Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that if $2a^3 + 27c = 9ab,$ then the roots of $x^3 + ax^2 + bx + c = 0$ form an arithmetic sequence. I am not sure how to begin this problem. Can someone help me?
I have a hint: Let $y = x + \frac{a}{3}$ and rewrite $x^3 + ax^2 + bx + c = 0$ in terms of $y.$
How do I do this?
| Isolating $c$ in what we are given, and then plugging it into the polynomial, we have, $$x^3+ax^2+bx+c=x^3+ax^2+bx+\frac{1}{27}(9ab-2a^3)=$$
$$=\frac{1}{27}(3x+a)(9x^2+6ax+9b-2a^2),$$
which says $$x_1=-\frac{a}{3}$$ and
$$x_2+x_3=-\frac{6a}{9}=-\frac{2a}{3}=2x_1$$ and since $$x_2-x_1=x_1-x_3,$$ we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is ...
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_1x-36y+k_2=0$, then find $k_1$ and $k_2... | $$\frac{(x-4)^2}{16}+\frac{(y-3)}{9}=1 \implies x=4 \cos t+4,y=3 \sin t+3 ~~~(1).$$ The image $(X,Y)$ of $(x,y)$ in the line $ax+by+c=0$ is given by:
$$\frac{X-x}{a}=\frac{Y-y}{b}=-2\frac{(ax+by+c)}{a^2+b^2}$$
So we get $$\frac{X-4\cos t-4}{1}=\frac{Y-3\sin t-3}{-1}=-2\frac{4\cos t+4-3\sin t -3-2}{2}$$, we get
$$X=-4\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that ge... | That is already written as partial fractions.
It seems you want:
$\begin{align*}
\int \frac{d x}{(x^2 + 2 x + 3)^2}
&= \frac{1}{\sqrt{2}} \int \frac{d u}{(u^2 + 1)^2} \qquad u = \sqrt{2} (x + 1)
\end{align*}$
This last one yields to a trigonometric substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate : $I=\int_0^{\infty}\frac{\ln (1+ax+x^{2})}{1+x^2}\,dx$ Does the following integral have a closed form :
$$I=\displaystyle\int\limits_0^{\infty}\frac{\ln (1+ax+x^{2})}{1+x^2}dx$$
Where $|a|≤1$ ,
I was trying using Feynman's trick.
Define
$$I(b)=\displaystyle\int\limits_0^{\infty}\frac{\ln (b(1+x^{2})+ax)}{1+x... | Parameterising $2\sin a$ instead of $a$ yields the integral
$$
I(a)=\int_{0}^{\infty} \frac{\ln \left(1+2 x \sin a+x^{2}\right)}{1+x^{2}} d x,
$$
where $a\in (0, \frac{\pi}{2}). $
Differentiating $I(a)$ w.r.t. $a$ yields
$$
\begin{aligned}
I^{\prime}(a) &=\int_{0}^{\infty} \frac{2 x \cos a}{\left(1+x^{2}\right)\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is: $f'(x) = x \sqrt{16-x^2}$ Section 5.2
Can somebody verify this solution for me?
Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is:
$f'(x) = x \sqrt{16-x^2}$
So ... | The correction is : $\frac{4}{3}=\frac{-1}{3}(16)^{3/2} +C$ then $ 4=-1(16)^{3/2} +3C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The largest area of the rectangle inscribed into an acute triangle
A triangle of base $b$ and height $h$ has acute base angles. A rectangle is inscribed in the triangle with one side on the base of the triangle. Show that the largest such rectangle has base $b/2$ and height $h/2$, so that its area is one-half the area... | The solution for the general case requires a bit more effort than the right triangle case. We will be using a lot of "Basic Probotionality theorem for triangles"
$$\frac{y}{h} = \frac{b_1 - x_1}{b_1}$$
Similarly,
$$\frac{y}{h} = \frac{b_2 - x_2}{b_2}$$
Giving us,
$$\frac{x_1}{b_1} = \frac{x_2}{b_2}$$
Using Compedendo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that: $\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}l{m \choose l}{2l \choose l}^{-1}{m+l \choose l}=\sum_{i=1}^{m}i^2$ By observing and alteration of the OP's question we found this sum which is equivalent to the natural square number, $\sum_{i=1}^{m}i^2$
$$\sum_{l=0}^{m}(-1)^{m+l}4^{l-1}l{m \choose l}{2l \choose l}^{-1}{m+l ... | We seek to evaluate
$$\sum_{q=0}^m (-1)^{m+q} 4^{q-1} q {m\choose q} {2q\choose q}^{-1}
{m+q\choose q}.$$
We get from the binomial coefficients
$$\frac{m!}{(m-q)! \times q!}
\frac{q! \times q!}{(2q)!} \frac{(m+q)!}{m! \times q!}
= \frac{(m+q)!}{(m-q)! \times (2q)!}
= {m+q\choose m-q}.$$
Our sum becomes
$$\sum_{q=0}^m (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Find positive integer solutions for $n^4(n+1)+1=7^m$ This is the statement:
Find all pairs of natural numbers $(n,m)$ that satisfies equation $$n^4(n+1)+1=7^m$$
I found one solution (2,2) and believe there is no more solutions. I'm beginner in number theory and don't know how to even start this. Can anyone help?
| Use $$n^5+n^4+1=n^5-n^2+n^4-n+n^2+n+1=(n^2+n+1)(n^3-n+1).$$
Indeed, we obtain that there are non-negative integers $p$ and $q$, for which
$$n^2+n+1=7^p$$ and $$n^3-n+1=7^q.$$
Now, if $p=0$ or $q=0$ we obtain solutions $(0,0),$ $(-1,0)$ in integer numbers.
Let $p\geq1$ and $q\geq1$.
Thus, from the first equation we obta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Please check my work! Question about cubic polynomials I need some help with this problem. Here is the link. Can you please tell me if there is an easier way to show that cubic polynomials have a real root? The question is in an analysis book from the continuity section so it has to use that. Here is the latex:
Show th... | Assume wlog $a=1$ by factoring $a$ out, which is doable since $a\ne0$.
Simpler inequalities can be deduced by taking $x\ge|b|+|c|+|d|+1$ so that we have
$$x+b,x+c,x+d\ge1$$
and noticing that we then have
\begin{align}f(x)&=x^3+bx^2+cx+d\\&=(x+b)x^2+cx+d\\&\ge(x+c)x+d\\&\ge x+d\\&\ge1\end{align}
and similarly that if we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.
My approach: Given that $I_n=\int_0^1\frac{x^... | You can still use Sandwich theorem, if you haven't yet seen dominated convergence theorem :
Let $ n $ be a positive integer.
As you said, using the substitution $ \left\lbrace\begin{aligned}y&=x^{n}\\ \mathrm{d}y &=n x^{n-1}\,\mathrm{d}x\end{aligned}\right. $, we get : $$\int_{0}^{1}{\frac{n x^{n}}{\sqrt{1+x^{2}}}\,\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
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How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T
\begin{pmatrix}
10 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25 \\
\en... | In our case we have two given symmetric matrices $B,C$,
$$
\begin{aligned}
B
&=
\begin{bmatrix}
10 &&\\&20&\\&&25
\end{bmatrix}=B_1^2 =B_1^TB_1=B_1B_1^T\ ,\text{ where }
\\
B_1
&=
\begin{bmatrix}
\sqrt{10} &&\\&2\sqrt 5&\\&&5
\end{bmatrix}=B_1^\ ,
\\[3mm]
C &=
\begin{bmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Calculate $ \int_0^1{\frac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx$
Given a function $f(x)$ that is differentiable on $\left[0; 1\right]$
satsifies: $$ f(1) = 1 $$ $$ f(x)f(1-x) = e^{x^2 - x} $$
Calculate: $$ \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx $$
Attempt number 1:
Using integration by parts... | From the given functional equation, we obtain that $f(0)=f(1)=1.$
Now setting $I=\int_0^1 \frac{(2x^3-3x^2)f'(x)}{f(x)}\mathrm dx$ and integrating by parts gives $$I=(2x^3-3x^2)\int \mathrm d\left(\log f(x)\right)-\int(6x^2-6x)\log f(x)\mathrm dx.$$ Substituting the limits gives that $$I=-\int(6x^2-6x)\log f(x)\mathrm ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that : $m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$ Let $m_{a},m_{b},m_{c}$ be the lengths of the medians and $a,b,c$ be the lengths of the sides of a given triangle , Prove the inequality :
$$m_{a}m_{b}m_{c}\leq\frac{Rs^{2}}{2}$$
Where :
$s : \operatorname{Semiperimeter}$
$R : \operatorname{circumradius}$
I know the... | In the standard notation we need to prove that:
$$\frac{1}{8}\sqrt{\prod_{cyc}(2a^2+2b^2-c^2)}\leq\frac{1}{2}\cdot\frac{abc}{4S}\cdot\frac{(a+b+c)^2}{4}$$ or
$$a^2b^2c^2(a+b+c)^3\geq\prod_{cyc}(2a^2+2b^2-c^2)\prod_{cyc}(a+b-c).$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $$\prod_{cyc}(2a^2+2b^2-c^2)=\pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Quadratic programming: KKT Optimality conditions I am struggling with an exercice with the following quadratic program:
$$min:x_{1}x_{2} + x^{2}_{1} + \frac{3}{2}x^{2}_{2} + 2x^{2}_{3} + 2x_{1} + x_{2} + 3x_{3}$$
subject to
$$x_{1} + x_{2} + x_{3} = 1$$
$$x_{1} − x_{2} = 0$$
$$x_{1}, x_{2}, x_{3} ≥ 0$$
Is the quadratic... | Let $x_1=x+2=a$. Thus, $c=1-2a$ and
$$x_{1}x_{2} + x^{2}_{1} + \frac{3}{2}x^{2}_{2} + 2x^{2}_{3} + 2x_{1} + x_{2} + 3x_{3}=$$
$$=\frac{23}{2}a^2-11a+5=\frac{23}{2}\left(a^2-\frac{22}{23}a+\frac{10}{23}\right)=$$
$$=\frac{23}{2}\left(a-\frac{11}{23}\right)^2+5-\frac{121}{46}\geq\frac{109}{46}.$$
The equality occurs for ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Suggestions for $ \lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2} $? I'm trying to evaluate $$
\lim_{(x,y)\to (0,0)} \frac{x-\sqrt{xy}}{x^2-y^2}
$$
over the domain $x>0$, $y>0$.
============
My attempt:
$f(x,x^2)\to +\infty$; so if the limit exists it must be $+\infty$.
I tried to evaluate the limits "near" $(x,... | In order to disprove your conjecture that $\lim_{(x, y) \to (0, 0)} f(x, y) = +\infty$, take the limit along the curve $(x, y) = (t^5, t)$ as $t \to 0^+$. Then we have:
$$f(t^5, t) = \frac{t^5 - t^3}{t^{10} - t^2} = \frac{t(1-t^2)}{1-t^8}$$
and from the last expression, we see that $f(t^5, t) \to 0$ as $t \to 0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find delta for the limit I'm having difficulty to solve this problem:
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$.
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displayst... | Firstable you can make $|1+x| < 1 \implies |x| -1 =|x| - |1| \le |1+x| < 1$. So:
$|x| < 2$. Now call the expression just before the $\epsilon$ in your work above $D$, then $D \le \dfrac{|1-x^2|}{2\cdot \frac{1}{\sqrt{2}}} \le |1-x^2|= |1+x||1-x|\le |1+x|(1+|x|) < 3|1+x|< \epsilon $ if $1+x| < \frac{\epsilon}{3}$. Thus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, ... | Hint: You can solve by showing $${1\over 1-x^2} \geq {27\over 32}(x+1)$$ for $x\in (0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n. If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.
I did and I got
$$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?
| Note that $m$ and $n$ have to be non-negative integers, and perhaps it is implied in the question that $m$ and $n$ are positive as well. The only positive solution to $3n+2m=12$ is $n=2,m=3$, so $m+n=5$.
If zero powers are allowed, then we also have $n=4,m=0$ and $n=0,m=6$, for sums of $4$ and $6$ respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straight... | My approach.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
$$=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\:+\int _1^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
Let $\displaystyle x=\frac{1}{t}$ for the $2$nd integral.
$$\int _0^1\frac{\ln \left(t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
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How to show the divergence of the improper integral $\int_{0}^{\infty} \frac{dx}{1+x^{2}\sin^{2}(x)}$? How can I show the given improper integral $\int_{0}^{\infty} \frac{dx}{1+x^{2}\sin^{2}(x)}$ is divergent?
Approach:
\begin{align*}
x^{2}\sin^{2}(x) \le x^2 \\
1+x^{2}\sin^{2}(x) \le 1+ x^2 \\
\frac{1}{1+x^{2}\sin^{... | Let $ n $ be a positive integer, we have : \begin{aligned} \int_{0}^{n\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}&=\sum_{k=0}^{n-1}{\int_{k\pi}^{\left(k+1\right)\pi}{\frac{\mathrm{d}x}{1+x^{2}\sin^{2}{x}}}}\\ &=\sum_{k=0}^{n-1}{\int_{0}^{\pi}{\frac{\mathrm{d}x}{1+\left(x+k\pi\right)^{2}\sin^{2}{x}}}}\\ &\geq\sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of ... | As $\{x^2,y^2, xy\}$ go to zero more quickly by one order degree than $\{x,y\}$ we have the near zero the dynamical system behaves as
$$
\dot x = y\\
\dot y = x
$$
The Jacobian at the origin is
$$
J = \left(
\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array}
\right)
$$
characterizing a saddle point which is unstable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Show $\sqrt{r_2^2-x^2}-\sqrt{r_1^2-x^2}\geq r_2-r_1$ for vertical chord distance between circles Let $0<r_1<r_2$.
Consider two circles centered at the same point, one with radius $r_1$ and the other with radius $r_2$. According to all of the pictures I have drawn, each vertical line from the smaller circle to the la... | Given $r_2 > r_1$ and $x\in [0,r_1]$, we have
$$r_1 \ge \sqrt{r_1^2-x^2}$$
$$ 2r_1(r_2-r_1) \ge 2(r_2-r_1)\sqrt{r_1^2-x^2}$$
$$ (r_2-r_1)^2 + 2r_1(r_2-r_1) + r_1^2-x^2 \ge (r_2-r_1)^2 + 2(r_2-r_1)\sqrt{r_1^2-x^2}+ (r_1^2-x^2) $$
$$ (r_2-r_1+r_1)^2-x^2 \ge \left[(r_2-r_1) + \sqrt{r_1^2-x^2}\right]^2$$
$$ \sqrt{r_2^2... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
simplify $\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$ -- if $x$ is negative? Was given this recreational problem: simplify $$\sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$$
The solution isn't hard. Let $y = \sqrt[3]{x \sqrt[3]{ x \sqrt[3]{x ...}} }$, then $y^3 = xy$, $y=\sqrt{x}$
The problem didn't specify, but if $x$ is ne... | Let $y=\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\dots}}}$
So,
$y=\sqrt[3]{xy}$
Now raise both side to the power $3$, we get:
$y^3=xy$
Now, $y\ne0$ because $x$ is negative as mentioned in the problem statement, we can divide both sides by $y$, we get:
$y^2=x$
However $y^2=x$ has no real solution since $x$ is negative. Hence the gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
System of congruences where $\gcd(m, n)\ne1$ I have to solve this system of congruences:
$$
\begin{cases}
x^2+2x+2\equiv 0\pmod{10}\\
7x\equiv 20\pmod{22}
\end{cases}
$$
after some calculations
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cas... | Solving the linear congruence gives $x\equiv 6\bmod 22$. Write $x=22n+6$ and substitute into the quadratic congruence:
$$
(22n+6)^2 +2(22n+6)+2 \equiv 0\bmod 10
$$
$$
\Rightarrow 4n^2+4n+6+4n+2+2 \equiv 0\bmod 10
$$
$$
\Rightarrow 4n^2+8n \equiv 0\bmod 10
$$Everything is even, so let's cancel 2:
$$
\Rightarrow 2n^2+4n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Linear Algebra: Looking at points M on a circle. Consider the points
$
M=\left\{
\begin{pmatrix}
0\\
2
\end{pmatrix}
,
\begin{pmatrix}
2\\
2
\end{pmatrix}
,
\begin{pmatrix}
1\\
0
\end{pmatrix}
,
\begin{pmatrix}
1\\
2
\end{pmatrix}
\right\}
$
in $\mathbb{R}^2$.
The points on a circle with radius r and center in $(a,b)^... | Take a generic circle $\mathcal C_{a,b,r}=\{(X,Y)\in \mathbb R^2 \ | \ 2aX+2bY+(r^2-a^2-b^2) = X^2+Y^2\}$.
We have:
\begin{equation}
M\subseteq \mathcal C_{a,b,r} \Longleftrightarrow
\begin{cases}
2a (0) +2b (2) + (r^2-a^2-b^2) = 0^2+2^2 = 4\\
2a (2) +2b (2) + (r^2-a^2-b^2) = 2^2+2^2 = 8\\
2a (1) +2b (0) + (r^2-a^2-b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$ I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities.
For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^... | Here is another approach:
Denote $t=a+b$ and $s=a-b$, then $f(x)$ becomes to be $(x^2-t^2)(x^2-s^2)$. Henece $f(x)$ is quintic which is very easy to work with, more specifically,
$$f(x)=x^4-(s^2+t^2)x^2+t^2s^2.$$
Now, you can make another substitution, namely $y=x^2$ to obtain a quadratic, which you know how to find a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x... | Put $g(x)=(x+3)^{1/3}$. We have $f(x)=g(x)(x-27)$. It suggests that if a Taylor series of $f$ at $x_0=27$ is $\sum_{n=0}^\infty \frac {f^{(n)}(x_0)}{n!}(x-x_0)^n$ and a Taylor series of $g$ at $x_0=27$ is $\sum_{n=0}^\infty \frac {g^{(n)}(x_0)}{n!}(x-x_0)^n$ then for each $n\ge 0$ we have $$\frac {g^{(n)}(x_0)}{n!}= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}dx$ as partial fraction solved using matrix equation in order to solve the integral $$\int \frac{2x^5-2x^4+2x^3+3}{2x^4-2x^3-x^2+1}\mathrm dx,$$
the expression inside the integral can be expressed as
$$(2x^5-2x^4+2x^3+3/2x^4-2x^3-x^2+1)= x+(A/(x-1))+(B/(x-1)^2)+(Cx... | Decompose the simplified integrand as,
$$\frac{3x^3-x+3}{(x-1)^2(2x^2+2x+1)}=\frac A{x-1}+\frac B{(x-1)^2}+\frac{Cx+D}{2x^2+2x+1}$$
The coefficients $A$, $B$, $C$ and $D$ are obtained successively as follows. Multiply both sides by $(x-1)^2(2x^2+2x+1)$
$$3x^3-x+3=A(x-1)(2x^2+2x+1)+B(2x^2+2x+1)+(Cx+D)(x-1)^2\tag 1$$
Fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a$ red balls and $b$ blue balls in an urn. Probability when a game lasts forever. In an urn there are $a$ indistinguishable red balls and $b$ indistinguishable blue balls. Every round, you take one ball in random from the urn. If this ball is blue, game over. If this ball is red, you put it back into the urn and put a... | This calculation of $E_{a,b}$ will get you started. On the $n-$th round, the urn contins $a+n-1$ red balls and $b$ blue balls. The probability that the $n-$ th round is the last is $\frac{b}{a+b+n-1}$. The only way of playing the $n-$th round is by surviving all the previous rounds.Thus the probability that the game en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimize $|a-1|^3+|b-1|^3$ with constant product $ab=s$ Let $0<s$, and define
$$ F(s):=\min_{a,b \in \mathbb{R}^+,ab=s} \left(|a-1|^3+|b-1|^3\right). $$
I would like to find proofs for the claim
$$
F(s)=\begin{cases}
1 - 3 s - 2s^{3/2}=F\big(a(s),b(s)\big), &\text{ if } 0<s\le1/9, \\
2 + 6 s - 2(3 + s)s^{1/2}=F(\sqrt ... | For $0 < s < 1$, we have
\begin{align}
F(s) &= \min_{a, b > 0;\ ab = s}\ |a-1|^3 + |b-1|^3 \\
&= \min_{a\ge b > 0;\ ab = s}\ |a-1|^3 + |b-1|^3\tag{1}\\
&= \min_{0 < b \le \sqrt{s}}\ \left|\frac{s}{b} - 1\right|^3 + (1-b)^3\\
&= \min_{s \le b \le \sqrt{s}}\ \left(1 - \frac{s}{b}\right)^3 + (1-b)^3. \tag{2}
\end{align}
E... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Quartic polynomials of a complex variable I want to answer the following:
The equation
$$
a \bigg(z + \frac{1}{z}\bigg)^2
+ b \bigg(z + \frac{1}{z}\bigg)
+ c = 0
$$
has four solutions. Find which quartics can be put in this form after apply a linear change of vars.
| The crucial step is the substitution:
$u = z + \frac{1}{z}$
Then, we obtain the quadratic:
$au^{2} + bu + c = 0$
By the Quadratic Formula:
$u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Let the roots of the above quadratic be $s$ and $t$. Then, for $s$:
$z + \frac{1}{z} = s$
$z^{2} - sz + 1 = 0$
By the Quadratic Formula:
$z = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The probability that a binomial coefficient is divisible by 12. I do not know how to solve the following question:
What is the probability that $12$ divides $\binom{n}{12}$ for a randomly chosen natural number n?
Context:
I set this as a question in a problem set. I was trying to write these problems from memory whil... | $$\binom{n}{12}=\frac{n(n-1)(n-2)\ldots(n-11)}{12!}=\frac{n(n-1)(n-2)\ldots(n-11)}{2^{10}\cdot 3^5\cdot 5^2\cdot 7\cdot 11}$$
Any $12$ consecutive integers include $4$ consecutive factors of $3$, at least one of which must be a multiple of $9$, and six consecutive multiples of $2$. Three of these must be multiples of $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Showing that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence. I need to show that $x_{n+1} = 1/2(x_n + 2/x_n)$ is a decreasing sequence, where $x_1 = 2$ and n = 1,2,3,....
I tried to show this with induction, where since $x_1= 2$ then $x_2 = 1.5$, hence the base case satisfies.
Assuming that $x_n \geq x_{n+1}$ , ... | At first one could show that $f(x)\geq 2$ for all $x>0$, where $f(x):=\frac12\left(x+\frac2{x}\right)$. This follows immediately from $f(x)^2-4= \frac14\left(x-\frac2{x}\right)^2\geq0$. But then $f(x)\leq x$ is equivalent to $\frac2{x}\leq x$ which is always true when $x^2\geq 2$. But this holde true for $x_0=2$ by as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Domain of $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$ $f(x,y) = \ln((16-x^2-y^2)(x^2+y^2-4))$
I'm stuck in this one because this can be rewritten as:
$$f(x,y) = \ln(16-x^2-y^2) + \ln(x^2+y^2-4)$$
Yet, the domain of the given function is $\{(16-x^2-y^2>0)\land(x^2+y^2-4>0)\} \lor \{(16-x^2-y^2<0)\land (x^2+y^2-4<0)\}$. But ... | When does $log(xy) = log(x) + log(y)$ hold?
Only when $x$ and $y$ are both greater than zero! Does that solve your doubt?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Question: Prove that for any integers $a,b,c,$ there exists a positive integer $n$ such that the number $n^3+an^2+bn+c$ is not a perfect square.
Solution: Let $f:\mathbb{N}\to\mathbb... | Let $a,b,c \in \mathbb Z$, and let $f(n)=n^3+an^2+bn+c$, $n \in \mathbb N$. We show that at least one of $f(1)$, $f(2)$, $f(3)$, $f(4)$ is not a perfect square. We use the fact that $m^2 \equiv 0\:\text{or}\:1\pmod{4}$ for $m \in \mathbb Z$.
Suppose $f(n)$ is a perfect square, $n \in \{1,2,3,4\}$. We note that
$$ \begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on bo... | A few proofs I came up with are:
Proof 1: This uses real analysis. Let $f(x) = \sin^2x + \cos^2x \implies f'(x) = 2\sin x \cos x - 2\sin x \cos x = 0$. Thus, $f(x)$ is a constant function. To find the value of $f(x)$, it is sufficient to find $f(c)$, where c is any convenient constant (say 0). Therefore, $f(x) = f(0) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Generating function for the sequence $(a_k) = (1,-1,2,-2,3,-3,4,-4,...).$ I'm trying to find the generating function for $(a_k)=(1,-1,2,-2,3,-3,4,-4,...)$. I know the answer is $$A(x) = \frac{1}{1-x} \cdot \frac{1}{(x+1)^2}$$ but I can't figure out how to get there.
First I broke down the sequence, $$(a_k) = 1-x+2x^2-2... | This is
$$\sum_{i=1}^\infty i(x^2)^{i-1}-x\sum_{i=1}^\infty i(x^2)^{i-1}$$
which is equal to
$$\frac1{(1-x^2)^2}-\frac x{(1-x^2)^2} = \frac{1-x}{(1-x)^2(1+x)^2}=\frac{1}{(1-x)(1+x)^2}$$
Here I have used the fact that $\sum_{i=1}^\infty{ix^{i-1}}$ is the derivative of a geometric series and used the substitution $x\maps... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha... | Here is another, slightly more sophisticated, take:
The other root is $\beta= \gamma ^2+2\gamma-4$, where $\gamma = \alpha^2+2\alpha-4$.
Expanding $\beta$ in terms of $\alpha$ and reducing it mod $\alpha^3+2\alpha^2-5\alpha+1$ gives $\beta=-\alpha^2 - 3 \alpha + 2$. Note that $\beta=g(\gamma)=g(g(\alpha))$, where $g(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Using definition of limit to prove that $\lim\limits_{x \to 1}\frac{2-x}{4-x}=\frac{1}{3}$ I hit a block when discovering a negative $\delta$. This is how:
I need to show that$$\forall \epsilon>0 \; \exists \delta>0 \text{ s.t. } \mid x-1 \mid < \delta \Rightarrow \Bigl| \frac{2-x}{4-x}-\frac{1}{3} \Bigr| < \epsilon$$... | If $\delta < 1$ then $|x-1| < \delta$ implies
$$x-1 \in [-1,1] \implies x-4 \in [-4,-2] \implies |x-4| \in [2,4] \implies \frac1{|x-4|} \in \left[\frac14,\frac12\right]$$
so $\frac1{|x-4|} \le \frac12$.
Therefore we have
$$\left|\frac{2(x-1)}{3(4-x)}\right| = \frac23 \cdot \frac1{|x-4|} \cdot |x-1| < \frac23 \cdot \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{al... | First Identity
$$
\begin{align}
&\sum_{j=n-k}^n\binom{n}{j}(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&=\sum_{j=n-k}^n\binom{n}{j}(1-x)^{n-j-1}x^{j-1}[j(1-x)-(n-j)x]\tag1\\
&=\sum_{j=n-k}^nn\binom{n-1}{j-1}(1-x)^{n-j}x^{j-1}-\sum_{j=n-k}^nn\binom{n-1}{j}(1-x)^{n-j-1}x^{j}\tag2\\
&=\sum_{j=n-k-1}^{n-1}n\binom{n-1}{j}(1-x)^{n-j-1}x^{j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Is there a way to show that $x^2=1$ is the only solution of $x^2+3=k^2$ for $x,k\in\mathbb{N}$ The question i'm trying to solve is: The sum of all natural numbers a such that $a^2-16a+67$ is a perfect square is (a)10, (b)12, (c)16, (4)22.
$a^2-16a+67$
= $a^2-16a+64+3$
=$(a-8)^2+3$
This is a perfect square when $(a-8)... | If $k,x \in \Bbb N$ and $3=k^2-x^2$
then $k^2>x^2\,$ so $k>x\,$ so $k\ge x+1\,$ so $k^2\ge (x+1)^2$
so $3=k^2-x^2\ge (x+1)^2-x^2=2x+1$
so $3\ge 2x+1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A function divisible by $p$ I want to ask about what is the intuition for making functions like these
*
*$4^n+2$ is divisible by $3$
*$ 2^{4n+2}+3^{n+2}$ is divisible by $13$
And if so, how can I make my own ones? Thank you all for your time.
| There is a simple explanation using the binomial theorem:
$$
4^n+2 = (3+1)^n+2 = 3a + 1 + 2
$$
and
$$\small
2^{4n+2}+3^{n+2}
=4 \cdot 16^n + 9 \cdot 3^n
=4 \cdot (13+3)^n + 9 \cdot 3^n
=13a + 4 \cdot 3^n + 9 \cdot 3^n
=13a + 13 \cdot 3^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Olympiad inequalities. I am trying to solve this problem but unable to. Can someone please give some hint or help.
I have to use holder inequality.
For $a,b,c$ positive real numbers prove.
$ \frac{a^6}{b^2+c^2} + \frac{b^6}{c^2+a^2} + \frac{c^6}{a^2+b^2} \ge \frac{abc(a+b+c)}{2}$
| By Cauchy-Schwarz inequality (Titu's lemma or Sedrakyan's inequality specifically):
$$F=\frac{a^6}{b^2+c^2}+\frac{b^6}{c^2+a^2}+\frac{c^6}{a^2+b^2} \ge
\frac{(a^3+b^3+c^3)^2}{2(a^2+b^2+c^2)}$$
Next use, $$(a^3+b^3+c^3) \ge \frac {(a^2+b^2+c^2)(a+b+c)}{3}~~\text{(Tchebechev)}$$
So,
$$F \ge \frac{1}{18} \ \underbrace{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Derivative of the inverse of a parametric function with respect to the parameter Given a function
$y = f(x)$
its inverse with respect to the argument x is
$x = F(y) = f^{-1}(y)$
Now, suppose that function has a parameter p
$y = f(x;p)$
whose inverse is
$x = F(y;p) = f^{-1}(y;p)$
How do I compute the parametric derivati... | We have the following relations
\begin{align}
y &= f(x; p), \tag{1}\\
x &= f^{-1}(y; p). \tag{2}
\end{align}
By taking derivative of (1) with respect to $p$ and by noting (2), we have
$$0 = \frac{\partial f}{\partial x} \frac{\partial x}{\partial p} + \frac{\partial f}{\partial p}$$
and hence
$$\frac{\partial x}{\parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\r... | Note
$$\frac{1}{\sqrt{1-x}}=\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}$$
and hence
$$\frac{1+x}{\sqrt{1-x}}=(1+x)\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}=\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}+\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^{n+1}}{2^n n!}.$$
So
$$ \frac{d^{100}}{dx^{100}}\frac{1+x}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose tha... | We can write
I) $z=\frac{a-yc}{b}$ from (1)
Now, multiplying $y$ on both sides of (3) we get $yc=y^2a+xyb$.
So, from I) we get $\frac{a-ay^2}{b}-xy=z$.....(1')
Similarly from equation (2) we get
II) $z=\frac{b-xc}{a}$ from (2)
Now, multiplying $x$ on both sides of (3) we get $xc=xya+x^2b$.
And we get from (2) $\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 1
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If $a^2 + b^2 + c^2$ is divisible by $16$, then show that$ a^3 + b^3 + c^3$ is divisible by $64$. Where $a, b, c \in \mathbb{Z}$. If $a^2 + b^2 + c^2$ is divisible by $16$, then show that $a^3 + b^3 + c^3$ is divisible by $64$; where $a, b, c \in \mathbb{Z}$.
I began by proving that if $(a^3+b^3+c^3) -(a^2+b^2+c^2)$ is... | I don't see any way offhand to continue with what you started to finish the proof.
Instead, note that perfect squares have remainders of only $0$, $1$, $4$ or $9$ when divided by $16$. As such, if $a^2$ had a remainder of $1$, the other $2$ would need to have remainders which sum to $15$. This is not possible as the on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that the value of $\Delta$ is an integer for the given determinant
$\Delta=\begin{vmatrix} \sqrt 6& 2i& 3+\sqrt 6 \\ \sqrt{12}&\sqrt 3 +\sqrt8i &3\sqrt 2 +\sqrt 6i \\ \sqrt{18} &\sqrt 2+ \sqrt {12}i &\sqrt {27}+2i \end{vmatrix}$
taking $\sqrt 6$ out from the first column and performing the following operations ... | The obvious way is to simply find the determinant right away. But the fact that you have asked this must mean that must not be allowed. So to prove this without finding the determinant at all:
Take a factor of $\sqrt3$ from $R_2$ and $\sqrt2$ from $R_3$.
$$\Delta=6\begin{vmatrix} 1 & 2i & 3+\sqrt6 \\
0 & 1 & \sqrt2 i-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3680882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$ For $a,b,c>0$$,$ prove$:$ $$(a+b+c)^3 (a+b-c)(b+c-a)(c+a-b) \leqq 27a^2 b^ 2 c^2$$
My proof by S-S method$,$ see here.
Another proof by $pqr$ method$:$
Let $p=a+b+c,\,q=ab+bc+ca,\, r=abc.$ This inequality equivalent to$:$ $${p}^{6}-4\,{p}^{4}q+8\,{p}^{3}r+27\... | Geometric approach :
Let $a,b,c$ be the side of an triangle $ABC$ then your inequality is :
$$T\leq\frac{3\sqrt{3}abc}{4(a+b+c)}$$
Or $$4\sqrt{3}T\leq \frac{9abc}{a+b+c}$$
Where $T$ is the area of the triangle $ABC$
For a proof see (found on Wikipedia):
Posamentier, Alfred S. and Lehmann, Ingmar. The S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$
My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we ge... |
Let $M$ the second intersection of $AD$ and the circumcircle of triangle $ABC$. By power of a point, $BD \cdot DC=AD\cdot DM$. Let $x=AD$ and $y=DM$. The inequality can be rewritten as $$x^2y\leq \dfrac{32}{27}.$$
It is clear that $$(x-2y)^2(4x+y)\geq 0$$ which is equivalent to $$x^2y\leq (x+y)^3\cdot \dfrac{4}{27}$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$x-\sin(x) \geq \dfrac{x^3}{(x+\pi)^2}$ Let $x \geq 0.$ I need to prove that $x-\sin(x)\geq\dfrac{x^3}{(\pi+x)^2}.$
I tried the derivative, of $f(x)=x-\sin(x)-\dfrac{x^3}{(\pi+x)^2}$ which is $1-\cos(x)-\dfrac{x^2(x+3\pi)}{(\pi+x)^3},$ but it has a complicated formula.
Any ideas, hints?
Edit: sorry, there was a mistak... | Put $x=z-\pi$; it suffices to show the inequality for for $z\geq \pi$.
$$
z-\pi -\sin(z-\pi)\geq \frac{(z-\pi)^3}{z^2}
$$
$$
z-\pi +\sin(z)\geq z -3\pi+\frac{3\pi^2}{z} -\frac{\pi^3}{z^2}
$$
$$
2\pi +\sin(z)\geq \frac{3\pi^2}{z} -\frac{\pi^3}{z^2}
$$Note that $\sin(z)\geq -1$. If we set
$$
2\pi- 1=\frac{3\pi^2}{z} -\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Sigmoid Function Typically sigmoid function is calculated as
1/(1 + exp(-x))
I see sometimes it is calculated as
1 - 1/(1 + exp(x))
or even
exp(x)/(1 + exp(x))
Could you clarify the difference please?
| $$
\frac
{1}{1+\exp(-x)}
=
\frac
{1}{1+e^{-x}}
=
\frac
{1}{1+e^{-x}}
\cdot
\frac
{e^x}{e^x}
=
\frac
{e^x}{e^x+1}
=
\frac
{e^x}{1+e^x}
=
\frac
{\exp(x)}{1+\exp{x}}
$$
and also
$$
1-\frac
{1}{1+\exp(x)}
=
1-\frac
{1}{1+e^{x}}
=
\frac{1+e^x}{1+e^x}
-
\frac{1}{1+e^x}
=
\frac{(1+e^x)-1}{1+e^x}
=
\frac
{e^x}{1+e^x}
=
\frac
{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3684762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integer solution of $a+b+c=15$ with restrictions on $a, b$ and $c$ I'm checking this problem:
Find the integer solutions of $a+b+c=15$ if $a$ is multiple of 3, $b$ is less than 10 and $c$ is multiple of 2. With $a, b, c ≥ 0$
We can make a series of polynomials with combinatorics. The polynomials would be:
$$P_a(x)=1+x... | Can't this simply be done by making use of the fact that $b + c = 15 - a$ is divisible by $3$ to conclude that $15-a$ can only equal one of $0,$ $3,$ $6,$ $9,$ $12,$ $15$? In the lists below, keep in mind that the values are nonnegative and $b = 0,\,1,\,\ldots,\,9$ and $c$ is even. Thus, for example, if $b+c = 9,$ then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3685534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Finding $\lim_{(x,y) \to (0,0)} \frac{x^5y^5}{|x|^9 + |y|^{11}}$ Setting $y = mx^k$ suggests the limit evaluates to $0$, and as far as the techniques I know, dismisses the possibility of its non-existence. But I'm having trouble using the Squeeze Theorem to prove that it does exist.
$$\lim_{(x,y) \to (0,0)} \frac{x^5y... | By weighted AM-GM inequality, we get
\begin{align*}
\lvert x \rvert^9 + \lvert y \rvert^{11}
&= \frac{11}{20} \cdot \frac{20}{11} \lvert x \rvert^9 + \frac{9}{20} \cdot \frac{20}{9} \lvert y \rvert^{11} \\
&\geq \left(\frac{20}{11} \lvert x \rvert^9\right)^{11/20} \left(\frac{20}{9} \lvert y \rvert^{11}\right)^{9/20} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Modular calculation high exponent? I want to show, that $5^{96}\equiv -1 \pmod{193}$, without using the formula for quadratic residue.
So far I have :
$5^{96}\equiv 5^{4\cdot24} \equiv 625^{24}\equiv 46^{24}\equiv 186^{12}\equiv -7^{12}\equiv 7^{12}\equiv 7^{3\cdot4}\equiv 150^4\equiv -43^4\equiv 43^4\equiv 112^2\equi... | Another idea: since $96$ is a multiple of $3$ and you know $(5^{96})^2\equiv 1$, you have $(5^{32})^3\pm 1\equiv 0$. Factor this using the familiar factorization for the sum or difference of cubes:
$5^{96}\pm1=(5^{32}\pm1)×(5^{64}\mp5^{32}+1)$
Now square $5$ six times to get
$5^2\equiv 25, 5^4\equiv 46, 5^8\equiv 186... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Help with complex contour integral involving a logarithm and a square root I've been trying to do the following integral using the 'keyhole' contour: $\int_0^{\infty} \frac{\ln(x)}{\sqrt{x}(1+x^3)}d x.$ I know the result is $-\frac{2\pi^2}{3\sqrt{3}},$ yet when I use the Residue Theorem the result amounts to something ... | We seek to compute using contour integration the integral
$$J = \int_0^\infty \frac{\log{x}}{\sqrt{x}(x^3+1)} \; dx$$
We work with
$$f(z) = \frac{\mathrm{Log}(z)}
{\exp(\mathrm{Log}(z)/2) (z^3+1)}$$
where $\mathrm{Log}(z)$ is the branch with argument in $[0,2\pi).$ We
use a keyhole contour wih radius $R$ and the slot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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integral of $x^{1/n}\sin x$ how to prove $\lim_{n \to \infty} \int_0^{\pi/2} x^{1/n}\sin x dx = 1$? I've tried it by using squeeze theorm. It worked on right side by putting $\pi/2$ into $x$, but i have no idea for left side
| Here's an elementary approach (without using any measure theory):
Since $\int_0^\frac{\pi}{2} \sin(x)dx=1$, we have that
$$
\left| \int_0^\frac{\pi}{2} x^\frac{1}{n}\sin(x)dx-1\right|=\left|\int_0^\frac{\pi}{2}\sin(x)\left(x^\frac{1}{n}-1\right)dx\right|\le \int_0^\frac{\pi}{2}\left|x^\frac{1}{n}-1\right|dx
$$
were th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is this series convergent? $1 + 1/2 + 1/2 + 1/4 + 1/4 + 1/4 + 1/4 + ...$ Is this series convergent? How to prove?
$$1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + ... + \frac{1}{8} \ (8 \times 1/8) + \frac{1}{16} + ...$$
It's equal to $\{1 + 1 + 1 + 1 + ...\}$, wh... | Observe that the $(2^n - 1)$th partial sum is given by $$s_{2^n - 1} = 1 + 2 \cdot \frac 1 2 + 4 \cdot \frac 1 4 + \cdots + 2^{n - 1} \cdot \frac 1 {2^{n - 1}} = \underbrace{1 + 1 + 1 + \cdots + 1}_{n \text{ summands}} = n.$$ Consequently, we have that $\lim_{n \to \infty} s_{2^n - 1} = \lim_{n \to \infty} n = \infty,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Leading behavior of a logarithmic integral The integral
$$\int_0^1 du \frac{\ln u}{u^2+b^2}$$
is governed by the divergence at $u=0$. I'm interested in the leading behavior of the integral, and it seems reasonable to expand around $u=0$ (as it is typically done in Laplace's method for instance), but what should be done... | Here's my attempt. First we start off by integrating by parts.
\begin{equation}
\int_0^1 \frac{\ln u}{u^2+b^2} du = \frac{1}{b} \arctan\left( \frac{u}{b} \right) \ln u \bigg |_0^1 - \frac{1}{b} \int_0^1 \frac{\arctan\left(\frac{u}{b} \right)}{u} du.
\end{equation}
Assuming $b\neq0$, we can evaluate the limit
\begin{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3700566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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I saw steps in an equation that added a 1 and a -1. Why is this okay? $(x^2+x)-(x+1)$=
$(x \cdot x + x \cdot 1)+(-1)\cdot(x+1)$
Does this have to be a multiplication by +1 and -1? I'm not sure why the - in the middle changed to a + when the (-1) appeared.
(Sorry I don't know the notation for the dot)
| It might be easier to keep track of everything if you use a number instead of the variable $x$. If we substitute $x = 3$, the equation becomes
$$
(3^2+3)-(3+1) = (3 \cdot 3 + 3 \cdot 1)+(-1)\cdot(3+1).
$$
It's easy to see that $(3^2 + 3)$ is the same as $(3 \cdot 3 + 3 \cdot 1)$. $3 \cdot 3$ and $3^2$ are both just $9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$F(x , y)=\cos ^{y}\left(\frac{\pi}{x}\right)+\cos ^{y}\left(\frac{3 \pi}{x}\right)+\cos ^{y}\left(\frac{5 \pi}{x}\right)$ I am stuck with this problem , may you help me , it is from Spanish math olympiad ....
Given
$F(x , y)=\cos ^{y}\left(\frac{\pi}{x}\right)+\cos ^{y}\left(\frac{3 \pi}{x}\right)+\cos ^{y}\left(\fra... | First let's define
$$p_n:=\left(2\cos\frac\pi7\right)^n+\left(2\cos\frac{5\pi}7\right)^n+\left(2\cos\frac{3\pi}7\right)^n\quad(n\in\mathbb{Z}) $$
$$p_0=3,\quad p_1=1,\quad p_2=5$$
$$\forall n\in\mathbb{Z}c\quad p_{n+3}-p_{n+2}-2p_{n+1}+p_n=0$$
Then your sum is :
$$M=\dfrac{p_2}{4}+\dfrac{p_3}{8}+\dfrac{p_5}{32}-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3701887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate $| P ( A )+2 I |$ Given $A=\left[\begin{array}{ccc}4 & 1 & 0 \\ 2 & -1 & 1 \\ 0 & 2 & 1\end{array}\right]$
And
$P(x)=x^{3}-3 x+4$
Calculate $| P ( A )+2 I |$
a) $1648$
b) $1624$
c) $1528$
d) $1728$
e) $1696$
My try
$P(A)+2I=A^{3}-3 A+4+2I$
$P(A)+2I=A^{3}-3 A+4I+2I$
$P(A)+2I=A^{3}-3 A+6I$
= ${\left[\be... | It's not a lot easier, but here is the calculation using the characteristic polynomial and eigenvalues of $A$. We calculate the characteristic polynomial
$$
|xI - A| = x^3 - 4x^2 - 5x + 14 = (x-2)(x-6x + 7).
$$
By the Cayley Hamilton theorem,
$$
P(A) + 2I = A^3 - 3A + 6I =
A^3 - 3A + 6I - 0
\\=
(A^3 - 3A + 6I) - (A^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Maximum ellipse inscribed in Witch of Agnesi curve An ellipse with variable $(2a,2b)$ axes parallel to the $(x,y)$ coordinate axes is inscribed inside fixed curve of equation.
$$ y=\pm\dfrac{1}{1+x^2}$$
Show that maximum ellipse area occurs when it touches the curve at its inflection point.
I am looking to generalizin... | Take $ f $ an even function and consider $ \mathcal E $ the ellipse that touches $ f $ in $ (c, f(c)) $. Suppose it has equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ (and thus an area of $ ab $). You get, from $ (c, f(c)) \in \mathcal E $ and from $ f $ tangent to $ \mathcal E $,
$$ \frac{c^2}{a^2} + \frac{f(c)^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Mistake in the proof of Binet's formula I want to prove Binet's formula $F_{k} = \dfrac{1}{\sqrt{5}}\times\left[\left(\dfrac{1+\sqrt{5}}{2}\right)^k-\left(\dfrac{1-\sqrt{5}}{2}\right)^k\right]$ for the $k_{th}$ fibonacci number.
I did as follows -
$F_{0}=0, F_{1} = 1$ .....
$F_{k} = F_{k-1}+F_{k-2}$
$F_kx^k=xF_{k-1}x... | Since the roots of $x^2+x-1$ are $\frac{-1\pm\sqrt{5}}{2}$, at the begining it should be
$$g(x) = \dfrac{-x}{\left(x+\frac{1+\sqrt{5}}{2}\right)\left(x+\frac{1-\sqrt{5}}{2}\right)}.$$
and therefore, at the end, we find
$$F_k = -\dfrac{1}{\sqrt{5}}\times \left[\left(\dfrac{-2}{1+\sqrt{5}}\right)^k-\left(\dfrac{-2}{1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the values $a$ and $b$ such that the function is differentiable at $x=0$ $\mathbf{Question:}$
Find the values $a$ and $b$ such that the function is differentiable at $x=0$
$$
f(x)=
\begin{cases}
x^{2}+1 &x≥0\\
a\sin x+b\cos x & x<0\\
\end{cases}
$$
$\mathbf{Solution:}$
$f(x)$ is differentiable at $x=0$ if $f'(0)$ ... | $$f(0)=1$$
$$\lim_{x\to 0^+}f(x)=\lim_{x\to0^+}(a\sin(x)+b\cos(x))=b$$
$f$ is continuous at $x=0$ if $b=f(0)=1$.
$$f'(0^-)=\lim_{x\to 0^-}\frac{f(x)-f(0)}{x-0}$$
$$=\lim_{x\to0^-}x=\color{red}{0}$$
$f$ is differentiable at $x=0$ if
$$\lim_{x\to0^+}\frac{f(x)-f(0)}{x-0}=\color{red}{0}$$
or
$$\lim_{x\to0^+}\frac{a\sin(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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inequality :$x^2+y^2+z^2=1$ , $(1-xy)(1-yz)(1-zx)\ge\frac{8} {27}$ and $a+b+c=1$,$a^2+b^2+c^2 +3abc \ge \frac {4}{9}$ 1.Let $x$,$y$,$z$ be nonnegetive reals such that $x^2+y^2+z^2=1$.
$(1-xy)(1-yz)(1-zx)\ge\frac{8} {27}$
*let $a$,$b$,$c$ be positive reals with $a+b+c=1$ . Prove that $a^2+b^2+c^2 +3abc \ge \frac {4}{... | *
*For the first one:
Let $p=x+y+z, q= xy+yz+zx, r=xyz$ Which are greater or equal to zero.
$x^2+y^2+z^2=p^2-2q \rightarrow p^2-2q=1 $ (1)
$\rightarrow (1-xy)(1-yz)(1-zx)=1-q+pr-r^2\geq \frac{8}{27}$ (2)
$p^3 - 4pq +9r \geq 0 \rightarrow p(p^2 -4q)+9r\geq 0 \rightarrow 9r \geq p(2q-1)$ (3)
Also $p^2 \geq 3q $ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove that $\sum_{i=1}^{n}\binom{a_{i}}{i}\ge\frac{\binom{2n+1}{n}}{2^{2n+1}}\sum_{i=1}^{n}2^{a_{i}}$ let $a_{1}<a_{2}<\cdots<a_{n}=2n$ be postive integers,Prove that
$$\sum_{i=1}^{n}\binom{a_{i}}{i}\ge\dfrac{\binom{2n+1}{n}}{2^{2n+1}}\sum_{i=1}^{n}2^{a_{i}}$$
from Yongxi wang
I want use induction to prove it.
Suppose ... | The induction approach is a good choice in this case, but we have to take into account the fact that, when passing from $n$ to $n+1$, only the first terms of the sequence can be left unchanged. For given $n$, the $n^{th}$ term is necessarily equal to $2n$, whereas for $n+1$ it has no fixed value and can range from $n$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3712767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$
Now if $(x+4y... | Another way:
Discriminant $$=(6y+3)^2-4(8y^2+6y-2)=4y^2+12y+17=(2y+3)^2+8$$ which needs to be perfect square $=z^2$(say)
WLOG $z\ge0$
so that $$x=\dfrac{-(6y+3)\pm z}2$$
Now $$(2y+3)^2-z^2=8$$
$$\iff\dfrac{2y+3-z}2\cdot\dfrac{2y+z+3}2=2=1\cdot2=(-2)(-1)$$
As $2y+3+z\ge2y+3-z,$
$\dfrac{2y+3+z}2=2\iff\dfrac{2y+3-z}2=-1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$ $$\lim_{n\to\infty} \left(\frac{n}{1+n^2}+\frac{n}{4+n^2}+\cdots+ \frac{n}{n^2+n^2}\right)$$
I will give my solution, which, to my surprise, turned out to be erroneous:
$S_n=\frac{a_1+a_n}{2}n=\frac{\frac{n}{1+n^2}+\frac{n}{... | Hint:
You can rewrite it as a Riemann sum:
$$\sum_{k=1}^n\frac{n}{n^2+k^2}=\sum_{k=1}^n\frac{n}{n^2\Bigl(1+\frac{k^2}{n^2}\Bigr)}=\frac1n\sum_{k=1}^n\frac{1}{1+\frac{k^2}{n^2}}.$$
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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what is the volume of a tilted glass I was wondering about the volume of a tilted glass (60$^o$).
Not sure, but my attempt:
$$y=\sqrt{x}, \space 0<x<9$$
$$x=y^2,\space 0<y<3$$
$$A(y)=\pi x^2=\pi y^4$$
$$V(y)=\int_0^3 \pi y^4\,dy$$
$$V=\frac{243 \pi}{5}$$
$$V(\text{adjusted})=\sin(60^o)*V=\frac{243\sqrt{3 \pi}}{10}$$
So... | The line of the $60$ degree angle can be described by $x = \frac{1}{\sqrt{3}}y + 9 - \sqrt{3}$
The area enclosed by the curve and the line can be calculated with an integral.
$\int{}{}{\frac{1}{\sqrt{3}}y + 9 - \sqrt{3}} - y^2$
You need to find the limits of integration. For the right side, it will be 3 since the limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding Inverse of a 4*4 matrix Find the inverse of the following matrix
$$
\begin{array}{c}
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
\text { where } c_{0}=\frac{1+\sqrt{3}}{4 \s... | The columns are pairwise orthogonal to each other.
So, with
$$A=
{\left[\begin{array}{cccc}
c_{0} & c_{1} & c_{2} & c_{3} \\
c_{2} & c_{3} & c_{0} & c_{1} \\
c_{3} & -c_{2} & c_{1} & -c_{0} \\
c_{1} & -c_{0} & c_{3} & -c_{2}
\end{array}\right]} \\
$$
You have
$$A^TA = qI_{4\times 4} \text{ with } q= \sum_{k=0}^3c_k^2 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find inverse element of $1+2\alpha$ in $\mathbb{F}_9$
Let $$\mathbb{F}_9 = \frac{\mathbb{F}_3[x]}{(x^2+1)}$$ and consider $\alpha = \bar{x}$. Compute $(1+2 \alpha)^{-1}$
I think I should use the extended Euclidean algorithm: so I divide $x^2 +1 $ by $(1+2x)$:
$$x^2 + 1 = (1+2x)(2x+2)+2$$
$$(2x+2)(1+2x) + 2(x^2+1) = ... | Your answer is correct.
Note that $\alpha^2=-1$, and the conjugate of $\alpha$ is $-\alpha$,
so $\dfrac1{1+2\alpha}=\dfrac1{1+2\alpha}\dfrac{1-2\alpha}{1-2\alpha}=\dfrac{1-2\alpha}{1-4\alpha^2}=\dfrac{1-2\alpha}{1-\alpha^2}=\dfrac{1-2\alpha}2=2^{-1}-\alpha=2+2\alpha$
in $\mathbb F_9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find all pairs of positive integers $(m, n)$ for which $X^m + X + 1$ divides $X^n + 1$ and pairs $(m,n)$ for which $X^m +X −1$ divides $X^n +1$?
The special case I can think of is when $n=3$, and $m=2$ for the first part. But I don't know if other cases exist. Any help would be appreciated. Thanks!
| I get that
there are no solutions
for $x^m+x+1$.
The roots of
$x^n+1$ are
$e^{(2k+1) \pi i/n}$
for $k = 0$ to $n-1$,
so
all the roots have magnitude $1$.
If $m = 1$ then
$x = -\frac12$
so
$\dfrac{(-1)^n}{2^n}+1 = 0$
which can't be.
If $m = 2$
the roots of
$x^2+x+1$
are
$x_{1, 2}
=\dfrac{-1\pm\sqrt{-3}}{2}
=e^{\pm 2\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim\limits_{n \to \infty}\sum\limits_{i=1}^n\frac{1-\cos \frac{\pi}{\sqrt{n}}}{1+\cos \frac{i\pi}{\sqrt{2n}}}$ We may have
\begin{align*}
\lim_{n \to \infty}\sum_{i=1}^n\frac{1-\cos \frac{\pi}{\sqrt{n}}}{1+\cos \frac{i\pi}{\sqrt{2n}}}&=\lim_{n \to \infty}\left(1-\cos \frac{\pi}{\sqrt{n}}\right)\sum_{i=1}^n\f... | The limit doesn't exist (i.e. it is infinite).
For $1\leqslant k\leqslant m_n:=\lfloor\sqrt{n/8}\rfloor$ (and $n$ large enough), let $$i_k:=\lfloor(2k-1)\sqrt{2n}\rfloor=(2k-1)\sqrt{2n}-\epsilon_k.$$ Then $$\cos^2\frac{i_k\pi}{2\sqrt{2n}}=\cos^2\left((2k-1)\frac{\pi}{2}-\frac{\epsilon_k\pi}{\sqrt{8n}}\right)=\sin^2\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$ For a fixed $k$ what is the value of $\sum_{l=1}^{5^m-1} \Big\lfloor \dfrac{l}{5^k}\Big \rfloor$
By dividing the numbers between $1$ and $5^m$ as intervals of $5^k$, I was getting the following expression:
$$\binom{5^{m-k}}... | Your original approach of doing casework on intervals works. The key (at least to my method) is to interpret $$\left\lfloor\frac{a}{b}\right\rfloor$$ as the number of positive multiples of $b$ that are less than or equal to $a,$ where $a$ and $b$ are positive integers. Your list of disjoint intervals that cover all of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick... | Solving for $c$, we get
$$ c=-{\frac {-18\,b \pm \sqrt {24\,{b}^{6}+36\,{b}^{4}+18\,{b}^{2}-159}
}{3 (2\,{b}^{2}+1)}}$$
If there is to be an integer solution, we need $24\,{b}^{6}+36\,{b}^{4}+18\,{b}^{2}-159$ to be a square. Taking $s = b^2$, let's look for integer solutions of
$t^2 = 24 s^3 + 36 s^2 + 18 s - 159$, wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Let z be a complex number such that $\frac{z-i}{z-1}$ is purely imaginary. Then the minimum value of $|z-(2+2 i)|$ Let
z be a complex number such that $\frac{z-i}{z-1}$ is purely imaginary. Then the minimum value of $|z-(2+2 i)|$ is ?
My approach:-
$$
\begin{array}{l}
\\
\left|z-z^{\prime}\right| \geqslant| \ |z|-\lef... | A geometric approach:
$\frac{z-i}{z-1}$ is purely imaginary signifies that $$\arg\left(\frac{z-i}{z-1}\right)={\pi \over 2}+k\pi$$ with $k\in \mathbb{Z}.$
Geometrically, locus of point $z$ is the circle with diameter joining $i$ and $1,$ except these two points.
The center is ${1\over2}+i{1\over2},$ the midpoint of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do I evaluate $\lim_{n\to\infty} \,\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}$? I came across the following problem recently in a problem sheet aimed at high school students:
Evaluate $$\lim_{n\to\infty} \,\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}.$$
I tried to rewrite the inner sum a... | We can rewrite the sum as
$$S = \lim_{n\to\infty}\sum_{k=1}^n \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}}\cdot\frac{k}{n}\cdot\frac{1}{n}$$
We also have that for $1\leq k \leq n$
$$\left(\frac{1}{n}\right)^{\frac{1}{n}} \leq \left(\frac{k}{n^2}\right)^{\frac{k}{n^2}} \leq \left(\frac{1}{n^2}\right)^{\frac{1}{n^2}}$$
for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The functional equation $\big(1 + yf(x)\big)\big(1 - yf(x + y)\big) = 1$ for $f:\mathbb R^+\to\mathbb R^+$ Functional equation from USAMO 2010 preparation session:
Find all functions $f:\mathbb R^+\to\mathbb R^+$ such that
$\big(1 + yf(x)\big)\big(1 - yf(x + y)\big) = 1$
for all $x, y \in \mathbb R^+$, where $\mathbb ... | Since $x,y \in \mathbb R^+$, we may without restriction assume $x,y \neq 0$.
Now from given,
$ \begin{align} (1 + yf(x))(1 − yf(x + y)) &= 1 \\
1 − yf(x + y)&= \frac{1}{1 + yf(x)} \\
1-\frac{1}{1 + yf(x)} &= yf(x + y) \\
\frac{1+yf(x)-1}{1 + yf(x)} &= yf(x + y) \\
\frac{f(x)}{1+yf(x)} &= f(x+y)= \frac{f(y)}{1+xf(y)} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Calculating $ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$ I need to find the limit of:
$$ \lim_{b \to a} \frac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}$$
I've tried throught "rationalization" and completing squares... This is my work so far (i... | Let
$b = a-x$
where $x > 0$.
$\begin{array}\\
f(a, b)
&=\dfrac{a \cdot (a + \sqrt{a^2-b^2}) - b^2}{a \cdot (a - \sqrt{a^2-b^2})-b^2}\\
&=\dfrac{a \cdot (a + \sqrt{a^2-(a-x)^2}) - (a-x)^2}{a \cdot (a - \sqrt{a^2-(a-x)^2})-(a-x)^2}\\
&=\dfrac{a(a + \sqrt{2ax-x^2}) - a^2+2ax-x^2}{a(a - \sqrt{2ax-x^2})-a^2+2ax-x^2}\\
&=\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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If $a+b+c=k$ and $a^2+b^2+c^2 =2k$ what is the maximum value of $k$? $a,b,c$ are real numbers and they satisfy the following equations.
$a+b+c=k$
$a^2+b^2+c^2=2k$
Find the maximum value of $k$.
I tried substituting for k in the second equation from the first and got
$a^2+b^2+c^2=2(a+b+c)$
Rearranging the terms I got
$a... | Your ideas are good. You obtained $(a-1)^2 + (b-1)^2 + (c-1)^2 = 3$. That means there is a sphere around the point $(1,1,1)$ with radius $\sqrt{3}$ and the point $(a,b,c)$ is situated on that sphere.
The condition $a^2+b^2+c^2 = 2k$ tells us that the point $(a,b,c)$ lives on a sphere around the origin at distance $\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Bound for $\sum_{k=0}^{n}(-1)^k{3n\choose k}{n\choose k}$. In the book Complex analysis by Bak J. & Newman J., chapter 11, talks about Sums Involving Binomial coefficients and find a bound $\frac{16}{9}\sqrt{3}$ for $|(z-1)^2(z+1)|$ in the "Example 3" on the unit circle, my way was using lagrange multipliers in this fo... | Using the "coefficient-of" notation, we have
$$S_n:=\sum_{k=0}^{n}(-1)^k\binom{3n}{k}\binom{n}{k}=\sum_{k=0}^{n}[z^k](1-z)^{3n}\times[z^{n-k}](1+z)^n=[z^n]\big((1-z)^3(1+z)\big)^n.$$ Let $f(z)=(1-z)^3(1+z)/z$; then, by Cauchy integral formula, for any $r>0$ we have $$S_n=\frac{1}{2\pi\mathrm{i}}\oint_{|z|=r}\big(f(z)\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove that for every integer $n$, $n^3$mod$6$=$n$mod$6$ I will use induction to prove this.
Firstly for $n=1$,
$1^3\text{mod}6=1\text{mod}6$
Now we assume that this holds for some $n=k$ and prove that if it holds for $n=k$ it should also hold for $n=k+1$.
$(k+1)^3\text{mod}6=(k^3+3k^2+3k+1)\text{mod}6$
As for $n=k$ bec... | An easy approach:
$$n^3-n=(n-1)n(n+1)$$
As in any two consecutive integers, one is even, and in any three consecutive, one is multiple of three. So the product of three consecutive numbers is a multiple of $6$. $$\begin{split}n^3-n&\equiv0\pmod6\\n^3&\equiv n\pmod 6\end{split}$$ Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Remainder when divided by $7$ What would be the remainder when
$12^1 + 12^2 + 12^3 +\cdots + 12^{100}$ is divided by $7$ ?
I tried cyclic approach (pattern method), but I couldn't solve this particular question.
| As $12\equiv-2\pmod7, 12^3\equiv(-2)^3\equiv-8\equiv-1,$ord$_712=6$
$\implies12^{6k+r}\equiv12^r\equiv(-2)^r\pmod7$
$$\sum_{r=1}^{100}12^r\equiv12^1+12^2+12^3+12^4+16\sum_{r=0}^512^r\pmod7$$
$$\equiv(-2)+(-2)^2+(-2)^3+(-2)^4+2\sum_{r=0}^5(-2)^r\pmod7$$
Finally $\displaystyle\sum_{r=0}^5(-2)^r\equiv\dfrac{(-2)^6-1}{-2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How to solve $\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}$?
How can i evaluate the following integral $$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=?$$
This is taken from a definite integral where $x$ varies from $0$ to $1$.
My attempt:
multiplied by conjugate
$$\int \frac{dx}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{(\sqrt{1+x}+\sqrt{... | $\begin{aligned} I &:= \int \frac{d x}{\sqrt{1+x}-\sqrt{1-x}}=\int \frac{\sqrt{1+x}+\sqrt{1-x}}{2 x} d x =\frac{1}{2}\left[ \underbrace{\int \frac{\sqrt{1+x}}{x}-d x}_{J}+\underbrace{\int \frac{\sqrt{1-x}}{x} d x}_{K}\right] \end{aligned}$
$$
\begin{aligned}
J &=\int \frac{\sqrt{1+x}}{x} d x \\
&=\int \frac{1+x}{x \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3741855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find every equation of the line that passes through the point $(5,13)$ "Find every equation of the line that passes through the point $(5,13)$ and passes both axis at non-negative, whole values."
Here's my attempt:
Finding first two equations, with $k=\pm1$ is fairly simple. After that, plugging in the $x=5$ and $y=13... | A few things to note: If $b$ is the $y$ intercept of line and $c$ is the $x$ intercept then the line goes through $(0,b), (c,0)$ and the slope is $m =-\frac bc$. As $b,c$ are positive the slope is negative.
Point $(5,13),(0,b),(c,0)$ is on the line and the line has a negative slope. So $\frac {13-b}5 =\frac {13}{c-5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3742603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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limit of the sequence $x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$ I was thinking what happens with the sequence $\{x_n\}_{n\in \Bbb N}$ where:
$$x_{n}:= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n\ldots}}}$$
When you look some terms, for example $x_{1}=1$, $x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}$, $x_{3}=\sqrt[3]{3 \s... | An Inductive idea is:
$$x_{2}=\sqrt[]{2 \sqrt[]{2 \sqrt[]{2 ...}}}\to 2\\
x_{3}=\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3 ...}}}\to \sqrt[2]3\\
x_{4}=\sqrt[4]{4 \sqrt[4]{4 \sqrt[4]{4 ...}}}\to \sqrt[3]4\\
x_{5}=\sqrt[5]{5 \sqrt[5]{5 \sqrt[5]{5 ...}}}\to \sqrt[4]5\\\vdots\\
x_{n}= \sqrt[n]{n \sqrt[n]{n \sqrt[n]{n ...}}}\to \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Simplifying $\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}$
Simplify $$\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{c^2+a^2-b^2}{(b-c)(b-a)}+\frac{a^2+b^2-c^2}{(c-a)(c-b)}\,.$$
I tried very hard but I am not being able to solve it easily I opened up everything and multiplied all... | First of all, we must have $$a\ne b\ne c$$
Take out the terms containing $b^2$ in the numerator
$$b^2\left(-\dfrac1{(a-b)(c-a)}+\dfrac1{(b-c)(a-b)}-\dfrac1{(c-a)(b-c)}\right)$$
$$=b^2\cdot\dfrac{-(b-c)+c-a-(a-b))}{(a-b)(b-c)(c-a}$$
$$=\dfrac{2b^2(c-a)}{(a-b)(b-c)(c-a)}$$
$$\implies\frac{b^2+c^2-a^2}{(a-b)(a-c)}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3743929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2... | Shortly: for $x\ne\sqrt 3$,
$$\frac{2x-6}{\sqrt x-\sqrt3}=2\frac{(\sqrt x-\sqrt3)(\sqrt x+\sqrt3)}{\sqrt x-\sqrt3}=2(\sqrt x+\sqrt3)\to4\sqrt3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How can i solve $\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$ How can i solve following $$\int \frac{x^3+2x-7}{\sqrt{x^2+1}}\ dx?$$
My work:
I substituted $x=\tan\theta$, $dx=\sec^2\theta d\theta $
integral becomes
$\int \dfrac{\tan^3\theta+2\tan \theta-7}{\sqrt{\tan^2\theta+1}}\ \sec^2\theta d\theta$
$\int \dfrac{\tan\the... | \begin{aligned}
\frac{x^{3}+2 x-7}{\sqrt{x^{2}+1}} d x &=\int \frac{x^{3}+2 x-7}{x} d \sqrt{x^{2}+1} \\
&=\int\left(x^{2}+2-\frac{7}{x}\right) d \sqrt{x^{2}+1} \\
&=\int\left(x^{2}+1\right) d \sqrt{x^{2}+1}+\sqrt{x^{2}+1}-7 \int \frac{d \sqrt{x^{2}+1}}{x}\\
&=\frac{\left(x^{2}+1\right)^{\frac{3}{2}}}{3}+\sqrt{x^{2}+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sums of powers of cosines and sines shifted by $2\pi/3$ I have stumbled across these two identities
$$
\begin{split}
\cos^2(x)+\cos^2(x+2\pi/3)+\cos^2(x+4\pi/3) &= 3/2,\\
\cos^4(x)+\cos^4(x+2\pi/3)+\cos^4(x+4\pi/3) &= 9/8.
\end{split}
$$
There is also the more intricate
$$
\begin{split}
\cos^2(x)\sin^2(x)+\cos^2(x+2\pi... | If $\cos3y=\cos3x$
$3y=2n\pi\pm3x$ where $n$ is any integer
$y=\dfrac{2n\pi}3+x$ where $n=0,1,2$
Again, $\cos3y=4\cos^3y-3\cos y$
So, the roots of $$4\cos^3y-3\cos y-\cos3x=0$$ are $p=\cos x,q=\cos\left(\dfrac{2\pi}3+x\right),r=\cos\left(\dfrac{4\pi}3+x\right)$
Using Vieta's formula, $$p+q+r=\dfrac04\ \ \ \ (1)\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
plane perpendicular to the straight line I have this problem to find the equation of the plane perpendicular to the straight line joining the points $(1, 3, 5)$ and $(4, 3 ,2)$ at its middle points. How would you solve this?
| The formula for it is: $(X-X_0)\cdot \vec{n}=0$ whereas $X = (x,y,z), X_0 = \left(\frac{1+4}{2}, \frac{3+3}{2}, \frac{2+5}{2}\right)=\left(\frac{5}{2}, 3, \frac{7}{2}\right)$ , and $\vec{n}= \dfrac{1}{\sqrt{3^2+0^2+(-3)^2}}\left(4-1,3-3,2-5\right)=\dfrac{1}{2\sqrt{3}}\left(3,0,-3\right)=\left(\frac{\sqrt{3}}{2}, 0,\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $\int\limits_0^{1/\sqrt{2}} \frac{au^2}{5(1-u^2)^2}du = 1$ for $a$ Problem:
The function $f_U(u) = \frac{au^2}{5(1-u^2)^2}$ is a probability density for the random variable $U$, which is non-zero on the interval $(0, \frac{1}{\sqrt{2}})$. I am supposed to find the value $a$.
I understand that that amounts to solv... | Start by pulling the $\frac{a}{5}$ out of the integral. You know how to factor the denominator into linear factors, so perform partial fraction decomposition.
$$ \frac{u^2}{(1-u^2)^2} = \frac{1/4}{u-1} + \frac{1/4}{(u-1)^2} + \frac{-1/4}{u+1} + \frac{1/4}{(u+1)^2} \text{.} $$
The antiderivatives of these terms are,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3752754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Derivative of $y = \log_{\sqrt[3]{x}}(7)$. Never dealt with a derivative of these type. My approach was $$y = \log_{\sqrt[3]{x}}(7) \iff 7 = (\sqrt[3]{x})^y.$$ Then,
$$\frac{d}{dx}(7) = \frac{d}{dx}\left(\sqrt[3]{x}\right)^y \Rightarrow (\sqrt[3]{x})^y = e^{\frac{y\ln(x)}{3}} $$
From here,
$0 = e^u\dfrac{du}{dx}$ and $... | $$\frac{\mathrm{d}}{\mathrm{d}x}\left(\log_{\sqrt[3]{x}}(7)\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\dfrac{3\ln\left(7\right)}{\ln\left(x\right)}\right]$$
$$=3\ln\left(7\right)\cdot\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{\ln\left(x\right)}\right]$$
$$=-\frac{3\ln\left(7\right)\cdot\dfrac{1}{x}}{\ln^2\left(x\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluating Sum at bounds I have to find an expression in terms of n using standard results for $$\sum_{r=n+1}^{2n} r(r+1)$$
And have found the general equation
$$\sum_{r=n+1}^{2n} r(r+1) = \frac{2n^3+6n^2+4n}{6}$$
However evaluating it as $$\frac{2(2n)^3+6(2n)^2+4(2n)}{6} - \frac{2(n+1)^3+6(n+1)^2+4(n+1)}{6}$$
does n... | Your general equation should be
$$\begin {align} \sum_{r=n+1}^{2n} r(r+1)&=\sum_{r=n+1}^{2n} (r^2+r)\\
&=\sum_{r=1}^{2n} (r^2+r)-\sum_{r=1}^{n} (r^2+r)\\
&=\frac 16\left((2n)(2n+1)(4n+1)\right)+\frac 12\left(2n(2n+1)\right)-\frac 16\left((n)(n+1)(2n+1)\right)+\frac 12\left(n(n+1)\right)\\
&=\frac 16\left(16n^3+12n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Simplify $\sqrt{8-\sqrt{63}}$ I simplified the expression into $$\sqrt{8-3\cdot \sqrt{7}}$$ but my tutor said it wasn't the answer he was looking for. Can someone help me?
| if $x = \sqrt{8-\sqrt{63}},$ then $0<x<1$ and $x^2 - 8 = - \sqrt{63},$ then $x^4 - 16 x^2 + 64= 63,$ then
$$ x^4 - 16 x^2 + 1 = 0. $$ Also
$$ x^2 - 16 + \frac{1}{x^2} = 0 $$
Taking $$ u = x + \frac{1}{x} $$
we get $u^2 - 18 = 0 $ and $$ u = \sqrt {18} $$
and
$$ x = \frac{3 \sqrt 2 \pm \sqrt{14}}{2} $$
and $x<1$ g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem with proving inequalities Question:
Prove that if $x,y,z$ are positive real numbers such that $x+y+z=a$ then $(a-x)(a-y)(a-z)>\frac8{27}a^3$ is not true.
My Approach:
$$\frac{a-x}{2}=\frac{y+z}2$$
$$\frac{a-y}{2}=\frac{x+z}2$$
$$\frac{a-z}{2}=\frac{x+y}2$$
Using $AM>GM$ we get $$\frac{x+y+z}{3}>\root 3 \of {x... | Use AM-GM of three items as
$$F=[(a-x)(a-y)(a-z)]^{1/3} \le \frac{3a-(x+y+z)}{3}$$
$$\implies (a-x)(a-y)(a-z) \le \frac{8a^3}{27}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find an inequality with more precise bound? In an other topic, i found that : $x-1\le\lfloor x\rfloor\le x$ which really easy to show.
However, i want to found an inequality with more precise bound.
Like i found that : $$\forall x\in \mathbb R\text{, }x-1\le x-1+\frac{1}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\l... | Let us prove that
$$x-1+\frac{1 + \frac{1}{6}(\sin \pi x)^2}{\pi}\vert\sin{\pi x}\vert\le
\lfloor x\rfloor\le x -\frac{1 + \frac{1}{6}(\sin \pi x)^2}{\pi}\vert\sin{\pi x}\vert .$$
By letting $y = \pi(x - \lfloor x\rfloor)\in [0, \pi)$ and $z = \pi (1 - (x - \lfloor x\rfloor))\in (0, \pi]$, it suffices to prove that
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the number of $6$ digit positive integers whose sum of the digits is at least $52$?
What is the number of $6$ digit positive integers whose sum of the digits is at least $52$?
My kind of approach was:
I thought to use Multinomial Theorem concept here.
So my primary aim was to find the coefficient of $x^{52}$... | You should have:
$$[x^{52}](1+x+x^2+\cdots +x^9)^6=[x^{52}]\left(\frac{1-x^{10}}{1-x}\right)^6=\\
[x^{52}](1-x^{10})^6\cdot (1-x)^{-6}=\\
[x^{52}]\sum_{i=0}^6 {6\choose i}(-x^{10})^i\cdot \sum_{j=0}^{\infty}{-6\choose j}(-x)^j=\\
[x^{52}]\sum_{i=0}^6 {6\choose i}(-x^{10})^i\cdot \sum_{j=0}^{\infty}{6+j-1\choose j}x^j=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many 6 digit numbers can be formed from two sets of digit? There are two sets of digit :
$ \text{set 1 :} \{~0,1,2,3,4~\}$
$ \text{set 2 :} \{~5,6,7,8,9~\}$
Now how many 6 digit number can we make by taking numbers from these two sets ? From $\text{set 1}$ repetition is permitted but from $\text{set 2}$ repetition ... | Yes, but I don't see you taking into account the positioning of the digits. For example, in combo 2, the digit from group 2 can be in one of the 6 places, while you let it be only in the last (it will affect also the 0 consideration).
Sanity check: assume we don't allow the usage of any digit twice. There are 9∗9∗7∗6∗5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
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.