,2,4,0,8,24,120,184,312,56,568,1592,...$ I'm having trouble finding a generating function for the sequence that has a closed form. The sequence can be deduced using two powers, with alternating negative and multiples of three as shown:
first term:
$1=1$
second term:
$2=1+2^{0... | Here is an idea:$$
-2^2 = 2^2-2\times 2^2$$ and $$3\times 2^5 = 2^5 + 2 \times 2^5.$$
So, for $n \ge 7$, we can split your sum as
$$f(n) = 1 + \sum_{i=0}^{n-2}2^i - 2\sum_{i=0}^{\lfloor\frac{n-4}{6}\rfloor}2^{6i+2} + 2\sum_{i=0}^{\lfloor\frac{n-7}{6}\rfloor}2^{6i+5}.$$
The last piece should be a matter of calculating t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is.
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$,... | $$\sinh x=\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}$$
differentiating and multiplying by $x$:
$$x\cosh x=\sum_{n=0}^\infty\frac{(2n+1)x^{2n+1}}{(2n+1)!}.$$
Again:
$$x^2\sinh x+x\cosh x=\sum_{n=0}^\infty\frac{(2n+1)^2x^{2n+1}}{(2n+1)!}.$$
Find $a$, $b$ and $c$ such that
$$a+b(2n+1)+c(2n+1)^2=(n+1)^2.$$
Then
$$a\sinh x+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve the complex equation $ω^2=-11/4+15i$ The question is stated as following:
"First, solve the equation:
$ω^2=-11/4+15i$
and after, with the help of that, solve:
$z^2-(3-2i)z+(4-18i)=0$"
The problem for me lies in solving the system of equations for ω;
$Re:a^2-b^2=-11/4$ and $Im:2ab=15$
Where I eventually e... | As you have noted, equating the real parts gives $a^2 - b^2 = -\frac{11}{4}$. Now since the modulus of $\omega^2$ is the squared modulus of $\omega$ (think in polar form), $|\omega|^2 = a^2 + b^2 = \sqrt{(-11/4)^2 + 15^2} = \frac{61}{4}$.
Thus adding and subtracting, $a^2 = \frac{1}{2} \left(\frac{61}{4} + -\frac{11}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $\frac{|z-a|}{|z-b|}=c$ for a constant $c$ not equal to $1$, then prove that the expression is a circle I saw a similar question which asked to show that the locus traced out by $arg(\frac{z-a}{z-b})=c$ is a circle. I know how to prove that but what is the connection between the $2$ questions. Can I describe the e... | Let $$\frac{|z-a|}{|z-b|}=c.$$
Let $z=x+iy$ and $a=a_1+ia_2$, $b=b_1+ib_2.$
We have $$(x-a_1)^2+(y-a_2)^2= c^2(x-b_1)^2+c^2(y-b_2)^2$$
$$ (1-c^2)x^2 + ( 1-c^2)y^2 -2(a_1-c^2 b_1)x-2(a_2-c^2 b_2)y+a_1^2+b_2^2-c^2b_1^2-c^2b_2^2=0$$
Which is the equation of a circle if the constant $ a_1^2+b_2^2-c^2 b_1^2-c^2 b_2^2$ is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2919737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A ring such that $(a+b)^2=a^2+b^2$ and $(a+b)^3=a^3+b^3$
Let $(A,+,\cdot)$ be a ring such that there are $a,b \in A$ which satisfy $$(a+b)^2=a^2+b^2, \quad (a+b)^3=a^3+b^3$$
Prove that $(a+b)^n=a^n+b^n,$ for all positive integers $n.$
I have found the following solution, but I am not quite satisfied with it.
From t... | It is no hard to see by using induction $a^nb=-b^na$ and $a^nb^n=-b^na^n$
Assume that $n$ is even.
$$(a+b)^n= \underbrace{(a+b)^2(a+b)^2(a+b)^2(a+b) ^2\cdots(a+b)^2}_{\text{$\frac{n}{2}$}\ \ \text{times}}$$
$$=(a^2+b^2)(a^2+b^2)\cdots(a^2+b^2)$$
$$=(a^4+a^2b^2+b^2a^2+b^4)\cdots (a^4+a^2b^2+b^2a^2+b^4)$$
Using $a^nb^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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"answer_id": 1
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Limiting Behavior of a Rational Function at Infinity Consider the following function:
$$f(x) = \frac{2x^2+x}{x^2-1}$$
I know there's a horizontal asymtote at $y = 2$. Without graphing, is there a way to tell that the $x\rightarrow \infty$ part is approaching from above and the $x \rightarrow -\infty$ is approaching fro... | For $x > 1$,
$$\frac{2x^2+x}{x^2-1} > \frac{2x^2+1}{x^2-1}>\frac{2x^2-2}{x^2-1}=2$$
Since the denominator is positive and $2x^2+x > 2x^2+1$. So $f(x) \to 2$ from above as $x \to \infty$.
Similarly
For $x< -2$,
$$\frac{2x^2+x}{x^2-1} < \frac{2x^2-2}{x^2-1}=2$$
Since the denominator is positive and $2x^2+x < 2x^2-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dimension of a vector space consisting of two $2$ by $2$ matrices that commute with all $2$ by $2$ matrices. What is the dimension of the vector space that consists of $2$ by $2$ matrices that commute with all $2$ by $2$ matrices?
Let $A$ be a matrix that commutes with all $2$ by $2$ matrices and let $B$ be any $2$ by ... | It appears that you’re looking for all $2\times2$ that commute with every $2\times2$ matrix, that is, matrices $A$ such that $[A,X]=AX-XA=0$ for all $X$. Let $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}.$$ Since this has to commute with every $2\times2$ matrix, it has to commute with the elements of the standard basis, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculating the sum of the infinite series $\frac{1}{5} + \frac{1}{3}\frac{1}{5^3} + \frac{1}{5} \frac{1}{5^5} +......$ How do I calculate the sum of the infinite series?
$$\frac{1}{5} + \frac{1}{3}.\frac{1}{5^3} + \frac{1}{5}. \frac{1}{5^5} +......$$
My attempt :
I know that $$\log (\frac{1+x}{1-x}) = 2 \, \left(x ... | Let $$f(x):=\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}.$$ The radius of convergence is $1$.
Then
$$f'(x):=\sum_{k=0}^\infty{x^{2k}}=\sum_{k=0}^\infty{(x^2)^k}=\frac1{1-x^2}.$$
By integration,
$$f(x)=\int_0^x\frac{dx}{1-x^2}=\frac12\int_0^x\left(\frac1{1+x}+\frac1{1-x}\right)dx=\frac12\log\left|\frac{1+x}{1-x}\right|$$ whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Cylotomic polynomial $\Phi_n(x) $: if $a$ is a root then so is $a^{k}, (k, n) =1$ It is well known that the $n$'th cylotomic polynomial $\Phi_n(x) \in\mathbb {Q} [x] $ has its roots as $\cos(2k\pi/n)+i\sin(2k\pi/n),(k,n)=1$.
I would like to establish the same fact without using any knowledge of complex numbers or circu... | Your statement about the remainder after polynomial long division is false. For instance, $$\Phi_{15}(x) = x^8 -x^7 + x^5-x^4+x^3-x+1$$
$$\Phi_{15}(x^2) = x^{16} -x^{14} + x^{10}-x^{8}+x^6-x^2+1$$
which has remainder $-x^{14} +x^{10} -x^8+x^6-x^2+x+1$ after division by $x^{15}-1$.
Instead, let us proceed via the defini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Transforming a matrix to diagonal matrix Show that the matrix $$A = \begin{bmatrix}a&h\\h&b\end{bmatrix} ,\quad a \ne b$$ is transformed to diagonal matrix $D = P^{-1}AP$, where $$P = \begin{bmatrix}\cos x& -\sin x\\\sin x& \cos x\end{bmatrix}$$ and $$\tan2x=\frac{2h}{(a-b)}$$
I understand that a $n \times n$ matrix $A... | You seem to know the theory behind the problem. So there is no reason you cannot find the eigenvalues of $A$ and the corresponding eigenvectors required for the diagonalisation. Note that since $A$ is symmetric, it is orthogonally diagonalisable.
To do this in a slightly different setting but essentially the same meth... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by... | Since $>,<$ are not defined concepts over the complex numbers, we can assume that $\sqrt{8-x^2}, \sqrt{25-x^2}$ are real
And $\sqrt{25-x^2} > \sqrt{8-x^2}$
Which implies $x\in [-2\sqrt 2,0]$
but for all $x$ in this interval, $\sqrt{8-x^2} - \sqrt{25-x^2} > x$
there is no solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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What are all of the negative integral solutions of $y^2+6xy-8x=0$? I got the answer as $(0,0)$ by making $D\ge 0$ (quadratic in $y$).
However, how do I know this is the only possible answer?
| $y^2+6xy-8x=0$ iff $y^2=2x(4-3y)$. Therefore, $\dfrac{y^2}{4-3y}=2x$ is an integer.
Computing the gcd of $y^2$ and $4-3y$ gives
$$
\frac{16}{4-3y} = 9 \frac{y^2}{4-3y} + (3y+4) = 18x + (3y+4)
$$
Therefore, $4-3y$ divides $16$ and so $4-3y \in \{ \pm 1, \pm 2, \pm 4, \pm 8,\pm 1 6 \}$. The only integer solutions $y$ for... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using complex numbers prove that $\sin(\frac {\pi}{m})\sin(\frac {2\pi}{m})\ldots\sin(\frac {(m-1)\pi}{m})= \frac{m}{2^{m -1}}$ Prove that for $m=2,3,\dots$
$$
\sin\left(\frac{\vphantom{1}\pi}m\right)\sin\left(\frac{2\pi}m\right)\sin\left(\frac{3\pi}m\right)\cdots\sin\left(\frac{(m-1)\pi}m\right)=\frac{m}{2^{m-1}}
$$
I... | $|\cos \theta + i\sin \theta - 1|\\
\sqrt {1 - 2\cos \theta} = 2\sin \frac {\theta}{2}\\
2\sin\theta = |e^{2\theta i} - 1|$
$\prod_\limits{n=1}^{m-1} \sin \frac{n\pi}{m} = \frac {1}{2^{m-1}}\prod_\limits{n=1}^{m-1}|e^{\frac{2n\pi}{m} i} - 1|$
The set $\{e^{\frac{2n\pi}{m} i}\}$ make up the roots of $z^m - 1 = 0$ exclu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Finding a particular coefficient in a polynomial I'm trying to get the coefficient of $x^6$ of this polynomial product:
$$x^2(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4).$$
I know with infinite series, you can use the closed form solution of the series to calculate the coefficient, but I haven't been able to figure out a ... | What is being asked is to find the coefficient of $x^4$ of
$$(1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4),$$
which, when multiplied by $x^2$ leads to the coefficient of $x^6$ overall.
This can be obtained by expansion:
\begin{align}
P &= (1+x+x^2+x^3+x^4+x^5)(1+x+x^2)(1+x^2+x^4) \\
&= (1+x+x^2+x^3+x^4+x^5)(1+x+2 x^2 + x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find $A = x - y + z$ if $3x + 5y + 7z = 29$ and $x , y , z \in \mathbb{Z}^+$ I've found the answer by trial and error $(x = 4 , y=2,z=1) \Rightarrow$ $A = 3$. I tried to solve it using modular arithmetic but it didn't work .
$$3x + 5y + 7z = 29 $$
$$ 3x + 5y+7z \equiv 29 \mod 3$$
$$ 2y +z \equiv 2 \mod 3 $$
$$ y= -k ,... | You don't get $z = 2k + 2$. you get $z \equiv 2k + 2 \mod 3$.
Or $z = 2k + 2 + 3M$.
Plugging that into the equation we get
$3x -5k + 2k + 14k + 14 + 21M = 29$ so
$3x = -9k + 15 + 21M$ or
$x = -3k + 5 + M$ for some integer $M$. As $M$ can be any integer that is pretty useless but let's continue.
Plugging $x = -3k + 5 +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Miscalculating the determinant I am learning linear algebra and am getting stuck when trying to calculate the determinant using elementary row operations. Consider the matrix A.
\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 1 & 1 & 1 \\
-2 & -2 & 3 & 3 \\
1 & 2 & -2 & -3 \\
\end{vmatrix}
According to the solution in my textbook... | You have a few mistakes in calculations. In step $2$ when you subtract row $1$ from row $2$ it should be $1-(-3)=4$ in the fourth column. In the same step when you add row $1$ twice to row $3$ it should be $3+2\times(-3)=-3$ in the fourth column.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solution for $\frac{3}{x-9} \gt \frac{2}{x+2}$ What I did:
$\frac{3}{x-9} \gt \frac{2}{x+2}$
$3(x+2) \gt 2(x-9)$
$3x+6 \gt 2x-18$
$x \gt -24$
When typing this in in symbolab, it showed me that the solution is $-24 \lt x \lt -2$ or $x \gt 9$
What did i do wrong ?
How come i didnt get the correct solution ?
| A shorter way
The domain of the inequality is $\mathbf R\smallsetminus\{-2,9\}$.
You can multiply both sides of the inequality by a positive number. To remove the denominators, we'll multiply by $(x+2)^2(x-9)^2$ and obtain
\begin{align}
3(x+2)^2(x-9)>2(x+2)(x-9)^2&\iff(x+2)(x-9)\bigl(3(x+2)-2(x-9)\bigr) >0\\
&\iff (x+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the relationship of the length of triangle's sides.
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
Now determine what kind of triangle $\triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
... | HINT:
Write the equality as $$a^2-10a+22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}=0$$ and since we know that $a$ is real, $$22+b+|\sqrt{c-1}-2|-2\sqrt{b-4}\le25\\b-2\sqrt{b-4}\le3-|\sqrt{c-1}-2|\le3.$$ But if $f(b)=b-2\sqrt{b-4}$, $f'(b)=1-\dfrac1{\sqrt{b-4}}=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Constructing a field with exactly $81$ elements I was thinking $\frac{\mathbb{Z_3}[x]}{(x^3+x+1)} \times \frac{\mathbb{Z_3}[x]}{(x^3+x+1)}$.
$(x^3+x+1)$ is irreducible in $\mathbb{Z_3[x]}$ so the quotient will be a field, and a field cross a field looks like it should be a field to me!
How would I do this if working ov... | Let's see. What are the irreducible quadratics: $x^2+x+2\,,x^2+2x+2\,,x^2+1$.
I count $3$ (thanks @Lubin).
So, there are ${3\choose 2}+3=6$ combinations to check.
These are the $6$ quartics I get: $x^4+1\,,x^4+x^3+x+1\,,x^4+2x^3+2x+2 \,,x^4+2x^3+x^2+x+1\,,x^4+x^3+x^2+2x+1\,,x^4+2x^2+1$.
But there are $2\cdot 3^4=162... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Divergence of infinite series $(\frac{3k-2}{4k+2})^{2k-3}$ I would like to ask if my solution for testing the divergence of the infinite series below is correct.
$$\sum_{k=1}^{\infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3}$$
I used the Cauchy ratio test.
$$
\lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3} =
\l... | $$
\left\lvert \frac{9^{k}}{9^{k+1}} \cdot \frac{16^{k+1}}{16^{k}} \right\rvert =
\left\lvert 9^{k-\color{red}{(k+1)}}\cdot16^{k+1-k} \right\rvert =
\left\lvert 9^{k-k-1}\cdot16^{k+1-k} \right\rvert =
\left\lvert \frac {16} 9 \right\rvert $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a Jordan Basis after finding the Jordan Canonical Form The question asked to find the Jordan Canonical Form and Jordan Basis of $\begin{bmatrix}1 & 1 & 0 & -1\\0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$, and after finding the characteristic polynomial, I used its eigenvalues and their multiplic... | let your matrix be $A$ and name $B = A - I.$ the characteristic polynomial says $A B^3 = 0.$ This is also the minimal polynomial.
The method I like is to find a column vector $w$ with $B^3 w = 0$ but $B^2 w \neq 0.$ I like zeros and ones, so I am choosing
$$
w =
\left(
\begin{array}{r}
0 \\
0 \\
0 \\
1 \\
\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove by the limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$ I was reviewing $\mathbb{R}-$analisys with a friend and I'm thinking about one of the questions...
Prove by the $\epsilon-\delta$ limit definition that $\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$.
My answer was very long, someone could do ... | I think (i) part is correct, and assume we choose $x$ from $|x-2|<\dfrac12$ then
$$\dfrac32<x<\dfrac52$$
$$-\dfrac32<x-3<-\dfrac12$$
$$\dfrac{13}{2}<x+5<\dfrac{15}{2}$$
these show
$$|\frac{(x-2)(x+5)}{x-3}|<\dfrac{15}{2}|x-2|.2<15\delta$$
so it is sufficient to have $\delta\leq\min\{\dfrac{1}{15}\varepsilon,\dfrac12\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Which is greater, $\left(\frac{e}{2}\right)^\sqrt{3}$ or $(\sqrt{2})^{\pi/2}$? (no calculators) From a math contest in 1985:
Determine which of the following is greater: (no calculators)
$$\left(\frac{e}{2}\right)^\sqrt{3} \, \hspace{3mm} \text{or} \hspace{3mm} \, (\sqrt{2})^{\pi/2}$$
Hints are welcome, but I'm totall... | This comparison is potentially pretty tricky inasmuch as the numerical values are pretty close, within $2\%$:
$$\left(\frac{e}{2}\right)^{\sqrt{3}} = 1.701\ldots, \qquad (\sqrt 2)^{\pi / 2} = 1.723\ldots$$
The following method is pretty quick-and-dirty, and I doubt that it's the most elegant way. But it doesn't use any... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Simplify the Dot Product in terms of $a$ and $b$ Where $a$ and $b$ are arbitrary vectors
$(a+2b) \cdot (2a-b)$
$$a\cdot(2a-b)+2b\cdot(2a-b) = 2(a\cdot a)-a\cdot b+1(b\cdot a)-2(b\cdot b)$$
$$=2(a)-ab+4ab-2(b)^2$$
$$=2a^2-2b^2$$
$$=2(a^2-b^2)$$
Where did i go wrong in simplifying this?
| If you denote dot product as $\cdot$, you get
$$
(a+2b) \cdot (2a - b)
= 2 a \cdot a + 4 b \cdot a - a \cdot b - 2 b \cdot b
= 2 a \cdot a + 3 b \cdot a - 2 b \cdot b
$$
and you can further simplify $a\cdot a = |a|$ if you like.
Your error is the 2nd step (3rd line), where you assumed $$2b\cdot(2a-b) = 1 b\cdot a - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving that the series with the general term $u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$ diverges.
$$u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$$ for $ n \in \mathbb{N}^*$.
*
*Prove that $(u_n)$ is convergent toward $0$.
*Prove that the series with the general term $(-1)^n u_n$ converges.
*Prove tha... | You can see the convergence of $u_n$ by induction:
$$\int^{\pi/2}_0 \cos x=[ \sin x]^{\pi/2}_0=1$$
$$\int^{\pi/2}_0 \cos^2 x=[ x/2+(1/4)\sin 2x]^{\pi/2}_0=\frac{\pi}{4}$$
$$\int^{\pi/2}_0 \cos^3 x=[ \sin x-(1/3)\sin^3 x]^{\pi/2}_0=1-\frac{1}{3}$$
$$\int^{\pi/2}_0 \cos^4 x=[(3/8)x+(1/32)\sin 4x+(1/16) \sin 2x]^{\pi/2}_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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$a = \lim_{n \rightarrow \infty} ( 1+ \frac{1}{n^2})^n$ and $b =\lim_{n \rightarrow \infty}( 1+\frac{1}{n})^{n^2}$
If $$a = \lim_{n \rightarrow \infty} \Big( 1+ \frac{1}{n^2}\Big)^n\;\;\;\; \text{and}\;\;\;\; b =\lim_{n \rightarrow \infty}\Big( 1+\frac{1}{n}\Big)^{n^2}$$
then choose the correct option
*
*$a= 1, b=\... | Generalize:
$$
a_{k,l} = \lim (1+\frac{1}{n^k})^{n^l}\\
$$
So $a=a_{2,1}$ and $b=a_{1,2}$
$$
\ln a_{k,l} = \lim n^l \ln (1+\frac{1}{n^k})\\
= \lim \frac{\ln (1+\frac{1}{n^k})}{n^{-l}}\\
= \lim \frac{\frac{-k/n^{k+1}}{(1+\frac{1}{n^k})}}{-l n^{-l-1}}\\
= \lim \frac{k}{l} \frac{n^{l+1}}{n^{k+1} (1+\frac{1}{n^k})}\\
= \li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2952181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How evaluate $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ How prove
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$
$\mathbf {My\,Attempt:}$
... | $$\iint_{(0,1)^2}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy = 2\iint_{0\leq y\leq x\leq 1}\frac{-\log(1-xy)}{(1+x)(1+y)}\,dx\,dy $$
equals
$$ 2\int_{0}^{1}\int_{0}^{1}\frac{-x\log(1-x^2 z)}{(1+x)(1+xz)}\,dx\,dz = 2\int_{0}^{1}\frac{\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2(1-x)-\log(x)\log(1-x^2)}{1+x}\,dx$$
and we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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How to apply induction to this formula? I want to justificate following equation:
$$\sum_{k=0}^n \frac{(-1)^k}{k!(n-k)!}\frac{1}{2k+1} = \frac{2^n}{(2n+1)!!}$$
I calculated the both sides for $n$ from 1 to 10 and it was true. How the mathematical induction can be applied to this equation? Or is there other way to justi... | There is a standard technique for this type of sum which has appeared
here several times. Introduce
$$f(z) = (-1)^n \frac{1}{2z+1} \prod_{q=0}^n \frac{1}{z-q}.$$
We use the fact that residues sum to zero and we have for the sum over
$0\le k\le n$
$$\sum_{k=0}^n \mathrm{Res}_{z=k} f(z)
= \sum_{k=0}^n (-1)^n \frac{1}{2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Is there an elegant way to determine $Av$ given $Au_1, Au_2$, and $Au_3$ for a $3\times3$ matrix $A$? Let A be a 3x3 matrix such that ${A} \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ -13 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \\ 4 \end{pm... | Try writing
$$\begin{pmatrix} 3\\ -11\\ -1\end{pmatrix}=a\begin{pmatrix} 3\\ 4\\ 5\end{pmatrix}+b\begin{pmatrix} 4\\ 5\\ 6\end{pmatrix}+c\begin{pmatrix} 5\\ -9\\ 1\end{pmatrix}$$
for some real numbers $a$, $b$ and $c$. Then
$$A\begin{pmatrix} 3\\ -11\\ -1\end{pmatrix}=aA\begin{pmatrix} 3\\ 4\\ 5\end{pmatrix}+bA\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$
My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\... | $$F(x)=\tan x\tan y=\dfrac{2\sin x\sin y}{2\cos x\cos y}=\dfrac{\cos(x-y)-\cos(x+y)}{\cos(x-y)+\cos(x+y)}=1-\dfrac{2\cos(x+y)}{\cos(x-y)+\cos(x+y)}$$
Method$\#1:$
For $x+y=\dfrac\pi3$
$$F(x)=1-\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$$ which will be maximum
if $-\dfrac2{2\cos\left(2x-\dfrac\pi3\right)+1}$ is maximu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first ... | From numerator of your solution
$$1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta$$
$$(1+2\sin\theta+\sin^2\theta-\cos^2\theta)+(2i\cos\theta+2i\cos\theta\sin\theta)$$
$$(1+2\sin\theta+\sin^2\theta-1+\sin^2\theta)+2i\cos\theta(1+\sin\theta)$$
$$(2\sin\theta)(1+\sin\theta)+2i\cos\theta(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Help with this proof by induction with inequalities. Show that mathematical induction can be used to
prove the stronger inequality $\frac{1}{2}\cdot ...\cdot \frac{2n-1}{2n} < \frac{1}{\sqrt{3n + 1}}$ for all integers greater than 1, which, together with a verification for the case where n = 1, establishes the weaker i... | Assuming $n\ge0$, we can show that
$$
\frac{2n+1}{2n+2}\le\frac{\sqrt{3n+1}}{\sqrt{3n+4}}\tag1
$$
by squaring both sides to get the equivalent
$$
\frac{4n^2+4n+1}{4n^2+8n+4}\le\frac{3n+1}{3n+4}\tag2
$$
and cross-multiplying to get the equivalent
$$
12n^3+28n^2+19n+4\le12n^3+28n^2+20n+4\tag3
$$
which is true since $n\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How do I evaluate the integral $\int_0^1\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$. stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was def... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
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Finding minimum value of $ \frac{x^2 +y^2}{y} $
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are
real
numbers satisfying $7x^2 + 3xy + 3y^2 = 1$
Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$
in $7... | Making $y = \lambda x$ we have
$$
\min f(x,\lambda) \ \ \mbox{s. t. }\ \ g(x,\lambda) = 0
$$
here
$$
\begin{cases}
f(x,\lambda) = \frac{1+\lambda^2}{\lambda}x\\
g(x,\lambda) = x^2(7+3\lambda+3\lambda^2)-1=0
\end{cases}
$$
this minimization problem is equivalent to
$$
\min F(\lambda) = \left(\frac{1+\lambda^2}{\lambda}... | {
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"url": "https://math.stackexchange.com/questions/2965865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculating the cross product of a cross product so I really can't see what I am doing wrong. I want to use this formula:
$a\times (b\times c) = b(a\cdot c) - c(a\cdot b)$
Calculate the rotation of $v(x,y,z)=(x,y,z)^T \times \omega$ with $\omega \in \mathbb R^3$
Solution:
$a\cdot c=\nabla\cdot \omega=0$
and
$a\cdot b =... | On Wikipedia you can see that the formula for the curl of a cross product is given by
$$
\nabla \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A}\ (\nabla \cdot \mathbf{B}) - \mathbf{B}\ (\nabla \cdot \mathbf{A}) + (\mathbf{B} \cdot \nabla) \mathbf{A} - (\mathbf{A} \cdot \nabla) \mathbf{B}
.
$$
Applying this on your c... | {
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Proof: For all n > 0, if $0I need to solve this problem without using induction, only with the following properties.
(1) Trichotomy law;
(2) if $a$ and $b$ are in $P$, then $a+b$ is in $P$;
(3) if $a$ and $b$ are in $P$, then $ab$ is in $P$.
Where P is the collection of all Positive Numbers.
My idea is to use the fac... | First a comment:
You don't need the assumption that
$x^2 < y^2$.
In general,
any statement of the form
$\forall n P(n)$
can only be proved by induction,
ultimately.
In your case,
$y^n - x^n = (y - x)(x^{n-1} + yx^{n-2} + ... + xy^{n-2} + y^{n-1})
$
has to be proved by induction
with $P(n)$ being
$y^n-x^n
=(y-x)\sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Need someone to explain answer to question without using Markov chains. Consider the following question here:
Probability of two consecutive head or tail or any one of them in a row?
Question. Fair coins are tossed and when either four consecutive heads and tails appear the process will be stopped. What is the probabi... | If I understand correctly, you want to compute the probability of getting 2 heads before 4 tails (ie ending the process). Let
*
*$p$ be the probability of landing heads (here $p=1/2$)
*$P(A \mid 0)$ be the probability of 2 consecutive heads at the beginning of the sequence
*$P(A\mid 1)$ be the probability of 2 co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Dimension of $W_{2}$? Let $A = \begin{bmatrix} 1 & -1 & -5 & 1 & 4\\ -1 & 2 & 8 & -3 & -4\\ 3 & -1 & -9 & 0 & 4 \\ 2 & 2 & 2 & -5 & -10\\ 0&-3&-9&5&13\end{bmatrix}$
Now we define the subspace $W_{1},W_{2}$ of $A$ as follows -
$W_{1} = \{X \in M_{5 \times 5}| AX = 0\}$
$W_{2} = \{Y \in M_{5 \times 5} | YA =0\}$
I can s... | Let us first look at
\begin{align*}
W'&= \{X\in M_{5\times 1}; AX=0\}\\
W''&= \{Y\in M_{1\times 5}; YA=0\}
\end{align*}
In the other words we look at similar equations but with column/row vectors instead of matrices.
By a direct computation you can get that $\operatorname{rank}A=4$, which implies that $\dim(W')=\dim(W... | {
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"source": "stackexchange",
"question_score": "5",
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Find Derivative of $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$?
Question. If $y=\tan^{-1}\left(\sqrt{\frac{x+1}{x-1}}\right)$ for $|x|>1$ then $\frac{d}{dx}=?$
Answer: $\displaystyle\frac{dy}{dx}=-\frac{1}{2|x|\sqrt{x^2-1}}$
My 1st attempt- I followed the simple method and started by taking darivati... | $$\dfrac{d\arctan\sqrt{\dfrac{x+1}{x-1}}}{dx}=\dfrac1{1+\dfrac{x+1}{x-1}}\cdot\dfrac{d\sqrt{\dfrac{x+1}{x-1}}}{dx}$$
For $x-1\ne0,$
$\sqrt{\dfrac{x+1}{x-1}}=\sqrt{\dfrac{\sqrt{x^2-1}}{(x-1)^2}}=\dfrac{\sqrt{x^2-1}}{|x-1|}$
$F=\dfrac{d\sqrt{\dfrac{x+1}{x-1}}}{dx}=\dfrac{d\dfrac{\sqrt{x^2-1}}{|x-1|}}{dx}$
If $x-1>0,F=\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The error for approximation of bessel function
The sixth degree polynomial
$1-{x^2\over 4}+{x^4\over 64}-{x^6\over 2304}$
Is sometimes used to approximate the Bessel function $J_0(x)$ of the first kind of order zero for $0 \leq x\leq 1$. Show that the error $E$ involved in this approximation is less than $0.00001$.
I... | Statement:
$E(x)=J_0(x)-\sum\limits_{n=0}^3\frac{(-1)^n}{(n!)^2} (\frac{x}{2})^{2n}\leq\frac{x^8}{2^8 (4!)^2}\tag{1}$
$E(x)=\sum\limits_{n=4}^\infty\frac{(-1)^n}{(n!)^2} ({\frac{x}{2}})^{2n}=\sum\limits_{n=0}^\infty\frac{(-1)^{n+4}}{(n+4)!^2} ({\frac{x}{2}})^{2(n+4)}=\frac{x^8}{2^8}\sum\limits_{n=0}^\infty\frac{(i)^{2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Verify the proof that $x_n = \ln^2(n+1) - \ln^2n$ is a bounded sequence.
Let $n\ \in \mathbb N$ and:
$$
x_n = \ln^2(n+1) - \ln^2n
$$
Prove that $x_n$ is a bounded sequence.
I've taken the following steps. Consider $x_n$
$$
\begin{align}
x_n &= \ln^2(n+1) - \ln^2n = \\
&= (\ln(n+1) + \ln n)(\ln (n+1) - \ln n) = \... | Another concise option: $\ln(n+1)<\ln n+\frac{1}{n}$ so $0<\ln^2(n+1)-\ln^2 n<2\frac{\ln n}{n}+\frac{1}{n^2}<2\cdot 1+1=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Sequence Inequality question from RMO 2018 Define a sequence {$a_n$} of real numbers by $a_1 = 2$ and $a_{n+1} = \dfrac{a_n^2+1}{2}$, for $n\ge 1$,
Prove that for every natural number $N, \sum_{j=1}^{N} \frac{1}{1+a_j} \lt 1$
I tried mathematical induction after coming to the step where
$Sum_n = \frac{1}{2}\left(\fra... | By computing $1-\sum_{k=1}^{N}\frac{1}{a_k+1}$ with the help of Mathematica it is not difficult to conjecture that
$$ 1-\sum_{k=1}^{N}\frac{1}{a_k+1} = \frac{2^{2^N-1}}{\prod_{k=1}^{N}b_k}\tag{1}$$
with $\{b_n\}_{n\geq 1}=\{3,7,37,1033,868177,701129422753,\ldots\}$.
It looks like $b_n = (a_n+1) 2^{2^{n-1}-1} $, hence i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Person with seven friends invites subset of three for one week Suppose that a person with seven friends invites a different subset of three friends to dinner every night for one week (seven days). How many ways can this be done so that all friends are included at least once?
I know the number of ways to do this with no... | This answer is essentially the same as Thomas Bladt's but I use a different format for counting. $$\frac {\binom{7}{3}!}{(\binom{7}{3}-7)!} - 7\cdot \frac{(\binom{7}{3}-\binom{6}{2})!}{(\binom{7}{3}-\binom{6}{2}-7)!}+ \binom{7}{2}\cdot \frac{(\binom{7}{3}-\binom{6}{2}-\binom{5}{2})!}{(\binom{7}{3}-\binom{6}{2}-\binom{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $df^{-1}(f(\begin{smallmatrix}x\\y\end{smallmatrix}))$?
$$f:\mathbb{R}\times[0,2\pi) \to \mathbb{R}^2,f\begin{pmatrix}x \\y\end{pmatrix}=\begin{pmatrix}e^x \cos y \\e^x \sin y\end{pmatrix}$$
which is $\exp(z)$
I know that $df(x,y)=\begin{pmatrix}e^x\cos y && -e^x\sin y\\e^x\sin y&&e^x\cos y\end{pmatrix}$.
How... | We have $df^{-1}\Bigg(f\begin{pmatrix}x \\y\end{pmatrix}\Bigg)=(df(x,y))^{-1}=\begin{pmatrix}e^x\cos y && -e^x\sin y\\e^x\sin y&&e^x\cos y\end{pmatrix}^{-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of picking two marbles each from two colors when selecting $4$ marbles out of $30$ marbles I have $\ 30 $ marbles. $\ 25 $ are white, $\ 3 $ are blue and $\ 2 $ are red. same color marbles are identical.
If I pick randomly and without replacement $\ 4 $ marbles, what is the probability that I'll pick two ea... | Correct answer is $\frac{401}{9135}$. It is calculated as follows:$\frac{\binom{25}{2}*\binom{3}{2}}{\binom{30}{4}}+\frac{\binom{25}{2}*\binom{2}{2}}{\binom{30}{4}}+\frac{\binom{3}{2}*\binom{2}{2}}{\binom{30}{4}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that $5|n^5-n$ an... | Method 1:We can try factoring but ... I don't want to.
Method 2: Euler's theorem. If $\gcd(n, 10) = 1$ then $n^{\phi(10)}=n^4 \equiv 1\pmod {10}$ So $n + 3n^3 + 7n^7 + 9n^9 \equiv n + 3n^3 + 7n^3 + 9n = 10n + 10n^3 \equiv 0 \mod 10$.
But if $\gcd(n,10) \ne 1$???
Well chinese remainder theorem.
$\mod 2$ we have $0^k \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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calculate $\lim_{x\to\infty} x + \sqrt[3]{1-x^3}$ So I multiplied by the conjugate and got $$\lim_{x\to\infty} \frac{x^2-(1-x^3)^\frac{2}{3} + x(1-x^3)^\frac{1}{3}-(1-x^3)}{x-(1-x^3)^\frac{2}{3}}$$
and this is where I got stuck.
|
So I multiplied by the conjugate and got
What conjugate expression was that exactly...?
You want to get rid of the cube root by using:
$$a+b=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2-ab+b^2}=\frac{a^3+b^3}{a^2-ab+b^2}$$
with, in your case, $a=x$ and $b=\sqrt[3]{1-x^3}$.
| {
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"source": "stackexchange",
"question_score": "1",
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Calculate the limit : $\lim_{x\rightarrow \infty}\tan ({\frac{\pi x}{2x+1}})^\frac{1}{x}$ Exercise :
Calculate the following limit
$$\lim_{x\rightarrow \infty}\tan \bigg({\frac{\pi x}{2x+1}}\bigg)^\frac{1}{x}$$
Attempt :
$$\lim_{x\rightarrow \infty} \frac{1}{x} = \frac {1}{\infty} = 0$$
$$\lim_{x\rightarrow \infty}... | As an alternative
$$\frac{\pi x}{2x+1}=\frac{\frac{\pi}2 (2x+1)-\frac{\pi}2}{2x+1}=\frac{\pi}2-\frac{\pi}{4x+2}$$
then
$$\left[\tan \bigg({\frac{\pi x}{2x+1}}\bigg)\right]^\frac{1}{x}=\left[\cot \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}=\frac{1}{\left[\tan \bigg(\frac{\pi}{4x+2}\bigg)\right]^\frac{1}{x}} \to 1$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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determine the error bound for the interpolation error at $x = \frac{\pi}{4}.$ I have been able to derive the interpolation polynomial $P_2(x)$ of degree two which interpolates $f(x) = \sin x$, given the points $(0,0), \left(\frac{\pi}{2}, 1\right), (\pi, 0).$
Solution:
$$P_2(x) = \frac{4}{\pi ^2}x(\pi - x)$$
Here is th... | For a polynomial interpolation of order $n$, the maximum error is given by
$${\rm err}=\frac{1}{(n+1)!}\max |f^{(n+1)}(x)|\max|\prod_{p=0}^n(x-x_p)|$$
Here $x_p$ are the roots of your polynomial.
Let's suppose that you are interested in finding the maximum only in the interval from $0$ to $\pi$.
The maximum of the deri... | {
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"source": "stackexchange",
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$\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx=$? Consider the convergent integral (I looked up difficult indefinite integrals on google images and then I saw this integrand and I was like hey let's see if it converges)
$$I=\int_{1}^{\infty}\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
We have the numerical approximatio... | $$
\begin{aligned}
\int_{1}^{\infty} \frac{x^{2}-1}{\left(x^{2}+1\right) \sqrt{x^{4}+1}} d x =& \int_{1}^{\infty} \frac{1-\frac{1}{x^{2}}}{\left(x+\frac{1}{x}\right) \sqrt{x^{2}+\frac{1}{x^{2}}}} d x \\
=& \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^{2}-2}} \\
=& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove the inequality: $\left(\frac{a}{a+2b}\right)^2+\left(\frac{b}{b+2c}\right)^2+\left(\frac{c}{c+2a}\right)^2 \geq \frac{1}{3}$
Let $a,b,c$ are all positive real numbers. Prove
$$\left(\frac{a}{a+2b}\right)^2+\left(\frac{b}{b+2c}\right)^2+\left(\frac{c}{c+2a}\right)^2 \geq \frac{1}{3}$$
Can anyone give a hint?
| By C-S and by the Vasc's inequality we obtain:
$$\sum_{cyc}\frac{a^2}{(a+2b)^2}=\sum_{cyc}\frac{a^4}{a^2(a+2b)^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+4a^3b+4a^2b^2)}=$$
$$=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}4a^3b+\sum\limits_{cyc}2a^2b^2+(a^2+b^2+c^2)^2}\geq\tfrac{(a^2+b^2+c^2)^2}{\frac{4}{3}(a^2+b^2+c... | {
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"url": "https://math.stackexchange.com/questions/2983036",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find the analytic form of expression for the following integral $$
\hspace{0.3cm} \large {\int_{0} ^{\infty} \frac{\frac{1}{x^4} \hspace{0.1cm} e^{- \frac{r}{x^2}}\hspace{0.1cm}e^{- \frac{r}{z^2}} }{ \frac{1}{x^2} \hspace{0.1cm} e^{- \frac{r}{x^2}}+ \frac{1}{y^2} \hspace{0.1cm} e^{- \frac{r}{y^2}}}} dr \hspace{.2c... | The integral is pretty awkward as written. Let's rename the parameters:
$$a=\frac{1}{x^2} \\ b= \frac{1}{y^2} \\ c= \frac{1}{z^2}$$
Then we have:
$$a^2 \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}$$
So we need to find the following integral:
$$I(a,b,c)= \int_0^\infty \frac{e^{-(a+c)r} ~dr}{ae^{-ar}+be^{-br}}... | {
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"url": "https://math.stackexchange.com/questions/2985319",
"timestamp": "2023-03-29T00:00:00",
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How to solve this radical expression I've been trying to solve this expression for at least two hours now... And I always get stuck towards the end, I don't know what I'm missing.
$\frac 1{xy} \times (\sqrt{xy} - \frac{xy}{x-\sqrt{xy}})\times (\sqrt{xy} + \frac{xy}{x+\sqrt{xy}})$
My first step was to rationalize the fr... | For simplicity, set $z=\sqrt{xy}$, with $z^2=xy$.
Then your expression becomes
\begin{align}
\frac {1}{xy}\left(z - \frac{xy}{x-z}\right)\left(z + \frac{xy}{x+z}\right)
&=\frac {1}{xy}\frac{xz-z^2-xy}{x-z}\frac{xz+z^2+xy}{x+z}\\[4px]
&=\frac {1}{xy}\frac{xz-2xy}{x-z}\frac{xz+2xy}{x+z}\\[4px]
&=\frac {1}{xy}\frac{x^2(z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
| Let $y:=x-\dfrac{3}{2}$. The equation becomes
$$\frac{1}{\left(y+\frac{3}{2}\right)^2}+\frac{1}{\left(y-\frac{3}{2}\right)^2}=\frac{104}{25}\,.$$
This is equivalent to
$$\frac{y^2+\frac{9}{4}}{\left(y^2-\frac{9}{4}\right)^2}=\frac{52}{25}\,.$$
Let $z:=\dfrac{1}{y^2-\frac{9}{4}}$, we have
$$\frac{9}{2}z^2+z=z^2\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2988342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
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Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tr... | For real $x$, we have $x^2 + x + 1 = (x+\frac12)^2 + \frac34 \ge \frac34$.
This implies for any $t \in (0,\frac34)$, following expansion in $t$ converges:
$$\frac{1}{x^2+x+1 - t} = \sum_{k=0}^\infty \frac{t^k}{(x^2 + x + 1)^{k+1}}$$
Since everything on RHS is non-negative, by Tonelli, we can integrate them term by te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Sum of a finite complex series
Let
$$C=\cos\theta+\cos(\theta+ \frac{2\pi}{n})+ \cos(\theta+ \frac{4\pi}{n})+...+\cos(\theta+ \frac{(2n-2)\pi}{n})$$
and
$$S=\sin\theta+\sin(\theta+ \frac{2\pi}{n})+ \sin(\theta+ \frac{4\pi}{n})+...+\sin(\theta+ \frac{(2n-2)\pi}{n})$$
Show that $C+iS$ forms a geometric series ... | Yes, you are close indeed:
\begin{align}
&\exp(i\theta)+ \exp(i(\theta+2\pi/n)) +\ldots+ \exp(i(\theta+2\pi(n-1)/n))\\
&= \exp(i\theta) \sum_{k=0}^{n-1} \exp(2\pi ik/n) = \exp(i\theta) \sum_{k=0}^{n-1} \exp(2\pi i/n)^k\\
&= \exp(i\theta) \frac{1-\exp(2\pi i/n)^n}{1-\exp(2\pi i/n)} =
\exp(i\theta) \frac{1-\exp(2\pi i)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2995828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\... | You can do it using the mean of the $m$ power for $m=\frac{3}{2}$ (or convexity of $x^\frac{3}{2}$ for that matter) :
We have that $\frac{x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^\frac{3}{2}}{3} \geq (\frac{x+y+z}{3})^\frac{3}{2}$ for any $x,y,z > 0$
Substituting $x=a^2, y=b^2, z=c^2$ we obtain $\frac{a^3+b^3+c^3}{3} \geq (\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theore... | Note that $4k+3$ is of the form $12k+3, 12k+7,\text { or } 12k+11,$ never of the form $12k+4,$ so there are no numbers in common in your sequences.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Domain of definition for $u_x + uu_y = 1$ How do i find the domain of definition for $u_x + uu_y = 1$
with $u = x/2$ on $y=x$ , $0 \leq x \leq 1$
I parametrise by letting $y=s$ , $x=s$ , $u=s/2$ , $0 \leq s \leq 1$ at $t=0$
The characteristic equations are:
$dx/dt = 1$, $dy/dt = u$, $du/dt = 1$
Solving $dx/dt = 1$ gi... | $$u_x+uu_y= \tag 1$$
Your three equations written on a equivalent form:
$$\frac{dx}{1}=\frac{dy}{u}=\frac{du}{1}=dt$$
A first family of characteristic equations comes from $\frac{dx}{1}=\frac{du}{1}$
$$u-x=c_1$$
A second family of characteristic equations comes from $\frac{dy}{u}=\frac{du}{1}$
$$\frac{u^2}{2}-y=c_2$$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $a$ is of order $3$ mod a prime $p$, then ... The question says:
Prove that if $a$ is of order $3$ modulo a prime $p$, then $1+a+a^2\equiv 0 \pmod p$. Moreover, $a+1$ is of order $6$.
For the First Part:
The typical idea is to start with $a^3 \equiv 1 \pmod p \to a^3 -1 \equiv 0 \pmod p$. Factoring the term on the... | For the second part, taking from the first part that $1+a = -a^2\pmod p\implies (1+a)^6= (-a^2)^6 = a^{12}\pmod p= (a^3)^4 =1^4\pmod p=1\pmod p$ . Thus $1+a$ is of order $6$ as claimed. And if $a$ is of order $3$ then $(-a)^3 = -a^3 = -1 = p-1\pmod p$. And from this we have: $(-a)^6 = ((-a)^3)^2 = (-1)^2 = 1\pmod p$. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^... | Observe that:
$$(k+2)^2+\cdots+(2k+2)^2=$$$$\left[(k+1)^2+(k+2)^2+\cdots+(2k)^2\right]+(2k+1)^2+(2k+2)^2-(k+1)^2=$$$$f(k)+(2k+1)^2+(2k+2)^2-(k+1)^2$$
where $f(k)$ denotes the RHS of 2).
If the equality indeed holds then the outcome must be $f(k+1)$ and that can be checked.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*... | Using Taylor's expansion:
When x approches $-\infty$:
$x + \sqrt{x^2-3x} = x - x(1 - \frac{3}{2x}+\text{o}(x)) = x - x + \frac{3}{2}+\text{o}(1) \rightarrow \frac{3}{2}$.
So, yeah
$
\begin{align}
\lim_{x\to-\infty}x + \sqrt{x^2-3x} =\frac{3}{2}.
\end{align}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle... Four squares, all of different areas, are cut from a rectangle, leaving a smaller rectangle of dimensions $1\times2$. If the largest square has area 64, and the other three squares have side lengths that are whole numbers no ... | Unless one side of the big rectangle is $8$, the $8\times 8$ must touch smaller parts on two edges, which already accounts for all parts. So one of the edges must touch two smaller squares - which leaves a gap at the smaller square that cannot be filled. We conclude that one side of the rectangle is $8$.
After removing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the remainder of the division of polynomials $x^{2007}$ divided by $x^2-x+1$.
I consider to solve this problem, should I break the $x^{2007}$ to find the formula $x^2-x+1$?
| Write: $$\boxed{x^{2007} = k(x)(x^2-x+1)+r(x)}$$ where $r(x)$ is a linear polynomial.
Say $a$ is zero of $x^2-x+1$, then $$ a^2-a+1=0\;\;\;/\cdot (a+1)$$ we get $$a^3+1 =0 \;\;\Longrightarrow \;\;a^3 = -1$$ and if we put $x=a$ in boxed equation we get $$-1= a^{2007} = k(a)\cdot 0+r(a)$$
and the same for other zero $b =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Explain this derivative identity: $ \frac{1}{2^n} \frac{d^n}{dy^n} \frac{(1+y)^{2n+3}(1-y)}{((1+y)^2 -2yx)^2} \bigg|_{y=0} = (n+1)! x^n $ I have the following result that I believe to be true:
$$
\frac{1}{2^n} \frac{d^n}{dy^n} \frac{(1+y)^{2n+3}(1-y)}{((1+y)^2 -2yx)^2} \bigg|_{y=0} = (n+1)! x^n
$$
The LHS is something ... | We have
$$\frac1{(1-t)^2}=\sum_{m=0}^\infty (m+1)t^{m},\qquad (|t|<1)$$
Put $t=\left(\dfrac{2y}{(1+y)^2}\right)x$, where $0<|y|<1$,
$$\frac1{\left(1-\left(\frac{2y}{(1+y)^2}\right)x\right)^2}=\sum_{m=0}^\infty (m+1)\left(\dfrac{2y}{(1+y)^2}\right)^mx^{m}.\qquad (|x|<\frac{(1+y)^2}{2|y|})$$
Multiply $(1+y)^{2n-1}(1-y)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3018164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate inverse of matrix with -1 on diagonal and 1 on the rest Calculate the inverse of the matrix
\begin{bmatrix}
-1& 1& ...& ...&1 \\
1& -1& 1& ... &1 \\
...& ...& ...& ...&1 \\
1&1 &1 & ...&1 \\
1& 1 &1 &1 &-1
\end{bmatrix}
$-1$ on the diagonal and $1$ on the rest.
The key I think is to per... | The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n \geq 3$
for $n=3:$
$$
\left(
\begin{array}{rrr}
-1 & 1&1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3020465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Laurent Series expansion about the point $z_0 = i$ of $\frac{z}{z^2+1}$ I am trying to construct the Laurent series expansion of $f(z) = \frac{z}{z^2+1}$ about $z_0 = i$ in the region $\{z \in \mathbb{C}: 0 < |z - i| < 2\}$ but I am stuck.
We can re-write $f(z) = \dfrac{z}{z^2+1} = \dfrac{z}{(z-i)(z+i)}$
This allows us... | Following your argument,
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right) = (z-i+i)\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)\\ = \left(\frac{i}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the Limit $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$ Find $\lim_{n \to \infty}\sum_{k=1}^{\infty}\frac{k^{n}}{1+k^{n+2}}$
My ideas: let $ n \in \mathbb N$ be constant, looking at $\frac{k^{n}}{1+k^{n+2}}$, we know
$$\frac{k^{n}}{1+k^{n+2}}\leq\frac{k^{n}}{k^{n+2}}=\frac{1}{k^{2}}$$
but this do... | I would do in this way
$$
\eqalign{
& \sum\limits_{1\, \le \,k} {{{k^{\,n} } \over {1 + k^{\,n + 2} }}}
= \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {{{k^{\,n + 2} } \over {1 + k^{\,n + 2} }}} \right)}
= \sum\limits_{1\, \le \,k} {{1 \over {k^{\,2} }}\left( {1 - {1 \over {1 + k^{\,n + 2} }}} \right)} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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