Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
| It is also true that $$ \rho \leq 1 $$
In the case that $\rho > 1,$ take $a=1, \; b = -1, \; c = 0.$ The polynomial becomes $2 - 2 \rho = 2(1 - \rho) < 0.$
In a similar style, we can do the original problem this way: if $\rho < -\frac{1}{2},$ take $a=1, \; b = 1, \; c = 1.$ The polynomial becomes $3 + 6 \rho < 3 - 3 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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inequality in geometry
Given triangle $ABC$ has $BC=a$, $CA=b$, $AB=c$ and $M$ is a point of the triangle plane. Prove that:
\begin{align*}
\cos \dfrac{A}{2} \cdot MA+\cos \dfrac{B}{2} \cdot MB+\cos \dfrac{C}{2} \cdot MC \geq \dfrac{a+b+c}{2}.
\end{align*}
My attempts:
\begin{eqnarray*}
a \cdot \overrightar... | Here is my proof, based on the main result that if $DEF$ is the pedal triangle of $M$ then we have: $$\cos\left(\frac{A}{2}\right)\cdot MA\ge \frac{AE+AF}{2}\qquad (1)$$
In short, we prove the above inequality by using the formula for $\sin\alpha -\sin\beta$ to show that $$\cos\left(\frac{A}{2}\right) \cdot MA = \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Let $s_{n} := 1 + 1/2 + \cdots + 1/n - \ln(n+1)$ and show that $s_{n} \leq 1 - 1/(n+1)$. This is in a Riemann integration section of a practice sheet from real analysis. I know that if $f(x) := \frac{1}{[x]} - \frac{1}{x+1}$, where $[*]$ is the floor function, then $\int_{1}^{n} f = s_{n}$. But I'm not sure where to go... | Induction does the trick
Assume it true for $n-1$, i.e.
$$1 + \ldots + \frac{1}{n-1} - \ln n < 1 - \frac{1}{n}$$
Then
$$1 + \ldots + \frac{1}{n-1} + \frac{1}{n} - \ln (n+1) < \ln n + 1 - \frac{1}{n} + \frac{1}{n} - \ln (n+1) = 1 + \ln (\frac{n}{n+1}) = 1 + \ln(1 - \frac{1}{n+1})$$
Note that $$\ln (1-x) < -x$$ for $0 < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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For positive number $a,b$, when $a,b$ satisfies $2a^2 +7ab+3b^2=7$, what is maximum value of $a+{\sqrt{3ab}}+2b$ Question is
For positive numbers $a,b$ such that $2a^2 +7ab+3b^2 = 7$, what is the maximum value of $a+{\sqrt{3ab}}+2b$?
I use AM-GM Inequality to do this
$$(2a+b)(a+3b)=7\text{ and }2a+b=\frac{7}{a+3b}.$... | The mistake is when you apply AM-GM in the last step. Applying it you get $\ge$ inequality, which corresponds to the minimal value of l.h.s, not maximal.
Also $m \le \frac{7}{2\sqrt{3ab}} + \sqrt{3ab}$, not equal.
UPD: You did the following
$m = \frac{7}{a + 3b} + \sqrt{3ab} \le \frac{7}{2\sqrt{3ab}} + \sqrt{3ab} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$ Evaluate $$I=\int \frac{dx}{(x^2-x+1)\sqrt{x^2+x+1}}$$
My book gave the substitution for $$\int \frac{dx}{P\sqrt{Q}}$$ as
$\frac{Q}{P}=t^2$ when $P$ and $Q$ are quadratic expressions
So accordingly i used
$$\frac{x^2+x+1}{x^2-x+1}=t^2 \tag{1}$$ we get
$$\frac{(1-x^2... | We have that $$\frac{x^2+1}{x}=\frac{t^2+1}{t^2-1}\tag{1}$$ so $$\frac{\sqrt{x^2-x+1}}{1-x^2}=\frac{\sqrt{x\left(\frac{t^2+1}{t^2-1}-1\right)}}{1-x\cdot\frac{t^2+1}{t^2-1}+1}=\frac{(t^2-1)\sqrt{\frac{2x}{t^2-1}}}{2(t^2-1)-x(t^2+1)}=\frac{\sqrt{2x(t^2-1)}}{2(t^2-1)-x(t^2+1)}$$ and you can further invoke $(1)$ to solve f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Polynomial problem with two conditions I have to find $P(0)$ from the polynomial with minimum degree given that
$$(x-1)^3|(P(x)+1)$$
$$(x+1)^3|(P(x)-1)$$
Plugging in $x=\pm 1$ gets something nice, also division by a polynomial of third order gives successively: $$P(1)+1 =0; \ P'(1)=0; \ P''(1)=0$$
$$P(-1)-1 =0; \ P... | Suppose $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that
\begin{align*}
(x+1)^3&{\,\mid\,}(P(x)-1)\\[4pt]
(x-1)^3&{\,\mid\,}(P(x)+1)\\[4pt]
\end{align*}
Equivalently, $P\in\mathbb{Q}[x]$ is a polynomial of least degree such that
\begin{align*}
P(x)&=A(x)(x+1)^3+1\\[4pt]
P(x)&=B(x)(x-1)^3-1\\[4pt]
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the $n$-th derivative of $f(x)=\frac{x}{\sqrt{1-x}}$ Find the $n$-th derivative of
$$f(x)=\frac{x}{\sqrt{1-x}}$$
First I just calculated the first, second and 3-th, 4-th derivatives and now I want to summarize the general formula. But it seems too complicated. Then I want to use binomial theorem or Taylor expansio... | We can prove that
$$f^{(n)}(x) = - \frac {(2n - 3)!!\, (x - 2n)} {2^n (1 - x)^{(2n + 1)/2}}$$
for $n \ge 2$ by induction on $n$.
The base case is easy. For the inductive step,
$$\begin{align*}
\frac d {dx} f^{(n)} (x) & = - \frac {(2n - 3)!!\, 2^n (1 - x)^{(2n + 1)/2} + (2n - 3)!!\, (x - 2n) 2^n \frac {2n+1} 2 (1 - x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Does there exist a right triangle with area 7 and perimeter 12? This question is really trivial. I can prove that there is no right triangle with area 7 and perimeter 12, but what I do is solve the following system: if $a$, $b$ and $c$ are, respectively, the two legs and hypotenuse of such a triangle, then
$$a^2 + b^2 ... | Given a general triangle with
inradius $r$, circumradius $R$
and semiperimeter $\rho=\tfrac12(a+b+c)$,
the lengths of its sides $a,b,c$ can be found as
three roots of the cubic equation
\begin{align}
x^3-2\rho\,x^2+(\rho^2+r^2+4\,r\,R)\,x-4\,r\rho\,R
&=0
\tag{1}\label{1}
.
\end{align}
For the right triangle
the equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 5
} |
Elliptic integral of a quartic I am trying to compute the following integral:
\begin{equation}
I(a) = \int_{-\infty}^{+\infty} \left[ 1-x^2 + \sqrt{(1-x^2)^2 + a} \right] dx
\end{equation}
where $a >0$.
I am pretty confident this integral is well defined, as it basically looks like a bell-shaped curve that behaves like... | With CAS help:
$$\int_{-\infty }^{\infty } \left(1-x^2+\sqrt{\left(1-x^2\right)^2+a}\right) \, dx=\\\frac{4}{3} \sqrt[4]{1+a} \left(2 E\left(\frac{1}{2}
\left(1+\frac{1}{\sqrt{1+a}}\right)\right)+\left(-1+\sqrt{1+a}\right) K\left(\frac{1}{2}
\left(1+\frac{1}{\sqrt{1+a}}\right)\right)\right)$$
for $a > 0 $.
where:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$dy/dx=\sqrt{x^2+y^2}$
$$\frac{dy}{dx}=\sqrt{x^2+y^2}$$
slope=distance from origin, should be simple and interesting. May have no solution!
I have tried several approaches, best:
$(\frac{dy}{dx}-y)(\frac{dy}{dx}+y)=x^2$ multiply by $e(-x) * e(+x)$ as integrating factor. Substitute $\frac{1}{2}x^2=t$.
Second approac... | Similar to To solve $ \frac {dy}{dx}=\frac 1{\sqrt{x^2+y^2}}$:
Apply the Euler substitution:
Let $u=y+\sqrt{x^2+y^2}$ ,
Then $y=\dfrac{u}{2}-\dfrac{x^2}{2u}$
$\dfrac{dy}{dx}=\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\... | We have that for $x>3$
$$\log{\left|\frac{x+2}{3-x}\right|}=\log{\left(\frac{x+2}{x-3}\right)}=\log{\left(1+\frac{5}{x-3}\right)}$$
and therefore
$$x\log{\left|\frac{x+2}{3-x}\right|}=\frac{5x}{x-3}\frac{\log{\left(1+\frac{5}{x-3}\right)}}{\frac{5}{x-3}}\to 5\cdot 1=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How to find the analytical representation of eigenvalues of the matrix $G$? I have the following matrix arising when I tried to discretize the Green function， now to show the convergence of my algorithm I need to find the eigenvalues of the matrix $G$ and show it has absolute value less than 1 for certain choices of $N... | Eigenvalues of matrix $G$ are
$$\lambda_k(G)=-\tfrac{1}{4}(N+1)\sin^{-2}\left(\frac{k\pi}{2(N+1)}\right)\,,\quad k=1,\ldots,N\,.$$
To prove the statement, we start by writing elements of matrix $G$ as $g_{ij}=ij-(N+1)\min(i,j)$.
Next, we define matrix $L$ as a lower triangular matrix with all elements in the lower tria... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\s... | Hint: It is $$\sqrt{4x^2+7x}=\sqrt{x^2\left(4+\frac{7}{x}\right)}=-x\sqrt{4+\frac{7}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. T... | Note that $\sqrt{25y^2+56}$ is an integer if and only if $(5y)^2+56=z^2$ for some integer $z$. This limits how big $5y$ can be. After all;
$$29^2=(28+1)^2=28^2+2\cdot28+1=28^2+57>28^2+56,$$
so surely $z^2<29^2$ and $(5y)^2<28^2$. This only leaves a few values of $y$ to try.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Finding $\binom{n}{4}$ if $\binom{n}{6}=\binom{n}{9}$ I was trying to solve a question related to combinatorics as follows :
A group contains $n$ persons. If the number of ways of selecting $6$ persons is equal to the number of ways of selecting $9$ persons, then find the number of ways of selecting $4$ persons from t... | If $n<6$ then the number of ways to select $6$ people is the same as the number of ways to select $9$ people; in both cases this is $0$. For $n<6$ we can easily compute $\binom{n}{4}$:
$$\binom{5}{4}=5,\qquad\binom{4}{4}=1,$$
and $\binom{n}{4}=0$ when $n<4$.
If $n>6$, then by selecting $6$ people we leave $n-6$ people ... | {
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"url": "https://math.stackexchange.com/questions/2893110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differentiate $\frac{x^3}{{(x-1)}^2}$
Find $\frac{d}{dx}\frac{x^3}{{(x-1)}^2}$
I start by finding the derivative of the denominator, since I have to use the chain rule.
Thus, I make $u=x-1$ and $g=u^{-2}$. I find that $u'=1$ and $g'=-2u^{-3}$. I then multiply the two together and substitute $u$ in to get:
$$\frac{d}... | You're mixing the product rule and the quotient rule. You can apply each of them, but not simultaneously.
*
*by the product rule: remember $\dfrac{\mathrm d}{\mathrm dx}\Bigl(\dfrac1{x^n}\Bigr)=-\dfrac n{x^{n+1}}$, so
\begin{align}
\dfrac{\mathrm d}{\mathrm dx}\biggl(\dfrac{x^3}{(1-x)^2}\biggr)&=\dfrac{\mathrm d}{\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895284",
"timestamp": "2023-03-29T00:00:00",
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What is the radius of the black circle tangent to all three of these circles? The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the ... | Radius is exactly $\dfrac{72}{23}$, pretty neat. See the figure below, where dotted lines from the center of the circumcribing circle passes through the midpoints of the sides of the triangle.
Let : $B$ be the origin $(0,0)$, and the center of circle of unknown radius $r$ be $(x,y)$. Then we solve the following three ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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How to determine continuity at origin for given 2 variable function Consider the function,
$f(x,y)=\dfrac{x^3\cos\left(\frac1y\right)+y^3\cos\left(\frac1x\right)}{ x^2+y^2},\qquad\forall x,y\neq0$
And $f(x,y)= 0,\qquad\text{ if } xy=0$
Clearly this function is not continuous at either $x-$ or $y-$ axis
As, when approa... | By inequalities and polar coordinates we have that
$$0\le \left|\frac{x^3\cos(1/y)+y^3\cos(1/x)}{ x^2+y^2}\right|\le \left|\frac{x^3\cos(1/y)}{ x^2+y^2}\right|+\left|\frac{y^3\cos(1/x)}{ x^2+y^2}\right|\le$$$$\le \left|\frac{x^3}{ x^2+y^2}\right|+\left|\frac{y^3}{ x^2+y^2}\right|=r\left(|\cos^3\theta|+|\sin^3\theta|\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding sum of a geometric series I am asked to find the summation of $1/3^n$ from $n=5$ to infinity.
I have done the calculation: $1/(1-r)$, for $r=1/3$, and received $1.5$. As this summation starts from $5$, I subtracted $3^0, 3^-1, 3^-2, 3^-3$ and $3^-4$ from $1.5$ and got $6.17e-3$. However, apparently this answer ... | $$1,\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\frac{1}{3^4} \ldots$$
You need to calculate $$S_{\infty}-S_5$$
$$=\frac{1}{1-\frac{1}{3}}-\frac{1(1-\frac{1}{3^5})}{1-\frac{1}{3}}$$
$$=\frac{\frac{1}{3^5}}{\frac{2}{3}}$$
$$=\frac{1}{2\cdot{3^4}}$$
$$=\frac{1}{162}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqr... | Check again it should be
$$=\int \left(x^{-\frac{1}{2}}-4x^{-\frac{3}{2}}+10x^{-\frac{5}{2}}\right)dx$$
indeed $\frac{1}{x^2\sqrt x}=x^{-\frac{5}{2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Evaluate: $ \int \frac{\sin x}{\sin x - \cos x} dx $ Consider
$$ \int \frac{\sin x}{\sin x - \cos x} dx $$
Well I tried taking integrand as $ \frac{\sin x - \cos x + \cos x}{\sin x - \cos x} $ so that it becomes,
$$ 1 + \frac{\cos x}{\sin x - \cos x} $$
But does not helps.
I want different techniques usable here.
| Hint:
In general for
$\displaystyle\dfrac{d\left(\dfrac{a\sin x+b\cos x}{c\sin x+d\cos x}\right)}{dx}$
write numerator$(a\sin x+b\cos x)$ as $A(c\sin x+d\cos x)+B\cdot\dfrac{d(c\sin x+d\cos x)}{dx}$
Compare the coefficients of $\sin x,\cos x$ to find $A,B$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term ... | We have that
$$a_r=\frac{r + 2}{r (r + 1) (r + 3)} = \frac2{3 r}-\frac1{2 (r + 1)} - \frac1{6 (r + 3)}=$$$$= \frac1{2r}-\frac1{2 (r + 1)}+ \frac1{6r} - \frac1{6 (r + 3)}=$$
$$= \frac12\left(\frac1{r}-\frac1{r + 1}\right)+ \frac16\left(\frac1{r} - \frac1{r + 3}\right)$$
and therefore
$$\sum_1^n a_r=\frac12 \sum_1^n \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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When The curvature is maximum of $x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$ QUESTION
Find Where The Curvature has an extremum ... | the equation $$\sqrt{x}+\sqrt{y}=\sqrt{a}$$ is equivalent to $$y=x-2\sqrt{ax}+a,~~~0\leq x \leq a.$$
Hence, $$y'=1-\frac{a}{\sqrt{ax}},~~~y''=\frac{a}{2x\sqrt{ax}}.$$
Therefore, $$k=\frac{|y'|}{(1+y''^2)^{3/2}}=\frac{1}{2}\sqrt{\frac{a}{(2x-2\sqrt{ax}+a)^3}}.$$
Notice that $$2x-2\sqrt{ax}+a=2\left(\sqrt{x}-\frac{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Sum of Lagrange polynomials: $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $. Given $\{x_0,...,x_n\}$ I am asked to show that $\sum_{i=0}^{n}L_i(0)x_i^{n+1} = (-1)^{n}x_0\cdot\cdots\cdot x_n $.
I already showed that $\sum_{i=0}^{n}L_i(x)x_i^{j} = x^j$ for $j=1,...,n$ and that $\sum_{i=0}^{n}L_i(x)=1... | It can be proven with determinant form of Lagrange polynomial that interpolates $(x_0;y_0)$, $\dots$, $(x_n;y_n)$
$$
P(x) = (-1)
\frac{
\det
\begin{pmatrix}
0 & y_0 & y_1 & \cdots & y_n \\
x^n & x_0^n & x_1^n & \cdots & x_n^n \\
x^{n-1} & x_0^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is it possible to find such integers $y,b$ that $\left|-\frac{x_{0}}{n}-\frac{y}{b}\right|<\frac{1}{b\sqrt{n}}$ Given integers $x_{0},n$ with $x_{0}^{2}\equiv -1$ (mod $n$) then there are integers $y,b$ with $(y,b)=1,0<b\le\sqrt{n}$ and
$$\left|-\frac{x_{0}}{n}-\frac{y}{b}\right|<\frac{1}{b\sqrt{n}}$$
I tried to solve... |
Lemma. If $m\mid(x^2+1)$, then there exists $y,d$ such that
*
*$m=y^2+d^2$
*$xy\equiv d\pmod m$.
Proof.
Let $x^2+1=mn$.
If $m$ is prime, then $m=2$ or $m\equiv 1\pmod 4$, hence $m$ is the sum of two squares $m=y^2+d^2$.
Since $x^2\equiv -1\pmod{m}$, we have
$$(xy+d)(xy-d)=x^2y^2-d^2\equiv 0\pmod{m}$$
h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the smallest four digit number which is divisible by $15,25,40$ and $75$ I'm stuck on this question. My working:
\begin{align*}
15 & = 3 \cdot 5\\
25 & = 5^2\\
40 & = 2^3 \cdot 5\\
75 & = 3 \cdot 5^2
\end{align*}
LCM $= 600$
And I'm not sure what to do after this (if the above steps are right).
| The answer is $\boxed{1200}$. Following your method, we have
$$15 = 3 \cdot 5 \\ $$
$$25 = 5^2 \\$$
$$40 = 5^1 \cdot 2^3 \\$$
$$75 = 3 \cdot 25 $$
Thus, to find the LCM, we take the maximum exponents for each of the prime factors, and we obtain $600$. But, since we need a four-digit number, we can multiply by $2$ to o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\,,\;x\geq 0$ is equicontinous on $[0,+\infty)$ Prove that $f_n(x)=\sin{\sqrt{x+4n^2\pi ^2}}\,,\;x\geq 0$
*
*is equicontinous on $[0,+\infty)$.
*converges pointwise to $0$ on $[0,+\infty)$.
MY TRIAL
*\begin{align}\sqrt{x+4n^2\pi ^2}&=\sqrt{x+4n^2\pi ^2}\left[\dfrac{\sq... | By mean value theorem：
$$\left|f_n(x)-f_n(y)\right|=\left|\sin{\sqrt{x+4n^2\pi ^2}}-\sin{\sqrt{y+4n^2\pi ^2}}\right|= |f_n'(\xi_n)||x-y|\leq|x-y|.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2913943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^a \lfloor x^n \rfloor \,\mathrm{d}x$
Evaluate $\int_0^a \lfloor x^n \rfloor \,\mathrm{d}x$ (where $ \lfloor \,\cdot\, \rfloor $ denotes greatest integer function).
Can anyone please give a detailed explanation of how to do this?
This is my first question on MathStack Exchange.
Thank You
| For a given $a\geq0$ there should $\exists k \in \mathbb{N}: \color{red}{k}\leq a^{n}<\color{red}{k}+1$ (i.e. $k=\left \lfloor a^n\right \rfloor$) then
$$\int\limits_0^a \lfloor x^n \rfloor dx=
\int\limits_0^1 \lfloor x^n \rfloor dx+
\int\limits_1^{\sqrt[n]{2}} \lfloor x^n \rfloor dx+
\int\limits_{\sqrt[n]{2}}^{\sqrt... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find this inverse matrix using Gauss-Jordan? I am trying to find the inverse matrix of
$$\begin{pmatrix} \ln\left(x\right) & -1\\ \:\:1 & \ln\left(x\right) \end{pmatrix}$$
using the Gauss-Jordan method. Using a different method I could already find that the inverse matrix is:
$$\frac{1}{\ln ^2\left(x\right)+1}... | Suppose that the matrix $A$ is given by
$$ \begin{bmatrix} \ln(x) & -1 \\ 1 & \ln(x) \end{bmatrix} \tag{1} $$
using Gaussian Elimination
$$ U=A, L=I \tag{2} $$
$$ \ell_{21} = \frac{u_{21}}{u_{11}} = \frac{1}{\ln(x)} \tag{3} $$
the point is to find the coefficient to zero the column
$$ u_{2,1:2} = u_{2,1:2} -\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite integral of $x\sin^n x$ from $0$ to $\pi/2$ How to find
\begin{equation*}
\int_0^{\pi/2} x\sin^n x dx
\end{equation*}
where $n$ is a positive integer? I tried $y=x-\pi/4$ and that gives
\begin{equation*}
\frac{1}{2^{n/2}}\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} (\sin y+\cos y)^n dy+\frac{1}{2^{n/2}}\int_{-\pi/4}^{\p... | $I_0=\frac{\pi^2}8$ and $I_1=1$ and for $n\ge2$,
$$
\begin{align}
I_n
&=\int_0^{\pi/2}x\sin^n(x)\,\mathrm{d}x\tag1\\
&=-\int_0^{\pi/2}x\sin^{n-1}(x)\,\mathrm{d}\cos(x)\tag2\\
&=(n-1)\int_0^{\pi/2}x\cos^2(x)\sin^{n-2}(x)\,\mathrm{d}x+\int_0^{\pi/2}\cos(x)\sin^{n-1}(x)\,\mathrm{d}x\tag3\\
&=(n-1)\int_0^{\pi/2}x\sin^{n-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Solution for Simultaneous Logarithmic Equations Following are the Equations:
\begin{align}
\log x +\frac {\log(xy^8)}{((\log x)^2+(\log y)^2)} &= 2 \\
\log y + \frac{\log(x^8/y)}{((\log x)^2 +(\log y)^2)} &= 0
\end{align}
I tried substituting $\log x$ and $\log y$ with $a$ and $b$ but it results in a cubic equation wit... | These are
$\log x +\dfrac{\log(x/y^8)}{(\log x)^2+(\log y)^2}
= 2
$
and
$\log y+\dfrac{\log((x^8)y)}{(\log x)^2 +(\log y)^2)}
=0
$.
Expanding the logs,
$\log x +\dfrac{\log x-8\log y}{(\log x)^2+(\log y)^2}
= 2
$
and
$\log y+\dfrac{8\log x+\log y}{(\log x)^2 +(\log y)^2)}
=0
$.
Letting
$\log x = a, \log y = b$,
$a +\df... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \b... | You found the solutions to the homogeneous system $Tv = 0$.
The solutions to the system $Tv = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ are of the form $$\left(\text{one particular solution to }Tv = \begin{bmatrix} 1 \\ 2 \end{bmatrix}\right) \quad + \quad\left(\text{ any solution of the homogeneous system }\right)$$
We ca... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve,... | Let $x = 2+\cos \theta$ where $0\leq \theta \leq \pi$. Then the equation is equivalent to
$$
(\sin\theta + 4)^{2} + (\cos\theta-3)^{2} = 26 + 8\sin\theta - 6\cos\theta = 26 + 10\sin(\theta+\alpha)
$$
where $-\pi/2 < \alpha <0$ satisfies $\tan \alpha = -3/4$. This attains minimum at $\theta = \pi/2 - \alpha$, which cor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$
$$\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx$$
I tried $2$ or $3$ trigonometric transformations but it did not work. One of them is as follows
$$\frac{\sin^2x\cos^2x}{(\sin x+\cos x)(1-\sin x\cos x)}$$ after ... | Nosrati's idea is good. However his second identity is wrong. Using identities
$$\sin x+\cos x=\sqrt{2}\cos(\dfrac{\pi}{4}-x), \sin x\cos x=\frac12\big[2\cos^2(\dfrac{\pi}{4}-x)-1\big]$$
and substitution $x\to\dfrac{\pi}{4}-x$ gives
\begin{eqnarray}
I&=&\int_{0}^{\frac{\pi}{4}} \frac{\sin^2x\cos^2x}{\sin^3x+\cos^3x}dx\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921300",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Consider a partition P Consider a partition $P= \{ [0,\frac{2}{7}]\}, \{ [\frac{2}{7}, \frac{1}{2}\}], \{ [\frac{1}{2}, \frac{2}{3}]\}, \{ [\frac{2}{3}, \frac{5}{7}]\}, \{ [\frac{5}{7},1]\} $ of $[0,1].$. Compute $\overline{I}_P(f)$ and $\underline{I}_P(f)$ for $f(x)=(x-\frac{1}{2})^2$
Now I know that:
A partition $P $... | First, let us find the $l(J_k)$'s.
\begin{align*}
l(J_1) &= 2/7-0=2/7 \\
l(J_2)&=1/2-2/7=3/14 \\
l(J_3) &= 2/3 - 1/2 = 1/6 \\
l(J_4)&=5/7-2/3=1/21 \\
l(J_5) &= 1 - 5/7 = 2/7 \\
\end{align*}
Then the suprememum's (Because this seems to be a computation problem, I will simply use Wolfram Alpha):
\begin{align*}
\sup_{x\... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
First derivative of $f(x)= \frac{2}{x+1} +3$ I'm struggling to find the first derivative of $f(x)= \frac{2}{x+1} +3$ using the limit definition of derivative. I keep coming up with $f'(x) = \frac{-5}
{(x+1)^2}$ but I should be getting $f'(x) = \frac{-2}{(x+1)^2}$
\begin{align}
\lim_{x\to 0} &= \frac{(\frac{2}{x+h+1}+3)... | In case you want to go by this definition.
In some cases it is more straightforward than the one you have used. If you put $x-a=h$ in the definition below you get the one you have.
There is an equivalent definition which says that function is differentiable at a point $a$ if $$\lim_{x\rightarrow a} \frac{f(x)-f(a)}{x-a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How do I find the equation of a tangent to a hyperbola whose centre is (h,k)? Given that $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$ is equation a hyperbola,
I have to find its tangent at the point $\left(-2,\frac{14}{3} \right)$.
I know about the equations $c^2=(am)^2-b^2$ and $\frac{xx1}{a^2} - \frac{yy1}{b^2} = 1$... | $$\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$$
$$ {d\over dx}\left(\frac{((x-3)^2}{9} - \frac{(y-2)^2}{4} \right)=0$$
$$ \frac{2}{9}(x- 3)- \frac{2}{4}(y- 2){dy\over dx}= 0 $$
$$ \frac{4(x- 3)}{9(y- 2)}= {dy\over dx} $$
$${dy\over dx}\biggr|_{(-2,{14\over3})}= -\frac{5}{6} $$
Thus the equation of the tangent is
$$(y-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit Problem
If there exists $a \in \mathbb{R}$ such that:
$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$
has limit in $-2$. If such $a$ exists what is limit in $-2$ ?
Attempt to solve
My idea was first to try factorize ... | When you substituted $x=-2$ in $3x^2+ax+a+3$ you incorrectly put $(-2)$ in place of some of the $a$-s. The remaining part appears to be the follow-up of said mistake of algebra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Difficulty Finding $A^k$ Let $A=
\begin{bmatrix}
1& -1 & 1\\
0 & 1 & 1 \\
0 & 0 & 1\\
\end{bmatrix}$. Compute $A^k$.
My attempt
I'm trying to compute $A^k$ using this approach as follows:
$$
A=I+N=
\begin{bmatrix}
1& 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix}+
\begin{bmatrix}
0& -1 & 1\\
0 & 0 & 1 \\
0 & 0 &... | How about this. Take the exponential function $e^{tA}$, where $t$ is some parameter
$$e^{tA}=\sum_{k=0}^\infty\frac{t^kA^k}{k!}=e^{t(I+N)}= e^{tI}e^{tN}=e^t\left[I+tN+\frac{(tN)^2}{2}\right]$$
where we used the matrix identity $e^{A+B}=e^Ae^B$ that is valid when
matrices $A$ and $B$ commute.
Since the functions $t^k$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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The tangent at $P$ to $y = x^2 - x^3$ meets the curve again at $Q$. Show that locus of midpoint of $PQ$ is $y=1-9x+28x^2-28x^3$
The tangent at a variable point $P$ of the curve $y = x^2 - x^3$ meets it again at $Q$. Show that the locus of the middle point of $PQ$ is $$y = 1 - 9x + 28x^2 - 28x^3$$
My approach
$$y^\pr... | Given $f(x) = x^2-x^3$ the tangent line at $x_0$ is
$$
y = f(x_0) + f'(x_0)(x-x_0)\to y = \left(2 x_0-3 x_0^2\right) (x-x_0)-x_0^3+x_0^2
$$
Solving now for the intersection points we have
$$
y = x^2-x^3\\
y = \left(2 x_0-3 x_0^2\right) (x-x_0)-x_0^3+x_0^2
$$
with the feasible solution $x = 1-2x_0$ so the sought locus i... | {
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"url": "https://math.stackexchange.com/questions/2932801",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\lef... | My first thought was to factor out the $\frac 6{x^2}$ but yours was to factor out the $1 + \frac 1x$. So we'll do it your way:
Factor out the $1 + \frac 1x$ and we get
$(1 + \frac 1x) (\frac {-6}{x^2} + \frac 6{x^3}(1 + \frac 1x))$.
Now factor out the $\frac 6{x^2}$ to get:
$\frac 6{x^2}(1 + \frac 1x) (-1 + \frac 1x(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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determining the common difference of an arithmetic sequence and common ratio of a geometric sequence with related terms Coming from a finance guy, I understand how AP and GP work. However, I came upon a problem that combines the two and was stuck. Here it goes.
Given first term of AP and GP$=4$, and common ratio of GP... | Here, assume that the AP is :
$$a, a+d, a+2d, ...$$
and, the GP is :
$$ a, ar, ar^2 ,...$$
Then from the given data -
$a = 4$, $ r = d - 8$ ( or $d = r + 8$) and $\frac {a + 2d}{2r^2} = \frac{7}{16}$.
The third equation above simplifies as,
$\frac{4+2d}{4r^2} = \frac{2+d}{2r^2} = {7\over 16}$
$\implies \frac{r+10}{2r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution!
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$
So, make the substitution
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x = \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\... | Hint:
If trigonometric substitution is not mandatory, for (where $a$ is an arbitrary constant )
$$\dfrac{\sqrt{4x^2+a}}{x^{2n-1}}=4^{n-1}\dfrac{\sqrt{4x^2+a}}{(4x^2)^n}\cdot4x$$
set $\sqrt{4x^2+a}=y,4x^2+a=y^2,y\ dy=4x\ dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Simplify $(3\log x) - (2\log x)$ How to simplify $(3\log x) - (2\log x)$? Would this become $(\log x )^ {\frac{3}{2}}$ or would this be just $3\log x-2\log x =\log x$? If so how to get $\log x$?
I was given this question: solve for $x$ if $\log x + \log x^2 +...+ \log x^n =n(n+1)$. But, the answer to my main question w... | $$3\log x-2\log x$$
Here, we use two identities.
$$\log a^b = b\cdot\log a$$
$$\log_a b-\log_a c = \log_a \big(\frac{b}{c}\big)$$
Use the first identity to reach the following expression.
$$\implies \log x^3-\log x^2$$
Use the second identity to simplify.
$$\implies \log \big(\frac{x^3}{x^2}\big) \implies \boxed{\log x... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove that $2^q+q^2$ is divisible by 3 where $q$ is a prime and $q\geq5$. I'm looking to prove that $2^q+q^2$ is divisible by $3$ where $q$ is a prime such that $q\geq5$.
I know that primes greater than five will be congruent to either $1\ (\text{mod}\ 3)$ or $2\ (\text{mod}\ 3)$, which means that the $q^2$-term will a... | $2^{1} \equiv 2 \pmod{3}$;
$2^{2} \equiv 1 \pmod{3}$;
$2^{3} \equiv 2 \pmod{3}$;
$2^{4} \equiv 1 \pmod{3}$;
$2^{5} \equiv 2 \pmod{3}$;
$2^{6} \equiv 1 \pmod{3}$;
$...$
Note that when the exponent is odd, the remainder equals to $2$ and when the exponent is even, the remainder is $1$.
This pattern keeps repeating becau... | {
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"url": "https://math.stackexchange.com/questions/2942329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Antiderivative of $x\sqrt{1+x^2}$ I am attempting this problem given to me, but the answer key does not explain the answer. The question asks me to find the antiderivative of $x\sqrt{1+x^2}$.
My attempt:
$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$
$g'(x)$ must be $x$, therefore $g(x)=\frac{1}{2}x^2$
$f'(x)$ must equal $\sqr... | $u$-substitution is more standard, but I can follow your method to see how it works out.
$$\begin{align*}
\frac d{dx}f(g(x)) &= g'(x) \cdot f'(g(x))\\
&= x\sqrt{1+x^2}
\end{align*}$$
Try $g'(x) = x$, then
$$g(x) = \frac12x^2+C_1$$
Try $C_1 = \frac12$ to get $g(x) = \frac12(1+x^2)$. Then
$$\begin{align*}
f'(x) &= \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Calculate variance for a probability mass function In a family of ten people, the probability mass function of the number of people
who contracted the flu is given by
$$P_X(x) = K(2x + 9); x = 0,1,..., 10$$
Calculate the variance of the number of people with flu in the family.
My attempt:
The first thing I needed to fi... | Note that$$\sum_{x=0}^{10}9=9(11)=99$$
$$10K(11)+99K=1$$
$$110K+99K=1$$
$$209K=1$$
$$K=\frac1{209}$$
\begin{align}
E(X^2) &= K \sum_{x=0}^{10} x^2(2x+9)\\
&=K\left( 2\left(\frac{(10)(11)}{2}\right)^2+9\cdot\frac{(10)(11)(21)}{6}\right) \\
&= 10(11)K\left(\frac{10(11)}{2} + \frac{3(21)}{2} \right)\\
&=\frac{10(11)}{2}K\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
2nd order differentiel equation I was wondering how I could solve
$$
f''=\left(1+x^4\right)f
$$
for some $f$.
I must admit that when it is 2nd order depending on $x$ I've more struggle to succeed, but here i've no clue ! Any hint please ?
| Hint:
Let $f=e^{ax^3}u$ ,
Then $f'=e^{ax^3}u'+3ax^2e^{ax^3}u$
$f''=e^{ax^3}u''+3ax^2e^{ax^3}u'+3ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u=e^{ax^3}u''+6ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u$
$\therefore e^{ax^3}u''+6ax^2e^{ax^3}u'+(9a^2x^4+6ax)e^{ax^3}u=(1+x^4)e^{ax^3}u$
$u''+6ax^2u'+(9a^2x^4+6ax)u=(x^4+1)u$
$u''+6ax^2u'+((... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose t... | Let's continue from your last step.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
$$=\frac{k^3}{3}-\frac{k^2}{2}+2\frac{k^2}{2}+\frac{k}{6}$$
$$=\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6}+k^2$$
By induction hypothesis, first three terms yileds an integer and $k^2$ is an inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
probability of selection without replacement An urn contains 20 black marbles and 20 white marbles. Three marbles are chosen without replacement. What is the probability that the first marble is white given that the third marble was black?
| You could look at the various probabilities for the eight possibilities for the first three marbles,
but a quicker way is to use symmetry (each marble can be in any position) and say this is the same as the probability that the second marble is white given that the first marble was black, and that is $$\frac{20}{39}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Convergence for series failed using "Ratio Test" $$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$
Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$
which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$
t... | Hint:
$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{1\cdot3\cdot5\cdots\cdot(2n-1)\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=2n+1$$
and not
$$\frac{1\cdot3\cdot5\cdots\cdot(2n+1)}{1\cdot3\cdot5\cdots\cdot(2n-1)}=\frac{2n+1}{2n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Maximum value on a circle I need to find the maximum value of a function on a circle: Let $C$ denote the circle of radius $6$ centered at the origin in the $xy$-plane. Find the maximum value of $x^2y$ on $C$. Where do I even start with this?
| What about polars? If we allow the circle $C(R)$ to be of arbitrary positive radius $R$, then for $(x, y) \in C(R)$,
$x = R\cos \theta, \tag 1$
$y = R \sin \theta, \tag 2$
so that
$x^2y = R^3 \cos^2 \theta \sin \theta; \tag 3$
this will take on a maximum value whenever $\cos^2 \theta \sin \theta$ does; we have
$\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integrate $\int\frac{x^3+1}{x^4+x^3+x^2+x}\,dx$ How do I integrate $\displaystyle\int\dfrac{x^3+1}{x^4+x^3+x^2+x}\,\mathrm dx$?
I tried by splitting the equation in two parts like $\dfrac{x^3}{x^4+x^3+x^2+x}$ and $\dfrac1{x^4+x^3+x^2+x}$ and then cancelling out the $x$ terms from the first part and then trying to integ... | We first realise that both our numerator and denominator are factorable.
We can factor the numerator and denominator as:
$$\frac{(x+1)(x^2-x+1)}{(x)(x^3+x^2+x+1)}.$$
We can now multiply both the numerator and denominator by $x-1$,
we will now get:
$$\frac{(x^2-1)(x^2-x+1)}{(x)(x^4-1)}.$$
We can also factor the denomina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2950939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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show that $\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{5n}+\frac{1}{5n+1} < \frac{2}{3}$ , $\forall n \mathbb \in{N}$
Show that $$\frac{1}{3n+1}+\frac{1}{3n+2}+...+\frac{1}{5n}+\frac{1}{5n+1} < \frac{2}{3}$$ for all $n \in \Bbb{N}$
I tried with induction method but I can not find any results.
| $$\frac{1}{3n+1}+\frac{1}{3n+2}+\cdots+\frac{1}{5n+1} <\frac{1}{3n+1}+\frac{1}{3n+2}+\frac{1}{3n+2}+\cdots+\frac{1}{3n+2}$$ $$=\frac{1}{3n+1}+\frac{2n}{3n+2}<\frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$.
Now, consider a cubic polynomial function with 3 distinct real roots,
$f(x)=A(x-a)(x-b)(x-c)$
It ... | I'm taking capital $A=1$ and then $a < b < c$ as the roots. we get the cubic
$$ x^3 - \sigma_1 x^2 + \sigma_2 x - \sigma_3 \; , $$
where
$$ \sigma_1 = a + b + c, $$
$$ \sigma_2 = bc + ca + ab, $$
$$ \sigma_3 = abc. $$
The first derivative is $3 x^2 - 2 \sigma_1 x + \sigma_2,$ with roots
$$ \frac{\sigma_1 \pm \sqrt{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2951681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Proving that $(a,b)\cong [1,2)$. I aim to prove the proposition indicated in the title.
Is the following argument correct?
We now show that the map $h:(a,b)\to [1,2)$ defined as follows is a bijection.
\begin{align*}h(x) = \begin{cases}\frac{x-a}{b-a}+1&\text{ if }x\neq \frac{a+b}{2}\\1&\text{ if }x = \frac{a+b}{2}\\\... | First, note that your function is not surjective, since there is no value in $(a,b)$ that maps to $\frac32$. We know this because $\frac{x-a}{b-a}+1$ is continuous and strictly increasing on the set $(a,b)$. As such, we know that there is only one value on the set that can map to $\frac32$, which is $x=\frac{a+b}2$. Ho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.
When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ Accordin... | Hint: Look at the real and imaginary parts of $(3+4i)^n$, mod 5.
If $(3+4i)^n$ never has real and imaginary parts a multiple of 5, then $\left(\dfrac{3+4i}5\right)\!^{\Large n}$ never has real and imaginary parts an integer. (For $n\ge2$, anyway.) In particular, it never equals $1$, and $\dfrac{3+4i}5$ is not a root of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
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What can I say of x given these conditions? I know that $x^2-x$ is integer and also $x^n-x$ is integer for some $ n > 2$. Can I say that $x$ is integer? How can I show it? ($x \in \mathbb{R}$)
| Suppose that $x^2 - x = k \in \mathbf{Z}$. In order for $x$ to be real, either $k = 0$ and $x \in \{0,1\}$ is an integer, or $k > 0$. For any $k$ and $n$, long division shows that
$$x^n - x = F_{n,k}(x)(x^2 - x - k) + A_{n,k} x + B_{n,k} = A_{n,k} x + B_{n,k}.$$
In particular, for the LHS to be an integer, either $x$ m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How can I simplify the derivative further to match the correct answer? I've been stuck on trying to simplify my expression in order to match the correct answer but I can't seem to get the correct solution. Guidance towards the proper steps would be greatly appreciated!
The original question:
Differentiate and simplify... | Just multiply by $1$ (to free the denominator from irrationality):
$$g^\prime(x) = \frac{2x^2}{\sqrt[3]{\left(x^3-1\right)^2}\sqrt[3]{\left(x^3+1\right)^4}}\cdot \frac{\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}{\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3]{x^3-1}}=\\
\frac{2x^2\;\sqrt[3]{\left(x^3+1\right)^2}\;\sqrt[3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2956152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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What can be said about the roots of $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2 )x^2 + b(a + c)x + ac$?
Let a, b and c be real numbers. Then the fourth degree polynomial in
$x$, $acx^4 + b(a + c)x^3 + (a^2 + b^2 + c^2 )x^2 + b(a + c)x + ac$
(a) Has four complex (non-real) roots
(b) Has either four real roots or four com... | Hint: It is $$ac\left(x^2+\frac{1}{x^2}\right)+b(a+c)\left(x+\frac{1}{x}\right)+a^2+b^2+c^2=0$$
and now substitute $$t=x+\frac{1}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Let $F(x)=x^4-bx^3-11x^2+4(b+1)x+a$.Find $a,b$ Let $F(x)=x^4-bx^3-11x^2+4(b+1)x+a$. Find $a,b$
It’s given that $F(x)$ is a complete square of a quadratic polynomial and $(x+2) $ is a factor of $F(x)$
My attempt :
I can write $F(x) = (x+2)^2(Ax+B)^2$
Then putting $x=-2$, I got $a=36$
And by putting $x=0$, I got $4B^2=3... | You proved $F(x) = (x+2)^2(x\pm 3)^2$, and if you multiply out the terms, the coefficient of $x^2$ in $F$ will be
$$
\left[x^2\right]F(x) = 4 \pm 24 + 9 = 13 \pm 24
$$
but you know this must be $-11$, so the correct sign is $-$ and you end up with $$F(x) = (x+2)^2(x-3)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to solve $(x \mod 7) - (x \mod 8) = 5$? I'm trying to solve $(x \mod 7) - (x \mod 8) = 5$ but no idea where to start. Help appreciated!
| I'm going to assume you are incorrectly assuming $\mod n$ refers to the remainder function and want to solve $(x \% 7) -(x\% 8) = 5$ where $a \% b$ is the unique remainder when $a$ is divided by $b$.
$x \% 7 = a$ means $x = 7k + a$ for some integer $k$ and integer $0 \le a < 7$. So $a = x - 7k$.
And $x \% 8 = b$ mean... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the exact values of the square roots of $z=1+i$ are... Show that the exact values of the square roots of $z=1+i$ are...
$w_0=\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{-1+\sqrt{2}}{2}}$
$w_1=-\sqrt{\frac{1+\sqrt{2}}{2}}-\sqrt{\frac{-1+\sqrt{2}}{2}}$
My attempt
Let $z=1+i\in \mathbb{C}$. Then
$r=|z|=\sqrt{2}$
... | Just to expand on Dr. Sonnhard Graubner's answer, square-rooting the sum of the squares of these equations gives $A^2+B^2=\sqrt{2}$. (The right-hand side won't be $-\sqrt{2}$, for obvious reasons.) So $A^2=\frac{\sqrt{2}+1}{2},\,B^2=\frac{\sqrt{2}-1}{2}$. Although square-rooting these equations introduces $\pm$ signs, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute the following limit if they exist: $\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$ Compute the following limit if they exist:
$\lim_{(x, y, z) \to (0,0,0)} \frac{2x^2 y cos(z)}{x^2 + y^2}$
Attempt:
Given $\epsilon > 0$. Choose $\delta = \text{Not sure yet}$. Then $||(x, y, z) - (0,0,0)|| = \sqrt... | To compute the limit we have that
$$\left|\frac{2x^2 y \cos z}{x^2 + y^2}\right|\le \frac{2x^2 |y| }{x^2 + y^2}=2r\cdot(\cos^2 \theta|\sin \theta|) \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find $A+2B$ if $3\cos^2A+2\cos^2B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$
$A$ and $B$ are positive acute angles satisfying $3\cos^2A+2\cos^2B=4$ and $\dfrac{3\sin A}{\sin B}=\dfrac{2\cos B}{\cos A}$, then find the value of $A+2B$ ?
My Attempt
$\cos2B=2\cos^2B-1=3-3\cos^2A$ and $\sin2B=2\sin B\cos B=3\s... | Note that:
$$\begin{cases}\cos2B=2\cos^2B-1=3-3\cos^2A\\ \sin2B=2\sin B\cos B=3\sin A\cos A\end{cases} \Rightarrow \\
\cos^22B+\sin^22B=9-18\cos^2A+9\cos^4A+9\sin^2A\cos^2A \Rightarrow \\
1=9-18\cos^2A+9\cos^4A+9(1-\cos^2A)\cos^2A \Rightarrow \\
1=9(1-\cos^2A) \Rightarrow \\
\sin^2A=\frac19 \Rightarrow \\
\sin A=\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is $\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ convergent?
$\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ : convergent?
My attempt
$$
(\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} = (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n}}) + (\frac{1}{\sqrt{2}} - \f... | Hint:
Note that for $1\leq j \leq \frac{n}{2}$ we have
$$ \frac{1}{\sqrt{j}} - \frac{1}{\sqrt{n}} \geq \frac{\sqrt{2}}{\sqrt{n}}- \frac{1}{\sqrt{n}} \geq \frac{1}{10} \cdot \frac{1}{\sqrt{n}} $$
Use this to find a lower bound of your sequence which divergeces.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplfy this indices question Grade 10 :$\frac{3^n-4 \cdot 3^{n-2}}{3^{n+1}-3^{n-1}}$ I am a student in grade 10 and i was given this question about indices
$$\frac{3^n-4 \cdot 3^{n-2}}{3^{n+1}-3^{n-1}}$$
All I am capable to do is expanding it out, i.e. from $3^{n-2}$ to $3 \cdot 3^{-2}$. I am unable proceed to... | Welcome. You can write:
$3^n-4\times 3^{n-2}=3^n-3^{n-2}-3\times 3^{n-2}$
$3^{n+1}-3^{n-1}=3(3^n-3^{n-2})$
Then the fraction(F) reduces as follows:
$F=\frac{1}{3}-\frac{1}{3^{2-n}}\times(\frac{1}{3^{n+1}-3^{n-1}})=\frac{1}{3}-\frac{1}{8}=\frac {5}{24}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2965783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integration-by-part Here is the given question
$$\int{\frac{x^3}{\sqrt{1+x^2}}dx}$$
I solved using integration by part as follow:
$$
\int\frac{x}{\sqrt{1+x^2}}x^2\,dx = x^2\int\frac{x}{\sqrt{1+x^2}}\,dx - \int\biggl(\frac{dx^2}{dx}\int\frac{x}{\sqrt{1+x^2}}dx\biggr)dx\tag{i}\label{i}
$$
Solving for $\int \frac{x}{\sqr... | \begin{align}\displaystyle\int\dfrac{x^3}{\sqrt{1+x^2}}\,\mathrm dx&=\dfrac12\displaystyle\int\dfrac{x^2\mathrm d(x^2)}{\sqrt{1+x^2}}\\&=\dfrac12\displaystyle\int\dfrac{(1+x^2)-1}{\sqrt{1+x^2}}\mathrm d(x^2)\\&=\dfrac12\displaystyle\int\sqrt{1+x^2}\mathrm d(x^2)-\dfrac12\displaystyle\int\dfrac{\mathrm d(x^2)}{\sqrt{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)\cdots\left(1+\frac{1}{n^3}\right) < 3$ I have this problem which says that for any positive integer $n$, $n \neq 0$ the following inequality is true: $$\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}... | We have that
$$\prod_{k=1}^\infty \left(1+\frac{1}{k^3}\right)<3\iff \sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 3$$
and since $\forall x>0\, \log(1+x)<x$
$$\sum_{k=1}^\infty \log\left(1+\frac{1}{k^3}\right)=\log 2+\sum_{k=2}^\infty \log\left(1+\frac{1}{k^3}\right)<\log 2+\sum_{k=2}^\infty \frac{1}{k^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
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Die is rolled until 1 appears. What is the probability of rolling it odd number of times? Problem: Die is rolled until 1 appears. What is the probability of rolling it odd number of times?
So, so far I have this:
$\frac{1}{6}$ - this is a probability of rolling "1" on first try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \f... | $p(1)=\frac 1 6 $ $P(not 1)=\frac 5 6$
1 on $1^{st}$ roll: $$\frac 1 6=\frac{5^0}{6^1}$$
1 on $3^{rd}$ roll: $$\frac 5 6 \cdot \frac 5 6 \cdot \frac 1 6 = \frac {5^2} {6^3} $$
1 on $5^{th}$ roll: $$\frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6 \cdot \frac 5 6\cdot \frac 1 6=\frac {5^4}{6^5}$$
keep going upto $\infty$: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Proof by induction in inequalities $$\sum^n_{k=1} \frac1{k^3} \le \frac 5 4 - \frac 1 {2n^2}$$ For all $n\ge 2$
Now this really a tough one for me.
The base case holds at $n = 2$
Then i replaced it with $p$ and then $p+1$
I got an inequality where i deduced that $p+1 = p$ and then plus a 1. But its not burging
| Assume that the inequality be equal for $n=p$ as you did
$$
\sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}} \le {5 \over 4} - {1 \over {2p^{\,2} }}
$$
Then it ia also valid for $n=p+1$
$$
\sum\limits_{k = 1}^{p + 1} {{1 \over {k^{\,3} }}} = \sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}} + {1 \over {\left( {p + 1} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n <... | An alternative for monotonicity:
Consider the corresponding (continuous) function $f(x) = \dfrac{x^2 + 3}{(x+1)^2}$. You can then check (using calc 1 methods) that $f'(x) > 0$ if $x \geq 3$. Hence $a_n$ is increasing for $n \geq 3$. You have already checked that $a_1 \leq a_2 \leq a_3$, so the sequence is monotone ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_... | Note that$$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})=F_n(F_n+2F_{n-1}).$$And if $2\mid F_n$, $4\mid F_n(F_n+2F_{n-1})$.
On the other hand\begin{align}F_{n+1}^3-F_{n-1}^3&=(F_{n+1}-F_{n-1})(F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2)\\&=F_n\bigl((F_n+F_{n-1})^2+F_n+F_{n-1}+F_{n-1}^2+F_{n-1}^2\bigr)\\&=F_n(F_n^2+3F_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+... | I would like to present you a geometric explanation of what is happening.
If $\xi = \cos x$ and $\eta=\sin x$, then you rewrite your equation as:
$$
a\xi +b\eta=c,\qquad \xi^2+\eta^2=1.
$$
So you are trying to find intersection points of a line and a circle.
Simple geometry tells us that bisector $AF$ is perpendicul... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Determine the image of $S$ under $f(z)=\frac{z^2+2}{z^2+1}.$
Let $S$ be the region $\{z:0<|z|<\sqrt{2}, \ 0 < \text{arg}(z) <
\pi/4\}$. Determine the image of $S$ under the transformation
$$f(z)=\frac{z^2+2}{z^2+1}.$$
I'm facing some difficulties on problems of this nature that includes Möbius transformations. First... | I think this might help:
$$f(z)=\frac{2+z^2}{1+z^2}=1+\frac{1}{1+z^2}$$
If $z=x+iy$:
$$\frac{1}{1+z^2}=\frac{1}{(x^2-y^2+1)+2ixy}=\frac{(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$
So:
$$f(z)=\frac{(x^2-y^2+1)^2+4x^2y^2+(x^2-y^2+1)-2ixy}{(x^2-y^2+1)^2+4x^2y^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Algebraic proof to show product of three numbers is zero from a system Let $a, b, c$ be real numbers satisfying: $$(a+b)(b+c)(c+a)=abc$$
$$(a^3+b^3)(b^3+c^3)(a^3+c^3)=a^3b^3c^3$$
Prove that $abc=0$
My work: I tried to prove it by contradiction, assuming $abc\neq0$ so $a,b,c\neq0.$
Then I factored second equation to be ... | Look at the first equation as a quadratic equation in $a$. It has discriminant $(b^2+bc+c^2)^2-4bc(b+c)^2=(b^2-bc+c^2)-8b^2c^2$. So we have
$$
\begin{align*}
a^4b^4c^4&=(a^2b^2c^2)^2\\
&=\prod_{cyc}(b^2-bc+c^2)^2\\
&\geq\prod_{cyc}8b^2c^2\\
&=512a^4b^4c^4
\end{align*}
$$
which only works if $abc=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2979726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How is $2(\ln \tan x + c)$ simplified to $A \tan^2 x$ where $A =2c$ I'm trying to follow this reasoning:
$1/2 \ln(4 +y^2) = \ln(\tan x) + C$
$\ln(4 +y^2) = 2\ln(\tan x) + \ln A$ ( constant 2C = A)
$4 + y^2 = A \tan^2 x$
| $\frac{1}{2}ln(4+ y^2)= ln(tan(x))+ C$
Multiply on both sides by 2:
$ln(4+ y^2)= 2ln(tan(x))+ 2C$
Taking $A= e^{2C}$ (NOT A= 2C) we have $2C= ln(A)$ so
$ln(4+ y^2)= ln(tan^2(x))+ ln(A)= ln(Atan^2(x))$
Now take the exponential of both sides:
$e^{ln(4+ y^2)}= e^{ln(Atan^2(x))}$
$4+ y^2= Atan^2(x)$
I have used:
1) $2ln(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$ As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$
But unfortunately I do not have skills in techniques to handle double summation .
Help appreciated.
I've made some re... |
The double summation is equal to
$$\frac{11\zeta(4)}{8}=\frac{11\pi^4}{720}.$$
Note that
$$\frac{m^2+n^2}{m n\left(m^2-n^2\right)^2}=
\frac{1}{2 mn(m+n)^2}+
\frac{1}{2 mn(m-n)^2}.$$
Now consider the Tornheim double sums:
$$T(a,b,c)= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}
\frac{1}{m^an^b(m+n)^c}.$$
Then
$$ \sum_{m... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Calculate $\sum_{n=1}^{\infty} \arctan\bigl(\frac{2\sqrt2}{n^2+1}\bigr) $ $$ \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{2\sqrt2}{k^2+1}= \lim_{n \to\infty} \sum_{k=1}^{n} \arctan\frac{(\sqrt{k^2+2}+\sqrt2)-\sqrt{k^2+2}-\sqrt2)}{(\sqrt{k^2+2}+\sqrt2)(\sqrt{k^2+2}-\sqrt2)+1}= $$
$$\lim_{n \to\infty} \sum_{k=1}^{n} \... | Note that
$$\arctan\left(\frac{2\sqrt{2}}{k^2+1}\right)=\arctan\left(\frac{\left(\frac{k+1}{\sqrt{2}}\right)-\left(\frac{k-1}{\sqrt{2}}\right)}{\left(\frac{k+1}{\sqrt{2}}\right)\left(\frac{k-1}{\sqrt{2}}\right)+1}\right)=\arctan\left(\frac{k+1}{\sqrt{2}}\right)-\arctan\left(\frac{k-1}{\sqrt{2}}\right)\,.$$
Therefore,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987659",
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"source": "stackexchange",
"question_score": "1",
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Minimize the expression This was a problem given in a no calculator math contest (HMMT):
What is the minimum value of $(xy)^2+(x+7)^2+(2y+7)^2$ where $x$ and $y$ are reals. My friends and I discussed about using calculus, but we reasoned that there must be a faster trick (calc would take too long). Any ideas?
| Hint:
\begin{align}
f(x,y) &= (xy)^2+(x+7)^2+(2y+7)^2 \\
&= (xy-2)^2+(x+2y+7)^2+45 \\
\end{align}
Details for the thoughts
By $(a+b+c)^2=a^2+b^2+c^2+2(bc+ca+ab)$
Hence,
\begin{align}
(x+7)^2+(2y+7)^2 &= x^2+14x+49+4y^2+28y+49 \\
&= (x^2+4y^2+49+14x+28y \color{red}{+4xy})+49 \color{red}{-4xy} \\
&= (x+2y+7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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The $1997$ IIT JEE problem
Let $S$ be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$
This problem appeared in IIT JEE $1997$ (re-exam).I really do not ... | The smallest $a^2+b^2+c^2+d^2$ is when the quadrilateral has its 4 vertices on the midpoint of each side of square S. $a=b=c=d=0.5$, and the sum of the squares is $2$. The largest is simply achieved when the vertices of the quadrilateral lies on the vertices of square S. ie, both squares are the same. The sum of th... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Lograthmic equation $ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}$ solutions $$ \frac {1}{\log_2(x-2)^2} + \frac{1}{\log_2(x+2) ^2} =\frac5{12}.$$
I made the graph using wolfram alpha it is giving answer as
6. But how to solve it algebraically?
base of logarithm is 2.
Tried using taking Lcm but the... | $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2} = \frac{5}{12}$$
Rewrite the logs using $$\log_a b^c = c\log_a b$$
and factor. Then, simplify both sides.
$$\frac{1}{2}\cdot\frac{1}{\log_2\vert x-2\vert}+\frac{1}{2}\cdot\frac{1}{\log_2\vert x+2\vert} = \frac{5}{12}$$
$$\frac{1}{2}\bigg(\frac{1}{\log_2\vert x-2\vert}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996775",
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"source": "stackexchange",
"question_score": "1",
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Solving characteristic equation for closed form solution from a given recurrence equation Provided a recurrence relationship, I converted the equation assuming $T(n) = r(n)$, and found the roots to the quadratic equation.
root 1: $r_1 = 2+\sqrt{3}$
root 2: $r_2 = 2-\sqrt{3}$
Resulting in the equation with the form $T... | For brevity, take
\begin{align}
\alpha &= 2+\sqrt{3} \\
\beta &= 2-\sqrt{3} \\
A &= c_1 \\
B &= c_2
\end{align}
Now,
\begin{align}
\alpha+\beta &= 4 \\
\alpha-\beta &= 2\sqrt{3} \\
\alpha \beta &= 1 \\
A\alpha+B\beta &= 1 \tag{1} \\
A\alpha^3+B\beta^3 &= 3 \tag{2}
\end{align}
$(2)-\beta^2 \times (1)$,... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Help calculating $\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$ I need some help calculating this limit:
$$\lim_{x \to \infty} \left( \sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} \right)$$
I know it's equal to 1 but I have no idea how to get there. Can anyone give me a tip? I can't use l'Ho... | EDIT
I guess you need a basic method that preceeds the l'Hospital's rule.
Set $x=a^2,\; a>0.$ The limit rewrites
$$ \begin{aligned}\sqrt{a^2+a} - \sqrt{a^2-a}=&\;\left(\sqrt{a^2+a} - \sqrt{a^2-a}\right)\frac{\sqrt{a^2+a}+\sqrt{a^2-a}}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\
=&\;\frac{2a}{\sqrt{a^2+a}+\sqrt{a^2-a}}\\=&\;
\frac{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3000336",
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"source": "stackexchange",
"question_score": "3",
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Solving $\int_{0}^{\infty} \frac{\sin(x)}{x^3}dx$ In my attempt to solve the this improper integral, I employed a well known improper integral (part of the Borwein family of integrals):
$$ \int_{0}^{\infty} \frac{\sin\left(\frac{x}{1}\right)\sin\left(\frac{x}{3}\right)\sin\left(\frac{x}{5}\right)}{\left(\frac{x}{1}\rig... | As the other answers have pointed out, the integral does indeed diverge. But if want to assign a finite value to it, there are a couple ways to see that in fact $-\pi/4$ is the "right" value.
One is to take the integral not quite down to zero, but instead to $\epsilon$. If we do, then expand in a series in $\epsilon$, ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Prove $\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)=\frac{\pi^3}{32}-2G\ln2$ How to prove
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}(2H_{2k}+H_k)\stackrel ?=\frac{\pi^3}{32}-2G\ln2,$$
where $G$ is the Catalan's constant.
Attempt
For the first sum,
$$\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}H_{2k}=\Re\... | this summation also has some relationship with integral identity, which asked almost the same time, see this, rewrite as
$$\frac3{2} \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x} \mathrm{d}x} = \int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \mathrm{d}x}$$
given some famous series
$$\arctan x = \sum_{n=0}^{\infty} {\frac{(-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Complex root of equation 1 let $a,b$ & $c$ are the roots of cubic $$x^3-3x^2+1=0$$
Find a cubic whose roots are $\frac a{a-2},\frac b{b-2}$ and $\frac c{c-2} $
hence or otherwise find value of $(a-2)(b-2)(c-2)$
| HINT
We have that $(x-a)(x-b)(x-c)=x^3-3x^2+1$ and then
*
*$a+b+c=3$
*$ab+bc+ca=0$
*$abc=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Probability that 2 heads do not come consecutively.
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?
Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{H... | You are missing the cases where the tails can be split into more than 2 groups.
i.e. 3 tails can be split into 1+1+1, yielding another 10 cases = $ {5 \times 4 \times 3\over 3 \times 2 \times 1} $ etc...
The final answer should be $144\over2^{10}$. There are 28 more cases for 3 heads and 15 more cases for 2 heads.
Key... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What is $\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$? What's the result of:
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$$
Is it
$$\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|x|}=0$$
or
$$\frac{1}{|{x}|}-\frac{x^2}{|x|^2x}=\frac{1}{|{x}|}-\frac{1}{x}\frac{x^2}{|x|^2}=\frac{1}{|{x}|}-\frac{1}{x}=\left\{\begin{matrix}\fr... | Your first conclusion is right since $$|x|^3=|x|^2\cdot |x|=x^2\cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3\ne x^3= |x|^2\cdot x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
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Show that these non-linear recursions produce integers only The recurrence is of third order:
Start with
\begin{align*}a_0(x)&=1\\ a_1(x)&=1\\ a_2(x)&=x \end{align*} and then
\begin{align*}a_{n+3}(x)&=\frac{a_{n+2}^2(x)-a_{n+1}^2(x)}{a_{n}(x)}.\end{align*}
The calculation and factorization of $a_n(x)$ for $3\le n... | Nice recursion!! However, let us change the name and indexing of your sequence. Notice that if $\,U_n(x)$ is the Chebyshev polynomial of the second kind, then $\, U_n(x/2)$ is a polynomial in $x$ with integer coefficients. Define $\, P_n(x) := U_{F_n - 1}(x/2).\,$ Your sequence $\, a_n(x) = P_{n+1}(x).\,$ Please note t... | {
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"source": "stackexchange",
"question_score": "1",
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approximation of trigonometric functions in form $a + bx + cx^2 +...$ Is there any way to express the trigonometric functions as infinitely long polinomials? If so, how? If not why?
Obviously doing $x(x-\pi)(x-2\pi)(x-3\pi)...$ does not work as it doesn't match for values in between the zeroes.
| You already received good answers to the question.
In this area, we can do other things which are amazing. If you look here, being myself fascinated by the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which wasproposed, more than $1400$ years ago by Mahabhaskariya... | {
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"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculating $\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{m-1} |mi-nj|$ I'd like to calculate $f(n,m)=\sum\limits_{i=1}^{n-1}\sum\limits_{j=1}^{m-1} |mi-nj|$ for all $1 \leq n \leq N,\ 1 \leq m \leq M$. Straightforward brute force method runs in $O(N^2M^2)$ which is too slow. How to calculate all values in $O(NM)$?
|
We show the following is valid for positive integers $n,m$:
\begin{align*}
\sum_{i=1}^{n-1}\sum_{j=1}^{m-1}|mi-nj|=\frac{1}{6}\left(2m^2n^2-3m^2n+m^2-3mn^2+3mn+n^2-\left(\gcd(m,n)\right)^2\right)
\end{align*}
In the following we denote with $d=\gcd(m,n)$.
We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^{n-1}}... | {
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How to find the upper Riemann integral of following function? $f(x)$ is defined on $[a,b]$ as -
$= 0$ if $x ∈ [a, b] ∩ Q$
$= x$ otherwise
Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.
| Since the irrationals are dense, we have $\sup_{x \in [x_{j-1},x_j]}f(x) = x_j$.
Hence, for any partition, $a = x_0 < x_1 < \ldots < x_n = b$, the upper Darboux sum is
$$U(P,f) = \sum_{j=1}^n x_j(x_j - x_{j-1}) = \frac{1}{2} \sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line ... | By letting $x=t+1$ we get an even function
$$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}.$$
Now we show that the minimum value is attained at $t=0$: we have to verify
$$\sqrt{2t^2+6t+5}+\sqrt{2t^2-6t+5}\geq \sqrt{20}$$
or, after squaring,
$$4t^2+10+2\sqrt{4t^4-16t^2+25}\geq 20$$
that is
$$\sqrt{(5-2t^2)^2+4t^2}\geq 5-2t^2$$
whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros? Does $Q(x) =1 -x + \frac{x^2}{2}- \frac{x^3}{3} + \frac{x^4}{4} -\frac{x^5}{5} +\frac{x^6}{6}$ have any real zeros?
\begin{align}
Q'(x) &= -1 + x -x^2 + x^3 -x^4 + x^5 \\
&= -(1-x+x^2)+x^3(1-x+x^2)\\
&=(x... | egreg's answer already resolves the problem, but here's a fun one: recall the maclaurin series for $\ln(x+1)$:
$$\ln(x+1)=x-\frac12x^2+\frac13x^3-\frac14x^4+\cdots=-(Q(x)-1)+O(x^7).$$
This expansion is only valid when the series converges, of course. In that range,
$$Q(x)=-\ln(x+1)+O(x^7)+1.$$
As $x\to-1$ (here, the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$ not correct? $\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x = \lim_{x\to -\infty}(-x)+x=\lim_{x\to -\infty}0 = 0$
Apparently, the 2nd step is illegal here. Probably because for $x=-\infty$ I'd get $(+\infty-\infty)$ which is not p... | To avoid confusion with sign, in these cases I suggest to take $y=-x\to \infty$ then
$$\lim_{x\to -\infty}\sqrt{x^2+5x+3}+x =\lim_{y\to \infty}\sqrt{y^2-5y+3}-y $$
from here we can clearly see that the expression is an indeterminate form $\infty-\infty$ then we can proceed by
$$\sqrt{y^2-5y+3}-y=\left(\sqrt{y^2-5y+3}-y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3024120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 4
} |
Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\... | $$(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)$$
$$=(2\sec^2x-1)(2\csc^2x-1)$$
$$=4\sec^2x\csc^2x-2(\sec^2x+\csc^2x)+1$$
Now use $\sec^2x+\csc^2x=\cdots=\sec^2x\csc^2x$
The second one has been solved by Taussig
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Put a matrix $A$ in Jordan Normal Form and find a $P$ such that $P^{-1}AP=J$ I have a linear algebra exam tomorrow and this is a frequent question.
$A=
\begin{pmatrix} 4 & 0 & 1 & 0 \\ 2 & 2 & 3 & 0 \\ -1 & 0 & 2 & 0 \\ 4 & 0 & 1 & 2
\end{pmatrix}$
$C_T(x)=(x-2)^2(x-3)^2$ so eigenvalues are $2$ and $3$.
$a_2=2$ so the... | The left two columns are just a basis of 2 eigenvectors. For 3, we take the far right vector as some $w$ such that $(A-3I)^2 w = 0 $ but $(A-3I) w \neq 0. $ Then the third column is $v = (A - 3I)w.$
$$
P =
\left(
\begin{array}{rrrr}
0 & 0 & 1&1 \\
1&0&-1&3 \\
0& 0 & -1& 0 \\
0&1&3&1
\end{array}
\right)
$$
determinant i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum value of $\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$ Given that $0\lt x\lt 2$ and $0\lt y\lt 2$ then find the minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$$
My try:
On factorisation we need minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {(y-2)... | You can still use Minkowski as:
$$\sqrt{(x+y)^2+(y-x)^2}+\sqrt{(2-y)^2+x^2}+\sqrt{(2-x)^2+(2-y)^2}\geq\sqrt{(x+y+2-y+2-x)^2+(y-x+x+2-y)^2}=\sqrt{20}.$$
One can check that the equality is attained when: $$(x,y) = \left(\frac 25,\frac 65\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - ... | $x_1 + x_2 + x_4 \le 8 + 4 + 5 = 17$ so $x_3=40 - x_1 + s_2 + x_4 \ge 40 -17 \ge 4$ so we can ignore the restriction on $x_3$.
$2 \le x_1 \le 6$ so there are $7$ values that $x_1$ can be, $x\le 4$ so there are $5$ values it can be. $x_4 \le 5$ so there are $6$ values it can be and $x_3$ must be $40 - x_1 - x_2 -x_4$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"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.