Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$
Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$
I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$
$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$
gathering everything together,
$... | prove: $(n+1)^{n−1}(n+2)^n>3^n(n!)^2=3^n(n!n!)$ for $n>1$
$n=2:3^14^2=48>3^2(2)(2)=36$
assume: $(n+1)^{n−1}(n+2)^n>3^n(n!n!)$
need to arrive at: $(n+2)^{n}(n+3)^{n+1}>3^{n+1}(n+1)!(n+1)!$
for lhs need to multiply by:
${{(n+2)^{n}(n+3)^{n+1}}\over {(n+1)^{n−1}(n+2)^n}}
={{(n+3)^n(n+3)(n+1)}\over {(n+1)^n}}=({{n+3}\over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How do I solve quadratic equations when the coefficients are complex and real? I needed to solve this: $$x^2 + (2i-3)x + 2-4i = 0 $$
I tried the quadratic formula but it didn't work. So how do I solve this without "guessing" roots? If I guess $x=2$ it works; then I can divide the polynomial and find the other root; but... | The quadratic formula is perfectly valid here for the same reason for which it is valid when working only with real numbers:
$$
\frac{-b\pm\sqrt{b^2 - 4ac}} 2 = \frac{-(2i-3) \pm\sqrt{-3+4i}} 2.
$$
The question now is how to find $\pm\sqrt{-3+4i}.$
In polar form, you have $-3+4i = \sqrt{3^3+4^2} \cdot(\cos\alpha+i\sin\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 0
} |
Given a positive number n, how many tuples $(a_1,...,a_k)$ are there such that $a_1+..+a_k=n$ with two extra constraints The problem was: Given a positive integer $n$, how many tuples $(a_1,...,a_k)$ of positive integers are there such that $a_1+a_2+...+a_k=n$. And $0< a_1 \le a_2 \le a_3 \le...\le a_k$. Also, $a_k-a_1... | This can also be done with a simple generating function. Call the
desired quantity $T_n.$ We first choose the value $a_1$ and then the
gaps between successive values among the $a_q$ which are either zero
or one. We will have $1\le k\le n$ but for $k=1$ there is just one
possibility so we may assume $2\le k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Method of solving an Associated Legendre Equation with m=0, and m not equal to 0 with explanation I am looking to find the Legendre polynomial for the following standard Associated Legendre equation:
$$\left(\frac{d}{dx}[1-x^2]\frac{d}{dx}+\lambda-\frac{m^2}{1-x^2}\right)\Theta(x)=0$$
I have found the indicial equation... | What you can show is that, if $y$ is a solution of
$$
\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)+\mu y =0,
$$
then
$$
w=(1-x^2)^{m/2}\frac{d^my}{dx^m}
$$
is a solution of
$$
\frac{d}{dx}\left((1-x^2)\frac{dy}{dx}\right)-\frac{m^2}{1-x^2}y+\mu y = 0. \tag{$\dagger$}
$$
Therefore, if $P_n(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2069695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find integral $\int\frac{\arcsin(x)}{x^{2}}dt$ Find integral $$\int\frac{\arcsin(x)}{x^{2}}dx$$
what I've done: $$\int\frac{\arcsin(x)}{x^{2}}dx=-\int\arcsin(x)d(\frac{1}{x})=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$ I got stuck with that
| $$\int\frac{\arcsin(x)}{x^{2}}dx=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^{2}}}$$
Let $$u=\sqrt{1-x^2}$$
Hence
$$\frac{du}{dx}=\frac{-x}{\sqrt{1-x^2}}$$
$$\int \frac{dx}{x\sqrt{1-x^2}}=-\int \frac{du}{x^2}=-\int\frac{du}{1-u^2}=-\tanh^{-1}(u)+C=-\tanh^{-1}\left(\sqrt{1-x^2}\right)+C$$
Hence
\begin{align}
\int\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Generators of the intersection of twelve ideals. I have to show that the intersection of the twelve ideals $$(X\pm 1, Y\pm 1), (X\pm 1, Z\pm 1), (Y\pm 1, Z\pm 1) \subset \mathbb{R}[X,Y,Z]$$ is the ideal $$\Big( (X^2 −1)(Y^2 −1),(X^2 −1)(Z^2 −1),(Y^2 −1)(Z^2 −1)\Big).$$ Is there an easy way to do it, maybe some combinat... | Let $m$ be an element of the intersection of the twelve ideals. At first we have $m \in (x\pm1,y\pm1)$, so minimally we must have:
$y-1|m \lor x-1|m ,y+1|m \lor x-1|m ,y-1|m \lor x+1|m ,y+1|m \lor x+1|m$
There is a symmetry among these conditions, so assume WLOG that $y-1|m$, which satisfies the first and third minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Explicit solution of the recursion $x_n = x_{n-1}^2 - 2$ with $x_0>2$ Let $m>2$ be an integer, $x_0 = m$ and $x_n = {(x_{n-1})}^2 - 2$ for $n > 0.$ Prove that $x_n=\lceil\tau(n) \rceil$, where $\tau(n) = α^{2^n}$ and $\alpha >1$ satisfies $\alpha + \frac{1}{\alpha} = m$.
This is the problem. The only thing I can think... | From $x_0=m=\alpha + \frac{1}{\alpha}$ we have
$$x_1=m^2-2=\alpha^2+\frac{1}{\alpha^2}$$
And from $\alpha^2-m\alpha+1=0$ we have $$\alpha=\frac{m+\sqrt{m^2-4}}{2}\Rightarrow \alpha^2=\frac{m^2-2+m\sqrt{m^2-4}}{2}$$
which satisfies, from (easy to check) $m^2-4<m\sqrt{m^2-4}<m^2-2$: $$m^2-3<\alpha^2<m^2-2$$
So, we conclu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?
| The symbol $\sqrt{a}$ stands for the principal square root of $a$ which is always positive. Thus, the value of $x$ in the first equation is $2$.
The solution of the second equation, $x^2-4=0$, is given by
\begin{align*}
&\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[2ex]
=\ &\frac{0\pm\sqrt{(0)^2-4(1)(-4)}}{2(1)}
\end{align*}
Simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 13,
"answer_id": 2
} |
How to simplify an expression that does not have a common factor I am trying to simplify this expression :
$$9a^4 + 12a^2b^2 + 4b^4$$
So I ended up having this :
$$(3a^2)^2 + 2(3a^2)(2b^2) + (2b^2)^2$$
However, after that I don't know how to keep on simplifying the equation, it is explained that the answer is $(3a^2 + ... | Using the following formula.
$x^2 + 2xy + y^2 = (x+y)^2$
Here $x = 3a^2$, $y = 2b^2$
On putting values in left hand side of above formula.
$(3a^2)^2+2(3a^2)(2b^2)+(2b^2)^2 = 9a^4+12a^2b^2+4b^4 = (3a^2+2b^2)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
$\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ proof
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ for all $n\geq 2$
Basecase n=2
$\sum_\limits{k=1}^{2-1}\frac{1}{\sqrt{k(2-k)}}=1\geq 1$
Assumption
$\sum_\limits{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ holds for some $n$
Claim
$\sum_\limits{k=1}^{n}... | By AM-GM $\sqrt{k(n-k)} \le \frac{1}{2}\big(k+(n-k)\big) = \frac{n}{2}\,$ Then:
$$\sum_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq \sum_{k=1}^{n-1}\frac{2}{n} = \frac{2(n-1)}{n} \ge 1 \;\;\text{for}\;\; n \ge 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Geometric (Trigonometric) inequality $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$ How can one prove/disprove that $\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$ where $R$ and $r$ denote the usual circum and inradii respectively.
I know that $R=\frac{abc}{4\Delta}$ and $r=\frac{\Delta}{s}$, where $\Delta$ denotes area of tri... | We can prove that $$\frac{(a+b+c)^3}{3abc}\leq1+\frac{4R}{r}$$
Indeed, we need to prove that
$$\frac{(a+b+c)^3}{3abc}\leq1+\frac{\frac{abc}{S}}{\frac{2S}{a+b+c}}$$ or
$$\frac{(a+b+c)^3}{3abc}\leq1+\frac{8abc(a+b+c)}{16S^2}$$ or
$$\frac{(a+b+c)^3}{3abc}\leq1+\frac{8abc}{\prod\limits_{cyc}(a+b-c)}$$ or
$$\frac{(a+b+c)^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the general solution to $2 \cos^2 x-\cos x=0$? What is the general solution to this trig equation?
$$2 \cos^2 x-\cos x=0$$
Thanks.
| $2\cos^2 x-\cos x=0$
$\implies \cos x(2\cos x-1)=0$
Either
$\cos x=0$
$\implies x=\dfrac{\pi}{2}\pm 2n\pi$
Or
$2\cos x-1=0$
$\implies \cos x=\dfrac{1}{2}$
$\implies x=\dfrac{\pi}{3}\pm 2n\pi$
where
$n \in Z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
If $ f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval If $\displaystyle f(x) = \frac{\sin^2 x+\sin x-1}{\sin^2 x-\sin x+2},$ then value of $f(x)$ lies in the interval
assume $\sin x= t$ where $|\sin x|\leq 1$
let $\displaystyle y = \frac{t^2+t-1}{t^2-t+1}$
$\displaystyle yt... | hint: $y = 1+\dfrac{2t-3}{t^2-t+2}, -1 \le t \le 1, $ and proceed to find critical values of $t$ for $y'(t) = 0$. Then consider the values of $y$ at $\pm 1$, and those of critical values of $t$ above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Use Mean value theorem to prove the following inequality A) Use the Mean value theorem to prove that
\begin{equation}
\sqrt{1+x} < 1 + \frac{1}{2}x \text{ if } x>0
\end{equation}
B) Use result in A) to prove that
\begin{equation}
\sqrt{1+x}>1+\frac{1}{2}x-\frac{1}{8}x^2 \text{ if } x>0
\end{equation}
Can someone give ... | From (A), for $x>0$ we have:
$0<\sqrt{x+1}-1<\frac{1}{2}x\Rightarrow {(\sqrt{x+1}-1)}^2<\frac{1}{4}x^2\Rightarrow {(\sqrt{x+1})}^2+1^2-2\sqrt{x+1}<\frac{1}{4}x^2 \Rightarrow \sqrt{x+1}>1+\frac{x}{2}-\frac{x^2}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Simple Polynomial Algebra question to complete the square $4y^2+32y = 0$
$4(y^2+8y+64-64) = 0$
$4(y+4)^2 = 64$
Is that correct?
| $4y^2+32y = 0$
$4(y^2+8y+16-16) = 0$
$4(y+4)^2 = 64$
$(y+4)^2 = 16$
$y+4= \pm4$
$y_1=-8$, $y_2=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| Take the cube of the expression and expand it. You get
\begin{multline}
\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)^3
\\= 2 + \sqrt{5} + 2 - \sqrt{5} + 3\sqrt[3]{(2 + \sqrt{5})(2 - \sqrt{5})}\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)
\\ = 4 - 3 \left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 5
} |
Calculate the antiderivative of a given function Consider the function $f : \left[ 0, \frac{\pi}{4} \right)$, $f(x) = \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}}$, where $n \in \mathbb{N}^*$. Find the antiderivative $F$ of $f$ such that $F(0) = \frac{1}{2(n + 1)}$.
I've noticed that $(\cos x - \sin x)' = -(\c... | Noting
$$ \cos x+\sin x=\sqrt2\sin(x+\frac{\pi}{4}),\cos x-\sin x=\sqrt2\cos(x+\frac{\pi}{4}) $$
one has
\begin{eqnarray}
F(x)&=&\int \frac{(\cos x + \sin x)^n}{(\cos x - \sin x)^{n + 2}} dx\\
&=&\int \frac{(\sqrt2\sin(x+\frac{\pi}{4}))^n}{(\sqrt2\cos(x+\frac{\pi}{4}))^{n + 2}} dx\\
&=&\int \tan^n(x+\frac{\pi}{4})\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Asymptotes and focus of a conic? This is the conic $$x^2+6xy+y^2+2x+y+\frac{1}{2}=0$$
the matrices associated with the conic are:
$$
A'=\left(\begin{array}{cccc}
\frac{1}{2} & 1 & \frac{1}{2} \\
1 & 1 & 3 \\
\frac{1}{2} & 3 & 1
\end{array}\right),
$$
$$
A=\left(\begin{array}{cccc}
1 & 3 \\
3 & 1
\end{array}\right),
$$
... | You've two principal axes:
\begin{align*}
x^2+6xy+y^2+2x+y+\frac{1}{2} & \equiv
A\left( \frac{4x-4y-1}{\sqrt{4^2+4^2}} \right)^2+
B\left( \frac{8x+8y+3}{\sqrt{8^2+8^2}} \right)^2+C \\
& \equiv
-2\left( \frac{4x-4y-1}{4\sqrt{2}} \right)^2+
4\left( \frac{8x+8y+3}{8\sqrt{2}} \right)^2+\frac{9}{32}
\end{align*}
Now
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Doing definite integration $\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}$ We have to solve the following integration
$$
\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}
$$
I divided both in Nr and Dr by $\cos^2 x$.
But after that I stuck.
| We have $$I=\int_{0}^{\pi/4}\frac{2x}{2\cos^2x+2\cos x\sin x}dx=\int_0^{\pi/4}\frac{2x}{\cos 2x+\sin 2x+1}dx=\frac{1}{2}\int_0^{\pi/2}\frac{xdx}{\cos x+\sin x+1}=\frac{1}{2}\int_0^{\pi/2}\frac{\pi/2-x}{\cos x+\sin x+1}dx\\I=\frac{1}{4}\int_0^{\pi/2}\frac{\pi/2}{\cos x+\sin x+1}dx$$Now substitute $x$ for $2x$ again$$=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
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Quadratic equation system $A^2 + B^2 = 5$ and $AB = 2$ Given a system of equations
$A^2 + B^2 = 5$
$AB = 2$
what is the correct way to solve it?
I see immediately that the answers are
*
*$A=1, B=2$
*$A=2, B=1$
*$A=-1, B=-2$
*$A=-2, B=-1$
but I don't understand the correct way of getting there. I have tried to is... | From $AB=2$ we know $A$ and $B$ have the same sign.
Moreover, if we find a solution for some values $A=A_1$ and $B=B_1$,
then $A=-A_1$ and $B=-B_1$ also is a solution.
So we can assume initially that $A$ and $B$ are both positive,
find all such solutions, and then simply change signs to produce all the remaining soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 3
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The le... | The proof is only valid for $0<x<\frac\pi2$. For a more general proof, let:
$$a\cos x+b\sin x\ \equiv\ R\cos(x-\alpha)=R\cos\alpha\cos x + R\sin\alpha\sin x$$
So $a=R\cos\alpha$ and $b=R\sin\alpha$, from which you can readily find $R$ and $\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Provide a different method of proving:$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}]^2}dx={1\over 2}$ Accidentally founded this particular integral producing a rational number
I can't be for sure it is correct, so can one provide a proof of it.
$$\int_{-\infty}^{\infty}{1\over [\pi(x+e^{\pi})^2+\pi^{1/3}... | To avoid confusion, $a = A = \pi$ and $b = B = \pi ^{1/3}$
Let $z=\frac{a u^{2}}{b}$
\begin{align}
\int\limits_{-\infty}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du &= 2 \int\limits_{0}^{\infty} \frac{1}{(a u^{2} + b)^{2}} du \\
&= \frac{2}{b^{2}} \int\limits_{0}^{\infty} \frac{1}{(\frac{a}{b} u^{2} + 1)^{2}} du \\
&= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n... | After breaking up the fraction using Partial Fractions, we see that $\frac3n$ is bigger than $\frac4{3n+1}$, so we give $\frac4{3n}$ of $\frac3n$ to $-\frac4{3n+1}$ to make it positive, but decreasing.
$$
\begin{align}
\frac{n^2+2n+1}{3n^2+n}
&=\frac13\left(1+\frac{5n+3}{3n^2+n}\right)\\
&=\frac13\left(1-\frac4{3n+1}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Integral $\int \frac{\mathrm{d}x}{\sin x+\sec x}$ In the following integral
$$\int \frac{\mathrm{d}x}{\sin x+\sec x}$$
My attempt is
I first multiplied and divided by $\cos^2x$
And then substitued $\tan x = t$
But after that got stuck .
| $\displaystyle I = \int\frac{1}{\sin x+\sec x}dx = \int\frac{\cos x}{\sin x\cos x+1}dx = \int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\sin 2x +2}dx$
$\displaystyle =\int\frac{\cos x+\sin x}{\sin 2x+2}dx+\int\frac{\cos x-\sin x}{\sin 2x+2}dx = I_{1}+I_{2}$
for $\displaystyle I_{1} = \int\frac{\cos x+\sin x}{3-(\sin x-\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the $\gcd[x+y+z; x^2+xy+z^2; y^2+yz+z^2; z^2+zx+x^2]$ What I have done:
There exists a non-zero integer $t$ such:
$$x+y+z=kt$$
$$x^2+xy+y^2=ut$$
$$y^2+yz+z^2=vt$$
$$z^2+zx+x^2=wt$$
$\implies$
$$(x-y)(x+y+z)=(u-v)t$$
$$(y-z)(x+y+z)=(v-w)t$$
$$(z-x)(x+y+z)=(w-u)t$$
$\implies$
$$\dfrac{x+y+z}{t}= \dfrac{u-v}{x-y}=\d... | If $x,y,z$ are not coprime, the result is trivial. So We can assume that $x,y,z$ are not all even. Then $t$ must be an odd integer since the bottom 3 numbers are always odd.
$$x^2+y^2+z^2+\dfrac{xy+xz+zx}{2}=\dfrac{t(u+v+w)}{2}$$
$$(x+y+z)^2-\dfrac{3(xy+xz+zx)}{2}=\dfrac{t(u+v+w)}{2}$$
$$(kt)^2-\dfrac{3(xy+xz+zx)}{2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of $n$ terms of the given series. Find the sum of $n$ terms of the series:
$$\frac{1}{x+1}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\frac{4x^3}{(x+1)(x+2)(x+3)(x+4)}+.......$$
Could someone give me slight hint to proceed in this question?
| Recalling that the lower incomplete gamma function has the power series $$\gamma\left(s,x\right)=x^{s-1}\Gamma\left(s\right)e^{-x}\sum_{k\geq1}\frac{x^{k}}{\Gamma\left(s+k\right)}\tag{1}$$ we have $$\frac{d}{dx}\left(\frac{\gamma\left(s,x\right)}{x^{s-1}\Gamma\left(s\right)e^{-x}}\right)=\sum_{k\geq1}\frac{kx^{k-1}}{\G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Problem with integrating this function I have the function $$f(u) = \frac{1}{\sqrt{1-bu^2 -cu}}$$ with $b,c$ positive. I want to integrate this function from $0$ to its first root. Within this interval, the function is not complex since the polynomial under the root is positive. But the function itself is obviously com... | As suggested in the comments section, you need to complete the square, i.e.
\begin{equation} \int f(u) du = \int \frac{1}{\sqrt{1 - bu^2 - cu}} du
= \int \frac{1}{\sqrt{-b \big( u^2 + \frac{cu}{b} + \frac{c^2}{4b^2} - \frac{c^2}{4b^2} - \frac{1}{b} \big) }} du \\
= \frac{1}{\sqrt{- b \Big( \big(u^2 + \frac{cu}{b} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Remarkable limit and L’Hôpital $\lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}$
Compute
$$
\lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2}
$$
where $a$ and $b$ are positive numbers.
I came to the two different forms of this limit, as $\lim_{x\to0}$
$$e^{\frac{\ln b-\ln b}{\ln a-\l... | Assuming that the desired limit is $L$ we can proceed as follows:
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A polynomial question. I came across this question in my weekly test paper and didn't know how to tackle it. The question goes as follows:
$a_0,a_1,a_2....a_{34}$ are the coefficients of $x^0,x^1,x^2......x^{34}$ of the polynomial obtained on opening the parenthesis of $(1+x+x^2)^{17}$, then which is true?
$1)~a_1+a_2+... | Here is some info regarding (2) - (4)
*
*Ad (2): $a_1$ is the coefficient of $x^1$ in $(1+x+x^2)^{17}$. Since there are $17$ factors of the form $1+x+x^2$ we have to choose precisely from one factor $x$ whereas from all other factors we have to choose $1$. This can be done in
\begin{align*}
a_1=\binom{17}{1}=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$?
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$.
I think this is a simple exercise, but I get this:
$(n+1)^2+1=n^2+2n+2$.
$n^2+2n+2 = (n^2+1)+(2n+1)$
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$
and $\displaystyle n^2+1 = \frac{n(2n+1)}{... | Your step that says $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$ is correct but not useful. We are working in the naturals and this threatens (if $n$ is not even) to takes us out of the naturals.
If we just make a spreadsheet and try the small numbers we find that the GCD is $1$ unless $n\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is the... | Since $\cos x = 1-2\sin^2 \frac{x}{2}$,
$$
f(x)=\left(2-2\sin^2\frac{x}{2}\right)\sin \frac{x}{2} = 2\sin \frac{x}{2} - 2\sin^3\frac{x}{2}.
$$
Thus you can find the maximum of $f(x)$ by finding the maximum of $g(x)=2x-2x^3$ defined on $(0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$ How to evaluate
$$\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx$$
My attempt:
\begin{align*}
\int_0^1 \frac{\ln \left(1+x+x^2+\cdots +x^{n} \right)}{x}dx &= \int_0^1 \frac{\ln \left(\dfrac{1-x^{n+1}}{1-x} \right)}{x}dx \\
&= \int_0^... | $$\int_0^1 \dfrac{\ln(1-x)}xdx = \int_0^1 \left(-\sum_{k=1}^{\infty} \dfrac{x^{k-1}}k\right)dx = -\sum_{k=1}^{\infty} \dfrac1k\int_0^1 x^{k-1}dx = - \sum_{k=1}^{\infty} \dfrac1{k^2} = -\zeta(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Number of positive integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$ for distinct primes $p$ and $q$
Let $p$ and $q$ be distinct primes. Then find the number of positive integer solutions of the equation $$\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$$
We get $pq=\frac{xy}{x+y}$
Now $x+y$ must divide $xy$ as L.H.S.... | Since
$$
\begin{align}
(x-pq)(y-pq)
&=xy-pq(x+y)+p^2q^2\\
&=p^2q^2
\end{align}
$$
By breaking up the $9$ factors of $p^2q^2$, and solving for $x$ and $y$, we get
$$
\begin{align}
\frac1{pq}
&=\frac1{pq+p^2q^2}+\frac1{pq+1}\\
&=\frac1{pq+pq^2}+\frac1{pq+p}\\
&=\frac1{pq+q^2}+\frac1{pq+p^2}\\
&=\frac1{pq+q}+\frac1{pq+p^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving the differential equation $7x(x-y)dy = 2(x^2+6xy-5y^2)dx$ How do I solve the differential equation $$7x(x-y)dy = 2(x^2+6xy-5y^2)dx$$
Is it homogeneous? I have tried taking the variables from the LHS and applying them to the RHS, making $\frac{dy}{dx}$ subject and ending up with:
$$\frac{dy}{dx} = \frac{2(x^2+6x... | Starting with your equation
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$
multiply the top and bottom of each fraction on the right, giving
$$\frac{dy}{dx} = \frac{2(x+5y)}{7x}\frac{1/x}{1/x} + \frac{4y}{7(x-y)}\frac{1/x}{1/x}$$
$$\frac{dy}{dx} = \frac{2(1+5y/x)}{7} + \frac{4y/x}{7(1-y/x)}$$
Since $y'=u'x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A Ramanujan infinite series $$ 1-5\left(\frac{1}{2}\right)^3+9\left(\frac{1 \cdot 3}{2 \cdot 4}\right)^3-13\left(\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6}\right)^3+\cdots $$
I went on evaluating the above series and encountered that solving $\displaystyle \sum_{n\ge 0}\left(\binom{2n}{n}\right)^3x^n$ would suffice.
B... | This is not an answer but just a result obtained using a CAS.
Let $$f_k=\sum_{n=0}^\infty \binom{2 n}{n}^kx^n$$ The following expressions have been obtained $$f_1=\frac{1}{\sqrt{1-4 x}}$$ $$f_2=\frac{2 }{\pi }K(16 x)$$ $$f_3=\frac{4 }{\pi ^2}K\left(\frac{1}{2} \left(1-\sqrt{1-64 x}\right)\right)^2$$
$$f_4=\, _4F_3\left... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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How to calculate $\lim\limits_{n\to \infty} \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}$ I have no idea how to find a limit of sequence
$$a_n= \frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}}$$. I think that it can be easily bounded by 1 from down, but this shows only $1≤a_n$. I think also it can be shown quite easily that... | shows that using conjugation that
\begin{split}
a_n=\frac{(n+3)^{\sqrt{n+1}}}{(n+1)^{\sqrt{n}}} &= &\exp \left( {\sqrt{n+1}}\ln(n+3)- {\sqrt{n}}\ln(n+1)\right)\\
&=& \exp \left( \ln(n+3)\color{red}{\left(\sqrt{n+1}-\sqrt{n}\right)}+\sqrt{n}[\color{blue}{\ln(n+3)-\ln(n+1)} ]\right)\\
&=& \exp \left( \frac{\ln(n+3)}{\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Evaluating $\quad\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}$ find limit :
$$\lim_{ x \to -\infty }\frac{\sqrt[n]{x+1}+\sqrt[n]{x+2}-2\sqrt[n]{x-3}}{\sqrt[n]{x-1}+\sqrt[n]{x-2}-2\sqrt[n]{x+3}}=\;\; ? \;\quad \text {given }\,n \in \mathbb{N}, n>2 ,\... | Substitute $x=-1/t$, so you get
$$
\lim_{t\to0^+}
\frac{\sqrt[n]{t-1}+\sqrt[n]{2t-1}-2\sqrt[n]{-3t-1}}
{\sqrt[n]{-t-1}+\sqrt[n]{-2t-1}-2\sqrt[n]{3t-1}}
=
\lim_{t\to0^+}
\frac{\sqrt[n]{1-t}+\sqrt[n]{1-2t}-2\sqrt[n]{1+3t}}
{\sqrt[n]{1+t}+\sqrt[n]{1+2t}-2\sqrt[n]{1-3t}}
$$
The Taylor expansion of the numerator i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$ The sum of series $(2^2-1)(6^2-1)+(4^2-1)(8^2-1)+\cdots \cdots +\cdots (100^2-1)(104^2-1)$
Attempt Assume $\displaystyle S = \sum^{50}_{r=1}((2r)^2-1)((2r+4)^2-1) = \sum^{50}_{r=1}(4r^2-1)(4r^2+16r+15)$
$\displaystyle S = \sum^{50}_{... | Being lazy, we could use the classical sums of powers of integer numbers as given here. This would lead to
$$ S_n = \sum^{n}_{r=1}(16r^4+64r^3+56r^2-16r-15)=\frac{1}{5} \left(16 n^5+120 n^4+280 n^3+180 n^2-71 n\right)$$
For sure, much more elegant is mathlove's solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Sum of given infinite series: $\frac14+\frac2{4 \cdot 7}+\frac3{4 \cdot 7 \cdot 10}+\frac4{4 \cdot 7 \cdot 10 \cdot 13 }+....$ Find the sum of infinite series
$$\frac{1}{4}+\frac{2}{4 \cdot 7}+\frac{3}{4 \cdot 7 \cdot 10}+\frac{4}{4 \cdot 7 \cdot 10 \cdot 13 }+....$$
Generally I do these questions by finding sum of $n$... | The partial sums, according to Maple, are
$$-{\frac {2\,{3}^{1/2-N}\pi}{27\,\Gamma \left( 4/3+N \right) \Gamma
\left( 2/3 \right) }}+\frac{1}{3}
$$
It should be possible to prove that by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Sum of real roots of two different cubic polynomials Let $\alpha$ be the only real root of $p(x)=x^3-3x^2+5x-17$ and $\beta$ the only real root of $q(x)=x^3-3x^2+5x+11$. Compute $\alpha + \beta$.
I've noticed that the two graphs are just the same shifted vertically (the first coefficients are equal except for the last)... | Re-arrange $p(x)$ and $q(x)$:
\begin{align*}
p(x) &= x^3-3x^2+5x-17 \\
&= (x-1)^3+2(x-1)-14 \\
q(x) &= x^3-3x^2+5x+11 \\
&= (x-1)^3+2(x-1)+14
\end{align*}
Note that $$p(1+x) = -q(1-x)$$
If $p(1+x)=-q(1-x)=0$, then $$
\left \{
\begin{align*}
1+x &= \alpha \\
1-x &= \beta
\end{align*}
\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Explanation of Digital Root/ Sum formula The formula to find the digital root/ sum is:
digital root of n = 1 + ( (n - 1) % 9 )
Can someone explain me the intuition behind this formula? Why does this result give the sum of digits?
| Numbers in base $10$ are "arranged" into boxes of $10$ entries. For example, $21$ looks like
$$
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare\\
\blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare \blacksquare... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Compute $\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot \frac{1}{n^2}$ I need to compute:
$\lim \limits_{n \to \infty} \frac{\sqrt[4]{n^4+4n}\,-\,\sqrt[3]{n^3+3n}}{\sqrt[5]{n^5+1}\,-\, \sqrt[5]{n^5+n}}\cdot\frac{1}{n^2}$
I tried this
$\lim \limits_{n ... | I know you are all "begeistert" with taylor series, but here is another method.
Note that $$\sqrt[3]{n^3+3n}-n=\frac{3n}{(\sqrt[3]{n^3+3n})^2+(\sqrt[3]{n^3+3n})n+n^2}$$ so it follows easily that $$n(\sqrt[3]{n^3+3n}-n)\to \frac{3}{3}=1$$
in a similar fashion one sees that
$$n^2(\sqrt[4]{n^4+4n}-n)\to 1$$
$$n^4(\sqrt[5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Does the vector belong to the column space? I have to check whether the vector
$$b = \begin{vmatrix}
-2\\
4\\
5\\
6\\
6\\
\end{vmatrix}$$
belongs to the column space of matrix
$$A = \begin{bmatrix}
1 & 1 & 1 & −2& 1\\
1 & 0 & 2 & 4 & 2\\
1 & 0 & 2 & ... | It obviously does, because the given matrix and the augmented matrix have the same rank. You even can read the solutions in the reduced row echelon form (I denote the unknowns $x, y,z,t,u$, in the order of the columns):
$$\begin{cases}x=-2z+2u,\\y=z-5u,\\t=1-u.\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $y'=-2-y+y^2$
$$y'=-2-y+y^2$$ and $y_{0}=2$ is a solution
So it is a riccati ODE:
$y=2+z\Rightarrow z=y-2$
$z'=y'$
substation into the original ODE gives:
$z'=-2-(2+z)+(z+2)^2=-4-z+z^2+4z+4=z^2+3z$
So we have:
$z'=z^2+3z$ which is bernoulli $z'=3z+z^{1+1}$ when $k=1$
$w=\frac{1}{z}$
$w'=-z'\cdot z^{-2}=-(... | Clearly $y=-1$ and $y=2$ are trivial solutions. Note that
$$ y^2-y-2=(y-2)(y+1)$$
and hence
$$ y'=(y-2)(y+1)$$
or
$$ \frac{dy}{(y-2)(y+1)}=dx.$$
Integrating both sides gives
$$ \int\frac{dy}{(y-2)(y+1)}=\int dx+C.$$
So
$$ \frac13\ln\left|\frac{y-2}{y+1}\right|=x+C$$
or
$$ \frac{y-2}{y+1}=Ce^{3x}$$
So the solution is
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve $x=\pm\frac{y'}{\sqrt{(y')^2+1}}$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x^2=\frac{(y')^2}{{(y')^2+1}}$$
$$x^2(y')^2+x^2=(y')^2$$
$$(y')^2[x^2-1]=-x^2$$
$$(y')^2=\frac{-x^2}{x^2-1}$$
$$y'=\pm \sqrt{\frac{-x^2}{x^2-1}}$$
Have I got it wrong? as there is no ODE
| It looks alright.
You're mistaken in saying there's no ODE. Every line in your posting is just an ODE.
From where you left off I'd go on to say
$$
\int\pm\sqrt{\frac{-x^2}{x^2-1}}\,dx = \int\pm\sqrt{\frac{x^2}{1-x^2}} = \int\pm \frac x {\sqrt{1-x^2}} \, dx = \int\pm \frac {du}{2\sqrt{u}} = \pm \sqrt u + C
$$
$$
= \pm \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve the equation: $\sin 3x=2\cos^3x$ Solve the equation :
$$\sin 3x=2\cos^3x$$
my try :
$\sin 3x=3\sin x-4\sin^3x$
$\cos^2x=1-\sin^2x$
so:
$$3\sin x-4\sin^3x=2((1-\sin^2x)(\cos x))$$
then ?
| Hint:
$$\sin 3x=2\cos^3x$$
$$3\sin x-4\sin^3x=2\cos^3x$$
$$3\sin x(\cos^2x+\sin^2x)-4\sin^3x=2\cos^3x$$
$$\sin^3x-3\sin x\cos^2x+2\cos^3x=0$$
Divide both sides by $\cos^3x$ and $\tan x=t$
$$t^3-3t+2=0$$
$$(t-1)^2(t+2)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve a system of 3 trigonometric equations How to solve system of three trigonometric equations:
$(\sin x)^2 (\cos y)^2 = 4 \cos x \sin y\tag1$
$(\sin y)^2 (\cos z)^2 = 4 \cos y \sin z \tag2$
$1- \sqrt{\sin z}(1+\sqrt{\cos x})=\sqrt{\frac{1-\sin y}{1+\sin y}}\tag3$
and to verify that
$\sin x=\sqrt{2}(\sqrt{2}... | Note that, if any of $\sin x$, $\sin y$, $\sin z$ are zero, then they all are, so we have the solution
$$\sin x = \sin y = \sin z = 0 \qquad \cos x = 1 \tag{0}$$
(where the latter condition allows $\sqrt{\cos x}$ to be real). On the other hand, if any of $\cos x$, $\cos y$, $\cos z$ are zero, then they all are, and we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to factorize this cubic equation? In one of the mathematics book, the author factorized following term
$$x^3 - 6x + 4 = 0$$
to
$$( x - 2) ( x^2 + 2x -2 ) = 0.$$
How did he do it?
| There is a neat trick called the rational roots theorem. All we have to do is factor the first and last numbers, put them over a fraction, and take $\pm$. This gives us the following possible rational roots:
$$x\stackrel?=\pm1,\pm2,\pm4$$
due to the factorization of $4$. Checking these, it is clear $x=2$ is the only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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How to simplify derivatives
The math problem asks to find the derivative of the function
$$y=(x+1)^4(x+5)^2$$
I get to the part $$(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3$$
How do they arrive at the answer
$$2(x+1)^3(x+5)(3x+11) ?$$
| $$\begin{align}(x+1)^4 \cdot 2(x+5) + (x+5)^2 \cdot 4(x+1)^3&=(x+1)^3(x+5)\cdot\big[(x+1)2+(x+5)4\big]\\
&=(x+1)^3(x+5)\cdot\big[2x+2+4x+20 \big]\\
&=(x+1)^3(x+5)\cdot(6x+22)\\
&=(x+1)^3(x+5)\cdot 2(3x+11)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$ Prove that; $8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})=1$
My Attempt,
$$L.H.S=8\cos^3(\frac {\pi}{9}) - 6\cos(\frac {\pi}{9})$$
$$=2\cos(\frac {\pi}{9}) [4\cos^2(\frac {\pi}{9}) - 3]$$
$$=2\cos(\frac {\pi}{9}) [2+2\cos(\frac {2\pi}{9}) - 3]$$
$$=2\co... | \begin{align}
8\cos^3 (\frac {\pi}{9})- 6\cos(\frac {\pi}{9})
&=8\cos(\frac{\pi}9)\left[1-\sin^2(\frac{\pi}9)\right]-6\cos^(\frac{\pi}9)\\
&=2\cos(\frac{\pi}9)-8\cos(\frac{\pi}9)\sin^2(\frac{\pi}9)\\
&=2\cos(\frac{\pi}9)-4\sin(\frac{2\pi}9)\sin(\frac{\pi}9)\tag1\\
&=2\cos(\frac{\pi}9)-2\left[\cos(\frac{\pi}9)-\cos(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the value of $x$ that satisfy the equation: $3^{11}+3^{11}+3^{11} = 3^x$ I have this question:
$$3^{11}+3^{11}+3^{11} = 3^x$$
Find the value of $x$
| Takking $3^{11}$ common from left hand side,
$3^{11}(1 + 1 + 1) = 3^x$
$3^{11}(3) = 3^x$
$3^{12} = 3^x$
$x = 12$
Or -
$3(3^{11})= 3^x$
$3^{12} = 3^x$
$x = 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the determinant of order $100$ Find the determinant of order $100$:
$$D=\begin{vmatrix}
5 &5 &5 &\ldots &5 &5 &-1\\
5 &5 &5 &\ldots &5 &-1 &5\\
5 &5 &5 &\ldots &-1 &5 &5\\
\vdots &\vdots &\vdots &\ddots &\vdots &\vdots &\vdots\\
5 &5 &-1 &\ldots &5 &5 &5\\
5 &-1 &5 &\ldots &5 &5 &5\\
-1 &5 &5 &\ldots &5 &5 &5
\en... | If we reverse the order of the rows, we end up with a matrix of the form $M = A-6I$, where $A$ has $5$ for every entry. This requires $50$ transpositions, so the determinant of this new matrix is the same. To find this determinant, we can multiply the eigenvalues of the matrix $A - 6I$, which can be found to be
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
finding limit with $\cos$ function occur $n$ times Finding $\displaystyle \lim_{x\rightarrow 0}\frac{1-\cos(1-\cos(1-\cos(1-\cdots \cdots (1-\cos x))))}{x^{2^n}}$
where number of $\cos$ is $n$ times
when $x\rightarrow 0$ then $\displaystyle 1-\cos x = 2\sin^2 \frac{x}{2} \rightarrow 2\frac{x}{2} = x$
so $1-\cos (1-\cos... | Let $f^{n} $ denote the composition of $f$ with itself $n$ times and let $f^{0}(x)=x$. Then the numerator of the given expression (whose limit is to be evaluated) is equal to $f^{n} (x) $ where $f(x) = 1-\cos x$. Note that as $x\to 0$ each of the functions $f^{n} (x) \to 0$ and also note that we have $$\lim_{x\to 0}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Finding function when dx is integrated $\int (5-x^2+\frac{18}{x^4})dx$ I have a question on how to evaluate this integral:
$\int (5-x^2+\frac{18}{x^4})dx$
Is this correct?
$\int (5-x^2+\frac{18}{x^4})dx$
$ \int 5 dx = 5x + C$
$ \int -x^2 = \frac {-x^3}{3}= -\frac{1}{3}x^3 +C $
$ \int \frac {18}{x^4} = \int 18x^{-4} dx... | This is how you would do this
$$\int (5-x^2+\frac{18}{x^4})dx$$
Integration as you have probably shown spits across addition so we have;
$$
\int5dx+\int-x^2dx+\int\frac{18}{x^4}dx
$$
now we can also take out constaints so we have:
$$
5\int1dx-\int x^2dx+18\int x^{-4}dx
$$
now by defintion we have
$$
5x-\frac{x^3}{3}+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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What real numbers do algebraic numbers cover? Hardy and Wright mention ( though don't give a proof ) that any finite combination of real quadratic surds is an algebraic number. For example
$\sqrt{11+2\sqrt{7}}$. Are all finite combinations of cube root, fourth root ... $n^{th}$ root also algebraic ? such as $\sqrt[3]{2... | Yes, you can unpack them into a polynomial that they satisfy. For your example, we can write $$x=\sqrt[3]{2+3\sqrt[7]{5+3\sqrt{6}}}\\x^3=2+3\sqrt[7]{5+3\sqrt{6}}\\\frac 13(x^3-2)=\sqrt[7]{5+3\sqrt{6}}\\\left(\frac 13(x^3-2)\right)^7=5+3\sqrt 6\\\frac 19\left(\left(\frac 13(x^3-2)\right)^7-5\right)^2-6=0$$
Note: this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 3
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If:$(\sqrt{x + 9})^{\frac{1}{3}} - (\sqrt{x-9})^{\frac{1}{3}} = 3$, Find $x^2$
If:$$\sqrt[3]{(x + 9)} - \sqrt[3]{(x-9)} = 3$$
Find $x^2$
I can't seem to solve this question. Any hints or solutions is welcomed.
| We have this:
$$(x + 9)^{\frac{1}{3}} - (x-9)^{\frac{1}{3}} = 3$$
Then:
$$(x + 9)^{\frac{1}{3}} -3 = (x-9)^{\frac{1}{3}}$$
$$((x + 9)^{\frac{1}{3}} -3)^3 = ((x-9)^{\frac{1}{3}})^3$$
$$x + 9 -9(x+9)^{\frac{2}{3}} + 27(x+9)^{\frac{1}{3}} -27 = x -9$$
$$-9(x + 9)^{\frac{2}{3}} +27(x +9)^{\frac{1}{3}} =-9-9 +27 $$
$$-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solving a univariable identity that satisfies the following relationship:
Question: Consider the following equations:$$\begin{align*}1^2+2^2+2^3 & =3^2\\2^2+3^2+6^2 & =7^2\\3^2+4^2+12^2 & =13^2\\4^2+5^2+20^2 & =21^2\end{align*}$$
State a one variable identity that is suggested by these examples.
Since the question ... | I believe the identity you are looking for contains a quadratic for $n$, $$n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2$$
Putting $n=1,2,3,4$ gives $$\begin{align*}1^2+2^2+2^3 & =3^2\\2^2+3^2+6^2 & =7^2\\3^2+4^2+12^2 & =13^2\\4^2+5^2+20^2 & =21^2\end{align*}$$
Which are the examples above. These identities are well known, and c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$? The question is how to show the identity
$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$
I checked this numerically for several cas... | Enforcing the substitution $x^{-1}-x=u$ in your last integral we get:
$$ \int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+u^2}}\right)\frac{du}{u\sqrt{u^2+a+2}} $$
and by setting $u=\sqrt{a+2}\sinh\theta$ we get:
$$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+(a+2)\sinh^2\theta}}\right)\frac{d\theta}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
} |
Vector Parameterisation of Ellipse How would you show that, if
$$\mathbf r = \mathbf A \sin \theta + \mathbf B \cos \theta$$
where $\mathbf A$ and $\mathbf B$ are arbitrary constant vectors in $\Bbb R^2$,
then $\mathbf r$ may be written as,
$$\mathbf r = \mathbf a \sin (\theta + \alpha) + \mathbf b \cos (\theta + \alph... | Let $\boldsymbol{x}=
\begin{pmatrix}
\cos \theta \\ \sin \theta
\end{pmatrix} \in S^1$
\begin{align*}
\boldsymbol{x}^T \boldsymbol{x} &= 1 \\
\boldsymbol{y} &=
\begin{pmatrix}
\boldsymbol{a} & \boldsymbol{b}
\end{pmatrix}
\begin{pmatrix}
\cos \theta \\ \sin \theta
\end{pmatrix} \\
\boldsymbol{y}^T \boldsymb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Inequality problem with 4th root. $ \frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 $
I've tried putting $t= \frac{6x}{x-2} $ and play algebraically, using square of sum, but still no luck, any help?
| Instead, let's put $$t = \sqrt[4]{\frac{12x}{x-2}}.$$
Then:
$$\frac{6x}{x-2} - \sqrt{\frac{12x}{x-2}} - 2\sqrt[4]{\frac{12x}{x-2}}>0 \Rightarrow \\
\frac{1}{2}t^4 - t^2 - 2t>0 \Rightarrow \\
t^4 - 2t^2 - 4t > 0 \Rightarrow \\
t(t-2)(t^2+2t+2) >0.
$$
Then:
$$t <0 ~ \vee t > 2, $$
or equivalently:
$$\sqrt[4]{\frac{12x}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort?
I will keep it short and take only an extract (most important part) of
the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$
What I have done is a lot work and time consuming, I have "simply" s... | The most general method: do all multiplications on both sides to transform both numerators to canonical form, then verify polynomials are the same degree and have respective coefficients equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
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Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$ Prove that: $\sec^2 20^\circ +\sec^2 40^\circ +\sec^2 80^\circ = \textrm 36$
My Attempt:
$$L.H.S=\sec^2 20^\circ + \sec^2 40^\circ +\sec^2 80^\circ$$
$$=\dfrac {1}{\cos^2 20°} +\dfrac {1}{\cos^2 40°} +\dfrac {1}{\cos^2 80°}$$
$$=\dfrac {\cos^2... | A clean start, not remotely obvious ( but easy enough to prove ), the roots of
$$ x^3 - 3 x + 1 $$ are
$$ 2 \cos \frac{2 \pi}{9}, \; \; 2 \cos \frac{4 \pi}{9}, \; \; 2 \cos \frac{8 \pi}{9}. $$
From page 174 in Reuschle (1875). The method used is due to Gauss.
I learned today that, in 1933, D. H. Lehmer published ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root. Proposition: Suppose that $a$, $b$, and $c$ are real numbers with $c \not = 0$. Prove, by contradiction, that, if $cx^2 + bx + a$ has no rational root, then $ax^2 + bx + c$ has no rational root.
Hypothesis... | The roots of
$ax^2+bx+c = 0
$
are
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$
and
the roots of
$cx^2+bx+a = 0
$
are
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2c}
$.
If the first are rational
and the second are not,
then their sum
and ratio
are irrational.
Since the roots of the first equation are rational,
their sum and product are rational... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If a,b and c are sides of a triangle, then prove that the following polynomial has no real roots This is the polynomial: $$a^2x^2+(b^2+a^2-c^2)x+b^2=0$$
Now this is my progress:
Assuming l,m, and n are sides of a triangle, then $$|m-n|\lt l\lt m+n$$
Also, if a second degree polynomial in the form $kx^2+px+q$ has real r... | Nice exercise. It is well known (and not difficult to prove) that a second-degree polynomial has no real roots iff its discriminant is negative. The discriminant of the given polynomial is
$$ (b^2+a^2-c^2)^2-4a^2 b^2 = (a^2+2ab+b^2-c^2)(a^2-2ab+b^2-c^2) $$
or
$$ (a+b+c)(a+b-c)\color{red}{(a-b-c)}(a-b+c) $$
that is clea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Computing: $\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$ Find the following limit:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^2\frac{1}{x}}\left(3^\frac{1}{x}-5^\frac{-1}{x}\right)}{\log_2(1+x^{-2}+x^{-3})}$$
I'm not sure whether my solution is corre... | Here is a solution without using Taylor series or applying L'Hôpital's rule:
$$\lim_{x\to\infty}\frac{\sqrt{1-\cos^{2}\left(\frac{1}{x}\right)}\left(3^{\frac{1}{x}}-5^{\frac{-1}{x}}\right)}{\log_{2}\left(1+x^{-2}+x^{-3}\right)}$$$$=\lim_{x\to\infty}\frac{\sin\left(\frac{1}{x}\right)}{\color{blue}{\frac{1}{x}}}\cdot\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$
Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$.
I expanded the brackets and applied AM-GM on all of the eight terms to get :
$$\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big) \geq 3\... | For $a=b=c=1$ we get the value $125$.
We'll prove that it's the minimal value.
Indeed, let $f(x)=-\ln(3+2x^2)$.
Hence, $f''(x)=\frac{4(2x^2-3)}{(2x^2+3)^2}<0$ for all $0<x<1$.
Thus, by Vasc's LCF Theorem it's enough to prove our inequality for $b=a$ and $c=3-2a.$
Hence, it's enough to prove that
$$(3+2a^2)^2(3+2(3-2a)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Angles in pyramid
We have to find the value of $\cos\theta$.
I tried it alot. But could not able to do it.
Solution given in the book
| This works out nicely with vectors.
Place $A, B, C, D$ on the $x$ and $y$ axes. Let $A = (1, 0, 0), B = (0, -1, 0), C = (-1, 0, 0), D = (0, 1, 0)$
$O$ must be on the $z$ axis by symmetry and $O = (0, 0, h)$ The only question is the value of $h$. Since angle $AOB = 45$ degrees, $OA \cdot OB = ||OA|| ||OB|| cos 45$ degre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Compute $\lim\limits_{x \to 0} \frac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Compute $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+\sin x} - \sqrt[3]{1- \sin x }}{x}$
Original question was to solve $\lim\limits_{x \to 0} \dfrac{ \sqrt[3]{1+ x } - \sqrt[3]{1-x }}{x}$ and it was solved by adding and subtracting 1 i... | Just another way using Taylor series.
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ Now, using the generalized binomial theorem
$$\sqrt[3]{1+\sin(x)}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{x^3}{162}+O\left(x^4\right)$$
$$\sqrt[3]{1-\sin(x)}=1-\frac{x}{3}-\frac{x^2}{9}-\frac{x^3}{162}+O\left(x^4\right)$$
Using the above,
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Solutions to a Cubic Root I'm looking for a sanity check to my work.
We are supposed to find the solutions of the cubic polynomial, $x^3+3x+5=0$.
Let $x=w-\frac{1}{w}$ and substitute.
Then $(w-\frac{1}{w})^3+3(w-\frac{1}{w})+5=0$.
Then $w^3+5w^3-1=0$.
Finally $w^3=\frac{-5+\sqrt{29}}{2}$ and $\frac{-5-\sqrt{29}}{2}$.
S... | In general Let be $f(x)=x^3+ax+b$
put $x=u-v$, then $f(u-v)=(u^3-v^3)-(3uv-a)(u-v)+b$ and if $f(u,v)=0$, then $(u^3-v^3)+b=0$ and $3uv-a=0$. From second equation we have $v=\frac{a}{3u}$ and then $u^3-(\frac{a}{3u})^3+b=0 \Rightarrow 3^3u^6-a^3+3^3u^3b=0\Rightarrow 3^3y^2+3^3by-a^3=0 ; y=u^3 $, we solve the equation fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How does one show that $\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=\pi$? Consider
$$\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=L\tag1$$
How does one show that $L=\pi$ for $k>0$?
An attempt:
For $k=2$
$$\lim_{n\to \infty}{n\over 2}\cdot\sq... | $$x\to 0 \\\cos x=1-\frac{1}2x^2+\frac{1}{4!}x^4+...\sim 1-\dfrac12x^2\\
\cos ^kx\sim (1-\frac{1}2x^2)^k \sim 1-\dfrac k2x^2\\ $$
now put into ,$\frac{2\pi}{n} \to 0$
$$\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-\cos^k\left({2\pi\over n}\right)}=\\
\lim_{n\to \infty}{n\over \sqrt{2k}}\cdot\sqrt{1-(1-\dfrac k2(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Condition of two hyperbolas do not intersect Given two hyperbolas h1(with foci and center) and h2(with foci and center), In what condition these hyperbolas will not intersect to each other?. I can get the condition when h1 and h2 are standard hyperbola (parallel to axis and the center is the origin). I want to find the... | Let $x^2-y^2=1$ be the first hyperbola.
*
*Re-scaling
$$x^2-y^2=c^2 \tag{$c^2 \ne 1$}$$
*Conjugate
$$y^2-x^2=c^2$$
*Translation
\begin{align*}
(x-h)^2-(y-k)^2 &= 1 \\
2(ky-hx)+h^2-k^2 &= 0 \\
x &= \frac{h^2-k^2+2ky}{2h} \\
(h^2-k^2+2ky)^2-4h^2 y^2 &= 4h^2 \\
4(h^2-k^2)y^2-4k(h^2-k^2)y+4h^2-(h^2-k^2)^2 &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}}$ I tried first without L'Hôpital's rule:
$$\lim_{x \rightarrow +\infty}\frac{\sqrt{x}+x}{x+\sqrt[3]{x}} =
\frac{\sqrt{x}}{\sqrt[3]{x}} \cdot \frac{1+\frac{x}{\sqrt x}}{1+ \sqrt[3] x} = \frac{\sqrt x}{\sqrt[3] x} \cdot \frac{\frac{\sqrt x+x}{\sqrt x}}{... | set ${ t }^{ 6 }=x$ then we have $$\lim _{ x\rightarrow +\infty } \frac { \sqrt { x } +x }{ x+\sqrt [ 3 ]{ x } } =\lim _{ t\rightarrow +\infty } \frac { { t }^{ 3 }+{ t }^{ 6 } }{ { t }^{ 6 }+{ t }^{ 2 } } =\lim _{ t\rightarrow +\infty } \frac { { \left( \frac { 1 }{ { t }^{ 3 } } +1 \right) } }{ \left( \frac { 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Geometric interpretation of complex number.
Let $z \in \mathbb{C}, z^2, z^3$ be the verticies of a right triangle. Find the geometric images of $z$.
I did not understand the question but I guess it want me to find the figure formed by $z$ under these constraints.
Let $z = x + iy$, then $z^2 = x^2 - y^2 +2xyi$ and $z... | Hint: let $r = |z| \in \mathbb{R}^+\,$, so that $\bar z = r^2 / z\,$.
It forms a triangle, So, $|z - z^2|^2 + |z^3 - z^2|^2 = |z - z^3|^2$ by Pythagoras theorem.
This assumes that the hypotenuse is $z\,z_3\,$, and the following works on this assumption. The other cases would follow similarly.
Rewriting it in terms o... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the coefficient of $x^{29}$ in the given polynomial. The polynomial is :
$$
\left(x-\frac{1}{1\cdot3}\right) \left(x-\frac{2}{1\cdot3\cdot5}\right) \left(x-\frac{3}{1\cdot3\cdot5\cdot7}\right) \cdots \left(x-\frac{30}{1\cdot3\cdot5\cdots61}\right)
$$
What I've done so far : The given polynomial is an expression of... | HINT: The sum $S_k $ given by
$$\sum_{n=1}^{k} \dfrac {n}{(2n+1)!!} $$
for $k=1,2,3....$ gives
$$S_1=\frac {1}{3} \\ =\frac {1}{2}-\frac {1/2}{ 1 \cdot 3}$$
$$S_2=\frac {1}{3} + \frac {2}{15} =\frac {7}{15} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 }$$
$$S_3=\frac {1}{3} + \frac {2}{15} +\frac {3}{105} =\frac {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2151030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use infinite series to prove Use infinite series to prove that
$$\arcsin{x}\lt \frac{x}{1-x^2},$$ for $0\lt x\lt1$.
| From AM-GM inequality we have, for a given $t$ in the interval $(0,1)$,
$$\frac12\left[\frac{1}{(1+t)^2}+\frac{1}{(1-t)^2}\right]\ge \frac1{(1+t)(1-t)}\tag{1}$$
Now, since $1>\sqrt{(1+t)(1-t)}$, it follows
$$\frac1{(1+t)(1-t)}>\frac{\sqrt{(1+t)(1-t)}}{(1+t)(1-t)}=\frac1{\sqrt{(1+t)(1-t)}}=\frac1{\sqrt{1-t^2}}\tag{2}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that $\sum_{n=1}^{\infty}\sum_{m=1}^{x-1} \frac{m}{n^2 x^2 - m^2}=\frac{1}{2}H_{x-1}$ Consider $(1)$, $H_n$ is the nth-harmonic number
$$\sum_{n=1}^{\infty}\left({1\over (nx)^2-1}+{2\over (nx)^2-2^2}+{3\over (nx)^2-3^2}+\cdots+{x-1\over (nx)^2-(x-1)^{2}}\right)=S\tag1$$
$x\ge2$
How does one show that $$\col... | From Cotangent identity (formula 18) and Digamma reflection formula:
$$
\begin{align}
\pi &\,\cot(\pi z)=\frac{1}{z}+2z\,\sum_{n=1}^{\infty}\,\frac{1}{z^2-n^2}=\psi(1-z)-\psi(z) \\[3mm]
S &= \sum_{n=1}^{\infty}\,\sum_{m=1}^{x-1}\,\frac{m}{(nx)^2-m^2} = \sum_{m=1}^{x-1}\,\frac{m}{x^2}\sum_{n=1}^{\infty}\,\frac{1}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Finding a quartic with integer coefficients such that $\sqrt{3} −\sqrt{5}$ is a zero Find integers $a,b,c,d$ and $e$ such that $\sqrt{3} −\sqrt{5}$ is a solution to the equation:
$$ax^4 + bx^3 + cx^2 + dx + e = 0.$$
Being new to quartic equations I wasn't sure how to solve this problem, must we use the quartic formula?... | Since $$\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a),$$
we have that $\sqrt3-\sqrt5$ is a root of the equation
$$2(3x^2+5x^2+15)-x^4-9-25=0$$ or
$$x^4-16x^2+4=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation : $\tan \theta + \tan 2\theta + \tan 3\theta = \tan \theta \tan 2\theta \tan 3\theta $ I've been having some trouble solving this equation. (The solution in my book is given as $ \frac {n \pi}{3}, n \in Z $)
Here is what I've done
$$\frac {\sin \theta}{\cos \theta} + \frac {\sin 2\theta} {\cos 2\thet... | In your simplified expression there is $\cos 3 \theta $ term in the denominator that goes to zero for solution you obtained. Should be checked before accepting or discarding it as a valid solution.
I cannot resist an elementary trig approach ..
If $ (A+B+C) = 2 \pi, $ then $ {(\tan A + \tan B + \tan C) = \tan A \ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{9a^2+b^2}}+\frac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \frac32.$
Prove that $$\dfrac{a}{\sqrt{a^2+b^2}}+\dfrac{b}{\sqrt{9a^2+b^2}}+\dfrac{2ab}{\sqrt{a^2+b^2}\times \sqrt{9a^2+b^2} }\leq \dfrac{3}{2}.$$
When is equality attained ?
My Attempt :
I could not t... | setting $$x=\sqrt{\frac{a^2}{a^2+b^2}}$$ and $$y=\sqrt{\frac{b^2}{9a^2+b^2}}$$
we get further $$1+\left(\frac{b}{a}\right)^2=\frac{1}{x^2}$$ and $$9\left(\frac{a}{b}\right)^2+1=\frac{1}{y^2}$$ we can eliminate $$\frac{a}{b}$$ and we get $$y^2=\frac{1-x^2}{8x^2+1}$$ thus our inequality is equivalent to
$$x+\sqrt{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154453",
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"source": "stackexchange",
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Considering $0Problem: Considering $0<s<1$ for the series $\sum_{i=1}^{\infty} \frac {1}{i^s}$, I want to show that $a_{2^{n+1} -1}$ $<\sum_{j=0}^{n}$ $({\frac {1} {2^{s-1}})}^j$ where $a_n$ is the partial sums of the aforementioned series..
So for $n=1$ this would yield the inequality:
$\frac {1}{1^s} +\frac {1}{2^s} ... | $$
\begin{align}
a_{2^{n+1}} &= \sum_{i=1}^{2^{n+1}} \frac 1{i^s} \\
&= \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{i^s} \\
& \leq \sum_{j=0}^{n} \sum_{i=2^j}^{2^{j+1}-1} \frac 1{2^{js}} \\
& = \sum_{j=0}^{n} \frac {2^j}{2^{js}} \\
& = \sum_{j=0}^{n} \left(\frac 1{2^{s-1}}\right)^j \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derivation of $1/(s^2-a^2)^{3/2}$ in Laurent Series So Given modified bessel equation (t/a)*I_1(at) using the general formula
I have arrived at the answer.
$$\frac{1}{s^3}+\frac{3a^2}{2s^5}+\frac{15a^4}{8s^7}+ \cdots$$
And according to wolfram alpha its laurent series is equal to $\frac{1}{(s^2-a^2)^{3/2}}$ equal to m... | Using
$$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$
we have
\begin{eqnarray}
\frac{1}{\sqrt{(s^2-a^2)^3}}&=&(s^2-a^2)^{-\frac32}\\
&=&s^{-3}\left(1-\left(\dfrac{a}{s}\right)^2\right)^{-\frac32}\\
&=&\frac{1}{s^3}\left(1
-\left(-\dfrac{3}{2}\right)\left(\dfrac{a}{s}\right)^2
+\frac{(-\frac{3}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2155636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partition of numbers : $1, 2, ..., 20$ Integers $1, 2, ..., 20$ are partitioned into $2$ groups. Sum of all integers in one group is equal to $n$ and the product of all integers in another group is also equal to $n$. Find the maximal $n$.
Since $1 + 2 + ... + 20 = 210$, so the product of all integers in another group ... | Here's a answer by trial and error.
Notice that the product of the smallest $5$ integers is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$ and the product of $6$ integers is bigger then $210$.
Now the sum will be bigger if we choose smaller numbers,lets try to make the sum the max possible value $<210$ by changing the digit $5... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $f_k(x)=\frac{1}{k}\left (\sin^kx +\cos^kx\right)$, then $f_4(x)-f_6(x)=\;?$ I arrived to this question while solving a question paper. The question is as follows:
If $f_k(x)=\frac{1}{k}\left(\sin^kx + \cos^kx\right)$, where $x$ belongs to $\mathbb{R}$ and $k>1$, then $f_4(x)-f_6(x)=?$
I started as
$$\begin{align... | Hint.
Let $x=\frac{\pi}{4}$
In your answer $f4(x)-f(6x) = 0$
By brute-forse, $f4(x)-f6(x) = \frac{0.5}{4} - \frac{0.25}{6} = \frac{1}{8}-\frac{1}{24} =\frac{1}{12}$, so, you make a typo :(
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find all primes $p$ for which there are integers $n, x, y$ such that $p^n = x^3 + y^3$ I want to find all primes $p$ for which there exist integers $n, x, y$ such that $p^n = x^3 + y^3$. One solution is $p = 2$ because $2^1 = 1^3 + 1^3$. Now I can assume that $p$ is odd, so exactly one of $x$ and $y$ must be odd and an... | Hint
Let solution exists => exists a solution with $gcd(x,y) = 1$ => exists a solution with $gcd(x,p) = gcd(y,p) = 1$
in this case: $x+y=p^k,x^2-xy+y^2=p^m => 3xy=p^{2k}-p^m = p^m(p^{2k-m}-1)$ =>$m=0$ or $m=1, p=3$, in other cases $p|xy$
$x^2-xy+y^2=1$ => $x=1, y=1$ (you was found this solution)
$x^2-xy+y^2=3$ => $(x-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Logarithmic differentiation - need help please Determine the following by using logarithmic differentiation.
$$y=\frac{\sqrt{x}(x^2-1)^5}{\cos(x))}$$
My answer is up to:
Take the ln on both sides so:
$\ln(y) = (1/2)\ln(x) + 5\ln(x^2-1) - \ln(\cos(x))$
Am I doing it right though?
| You have everything right so far.
$\ln(y) = (1/2)\ln(x) + 5\ln(x^2-1) - \ln(\cos(x))$
Now remember, that the derivative of $\ln(y) = \frac{1}{y}\times y'$
This is because $y$ is a function, whereas $x$ is just a variable.
The derivative of $\ln(x)$ in this case is just $\frac{1}{x}$
Also remember that if you have some... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2163588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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If $abc=1$ so $\sum\limits_{cyc}\frac{a}{a^2+b^2+4}\leq\frac{1}{2}$
Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{a^2+b^2+4}+\frac{b}{b^2+c^2+4}+\frac{c}{c^2+a^2+4}\leq\frac{1}{2}$$
This inequality is a similar to the following.
Let $a$, $b$ and $c$ be non-negative numbers. Prov... | The Buffalo Way works.
With the substitutions $a = \frac{x^2}{yz}, \ b = \frac{y^2}{zx}, \ c = \frac{z^2}{xy}; \ x, y, z > 0$, after clearing denominators,
it suffices to prove that $f(x, y, z) \ge 0$ where $f(x,y,z)$ is a homogeneous polynomial.
WLOG, assume that $z = \min(x, y, z) = 1$.
There are two possible cases:
... | {
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"url": "https://math.stackexchange.com/questions/2166910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Equation in radicals $x^7+7x^3-7x^2+7x+1=0$ Solve the following equation in radicals.
$$x^7+7x^3-7x^2+7x+1=0$$
It's obvious that this equation has no rational roots.
And what is the rest?
| The Galois group of the polynomial is $D_{14}$, thus it is solvable.
RadiRoot is an implementation of the algorithm described in this paper:
http://www.icm.tu-bs.de/ag_algebra/software/distler/Diplom.pdf. This is the output of RadiRoot:
Let $\zeta_7$ be a primitive $7$-th root of unity and
$$\omega_1 = \sqrt[7]{ - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Infinitely many $n$ such that $n, n+1, n+2$ are each the sum of two perfect squares. Prove that there exist infinitely many integers $n$ such that $n$, $n+1$, $n+2$ are all the sum of two perfect squares. Induction does not seem to be yielding any results.
| One such infinite family is
$$(2a^2 + 1)^2 - 1, (2a^2 + 1)^2 + 0^2, (2a^2 + 1)^2 + 1^2$$ for all integer $a$.
This works because
$$(2a^2 + 1)^2 - 1 = 4a^4 + 4a^2 = (2a^2)^2 + (2a)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
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Prove by induction that $n^4+2n^3+n^2$ is divisible by 4
I'm trying to prove by induction that $n^4+2n^3+n^2$ is divisible by 4.
I know that P(1) it's true. Then $ n=k, P(k):k^4+2k^3+k^2=4w$ it's true by the hypothesis of induction.
When I tried to prove $n=k+1, P(k+1):(k+1)^4+(k+1)^3+(k+1)^2 = 4t$,
$$k^4+4k^3+6k^2+... | You don't need induction to prove this.
$f(n)=n^4+2n^3+n^2 = n^2\cdot(n^2+2n+1) = n^2\cdot(n+1)^2$
Case 1: If $n$ is even, $n=2k, n^2=2k\cdot2k=4k^2$, now $4k^2\cdot(n+1)^2$, which is obvious that is divisible by 4
Case 2: If $n$ is odd then $n+1$ is even, let $m=n+1$, $m=2k, \ m^2=2k\cdot2k = 4k^2$, if we change in ou... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$ Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$
Wolfram Alpha gave this result.
But may be there is simple way to get all a?
| Rewrite the equation by completing the squares
\begin{align}
a^4-6a^2b-8ac-3b^2
=(a^2+s)^2 -( 2s+6b)\left( a +\frac{4c}{2s+6b} \right)^2\tag1\\
\end{align}
where $s$ happens to satisfy $(s+b)^3=8(c^2-b^3)$, or
$s=2\sqrt[3]{c^2-b^3}-b$.
For notational convenience, denote
$$p=\frac12\sqrt{2s+6b}= \sqrt{\sqrt[3]{c^2-b^3}+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Does series $\sum _{n=1}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ converge? Does series
$\sum _{n=2}^{\infty \:}\ln\left(\frac{n\left(n+2\right)}{\left(n+1\right)^2}\right)$ converge?
My idea is to apply the Cauchy test, but I dont know how to simplify it next.
Thanks
| Using the identity $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\,a+b=c+d$$ we have $$\prod_{n\geq2}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}=\prod_{n\geq0}\frac{\left(n+2\right)\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
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Number of real zeros of a polynomial Given a $n$-th degree polynomial with real coefficients
$$c_nx^n + c_{n-1}x^{n-1} + ... + c_1x+c_0$$
Is there any theorem that tells me how many zeros are real (zero imaginary part). Anything related would help. Thanks!
| Sturm's theorem might help!
Sturm's Theorem: If $f(x)$, freed from equal roots, be divided by $f'(x)$, and the last divisor by the last remainder, changing the sign of each remainder before dividing by it, until a remainder independent of $x$ is obtained, or else a remainder which cannot change its sign; then $f(x)$, $... | {
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"timestamp": "2023-03-29T00:00:00",
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A rigorous proof of a trigonometric inequality in $(0,\pi/2)$ I want to show
$$f(x)=\sin \left(\frac{2 x}{17}\right)-8 \sin
\left(\frac{4 x}{17}\right)+27 \sin
\left(\frac{6 x}{17}\right)-64 \sin
\left(\frac{8 x}{17}\right)+125 \sin
\left(\frac{10 x}{17}\right)-216 \sin
\left(\frac{12 x}{17}\right)+343 \... | All right, that is easier than expected. If we define
$$ C_n(x) = \sum_{k=1}^{n}(-1)^k k^3\sin\left(\frac{2kx}{2n+1}\right)\tag{1} $$
through $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ we have:
$$\small{ C_n(x) = -\frac{i (-1)^n}{2z^n (1+z)^4}\left(3 n^2 z (1+z)^2 \left(-1+z^{2 n}\right)+z (1-4z+z^2)\left(-1+z^... | {
"language": "en",
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How do we find the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}?$ We would like to find out the closed form for integral $(1)$
$$\int_{0}^{\pi/2}{\mathrm dx\over \sqrt{\sin^6x+\cos^6x}}\tag1$$
An attempt:
We may write
$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$
Let $x=\sin x$ and $y=\cos x$
$x^... | Using this,$$\sin^6(x)+\cos^6(x)=1-\frac 34 \sin^2(2x)$$ makes $$\int{ dx\over \sqrt{\sin^6(x)+\cos^6(x)}}=\int{ dx\over \sqrt{1-\frac 34\sin^2(2x)}}=\frac{1}{2} F\left(2 x\left|\frac{3}{4}\right.\right)$$ where appears the elliptic integral of the first kind.
Using the bounds, the result simplifies to $$K\left(\fr... | {
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Finding a basis for a set of $2\times2$ matrices
Find a basis for $M_{2\times2}$ containing the matrices $$\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}$$ and $$\begin{pmatrix} 1 & 1 \\ 3 & 2 \end{pmatrix}$$
I know that every $2\times2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ can be written as:
$$a\begi... | The two matrices are linearly independent, so they are a basis for the two dimensional vector space spanned by them.
If you want a basis for $M_{2\times2}$ add two linearly independent matrices that are linearly independent from them. As an example you can chose:
$$
\begin{pmatrix} 0 & 0 \\ 2 & 3 \end{pmatrix}
\qquad
... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove an identity which uses Drazin inverses and Moore-Penrose inverses. Let $A$ be $m\times n$ complex matrix. Then how can we prove the following.
$$A^+=(A^*A)^DA^*=A^*(AA^*)^D$$ where $D$ denotes the Drazin inverse and $A^+$ is the Moore-Penrose Pseudoinverse of $A$. I found this question at http://planetmath.org/d... | Core-Nilpotent Decomposition
Let $\mathbf{A}$ be a square matrix of rank $\rho<n$, with index $k$ such that $\text{rank} \left( \mathbf{A}^{k} \right) = \rho$:
$$
\mathbf{A} \in \mathbb{C}^{n\times n}
$$
then there exists a nonsingular matrix $\mathbf{Q}$ such that
$$
\mathbf{A} =
\mathbf{Q}
\left[ \begin{arra... | {
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Does $\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ \operatorname dx$ converge? I have been trying to prove the following integral:
$$\int _0^{\infty }\:\frac{1}{1+x^2\left(\sin x\right)^2}\ dx$$
diverges (please correct me if I am mistaken).
I have tried to use different comparison tests (as this is an ... | The integral would diverge if $x^2\sin^2(x)$ was "quite small" when $x$ is near a multiple of $\pi$. Using the periodicity of $\sin(x)$, we can examine the behavior near $k\pi$ by shifting and looking at
$$ (x+k\pi)^2 \sin^2(x) \approx (x+k\pi)^2 x^2$$
for $x$ near $0$. This is valid for very small $x$ through the Tayl... | {
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Inequality with 4 variables, powers. $\{ a,b,c,d \in\Bbb R_+\ \}$
$a^6 + b^3 + c^2 + d \ge 2 \sqrt[3]{2} \sqrt[4]{3} \sqrt {abcd} $
I have no idea how to do this sort of inequalities, never seen it before.
| By the AM-GM inequality, one has
$$ a^6+b^3+c^2+d=a^6+\frac{1}{2}b^3+\frac{1}{2}b^3+\frac{1}{3}c^2+\frac{1}{3}c^2+\frac{1}{3}c^2+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d+\frac{1}{6}d\ge 12\sqrt[12]{a^6(\frac12b^3)^2(\frac13c^2)^3(\frac16d)^6}$$
After simple calculation you will get the answer.
| {
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Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}$ For $a\geq b\geq c >0$. Prove that $$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$$
$a=100;b=1;c=1/100$ it's wrong ???
$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}$
$<=>\sum\frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac... | For $a\geq b\geq c>0$ we obtain:
$$\sum_{cyc}\frac{a}{a+b}-\frac{3}{2}=\sum_{cyc}\left(\frac{a}{a+b}-\frac{1}{2}\right)=\sum_{cyc}\frac{a-b}{2(a+b)}=$$
$$=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{2\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{c^2a-c^2b}{2\prod\limits_{cyc}(a+b)}=$$
$$=\sum_{cyc}\frac{a^2b-a^2c}{2\prod\limits_{c... | {
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"timestamp": "2023-03-29T00:00:00",
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Find value or $W=\sqrt[4x+4]{3^{x}\left ( \sqrt{5}-1 \right )}$ according to condition $9^{x}=6^x +4^x$ If $$9^{x}=6^x +4^x$$
Find the value of:
$$W=\sqrt[4x+4]{3^{x}\left ( \sqrt{5}-1 \right )}$$
Solving the equation arrives at:
$x = \frac{\log\left ( \sqrt{5}-1 \right ) - \log(2)}{\log(2) - \log(3)}$
But W yields... | Note that
$$1=\left(\frac23\right)^x+\left(\frac49\right)^x$$
Let $a=\left(\frac23\right)^x$
$$a^2+a-1=0$$
Solving gives
$$a=\left(\frac{\sqrt5 -1}{2}\right)$$
$$\left(\frac23\right)^x=\frac{\sqrt5 -1}{2}$$
$$\left(\frac32\right)^x=\frac{2}{\sqrt5-1}$$
$$3^x=\frac{2^{x+1}}{\sqrt5-1}$$
Substitute for this in question
$... | {
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