Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Showing that the inflection points of $x\sin x$ lie on a certain curve
Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$
I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?
| $f(x)$ has an inflection point only if $f''(x)=0.$
$$f'(x) = \sin x + x \cos x$$
$$f''(x) = \cos x + \cos x -x \sin x=2\cos x - f(x)$$
$$f''(x) = 2\sqrt{1-\sin^2 (x)} - f(x)$$
$$f''(x) = 2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$0=2\sqrt{1-f(x)^2/x^2} - f(x)$$
$$y=2\sqrt{1-y^2/x^2}$$
$$y^2 = 4-4y^2/x^2$$
$$...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Inequality $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$
prove that for $a,b,c$ being positives and $a+b+c=1$:$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}-\frac{c}{a+b}-\frac{a}{b+c}-\frac{b}{a+c}\ge 3/2$$
This is a very interesting inequality which i came upon accidentally.We also s... | First inequality:
$$ LHS = \sum_{cyc} \dfrac{bc}{a(a+c)} = \sum_{cyc} \dfrac{(bc)^2}{a^2bc+abc^2} \ge \dfrac{(ab+bc+ca)^2}{ 2abc(a+b+c)} \ge \dfrac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Finding $\frac{\sum_{r=1}^8 \tan^2(r\pi/17)}{\prod_{r=1}^8 \tan^2(r\pi/17)}$ I have tried to wrap my head around this for some time now, and quite frankly I am stuck.
Given is that :
$$a=\sum_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right) \qquad\qquad b=\prod_{r=1}^8 \tan^2\left(\frac{r\pi}{17}\right)$$
Then what is the va... | Here is a solution using the observation in Cauchy's proof of Basel problem.
Lemma. We have
$$ \prod_{k=1}^{n} \left( t - \tan^2\left(\frac{k\pi}{2n+1}\right)\right) = \sum_{k=0}^{n} (-1)^{n-k}\binom{2n+1}{2k+1} t^k. \tag{*} $$
Using this, we immediately know that
$$ a = \sum_{k=1}^{n} \tan^2\left(\frac{k\pi}{2n+1}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can you find the limit of this: $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$? $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$
I would start with expanding it with $* \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\co... | I'm going to write it out.
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}} * \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}$
$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Limit of a product with growing number of factors I'm trying to solve the following limit:
$$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$
Since originally, the limit yields an $1^\infty$ indeterminate expression, my first id... | $\require{cancel}$
$$L=\lim_{n\to\infty}\frac{\cancel{n^{2n}}(1-\frac1{n^2})(1-\frac2{n^2})\cdots(1-\frac{n}{n^2})}{\cancel{n^{2n}}(1+\frac1{n^2})(1+\frac3{n^2})\cdots(1+\frac2n-\frac1{n^2})}=\lim_{n\to\infty}\frac{e^{\frac{-1}{n^2}}e^{\frac{-2}{n^2}}...e^{\frac{-n}{n^2}}}{e^{\frac{1}{n^2}}e^{\frac{3}{n^2}}...e^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove that $n!>2^n$ holds for all $n>3$? I suspect something with sets since one with cardinality $n$ has $n!$ permutations and its powerset contains $2^n$ elements. It could also involve binomial coefficients because of$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n\\1\end{pmatrix}+\begin{pmatrix}n\\2\end{p... | The number $n!$ is the product of the $n-1$ numbers $2,3,\ldots,n$, each of which is greater than or equal to $2$ (and all of them but $2$ is actually greater than $2$) and, since $n>3$, at least one of them is greater than or equal to $4$. So, $n!>2^{n-2}\times4=2^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$ $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.
The problem is to actually find... | $$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}=\\ \frac{\sqrt{1+a_n^2}-1}{a_n}.\frac{\sqrt{1+a_n^2}+1}{\sqrt{1+a_n^2}+1}=\\\frac{1+a_n^2-1}{a_n(\sqrt{1+a_n^2}+1)}=\\
\frac{a_n}{(\sqrt{1+a_n^2}+1)}\leq \frac{a_n}{(\sqrt{1}+1)}=\frac{a_n}{2} $$so $$0<a_{n+1}\leq \frac 12a_n\leq\frac14 a_{n-1}\leq\frac18a_{n-2}\cdots<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Minimizing $\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$ over positive reals with $a+b=1$. Why is the minimum not $18$?
If $a,b \in R^+$ such that $a+b=1$, then find the minimum value of $$\left(a+\frac{1}{a}\right)^{2}+\left(b+\frac{1}{b}\right)^{2}$$
We can write $$\left(a+\frac{1}{a}\right)^{2}+\... | The issue is in the first line of simplification.
$$a+\frac{a+b}{a}=a+1+\frac{b}{a}\neq 2+\frac{b}{a}.$$
If you make the correct simplification, you can continue
$$\sqrt{\frac{\left(a+1+\frac ba\right)^2+\left(b+1+\frac ab\right)^2}{2}}\geq \frac{a+b+2+\frac ab+\frac ba}{2}=\frac{3+\frac ab+\frac ba}{2}\geq \frac 52,$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is this method right to solve some equations of degree 2? Given this equation
$-x^2+x+6=4$
We can write it as
$(3-x)(x+2) = 4$
So weridly if I take
$(3-x) = 4$ and $(x+2) = 4$, I can get $x=-1$ and $x=2$ that are indeed the correct solutions.
Why does this happen? And in which cases can this occur?
| Consider $(a-x)(x-b) = c$
With roots $r_1$ and $r_2$.
We need four equalities to be true
$$a-r_1 = c, \quad r_1-b = 1, \quad a-r_2 = 1, \quad r_2-b = c$$
We get
$$\text{$r_1=a-c=b+1$ and $r_2=a-1 = b+c$}$$
Which implies $c=a-b-1$
So for any $a$ and $b$
$$(a-x)(x-b) = a-b-1$$
will behave as you described.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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find the range of $x$ on which $f$ is decreasing, where $f(x)=\int_0^{x^2-x}e^{t^2-1}dt$ I want to find the range of $x$ on which $f$ is decreasing, where
$$f(x)=\int_0^{x^2-x}e^{t^2-1}dt$$
Let $u=x^2-x$, then $\frac{du}{dx}=2x-1$, then $$f'(x)=\frac{d}{dx}\int_0^{x^2-x}e^{t^2-1}dt=\frac{du}{dx}\frac{d}{du}\int_0^{x^2-... | $f(x)=\int_{0}^{x^2-x} e^{t^2-1} dt \implies f'(x)= (2x-1) e^{(x^2-x)^2-1} >0 ~if~ x>1/2$. Hence $f(x)$ in increasing for $x>1/2$ and decreasing for $x<1/2$. Yes you are right. there is a min at $x=1/2$. This one point does not matter, you may also say that $f(x)$ is increasing in $[1/2,\infty)]$ and decreasing on 4(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3931246",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 0
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Find maxima and minima of $f(x,y) = x^3 + y^3 -3x -3y$ in $x + 2y=3$ I need to find max and min of $f(x,y)=x^3 + y^3 -3x -3y$ with the following restriction: $x + 2y = 3$.
I used the multiplier's Lagrange theorem and found $(1,1)$ is the minima of $f$. Apparently, the maxima is $(-13/7, 17/7)$ but I could not find it v... | This is from your working -
$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$
$3x^2 - 3 = \lambda, 3y^2-3 = 2\lambda$
Equating $\lambda$ from both equations,
$6x^2-6 = 3y^2-3 \implies 2x^2 - y^2 = 1$
Substitute $x$ from $x+2y = 3$
$2(3-2y)^2 - y^2 = 1$
$\implies 7y^2 - 24y + 17 = 0 \, $ or $(7y-17)(y-1) = 0$
Can you take it from he... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3931807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Determine the linear function $f: R \rightarrow R$ There's a linear function $f$ whose correspondence rule is $f(x)=|ax^2-3ax+a-2|+ax^2-ax+3$. State the values of the parameter $a$ that fully define the function $f$.
According to the problem condition the function $ f (x) $ is linear by
so the coefficient that accompan... | If there's some $x_0$ such that $ax_0^2-3ax_0+a-2>0$, then there's some neighborhood around $x_0$ such that $\forall x \in (x_0-\delta, x_0+ \delta)$, $ax^2-3ax+a-2>0$, $f(x)$ is quadratic unless $a=0$. On the other hand if $a=0$ then $f(x)$ is indeed linear.
If $a\ne 0$, it follows that $ax^2-3ax+a-2 \le 0, \forall x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3933777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
| According to the Vieta's formulas and the proposed relation, one has that
\begin{align*}
\begin{cases}
x_{1} + x_{2} + x_{3} = 3\\\\
2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3} = k\\\\
2x_{1}x_{2}x_{3} = 1\\\\
x^{2}_{1} + x^{2}_{2} + x^{2}_{3} = 6
\end{cases}
\end{align*}
Consequently, we have that
\begin{align*}
3^{2} = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel.
Two identical squares $ABCD$ and $PQRS$, overlapping each other in such a way that their edges are parallel, and a circle of radius $(2 - \sqrt{2})$ cm covered within these squares. Find the length of $AD$.
Wh... | Using the fact that the side of the outer square and diagonal of the inner square are equal to the diameter of the circle,
$AD = \frac{AL + PE}{2} = \frac{2r + r\sqrt2}{2} = \frac{(2 + \sqrt2)(2 - \sqrt2)}{2} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A question of Roots of unity
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
\cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.
I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.
| Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
The roots of $$\dfrac{z^9-1}{z-1}=0$$ are $e^{2\pi ir/9}; r=1,2,3,4,5,6,7,8$
Dividing both sides $z^4,$
$$z^4+\dfrac1{z^4}+z^3+\dfrac1{z^3}+z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$
$z+\dfrac1z=2\cos\dfrac{2\pi r}9=2y$(say)
Use
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding recursive formula for repeated term on sequence I have a sequence
$$3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,...$$
The pattern is so difficult. The best i can do is knowing that after jumping to the "next number" it has length $2+2n$. I mean i have $3$ as initial value. And first repeated term has lenght $2+2(1)=4$.... | $$a_0=3;\;a_n=a_{n-1}+\left\lfloor -\text{frac}\left(\frac{1}{2} \left(\sqrt{4 n+9}-3\right)\right)\right\rfloor +1$$
where $\text{frac}(x)$ is the fractional part of $x$.
From $0$ to $20$ we have
$$\{3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,6,6,6\}$$
This works because $\frac{1}{2} \left(\sqrt{4 n+9}-3\right)$ is integer w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$... | You can correct the error that Ryan points out. Instead, note that
$$a^3 + b^3 = (a + b)^3 - 3ab(a + b) = 8 - 6ab.$$
So,
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
becomes
$$a^5 + b^5 + 5ab(8 - 6ab) + 20a^2b^2 = 32.$$
Using $a^5 + b^5 = 32$, then expanding and collecting like terms,
$$40a^2b^2 - 10ab = 0 \implies ab = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 2
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Evaluating $\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} dx$ This question is not a homework question, but I just searched it on WolframAlpha and just want to know how it's done. WolframAlpha lists the following:
$$\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} \mathrm dx = \frac{\sqrt{\pi}\Gamma (\frac{4}{3})}{\Gamma (\frac{5}{6})}$$
H... | Substituting $v=-x$ and $u=v^3$ gives
$$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}=\int_0^1\frac{dv}{\sqrt{1-v^3}}=\frac{1}{3}\int_0^1 u^{-\frac{2}{3}}(1-u)^{-\frac{1}{2}}du=\frac{1}{3}\beta\left(\frac{1}{3},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{2}\right)}{3\Gamma\left(\frac{5}{6}\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $x^d \equiv a \pmod{29}$ using primitive roots Can anyone please detail the general approach to questions of the form $x^d \equiv a \pmod{29}$? For example, Wolfram Alpha states that $x^5 \equiv 8 \pmod{29}$ has one solution, $x^4 \equiv 4 \pmod{29}$ has no solution, and $x^7 \equiv 12 \pmod{29}$ has many solut... | $(x^5)^{17}\equiv x^{85}\equiv (x^{28})^3x\equiv x\bmod 29$,
so if $x^5\equiv8\bmod29$, then $x\equiv8^{17}=2^{51}=2^{28}2^{23}\equiv2^{23}\equiv10\bmod29$.
If $x^4\equiv4=2^2\bmod29$ had solutions, then $1\equiv x^{28}\equiv (x^4)^7\equiv(2^2)^7=2^{14}\bmod29$, so $2$ would not be a primitive root.
$2^7\equiv12\bmod29... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$ Let $z=e^{i\frac{2\pi}{7}}$
Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$
How do I solve from here?
| Since $$100\equiv _7 2$$ and $$300\equiv _7 -1$$ and $$500\equiv _7 3$$ We have \begin{align} &=z^{2}+z^{-2} + z^{-1}+z^{1} + z^{3}+z^{-3} \\
&= {z^5+z+z^2+z^4+z^6+1\over z^3}\\
& ={z^6+z^5+z^4+\color{red}{z^3}+z^2+z+1-\color{red}{z^3}\over z^3}\\ &= {{z^7-1\over z-1} -z^3\over z^3} \\
&= {0-z^3\over z^3} =-1\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$ Got limitation need superemum. I need to find supremum of:
$$ A: \{\frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2m)^2+(n^3 + 3n^2)^2}, n, m \in \mathbb{N} \}$$
I found out that:
$$ \frac{m^2n^3 + 3m^2n^2+2mn^3+6mn^2}{(m^2 + 2... | This is not an answer, but something you could try. You could still use differentiation. Replace the function with a continuous one,
$$f(x,y)=\dfrac{x^2y^3+3x^2y^2+2xy^3+6xy^2}{(x^2+2x)^2+(y^3+3y^2)^2},$$
Calculate $\max_{(x,y)\in\mathbb{R}^2}f(x,y)$. Then calculate the values near it which correspond to integers and p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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3D transform cylinder to make center line z axis I have an array of points that represent the surface of a cylinder, I was able to calculate the center line of this cylinder but it is translated and rotated away from origin. I want to transform this cylinder to make the center line become the Z axis, I remember it was ... | With the rotation matrix
$$R=\left(
\begin{array}{ccc}
\frac{1}{6} \left(\sqrt{3}+3\right) & \frac{1}{6} \left(\sqrt{3}-3\right) & -\frac{1}{\sqrt{3}} \\
\frac{1}{6} \left(\sqrt{3}-3\right) & \frac{1}{6} \left(\sqrt{3}+3\right) & -\frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$ I found the following exercise in a problem book (with no solutions):
Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$
I tried AM-GM for the fraction on the LHS but got stuck ... | By Schur's inequality and AM-GM $$\begin{align*}x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(x+z)&\\ \ge 2( {(xy)}^{3/2}+{(yz)}^{3/2}+{(zx)}^{3/2})\end{align*}\tag S$$
Now as $$\frac{1}{1+x}\le \frac{1}{2\sqrt{x}}$$ we have to prove $$a+b+c+3\sqrt[3]{abc}\ge 4\left(\frac{1}{2\sqrt{a}}+\frac{1}{2\sqrt{b}}+\frac{1}{2\sqrt{c}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Application of Induction in the analyze of the convergence a sequence defined recursive.
Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.
My approach: Note that $$a_{n ... | Update: Thanks Brian M. Scott for your insight.
I'll add the case where some $a_k=\frac 43$. Per Brian, we need to solve for the sequence $b_k$ such that $b_1=\frac 43$, $b_{k+1}=\frac{4b_k-1}{3b_k}$. This can be solved in a similar fashion, but easier because $b_1$ is given.
Note that
$$
b_{k+1} - 1 = \frac{b_k-1}{3b_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Surface integral on $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$ Let $S=\{(x,y,z)|x^2+y^2+z^2=1,x+y+z\leq 1\}$, $F(x,y,z)=(x,0,-x)$ and $n(x,y,z)$ be the unit normal vector of $S$ such that $n(0,0,-1)=(0,0,-1)$.
I want to evaluate $\displaystyle \iint_{S}F(x,y,z)\cdot n(x,y,z)dS$.
My Attempt
Let $f(x,y,z)=x^2+y^2+z^2-1$. ... | I will present three ways of tackling this problem.
$\textbf{Option 1}$: Directly
Parametrizing spherical coordinates as usual we can find the bounds by examining the plane equation
$$x+y+z = 1 \implies \sin\phi\cos\theta+\sin\phi\sin\theta+\cos\phi = 1$$
which after a little manipulation becomes
$$\cos\theta+\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the coefficient of ${x^9}$ in the expansion of $ (1 + x)( 1 + x^2)( 1 + x^3)..(1 + x^{100})$ Find the coefficient of ${x^9}$ in the expansion of $\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^3}} \right)..\left( {1 + {x^{100}}} \right)$.
The official answer is 8.
How do I find the general term,
Di... | Hint:
$$\begin{align}9=9+0\\=8+1\\=7+2\\=6+3\\=6+2+1\\=5+4\\=5+3+1\\=4+3+2\end{align}$$
We don't have to worry about 4 summands, since $1+2+3+4>9$. There is no known closed form for the general term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Nice geometry question to prove tangency Let $\triangle ABC$ be a scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ meet at a point $P$. Let $AP$ meet $BC$ at a point $K$ and let $M$ be the midpoint of $BC$. Let $X$ be a point on $AC$ such that $AB$ is tangent to the circumcircle of $... | Here you have an approach employing barycentric coordinates. For more information, have a look at this awesome handout by Evan Chen. Following his advice, I will use , when referring to normalized coordinates, and : for homogeneous coordinates.
Let $\triangle ABC$ be the reference triangle, i.e. $A=(1,0,0); B=(0,1,0);C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, u... | As another approach, factor the two quadratics as
$$x^2 + x + 1 = (x-\phi)(x+\psi)$$ $$x^2 - x + 1 = (x + \phi)(x - \psi)$$
Where $\phi$ and $\psi$ are solutions to $\phi\psi=-1$, $\psi - \phi=1$
So problem reduces to solving for $A$ and $B$
$$(x-\phi)^n(x+\psi)^n = P(x)(x+\phi)(x-\psi) + Ax + B$$
Evaluate this for $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Dividing a polynomial with integer coefficients by a quadratic In the following question, one of the coefficients of the divisor quadratic polynomial is unknown; so, it cant be factored for performing synthetic division. I tried to perform long division hoping to get some clues but ran into a conflict.
Appreciate your ... | Some quick observations : if $x$ is a root of $P(x)$ then so is $-x$. Thus if $(x^2-x+a)$ is a factor of $P(x)$, so is $(-x)^2-(-x)+a=x^2+x+a.$
Then $(x^2-x+a)(x^2+x+a)=x^4+(2a-1)x^2+a^2$ is a factor of $P(x)$. $\, a^2|36 \,$ $\Rightarrow a \in \{1,2,3,6\}$. (Here @Anwesha1729 observe in their answer that as $P(x)$ tak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Help solving a recurrence relation I'm quite stuck on how to solve recurrence/difference equations.
For example I have the following linear inhomogeneous equation:
$a_n = 2a_{n-1} + 2^n n$, and $a_0 = 1/2$
I know that $2a_{n-1} = 1/2(2^n)$, and then we have $a_n = 1/2(2^n) + n(2^n)$. This is where I'm stuck though and ... | You didn't use the generating-functions tag, so here is more than a nudge for that approach. Let $A(z)=\sum_{n=0}^\infty a_n z^n$ be the ordinary generating function for $(a_n)$. Then the recurrence and initial condition imply that
\begin{align}
A(z)
&= \frac{1}{2} z^0 + \sum_{n=1}^\infty \left(2 a_{n-1} + 2^n n\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$24ml+1=k^2$ has no solution for all $l=1 \dots m$ Investigating solutions of $$24ml+1=k^2$$ for $l=1\dots m$
The question is to find the $m$-s for which the above equation has no solution for all $l=1..m$-s.
The first few $m$-s are:
$$3, 9, 24, 27, 81, 192, 243, 432, 729$$
Actually have found that the $m$ should be of... | The OP also conjectures that $12ml+1=k^2$ has no solutions for $1\le l\le m$ if and only if $ m=3^a$. This is true and can be proved as follows.
Only if
For $12ml=(k-1)(k+1)$ we must have $k=6x\pm 1$. Then $ml=x(3x\pm 1).$
Let $m=3^aM$, where $M>1$ is coprime to $3$. Let $x=3^aX$ and then $$Ml=X(3^{a+1}X\pm 1).$$
We ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proof Check: $ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} $ The limit can be rewritten as
$$ \lim_{x \rightarrow 0} \frac{x-\sin x}{x^2} = \lim_{x \rightarrow 0} \frac{1}{x} \left[ 1- \frac{\sin x}{x} \right] $$
Recall the inequality,
$$ \cos x < \frac{\sin x}{x} < 1$$
holds for $ x\in (-\pi/2, \pi/2) $.
This provides... | Your proof is right. But I would have used Taylor developments :
$$\frac{x-\sin x}{x^2} \underset{x\rightarrow 0}{=} \frac{x-(x-x^3/6+o(x^3))}{x^2}\underset{x\rightarrow 0}{=}\frac{x}{6}+o(x)\underset{x\rightarrow 0}{\longrightarrow} 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both ... | By drawing triangles it should be easy to see that \begin{align*}
\cos \arcsin \left(\frac{3}{5}\right)&=\frac{4}{5}\\
\cos \arctan \left(\frac{7}{24}\right)&=\frac{24}{25}\\
\sin \arcsin\left(\frac{3}{5}\right)&=\frac{3}{5}\\
\sin \arctan\left(\frac{7}{24}\right)&=\frac{7}{25}
\end{align*}
(note the Pythagorean tripl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Approach ideas for the integral $\int\frac{dx}{(x^4-16)^2}$ Well, the title sums it up pretty well. I'm in search for some smart approach ideas for solving this indefinite integral:
$$\int\frac{dx}{(x^4-16)^2}$$
I know one that would work for sure, namely partial fraction decomposition, but it gets really heavy when re... | As $x^4-2^4=(x^2-2^2)(x^2+2^2)$ and $(x^2+2^2)-(x^2-2^2)=8$
$$\dfrac{8^2}{(x^4-16)^2}=\dfrac{(x^2+4-(x^2-4))^2}{(x^2-4)^2(x^2+4)^2}=\dfrac1{(x^2-4)^2}+\dfrac1{(x^2+4)^2}-\dfrac2{(x^2-4)(x^2+4)}$$
For the first, write the numerator as $\dfrac{(x+2-(x-2))^2}{16}$ and expand
For the second either $x=2\tan t$
or integrate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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How to solve the recurrence $()=5(/2)+^3()$ using iteration method How do we solve the recurrence $()=5(/2)+^3()$ using the Iteration method?
I solved the recurrence using a master method - master method
Now using the iteration method
$()=5(/2)+3() = 5(5T(/4)+(n/2)^3)*log(n/2))+n^3*logn$ $=$ ... $= 5^i(n/2^i)+n^3*log(n... | Do the substitution:
$$\begin{array} {rcl}
T(n) &=& n^3\log(n) \\
&+& 5(n/2)^3\log(n/2)\\
&+& 5^2(n/2^2)^3\log(n/2^2)\\
&+& 5^3(n/2^3)^3\log(n/2^3)\\
&+& \dots \\
\\
&=& n^3\log(n) \\
&+& 5/2^3 n^3 (\log n - \log 2)\\
&+& 5^2/2^6 n^3 (\log n - 2\log 2)\\
&+& 5^3/2^9 n^3 (\log n - 3 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is the solution that Wolfram Alpha gives really the only solution to this problem? My formula
$$(z-y)(z^2+zy+y^2)=(y-x)(y^2+yx+x^2)$$
$$z>=y+2$$
$$y>=x+2$$
$$x>=2$$
I am trying to understand a geometric problem better by plugging my formula into Wolfram Alpha. Normally I get a result which helps me understand the issue... | Your equation says $x^3 + z^3 = 2 y^3$. Note that you want $x,y,z$ to be all $\ge 2$.
With $y = ((x^3 + z^3)/2)^{1/3}$, your inequalities are equivalent to
$$ \eqalign{\frac{x^3}{2} + \frac{z^3}{2} &\le (z-2)^3\cr
(x+2)^3 &\le \frac{x^3}{2} + \frac{z^3}{2}\cr
x \ge 2\cr}$$
Now the resultant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3983913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate $x^3 + \frac{1}{x^3}$ Question
$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 + \frac{1}{x^3}$ and choose the correct answer from the following options:
*
*198
*216
*200
*186
What I have did yet
I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$... | Let have a look at this: finding $x^6+y^6$ given $x+y$ and $xy$
$$U_n=x^n+\frac 1{x^n}$$
$s=x+\frac 1x=U_1$ and $p=x\times\frac 1x=1$
also we have $\begin{cases}U_2=34\\U_0=x^0+\frac 1{x^0}=1+1=2\end{cases}$
We have the relation $$U_{n+1}=U_1U_n-U_{n-1}$$
$U_2=U_1^2-U_0\implies U_1^2=34+2=36\implies U_1=\pm 6$
And $U_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $2\le U_n \le 3$ for all $n$ Given $U_n = (1 + 1/n)^n$ , $n = 1,2,3,\ldots\;.$
Show that $2 \le U_n \le 3$ for all n
This is what I've done. Can anyone help?
$$\begin{align*}
a_n=\left(1+\frac1n\right)^n&=\sum_{r=0}^n{^nC_r}(1)^r\left(\frac1n\right)^{n-r}\\
&=\sum_{r=0}^n{^nC_r}\left(\frac1n\right)^{n-r}\\
... | Here is another way to look at it that might be helpful, if the question is why for all $n$ it holds. Consider $f(x)=(1+\frac{1}{x})^x$. It is an increasing function of $x$:
$$ f'(x)=\frac{d}{dx}e^{x \ln(1+\frac{1}{x})}
= \left[\ln\left(1+\frac{1}{x}\right) - \frac{1}{1+x}\right]\left(1+\frac{1}{x}\right)^x >0, \foral... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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proving limits exist Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.
I basically need to determine how large $n$ has to be to imply
$$\frac{3n+24}{n^3-6}<\epsilon$$
the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.
it i... |
I basically need to determine how large $n$ has to be to imply
$$\frac{3n+24}{n^3-6}<\epsilon$$
the idea is to upper bound the numerator and lower bound the denominator.
For $n > 9$, $0 < (n^3 - 64n),$ which is a lower bound for the denominator.
For $n > 9$, $(n-8) > 1 \implies$ that the numerator is
less than $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Finding unique number and value of p. Function = f(x) = $\frac{x^2+px+1}{x^2+x+1}.$
Q 1For any value of p , a unique number will be there in the range . Find it.
Q 2 If nine integers are in the range of f(x) , then solve for p.
Q3If range of function is non negative , then find maximum no of integers in the range can ... | Question 1
let $x=0$. No matter what the value of $p$ is, this value will be in common with all possible functions.
Question 2
Question 2 is asking the following:
If we have the function $f(x)=\frac{x^2+px+1}{x^2+x+1}$ and we look at range of $f(x)$, for some value of $p$ the range of $f(x)$ includes exactly $9$ integ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Given the minimal polynomial of a matrix $A^2$, what could the minimal polynomial of $A$ be? It is given that the minimal polynomial of $A^2$ is $\phi_{A^2}(x) = (x-1)^2$, where $A$ is a complex $4\times4$ matrix. The question is, what are the possible minimal polynomials for $A$?
From the given that $\phi_{A^2}(x) = (... | HINT
You know a polynomial satisfied by $A$, namely $(A^2-1)^2$. So the minimal polynomial has to be a factor of $(x-1)^2(x+1)^2$.
Which of the possible factors could give the correct minimal polynomial for $A^2$?
If neither power of $(x-1)^i(x+1)^j$ is $2$ then we would have one of $A-I=0,A+I=0,(A+I)(A-I)=0$. In all t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$
I tried the most rudimentary method ie. differentiating w... | I think that the derivative gives a good indication.
$$f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$$
$$f'(x)=4 \sqrt{\frac{a}{x (a x+8)^3}}-\frac{1}{2 (x+1)^{3/2}}$$ Without any constraint, the derivative cancels at four points
$$x_{1,2}=\pm \frac{2 \sqrt{2}}{\sqrt{a}}\qquad x_{3,4}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $ Prove by induction $2^n \geq n^3 \ \ \forall n\geq 10 $
I did these steps:
*
*Basis step
$$P(10): \ \ 1024 \geq 1000 \ (True)$$
*Inductive step
$$P(n) \implies P(n+1) \\P(n+1) = 2^{n+1} \geq (n+1)^3$$
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1... |
so $$2^n \geq n^3 \\ 2^n \cdot 2 \geq n^3 \cdot 2 \\ 2^{n+1} \geq 2n^3 \\ 2^{n+1} \geq (n+1)^3$$
taking advantage of the fact that $(n+1)^3 \geq 2n^3$ and less than $2^{n+1}$.
does this demonstration work?
No, your analysis (in and of itself) hasn't shown that
$(n+1)^3 \leq 2^{n+1}$, assuming that I am not overlo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3996557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential equation with partial fraction How do you separate this differential equation into a partial fraction?
Solve the following differential equation:
$$\frac{dy}{dx}=\frac{2y^2-xy+x^2}{xy-x^2}$$
| Let $\frac{y}{x}=v$
$$\frac{{\rm d}y}{{\rm d}x} = v + x\frac{{\rm d}v}{{\rm d}x} =\frac{2v^2-v+1}{v-1} \\
x\frac{{\rm d}v}{{\rm d}x} = \frac{v^2+1}{v-1} \\
\frac{v-1}{v^2+1}dv = \frac {{\rm d}x}{x} \\
\text{Integrating both sides} \\
\int \frac{v-1}{v^2+1}dv =\int \frac {{\rm d}x}{x} \\
\int \frac{v}{v^2+1}dv - \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}$ Solve $$\begin{cases}x^2+y^4=20\\x^4+y^2=20\end{cases}.$$ I was thinking about letting $x^2=u,y^2=v.$ Then we will have $$\begin{cases}u+v^2=20\Rightarrow u=20-v^2\\u^2+v=20\end{cases}.$$ If we substitute $u=20-v^2$ into the second equation, we will get $$v^4-40v^2... | You have the following problem: $$\begin{bmatrix}x^2+y^4=20\\ x^4+y^2=20\end{bmatrix}$$
In the first equation, you can isolate $x$ to get $x=\sqrt{20-y^4}$ and $\:x=-\sqrt{20-y^4}$. You then substitute the $x$ values into the second equation to get:
$$\left(\sqrt{20-y^4} \right)^4 +y^2=20, \left(-\sqrt{20-y^4} \right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$ Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.
Here is what I have done so far
$$ \cos x + \cos y = -\cos z $$
$$ (\... | Continuing from your work, we have $y-z\equiv z-x\equiv x-y\equiv\frac23 r\pi\;\,(\!\!\!\!\mod2\pi)$, with $r\in\{1,2\}$ ($r$ is constant, since the three differences sum to zero, modulo $2\pi$). So $$\cos2x+\cos2y+\cos2z=\cos2x+\cos(2x-\tfrac43r\pi)+\cos(2x+\tfrac43r\pi)\qquad\qquad$$
$$=\cos2x+2\cos2x\cos\tfrac43r\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\fra... | Working a little differently, you get an exact result:
$$
\begin{align}
x &= \left(\frac 21 \times \frac 22\right) \times \left(\frac 43 \times \frac 44\right) \times \cdots \times \left(\frac{10000}{9999} \times \frac {10000}{10000}\right) \\
&= \frac{1}{10000\, !} [2\times 4 \times \cdots \times 10000]^2\\
&= \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Integration by substitution: $\displaystyle\int\frac{dx}{(1+x^2)^2}$ I came across this solution of a tricky integral using implicit substitution (i.e., manipulating differentials without actually declaring $u=\cdots$).
\begin{align*}
\int \frac{dx}{(1+x^2)^2} &= \int \frac{1 + x^2 - x^2}{(1+x^2)^2}\,dx... | Here's something you might like.
Writing the indefinite integral as a definite one and applying Feynman's method
\begin{align}I(a)&=\int_c^x\dfrac{1}{a^2+x^2}\,\mathrm dx=\dfrac1a\arctan\left(\dfrac xa\right)\bigg|_c^x=\dfrac1a\arctan\left(\dfrac xa\right)+C\\I'(a)&=\int_c^x-\dfrac{2a}{(a^2+x^2)^2}\,\mathrm dx=-\dfrac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Expressing a projective conic in standard form I was given the task to find a projective transformation $\phi: \mathbb{P}_\mathbb{C} \rightarrow \mathbb{P}_\mathbb{C}$ that transforms the following conic:
$$2\,{x}^{2}+2\,xy-3\,xz+4\,yz+{z}^{2}=0$$
into the form:
$$X^2 + Y^2 + Z^2 = 0$$
Then my attempt to do so was to f... | I found some notes online, the task is to begin with symmetric $H$ and solve $P^T HP = D$ diagonal, which can be done over the reals. Then forcing the identity over the complexes requires multiplying by an additional diagonal complex matrix on both sides.
Algorithm discussed at reference for linear algebra books that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral question with cube roots Evaluate the following integral:
$$\int\sqrt[3]{x^3+\frac 1 {x^3}} dx$$
For $x^3=u$ and $1+\frac 1 {u^2}=v$, getting $$-\frac 1 6 \int(v-1)^{-\frac 4 3}v^{\frac 1 3} dv$$, what is the next step?
| I would chose a different substitution:
$$I =\int \big(x^3 +\frac{1}{x^3} \big)^{\frac{1}{3}}dx$$
$$=\int \frac{(x^6 +1)^{\frac{1}{3}}}{x} dx $$
Let: $ u=(x^6+1)^{\frac{1}{3}} $, then:
$x^6=u^3-1 \text{ , and } dx= \frac{(x^6 +1)^{\frac{2}{3}}}{2x^5} du$ .
$$I=\frac{1}{2} \int \frac{u^3}{u^3-1}du$$
$$=\frac{1}{2} \int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to ... | $b-a=4^{2\cdot2^k}-4^{2^k}+2^{2\cdot2^k}-2^{2^k}$
$=4^{2^k}(4^{2^k}-1)+2^{2^k}(2^{2^k}-1)$
$=4^{2^k}(2^{2^k}-1)(2^{2^k}+1)+2^{2^k}(2^{2^k}-1)$
$=(2^{2^k}-1)(8^{2^k}+4^{2^k}+2^{2^k})$
$=(2^{2^k}-1)2^{2^k}\color{blue}{(4^{2^k}+2^{2^k}+1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
Asymptotic Expansion of Digamma Function While reading the wikipedia page of the Digamma function (https://en.wikipedia.org/wiki/Digamma_function#Asymptotic_expansion)
I noticed that it said the asymptotic expansion for the digamma function ($\psi(z)$) can be obtained from using
\begin{equation} \psi(z+1) = -\gamma + \... | I shall assume that $|\arg (z+1)|<\pi$. By the Euler–Maclaurin formula,
\begin{align*}
\psi (z + 1) = & - \gamma + \int_1^{ + \infty } {\left( {\frac{1}{t} - \frac{1}{{t + z}}} \right)dt} + \frac{1}{2}\left( {1 - \frac{1}{{1 + z}}} \right) \\ & + \int_1^{ + \infty } {\left( {\left\{ {1 - t} \right\} - \frac{1}{2}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4022702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Set of points that satisfy the sum of the distances between a point P and points A(-2,0) and B(2,0) = 7 So, I have to find the set of points whose distance to point $A(-2, 0)$ plus the distance to point $B(2, 0)$ is equal to $7$.
I am solving it this way:
$ d(P, A) + d(P, B) = 7 \Leftrightarrow $
$ \sqrt{(x+2)^2+y^2}+... | When you expand $(a+b)^2$, you get $\color{blue}{\text{cross terms}}$ also as
$$(a+b)^2=a^2+b^2+\color{blue}{2ab}$$
You have missed
$$(x+2)^2+y^2+(x-2)^2+y^2+\color{blue}{2\cdot \sqrt{(x+2)^2+y^2}\cdot\sqrt{(x-2)^2+y^2}}=49$$
One more squaring will clear the square roots and finally give correct equation (of an ellipse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Moment of inertia of hollow paraboloid A hollow paraboloid of length $l$ and radius $r$ with mass $m$ and uniform density $\rho$ has the origin at the base of its circular cross section, and at the center of its cross section. I am trying to find the moment of inertia for the axis that cuts through the paraboloid's cen... | We have a hollow paraboloid of length $L$ and radius $R$ along $x-$axis. If the equation of the paraboloid is $y^2 + z^2 = r^2 = ax$,
Then $R^2 = aL \implies a = \frac{R^2}{L}$. We will continue our working with $a$ and substitute this value of $a$ in the end.
As we know, moment of inertia of mass $dm$ around an axis =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem by proving a statement by induction I want to prove the following statement by induction for $n\leq 1$:
\begin{align*}
\sum_{k=n}^{2n-1}\frac{1}{k}= \sum_{k=1}^{2n-1}\frac{(-1)^{k+1}}{k}
\end{align*}
For $n=1$ its easy to see:
\begin{align*}
\sum_{k=1}^{2-1}\frac{1}{k}= \sum_{k=1}^{1}\frac{1}{k}= 1 = \sum_{k=1}... | Call the left sum $L(n)$ and the right sum $R(n)$. Then we have
$$L(n + 1) - L(n) = \frac{1}{2n}+\frac{1}{2n+1} - \frac{1}{n},$$
and
\begin{align}
R(n + 1) - R(n) &= \frac{(-1)^{2n}}{2n+1} + \frac{(-1)^{2n+2}}{2n+1} \\
&= - \frac{1}{2n} + \frac{1}{2n+1}.
\end{align}
So by comparison, we need to check that
$$\frac{1}{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$ For $a,b,c>0;abc=1.$ Prove that $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$$
I will post my solution in the answer. Now I'm looking forward to another solution.
| My solution. Assume $(b-1)(c-1)\ge 0 \rightarrow 1=abc\ge (a+b-1)c\rightarrow a+b \le \dfrac{1}{c}+1.$
\begin{align*} \text{LHS}&=\sqrt{\left(a^2+b^2\right)-(a+b)+2+2\sqrt{\left(a^2-a+1\right)\left(b^2-b+1\right)}}+\sqrt{c^2-c+1}\\&\ge \sqrt{t^2-2t+4}+\sqrt{c^2-c+1}=\text{P}\,(\text{where}\,t=a+b)\end{align*}
Let $f(t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4028714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Any neat way to solve the integral $\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$? Straight to the point: given the integral
$$\iint_Q \frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy$$
where $Q=[-a,a]\times[-b,b]$, can you think of any neat way to solve it?
At a first glance it looked quite inno... | Substitute $x = \sqrt{b^2+z^2}\sinh t$ after integrating over $y$
\begin{align}
&\int_{-a}^a \int_{-b}^b\frac{1}{\left(x^2+y^2+z^2\right)^{3/2}}\,dxdy \\
= &2b \int_{-a}^a \frac{1}{\left(x^2+z^2\right)\sqrt{x^2+b^2+z^2}}\,dx \\
=&4b \int_0^{\sinh^{-1}\frac a{\sqrt{b^2+z^2}}}\frac1{(b^2+z^2)\sinh^2 t +z^2}dt\\
=&4b \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Partitioning 18 into odd number of even components I need to find a total number of ways $18$ can be partitioned into odd number of even components.
Does the function below satisfy the "odd number of components" requirement?
$$
P_{1,3,5,7,9}(X) = \frac{1}{1-x} \cdot \frac{1}{1-x^3} \cdot \frac{1}{1-x^5} \cdot \frac{1}{... | It may be quickest to just write the numbers down: you should be able to find $16$ cases.
$$18 =2+2+14=2+4+12=2+6+10=2+8+8=4+4+10 \\=4+6+8=6+6+6 =2+2+2+2+10=2+2+2+4+8\\=2+2+2+6+6=2+2+4+4+6
=2+4+4+4+4 \\= 2+2+2+2+2+2+6=2+2+2+2+2+4+4 \\
=2+2+2+2+2+2+2+2+2$$
One approach could be to say that, by dividing each part by $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4032487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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confirming solution to the series $\frac{n(n+1)}{(n+3)^3}$ with ratio test I want to confirm my solution to this series using the ratio test correct, I tested to show the series is divergent.
$$\frac{n(n+1)}{(n+3)^3}$$
Using the ratio test, then simplifying in stages:
$$\frac{(n+1)(n+1+1)}{(n+1+3)^3}\frac{(n+3)^3}{n(n+... | Ratio test is not effective because
$$\lim_{n\to \infty }\frac{(n+2) (n+3)^3}{n (n+4)^3}=1$$
You can use comparison test as
$$\frac{n (n+1)}{(n+3)^3}\sim \frac1n ;\text{ as }n\to\infty$$
and $\sum\frac1n $ diverges
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the fami... | Let $\alpha$ denote the real root of $X^3+X^2-1$. Then $\alpha^3 = 1-\alpha^2$. If we plug in $\alpha^2$ into $X^3-X^2+2X-1$, we obtain
\begin{align*}
(\alpha^2)^3-(\alpha^2)^2+2(\alpha^2)-1 &= (1-\alpha^2)^2-\alpha^4+2\alpha^2-1\\
&=(1-\alpha^2)^2-(1-\alpha^2)^2=0,
\end{align*}
so $\alpha^2$ is a root of $X^3-X^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.
The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
x &= 1\\
x &= -\sqr... | Your elements are basically $(e^{ik\pi/9})_{k=0}^8$. They form the cyclic group of $9$ elements under ordinary multiplication. The most obvious generator is $r=e^{i\pi/9}$ (with $k=1$).
However, any such element with $k$ relatively prime to $9$ (e.g. $4,5,8$) would generate the group as well. For example, starting with... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Proof by induction: Inductive step struggles Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$
where $ n $ is a nonnegative integer.
Preforming the basis step where $ n $ is equal to 0
$$
1 = \frac{2^{1}+(-1)^{0}}{... | \begin{align}
\frac{2^{k+1}+(-1)^k}{3 \times 2^k}+\left(-\frac{1}{2}\right)^{k+1}
&=\frac{2 \times 2^{k+1}+2 \times(-1)^k}{3 \times 2^{k+1}}+\frac{3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}+2 \times (-1)^k+3 \times (-1)^{k+1}}{3 \times 2^{k+1}} \\
&=\frac{2^{k+2}-2 \times (-1)^{k+1}+3 \times (-1)^{k+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If $\tan x\tan y=\frac{b}{a},\,a,\,b\ne 0$, prove that $\frac{\sec^2x}{a\tan^2x+b}+\frac{\sec^2y}{a\tan^2y+b}=\frac{a+b}{ab}$ If $\tan x\tan y=\frac{b}{a},\,a,\,b\ne 0$, prove that $\frac{\sec^2x}{a\tan^2x+b}+\frac{\sec^2y}{a\tan^2y+b}=\frac{a+b}{ab}$
I solved it in the following way:
$\frac{1+\tan^2x}{a\tan^2x+b}+\fra... | Let $t:=\tan x$ so $\tan y=\frac{b}{at}$ and we want$$\frac{1+t^2}{at^2+b}+\frac{1+b^2/(a^2t^2)}{b^2/(at^2)+b}=\frac{b(1+t^2)+at^2+b^2/a}{b(at^2+b)}=\frac{a+b}{ab},$$taking out a factor of $t^2+b/a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4038495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Question regarding a proof of a inequality Given $a,b,c, \in (0,\infty)$, then the following inequality holds
$$\sqrt{5a^2+12ab+7b^2}+\sqrt{5b^2+12bc+7c^2}+\sqrt{5c^2+12ca+7a^2} \leq 2 \sqrt6 (a+b+c)$$
What I've tried:
First, I noticed that we can factor
$$5a^2+12ab+7b^2 = (a+b)(5a+7b)$$
1.) Therefore we can use AM-GM ... | Your inequality become the equality when $a=b=c.$ But for $a=b,$ then
$$\sqrt{(a+b)(5a+7b)} \ne \frac{(a+b)+(5a+7b)}{2}.$$
By the AM-GM inequality, we have
$$\sqrt{(a+b)(5a+7b)} = \frac{1}{\sqrt6} \cdot \sqrt{6(a+b) \cdot (5a+7b)} \leqslant \frac{6(a+b)+(5a+7b)}{2\sqrt6} = \frac{11a+13b}{2\sqrt6}.$$
Equality hold for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about Sui Zhen Lin's proof for inequality $\sqrt{\frac{a^2}{9b^2-8b+4}}+\sqrt{\frac{4b}{a+4}}\leq 1$ with positive numbers $a,b$ so $a+b=1$ given two positive numbers $a, b$ so that $a+ b= 1$
Sui Zhen Lin ; @szl6208 gave a very beautiful proof for the following inequality
$$\sqrt{\frac{a^{2}}{9b^{2}- 8b+ 4}}+... | Using algebra.
From the constraint $b=1-a$. So, we face
$$f(a)=\sqrt{\frac{a^2}{9 a^2-10 a+5}}+2 \sqrt{\frac{1-a}{a+4}}$$
$$f'(a)=\frac{5-5 a}{(9 a^2-10 a+5)^{3/2}}-\frac{5}{\sqrt{1-a} (a+4)^{3/2}}$$
Square and factor; the numerator is just
$$(2 a-1) (5 a-1) \left(a^2-a+1\right) \left(73 a^2-118 a+61\right)$$ So, the d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculation in Routh's theorem The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it... | A variant on my option 2: as @RiverLi notes, the numerator is of at most degree $2$ in $z$, so is of the form $(c-1)(Aa+Bb+Cc+D)$. That $A=B=0,\,C=D=1$ follows from the case $x=y=z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000... | You can distinguish between the two possibilities you found by starting modulo $2000$:
because the Carmichael function of $2000$ is $100$, $7^{2000}\equiv(7^{100})^{20}\equiv1\bmod2000$,
so $7^{2005}\equiv7^5\equiv807\bmod 2000$.
Therefore, $7^{2005}-1\equiv806\bmod2000$.
Therefore, $\dfrac{7^{2005}-1}2\equiv403\bmod10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the limit: What does it mean when one obtains zero in the calculation? Consider the sequence $a_n = \sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\\$. To determine the limit I did the following:
\begin{aligned}
a_{n} &=\left(\sqrt{n+\sqrt{n}}-\sqrt{n-\sqrt{n}}\right) \frac{\sqrt{n+\sqrt{n}}+\sqrt{n-\sqrt{n}}}{\sqrt{n+... | Your second approach is wrong.
To save writing let $x=\sqrt{n}$, and expand square roots in power series in $\frac{1}{x}$, so expression becomes $x(\sqrt{1+\frac{1}{x}}-\sqrt{1-\frac{1}{x}})$
$=x(1+\frac{1}{2x}+O(\frac{1}{x^2})-1+\frac{1}{2x}+O(\frac{1}{x^2}))$
$=1+O(\frac{1}{x})\to 1$ as $x\to \infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplify the sum of the square of a polynomial having a fourth degree. Today I learn about polynomial. Because I want to improve my knowledge. Thank you for your support and time for sharing information and experience.
From question :
If $a, b, c$ and $d$ are the roots of polynomial $Ax^4+Bx^3+Cx^2+Dx+E$ then find the ... | Indeed, we know that
$$(a+b+c+d)^2=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd$$ and you find these sums from Vieta's formulas.
Solving from $abcd$ alone is obviously impossible, because there are infinitely many ways to obtain the same value from different $a,b,c,d$, and these result in different sums $a^2+b^2+c^2+d^2$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I evaluate $\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx$? I need to calculate the following definite integral:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx.$$
The only thing that I've found is:
$$\int_{1/3}^3 \frac{\arctan x}{x^2 - x + 1} \; dx = \int_{1/3}^3 \frac{\arctan \frac{1}{x}}{x^2 - x + 1} \;... | $$\begin{eqnarray*}\int_{1/3}^{3}\frac{\arctan x}{x^2-x+1}\,dx&=&\int_{1/3}^{1}\frac{\arctan x}{x^2-x+1}\,dx+\int_{1}^{3}\frac{\arctan x}{x^2-x+1}\,dx\\&=&\int_{1}^{3}\frac{\arctan x+\arctan\frac{1}{x}}{x^2-x+1}\,dx\\&=&\frac{\pi}{2}\int_{1}^{3}\frac{dx}{x^2-x+1}\\&=&2\pi\int_{1}^{3}\frac{dx}{(2x-1)^2+3}\\&=&\pi\int_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare ... | Answer
$$
\begin{align}
0
&\le\tfrac34((x-1)-(y-1))^2+\tfrac14((x-1)+(y-1))^2\\
&=(x-1)^2+(y-1)^2-(x-1)(y-1)\\
&=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)-(xy-x-y+1)\\
&=\left(x^2+y^2+1\right)-(xy+x+y)
\end{align}
$$
Motivation
To prove $x^2+y^2+1\ge xy+x+y$, look at the difference
$$
\begin{align}
\left(x^2+y^2+1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 7
} |
Quick questions of infinite series Are these true:
$$1.\sum_{n=1}^{\infty}\frac{1}{n^2+88n}=\frac{1}{88}\sum_{n=1}^{88}\frac{1}{n}$$
$$2.\sum_{n=1}^{\infty}\frac{1}{n^2+4n}=\frac{25}{48}$$
For first one, it's factored. ¿It's not factored correctly? Because the bound is from $\infty$ to $88$ and if you multiply $\frac{1... | $$\frac{1}{n^2 + 88n} = \frac{1}{n(n+88)} = \frac{1}{88}\left( \frac{1}{n} - \frac{1}{n+88} \right).$$ Therefore, $$\sum_{n=1}^\infty \frac{1}{n^2 + 88n} = \frac{1}{88} \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+88} = \frac{1}{88} \sum_{n=1}^{88} \frac{1}{n}.$$
The other sum is handled in a similar fashion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$? What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)
I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect e... | First of all $x > 0$.
If $x \le 0$ then $x^3 < 0$ so $[x] = x^3 -3 \le -3$ so $[x] \le -3$ and $x \le -2$. (Obviously I'm being overly cautious. $[x]\ne x$ unless $x$ is on integer so $x \le -3$ and obviously $x$ can't be $0$ so $[x]\le -4$ and ... so on but... this just stick with $x \le -2$.)
This means that $|x^3| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Epsilon-delta proofs $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I am getting back to $\delta-\epsilon$ arguments after a break, so I'd like ask if my $\delta$-response to the below $\epsilon$-challenges is technically correct and... | all good except part (b), which is wrong because you are requested to find the largest $\delta$, which must be $(\delta = 3)$, since
$|\sqrt{4-3} - 2| = 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4058162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Rank inequality
Let $ A, B \in M_n (\mathbb{C})$ such that $(A-B)^2 = A -B$. Then $\mathrm{rank}(A^2 - B^2) \geq \mathrm{rank}( AB -BA)$.
I tried to apply the basic inequalities without results. How to start? Thank you.
| $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
$\newcommand{\M}{\begin{pmatrix} I_{(r)} & 0 \\ 0 & 0 \end{pmatrix}}$
The key to this problem is to use the property of idempotent matrix $A - B$. As @Berci suggested, denote $A - B = P$, then $P^2 = P$, which implies there exists an order $n$ inve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to deal with subtraction of sigma? How to solve this?
$$\displaystyle\sum_{n=1}^\infty
\frac{n}{3^{n-1}}-\displaystyle\sum_{n=1}^\infty \frac{n}{3^n}$$
My work
$$S=\displaystyle\sum_{n=1}^\infty \left(\frac{n}{3^{n-1}}-\frac{n}{3^n}\right)$$
$$S=\displaystyle\sum_{n=1}^\infty\frac{3n-n}{3^n}=\displaystyle\sum_{n... | I suspect that you're supposed to shift the index of one of the series. What you've done is 100% valid, but as you said, it's not yielding something "useful". Instead, take the series
$$\sum_{n=1}^\infty \frac{n}{3^{n-1}},$$
and shift the index. Let $m = n - 1$. Then when $n = 1$, $m = 0$, so
$$\sum_{n=1}^\infty \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Reduction formula for $\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $ I'm struggling to go any further. Anyone have any hints?
$$ I _ n = \int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } \ \mathrm d x $$
I used integration by parts where I differentiated $ x ^ n $, but it resulted in an $ \arcsin x $ term whi... | Integrate by parts as follows
\begin{align}
\int \frac { x ^ n } { \sqrt { 1 - x ^ 2 } } d x
&= -\int \frac { x ^{n-1} } { (1 - x ^ 2 )^{\frac{n-1}2}} d \left(\frac{(1-x^2)^{\frac n2}}n\right)\\
&=-\frac1n \frac{x^{n-1}}{\sqrt { 1 - x ^ 2 } }+ \frac{n-1}n\int \frac { x ^ {n-2} } { \sqrt { 1 - x ^ 2 } } d x \\
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Intuition for polynomial long division So I am trying to learn some basic algebra and go over some precalc material and I am terribly confused on why or how polynomial long division works.
I have looked at many sources online and they seem to all suggest that division of numbers is identical to division of polynomials ... | If you really want to understand the mathematics, do not do the kind of long division that you were taught. Instead, you must learn to write down true equalities. For example:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
$\lfrac{4x^3-7x^2-6}{2x^2+5}$
$ = \lfrac{\color{blue}{2x}(2x^2+5)-10x-7x^2-6}{2x^2+5}$
$ = 2x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
$b=a^p+1$ is a perfect square. Show that $p|(b-9)$ $p$ is a prime, and $a$ is a positive integer, $b=a^p+1$ is a perfect square. Show that $p|(b-9)$
It seem very interesting problem.if let $a^p+1=x^2$,it is clear $p\neq 2$
I have prove : $x$ is odd
proof:if $x$ is even number,then $(x+1,x-1)=1$,and note $a^p=(x+1)(x-... | From the comments we have the elementary proofs that $x^2 = a^p + 1$ has no solutions for prime $p > 3$, so it suffices to prove that when $p = 3$, $x$ is divisible by $3$.
We have $\gcd(a+1, a^2-a+1) \mid 3$, so we can assume that they are coprime and hope for a contradiction. If they are coprime, then both are square... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$? If for the real numbers $a,b(a\ne b)$ it is true that $a^2=2b+15$ and $b^2=2a+15$, then what is the value of the product $ab$?
I tried to solve it as follows:
I state that $p=ab$
$p^2=(2b+15)(2... | This is not a straightforward way to answer the question, but illustrates a matter that the problem touches on.
We can treat the given equations as "solutions" to the problem: if $ \ y^2 \ = \ 2x + 15 \ $ describes a transformation such that $ \ T(a) \ = \ b \ $ and $ \ T(b) \ = \ a \ \ , $ and so $ \ T( \ T(a) \ ) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In p... | The case where $p \leqslant 0$ is quite simple to establish, so I will look only at the case where $p > 0$. One way you can establish the convergence/divergence result is by using upper and lower bounds on the summation. For all values $n \geqslant 1$ and $p>0$ we can easily establish the bounds:
$$\int \limits_{n}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Indefinite integral $\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$ I'm a bit lost in this integral: $$\int \frac{1}{1+\sin^4(x)} \, \mathrm dx$$
I have tried solving with Wolfram, but I was getting a cosecant solution which doesn't seem as the correct method.
Do you have any ideas? :)
EDIT:
Do you please have step-by-step ... | Double-check my arithmetic, but here's the strategy.
@Bernard's suggested substitution $t=\tan x$ gives
\begin{align}
& \int\left(1-\frac{t^4}{(t^2\sqrt{2}+1)^2-2(\sqrt{2}-1)t^2}\right) dt \\
= {} & t-\int\frac{t^4}{(t^2\sqrt{2}+ct+1)(t^2\sqrt{2}-ct+1)} \, dt\end{align}
with $c:=\sqrt{2\sqrt{2}-1}$. You can do the rest... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When i... | You can use induction to prove
$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$
is an integer when $n = 3^k$. The base case of $k = 0$ gives $3^k = 1$, with \eqref{eq1A} becoming $\frac{10-1}{9} = 1$. Next, assume \eqref{eq1A} is an integer for $k = m$ for some integer $m \ge 0$. Then, for $k = m + 1$,
$$\begin{equation}\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Let x and y be real numbers such that $6x^2 + 2xy + 6y^2 = 9 $.Find the maximum value of $x^2+y^2$ I tried to re-arrange the terms
$6x^2 + 2xy + 6y^2 = 9 $
$6x^2 + 6y^2 = 9 - 2xy $
$6 (x^2 + y^2) = 9 - 2xy $
$x^2 + y^2 = \frac{9 - 2xy}{6} $
Using A.M $\geq$ G.M
$\frac{x^2 + y^2}{2} \geq xy $
Can ayone help me from here... | Without AM-GM inequality:
Let, $x=a+b$ and $y=a-b$, then you get
$$6(a+b)^2+2(a+b)(a-b)+6(a-b)^2=9$$
$$14a^2+10b^2=9$$
$$x^2+y^2=(a+b)^2+(a-b)^2=2a^2+2b^2$$
$$x^2+y^2=2a^2+2 \times \frac{9-14a^2}{10}$$
$$x^2+y^2=\frac{9-4a^2}{5}≤\frac95$$
$$\color {gold}{\boxed {\color{black} {\text{max}[x^2+y^2]=\frac 95.}}}$$
If you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4073042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$ How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$
What I tried
Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamm... | $$a_n=2^{4 n}\frac{ \Gamma^4 \left(n+\frac{3}{4}\right)}{\Gamma^2 (2 n+1)
}$$
$$\log(a_n)=4 n \log (2)+4 \log \left(\Gamma \left(n+\frac{3}{4}\right)\right)-2 \log (\Gamma (2 n+1))$$
Applying Stirling approximation twice and continuing with Taylor series
$$\log(a_n)=\log (\pi )-\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4074052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Series $A= \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$ Consider convergence of series:
$A=\displaystyle \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$
I have a idea
\begin{align*}
a_n&=n^\frac{1}{n^2+1}-1
\\&=e^{\frac{\ln n}{n^2+1}}-1 \sim b_n= \dfrac{\ln\,n}{n^2+1} \text{ when } n \to \infty
\end{align... | OP -- Consider convergence of series:
$A=\displaystyle \sum_{n=1}^{\infty}\left(n^\frac{1}{n^2+1}-1\right)$
$$ n^\frac{1}{n^2+1}-1\ =
\ \frac{n-1}{\sum_{k=0}^{n^2}\,n^{\frac k{n^2+1}}\ }\ < $$
$$ \frac{n-1}{\left(n^2-\lfloor\frac{n^2+1}2\rfloor\right)
\cdot n^{\frac 12}}\ <
\ \frac{n^{\frac 12}}{n^2-\lfloor\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$ I am trying to prove a statement about the decomposition field of a polynomial that has both $\sqrt{2-\sqrt{2}}$ and $\sqrt{2+\sqrt{2}}$ as roots. I cannot find a way to prove that $\sqrt{2-\sqrt{2}} \in \mathbb{Q}(\sqrt{2+\sqrt{2}})$. I have tried... | Alternatively (without using conjugates),
Observe that,
*
*If $\sqrt{2-\sqrt{2}}=x$, then
$$(2-x^2)^2=2$$
*
*If $\sqrt{2+\sqrt{2}}=x$, then
$$(x^2-2)^2=2$$
Finally, we can observe that both polynomial equations are equal to each other:
$$(2-x^2)^2=2 \iff (x^2-2)^2=2$$
This implies, if $\sqrt{2-\sqrt{2}}$ is the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4086523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Find all odd functions of the form $f(x) = \frac{ax + b}{x + c}$ I am working through a pure maths book and am stuck on odd and even functions.
Let $f(x) = \frac{ax + b}{x + c}$ where x, a, b, c are real and $x \ne \pm c$. Show that if $f$ is an even function then $ac = b$. Deduce that if $f$ is an even function then ... | A polynomial is odd iff all its exponents are odd and even iff all its exponents are even (including the exponent $0$).
A fraction is odd iff the numerator is odd an the denominator is even or vice versa.
The denominator in question can't be even; it's odd iff $c=0$. Now the numerator must be even, that is $a=0$. Henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4087772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $P=\sqrt{a^2-2ab+b^2}+\left(\frac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\frac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$ is rational Show that the number $$P=\sqrt{a^2-2ab+b^2}+\left(\dfrac{a}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\right):\left(\dfrac{b\sqrt{a}}{a-\sqrt{ab}}+\sqrt{b}\right)$$ is a rational number ($... | Just keep chewing on it until it falls apart
$|a-b|+\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}.\dfrac{a-\sqrt{ab}}{b\sqrt{a}+a\sqrt{b}-\sqrt{b^2a}}=$
$|a-b| +\frac {\sqrt {ab}}{\sqrt a - \sqrt b}\cdot \frac {\sqrt a(\sqrt a -\sqrt b)}{b\sqrt a + a\sqrt b - b\sqrt a}=$
$|a-b| + \frac {\sqrt{ab}}{\sqrt a-\sqrt b}\cdot \frac {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4090748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find value of $x^3+y^3+z^3$ if $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$ If $x,y$ and $z$ are positive real numbers such that $x+y+z=12$ and $(xyz)^3(yz)(z)=(0.1)(600)^3$, then what is the value of $x^3+y^3+z^3$?
I first thought of making them all equal because in that case the product is maximum, but obviously that... | Applying the AM-GM inequality for 3 times $\frac{x}{3}$, 4 times $\frac{y}{4}$ and 5 times $\frac{z}{5}$, we have
\begin{align}
12 &= x+y+z \\ & = 3\times \frac{x}{3} + 4 \times \frac{y}{4} +5 \times \frac{z}{5} \ge (3+4+5)\sqrt[12]{\left(\frac{x}{3}\right)^3\left(\frac{y}{4}\right)^4\left(\frac{z}{5}\right)^5} = 12
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4095618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Showing $\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)$, given $\sin z=z\prod_{n=1}^{ \infty}\left(1 - \frac{z^2}{n^2\pi^2}\right)$ How to show that
$$\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)$$
using the well-know representation of sine as infinite product that is
$$\sin z=z\prod_... | Since $\sinh z = \frac{1}{i}\sin(iz)$, we have
$$\begin{align*}
\sinh z &= \frac{1}{i}\sin(iz) \\
&= \frac{1}{i}\times (iz)\prod\limits_{n=1}^{\infty}\left(1 - \frac{(iz)^2}{n^2\pi^2}\right) \quad (\text{using the product formula for }\sin(\cdot))\\
&= z\prod\limits_{n=1}^{\infty}\left(1 -\frac{-z^2}{n^2\pi^2}\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Combinatorial Summation I'm trying to solve the following question:
If $s_n$ is the sum of first $n$ natural numbers, then prove that
$$2(s_1s_{2n}+s_2s_{2n-1}+\dots+s_ns_{n+1})=\frac{(2n+4)!}{5!(2n-1)!}$$
This is where I've come so far:
The general term of $(1-x)^{-2n}$ happens to be $t_{r+1}=\frac{(2n+r-1)!}{(2n-1)!... | Here is a rather detailed algebraic derivation. We recall that multiplication of a generating function $A(x)$ with $\frac{1}{1-x}$ gives the sum of the coefficients of $A(x)$:
\begin{align*}
A(x)=\sum_{n=0}^\infty \color{blue}{a_n}x^n\qquad\qquad \frac{1}{1-x}A(x)=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^na_k}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Let $a_1=\sqrt{6},a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to \infty} (a_n-3)6^n$.
Let $a_1=\sqrt{6}$, $a_{n+1}=\sqrt{6+a_n}$. Find $\lim\limits_{n \to
\infty} (a_n-3)6^n$.
First, we may obtain $\lim\limits_{n\to \infty}a_n=3$. Hence, $\lim\limits_{n \to \infty}(a_n-3)b^n$ belongs to a type of limit with the for... | Here's a proof that $b_n:=(3-a_n)6^n$ converges to a finite limit. This is merely a proof of existence of a finite limit (which I don't think is trivial).
It is shown here that $a_n$ is strictly increasing and has limit $3$.
Note that
$$3-a_{n+1} = \frac{3-a_n}{3+\sqrt{6+a_n}}$$
hence
$$(3-a_{n+1})6^{n+1} = \frac{6}{3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 7,
"answer_id": 0
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Show it is impossible to find a $+ve$ integer such that sum of its sq. and its cube is an integral multiple of the sq. of the next highest integer. Based on the problem statement, we can form the equation as :-
$x^2+x^3 = k(x+1)^2$
where $x$ and $x+1$ are positive integers and $k$ will be the integral multiple.
One ... | An easier approach might be to say
$$k = \frac{x^2+x^3}{(x+1)^2} = \frac{x^2(1+x)}{(x+1)^2}= \frac{x^2}{x+1}= \frac{x^2-1}{x+1}+\frac{1}{x+1} =x-1+\frac{1}{x+1}$$
and, for positive integer $x$, that is strictly between $x-1$ and $x$, so not an integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluate:- $\frac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
Evaluate:- $\dfrac{[4 + \sqrt{15}]^{3/2} + [4 - \sqrt{15}]^{3/2}}{[6 + \sqrt{35}]^{3/2} - [6 - \sqrt{35}]^{3/2}}$
What I Tried:- Let $a = 4 , b = \sqrt{15} , c = 6, d= \sqrt{35}$ . Then I get :-
$$\rightar... | Simply square the expression. A lot of cancelation will happen and you'll end up in $490/1690$. Now take square root to get the final answer $7/13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4109671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For small $|x|<1$, $\alpha\in \mathbb{R}$, is it true that $e^{\alpha t}-(1+\alpha x+\alpha^2 x^2)^{\frac{t}{x}}=O(x)$? So expanding out the second term gives:
$$(1+\alpha x+\alpha^2 x^2)^{\frac{t}{x}}=e^{\frac{t}{x}\ln(1+\alpha x+\alpha^2 x^2)}$$
$$=e^{\frac{t}{x}((\alpha x+\alpha^2 x^2)-\frac{1}{2}(\alpha^2 x^2+2\alp... | Your work is perfectly correct but, if I may suggest, it could be done faster.
Considering $$\Delta=e^{\alpha t}-(1+\alpha x+\alpha^2 x^2)^{\frac{t}{x}}$$ let $\alpha t= T$ and $\alpha x= X$ to make
$$\Delta=e^{T}-(1+X+X^2)^{\frac{T}{X}}$$
$$y=(1+X+X^2)^{\frac{T}{X}}\implies \log(y)=\frac T X \log(1+X+X^2)$$ By Taylor
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4114933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $f(x,y) =\sqrt{16-x^2-y^2}$ is continuous using $\epsilon$, $\delta$? I want to prove that for every $\epsilon > 0$ there exists a $\delta > 0$ such that
$$(x-x_0)^2 + (y-y_0)^2 < \delta^2 \Rightarrow \left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right| < \epsilon.$$
I note that $|x-x_0| \leq \delta$ and $|y-y_... | First note that if $|x-x_0|+|y-y_0|<1$ then each of $|x+x_0|,\ |y+y_0|$ is bounded by some $M>0\ $ (Proof: triangle inequality). So assume $\delta<1.$ There are two cases:
$1).\ $ If $x_0^2+y_0^2\neq16,\ $ set $N=\sqrt{16-x_0^2-y_0^2}.$ Then
$\displaystyle\left|\sqrt{16-x^2-y^2}-\sqrt{16-x_0^2-y_0^2} \right|\le \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Induction on $S(n,k)$ Consider $$S(n,k):= 1^{k}+\ldots+n^{k}.$$
I have to find a formula for $$S(n,3).$$
After trying for some n values I found out that
$$S(1,3)=1^{3}=1^{2}$$
$$S(2,3)=1^{2} +2^{2}=3^{2}$$
$$S(3,3)=1^{2} +2^{2} +3^{3}=6^{2}$$
$$S(4,3)=1^{2} +2^{2} +3^{3}+4^{3}=10^{2}$$
$$S(5,3)=1^{2} +2^{2} +3^{3}+4^{... | On inductive step we can use
$$1^3+2^3+\cdots +k^3+(k+1)^3 = (1+2+\cdots +k)^2 +(k+1)^3 =(1+2+\cdots +k+(k+1))^2$$
Last equality is same with
$$\left( \frac{k(k+1)}{2} \right)^2 +(k+1)^3 = \left( \frac{(2+k)(k+1)}{2} \right)^2$$
and can be easy checked.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4121617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Right solution to an integration problem One more from me.
Please help me understand the following. How do you get from here:
$\frac{1}{8}\int{1dx}-\frac{1}{8}\int{\cos(2x)dx}-\frac{1}{8}\int{\cos^2(2x)dx}+\frac{1}{8}\int{\cos^3(2x)dx}$
to here:
$=\frac{1}{16}x-\frac{1}{64}\sin(4x)-\frac{1}{48}\sin^3(2x)+C$
Is the abov... | We know that $$\cos^2(x) = \frac {1 + \cos(2x)}{2}$$ and $$\sin^2(x) = \frac {1 - \cos(2x)}{2}$$. So let us convert the integrand into something we can integrate:
$$\sin^4(x)\cos^2(x) = (\sin^2(x))^2\cos^2(x)$$ $$=\frac {1-\cos(2x)}{4}^2 \times \frac {1+\cos(2x)}{2}$$ $$=\frac {1 - 2\cos(2x) + \cos^2(2x)}{4} \times \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1}{1⋅2}+\frac{1}{3⋅4}+\frac{1}{5⋅6}+....+\frac{1}{199⋅200}$ = $\frac{1}{101}+\frac{1}{102}+\frac{1}{103}...+\frac{1}{200}$
Prove that $$\frac{1}{1⋅2}+\frac{1}{3⋅4}+\frac{1}{5⋅6}+....+\frac{1}{199⋅200}= \frac{1}{101}+\frac{1}{102}+\frac{1}{103}...+\frac{1}{200}$$
My Approach:
$T_{r}=\frac{1}{\left(2r... | $S=1-\frac{1}{2}+\frac{1}{3}...-\frac{1}{200}=\sum_{n=1}^{200}\frac{1}{n}-2\sum_{n=1}^{100}\frac{1}{2n}=\sum_{n=101}^{200}\frac{1}{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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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.