Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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minimal value of $x^2+2y^2+5z^2$ with constraint. $x,y,z>0$, and $xy+yz+zx=1$.
I need to find the minimum value of
$x^2+2y^2+5z^2$
In general what can we say about the minimal value of $\frac{ax^2+by^2+cz^2}{xy+xz+yz}$, over all positive numbers with $a,b,c>0 $
| Using AM-GM, we have:
$$4x^2+9y^2 \ge 12xy$$
$$3y^2+12z^2 \ge 12yz$$
$$18z^2+2x^2\ge 12zx$$
Combining and using the constraint,
$$6x^2+12y^2+30z^2 \ge 12(xy+yz+zx) \implies x^2+2y^2+5z^2 \ge 2$$
Equality is when $x:y:z = 3:2:1$ and satisfying the constraint, i.e. when $x = \frac3{\sqrt{11}}, y = \frac2{\sqrt{11}}, z =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Logarithmic inequality Solve the inequality:
$$
\log_8(x^2-4x+3) < 1
$$
$$
\log_8(x^2-4x+3) < \log_8(8)
$$
$$
\log_8(x^2-4x+3) - \log_8(8) <0
$$
$$
\log_8 [(x^2-4x+3)/8] < 0
$$
Thats what I did for the question so far... and I'm confused as to what to do next. Can someone verify this for me?
| Firstly, $\log(n)$ is defined only if $n \in (0,\infty)$
$$
\begin{align}
\Rightarrow\quad x^2 - 4x + 3 &\in (0,\infty)\\
\Rightarrow\quad x^2 - 4x + 3 &> 0\\
\Rightarrow (x-3)(x-1) &> 0\\
\end{align}
$$
$$
\boxed{\therefore\quad x \in (-\infty, 1)\cup(3,\infty)
\quad\text{or}\quad x\in \mathbb R \sim [1,3]}\tag{1}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How would you integrate $\sqrt{1+\frac{1}{x^2}}$ I need to integrate $\sqrt{1+\frac{1}{x^2}}$
I've tried to let $u=1/x^2$ but end up with, $\int \frac{\sqrt{1+u^2}}{u^{3/2}}du$ . I attempted to then substitute $u=\tan\theta$ and lead me to $-\frac{1}{2}\int \frac{1}{(\sin\theta\cos\theta)^{3/2}}d\theta$.
Not sure how t... | $$\int\sqrt{1+\frac{1}{x^2}}\,dx=\int \frac{\sqrt{1+x^2}}{x}\,dx=\int \frac{x\sqrt{1+x^2}}{x^2}\,dx$$ Let $1+x^2=u^2 $ so that$$\int \frac{x\sqrt{1+x^2}}{x^2}\,dx=\int \frac{u^2}{u^2-1}\,du$$ hence we have that $$\int \frac{u^2}{u^2-1}\,du=u+\frac{1}{2}\ln |\frac{u-1}{u+1}|+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solving for $x$ given $y = 2x - 9$ and $y = 5$. When does the 5 come into play? I am doing some problems. It states "Isolate the left side for each of the following equations. Then solve for x, assuming the value of y is 5 in a all cases." I noticed that I can get two different answers, depending on when I decide to pl... | The first is wrong because you forgot to multiply $y$ by $\tfrac12$. When you got into
$$
9 + y = 2x
$$
By the law of equations, you should do as follows:
$$
(9 + y) = 2x \\
\frac12(9+y) = \frac12 2x \\
\frac12(9+y) = x
$$
The you plug the value of $y$
$$
\frac12(9 + 5) = x \\
\frac12 14 = x \\
x = 7
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Am I differentiating this wrong? Differentiation is the opposite of Integration
$$\begin{align}\int \cos^2x dx\end{align}$$
$$\begin{align}-\frac{\cos^3x}{3\sin x}\end{align}$$
Now if we differentiate $-\frac{\cos^3x}{3\sin x}$ we should get $\cos^2x$. But I have always thought the differentiation goes like (and works ... | You are wrong in $\int \cos^2 x dx = -\frac{\cos^3 x}{3\sin x}$. You seem to be confused with power rule ($\int x^n dx = \frac{x^{n+1}}{n+1} + c $, for $n \neq -1$), which is only valid for powers of $x$, not $f(x)$. Also, there's an extra $\sin x$ in denominator...
The correct way to evaluate $\int \cos^2 x dx$ is:
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
computing integral without softwares: $\int \frac{2x+3}{x^2+\sqrt{1-x^2}}dx$ I was wondering if this integral can be solve without wolfram and others:
$$\int \frac{2x+3}{x^2+\sqrt{1-x^2}}dx$$
Thanks.
| This is a plan of attack, at least:
1) Separate into two terms, and to find $\displaystyle\int\frac{2x}{x^2+\sqrt{1-x^2}} dx$, let $u=x^2$ to get $\displaystyle\int\frac{1}{u+\sqrt{1-u}} du$, and then let $t=\sqrt{1-u}$ to obtain $$ 2\int\frac{t}{t^2-t-1} dt.$$
Now use partial fractions.
2) For the second term, let $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the volume of the area between two curves when rotated about the y-axis I keep going down a rabbit hole when answering this question:
Find the volume of the area lying in the first quadrant and bounded by the $y$-axis, the curve $y = x^3$ and the line $y = 3x + 2$, when it performs one revolution about the $y$... | When calculating $V_2$, you need to integrate from $2$ to $8$ because for $0 \le y < 2$, we have $x=(y-2)/3<0$, which is left of the $y$-axis (plot the graph to see this more clearly).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/940602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int_{-\pi}^{\pi} \frac{ d\theta}{w - sin \theta}$ I know that the integral
$$\int_{-\pi}^{\pi} \frac{ d\theta}{w - sin \theta} = \frac{2\pi}{\sqrt{w^2-1}}$$
where w, is an arbitrary constant and at some point you must do the substitution $$u = tan( \frac{\theta}{2} )$$ Does any one know how to do the inte... | This is a partial answer, assuming that $w>1$:
Let $u=\tan\left(\frac{\theta}{2}\right)$, so $\theta=2\arctan u$, $d\theta=\frac{2}{1+u^2}du$, $\sin\theta=\frac{2u}{1+u^2}$.
Then $\displaystyle\int\frac{d\theta}{w-\sin\theta}=\int\frac{1}{w-\frac{2u}{1+u^2}}\cdot\frac{2}{1+u^2}du=2\int\frac{1}{wu^2-2u+w}du$
$=\displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Under what conditions will one solution of $ax^2+bx+c = 0$ be the reciprocal of the other? Under what conditions will one solution of
$ax^2+bx+c = 0$ be the reciprocal of the other?
| Here's one way using the quadratic formula:
The roots of $ax^2+bx+c=0$ can be written as $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$.
Let $x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ and let $x_2$ be the reciprocal of this, so $$x_2 = \frac{2a}{-b+\sqrt{b^2-4ac}}.$$
However, since $x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}$, we can set th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Tough trigonometric identity: $\cot 13^\circ\cot 23^\circ \tan 31^\circ\tan35^\circ\tan41^\circ = \tan 75^\circ$ Prove that $$\cot 13^o\cot 23^o \tan 31^o\tan35^o\tan41^o = \tan 75^o$$
I managed to rearrange it to the form $$\tan 31^o\tan 35^o\cot 49^o = \cot 15^o\tan 23^o\cot 77^o$$
and in this form we have the intere... | First convert all ratios of the left hand side in tangent
Observe that $\tan(5\cdot77^\circ)=\cdots=\tan25^\circ$ etc.
So, let $\tan5x=\tan25^\circ\implies5x=180^\circ n+25^\circ\iff x=36^\circ n+5^\circ$ where $n$ is any integer
Like Sum of tangent functions where arguments are in specific arithmetic series or this
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Find the sum to n terms of the series Find the sum to n terms of the series
$$\frac {\sin x}{\cos x+\cos2x} + \frac {\sin2x}{\cos x+\cos4x} +
\frac {\sin3x}{\cos x + \cos6x} +\dotsb $$
How can I solve this?
Here is what I did for the first term:
$$\frac {\sin\bigl(\frac{3x}{2} - \frac x2\bigr)}
{2\cos\bigl(\frac{x}{... | So I figured out the answer :
$ S = \frac {sinx}{cosx+cos2x} + \frac {sin2x}{cosx+cos4x}+ \frac {sin3x}{cosx+cos6x} + \quad ...$
Multiply both sides by $2sin\frac x2$
Take the first term :
$$\frac {2sin\frac x2sinx}{cosx+cos2x}$$
$$\frac {cos\frac{x}2-cos\frac {3x}2} {2cos\frac {3x}2cos\frac x2}$$
$$\frac 12(sec \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expressing $1 + \cos(x) + \cos(2x) +... + \cos(nx)$ as a sum of two terms Question in title, my progress:
let $z = \cos(x) + i\sin(x)$
then $1 + \cos(x) + \cos(2x) +\dots + \cos(nx) = Re(1 + z + z^2 +\dots + z^n) = Re\left (\dfrac{1-z^{n+1}}{1-z} \right)$
by geometric series;
multiplying $\dfrac{1-z^{n+1}}{1-z}$ by $\o... | Use $\sin(a) \cos(b) = \frac{1}{2} \sin(b+a) - \frac{1}{2} \sin(b-a)$ and multiply your sum with $\sin\left(x/2\right)$.
$$
\sum_{m=0}^{n} \sin\left(\frac{x}{2}\right) \cos(m x) = \frac{1}{2} \sum_{m=0}^n \left\{\sin \left(\left(m+1-\frac{1}{2}\right)x \right) - \sin\left( \left(m-\frac{1}{2}\right)x \right) \right\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\p... | Using $e^{i x} = \cos(x) + i \sin(x)$ then for $z_{1} = e^{2 \pi i/3}$ it is seen that
\begin{align}
z_{1}^{10} + \frac{1}{z_{1}^{10}} &= e^{20 \pi i/3} + e^{- 20 \pi i/3} = e^{6 \pi i + 2 \pi i/3} + e^{-6 \pi i - 2\pi i/3} \\
&= e^{2\pi i/3} + e^{-2\pi i/3} = 2 \cos(2\pi /3) = -1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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How to solve this differential equation: $x^2dy-y^2dx+xy^2(x-y)dy=0$
$$x^2dy-y^2dx+xy^2(x-y)dy=0$$
What I tried:
$$\frac{x^2}{y^2} \frac{dy}{dx}+x(x-y)\frac{dy}{dx}=1\\$$
Let $h=-1/x, \; k=-1/y,\; dh=1/x^2 \, dx, \; dk=1/y^2 \,dy$
$$\frac{dk}{dh}+\frac{(k-h)}{k^2} \frac{dk}{dh}=1\\
\frac{dk}{dh}+\frac hk=1+\frac1k\... | $\left(\dfrac{x^2}{y^2} +x(x-y)\right)=\dfrac{dx}{dy}$
$x=vy\implies \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$
$\therefore \left(\dfrac{x^2}{y^2} +x(x-y)\right)=\dfrac{dx}{dy} \\\implies v+y\dfrac{dv}{dy}=v^2+v^2y^2-vy^2 \\ \implies y\dfrac{dv}{dy}=v^2+v^2y^2-vy^2-v\\ \implies y\dfrac{dv}{dy}=(1+y^2)v(v-1) \\\implies \displays... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Find the last two digit of $3^{3^{100}}$. Find the last two digit of $3^{3^{100}}$.
I know how to calculate if I have $3^{100}$. That I will use euler's theorem. which gives me $3^{40}\equiv 1 \pmod{100}$. And so on... but if I have $3^{3^{100}}$ what should I do?
I Tried:
Infact, i need $3^{3^{100}}\equiv x\pmod{ 100}... | If we can find $3^{3^{100}}$ in mods 4 and 25 we're done by the Chinese Remainder Theorem.
Mod 4: Note that $3\equiv -1\pmod 4$, hence $3^{3^{100}}\equiv (-1)^{3^{100}}\equiv (-1)^{2k+1}\equiv -1\equiv 3\pmod 4$.
Mod 25: Since $\varphi(25)=20$ we have $3^{3^{100}}\equiv 3^{3^{100}\pmod {20}}\pmod {25}$. Similarly, $3^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/955355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty... | If $x\lt 0$, then
$$\frac{\sqrt{6x^2-2}}{x}\not=\sqrt{\frac{6x^2-2}{x^2}}$$
because the left-hand-side is negative and the right-hand-side is positive.
Since $-x\gt 0\Rightarrow -x=\sqrt{(-x)^2}=\sqrt{x^2}$, we have
$$\frac{\sqrt{6x^2-2}}{x}=\frac{\sqrt{6x^2-2}}{-(-x)}=\frac{\sqrt{6x^2-2}}{-\sqrt{(-x)^2}}=-\sqrt{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Are there any positive integers $a, b, c, d$ such that both $(a, b, c)$ and $(b, c, d)$ are Pythagorean triples? Pythagorean triple is a triple of integers $(a, b, c)$ such that $a^2+b^2=c^2$. Is there any Pythagorean triple such that, not only $a^2+b^2$, but also $b^2+c^2$ is a square number? If not, how to prove it?
... | For the $(a,b,c)$ and $(b,c,d)$ Pythagorean triples to work, we note:
*
*$a,b,c$ is primitive (otherwise reduce by the gcd)
*likewise $b,c,d$ is primitive
*Using the traditional solution for a triple:
$2mn$ ; $m^2-n^2$ ; $m^2+n^2$, then $c$ is odd
*if $c$ is odd in the $(b,c,d)$ triple, then $b$ must be even = $2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Diophantine equation $(x^2-1)^2-4y^2=0$ We have the diophantine equation $(x^2-1)^2-4y^2=0$, where $x$ and $y$ are positive integers.
Is the only solution $x=1$, $y=0$ or can there be infinitely many solutions?
| Note that $$\begin{align}(x^2-1)^2-4y^2=0&\iff (x^2-1-2y)(x^2-1+2y)=0\\&\iff x^2-1-2y=0\ \ \text{or}\ \ x^2-1+2y=0.\end{align}$$
So, considering $x^2-1-2y=0\iff x^2-1=2y$ gives us, for example,
$$(x,y)=(2a+1,2a^2+2a)$$
for positive integers $a$.
Hence, we know that there are infinitely many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof by Induction that $3^n ≥ 1+2^n$ Use the PMI to prove the following for all natural numbers:
$3^n ≥ 1+2^n$.
I have already verified the base case but am having trouble doing so with the inductive case. Thanks!
| Suppose the result holds for the case $n$ : $3^n \geq 1 + 2^n$. Now, to complete the induction proof, we must show $ 3^{n+1} \geq 1 + 2^{n+1} $.
$$ 3^{n+1} = 3^n \cdot 3 \geq(1 + 2^n)3 = 3 + 2^n \cdot 3 = 1 + 2 + 2^n \cdot 3 = 1 + 2 + (1+2)2^n = 1 + 2 + 2^{n+1} + 2 > 1 + 2^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/966128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do I solve this recurrence relation? Given a recursive relation
$$a_n = \begin{cases}
(1 - 2b_n)a_{n-1} + b_n, & n > 1 \\
\frac{1}{2}, & n =1
\end{cases}
$$, how can I expression $a_n$ in term of $b_i, i \in \{1, 2, \dots n\}$?
| Suppose that $a_n=\frac{1}{2}$ for some $n$. Then according to the recurrence:
\begin{align*}
a_{n+1}&=(1-2b_{n+1})a_n+b_{n+1} \\
&= \frac{1}{2}-b_{n+1}+b_{n+1} \\
&= \frac{1}{2}
\end{align*}
Since $a_1=\frac{1}{2}$, by induction this shows that $a_n=\frac{1}{2}$ for all $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $10^{x^2+x}+\log{x} = 10^{x+1}$? In one of my recent exam, I was ask to solve this:
$$
10^{x^2+x}+\log{x} = 10^{x+1}
$$
My attempt to solve it was:
$$
10^{x^2+x}+\log{x} = 10^{x+1} \\
\log{x}=10^{x+1}-10^{x^2+x} \\
\log{x} = 10^{x+1}(1-10^x) \\
\log(\log{x})=(x+1)+\log(1-10^x) \\
$$
At ... | In your manipulation of the equation you factored the powers of $10$ incorrectly. Correction:
$$10^{x^2+x} + \log(x) = 10^{x+1}\\\log(x) = 10^{x+1}(1-10^{x^2 - 1})$$
We know that $x > 0$ for $\log(x)$ to be defined.
Case ($0 < x < 1$): $\log(x) < 0 < 10^{x+1}(1-10^{x^2-1})$
Case ($x = 1$): $\log(1) = 0 = 10^2(1-10^0)$... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 2
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Minimum volume cone. What would be the radius and the altitude of a right circular cone that circumscribes a sphere with a radius 8 cm if the volume of the cone is to be minimized?
Here is my rough sketch;
My idea is to write some characteristics of the cone as a function of the radius of the circle, minimize it with ... | Let $y=AC$ and $r=BD$, so $h=y+8$.
By similar triangles, $\displaystyle\frac{r}{y+8}=\frac{8}{\sqrt{y^2-64}}$ $\;\;$so $\displaystyle r^2=\frac{64(y+8)^2}{y^2-64}=\frac{64(y+8)}{y-8}$ and
$V=\displaystyle\frac{1}{3}\pi r^2h=\frac{64\pi}{3}\cdot\frac{(y+8)^2}{y-8}$.
Then $\displaystyle V^{\prime}(y)=\frac{64\pi}{3}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/967010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving by induction $5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$ is divisible by $4$ I want to prove the following twice. Once by induction then again by any other method.
$$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$$ is a multiple of 4 for all nonnegative integers n.
Let n=0 , since it is the first nonnegative integer
$$5^{3(0)}+... | $$5^{3n} + 2 \cdot5^{2n} - 5^{n} - 2=5^n\cdot5^{2n}+2 \cdot5^{2n} - 5^{n} - 2=$$
$$5^{2n}( 5^{n} +2)-( 5^{n}+2)=( 5^{n} + 2)(5^{2n} - 1)=$$
$$=( 5^{n} + 2)(5^{n} + 1)(5^{n} -1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the largest value of x given the equation... Find the largest value of $x$ for which $x^2 + y^2 + z^2 = x + y + z$.
What I did was subtract the RHS, to get $$x^2 - x + y^2 - y + z^2 - z = 0$$
$$x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} + z^2 - z + \frac{1}{4} = \frac{3}{4}$$
$$(x-\frac{1}{2})^2 + (y-\frac{1}{2... | Yes, you are correct. The point where $(y - a)^2$ and $(z - b)^2$ are zero is where we would find the maximum $x$. Anything greater than zero would make the $x$ part of the equation smaller.
Nice thinking!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum\limits_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$ This sum $\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} $showed up as I was computing the expected value of a random variable.
My calculator tells me that $\,\,\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$.
How can I show that?
I know... | As hinted by @AlexandreHalm,
$$\sum_{k=0}^\infty x^k=\frac1{1-x}.$$
Then deriving on $x$ and multiplying by $x$,
$$\sum_{k=0}^\infty kx^k=\frac x{(1-x)^2},$$
$$\sum_{k=0}^\infty k^2x^k=\frac{2x^2}{(1-x)^3}+\frac x{(1-x)^2},$$
$$\sum_{k=0}^\infty k^3x^k=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac x{(1-x)^2}.$$
Ev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/970133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
} |
Evaluate a linear system of three equations Solve for $x, y\ \text{and}\ z\ $:
$x-3z=10\\
-x+y+2z=7\\
2x+2y-5z=-8$
My working:
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
-1 & 1 & 2 & 7 \\
2 & 2 & -5 & -8
\end{array}\right) =
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 3 & 0 & -11
\end{a... | The last line should read
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 0 & 3 & -62
\end{array}\right) \begin{array}{l}
\\
\\
R_3 - 3R_2
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -52 \\
0 & 1 & 0 & -11/3 \\
0 & 0 & 1 & -62/3
\end{array}\right) \begin{array}{l}
R_1 + R_3 \\
R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does the series $\sum_{n=1}^\infty \frac{n+1}{n^3+10n}$ converge? Using the ratio test, we evaluate:
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty}\left| \frac{(n+1) + 1}{(n+1)^3 + 10(n+1)} \cdot \frac{n^3+10n}{n+1} \right| = \lim_{n \to \infty} \left| \frac{n^4 + 2n^3 + 10n^2 + 20n}{(n+... | $$
\sum_{n=1}^\infty \frac{n+1}{n^3+10n} \leq \sum_{n=1}^\infty \frac{n+n}{n^3} = 2 \sum_{n=1}^\infty \frac{1}{n^2} < \infty.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
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Differentiate the following function: $y = \frac{2x^2 + 6\sqrt{x} }{9x}$ My answer is $- \dfrac{1}{9 \sqrt{x} }$, however, the program I am using states that I am wrong. Where have I went wrong?
| $y = \frac{2x^{2}}{9x} + \frac{6x^\frac{1}{2}}{9x}=\frac{2x}{9} + \frac{2}{3}x^{-\frac{1}{2}}$.
Then $\frac{dy}{dx}=\frac{2}{9} -\frac{2}{3}⋅-\frac{1}{2}x^{-\frac{3}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/973580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Probability of a random triangle containing the center of a polygon Consider a regular polygon of $2n+1$ sides. Let three random vertices be chosen at random to get a triangle. The probability that the chosen triangle contains the center of the polygon is 5/14.
What is n?
I'm confused as to how to proceed on the cases ... | The probability seems to work out for a regular $9$-gon (so $n=4$).
Consider a circle with $9$ equally spaced points marked. Let's label the points $0, 1, 2, \dots, 8$ going clockwise around the circle. These points form our regular $9$-gon. When we choose $3$ of these vertices and form a triangle, the triangle cont... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$ I need some starting ideas, hints for proving that
$$2 \le \int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \le 3$$
I already checked that with Mathematica that numerically says that
$$\int_0^1 \ \frac{{(1+x)^{1+x}}}{x^x} \ dx \approx 2.577632915067858 $$
| This can be done with 4 observations:
$$
\lim_{x\rightarrow 0^+} \frac{(1+x)^{1+x}}{x^x} = 1
$$
$\frac{(1+x)^{1+x}}{x^x}$ has value $4$ at $x=1$.
$\frac{(1+x)^{1+x}}{x^x}$ has value $\frac{\sqrt{27}}{2}$ at $x=\frac{1}{2}$.
At $x=1$, $\frac{d^2}{dx^2} \frac{(1+x)^{1+x}}{x^x} = 4 \ln 2$.
And $\frac{d^2}{dx^2} \frac{(1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Proof of: if $x^2+y^2=2xy$ then $x=y$ I am trying to prove $x^2+y^2=2xy$ then $x=y$
What I have done is suppose $x^2+y^2=2xy$
then $x^2+y^2+(-2xy)=0\iff
x^2+(-xy)+(-xy)+y^2=0
\iff (x+(-y))\cdot x+(x+(-y))\cdot-y=0
\iff (x+(-y))^2=0$
i then square root both sides but i'm not sure if that's mathematically correct?
which ... | The claim as written is incorrect:
$$
y=\sqrt{2},x=2+\sqrt{2}\implies x^2-y^2=4+4\sqrt{2}+2-2=2\times\sqrt{2}(2+\sqrt{2})=2xy.
$$
Perhaps the problem was meant to be $x^2+y^2=2xy$ implies $x=y$. In this case
$$
0=x^2+y^2-2xy=(x-y)^2\implies x=y.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $2^{10}+5^{12}$ is composite
Prove that $2^{10}+5^{12}$ is composite
I need to solve this using only high school mathematics.
Any ideas?
| $2^{10}+5^{12}=(2^5)^2+(5^6)^2$ $=(2^5+5^6)^2 - 2\cdot2^5\cdot5^6$ $=(2^5+5^6)^2 - (2\cdot5)^6$ $=(2^5+5^6)^2 - (10^3)^2$ $=(2^5+5^6 - 10^3)(2^5+5^6 + 10^3)$ $=(15657 - 1000)(15657 + 1000)$ $=14657\times 16657$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/978222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Closed-form of $\int_0^1 B_n(x)\psi(x+1)\,dx$ Is there a closed-form of the following integral?
$$I_n = \int_0^1 B_n(x)\psi(x+1)\,dx,$$
where $B_n(x)$ are the Bernoulli polynomials and $\psi(x)$ is the digamma function.
The motivation of the problem was this question. In this answer achille hui showed that for all $n>0... | The complete answer for the question with $\,n\ge 1\,$ is the union of the answers of Start wearing purple and Olivier Oloa (the sign of the right side of the equation $(*)$ is wrong there), we get:
$$\int\limits_0^1 B_n(x)\psi(x+1)dx = (-1)^{n-1} \left(n \,\zeta’(1-n)- B_n H_{n-1}\right) + \sum\limits_{k=1}^n {\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$
*
*The below problem is taken from Joseph Edwards book Integral Calculus for beginners.
How does one show:
$$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2... | Note
\begin{align}
&\frac{\sin x}{\sin5x}=\frac1{4\left(\cos2x- \cos\frac{2\pi}{5}\right) \left(\cos2x- \cos \frac{4\pi}{5}\right)}\\=&\ \frac1{2\sqrt5}\bigg(\frac1{\cos2x- \cos\frac{2\pi}{5} }
-\frac1{\cos2x- \cos\frac{4\pi}{5} }
\bigg)\\
=&\ \frac1{4\sqrt5}\bigg(\frac1{\sin(\frac\pi5+x)\sin(\frac\pi5-x) }-\frac1{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $... | You are wrong considering $x,y\in\left[0,6\right]$.
$30+x+y$ is divisible by $9$. That's correct, but it may be $30+x+y=36$ or $30+x+y=45$. It cannot be $30+x+y=54$, since the greatest value of $\left(x+y\right)$ is $9+9=18$.
From these two options you have $x+y=6$ or $x+y=15$. Remember this.
From the division by $11$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/981544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Floor function inequality $\lfloor a\sqrt{2}\rfloor\lfloor b\sqrt{7}\rfloor <\lfloor ab\sqrt{14}\rfloor$ Let $a,b$ be positive integers. Show that $\lfloor a\sqrt{2}\rfloor\lfloor b\sqrt{7}\rfloor <\lfloor ab\sqrt{14}\rfloor$.
[Source: Russian competition problem]
| Let $a\sqrt{2} = m + \alpha$ where $m$ is an integer and $0 < \alpha < 1$, and $b\sqrt{7} = n + \beta$ where $n$ is an integer and $0 < \beta < 1$. (The fractional parts are positive since $\sqrt{2}$ and $\sqrt{7}$ are irrational.) We are to prove that
$$ (m+ \alpha)(n + \beta) - mn \geq 1.$$
We have
$$\alpha = a\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Finding the square roots of a complex number.
Express $z=4\sqrt2(1+i)$ in modulus/argument form. Hence find the two
square roots of $z$ and mark their representations on an Argand
Diagram.
So far I've worked out the mod/arg form of the complex number which is just
$$
z = 8(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) ... | Both are correct. In fact, they are the same.
$$ 2\sqrt 2 (\cos\frac{-7\pi}{8} + i\sin\frac{-7\pi}{8}) = -2 \sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8}) $$
Moreover, $-2\sqrt 2 (\cos\frac{\pi}{8} + i\sin\frac{\pi}{8})$ looks better.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Dividing a polynomial with $(x^2+1)^2$ I have been given that a polynomial $f(x)$ with real coefficients is divisible by $(x^2+1)$, and that when $f'(x)$ is divided by $(x^2+1)$, we get a remainder of $(x+1)$. I need to prove that $2f(x)+(x-1)(x^2+1)$ is divisible by $(x^2+1)^2$.
What I did was to express $f(x)$ as $f... | You can use the rule for a linear factor that $a$ is a double-root, and $(x-a)^2$ a factor, of $p(x)=0$ if and only if $p(a)=p'(a)=0$.
Use $a=i$ and $a=-i$. $(x-i)^2(x+i)^2=(x^2+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
maximum, complex quadratic function, Is my solutions correct? I'm trying to compute $\max_{|z| \le 1} |(z+2)(z-1)|$.
Here's how I do it:
$\{z \in \mathbb{C} \ | \ |z| \le 1 \}$ is compact and $f(z) = (z+2)(z-1)$ is continuous, so it suffices to look for $\max_{|z| = 1} |(z+2)(z-1)|$.
Let $z = \cos \phi + i \sin \phi, \... | setting $z=x+yi$ then we get
$|(x+2+yi)(x-1+yi)|=\sqrt{(x+2)^2+y^2}\sqrt{(x-1)^2+y^2}$ where $y^2=1-x^2$ thus you will have only one variable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/988930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Inequality of area of two triangles Let $ABC$ be a triangle with sides $a,b,c$ and $A_1B_1C_1$ be another triangle with sides $a+\frac{b}2$, $b+\frac{c}2$, $c+\frac{a}2$. Prove that:
$$\frac94[ABC]\le[A_1B_1C_1]$$
I tried using Heron's formula, but it is not getting me to the answer. Can anyone help? :)
| By Heron's formula,
$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$[A_1B_1C_1]=\frac{1}{4}\sqrt{\frac{3(a+b+c)}{2}(b+c-a+\frac{a+c-b}{2})(a+c-b+\frac{a+b-c}{2})(a+b-c+\frac{b+c-a}{2})}$
Set $b+c-a=x$, $a+c-b=y$, $a+b-c=z$,
Enough to show:
$\frac{27}{8}xyz\leq(x+\frac{y}{2})(y+\frac{z}{2})(z+\frac{x}{2}) $,
i.e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $ Prove $\lim_{n \rightarrow \infty} \frac{1}{n}\cdot \frac{3 + \frac{1}{n}}{4 - \frac{1}{n}} = 0 $
Let $\epsilon > 0$ be arbitrary. I want to find $N$ such that $n \in \mathbb{N}$ guarantees $ \left | \frac{1}{n}\cdot \f... | Since $n \in \mathbb N \implies n \geq 1$, we know that:
$$
\frac{1}{n} \leq 1 \iff 3 + \frac{1}{n} \leq 4
$$
as well as that:
$$
\frac{-1}{n} \geq -1 \iff 4 - \frac{1}{n} \geq 3
$$
Now to make a fraction smaller, we can either decrease the numerator or increase the denominator. Thus, it follows that:
$$
\epsilon > \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/991315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Using differentiation under integral sign to calculate $\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi$ I want to calculate the integral $$\int^{\pi/2}_0\frac{\log(1+\sin\phi)}{\sin\phi}d\phi$$ using differentiation with respect to parameter in the integral $$\int^{\pi/2}_0\frac{\log(1+a\sin\phi)}{\sin\phi}d\phi... | Let's consider the integral
\begin{align}I(\alpha)&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\sin\,\phi)}{\sin\,\phi}\;d\phi\quad\Rightarrow\quad\phi\mapsto \frac{\pi}{2}-\phi\\
&=\int_0^{\Large\frac{\pi}{2}}\frac{\ln\,(1+\cos\alpha\,\cos\,\phi)}{\cos\,\phi}\;d\phi, \qquad 0 < \alpha < \pi.\end{align}
Dif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 2
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Equivalence of summations Show that $$\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2}\equiv \frac{1}{n}\sum^{n}_{i=1}x_{i}^{2} - \bar{x}^{2}.$$ Note that $\bar{x} = \frac{1}{n}\sum^{n}_{i=1} x_{i}$. So I have started by:
\begin{align}
\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2} &= \frac{1}{n}\sum^{n}_{i=1} (x_{i}^{... | When you go from the 3rd line of your calculation to the last line, you made a mistake. Thus the $\frac{1}{n^3}$ should really be $\frac{1}{n^2}$.
\begin{align}
\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2} &= \frac{1}{n}\sum^{n}_{i=1} (x_{i}^{2} - 2x_{i}\bar{x} + \bar{x}^{2}) \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd. Prove: If $n = 2^k - 1$ for $k \in \mathbb{N}$, then every entry in row $n$ of pascal's triangle is odd.
I know that the $n$th row in pascal's triangle correspond to the coefficients of $(x+y)^n$:
$$\begin{align}(x+y)^n = ... | As the answer above says, $\displaystyle{2^{n}-1\choose k-1}+{2^{n}-1\choose k} = {2^n \choose k}$. So if ${2^n \choose k}$ is even for $k \neq 0,2^n$, both terms are even or both terms are odd. First, let's prove that ${2^n \choose k}$ is even.
$$\displaystyle{2^n \choose k} = \frac{2^n!}{k!(2^n-k)!}$$
For $k = 1, {2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Rewrite the expression in the form $A \sin(x+C)$ Rewrite the following expression in the form $A \sin(x+C)$
$$4 \sin x + 4\sqrt{3} \cos x$$
This is what I have so far, and I'm not even sure it's the right approach. I just dont understand this concept as a whole:
$$A \cos(c)\sin(x) + A\sin(c)\cos(x)$$
$$A\cos (c)=4$$
$... | rewrite your term in the form
$\sqrt{4^2+4^2\cdot 3}\left(\frac{4}{\sqrt{4^2+4^2\cdot3}}\sin(x)+\frac{4\sqrt{3}}{\sqrt{4^2+4^2\cdot3}}\cos(x)\right)$
now we have
$\cos(\phi)=\frac{4}{\sqrt{4^4+4^2\cdot 3}}$ and
$\sin(\phi)=\frac{4\sqrt{3}}{\sqrt{4^2+4^2\cdot3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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When do we specify restrictions in rational expressions or equations In the following equation $x \neq 0$ apply's according to wolframalpha (expanded form).
$$\frac{x^3 + x^2}{2x} = \frac{x^2 + x}{2}$$
Yet in the following equation it does not...
$$\frac{ab^2}{b} = 5 \iff ab = 5$$
Why don't we specify the restriction $... | I think that
$$\dfrac{x^3 + x^2}{2x} = \dfrac{x^2 + x}{2} \Leftrightarrow
\begin{cases}
2x\cdot \dfrac{x^3 + x^2}{2x}=2x\cdot \dfrac{x^2 + x}{2}\\
2x\neq 0
\end{cases}$$
Also $$\dfrac{ab^2}{b} = 5 \iff
\begin{cases}
b\cdot\dfrac{ab^2}{b}=b\cdot 5\\
b\neq 0 \end{cases}\iff
\begin{cases}
ab^2=5b\\
b\neq 0 \end{cases}\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$ - is there any other way to solve this? Here's a problem that I think I've managed to solve:
$$\lim_{n\rightarrow\infty} \sqrt{2^n+n^2}-\sqrt{2^n+1}$$
Here's how I did it:
$(\sqrt{2^n+n^2}-\sqrt{2^n+1})\frac{\sqrt{2^n+n^2}+\sqrt{2^n+1}}{\sqrt{2^n+n^2}+\sqrt{2^n+1}... | If you know about Taylor expansions, you can do the following: as long as $\frac{a_n}{2^n}\xrightarrow[n\to\infty]{}0$,
$$
\sqrt{2^n + a_n} = \sqrt{2^n}\sqrt{1 + \frac{a_n}{2^n}} = \sqrt{2^n}\left(1+\frac{a_n}{2\cdot2^n}+o\!\left(\frac{a_n}{2^n}\right)\right) = 2^{\frac{n}{2}}+\frac{a_n}{2^{\frac{n}{2}+1}}+o\!\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
| Denote $\sqrt{14+4\sqrt{10}}=a, \sqrt{14-4\sqrt{10}}=b$, then
$$a^2+b^2=28, ab=6$$
So $(a-b)^2=16$ and $a-b=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Convergence of a summation How do I find out what this summation converges to? I dont even know how I'd start o_o
I assume the first part converges to infinity but dont know how the cos works in this case.
$$\sum^{\infty}_{k=1} 3^{-k}\cos{k}$$
| What you have is:
$$
\sum_{k=1}^\infty \frac{\cos k}{3^k}
$$
As Daniel mentioned, Euler's formula will be useful here:
You will notice that $\cos x = \frac{1}{2}(e^{ix}+e^{-ix})$. Thus:
\begin{align}
\sum_{k=1}^\infty \frac{\cos k}{3^k} &= \frac{1}{2}\sum_{k=1}^\infty \frac{e^{ik}+e^{-ik}}{3^k} \\
&= \frac{1}{2}\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Please the box method for factoring trinomial of the form $ax^2+bx=c$ I was given this method for factoring trinomials of the form $ax^2 + bx + c$
This is the method:
Find numbers $p$ and $q$ such as $a\cdot c=p\cdot q$ and $b=p+q$
With $p$ and $q$ $(GCF(a,q)x + GCF(c,p))(GCF(a,p)x+GCF(c,q))= ax^2+bx+c$
I have no ide... | The greatest common factor is also called the greatest common divisor. I will call it the gcd for greatest common divisor.
There is an implicit assumption here that $\gcd(a, b, c) = 1$. For instance,
\begin{align*}
8x^2 + 24x + 18 & = 8x^2 + 12x + 12x + 18\\
& = 4x(2x + 3) + 6(2x + 3)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1006720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| rewrite it in the form $(\sqrt{x}-\sqrt{y})\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}=y-x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? While Solving some integral problem, I encountered the following infinite series:
$$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$
I have tried many methods including partial fractions... I s... | We have, $\displaystyle B_{2n}(x) = (-1)^{n-1}\frac{2(2n)!}{(2\pi)^{2n}} \sum\limits_{k=1}^{\infty} \frac{\cos \left(2\pi kx \right)}{k^{2n}}$
Cubing and integrating both sides,
\begin{align*}&\int_0^1 \left((-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}(x)\right)^3\,dx \\&= \int_0^1 \left(\sum\limits_{k=1}^{\infty} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 2,
"answer_id": 1
} |
Integration question does this substitution work? $$\int_{0}^{3} \frac{1}{\sqrt{3-x}}dx$$
I'm a little lost here do i let $u = 3-x$, so that $du/dx = 1$, $du=dx$
$$\int_0^3 \frac{1}{\sqrt{u}}du = u^{-\frac{1}{2}} du$$
$$ \left[\frac{1}{\frac{1}{2}} u^{\frac{1}{2}}\right]_0^3 = \left[2(3-x)^{\frac{1}{2}}\right]_0^3$$
$... | If you substitute $u = 3-x$, $du = -dx$. The bounds change as well so $x =0$ means $u=3$ and $x=3$ means $u=0$
$$\int_0^3 \frac{dx}{\sqrt{3-x}} = \int_3^0 \frac{-du}{\sqrt{u}} = \int_0^3 \frac{du}{\sqrt{u}} = \left[ 2\sqrt{u} \right]_0^3 = 2(\sqrt{3} - 0) = 2\sqrt{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving Trigonometric Limits compute the following limit:
$$\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\ \mbox{as}\ x\to0.$$
How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
but without ... | Consider first $$\cos (y+x)-\cos y=\cos y\cdot(\cos x-1)-\sin x\sin y=-2\cos y\cdot\sin^2\frac x2-\sin x \sin y$$
And $$\sin(a+x)-\sin a=\sin a\cdot(\cos x-1)+\cos a\sin x=-2\sin a\cdot\sin^2 \frac x2+\cos a\sin x$$
Then (expressing the numerator as a double-difference ie a difference of differences) $$\cos (a+2x)-2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+a)}=\frac{1}{aa!}$ $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)\cdots(n+v)}=\frac{1}{vv!}$$
I am struggling to find a solution for this but no luck yet. How can I analyze it to get to second part?
| \begin{eqnarray}
\frac{1}{n(n+1)...(n+v)}&=&\frac{1}{v}\frac{(n+v)-n}{n(n+1)...(n+v)}\\
&=&\frac{1}{v}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]
\end{eqnarray}
$$\sum_{n=1}^{\infty}\frac{1}{n(n+1)...(n+v)}
=\frac{1}{v}\sum_{n=1}^{\infty}[\frac{1}{n(n+1)...(n+v-1)}-\frac{1}{(n+1)...(n+v)}]$$
$$\frac{1}{v}\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine whether this set of vectors is linearly independent For $n \geq 1$ determine if the set $S$ is linearly independent.
The set $S$ is these vectors:
$$
v_1 \equiv(\,1,2,\ldots,n\,)\,,\quad
v_2 \equiv(\,1,2^2,3^2,\ldots,n^2\,)\,,\ldots\,
v_n \equiv(\,1,2^n,3^n,\ldots,n^n\,)
$$
I know for vectors to be linearly ... | If $n=3$ it is
$$\left|\begin{array}{rrr} 1 & 1 & 1 \\ 2 & 2^2 & 2^3 \\ 3 & 3^2 & 3^3 \end{array}\right|=\left|\begin{array}{rrr} 1 & 1 & 1 \\ 0 & 2 & 2^2 \\ 0 & 2\cdot 3 & 2\cdot 3^2 \end{array}\right|=\left|\begin{array}{rr} 2 & 2^2 \\ 2\cdot 3 & 2\cdot 3^2 \end{array}\right|=2^2\cdot 3 \left|\begin{array}{rr} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the exact closed from expression of $1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2$ I know the above expression equals to $\frac{n(2n−1)(2n+1)}{3}$, but how exactly can i come up with something from scratch?
| Most basic way in my opinion:
$$f(n)=1^2 + 3^2 + 5^2 + · · · (2n-1)^2+ (2n + 1)^ 2$$
$$f(n-1)=1^2 + 3^2 + 5^2 + · · · + (2(n -1)+ 1)^ 2=1^2 + 3^2 + 5^2 + · · · + (2n -1)^ 2$$
$$f(n)-f(n-1)=(2n + 1)^ 2$$
$$f(n)=an^3+bn^2+cn$$
$$f(n-1)=a(n-1)^3+b(n-1)^2+c(n-1)$$
$$an^3+bn^2+cn-a(n-1)^3-b(n-1)^2-c(n-1)=4n^2+4n+1$$
If you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Confusion about the expansion of $\;x(2x - 3)(3x + 4)$ For this algebraic product:
$\;x(2x - 3)(3x + 4)$
What I expected to do is:
$$\begin{align} x(2x - 3)(3x + 4) & = 2x^2 -3x(3x + 4) \\ & = 2x^2 -9x^2 -12x \\ & = -7x^2 -12x\end{align}$$
However, what the book I'm reading does is:
$$\begin{align} x(2x - 3)(3x + 4) &... | First off, to address a misunderstanding expressed in your comment:
$$a - b(c + d) \neq (a - b)(c + d)$$ Rather $$(a-b)(c+d) = a(c+d) - b(c+d) = ac + ad -bc -bd$$
Now, we address the posted problem. If proceeding left to right, $x$ needs to multiply both $2x$ and $-3$, to give $(2x^2 - 3x)$, all of which needs to then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculation of $ \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ Calculation of $\displaystyle \cos\left(\frac{2\pi}{7}\right)+\cos \left(\frac{4\pi}{7}\right)+\cos \left(\frac{6\pi}{7}\right)$ and $\displaystyle \cos\left(\frac{2\pi}{7}\right)\times \cos\left(\frac{4\... | To complete Tim Raczkowski's solution we will the address the second part of the problem. Using the identical notation as above, we have
$$\cos(2\pi/7)\cdot \cos(4\pi/7)\cdot \cos(6\pi/7)=$$$$(\omega+1/\omega)(\omega^{2}+1/\omega^{2})(\omega^{3}+1/\omega^{3})/8=\omega(\omega^2+1)(\omega^4+1)(\omega^6+1)/8=$$$$(\omega^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Prove a limit for $g(x)$ with the definition of limit only We have the function $g(x) = x^3+1$
Prove, with the definition of limit only, that the limit $L=9$ is indeed the limit of the function when $x=2$.
I started with: $\lvert g(x) - L \rvert$ : $\lvert x^3+1-9 \rvert = \lvert x^3-8 \rvert = \lvert (x-2)(x^2+2x+4) ... | HINT: $x^2+2x+4<19$ for $1<x<3$, hence we can take $\delta=\varepsilon/19$ for $\varepsilon<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$
Show
$\forall \ x \in \mathbb{R}:\quad x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$
My attemps:
Case $x=-1$
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 +... | You can write your expression as
$$x(x-1)(x^4+x^2+1)+\frac{3}{4}\ .$$
Now consider $x(x-1)(x^4+x^2+1)$: the third factor is always positive, hence the whole assume negative values only for $x\in(0,1)$. In this interval $x(x-1)$ is negative and $x^4+x^2+1<3$. The minimum of $x(x-1)$ in $(0,1)$ is $-\frac{1}{4}$ for $x=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{d^n}{dx^n} (\sin^4 x + \cos^4 x) = 4^{n-1}\cos (4x + \frac{n\pi}{2})$
*
*Question
Prove that $\frac{d^n}{dx^n} (\sin^4 x + \cos^4 x) = 4^{n-1}\cos (4x + \frac{n\pi}{2})$
*
*My attempt
*
*First calculate $\frac{d}{dx} (\sin^4 x + \cos^4 x)$, that is,
$$\frac{d}{dx} (\sin^4 x + \cos^4 x) =4\si... | Hint: Note that if
$$\cos (4x + \frac{n\pi}{2})=\cos 4x \cos \frac{n\pi}{2} - \sin 4x \sin \frac{n\pi}{2}$$
If $n$ is odd, then $n=4k-3$ or $n=4k-1$, $\forall k\in\mathbb{N}$, then $\cos\frac{n\pi}{2}=0$: $$\sin{\frac{(4k-3)\pi}{2}}=1$$ and
$$\sin{\frac{(4k-1)\pi}{2}}=-1$$
Now If $n$ is even, then $n=4k$ or $n=4k-2$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $((x+y i)-\frac{1}{x+y i})/{(2 i)} = 2$
Solve the equation $$\frac{\left((x+y i)-\frac{1}{x+y i}\right)}{2 i} = 2$$
So far, I got $(0, 2-\sqrt{3}i)$ and $(0, 2+ \sqrt{3}i)$ as solutions for $x$ and $y$. Do I require $2$ more solutions?
| $$\frac{(x+iy)-\dfrac1{(x+iy)}}{2i}=2$$
$$\iff(x+iy)-\dfrac1{(x+iy)}=2(2i)$$
$$\iff(x+iy)^2-1=4i(x+iy)$$
$$\iff x^2-y^2-1+(2xy)i=(4x)i-4y$$
Equating the real parts, $x^2-y^2-1=-4y$
Equating the imaginary parts, $2xy=4x\iff2x(y-2)=0$
If $x=0, -y^2-1=-4y\iff y^2-4y+1=0,\implies y=?$
If $y-2=0\iff y=2, x^2-2^2-1=-4(2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1025494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Indefinite integral evaluation of fraction Here's the indefinite integral to evaluate:
$F(x) = \int \left(\frac{3x + 4}{x + 3}\right)dx$
Here are my steps (verbose):
$$\begin{align}
u & = x + 3 \\
(x + 3)dx & = du \\
\left(\frac{d}{dx}(x + 3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + \frac{d}{dx}(3)\right)dx & = du \\... | Nice work!
About your misgivings: Evaluating the definite integral $F(x) \Big|_b^a$ will yield $$ 3x - 5\ln\lvert x + 3\rvert + 9 + C\Big|_b^a = \Big(3a - 5\ln|a+3| + 9 +C\Big) - \Big( 3b - 5\ln |b + 3| + 9 + C \Big)$$
$$= 3a - 5\ln|a+3| +9 +C - 3b + 5\ln |b + 3| - 9 - C]$$
You can see that upon simplification, we e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I hav... | Here is what I have so far, so hopefully someone can take it further..
The above integral has a form of
$$
-\frac{1}{8}\int \frac{df}{dx}\frac{dg}{dx}dx
$$
Where
$$
f(x) = \ln\left(1-2x^2\right)\\
g(x) = \ln\left(\frac{2}{3}\sqrt{\left(2-x^2\right)\left(1+x^2\right)} +1\right)
$$
But as always double check the manipula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
} |
GCD of polynomials by using Euclid's algorithm
Let $g = x^2 +6x -7$ and $f = x^4 - 1$. Find the GCD of $f$ and $g$.
So I started by evaluating $f/g$ and the result is $q = x^2-6x+43, r = -300x+300$.
I tried to follow the algorithm one step further but I can't see how the result ends up to be $x-1$.
| As you say, euclidean division yields
\begin{align*}
(x^4-1)
&= (x^2 - 6x + 43)(x^2 + 6x - 7) + (-300x+300)\\
&= q(x)\cdot(x^2 + 6x - 7) + r(x)
\end{align*}
Then $\gcd(x^4-1,\,x^2+6x-7)=\gcd(x^2+6x-7, -300x+300)$. You can apply again euclidean division to conclude
$$\gcd(x^4-1,\,x^2+6x-7) = -300x+300$$
But you alrea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{a+b+c}{2}$ If $a,b,c$ are positive real numbers, not all equal, prove that $$\dfrac{bc}{b+c}+\dfrac{ca}{c+a}+\dfrac{ab}{a+b}<\dfrac{a+b+c}{2}$$
Progress:
$\frac{bc}{b+c}-\frac{b+c}{2}=$
| Since $(b+c)^2 \geq 4bc$, we have
$$\dfrac{bc}{b+c} \leq \dfrac{b+c}4$$
Use the same inequality for the rest and obtain what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Range of values of $t$ for which $ 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ Calculation of Range of values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ where $\displaystyle t \in \left[-\frac{-\pi}{2}\;,\frac{\pi}{2}\right]$
$\bf{My\; Try::}$ Given $\displaystyle 2\sin t = \frac{1-2x+5x^2}... | Hint: If we assume that $x$ can be any real number, then we know that $2\sin t$ can take values between -2 and 2. So the problem is then to find what possible values of the right hand side are, assuming it can't take all possible values. Given those, work out the arcsin of the values and the possible range of $t$ from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse
(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\qu... | You can rewrite the definition of $v_n$ as
$$
v_1=2\\[2ex]
v_{n+1}=\frac{(2n+1)(2n+2)}{n}v_n
$$
so
$$
\frac{v_{n+1}}{v_n}=\frac{(2n+1)(2n+2)}{n}
$$
Another way to see this is noting that
$$
v_n=\frac{(2n)!}{(n-1)!}
$$
For $t_n$ you have
\begin{align}
t_n
&=\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{n^n}\\
&=\frac{1}{n^n}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proof that $ \forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0 $ How to proof simply that
$$
\forall x,y \in \mathbb{R} \qquad x^2+y^2+(x-1)(y-1)>0
$$
| $x^2+y^2+(x-1)(y-1)>0\iff x^2+y^2+xy-x-y+1>0\iff2x^2+2y^2+2xy-2x-2y+2>0\iff(x-1)^2+(y-1)^2+(x+y)^2>0$ which is obviously true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1037503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Eisenstein integers and applications to Diophantine equations Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?
I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.
| It is known that the Eisenstein integers $\mathbb{Z}[\omega]$ is an unique factorization domain and it has six units
$$\pm 1, \pm \omega, \pm \omega^2$$
Over $\mathbb{Z}[\omega]$, the numbers $7, 13, 19$ factorize into its prime factors as
$$\begin{cases}
7 &= (3 + \omega)(3 + \omega^2)\\
13 &= (4 + \omega)(4 + \omega... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the determinants of the given matrices Consider scalars $a,b,c,d,e,f$ such that
$\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 1 & e\\
c & 1 & f\\ \end{array} \right) = 7$ and $\det\left( \begin{array}{ccc}
a & 1 & d\\
b & 2 & e\\
c & 3 & f\\ \end{array} \right) = 11$
What is $\det\left( \begin{array}{ccc}
a & 3 &... | hint: $\left(\begin{array}{l} 3 \cr 4 \cr 5\end{array} \right) = 2 \left( \begin{array}{l} 1 \cr 1 \cr 1 \end{array} \right) + \left( \begin{array}{l} 1 \cr 2 \cr 3\end{array}\right).$
determinant is a linear function of every column.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1039042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all solutions in positive integers of the diophantine equation $w^2+x^2+y^2=z^2$ It's an exercises of the text book Elementary Number Theory and It's Applications 6th Edition by Kenneth H.Rosen. I wanted to solve it using the method in solving the diophantine equation $x^2+y^2=z^2$. But some difficult gaps showed ... | In General it is possible in various ways to write the solution to this equation:
$$x^2+y^2+z^2=q^2$$
I like such kind.
$$x=2a^2s^2-2abs^2\pm{2apbs}$$
$$y=2a^2s^2+2abs^2\pm2apbs$$
$$z=p^2b^2-a^2s^2+s^2b^2\pm2apbs$$
$$q=p^2b^2+3a^2s^2+s^2b^2\pm2apbs$$
If you want you can write infinitely many formulas are not the proble... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Finding $\lim_{x\rightarrow 0}\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}$ We know that $$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\tan({x}/{2})$$ so we can change the above function to another form as follow $$\frac{x}{2}\sqrt{\frac{1+\cos(x)}{1-\cos(x)}}=\frac{(x/2)}{\tan(x/2)}$$ If we use the Taylor series of $\tan(x/2)... | You are comparing two different functions because
$$\sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\sqrt{\tan^2\left(\frac{x}{2}\right)} = \Bigg|\tan\left(\frac{x}{2}\right)\Bigg| \not = \tan\left(\frac{x}{2}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
| $$\left(\frac{1}{2}\right)^{\frac{1}{3}}=\frac{\sqrt[3]1}{\sqrt[3]2}=\frac1{\sqrt[3]2}$$
$$\left(\frac{1}{3}\right)^{\frac{1}{2}}=\frac{\sqrt1}{\sqrt3}=\frac1{\sqrt3}$$
Now it is obvious that
$$\sqrt[3]2<\sqrt3$$
Thus
$$\frac1{\sqrt[3]2}>\frac1{\sqrt3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 9,
"answer_id": 3
} |
Show that $\sum\limits_{i=0}^{n/2} {n-i\choose i}2^i = \frac13(2^{n+1}+(-1)^n)$ While doing a combinatorial problem, with $n$ being even, I came up with the expression
$$\sum_{i=0}^{n/2} {n-i\choose i}2^i$$
for which I used wolfram to get a closed form expression of $\dfrac{1}{3}\left(2^{n+1}+(-1)^n\right)$.
Is there ... | One way is by considering generating functions.
Denoting $\displaystyle a_n = \sum_{k=0}^{n} 2^k{n-k\choose k}$
(where, we use the notation $\displaystyle \binom{n}{m} = 0$ for $m > n$)
$\displaystyle \begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty x^n\sum_{k=0}^n2^k\binom{n-k}{k} =\sum_{k=0}^\infty\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Number theory: $x^y + 1 = y^x$ Today a friend told me the equality:
$2^3 + 1 = 3^2$, and I wondered if there exist more solutions to the general problem $$x^y + 1 = y^x$$ where $x$ and $y$ are integers. Some research led me to the result for $x^y = y^x$, which has no integer solutions except $x = 2$ and $y = 4$ (assume... | Consider function $f:(0,\infty)\rightarrow \mathbb{R}, f(x)=(1+\frac{n}{x})^x$ which is strictly increasing and $f(x)<e^n$.
I. $x<y, y=x+n, n>0$, integer
$$\begin{align} x^y+1=y^x
\implies & x ^{x+n}+1=(x+n)^x \\
\implies & x^n+\frac{1}{x^x}=(1+\frac{n}{x})^x<e^n \\
\implies & x<e \\
\implies & x=1, x=2\end{align}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1042371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Investigate convergence of the following series Investigate the corvergence of the following series?
$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13}+\frac{1}{15}-\frac{1}{8}+ \ldots$
$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{... | For the first one, you have consecutive triples
$$\frac1{4n-1}-\frac1{2n}+\frac1{4n+1}=\frac{1}{2n(4n-1)(4n+1)}\sim\frac1{32n^3}$$
The sum is thus convergent.
For the record, the denominators of the terms in the sum are, up to sign, this sequence in OEIS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1046616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
I need help with a limit problem I need help with the following limit:
$\lim_{n \to \infty}\left [ (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right ) \right ],a\in \mathbb{N}$
I am completely lost.
| Let us consider $$A= (n^2+1)\left ( \left ( 1+\frac{2}{n^2} \right )^a - 1 \right )$$ In a first time, let us focus on $(1+x)^a$ and use the generalized binomial theorem for generating a Taylor-like series when $x$ is small. So $$(1+x)^a=1+a x+\frac{1}{2} (a-1) a x^2+\frac{1}{6} (a-2) (a-1) a x^3+O\left(x^4\right)$$ No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$ How can I prove this equality?
$$\left(\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}\right)\left(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x+\cdots}}}}\right)=x$$
| Let $\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}=y\implies x+y=y^2\implies y^2-y-x=0\implies y=\dfrac{1\pm\sqrt{1+4x}}2$
As $x>1,y>1\implies y=\dfrac{1+\sqrt{1+4x}}2$
$\sqrt{x-\sqrt{x-\sqrt{x-\cdots}}}=z\implies x-z=z^2\iff z^2+z-x=0\implies z=\dfrac{-1\pm\sqrt{1+4x}}2$
As $x>1, z<1\implies z=\dfrac{-1+\sqrt{1+4x}}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1051744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
contractive and find its limit The real sequence $\{P_n\}$ is defined as $p_1$=2, $p_{n+1}$=$\frac{2}{1+p_n}$, n $\in N$. prove that $\{p_n\}$ is contractive, deduce that it converges, and find its limit.
first we need to show that $\frac{2}{3}$ $\leq$ $p_n$$\leq$2
pf:
as $p_1$ = 2, $p_2$=$\frac{2}{3}$,.. so that $\fr... | You are quite close. If we can show that $p_k\ge \frac{2}{3}$ for all $k$, we will have, by a calculation similar to yours, that
$$|p_{n+1}-p_n|=\left|\frac{2}{1+p_n}-\frac{2}{1+p_{n-1}}\right|=\frac{2|p_n-p_{n-1}|}{(1+p_n)(1+p_{n-1})}.$$
But since $p_k\ge \frac{2}{3}$ for all $k$, it follows that $(1+p_k)(1+p_{n-1})\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\int \frac{x\ln(x)}{\sqrt{x^2 - 1}}$ My Work
$x = \tan\theta$
$dx = \sec^2\theta d\theta$
$\int \frac{\tan\theta\ln(\tan\theta)}{\sqrt{\tan^2\theta - 1}}\sec^2\theta d\theta$
$\int \sec\theta \tan\theta ln(tan\theta)$
$u = \ln(\tan\theta)$
$du = \frac{\sec^2 \theta}{\tan \theta}$
$dv = \sec \theta \tan \theta d... | You can evaluate the integral without using secant substitution. Here is the way.
Since we know that
\begin{gather*}
\mathrm{d}\sqrt{x^2-1}=\frac{x}{\sqrt{x^2-1}}\mathrm{d} x,
\end{gather*}
we can calculate, by differentiation by parts, as follows,
\begin{align*}
&\quad \frac{x\ln(x)}{\sqrt{x^2-1}} \mathrm{d} x=\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1053868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Is $SL(2, 3) $ a subgroup of $SL(2, p)$ for $ p>3$? As the title says, I was wondering whether $SL(2,3)$ is a subgroup of $SL(2,p)$ for $p>3$. I know that it is for $p=5$ (it can be found explicitly using the quaternionic representation), and I have some evidence that it is for the other $p$'s, but I'm not sure how I ... | $SL_2(3)$ is the semidirect product of the quaternion group
$Q = \langle i,j\mid i^4, j^4, i^2=j^2, i^{-1}ji=j^{-1}\rangle$
with a cyclic group $Z_3 = \langle z\rangle$ of order $3$ via
$i^z = j$ and $j^z = ij$.
Hint: Find a subgroup of $SL_2(p)$ ($p$ any odd prime) isomorphic to
$Q$, and then extend it to full group $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1054917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
How to find the coefficient of a power in a power series?? How can I find the coefficient of $x^{80}$ in the power series $$(1+x+x^{2}+x^{3}+x^{4}+\cdots)(x^{2}+x^{4}+x^{6}+x^{8}+\cdots)(1+x^{3}+x^{5})\,?$$
Is there a general method to this?
| We can use the fact that
\begin{align}
1+x+x^2+\dotsb&=\frac{1}{1-x}\\
x^2+x^4+x^6+\dotsb&=\frac{x^2}{1-x^2}
\end{align}
so the expression is
$$
\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}
$$
and we can try doing partial fraction decomposition:
$$
\frac{x^2(1+x^3+x^5)}{(1-x)^2(1+x)}=
x^4+x^3+3x^2+3x+5+\frac{6x^2+2x-5}{(1-x)^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
fixed points for the following system I'm trying to find the fixed point for the system (see document attached) but it seems so hard and I don't know what Im doing wrong. Can somebody help me with this. I need to find the to look for the value where the bifurcation occur (for that process i need to evaluate the jacobia... | We need to simultaneously solve:
$$\begin{align*} a(1 - u) - u v^2 = 0 \\ u v^2 -(a + k) v = 0 \end{align*}$$
From the first equation we get:
$$u = \dfrac{a}{a+v^2}$$
Substituting this into the second equation yields:
$$-(a+k) v + \dfrac{a v^2}{a + v^2} = -(a+k)(v)(a + v^2) + av^2 = 0$$
Expanding and simplifying, yield... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Formula for a periodic sequence of 1s and -1s with period 5 I've been playing with periodic sequences of 1s and -1s lately. This is what I came up with:
\begin{eqnarray*}
-(-1)^n& = &1, -1, 1, -1,\ldots\quad\textrm{(Period 2)}\\
\left(-(-1)^n\right)^{\frac{n+2}{2}} & = &1, 1, 1, -1, 1, 1, 1, -1,\ldots\quad\textrm{(Peri... | We have
$$
\frac{4\cos\left(\frac{2\pi k}5\right)+4\cos\left(\frac{4\pi k}5\right)-3}5=
\left\{\begin{array}{}
-1&\text{if }k\not\equiv0\pmod5\\
+1&\text{if }k\equiv0\pmod5\\
\end{array}\right.\tag{1}
$$
Explanation
The roots of $z^5-1$ are $z=e^{2\pi ik/5}$ for $k\in\{0,1,2,3,4\}$. Vieta says that the coefficient of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Determining the last two digits of $229^{10} +37^{10}$ Determine the last two digits of: $229^{10}+37^{10}.$
I do not want to use the Euler-totient function or the carmichael function please!
Thanks
| We need $229^{10}+37^{10}\pmod{100}$
As $100=4\cdot25$ where $(4,25)=1$
$229\equiv1\pmod4\implies229^{10}\equiv1;37\equiv1\pmod4\implies37^{10}\equiv1$
$\implies229^{10}+37^{10}\equiv1+1\pmod4\equiv2\ \ \ \ (1)$
Again, $229\equiv4\pmod{25}\implies229^{10}\equiv4^{10}$
But $4^{10}=(2^2)^{10}=2^{20}$
and $37\equiv12\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the limit of function using Taylor series Good evening,
I'm somehow stuck on solving some easy exercises :
$$\lim_{x\to\infty} x^{3/2}\bigl(\sqrt{x+1}+\sqrt{x-1}-2\,\sqrt{x}\bigr)$$
| $$\sqrt{x+1}-\sqrt{x}=\frac{1}{\sqrt{x}+\sqrt{x+1}},\qquad \sqrt{x-1}-\sqrt{x}=-\frac{1}{\sqrt{x}+\sqrt{x-1}},$$
hence:
$$\sqrt{x+1}+\sqrt{x-1}-2\sqrt{x}=\frac{\sqrt{x-1}-\sqrt{x+1}}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x-1})}=\frac{-2}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}+\sqrt{x-1})(\sqrt{x-1}+\sqrt{x+1})}\approx\frac{-2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1068119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A faster way of calculating this determinant? I'm doing a problem involving Cramer's rule, and one of the determinants I have to work with is as follows:
\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3
\end{vmatrix}
So I started off by getting the matrix to a triangular matrix so I can just take the product of the diagonal... | You can compute the determinant of a generic $3\times 3$ matrix using a neat trick, if we have:
$$\mathbf{A}=\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$
Then we have the sum of the diagonals (highlighted in green) minus the sum of the antidiagonals (highlighted in red) as follows:
$$\det(\mathbf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$ Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$. For which $x$ does it converge to $f(x)$?
This is what I have been able to do so far
$$f(x)=\arctan(x)=\int \frac{1}{1+x^2}\, dx=\int \frac{1}{1+x^2+1-1}\,dx$$
Any hints please???
| I bet the OP was asking to find the values of the derivatives in $x=1$.
This is clearly the same as finding the derivatives in $x=1$ for $f(x)=\frac{1}{x^2+1}$. Since:
$$\frac{1}{x^2+1}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right),\tag{1}$$
we have that:
$$\frac{d^n}{dx^n}\frac{1}{x^2+1}=\frac{n!(-1)^n}{2i}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1069869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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More rigorous method for this elementary problem? The problem is:
Find all real values of $x$ such that $$(5+2\sqrt{6})^x+(5-2\sqrt{6})^x=2\sqrt{3}$$
One solution I received was as follows:
$5+2\sqrt{6}$ can be expressed as $(\sqrt{3}+\sqrt{2})^2$, and $5-2\sqrt{6}$ as $(\sqrt{3}-\sqrt{2})^2$. This means we want to ... | Answer is attributed to ah-huh-moment's answer.
Since $5+2\sqrt{6}=\dfrac{1}{5-2\sqrt{6}}$ letting $y=5-2\sqrt{6}$ gives
$$\frac{1}{y^x}+y^x=2\sqrt{3}$$
which becomes
$$y^{2x}-2\sqrt{3}y^x+1=0$$
so
$$y^x=\sqrt{3}\pm \sqrt{2}$$
Plugging $y=5-2\sqrt{6}$ back in and taking $\log$ of both sides give
$$x=\log_{5-2\sqrt{6}}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1071625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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How to solve these equations for x and y.. equations are
$(x-y)(x+2y)(2x+y) = 20$
and
$x^2+xy+y^2 = 7$
i want the METHOD not the solutions
| Equation 2 is the equation of an ellipse. Basically, any equation
\begin{eqnarray}
Ax^2 + Bxy + Cy^2 + Dx + Ey + C = 0
\end{eqnarray}
can be translated, by a change of variable, into:
\begin{eqnarray}
\left(\frac{\tilde{x}}{a}\right)^2 + \left(\frac{\tilde{y}}{b}\right)^2 = 1
\end{eqnarray}
Leading to:
\begin{eqnarray}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1073177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
If we know $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, how to find $x^4+y^4+z^4$? Let $x$, $y$, and $z$ be such that
$$\begin{align*}
x+y+z&=1\\
x^2+y^2+z^2&=2 \\
x^3+y^3+z^3&=3
\end{align*}$$
Then $x^4+y^4+z^4=?$
| We can use $$(x+y+z)^2=2\sum xy+(x^2+y^2+z^2)\ \ \ \ (1)$$
$$\sum x^3-3xyz=(x+y+z)(\sum x^2-\sum yz)$$
$$\iff\sum x^3-3xyz=(x+y+z)\{(x+y+z)^2-3\sum xy\}\ \ \ \ (2)$$
to find $xyz=c$(say)$,\sum xy=b$(say)
Then $x,y,z$ are the roots of $t^3-(1)t^2+bt-c=0$
$\implies t^4=t^3-bt^2+ct$
$\implies\sum x^4=\sum x^3-b\sum x^2+c\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How find this integral $I=\int_{-1}^{1}\frac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}$ Show this integral
$$I=\int_{-1}^{1}\dfrac{dx}{\sqrt{a^2+1-2ax}\sqrt{b^2+1-2bx}}=\dfrac{1}{\sqrt{ab}}\ln{\dfrac{1+\sqrt{ab}}{1-\sqrt{ab}}}$$
where $0<a,b<1$
my idea:
Let
\begin{align*}&(-2ax+a^2+1)(-2bx+b^2+1)=4abx^2-2(a+b+a^2b+b^2a)x+... | Notice for $|t| < 1$, $\frac{1}{\sqrt{1 - 2xt + t^2}}$ is the generating function for the Legendre polynomials:
$$\frac{1}{\sqrt{1 - 2xt + t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{*1}$$
It is known that for $x \in [-1,1]$, $|P_n(x)| \le P_n(1) = 1$. This means as long as $t$ is fixed and $|t| < 1$, the absolute values... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1075624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Spaces of polynomials
Let $p_{1}(x)=-2x+2$, $p_{2}(x)=x+2$, $p_{3}(x)=x^{2}+2x+3$, $p_{4}(x)=x^{2}-x+3$.
a) From the above four polynomials, determine a linearly independent subset that spans the polynomials.
b) What is the dimension of the (sub)space that this basis spans?
c) Does this basis span the space of the sec... | (a) Let $\vec p_i$ be the coefficient vector of $p_i(x)$. Then,
$$M=
\begin{bmatrix}
\vec p_1 & \vec p_2 & \vec p_3 & \vec p_4
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 1 & 1 \\
-2 & 1 & 2 & -1 \\
2 & 2 & 3 & 3
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 1
\end{bmatrix}
.$$
Colum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1081024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do I get $0.098765432098765432...$ when I divide $8$ by $81$? I got this remarkable thing when I divided $16$ by $162$, or, in a simplified version, $8$ by $81$. It's $0.098765432098765432\cdots$, or more commonly known as $0.\overline{098765432}$, with all the one-digit numbers going backwards...except for $1$. ... | More generally, in base $b$ (any integer $>1$)
$$ \dfrac{b-1}{b^2} + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} =
\dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}$$
so that
$$ \eqalign{
\dfrac{b-1}{b^2}& + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} +
\dfrac{0}{b^{b}} + \dfrac{b-1}{b^{b+1}} + \dfrac{b-2}{b^{b+3}} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1083079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
Find $\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$ $$\int\frac1x\sqrt\frac{1-x}{1+x}\ dx$$
How to integrate?
I tried the substitution $x=\sin\theta$ but didn't work.
| Let $ x = \sin \theta$
$$
\int \frac{1}{\sin \theta} \cdot \sqrt \frac{1-\sin \theta}{1 + \sin \theta} \cdot \cos \theta \; d \theta \;\;= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \sqrt\frac{(1-\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)} \; d\theta
\\
= \;\; \int \frac{\cos\theta}{\sin\theta}\cdot \frac{1-\sin\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator
It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$?
I did the following:
$$\begin{array}
& 5 \ \text{mod} \ 7=& &5 \\
55 ... | $$\frac{5}{9}\left(10^{1000}-1\right) \equiv \frac{-2}{2}\left(10^{4}-1\right) \equiv 1-3^4 \equiv -80 \equiv 4\pmod{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 5
} |
Solve $\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\frac 3 2$ Solve the following trigonometric eqation where $\alpha, \beta, \gamma$ are angles in a triangle ($\alpha + \beta + \gamma = 180$):
$$\sin \frac{\alpha - \beta}2 + \sin \frac{\alpha - \gamma}2 + \sin \frac{3\alpha}2 =\f... | i think the answer is $\alpha = 100^\circ, \beta = 40^\circ$ and $\gamma = 40^\circ$
here is my attempt at the solution. i will use a change of variable
$$b = \beta+{1 \over 2}\gamma,\ c = \gamma + {1 \over 2}\beta$$ and the following inverse relations
$$\alpha = 180^\circ - {2 \over 3}(b+c), \ \beta = {4 \over 3}(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.
| Since $x$ is the sum of three square roots, $x$ is positive. Thus there exists a positive real $y$ such that $x=y^2-2$.
The equation then becomes
$$y^2-2=\sqrt{y^2-1}+\sqrt{y^2}+\sqrt{y^2+1}$$
$$y^2-y-2=\sqrt{y^2-1}+\sqrt{y^2+1}$$
Squaring gives
$$y^4-2y^3-3y^2+4y+4=2y^2 + 2\sqrt{y^4-1}$$
$$y^4-2y^3-5y^2+4y+4=\sqrt{4y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1089589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
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.