Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the integral of $\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$
Evaluate $\displaystyle\int \ln\left(\sqrt{x-b}+\sqrt{x-a}\right)\,dx$.
I am tryed to integrate it by parts by taking $du = 1$ and $v=\ln\left(\sqrt{x-b}+\sqrt{x-a}\right)$
Therefore,
$vu - \displaystyle\int v du =x\ln\left(\sqrt{x-b}+\sqrt{x-a}\... | $$I=\int\frac x{\sqrt{x^2-(a+b)x+ab}}dx$$You can write $x=0.5[2x-(a+b)+(a+b)]$,$$I=\int\frac{0.5[2x-(a+b)+(a+b)]}{\sqrt{x^2-(a+b)x+ab}}dx\\=0.5\left[\int\frac{2x-(a+b)}{\sqrt{x^2-(a+b)x+ab}}dx+\int\frac{(a+b)}{\sqrt{\left[x-\frac{(a+b)}2\right]^2-(a+b)^2/4+ab}}dx\right]$$The first integral reduces to $\int\frac{du}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Conditional probability of min and max of two dice Consider the following problem, from Tijms's Understanding Probability:
Two dice are rolled. Let the random variable $X$ be the smallest of the two outcomes and let $Y$ be the largest of the two outcomes. What are the conditional mass functions $P (X = x | Y = y)$ and... | Partial answer due lack of time
For $P(Y=y)$ I have something slightly different. I´ve worked with the table below:
$P(Y=y|X=x)$
$$ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{ X/Y } & 1 &2 &3 &4 &5 &6 \\ \hline \hline \hline 1 &\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3225706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in... Let $\{a_n\}$ be a sequence of positive real numbers such that
$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.
Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
(A) $... | Following your calculations and according to the ratio test
$$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$
thus
$$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$
Now, applying the same technique from the proof on the ratio test
$$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Evaluate the limit $ \lim_\limits{n \to \infty} \frac{1^a+2^a\cdots+n^a}{n^{a+1}} $ The exact question is-
Find the real value(s) of $a (a \ne -1)$ for which the limit $$ \lim_{ n \to \infty} \frac{ 1^a+2^a\cdots+n^a}{(n+1)^{a-1}[ (na+1)+(na+2) \cdots+(na+n)]} = \frac{1}{60}$$
I simplified it a bit to get that limit. H... | Useful link related to the first equation;
Evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}$
Limit of the sequence $\frac{1^k+2^k+...+n^k}{n^{k+1}}$
$\lim_{{n}\to {\infty}}\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=?$
Showing that $\lim_{n \rightarrow \infty}\frac{1^k+2^k+...+n^k}{n^{k+1}}=\frac{1}{k+1}$ usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hard limit involving different order radicals $\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$ Please help me to calculate the following limit
$$\lim_{n \to \infty} (\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} )$$
I factored out $n$ from both radicals but it didn't work and tried to use the identity $a^2-b^2$ and $a^3-b^... | There is a way to use directly the two binomial formulas you mentioned. It only needs a little trick:
$$\sqrt[3]{n^3+3n^2}-\sqrt{n^2+2n} = \sqrt[3]{n^3+3n^2}\color{blue}{-n + n}-\sqrt{n^2+2n}$$
Now, consider
\begin{eqnarray*} \sqrt[3]{n^3+3n^2}-n
& \stackrel{n=\sqrt[3]{n^3}}{=} & \frac{n^3+3n^2 - n^3}{(\sqrt[3]{n^3+3n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Given three non-negative $a,b,c$ so that $c$ between $a$ and $b$. Prove that $2\sum\limits_{cyc}a^{3}+3abc\geqq3\sum\limits_{cyc}a^{2}b$ .
Given three non-negative $a, b, c$ so that $c$ between $a$ and $b$. Prove that $2\sum\limits_{cyc}a^{3}+ 3abc\geqq 3\sum\limits_{cyc}a^{2}b$
Inspried from: https://math.stackexcha... | By Schur's inequality, the basic inequality $3(a^2+b^2+c^2)\geq (a+b+c)^2$ (which can be proved by Cauchy-Schwarz if we write $3=1^2+1^2+1^2$) and the non-negativity of $a,b$ and $c$ we have that
$$a^3+b^3+c^3+3abc\geq a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\Rightarrow\\ 2(a^3+b^3+c^3)+3abc\geq (a+b+c)(a^2+b^2+c^2)\geq \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $2 - \sqrt{2}$ is irrational I suppose $2 - \sqrt{2} $ is rational.
so $$2- \sqrt{2} = {a/b} $$
where a,b are integers and gcd(a,b) = 1.
$$\text{Step 1. } 2 = (a/b)^2 \text{ //squared both sides }$$
$$\text{Step 2. } 2b^2 = a^2 \text{ //We see $a^2$ is even }$$
$$\text{Step 3. }2b^2 = (2k)^2$$
Step 3 since $a... | As an alternate approach, note that $2-\sqrt 2$ is a root of $x^2-4x+2$ (the conjugate root is $2+\sqrt 2$ so just compute $(x-(2-\sqrt 2))(x-(2+\sqrt 2))$.
By the Rational Root Theorem, this quadratic has no rational roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Integrating double integral with spherical coordinates problem with interpret a domain Hello i have the following problem i am solving integral with spherical coordinates but i am getting wrong answer - i think i am integrating correct so i think the problem is coming from the constraints. So i have $$\iint_Dxdx$$ and ... | The limits are $x^2+y^2\le 2y \ ,\ x\ge0 \ , y\ge\frac{1}{2} $
You've found the upper limit i.e, $r = 2\sin\theta$ (upper limit).
For $y\ge\frac{1}{2} \ , \ r\sin\theta \ge\frac{1}{2} $
So, the lower limit of $r$ is $\frac{1}{2\sin\theta}$
Also, $x\ge0$ , $r\cos\theta\ge0$ , $\cos\theta\ge0$ So, $\theta\le\pi/2$
At $y=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$ The inequality:
$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$$
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an e... | $$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \\
\left( \sqrt{a^2+c^2}+\sqrt{b^2+d^2} \right)^2 \ge \: \left( \sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \right)^2 \\ a^2+c^2+b^2+d^2+2\left( \sqrt{a^2+c^2}\sqrt{b^2+d^2} \right) \ge \: \left(a+b\right)^2+\left(c+d\right)^2 \\
a^2+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum using lagrange I've got the formula $$x^3+y^3+z^3$$
With the constraint $ax+by+cz = 1$
I tried to solve this using lagrange but every possible way I try to use does not get me to the right answer
Using the lagrange I got to
$$3x^2+ka = 0 $$
$$3y^2+kb = 0 $$
$$3z^2+kc = 0 $$
=
$$3x^2bc+kabc = 0 $$
$$3y^... | Assuming $x \ge 0, y \ge 0, z \ge 0$ to assure a bounded solution, we have from the stationary conditions
$$
x = \sqrt\frac{\lambda a}{3}\\
y = \sqrt\frac{\lambda b}{3}\\
z = \sqrt\frac{\lambda c}{3}\\
$$
and substituting into the restriction
$$
\frac{\lambda}{3} = \frac{1}{(a\sqrt a+b\sqrt b+c\sqrt c)^2}
$$
but
$$
x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
complete set of values of $a$ having modulus and linear terms
If $(9-x^2)>|x-a|$ has at least one negative real solution for $a\in\mathbb{R}.$ Then complete set of values of $a$ is
Plan
If $x>a$ Then $9-x^2>x-a\Rightarrow x^2+x-(a+9)<0$
If $x\leq a$ Then $9-x^2>a-x\Rightarrow x^2-x+a-9<0$
How do i solve these inequ... | Sketch a graph!
$k=5$">
Now you can see it more clearly. You want the red curve to be above the blue one.
For $a= 3$, the two curves intersect at $(3,0)$ and $(-2,5)$. (Solve the equation $9-x^2=-x+3$) So the solution set is $x\in (-2,3)$. The solution set is $x\in (-3,2)$ for $a=-3$.
For $a \leq 3$, we need to solve t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(... | Let $p = A+B = \sqrt{1+x}+\sqrt{1-x}$. We have
$$ p^2 = 2+2\sqrt{1-x^2}$$
$$ p^3 = \sqrt{(1-x)^3}+\sqrt{(1+x)^3} + 3p\sqrt{1-x^2}$$
so
$$ 1+ \sqrt{1-x^2} = \frac12 p^2$$
$$ \sqrt{(1-x)^3}+\sqrt{(1+x)^3} = -\frac12 p^3 +3p$$
and we get
$$ \frac{p}{\sqrt{2}} (-\frac12 p^3 +3p) = 1 + \frac12 p^2$$
$$ - \frac{1}{2\sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
} |
positive root of the equation $x^2+x-3-\sqrt{3}=0$ $x^2+x-3-\sqrt{3}=0$
using the quadratic formula we get
$$x=\frac{-1+\sqrt{13+4\sqrt{3}}}{2}$$ for the positive root
but the actual answer is simply $x=\sqrt3$
I am unable to perform the simplification any help would we helpful
| $13+4\sqrt{3}=(2\sqrt{3})^2+2(2\sqrt{3})(1)+1=(2\sqrt{3}+1)^2$
So it is just $\dfrac{-1+2\sqrt{3}+1}{2}$.
We may also start from the original equation.
\begin{align*}
x^2+x-3-\sqrt{3}&=0\\
x^2-3+x-\sqrt{3}&=0\\
(x-\sqrt{3})(x+\sqrt{3})+x-\sqrt{3}&=0\\
(x-\sqrt{3})(x+\sqrt{3}+1)&=0
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find the amplitude of a complex number
If
$$z = (1-\cos{\theta}) + i \sin{\theta},$$
then the amplitude of $z$ is:
$$a) \frac{\pi}{4}-\frac{\theta}{2}$$ $$b)
\frac{\pi}{2}-\frac{\theta}{2}$$ $$c) \frac{\pi}{2}-\theta$$ $$d)
\frac{\pi}{2}-\frac{\theta}{4}$$ where 0<$\theta$ <$\pi$.
I tried :
$x= 1-\cos{\theta}, y= ... | Your $z$ can be nicely written as
$$z=1-e^{-i \theta}$$
Now,
\begin{align}
|z| &= |1-e^{-i \theta}| \\
&=\bigg |e^{-i \frac{\theta}{2}}\left(e^{i \frac{i\theta}{2}} -e^{-i \frac{\theta}{2}} \right)\bigg | \\
&=\bigg |e^{-i \frac{i\theta}{2}} \left( \cos \left( \frac{\theta}{2}\right) + i \sin \left( \frac{\theta}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3242500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?
Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing aro... | $$2^a+2^b+2^c=2^c(2^{a-c}+2^{b-c}+1)$$
Since the second factor must be even, we see that $a>b=c$, and then it becomes
$$2^c(2^{a-c}+2)=2^{c+1}(2^{a-c-1}+1)$$
And now we see that $a=c+1$.
Now,
$$2^a+2^b+2^c=4\cdot2^c$$
so $c$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Find $ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$ How to compute the limit
$$ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$$
Does exist an explicit formula for finding the numerator?
| By the Stolz-Cesro Theorem , one has
\begin{eqnarray}
&&\lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}\\
&=&\lim_{k \rightarrow \infty } \frac{{(k+1)^{n + 1}}} {(k + 2)(k+1)^{n + 1}-(k + 1)k^{n + 1}}\\
&=&\lim_{k \rightarrow \infty } \frac{1} {(k + 2)-(k + 1)\bigg(\frac{k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3248353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find Maxima and Minima of $f( \theta) = a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta$ Show that, whatever the value of $\theta$, the expression
$$a \sin^2 \theta + b \sin \theta \cos \theta + c \cos^2 \theta\ $$
Lies between
$$\dfrac{a+c}{2} \pm \dfrac 12\sqrt{ b^2 + (a-c)^2} $$
My try:
The given expr... | By C-S we obtain:
$$a\sin^2\theta+b\sin\theta\cos\theta+c\cos^2\theta=a\cdot\frac{1-\cos2\theta}{2}+b\cdot\frac{\sin2\theta}{2}+c\cdot\frac{1+\cos2\theta}{2}=$$
$$=\frac{1}{2}\left(a+c+b\sin2\theta-(a-c)\cos2\theta\right)\leq\frac{1}{2}\left(a+c+\sqrt{(b^2+(-a+c)^2)\left(\sin^22\theta+\cos^22\theta\right)}\right)=$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Determining the coefficients of $(1 + x + x^2 +\cdots+x^n)^{n-1}$ Suppose we have the following polynomials:
$$f_1(x)=(1 + x + x^2)$$
$$f_2(x)=(1 + x + x^2 + x^3)^2$$
$$f_3(x)=(1 + x + x^2 + x^3 + x^4)^3$$
$$f_4(x)=(1 + x + x^2 + x^3 + x^4 + x^5)^4$$
$$\vdots$$
$$f_{n-1}(x)=(1 + x + x^2 + x^3 +x^4+ x^5+\cdots+x^n)^{n-1... | We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series.
We obtain for $0\leq k\leq n(n-1)$:
\begin{align*}
\color{blue}{[x^{k}]}&\color{blue}{\left(1+x+x^2+\cdots+x^n\right)^{n-1}}\\
&=[x^k]\left(\frac{1-x^{n+1}}{1-x}\right)^{n-1}\tag{1}\\
&=[x^k]\sum_{j=0}^\infty\binom{-(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\sum_{cyc}\frac{1}{x^2 + 1} \ge \frac{2}{3}\bigl(\sum_{cyc}\frac{x}{\sqrt{x^2 + 1}}\bigr)^3$ where $x, y, z > 0$ and $xy + yz + zx = 1$.
$x$, $y$ and $z$ are positives such that $xy + yz + zx = 1$. Prove that $$\large \frac{1}{x^2 + 1} + \frac{1}{y^2 + 1} + \frac{1}{z^2 + 1} \ge \frac{2}{3}\left(\frac{x}{\... | We need to prove that:
$$\sum_{cyc}\frac{1}{x^2+1}\geq\frac{2}{3}\left(\sum_{cyc}\frac{x}{\sqrt{x^2+1}}\right)^3$$ or
$$\sum_{cyc}xy\sum_{cyc}\frac{1}{(x+y)(x+z)}\geq\frac{2}{3}\left(\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\right)^3.$$
Now, by C-S
$$\sum_{cyc}xy\sum_{cyc}\frac{1}{(x+y)(x+z)}=\frac{\sum\limits_{cyc}x(y+z)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$ Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$
I proved it by induction:
"$n=1$" $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad \checkmark$
"$n \implies n+1$": $$\quad \sum_{k=0}^{n+1}{\frac{1}{k!}}\leq 3 \\ \left(... | $$\sum_{k=0}^{\infty}\frac{1}{k!} = 1 + 1 + \frac{1}{2!}+\frac{1}{3!}+ \cdots$$ $$ < 1+1+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5}+ \cdots$$ $$ = 1 + 1 +\left (1-\frac{1}{2}\right) + \left (\frac{1}{2}-\frac{1}{3}\right) + \left (\frac{1}{3}-\frac{1}{4}\right)+\cdots$$ $$ = 1+1 +1 =3.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
$\frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$ given that where $x, y, z > 0$ and $xyz = \frac{1}{2}$.
$x$, $y$ and $z$ are positives such that $xyz = \dfrac{1}{2}$. Prove that $$ \frac{xy}{z^2(x + y)} + \frac{yz}{x^2(y + z)} + \frac{zx}{y^2(z + x)} \ge xy + yz + zx$$
Before ... | Let $x=\frac{1}{a},$ $y=\frac{1}{b}$ and $z=\frac{1}{c}.$
Thus, $abc=2$ and by C-S we obtain:
$$\sum_{cyc}\frac{xy}{z^2(x+y)}=\sum_{cyc}\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}=xy+xz+yz.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find all ordered pairs $(x,y)$ that satisfy both $\frac{3x-4y}{xy} = -8$ and $\frac{2x+7y}{xy} = 43$ The following is listed under the "multiple variable" category of my Algebra I homework.
Find all ordered pairs $(x,y)$ that satisfy both $\frac{3x-4y}{xy} = -8$ and $\frac{2x+7y}{xy} = 43$
I can't wrap my head around... | $$\frac{3x-4y}{xy} = \frac{3}{y} - \frac{4}{x} = -8$$
$$\frac{2x+7y}{xy} = \frac{2}{y}+\frac{7}{x} = 43$$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$.
Then,
$3v - 4u = -8$ and $2v + 7u = 43$
Solve these linear equations and replace $x$ and $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the sum of 2 discrete random variables Let X and Y be independent random discrete variables and
$X = \begin{pmatrix}-1&0&1\\\frac13&\frac13&\frac13\end{pmatrix}$ and $Y = \begin{pmatrix}0&1\\\frac13&\frac23\end{pmatrix}$
Then what is $X + Y$? As a discrete random variable.. I understood that I have to add ... | Hint: Can you enumerate all the cases? For example, if $X+Y=1$ it means either $X=0$, $Y=1$ (probability $\frac{1}{3}\cdot\frac{2}{3}=\frac{2}{9}$) or $X=1$, $Y=0$ (probability $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$) $\Longrightarrow P\left(X+Y=1\right)=\frac{2}{9}+\frac{1}{9}=\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve system of congruences $k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$
Solve system of congruences
$$k^2 + l^2 \equiv 0 \mod 17 \wedge k^3+l^3 \equiv 0 \mod 17$$
Is there any faster way to solve that congurence than looking at table and finding such pairs of $i$ that in both cases it gives $0$? ... | From $k^3 + l^3 \equiv 0 \pmod{17}$, we have
$$k^3 \equiv (-l)^3 \pmod {17}.$$
But, as you can see from your table, cubes are distinct modulo $17$, which effectively lets us take cube roots: thus, $k \equiv -l \pmod {17}$. Therefore
$$k^2 + (-k)^2 \equiv 0 \pmod {17},$$
so $2k^2 \equiv 0 \pmod {17}$. Divide by $2$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find K such that the line is tangent to $y=-\frac{3}{4}x+4$ The function is $f(x)= \frac{k}{x}$ and the line is $y =-\frac{3}{4}x+3$
So the derivative of $f(x)$ should equal $-\frac{3}{4}$
If I take $k$ to be a constant I can remove it from the fraction like so
$$f'(x)=k \frac{d}{dx} \frac{1}{x}$$
to get $-\frac{k}{x^... | If the tangent occurs at $x = t$, we have $\displaystyle\frac{k}{t^2} = \frac{3}{4}$, as you've already found, but we also must have $\displaystyle\frac{k}{t} = -\frac{3}{4}t + 3$.
So we obtain $\displaystyle k = -\frac{3t^2}{4} + 3t$, and subbing into the first equation gives $\displaystyle -\frac{3}{4} + \frac{3}{t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2... | There is also the following: $$\tan\frac{\pi}{19}+4\sin\frac{7\pi}{19}+4\sin\frac{8\pi}{19}-4\sin\frac{6\pi}{19}=\sqrt{19},$$
$$-\tan\frac{3\pi}{19}+4\sin\frac{\pi}{19}+4\sin\frac{2\pi}{19}+4\sin\frac{5\pi}{19}=\sqrt{19},$$
$$\tan\frac{5\pi}{19}+4\sin\frac{2\pi}{19}-4\sin\frac{3\pi}{19}-4\sin\frac{8\pi}{19}=\sqrt{19},$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}$ How to prove that
$$\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}=\frac{7\pi}{16}\zeta(3)+\frac{\pi^3}{16}\ln2+\frac{\pi^4}{32}-\frac1{256}\psi^{(3)}\left(\frac14\right)$$
where $H_n=1+\frac1{2}+\frac1{3}+...+\frac1{n}$ is the $n$th harmonic number.
This s... | \begin{align}
\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}&=\sum_{n=1}^\infty(-1)^{n-1} H_n\int_0^1\frac12x^{2n}\ln^2 x\ dx\\
&=-\frac12\int_0^1\ln^2x\sum_{n=1}^\infty(-x^2)H_n\\
&=\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\tag{1}
\end{align}
\begin{align}
I&=\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\\
&=\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to get the value of the root? I have this statement:
If $\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3}$, Which of
the following values are the closest to $\sqrt{21}$ ?
A) 68/15 B) 14/3 C) 19/4 D) 55/12 E) 9/2
My development was:
$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} = \frac{\sqrt{35}-\sqrt{21}}{2} \appr... | Rearrange:
$$\frac{\sqrt{7}}{\sqrt{5}+\sqrt{3}} \approx \frac{2}{3} \iff \frac{7}{\sqrt{35}+\sqrt{21}} \approx \frac{2}{3} \Rightarrow \sqrt{21}\approx \frac{21}{2}-\sqrt{35}$$
Use linear approximation to find $f(35)=\sqrt{35}$:
$$f(x_0+\Delta x)\approx f(x_0)+f'(x_0)\cdot \Delta x \ ;\\
f(36-1)\approx f(36)+\frac1{2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.
I have provided a (dumbfounding) solution down be... | If $5\mid x^2-2xy-y$ and $5\mid xy-2y^2-x$ then $5$ also divides
$$(3y-1)(x^2-2xy-y)-(3x+2)(xy-2y^2-x)=2x^2+y^2+2x+y.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Arithmetic sequences. Writing the general term differently So I'm dealing with this problem that has to do with arithmetic sequences and I was wondering if I could write the general term of this kind of sequences (finite) like $$a_{k-1}=a_1+(k-\lfloor \frac{n}2 \rfloor)d$$ where $k$ can take values from $2$ to $n+1$
$... | If you really want to, you can, but $n$ isn't a free parameter. If you want the formula $$ a_{k-1} = a_1 + (k-\lfloor \frac{n}{2}\rfloor)d $$
to hold for $k=2$, you get a condition
$$ a_1 = a_1 + (2-\lfloor \frac{n}{2}\rfloor)d$$
that means that for $d\neq 0$ you'll need
$$\lfloor \frac{n}{2}\rfloor = 2$$
So, for natu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What am I doing wrong solving $\sqrt{x^2+1}- 2x+1>0$? First, I started looking at where $\sqrt{x^2+1}$ is defined: $\sqrt{x^2+1}>0$ is defined everywhere. Next I
$\sqrt{x^2+1}- 2x+1>0$
$\sqrt{x^2+1}> 2x-1$
$0> 3x^2-4x$
$0> x(3x-4)$ and I solve this for $x\in (0,\frac{4}{3}]$.
I know that this solution is wrong, becaus... | If $A,B\in \Bbb R$,$$\sqrt A>B\iff \begin{cases}A\ge 0\\ B<0\end{cases}\lor\begin{cases}A\ge 0\\ B\ge 0\\ A>B^2\end{cases}$$ In this instance $A=x^2+1$, $B=2x-1$ $$\begin{cases}x^2+1\ge 0\\ 2x-1<0\end{cases}\lor\begin{cases}x^2+1\ge 0\\ 2x-1\ge 0\\ x^2+1>4x^2-4x+1\end{cases}\iff x<\frac12\lor\begin{cases}x\ge \frac12\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$ In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.
This is my attempt:
$$ \begin{align} \frac{1}{(n+1)^... | Multiply through by $n(n+1)^2$ (a positive number since $n \ge 1$ ) to get $$n+n(n+1) < (n+1)^2$$
$$n < (n+1)^2 - n(n+1)$$
$$n < (n+1)(n+1-n)$$
$$n < n+1$$
Which is true of course for all $n$; in particular $n\ge 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
sum of series $\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $
The sum of series
$$\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $$
My attempt $$\displaystyle \sum^{n}_{k=1}\frac{1}{(3k-2)(2k+1)}=\frac{1}{7}\sum^{n}_{k=1}\bigg[\frac{3}{3k-2}-... | (CONSIDERING INFINITE SUM)
We have $$I_n=\frac 37\int_0^1 \frac {1-x^{3n}}{1-x^3} dx-\frac 27\int_0^1 \left(\frac {1-x^{2n+2}}{1-x^2} +\frac {x^2-1}{1-x^2}\right) dx$$
Let the first integral be denoted by $I_{1,n}$ and the second be denoted by $I_{2,n}$
Hence for the sum we need $$\lim_{n\to\infty} \frac 37 I_{1,n}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the n... | $f(x)\equiv{x+3}\pmod{(x-1)^2}\wedge f(x)\equiv{2x+4}\pmod{x^2} \overset{CRT}{\implies} f(x)\equiv{3x^3 - 5x^2 + 2x + 4}\pmod{(x-1)^2x^2}$
? chinese(Mod(x+3,(x-1)^2),Mod(2*x+4,x^2))
%1 = Mod(3*x^3 - 5*x^2 + 2*x + 4, x^4 - 2*x^3 + x^2)
Then $f(x)\equiv{-2x^2 + 2x + 4}\pmod{(x-1)x^2}$
? Mod(3*x^3 - 5*x^2 + 2*x + 4,(x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Fresnel integral $\int\limits_0^\infty\sin(x^2) dx$ calculation I'm trying to calculate the improper Fresnel integral $\int\limits_0^\infty\sin(x^2)dx$ calculation.
It uses several substitutions. There's one substitution that is not clear for me.
I could not understand how to get the right side from the left one. What ... | Well, if one puts $v=\frac{1}{u}$ then:
$$I=\int_0^\infty\frac{v^2}{1+v^4} dv =\int_0^\infty\frac{1}{1+u^4} du$$
So summing up the two integrals from above gives:
$$2I=\int_0^\infty\frac{1+u^2}{1+u^4} du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ . Problem. Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove:
$$a^{3}b+ b^{3}c+ c^{3}a\leqq 8$$
My solution in M&Y : (and I'm looking forward to s... | Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Thus, since $x+y+z=3$ and $xy+xz+yz=2,$ we obtain:
$$x+z=3-y,$$
$$xz=2-y(x+z)=2-y(3-y)=y^2-3y+2,$$
which gives $0\leq y\leq1$ because $y\geq2$ is impossible.
Thus, by Rearrangement we obtain:
$$a^3b+b^3c+c^3a=a^2\cdot ab+b^2\cdot bc+c^2\cdot ca\leq x^2\cdot xy+y^2\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$ Show that $$ \lim_{n\to\infty}[\frac{1}{\sqrt n}+\frac{1}{\sqrt {n+1}}+\frac{1}{\sqrt {n+2}}.......\frac{1}{\sqrt {2n}}] = \infty$$
LHS : $ \lim_{n\to\infty}\frac{1}{n}[\frac{n}{\sqrt n}+\fr... | Here is an
elementary proof that
$\dfrac{\sqrt{2}}{3n}
\lt \dfrac{s_n}{\sqrt{n}}-2(\sqrt{2}-1)
\lt \dfrac{2\sqrt{2}}{n}
$.
$\begin{array}\\
\sqrt{m+1}-\sqrt{m}
&=(\sqrt{m+1}-\sqrt{m})\dfrac{\sqrt{m+1}+\sqrt{m}}{\sqrt{m+1}+\sqrt{m}}\\
&=\dfrac{1}{\sqrt{m+1}+\sqrt{m}}\\
&>\dfrac{1}{2\sqrt{m+1}}\\
\text{and}\\
\sqrt{m+1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
AM-GM inequality in $3$ variables I am trying to prove AM-GM inequality in $3$ variables.
Prove that $a^3 + b^3 + c^3 \ge 3abc$ for all $a,b,c \in \mathbb{R}^+$.
Could you please verify if my attempt contains logical gaps/errors?
My attempt:
Lemma: $a^2 + b^2 \ge 2ab$ for all $a,b \in \mathbb{R}^+$.
It follows from ... | Hint: Better is to use that $$a^3+b^3+c^3-3abc= \left( a+b+c \right) \left( {a}^{2}-ab-ac+{b}^{2}-bc+{c}^{2}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
General solution of $\tan (2x)\tan (x)=1$ For the question, $\tan (2x) \tan x=1$, I divided it by $\tan x$, and got the solution as $\frac{(2n+1)\pi}{6}$.
$\tan 2x= \cot x= \tan\left(\frac{\pi}{2}-x\right)$. So, $2x=n\pi+ \frac{\pi}{2}-x$. So, $3x= \frac{(2n+1)\pi}{2}$
But the book solved using the formula of $\tan (2... | Your way is better, I think:
The domain gives $\cos2x\neq0$,$\cos{x}\neq0$ and $\sin{x}\neq0$ and by your idea we obtain:
$$\tan2x=\cot{x}$$ or
$$\frac{\sin2x\sin{x}-\cos2x\cos{x}}{\cos2x\sin{x}}=0$$ or
$$\cos3x=0$$ or
$$x=\frac{\pi}{6}+\frac{\pi}{3}k,$$ where $k\in\mathbb Z$.
Now, delete from these roots, roots of $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3276826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$ Show that $3^{7^n}+5^{7^n}\equiv 1 \pmod{7^{n+1}}$
I tried to induct on n:
For $n = 0$ we have $3+5 = 8$ and $8 \equiv 1 \pmod{7^{n+1}}$.
Suppose it is true for $n = k$:
$$3^{7^k}+5^{7^k}\equiv 1 \pmod{7^{k+1}}$$
so $3^{7^k}+5^{7^k}=7^{k+1}*q_1+1$
For $n = k+1$:
$$3^{... | Lemma: Suppose $n \geq 0$. Then ${7^n\choose k}7^k$ is divisible by $7^{n+1}$ for all $1 \leq k \leq 7^n$.
Proof: Legendre's formula says that the largest power of $7$ dividing $m!$ is $\sum\limits_{i=0}^\infty\left\lfloor\frac{m}{7^i}\right\rfloor$ (this is really a finite sum). Now use the formula ${m\choose k} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$ I have to simplify the following expression:
$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$
Answer: $\sqrt{a+b}-\sqrt{a-b}$
I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)
| 1) You have $\sqrt{a + b}$ in the denominator before simplification, so you can not have $\sqrt{a+ b} = 0$. So $a+b \ne 0$ and $a \ne -b$.
2) You have $\sqrt{a+b}$ before simplification so $a+b \ge 0$ and by 1) you have $a+b \ne 0$ so $a+ b > 0$ or $a > -b$.
3) You have $\sqrt{a^2 - b^2}$ before simplification so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solve $\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx$ using $\sinh x$
How to solve definite Integral
$$\int_0^\infty \frac{1}{\sqrt{x^2+1}(x^2+x+1)}dx=\frac{\ln(2+\sqrt{3})}{\sqrt{3}}$$
with hyperbolic function $\sinh(x)$?
Below is using $\tan x$ to do it. I thought it is same with this because $\sinh^2(x)+1=\cosh^... | Hint: Substituting $$\sqrt{x^2+1}=x+t$$ we get $$x=\frac{1-t^2}{2t}$$ so $$dx=-\frac{t^2+1}{2 t^2}dt$$
and $$\sqrt{x^2+1}=\frac{1+t^2}{2t}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove Pascal's theorem by homogeneous coordinates I was trying to prove Pascal's theorem by using homogeneous coordinates with the following configurations (interactive graph at Desmos):
*
*A,B,C,D,E,F (homogeneous coordinates) are on a conic.
*G,H,I are the intersection points of pairs of lines (AB,DE), (CD,FA), ... | Define $$\phi: X \longmapsto
\begin{vmatrix} X \\ D\\E\end{vmatrix}
\begin{vmatrix} C \\ F\\X\end{vmatrix}
\begin{vmatrix} E \\ B \\C\end{vmatrix}
\begin{vmatrix} B \\ D \\F\end{vmatrix}-\begin{vmatrix} B \\ D \\E\end{vmatrix}
\begin{vmatrix} D \\ F \\X\end{vmatrix}
\begin{vmatrix} F \\ B \\C\end{vmatrix}
\begin{vmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Other methods for a limit I know that I can compute the limit
$$
\lim_{x\to1}\frac{Nx^{N+1}-(N+1)x^N+1}{(x-1)^2}=\frac{N(N+1)}{2}
$$
using L'Hospital's rule (not one but two times) but I am looking for other ways. Are there any of them?
p.s.: the limit follows from a shortcut used in order to find the value of $$
\sum_... | $$Nx^{N+1}-(N+1)x^N+1 = Nx^N(x-1)-(x^N-1)$$
$$ = Nx^N(x-1)-(x-1)(x^{N-1}+...x^2+x+1)$$
$$ = (x-1)\Big(\underbrace{Nx^N-(x^{N-1}+...x^2+x+1)}_{p(x)}\Big)$$
Now $$p(x) = (x^N-x^{N-1})+...+\color{red}{(x^N-x^2)}+\color{blue}{(x^N-x)}+\color{green}{(x^N-1)}$$
$$= x^{N-1}(x-1)+...+\color{red}{ x^2(x^{N-2}-1)}+\color{blue}{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show That the Series $\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}$ Converges to a Number Between $\frac{1}{2}$ and $1$ $a_n = \frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}$ converges to a number between $\frac{1}{2}$ and $1$.
I know that it converges to Ln(2) but we still haven't learned about integrals in my c... | Hint 1:
$$a_{n+1}-a_n=\left( \frac{1}{n+1}+ \frac{1}{n+1} +\dots + \frac{1}{2n+2} \right)-\left(\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n}\right)\\ = \frac{1}{2n+1}+\frac{1}{2n+2}- \frac{1}{n}<\frac{1}{2n}+\frac{1}{2n}- \frac{1}{n}=0$$
Hint 2:
$$a_n =\frac{1}{n}+ \frac{1}{n+1} +\dots + \frac{1}{2n} \geq \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Study the sign of the quadratic matrix. Let $Q$ be the quadratic form associated with the matrix :
$$
\begin{pmatrix}
2 & 1 & 1 \\
1 & k & 0 \\
1 & 0 & 1 \\
\end{pmatrix}
$$
I reduced the matrix to the row echelon form and obtain :
$$
\begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & -1 \\
0 & 0 ... | Careful, eigenvalues of row echelon form are in general not equal to the eigenvalues of the original matrix. In this case the eigenvalues are not easy to calculate so we can directly consider the form $Q$ itself.
We have
$$Q(x,y,z) = \left\langle \begin{pmatrix}
2 & 1 & 1 \\
1 & k & 0 \\
1 & 0 & 1 \\
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3287807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On the Sum of Some of the Factors of $2310$
Given 3 natural numbers a,b,c such that $a\times b\times c=2310$. Find the sum of $\sum a+b+c$
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
So now since $a,b,c$ are symmetric , their sum should be the same , and I should get , $\sum a+b+c= ... | Silly overkill method:
For any (positive) natural number $N$, set
$$d_N=\sum_{abc=N}(a+b+c)\text{.}$$
Consider the Dirichlet series
$$f(s)=\sum_{N}\frac{d_N}{N^s}\text{.}$$
Then
$$\begin{split}f(s)&=\sum_{a,b,c}\frac{a+b+c}{a^sb^sc^s}\\
&=3\zeta(s-1)\zeta(s)^2\end{split}$$
so that $f(s)$ admits the Euler product
$$\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rational points on $2x^2+2y^2=1$ and integral solutions to $2X^2+2Y^2=Z^2$
(a) Find a solution to the diophantine equation $2X^2+2Y^2=Z^2$; hence find a solution for rational numbers of the form $2x^2+2y^2=1$.
We have
$2X^2+2Y^2 \equiv Z^2\pmod{2} \implies (X,Y,Z) =c(1,1,2), ~c \in\mathbb{Z}$ is a solution. And $(1... | You have found one rational solution $(x,y)=(\frac12,\frac12)$ to
$$
2|x+iy|^2=1
$$
This you can now use to transform any other solution as in
$$
2=|1-i|^2\implies |(1-i)(x+iy)|^2=1,
$$
that is, $(x+y, y-x)$ is a rational point on the unit circle. Now these are all parametrized by Pythagorean triples $a+ib=(p+iq)^2$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Logarithmic integral $\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$
Evaluate the integral in a closed-form :
$$I=\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$$
My attempt :
After put $x=\tan y$ we obtain:
$$I=\int_0^{\frac{π}{4}}\ln (\tan x)\ln (1-\tan^{2} x)dx$$
$$1-\tan^{2} x=(1+\tan x)(1-\tan x... | Presented below is a self-contained evaluation. Let $J=\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx$
\begin{align}
&\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx
\overset{x\to\frac{1-x}{1+x}}= J
-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx-2G\ln2+\frac{\pi^3}{16}\\
&\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx
\overset{(0,1)+\overset{x\to 1/x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$. I am getting x = 4 but the book gets x = 81. What am I doing wrong?
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}... | The mistake:
It should be $$2\log_3x=8$$ or
$$\log_3x=4$$ or
$$\log_3x=\log_3{3^4},$$
which gives $x=81$.
Actually, in the first step you can use the following property.
$$\log_{a^{\beta}}x=\frac{1}{\beta}\log_ax,$$ where $a>0$, $a\neq1$, $x>0$ and $\beta\neq0$.
Since $\frac{1}{3}=3^{-1},$ we obtain $$\log_3x=-\log_3x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
What is the remainder of polynomial $f(x) g(x)$ divided by $x^{2} + x - 6$? $F(x),g(x)$ be polynomials. $f(x)$ divided by $x^{2}-4$ has remainder $ax+a$. $g(x)$ divided by $x^{2}-9$ has remainder $ax+a-5$. It is known that the remainder
$$ Remainder \: f(x) \: divided \: by \: x+2 = Remainder \: g(x) \: divided \: by ... | Since you're dividing by a quadratic polynomial, the remainder has to be a linear polynomial, because its degree must be strictly less than that of the divisor. So you can write your remainder as $R(x)=px+q$. Knowing that $R(2)=-6$ and $R(-3)=14$, you can set up and solve a system of two equations with two unknowns $p$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimizing distance between an ellipse and a point Problem
An ellipse has the formula:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
What is the shortest distance from the ellipse to the point $P = (a,0)?$
Attempted solution:
I have previously solved the same problem with P = (1,0) and P = (2,0). For P = (1,0), the minimum whe... | The key is to remember $-3\leq x\leq 3.$
It is much easier, without even using calculus, to minimize $D^2$ than the minimize $D.$
We have that the distance $D$ satisfying:
$$\begin{align}D^2&=(x-a)^2+y^2\\
&=x^2-2ax+a^2+4\left(1-\frac{x^2}{9}\right)\\
&=\frac{5}{9}x^2-2ax+a^2+4\\
&=\frac{5}{9}\left(x-\frac{9}{5}a\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to determine the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$ I have a question
Find the coefficient of $x^{10}$ in the expansion $(1+x+x^2+x^3+.....+x^{10})^4$
There ARE questions like this on stack exchange already I know, but I'm not able to formulate a pattern or know how to apply t... | The standard geometric series tells us that
$$1 +x + x^2 + x^3 + \ldots +x^{10}= \frac{1-x^{11}}{1-x}$$
So taking this to the fourth power we get
$$\left( \frac{1-x^{11}}{1-x} \right)^4 = (1-x^{11})^4 (1-x)^{-4}\tag{1}$$
This would not seem to really help much, except that for negative powers of $n$ there
is a general... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find all pairs $(k, n)$ of positive integers such that $k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$ Find all pairs $(k, n)$ of positive integers such that
$$k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$$
I tried to solve this problem but only found one solution $(1,1)$. Please help me to solve thi... | Find the exponent of the largest power of 2 that divides both sides. In RHS it is $0 + 1 + \ldots + (n-1) = \frac{(n-1)n}{2}$. In LHS it can be found with Legendre's formula, which gives $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor$. Since $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor < \sum_{i=1}^{\infty} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?
| $x-\dfrac {\sqrt {8x}}{x}=9,\Rightarrow x^{2}-\sqrt {8x}=9x$
$\begin{aligned}x^{2}-9x=\sqrt {8x}\\ x^{2}-8x=\sqrt {8x}+x\\ \left( x-\sqrt {8x}\right) \left( x+\sqrt {8x}\right) =\sqrt {8x}+x\\ x -\sqrt {8x}=1\end{aligned}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving non-trivial integer solutions don't exist for $z^2=10x^2-5y^2$ Given $z^2=10x^2-5y^2$.
I am wanting to show this equation has integer solutions only $0=x=y=z$.
I attempted by considering modulo $2$, however this didn't give me a contradiction. Any help or hints would be appreciated.
Solution:
Consider modulo... | Outline of proof:
*
*If there is a non-trivial solution, there is a solution with $x$ odd.
*There are no solutions when $x$ odd.
We'll look entirely modulo $8.$
First note that if $x,y,z$ are solutions with $x=0$, then $y=0,z=0.$
So if there is any solution $(x,y,z)\neq (0,0,0),$ there must be one with $x>0.$ Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{2... | Your idea is correct.
You need to redo your calculation of intersects.
You have the point $(\frac {6}{5}, \frac {6}{5})$ on the line $y=\frac {4}{3} x$ which does not make sense.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Variance of a Fair Coin Consider Vamshi decides to toss a fair coin repeatedly until he gets a tail. He makes atmost $4$ tosses.The value of variance $T$ is ($T$ denoted number of tosses) ______
I tried like this Standard Deviation on marks of students
$x-----1 ----- 2 ------3-----4$
$P(x)----\frac{1}{2}-----\frac{1}{... | $$P(X=4)=1-P(X\le3)=\frac18\ne\frac1{16}$$ It is clear that one of the probabilities must be wrong as you currently don't have $\sum_{\forall x} P(X=x)=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding maximum of a given function
Show that $f(x)=\sin x(1+\cos x)$ attains its maximum at $x= \pi/3$.
I differentiated the function $f$ and got $f'(x)=\cos x(1+2\cos x)$. After equating with $0$, I got $x=\pi/2$ and $x =\pi/3 + n\pi$ with $n\neq 0$. So I did not get $x= \pi/3$ even as an extreme value.
| Probably your differentiation is not correct,
$$f(x)=\sin x(1+\cos x)$$
$$f'(x)=\sin x\frac{d}{dx}(1+\cos x)+(1+\cos x)\frac{d}{dx}\sin x$$
$$=\sin x(-\sin x)+(1+\cos x)\cos x$$
$$=-\sin^2 x+\cos x+\cos^2 x$$
$$=2\cos ^2x+\cos x-1$$
$$=(2\cos x-1)(\cos x+1)$$
$$\implies f''(x)=-4\sin x\cos x-\sin-1$$
for maximum value,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A lucky proof for the Basel problem. I'll modify this part since I want the proof to be here.
$$\sum_{n=1}^\infty \frac{1}{n^2}=\frac43\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=-\frac43\sum_{n=0}^\infty \int_0^1 x^{2n}\ln x dx=\frac43\int_0^1 \frac{\ln x}{x^2-1}dx$$
$$\int_0^1 \frac{\ln x}{x^2-1}dx\overset{x\rightarrow \fra... | Here is an 'elementary' proof of its validity. Note that
$$ \frac{c+1}{(x+c)(x-1)}= \frac{c-1}{(x+c)(x+1)} +\frac{2}{x^2-1} $$
Then
$$ \int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx = \frac{c-1}{c+1}I_1 + \frac{2}{c+1}I_2 \tag{1}$$
where
$$ I_1 =\int_0^\infty \frac{\ln x }{(x+c)(x+1)}
\overset{t=c/x}{dx}=\int_0^\infty \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
} |
Order of elements in $\Bbb Z_{24}$ What elements of $\Bbb Z_{24}$ are order $2$? Order $3$? Order $4$? Order $6$?
Order $2: 12$
Order $3: 8,16$
Order $4: 6,18$
Order $6: 4$
I had listed element $20$ as order $6$, but erased it late at night. A week later, I can't for the life of me remember why I didn't include it.
Am ... | You can brute force calculate:
\begin{align*}
20 \cdot 1 \equiv 20 \mod 24 \\
20 \cdot 2 \equiv 16 \mod 24 \\
20 \cdot 3 \equiv 12 \mod 24 \\
20 \cdot 4 \equiv 8 \mod 24 \\
20 \cdot 5 \equiv 4 \mod 24 \\
20 \cdot 6 \equiv 0 \mod 24 \\
\end{align*}
So the first (positive) multiple of $20$ that produces the group identit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac... | I think the most straightforward approach is computing $a$ in terms of $b$.
$3^a=4^b$ so $a = \log_34^b=b\log_34$
$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 9^{ \log_34}+16^{\log_43}=3^{ 2\log_34}+4^{2\log_43}=16+9=25$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Splitting field in math software. Can anyone help me to find a method for calculating the splitting field for a polynomial over a function field?
I think this feature is not currently supported..
| You can find the splitting field with the Isprime and PrimaryDecomposition function
P<x,y,z> := PolynomialRing(Rationals(),3);
I := ideal<P| x^2+y^3+y, z^3+z*(x^3+y)+1>;
IsPrime(I); // true
R := quo<P|I>; x:= R.1; y := R.2; z := R.3; Q<w> := PolynomialRing(R,1);
f := w^3+w*(x^3+y)+1; g:= f ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solutions of bivariate cubic Diophantine equation The problem is what are the integer solutions to
$$7x^2 - 40xy + 7y^2 = (|(x - y)| + 2)^3$$
I only got $x - y = 4 (\mod 13)$. Also, how to solve it?
| Let $(x,y)$ be an integral solution to the equation. Both sides of the equation are invariant under the transformation $(x,y)\ \mapsto\ (-x,-y)$, so without loss of generality $x\geq y$ and hence
\begin{eqnarray*}
7x^2 - 40xy + 7y^2&=& (|(x - y)| + 2)^3=(x-y+2)^3.
\end{eqnarray*}
Expanding the right hand side and bit ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Tips on solving $a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$ $$a^2 (a + 2 b) (a + b^2 + b)^{\frac{3}{2}}>b^3 (a^2 + a + b)^{\frac{3}{2}}$$
This is true with $a>b>0$, according to Wolfram Alpha, but I am not able to prove this.
I to simplify using the fact that $a^2(a+2b)=a^3+2ba^2>3b^3$,... | We prove the stronger result that comes from replacing $a+2b$ on the left by $a$
Note that for $0 \lt b \lt a, \frac 1b \gt \frac 1a$ and $\frac a{b^2} \gt \frac b{a^2}$ Then
$$1+\frac 1b+\frac a{b^2} \gt 1+\frac 1a+\frac b{a^2}\\
\sqrt{1+\frac 1b+\frac a{b^2}} \gt \sqrt{1+\frac 1a+\frac b{a^2}}\\
ab\sqrt{1+\frac 1b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find value of $k$? I have a question. How can I find value of $k$?
The given below
$$x + 4$$ is a factor of $$x^4+kx^3-4x^2$$
What is value of $k$?
Thanks for the solutions.
| Two ways.
1) Hard but straightforward way:
$x^4 + kx^3 -4x^2 = (x+4)(x^3 + ax^2 + bx + c)$ so
$x^4 + kx^3 - 4x^2 = x^4 + (a+4)x^3 + (b+4a)x^2 + (c + 4b)x + 4c$
So $4c = 0; c+4b=0; b+4a=-4; a+4 =k$
So $c = 0$ and $b = 0$ and $a =-1$ and $-1 +4 = 3 = k$.
$x^4 + 3x^3 -4x^2 = (x+4)(x^3 - x^2)$
2) Very easy way
If $x + 4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3313805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Max of $(a-x^2)(b-y^2)(c-z^2)$ when $x+y+z=a+b+c=1$, $x,y,z,a,b,c \geq 0$ What's the maximum value of
$$(a-x^2)(b-y^2)(c-z^2)$$
given $x+y+z=a+b+c=1$, $x,y,z,a,b,c \geq 0$
The tricky part, as you could see in one of the attempted answer, is how to handle the case when two of the 3 factors are negative
| Partial solution
It is enough to consider the case where the product is positive.
1) If there are two negatives and one positive amongst $a-x^2, b-y^2, c-z^2$: I'm working on this.
2) If all $a-x^2, b-y^2, c-z^2 \geq 0$, then by applying AM-GM, we have
$$(a-x^2)(b-y^2)(c-z^2) \leq \Big( \frac{a+b+c-(x^2+y^2+z^2)}{3} \B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute 100th derivative A friend suggested me a rather tricky problem, namely find the $100^{th}$ derivative of
$$
f(x)=\frac{x^2+1}{x^3-x}.
$$
I have computed the zeroth derivative
$$
\frac{x^2+1}{x^3-x}
$$
and the first derivative
$$
\frac{2x(x^3-x)-(3x^2-1)(x^2+1)}{(x^3-x)^2}=\frac{1-x^4-4x^2}{(x^3-x)^2}
$$
but I ... | Write your function as
$$
\frac{x^2 + 1}{x^3 - x} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{n\rightarrow \infty}\int_0^\infty \frac{x^{n-2}\cos(n\pi x)}{1+x^n} dx$ I feel it goes to zero I separate the integral $\int_0^1 \frac{x^{n-2}\cos(n\pi x)}{1+x^n}$ the sequence goes to zero and it bounded by $0.5$ so by bounded convergence theorem the limit is zero. Now for the $\int_1^\infty \frac{x^{n-2}\... | Hint: If $x \geq 1$ then
$$ \frac{x^{n-2}}{1+x^n} \leq \frac{1}{x^2}$$
and if $0\leq x \leq 1$ then
$$ \frac{x^{n-2}}{1+x^n} \leq x^{n-2} \leq 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof by induction - stuck on simple question! Question:
(Part 1) Show that the inequality
$$
\frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \ldots + \frac{1}{(n-1)\cdot n} < \frac{1}{2}
$$
works for all natural numbers $n > 2$.
(Part 2) Deduce that for all natural numbers $n$, the following inequality holds:
$$
\frac{1}{... | $\require{cancel}$No need for induction here. Just note that\begin{align}\frac1{2\times3}+\frac1{3\times3}+\cdots+\frac1{(n-1)n}&=\frac12-\cancel{\frac13}+\cancel{\frac13}+\cdots-\cancel{\frac1{n-1}}+\cancel{\frac1{n-1}}-\frac1n\\&=\frac12-\frac1n\\&<\frac12.\end{align}And then\begin{align}\frac1{1^2}+\frac1{2^2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Separate equations of the pair of lines $x^2+2xy\sec\theta+y^2=0$
Find the separate equations of the lines $x^2+2xy\sec\theta+y^2=0$
Attempt 1
$$
x^2+2xy\sec\theta+y^2=0\\
\frac{x^2}{y^2}+2\frac{x}{y}\sec\theta+1=0\\
\frac{x}{y}=\frac{-2\sec\theta\pm\sqrt{4\sec^2\theta-4}}{2}=-\sec\theta\pm\tan\theta=\frac{-1\pm\sin\... | \begin{align}
\dfrac{-1 \pm \sin \theta}{\cos \theta}
&= \dfrac{-1 \pm \sin \theta}{\cos \theta} \cdot
\dfrac{-1 \mp \sin \theta}{-1 \mp \sin \theta} \\
&= \dfrac{1-\sin^2 \theta}{\cos \theta (-1 \mp \sin \theta)} \\
&= \dfrac{\cos^2 \theta}{\cos \theta (-1 \mp \sin \theta)} \\
&= \dfrac{\cos \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3319175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $y$ in $\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$ I would like to solve this equation for $y$. Any tips? It seems like you cant really do it analytically?
$$\sqrt{4+(y-6)^2}+\sqrt{16+(y-3)^2}=3\cdot\sqrt{5}$$
| Set $y-3=z,$ then square both sides and simplify to get $$z^2-3z-8=-\sqrt{4+(z-3)^2}\sqrt{16+z^2}.$$ Squaring again and simplifying the right hand side gives $$(z^2-3z-8)^2=(z^2-6z+13)(16+z^2).$$ Now let $16+z^2=w,$ then expand and simplify to get $$z^2+16z+64=5w.$$ Substituting back for $w,$ simplifying and factoring ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$
*
*Show that the sequence $\left\{\frac{n^2}{9^n}\right\}_{n=1}^{\infty}$ is monotone decreasing and bounded below.
Let $a_n = \frac{n^2}{9^n}$. For $n\geq 1$ we have
\begin{equation*}
a_n-a_{n+1} = \frac{n^2}{9^n}-\frac{(n+1)^2}{9^{n+1}}.
\end{equation*}
T... | *
*First note that $a_n > 0$ for all $n.$ Therefore $$a_{n+1} \leq a_n \iff \frac{a_{n+1}}{a_n} \leq 1$$ and in our case, we have $$\frac{a_{n+1}}{a_n} = \frac{1}{9}\left(1 + \frac{1}{n}\right)^2 \leq \frac{1}{9}\left(1 + \frac{1}{1}\right)^2 = \frac{4}{9} < 1$$
*We know that the limit exists, so write $L = \display... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove inequality for a sum to $2^{n+1}$ Problem:I realize I made a mistake on one of the last questions I posted.Here is the problem:
Prove for all $n \in \mathbb{N}$
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$
Attempt:
I'm having trouble recognizing the pattern for the general term of the inequality, can s... | $\int_k^{k+1} \frac 1 t dt <\frac 1 k$. Summing over $k$ we see that the given sum exceeds $\int_1^{2^{n+1}+1} \frac 1 t dt=\ln (2^{n+1}+1) >ln (2^{n+1})=(n+1)ln 2>nln 2 >\frac n 2$ since $e <4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Verify the following limit using epsilon-delta definition: $ \lim_{(x,y)\to(0,0)}\frac{x^2y^2}{x^2+y^2}=0$ Show that $$ \lim\limits_{(x,y)\to(0,0)}\dfrac{x^2y^2}{x^2+y^2}=0$$
My try:
We know that, $$ x^2\leq x^2+y^2 \implies x^2y^2\leq (x^2+y^2)y^2 \implies x^2y^2\leq (x^2+y^2)^2$$
Then, $$\dfrac{x^2y^2}{x^2+y^2}\leq x... | Or alternatively, by AM-GM we get
$${x^2+y^2}\geq 2|xy|$$ so
$$\frac{x^2y^2}{x^2+y^2}\le \frac{x^2y^2}{2|xy|}=\frac{1}{2}|xy|$$ and this tends to zero if $x,y$ tend to zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
In simplifying the formula that I've derived for finding the square root of a complex number to the standard formula. So by easy means, I derived
$\sqrt{a+ib} = \sqrt{\frac{a(a+1)+b^2}{2}}+i(\frac{b}{\sqrt{2}\sqrt{a(a+1)+b^2}})$
But then I checked for the actual formula it is this;
$\sqrt{a+ib} = \sqrt{\frac{\sqrt{a^2... | One of the approache is use Euler form if Z is a complex numbers such that $ Z\, = \,r.e^{ix}$ then
Square root of Z will be $Z_{1}= \, \sqrt{r} . e^{iy} $
Where angle y will be x/2 and (x + 2π)/2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\frac {\cos 81^{\circ}}{\sin3^{\circ}\sin57^{\circ}\sin63^{\circ}} $
Find $$\frac {\cos 81^{\circ}}{\sin3^{\circ}\sin57^{\circ}\sin63^{\circ}} $$
This expression seemed quite easy to solve, but now the $63$ and $57$ in the equation are posing me with difficulties. I tried multiplying by $\cos9$ but I was ... | Let $\alpha = 60^\circ$ and $\theta = 3^\circ$. The numerator is
$$\begin{align}
\cos(81^\circ) = \sin(9^\circ) &= \sin(3\theta)\\
&= \sin\theta\cos(2\theta) + \cos\theta\sin(2\theta)\\
&= \sin\theta(\cos(2\theta) + 2\cos^2\theta)\\
&= \sin\theta(2\cos(2\theta) + 1)\end{align}$$
while the denominator equals to
$$\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means ... | This can be solved as a tangency problem.
Given the non degenerated conic $C(x,y) = 0$ determine the value for $\lambda$ such that $a x + b y = \lambda$ is tangent to $C(x,y)=0$ so we proceed as follows:
first we substitute $y = \frac 1b(\lambda-a x)$ into $C(x,y) = 0$ giving the relationship
$$
C\left(x, \frac 1b(\lam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Solve System of equation using elimination? \begin{align}
I:&& ~~ x+\frac12y &= 6
\\[.5em]
II:&& ~~ \frac32x + \frac{3}{2}y &= {17 \over 2}
\end{align}
when $x$ was multiplied by $(-3/2)$ in first equation the $x$ will be canceled and the resulting $y = -2/3$ and $x = 19/3$.
But when fractions were simplified first the... | Multiplying the second equation by $\frac{2}{3}$ we get
$$x+\frac{1}{2}y=6,$$
and
$$x+y=\frac{17}{3}.$$
Now multiplying the first equation by $-1$ and adding to the second
$$\frac{1}{2}y=-6+\frac{17}{3}$$
Can you finish?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Calculate in how many different ways can rotated square
We put a square $3 \text{ x } 3$ from $9$ square tiles in two types: or which can be freely rotated (we allow also symmetries). Calculate in how many different ways you can do that if we identify such systems that one goes to the other with some rotation or sym... | Let the small square with 1 shaded triangle be called A and the one with 2 shaded triangles be called B. We shall apply Burnside's Lemma to the group of symmetries of the large square acting on the $6^9$ patterns.
The identity fixes all $6^9$ patterns.
The rotations of 90 and 270 cannot fix the centre square and so fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Locus of midpoint of the line cutting the other two lines Given two lines $L_1: y = 2x-2$ and $L_2: y = -2x-2$. The variable line $L$ passes through the point $R(0,1)$ and meets $L_1$ and $L_2$ at $A$ and $B$, respectively. The slope of variable line is $m$. Find the equation of the locus of the midpoint $M$ of $AB$.
S... | Let $(h,k)$ be the midpoint $M$ of $AB$.
\begin{align*}
h&=\frac{x_A+x_B}{2}=\frac{3m}{4-m^2}
\end{align*}
But $R(0,1)$ also lies on the line $L$, so $m=\frac{k-1}{h-0}=\frac{k-1}{h}$. Thus
$$h=\frac{3m}{4-m^2}=\frac{3\left(\frac{k-1}{h}\right)}{\left(4-\left(\frac{k-1}{h}\right)^2\right)}.$$
Simplify this and you get
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to determine algebraically whether an equation has an infinite solutions or not? I was learning for the first time about partial fraction decomposition. Whoever explains it, emphasises that the fraction should be proper in order to be able to decompose the fraction. I was curious about knowing what happens if I try... | Here's a straightforward way to see that this particular equation ($(Ax+B)(x−3)+(Cx+D)(x+5)=^2−4$) has infinitely many solutions once you've found one: suppose we have some solution $\langle A, B, C, D\rangle$. We know that $(x+5)(x-3)+(3-x)(x+5)=0$ (since after all this is just saying that $(x+5)(x-3)-(x-3)(x+5)=0$!)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 2
} |
A double integral for $\frac{\pi}{2} \ln 2$.
Show that
\begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}} = \frac{\pi}{2} \ln 2.
\end{eqnarray*}
My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from th... | $$\int_0^1 \int_0^1 \frac{dxdy}{\sqrt{1-x^2y^2}}\overset{xy=t}=\int_0^1 \frac{1}{y}\int_0^y \frac{1}{\sqrt{1-t^2}}dtdy=\int_0^1 \frac{\arcsin y}{y}dy
$$
$$\overset{IBP}=-\int_0^1 \frac{
\ln y}{\sqrt{1-y^2}}dy\overset{y=\sin x}=-\int_0^\frac{\pi}{2}\ln(
\sin x)dx=\frac{\pi}{2}\ln 2$$
See here for the above integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find minimum of function $\frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$
Find minimum of function $f(x) = \frac{\left| x-12\right|
}{5}+\frac{\sqrt{x^2+25}}{3}$
I tried compute min using by definition of abs function.
I consider two cases:
*
*when $x > 12$ we have:
$f_{1}(x) = \frac{x-12}{5}+\frac{\sqrt{x... | We can use C-S and the triangle inequality:
$$\frac{|x-12|}{5}+\frac{\sqrt{x^2+25}}{3}=\frac{|x-12|}{5}+\frac{\sqrt{(3^2+4^2)(x^2+5^2)}}{15}\geq$$
$$\geq \frac{|x-12|}{5}+\frac{|3x+20|}{15}=\left|\frac{12}{5}-\frac{x}{5}\right|+\left|\frac{4}{3}+\frac{x}{5}\right|\geq $$
$$\geq \left|\frac{12}{5}-\frac{x}{5}+\frac{4}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding the length of the arc for function $8y = x^4 +2x^{-2}$
Find the length of the arc of curve $8y = x^4 +2x^{-2}$ from $x=1$ to $x=2$
I first isolated for $y$ and derived:
$$ f(x) = {1 \over 8 }x^{4} + {1 \over 4}x^{-2} $$
$$f'(x) = {1 \over 2} x^3 - {1 \over 2}x^{-3}$$
Then found the arc length:
$$L = \int ^2 _... | Hint
$$u = {1 \over 4} x^6 + {1\over 4}x^{-6} +{1 \over 2}= \left({1 \over 2} x^3 + {1\over 2}x^{-3} \right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Difficult integral involving trigonometric and hypertrigonometric functions This is definitely the most difficult integral that I've ever seen.
Of course, I'm not able to solve this.
Could you help me?
$$\int { \sin { x\cos { x } \cosh { \left( \ln { \sqrt { \frac { 1 }{ 1-\sin { x } } } +\tanh ^{ -1 }{ \left( \sin ... | Using the fact that $\cosh t = \frac{1}{\sqrt{1-\tanh^2 t}}$ and $\sinh t = \frac{\tanh t}{\sqrt{1-\tanh^2 t}} $ $$\cosh(a+b+c) = \cosh a \cosh b \cosh c + \sinh a \sinh b \cosh c + \sinh a \cosh b \sinh c + \cosh a \sinh b \sinh c$$
the integrand simplifies to
$$\int dx \left[\cosh\left(\log\left(\frac{1}{\sqrt{1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3337475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{i=0}^n (-1)^i\frac{C_i}{i+2}$ $C_i$ being the coefficient of $x^i$ in $(1+x)^n$ Evaluate: $\displaystyle\sum_{i=0}^n (-1)^i\frac{C_i}{i+2}$ where $C_i$ is the coefficient of $x^i$ in $(1+x)^n$
Attempt: We consider $f(x)=x(1-x)^n$
$\displaystyle\int_0^1x(1-x)^n dx= \int_0^1x^{(2-1)}(1-x)^{(n+1)-1} dx=B(2,n+1)=... | Hint:
Another way without calculus
$$\dfrac{\binom nr}{r+2}=\dfrac{(r+2-1)\binom nr}{(r+1)(r+2)}=\dfrac1{n+1}\binom{n+1}{r+1}-\dfrac1{(n+2)(n+1)}\binom{n+2}{r+2}$$
Now set $m=n+1,n+2$ in $$(1-1)^m=\sum_{r=0}^m(-1)^r\binom mr$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the reflection of the point $(4,-13)$ in the line $5x+y+6=0$
Find The image(or reflection) of the point $(4,-13)$ in the line $5x+y+6=0$
Method 1
$$
y+13=\frac{1}{5}(x-4)\implies x-5y-69=0\quad\&\quad 5x+y+6=0\implies (3/2,-27/2)\\
(3/2,-27/2)=(\frac{x+4}{2},\frac{y-13}{2})\implies(x,y)=(-1,-14)
$$
Method 2
$m=\... | The simplest (and shortest) way is to do some affine geometry:
Find first the projection of the point $A(4,-13)$ onto the line $5x+y+6=0$. As a directing vector for this projection is $\vec n (5,1)$, you have a parametric equation of the line of projection:
$$\overrightarrow{OM}=A+t\mkern 1.5mu\vec n, $$it suffices t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Proof: if $x^5-4x^4+3x^3-x^2+3x-4≥0$ then $x≥0$ I have proved this question by finding the solution of the polynomial using 17 steps Newthon Rhapson and found out that x ≥ 3,0735. I am wondering is there any more simple way to solve this?
| You have that
$$
\begin{gathered}
p(x) = x^5 - 4x^4 + 3x^3 - x^2 + 3x - 4 = \hfill \\
\hfill \\
= x^3 \left( {x^2 - 4x + 3} \right) - \left( {x^2 - 3x + 4} \right) \hfill \\
\end{gathered}
$$
Now:
*
*$$-\left( {x^2 - 3x + 4} \right)<0 \;\;\forall x\;\; \in \mathbb R$$
*$$
{x^2 - 4x + 3}>0\;\; \fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
A family of irreducible polynomials Does someone has an idea how to prove that the polynomials
\begin{align*} P_a=(a+2)(a+1)X^{a+4}-2(a+4)(a+1)X^{a+3}+(a+4)(a+3)X^{a+2}-2(a+4)X+2(a+1)
\end{align*}
are all of the form $(X-1)^4Q_a$ with $Q_a$ irreducible (over $\mathbb{Q}$)? Sage tell me it's true until 100 but I can't ... | $$P_a=(a+2)(a+1)x^{a+4}-2(a+4)(a+1)x^{a+3}+(a+4)(a+3)x^{a+2}-2(a+4)x+2(a+1)$$
Offset $a$ down by one:
$$P'_a=(a+1)ax^{a+3}-2(a+3)ax^{a+2}+(a+3)(a+2)x^{a+1}-2(a+3)x+2a$$
Subst $z = x-1$
$$P'_a=(a+1)a(z+1)^{a+3}-2(a+3)a(z+1)^{a+2}+(a+3)(a+2)(z+1)^{a+1}-2(a+3)(z+1)+2a \\
= \sum_i \left[ (a+1)a\binom{a+3}{i} - 2(a+3)a\bino... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding an integrating factor for $(3t + \frac{6}{y})dt + (\frac{t^2}{y} + \frac{3y}{t})dy = 0$ I want to solve the differential equation
$$
\begin{equation}
(3t + \frac{6}{y})dt + (\frac{t^2}{y} + \frac{3y}{t})dy = 0\
\end{equation}
$$
using an integrating factor of the form $t^a y^b$ in order to make it exact, where... | Here is a different approach to this problem. In the first place, notice that
\begin{align*}
3t + \frac{6}{y} + \left(\frac{t^{2}}{y} + \frac{3y}{t}\right)y^{\prime} = 0
\end{align*}
As suggested by Moo, multiply it by $ty$ in order to obtain
\begin{align*}
3t^{2}y + 6t + (t^{3} + 3y^{2})y^{\prime} = 0
\end{align*}
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3345642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
We have the equation $2x^2-\sqrt{3}x-1=0$ and have to find $|x_1-x_2|$
We have the following quadratic equation:
$2x^2-\sqrt{3}x-1=0$ with roots $x_1$ and $x_2$.
I have to find $x_1^2+x_2^2$ and $|x_1-x_2|$.
First we have: $x_1+x_2=\dfrac{\sqrt{3}}{2}$ and $x_1x_2=-\dfrac{1}{2}$
So $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\d... | Well, you know that, by the Quadratic Formula, $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, so the difference between the two roots is $\frac{1}{a}\sqrt{b^2-4ac}=\frac{1}{a}\sqrt{3+4(2)(1)}=\frac{1}{2}\sqrt{11}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove a trigonometric identity: $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$ when $A+B+C=\pi$ There is a trigonometric identity:
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C\equiv 1\text{ when }A+B+C=\pi$$
It is easy to prove it in an algebraic way, just like that:
$\quad\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos... | Since the accent in the OP is put on a purely geometric solution, i can not even consider the chance to write $\cos^2 =1-\sin^2$, and rephrase the wanted equality, thus having a trigonometric function which is better suited to geometrical interpretations.
So this answer has two steps, first we reformulate the given ide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3350887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | The remainder is a quadratic polynomial $g(x)$ that is divisible by $x$ (since $f(x)$ is divisble by $x$ and has remainders $f(1)$ and $f(-1)$ from division on $(x-1)$ and $(x+1)$ respectively.
One may argue that:
$$
g(x)=f(1)\frac{x+1}{2}+f(-1)\frac{x-1}{-2} + \gamma(x^2-1)
$$
From the fact $g(0) = 0$, one can find th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
complex series summation up to infinity
Sum of series
$$S(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots , |x|<1$$
What I tried.
Let $$A=\ln\bigg[(1+x)(1+x^2)(1+x^3)(1+x^4)\cdots \cdots \bigg]$$
$$A = \ln\bigg[(1+x+x^2+2x^3+2x^4+3x^5+\cdots+ )\bigg]$$
How do is solve it help me ple... | Regarding the relationship with the $A$ term
$$
S_n(x) = \sum_{k=1}^n\frac{g_k'(x)}{g_k(x)} = \frac{d}{dx}\sum_{k=1}^n\ln g_k(x) = \frac{d}{dx}\ln\prod_{k=1}^ng_k(x) = \frac{d}{dx}A
$$
here $g_k(x) = 1+x^k$. Also $\prod_{k=1}^n g_k(x) = \frac 12(-1;x)_{n+1}$ (see) and
$$
S_n(x) = \frac{d}{dx}\ln\left(\frac 12(-1;x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the maximum value of $\sum_{cyc}\frac{bc}{(b + c)^3(a^2 + 1)} $ where $a, b, c \in \mathbb R^+$ satisfying $abc = 1$.
Calculate the maximum value of $$\large \frac{bc}{(b + c)^3(a^2 + 1)} + \frac{ca}{(c + a)^3(b^2 + 1)} + \frac{ab}{(a + b)^3(c^2 + 1)}$$ where $a, b, c$ are positives satisfying $abc = 1$.
We... | The inequality $$\sum_{cyc}\frac{x}{x^4+1}\leq\frac{3}{2}$$ for positives $x$, $y$ and $z$ we can prove also by the following way.
We'll prove that
$$\frac{x}{x^4+1}\leq\frac{3(x^2+1)}{4(x^4+x^2+1)}$$ is true for all positive $x$.
Indeed, let $x^2+1=2ux.$
Thus, by AM-GM $u\geq1$ and it's enough to prove that
$$3\cdot2u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is the series $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$ convergent or divergent $\sum_{n=1}^{\infty} \frac{4+3^n}{2^n}$
$\begin{align}
\sum_{n=1}^{\infty} \frac{4+3^n}{2^n} &= \sum_{n=1}^{\infty} \frac{4}{2^n} + \sum_{n=1}^{\infty} \frac{3^n}{2^n} \\
&= \sum_{n=1}^{\infty} \frac{4}{2 \cdot 2^{n-1}} + \sum_{n=1}^{\infty} ... | Your answer is correct!
You can also argue that $
\frac{4+3^{n}}{2^{n}} \underset{n \rightarrow \infty}{\longrightarrow} \infty
$ that is the necessary condition for a series to converge doesn't hold in our case that is- the limit of a sequence defining a convergent series is $0$. Or the limit of a general term in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Definite integral of rational expression involving quadratics I was given the following exercise on my Calculus class:
$$
\int_2^4 \frac{x^2+4x+24}{x^2-4x+8}dx
$$
I studied from a book (author, Stewart) various methods to solve integrals of the form $\int \frac{P(x)}{Q(x)}dx$; a basic idea common to all of this methods... | It appears you have written your integral in the wrong manner, as $8+32\neq 24$.
\begin{align}
\int_2^4 \frac{x^2+4x+24}{x^2-4x+8}\ dx&=\int_2^4 \frac{(x^2-4x+8)+(8x+16)}{x^2-4x+8}\ dx\\
&=\int_2^4 dx + 8\int_2^4\frac{x+2}{x^2-4x+8}\ dx\\
&=x\big]_2^4+8\int_2^4 \frac{x+2}{(x-2)^2+4}\ dx\\
&=2+8\int_0^2 \frac{u+4}{u^2+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the exact value of trigonometric expression: $ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$? How can I simplify this trigonometric expression?
$$ \arctan\frac{2\cos \frac{\pi}{30} \cos\frac{\pi}{15}}{1+2\cos \frac{\pi}{30} \sin\frac{\pi}{15}}$$
I used
$$\sin... | $\begin{align}
T &= \large \left(\frac{2\cos 6° \cos 12°}{1\;+\;2\cos 6° \sin 12°}\right)
\left({2\sin 6° \over 2\sin 6°}\right)\cr
&= \large \frac{\sin 24°}{2\sin 6°+\;2\sin^2 12°}\cr\cr
{1\over T}&= \large \frac{2\sin(30°-24°)+\;(1-\cos 24°)}{\sin 24°}\cr
&=\large {(\cos 24° - \sqrt3\sin24°) + (1-\cos 24°) \over \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Numerical bases and prime number theory Let $N = 3^x$. $5^y$. $7^z$. Find $N$ such that $5N$ and $27N$ have $8$ and $18$ dividers, respectively, more than $N$.
I did what the statement asks and it was here:
dividers: $(x + 1) (y + 2) (z + 1) = (x + 1) (y + 1) (z + 1) + 8; (x + 4) (y + 1) (z + 1) = (x + 1) (y + 1) (z + ... | If $N=3^x5^y7^z$ then $N$ has $(x+1)(y+1)(z+1)$ divisors.
$5N = 3^x5^{y+1}7^z$ has $(x+1)(y+2)(z+1)$ divisors
And $27N = 3^{x+4}5y7^z$ has $(x+5)(y+1)(z+1)$ divisors.
It may be easier to replace $x+1 =j; y+1=k; z+1=m$
So
So $(x+1)(y+2)(z+1)-(x+1)(y+1)(z+1)= 8$. So $j(k+1)m - jkm=jkm +jm - jkm = jm=8$.
And $(x+4)(y+1)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.