Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove: $\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$ This is not a homework question, its from sl loney I'm just practicing.
To prove :
$$\arcsin\left(\frac 35\right) - \arccos\left(\frac {12}{13}\right) = \arcsin\left(\frac {16}{65}\right)$$
So I changed all t... | How exactly did you convert to arctan? Careful:
$$\arccos\left(\frac {12}{13}\right) = \arctan\left(\frac {5}{12}\right)
\ne \arctan\left(\frac {12}{5}\right)$$
Draw a right triangle with hypotenuse of length 13, adjacent side (from an angle $\alpha$) with length 12 and opposite side with length 5; then $\cos\alpha = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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A human way to simplify $ \frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^2 - 1}} - 2 a $ I end up with simplifying the following fraction when I tried to calculate an integral(*) with the residue theory in complex analysis:
$$
\frac{((\sqrt{a^2 - 1} - a)^2 - 1)^2}{(\sqrt{a^2 - 1} - a)^22 \sqrt{a^... | Start with
$$\begin{align}\left(\sqrt{a^2-1}-a\right)^2-1&=a^2-1-2a\sqrt{a^2-1}+a^2-1\\
&=\sqrt{a^2-1}\left(2\sqrt{a^2-1}-2a\right)\\
&=2\sqrt{a^2-1}\left(\sqrt{a^2-1}-a\right)\end{align}$$
So you are now down to
$$\frac{\left(2\sqrt{a^2-1}\left(\sqrt{a^2-1}-a\right)\right)^2}{\left(\sqrt{a^2-1}-a\right)^2\cdot2\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$ for $a, b, c > 0$ Prove for $a, b, c > 0$ that
$$a \sqrt{b + c} + b \sqrt{c + a} + c \sqrt{a + b} \le \sqrt{2(a+b+c)(bc + ac + ab)}$$
Could you give me some hints on this?
I thought that Jensen's inequality might be of use ... | Use the Cauchy-Schwarz inequality on the two vectors $(\sqrt a, \sqrt b, \sqrt c)$ and $(\sqrt{a(b+c)}, \sqrt{b(a+c)}, \sqrt{c(a+b)})$ (then take the square root on both sides or not, depending on which version of the CS inequality you use), and lastly note that we have:
$$
a(b+c) + b(a + c) + c(a + b) = 2(bc + ac + a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Function that is second differential continuous Let $f:[0,1]\rightarrow\mathbb{R}$ be a function whose second derivative $f''(x)$ is continuous on $[0,1]$. Suppose that f(0)=f(1)=0 and that $|f''(x)|<1$ for any $x\in [0,1]$. Then $$|f'(\frac{1}{2})|\leq\frac{1}{4}.$$
I tried to use mean value theorem to prove it, but I... | A bit more tricky than I thought at first. The idea is easy, the calculation may look complicated. The idea is to find a second order polynomial with second derivate $=1$ which has the same values as $f$ for $x=0 $ and $x= \frac{1}{2}$, and then to show that this function $-f$ is convex, which allows to get an estimate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\int_D \rvert x-y^2 \rvert dx \ dy $ $$\int_D \rvert x-y^2 \rvert dx \ dy $$
$D$ is the shape that is delimited from the lines:
$$
y=x \\
y=0 \\
x=1 \\$$
$$D=\{ (x,y) \in \mathbb{R}^2: 0 \le x \le 1 \ , \ 0 \le y \le x \}$$
$$\rvert x-y^2 \rvert=x-y^2 \qquad \forall (x,y) \in D $$
$$\int_0^1 \Big( \int_0^x... | This is correct, you are correct that the absolute value integrand is equal to x-y^2 for all x and y in your region of integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum of $xy+y^2$ subject to right-semicircle $x\ge 0,x^2+y^2\le 1$ Maximum of:
$$
xy+y^2
$$
Domain:
$$
x \ge 0, x^2+y^2 \le1
$$
I know that the result is:
$$
\frac{1}{2}+\frac{1}{\sqrt{2}}
$$
for
$$
(x,y)=\left(\frac{1}{\sqrt{2(2+\sqrt{2})}},\frac{\sqrt{2+\sqrt{2}}}{2}\right)
$$
But I don't know how to get this resu... | Hint$$x^2+y^2=x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 $$
Now notice $$x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 \ge 2(3-2\sqrt{2})^{\frac{1}{2}}xy+(2\sqrt{2}-2)y^2 (\because \text{AM-GM})$$Now note $(\sqrt{2}-1)^2=3-2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Given $\tan a = -7/24$ in $2$nd quadrant and $\cot b = 3/4$ in $3$rd quadrant find $\sin (a + b)$. Say $\tan a = -7/24$ (second quadrant) and $\cot b = 3/4$ (third quadrant), how would I find $\sin (a + b)$?
I figured I could solve for the $\sin/\cos$ of $a$ & $b$, and use the add/sub identities, but I got massive unwi... | $\tan(a) = -7/24$
Opposite side $= 7$ and adjacent side $= 24$
Pythagorean theorem
$\Rightarrow$ hypotenuse $= \sqrt{49+576} = 25$
$\sin(a) = 7/25$ (sin is positive in second quadrant)
$\cos(a) = - 24/25$ (cos is negative in second quadrant)
$\cot(b) = 3/4 \Rightarrow \tan(b) = 4/3$
Opposite side $= 4$ and adjace... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765720",
"timestamp": "2023-03-29T00:00:00",
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Prove that the sum of the squares of two odd integers cannot be the square of an integer. Prove that the sum of the squares of two odd integers cannot be the square of an integer.
My method:
Assume to the contrary that the sum of the squares of two odd integers can be the square of an integer. Suppose that $x, y, z \in... | Let $a=2n+1$, $b=2m+1$. Then $a^2 + b^2=4n^2 + 4n +4m^2 +4m+2$. This is divisible by $2$, a prime number, but not by $4=2^2$. Hence it cannot be the square of an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Finding the Exponential of a Matrix that is not Diagonalizable Consider the $3 \times 3$ matrix
$$A =
\begin{pmatrix}
1 & 1 & 2 \\
0 & 1 & -4 \\
0 & 0 & 1
\end{pmatrix}.$$
I am trying to find $e^{At}$.
The only tool I have to find the exponential of a matrix is to diagonalize it. $A$'s eigenvalue is 1. Therefore, ... | this is my first answer on this site so if anyone can help to improve the quality of this answer, thanks in advance.
That said, let us get to business.
*
*Compute the Jordan form of this matrix, you can do it by hand or check this link. (or both).
*Now, we have the following case: $$ A = S J S^{-1}.$$ You will fin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 3
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Calculation of total number of real ordered pairs $(x,y)$
Calculation of total number of real ordered pairs $(x,y)$
in $x^2-4x+2=\sin^2 y$ and $x^2+y^2\leq 3$
$\bf{My\; Try::}$ Given $x^2-4x+2=x^2-4x+4-2=\sin^2 y\Rightarrow (x-2)^2-2=\sin^2 y$
Now Using $0 \leq \sin^2 y\leq 1$. So we get $0\leq (x-2)^2-2\leq 1\Righta... |
Now How can I solve it after that
I don't know how to continue from that. So, let us take another approach.
Solving $x^2-4x+2-\sin^2 y=0$ for $x$ gives
$$x=2\pm\sqrt{2+\sin^2y}$$
Now $x=2+\sqrt{2+\sin^2y}$ does not satisfy $x^2\le 3$. So, we have $x=2-\sqrt{2+\sin^2y}$.
Now
$$\left(2-\sqrt{2+\sin^2y}\right)^2+y^2\le ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Curious inequality: $(1+a)(1+a+b)\geq\sqrt{27ab}$ I was recently trying to play with mean inequalities and Jensen inequality. My question is, if the following relation holds for any positive real numbers $a$ and $b$
$$(1+a)(1+a+b)\geq\sqrt{27ab}$$
and if it does, then how to prove it. By AM-GM inequality we could obtai... | Use AM-GM:
$$\frac{1}{2}+\frac{1}{2}+a\ge3\sqrt[3]{\frac{a}{4}}\\
\frac{1}{2}+\frac{1}{2}+a+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}\ge6\sqrt[6]{\frac{ab^3}{108}}\\\therefore(1+a)(1+a+b)\ge\left(3\sqrt[3]{\frac{a}{4}}\right)\left(6\sqrt[6]{\frac{ab^3}{108}}\right)=\sqrt{27ab}$$
Equality holds iff $\frac{1}{2}=a=\frac{b}{3}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of squares of integers divisible by 3 Suppose that $n$ is a sum of squares of three integers divisible by $3$. Prove that it is also a sum of squares of three integers not divisible by $3$.
From the condition, $n=(3a)^2+(3b)^2+(3c)^2=9(a^2+b^2+c^2)$. As long as the three numbers inside are divisible by $3$, we can ... | Induction on the power of $9$ dividing the number. Begin with any number not divisible by $9,$ although it is allowed to be divisible by $3.$ The hypothesis at this stage is just that this number is the sum of three squares, say $n = a^2 + b^2 + c^2.$ Since this $n$ is not divisible by $9,$ it follows that at least one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Calculate $ \lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx}$ My attempt:
\begin{align*}
\lim_{x\to0}\frac{\ln(1+x+x^2+\dots +x^n)}{nx} &= (\frac{\ln1}{0}) \text{ (we apply L'Hopital's rule)} \\
&= \lim_{x \to0}\frac{\frac{nx^{n-1}+(n-1)x^{n-2}+\dots+2x+1}{x^n+x^{n-1}+\dots+1}}{n} \\
&= \lim_{x \to0}\frac{nx^{n-1}+(n-1)x... | Besides using L'Hospital's Rule,
By the definition of derivative,
$\displaystyle\lim \limits_{x\to0}\frac{\ln(1+x+x^2+...+x^n)}{nx}=\lim \limits_{x\to0} \frac{\ln(1+x+x^2+...+x^n)-\ln1}{n(x-0)}=\left.\frac{1}{n}\frac{d}{dx}(ln(1+x+x^2+...+x^n))\right|_{x=0}$
$=\displaystyle\left.\frac{1}{n}\frac{1+2x+...+nx^{n-1}}{1+x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integral solutions of the equation $x^n+y^n+z^n=2016$
Find all integral solutions of equation
$$x^n+y^n+z^n=2016,$$
where $x,y,z,n -$ integers and $n\ge 2$
My work so far:
1) $n=2$ $$x^2+y^2+z^2=2016$$
I used wolframalpha n=2 and I received the answer to the problem (Number of integer solutions: 144)
2) ... | An approach that can sometimes help to find solutions to this type of equation is to consider the prime factors of the given integer. In this case:
$$2016 = 2^5.3^2.7 = 2^5(2^6-1) = 2^5((2.2^5)-1)$$
Hence:
$$\mathbf{2016 = 4^5 + 4^5 + (-2)^5}$$
And also (finding a solution of $x^3+y^3+z^3=252$ by trial and error or fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Integrate $\frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}$ Evaluate $$\int \frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}dx$$ I saw terms like $1+t+t^2$ in the denominator , so I thought of $t^3-1$ and then converting back into half angle but it doesn't help me.... | Let $$\displaystyle I = \int \frac{\sin ^3(x/2)}{\cos(x/2)\sqrt{\cos^3 x+\cos^2x+\cos x}}dx = \frac{1}{2}\int\frac{2\sin^2 \frac{x}{2}\cdot 2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}{2\cos^2 \frac{x}{2}\sqrt{\cos^3 x+\cos^2 x+\cos x}}dx$$
So we get $$\displaystyle I = \frac{1}{2}\int\frac{(1-\cos x)\cdot \sin x}{(1+\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Evaluate $\int\frac{\sqrt{x^2+2x-3}}{x+1}d\,x$ by trig substitution I am preparing for an exam and found this integral in a previous test. Did I do it correctly?
My attempt.
$$
\int\frac{\sqrt{x^2+2x-3}}{x+1}\,dx
$$
Complete the square of $x^2+2x-3$; I changed the integral to
$$
\int\frac{\sqrt{(x-1)^2-4}}{x+1}\,dx
$$
... |
You've made a mistake. I hope you can find it using my answer
$$\int\frac{\sqrt{x^2+2x-3}}{x+1}\space\text{d}x=\int\frac{\sqrt{(x+1)^2-4}}{x+1}\space\text{d}x=$$
Substitute $u=x+1$ and $\text{d}u=\text{d}x$:
$$\int\frac{\sqrt{u^2-4}}{u}\space\text{d}u=$$
Substitute $u=2\sec(s)$ and $\text{d}u=2\tan(s)\sec(s)\space... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$
Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$
$\bf{My\; Try::}$ I have solved Using Direct formula::
$$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{... | Using $2\sin a\sin b=\cos(a-b)-\cos(a+b)$ and $2\sin a\cos b=\sin(a+b)+\sin(a-b)$, write
$$\sin \frac{\pi}7\cdot\sin \frac{2\pi}7\cdot \sin \frac{3\pi}7 = \frac12\left(\cos\frac{\pi}7-\cos\frac{3\pi}7\right)\sin\frac{3\pi}7=\frac14\left(\sin\frac{4\pi}7+\sin\frac{2\pi}7-\sin\frac{\pi}7\right)\\=\frac14\left(\sin\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove ths sum of $\small\sqrt{x^2-2x+16}+\sqrt{y^2-14y+64}+\sqrt{x^2-16x+y^2-14y+\frac{7}{4}xy+64}\ge 11$ Let $x,y\in R$.show that
$$\color{crimson}{f(x,y)=\sqrt{x^2-2x+16}+\sqrt{y^2-14y+64} + \sqrt{x^2-16x+y^2-14y+\frac{7}{4}xy+64} \ge 11}$$
Everything I tried has failed so far.use Computer found this inequality $\c... | For convenience, we make the translation $x=2+a$ and $y=6+b$, so that the equality case is $a=b=0$. Then the expression to bound is:
$$\sqrt{(a+1)^2+15}+\sqrt{(b-1)^2+15}+\sqrt{\frac{7}{8}(a+b)^2+\frac{1}{8}(a-6)^2+\frac{1}{8}(b+6)^2} $$
Now recall the following form of Cauchy-Schwarz for $n$ nonnegative variables $x_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a$ and $b$ are roots of $x^4+x^3-1=0$, $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I have to prove that:
If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.
I tried this :
$a$ and $b$ are root of $x^4+x^3-1=0$ means :
$\begin{cases}
a^4+a^3-1=0\\
b^4+b^3-1=0
\end{cases}$
whic... | Let $a,b,c,d$ be the roots of $x^4+x^3-1=0$.
By Vieta's formula,
$$a+b+c+d=-1\quad\Rightarrow\quad c+d=-1-(a+b)\tag1$$
$$abcd=-1\quad\Rightarrow \quad cd=-\frac{1}{ab}\tag2$$
Since we have
$$a^4+a^3=1\quad\text{and}\quad b^4+b^3=1$$
we can have
$$1=(a^4+a^3)(b^4+b^3)$$
$$(ab)^4+(ab)^3(a+b+1)=1,$$
i.e.
$$a+b=\frac{1-(ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
| You could also consider that $$A(x)=\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}x^n=\sum\limits_{n=0}^∞ \frac{n(n+1)}{n!}x^n=\sum\limits_{n=0}^∞ \frac{n(n-1)+2n}{n!}x^n$$ $$A(x)=\sum\limits_{n=0}^∞ \frac{n(n-1)}{n!}x^n+2\sum\limits_{n=0}^∞ \frac{n}{n!}x^n=x^2\sum\limits_{n=0}^∞ \frac{n(n-1)}{n!}x^{n-2}+2x\sum\limits_{n=0}^∞ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Another of $\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\cdots=\frac{\pi}{4}$ type expressable in cube? Gregory and Leibniz formula
(1)
$$-\sum_{m=1}^{\infty}\frac{(-1)^m}{2m-1}=\frac{\pi}{4}$$
We found another series equivalent to (1)
This is expressed in term of square numbers
$$-\sum_{m}^{\infty}\frac{\sum_{... | A proof of mahdi's result:
$$ \sum_{k=1}^{n}k^3 = \frac{n^2(n+1)^2}{4}\tag{1}$$
$$ \sum_{k=1}^{2n}(-1)^{k+1} k^3 = -n^2(4n+3),\qquad \sum_{k=1}^{2n-1}(-1)^{k+1} k^3 = n^2(4n-3)\tag{2}$$
lead to:
$$\begin{eqnarray*}\sum_{n\geq 1}\frac{1^3-2^3+\ldots}{1^3+2^3+\ldots+n^3} &=& \sum_{m\geq 1}\frac{-(4m+3)}{(2m+1)^2}+\sum_{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Recurrent sequence limit Let $a_n$ be a sequence defined: $a_1=3; a_{n+1}=a_n^2-2$
We must find the limit: $$\lim_{n\to\infty}\frac{a_n}{a_1a_2...a_{n-1}}$$
My attempt
The sequence is increasing and does not have an upper bound. Let $b_n=\frac{a_n}{a_1a_2...a_{n-1}},n\geq2$. This sequence is decreasing(we have $b_{n+1}... | since take $a_{1}=x+\dfrac{1}{x},x>1$,then
$$a_{2}=x^2+\dfrac{1}{x^2},a_{3}=x^4+\dfrac{1}{x^4}\cdots,a_{n}=x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}$$
so we have
$$a_{1}a_{2}\cdots a_{n}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots\left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)=\dfrac{x^{2^n}-\dfrac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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Minimize $a^5+b^5+c^5+d^5+e^5 = p^4+q^4+r^4+s^4 = x^3+y^3+z^3 = m^2 + n^2$ with distinct positive integers
Find the minimum value of the following:
$$a^5+b^5+c^5+d^5+e^5 = p^4+q^4+r^4+s^4 = x^3+y^3+z^3 = m^2 + n^2$$
where all numbers are different/distinct positive integers.
I know the answer (see below), but wan... | the smallest is
76913 squares of: 263 88
76913 cubes of: 40 17 20
76913 fourth powers of: 16 3 6 10
76913 fifth powers of: 9 1 2 4 7
76913
the second smallest is
1560402 squares of: 1239 159
1560402 cubes of: 101 45 76
1560402 fourth powers of: 35 5 12 14
1560402 fifth powers of: 17 1 6 8 10
156040... | {
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Cardano's method returns incorrect answer for $x = u + v$ I'm trying to use Cardano's method to solve this equation:
$$x^3+6x=20 \tag{1}$$
As described on Wikipedia, I let $x = u + v$ and expand in $(1)$:
$$(u+v)^3+6(u+v)=20$$
$$u^3 + v^3 + (3uv+6)(u+v)-20=0 \tag{2}$$
I then let $3uv + 6 = 0$ and substitute in $(2)$:
$... | You forgot the condition $u^3v^3=-8$. The correct solution is the first you enunciated.
That said, your way of solving the system of equations $\; \begin{cases}u^3+v^3=20\\u^3v^3=-8\end{cases}\;$ is over complicated.
Just use what any high-school student knows to solve the problem of finding two numbers the sum $s$ and... | {
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Factor out (m+1) in the following so that the final answer is $\frac{(2m+1) (m+2) (m+1)} {6}$ Question: $\frac{m (m+1) (2m+1) + 6(m+1)^2}{6}$=$\frac{(2m+3)(m+2)(m+1)}{6}$
I must multiply by 6 on both sides and expand the brackets and collect like terms. I'm I correct?
Edit notes: The original problem was posed as:
$$ \... | The description as given is correct. Multiplying both sides by $6$ yields
$$(2m+1) (m+1) + 6(m+1)^2= (2m+3)+(m+2)(m+1)$$
Now what remains is to determine the value of both sides. Let $L_{m}$ be the left and $R_{m}$ be the right.
\begin{align}
L_{m} &= (2m+1)(m+1)m + 6(m+1)^2 \\
&= (2m^2 + 3m + 1)m + 6(m^2 + 2m +1) \\
... | {
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Power Diophantine equation: $(a+1)^n=a^{n+2}+(2a+1)^{n-1}$ How to solve following power Diophantine equation in positive integers with $n>1$:$$(a+1)^n=a^{n+2}+(2a+1)^{n-1}$$
| If the equation holds, then $(a+1)^n>(2a+1)^{n-1}$ and $(a+1)^n>a^{n+2}$.
Multiplying gives $(a+1)^{2n} > (2a+1)^{n-1}a^{n+2}$. This can be rewritten as $(a^2+2a+1)^n > (2a^2+a)^{n-1} a^3$. However, if $a\geq 3$, then $a^2+2a+1 < 2a^2+a$ and $a^2+2a+1 <a^3$, which gives a contradiction.
Hence $a=1$ or $a=2$. The first... | {
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Evaluation of Definite Integral
Evaluation of $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x\sin 2x \sin 3x}{x}dx$
$\bf{My\;Try::}$ Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\cos x\sin 2x \sin 3x}{x}dx = \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{2\sin 3x\cos x\sin 2x}{x}dx$
So we get $$I = \frac{1}{2}\int_... | Hint 1: $$ \int f(x) = F(X) + C \Longrightarrow \int f(ax+b) = \frac{1}{a} \cdot F(ax+b) + C$$
Hint 2: $$ \int \left(f(x) + g(x)\right) = \int f(x) + \int g(x) $$
Hint 3: $$ \int \frac{\cos x}{x} = Ci(x) + C$$
First step:
(use hint 2)
$$
4I=\int_{0}^{\frac{\pi}{2}}\frac{\cos2x-\cos6x+1-\cos 4x}{x}dx =\\
=2\int_{0}^{\fr... | {
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Mean-value Theorem $f(x)=\sqrt{x+2}; [4,6]$
Verify that the hypothesis of the mean-value theorem is satisfied for the given function on the indicated interval. Then find a suitable value for $c$ that satisfies the conclusion of the mean-value theorem.
$$f(x)=\sqrt{x+2}; [4,6]$$
So,
$$f'(x) = {1 \over 2} (x+2)^{-{1\ov... | You did everything correctly, let's solve for $c$ together.
You have
$$
\frac{1}{2} (c+2)^{-1/2} = \frac{a}{2}\\
(c+2)^{-1/2} = a \\
\frac{1}{\sqrt{c+2}} = a \\
c+2 = \frac{1}{a^2}
$$
so
$$
\begin{split}
c &= \frac{1}{a^2} - 2 = \frac{1}{\left(2 \sqrt{2} - \sqrt{6}\right)^2} - 2 \\
&= \frac{1}{8 + 6 - 4\sqrt{2}\sqrt{... | {
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find the maximum of the function $f(x)=a+b\sqrt{2}\sin{x}+c\sin{2x}$ let $a,b,c\in R$,and such $a^2+b^2+c^2=100$, find the maximum value and minimum value of the function
$$f(x)=a+b\sqrt{2}\sin{x}+c\sin{2x},0<x<\dfrac{\pi}{2}$$
Use Cauchy-Schwarz inequality?
| Use Cauchy-Schwarz inequality:
$$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le (a^2+b^2+c^2)(1+2\sin^2x+\sin^22x)$$
$$\left(a+b\sqrt{2}\sin{x}+c\sin{2x}\right)^2\le 100\cdot(1+2\sin^2x+\sin^22x)$$
$$|a+b\sqrt{2}\sin{x}+c\sin{2x}|\le 10\cdot\sqrt{1+2\sin^2x+\sin^22x}$$
$$1\le1+2\sin^2x+\sin^22x\le \frac{13}{4}$$
Then $... | {
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Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the prob... | From the first equation (inverted),
$$\frac{a^2+a+1}a=6$$ or $$\frac{a^2+1}a=5.$$
Then squaring,
$$\frac{a^4+2a^2+1}{a^2}=25$$
or
$$\frac{a^4+a^2+1}{a^2}=24.$$
| {
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Geometrical Description of $ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2} $
The question is in an Argand Diagram, $P$ is a point represented by the complex number. Give a geometrical description of the locus of $P$ as $z$ satisfies the equation:
$$ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2... | Let the points $z_A = -1-i$ and $z_B=1+i$. Then we look for the locus of all points Pwith $z_P=z$ such that $\vert\arg \frac{z-z_A}{A-z_B} \vert = \pi$ (in other words, the angle $\angle APB=\pi$). This is the circle in the complex plane with diameter $AB$, since we know that the angle under which the segment $AB$ is ... | {
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$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n}$ converges for all $a\in \mathbb{R}$ $$\sum_{n=0}^{\infty}\frac{a^2}{(1+a^2)^n}$$
Can I just see this series as a geometric series? Since $c = \frac{1}{1+a^2}<1$, we can see this as the geometric series:
$$\sum_{n=0}^{\infty}bc^n = \sum_{n=0}^{\infty}a^2\left(\frac{1}{1+a^2}\ri... | Let $a=0$. Then the series obviously converges to $0$.
Now suppose that $a\ne 0$. Then our series is the geometric series
$$a^2+a^2r+a^2r^2+a^2r^3+\cdots,$$
where $r=\frac{1}{1+a^2}\lt 1$. Since $|r|\lt 1$, the series converges. It is probably by now a familiar fact that when $|r|\lt 1$ the series $1+r+r^2+r^3+\cdots$... | {
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Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation:
$$4 \cos x(\cos 2x+\cos 3x)+1=0$$
Applying trigonometric identity leads to
$$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$
But I can't understand what to do from here. Could some suggest how to proceed... | Thinking about the answer, we might notice that if $\theta=\frac{2\pi k}9$, then $\cos9\theta=1$. We can write this as
$$\begin{align}\cos9\theta-1&=4(4\cos^3\theta-3\cos\theta)^3-3(4\cos^3\theta-3\cos\theta)-1\\
&=(16\cos^4\theta+8\cos^3\theta-12\cos^2\theta-4\cos\theta+1)^2(\cos\theta-1)=0\end{align}$$
From this we c... | {
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A curious approximation to $\cos (\alpha/3)$ The following curious approximation
$\cos\left ( \frac{\alpha}{3} \right ) \approx \frac{1}{2}\sqrt{\frac{2\cos\alpha}{\sqrt{\cos\alpha+3}}+3}$
is accurate for an angle $\alpha$ between $0^\circ$ and $120^\circ$
In fact, for $\alpha = 90^\circ$, the result is exact.
How can ... | Close but not exact even with regard to the linear term. With $x$ and $y$ as per Paramanand Singh's answer, if $x=1-\delta$, then, ignoring $o(\delta)$ throughout,
$$\begin{align}
y&=(1-\delta)[4(1-2\delta)-3]\\
&=(1-\delta)(4-8\delta-3)\\
&=(1-\delta)(1-8\delta)\\
&=1-9\delta.
\end{align}$$
By Paramanand Singh's equat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx\stackrel?=\frac{\pi}{4}\sqrt{5\sqrt2-7}$ $$\int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx=\frac{\pi}{15}$$
$$\int_{0}^{\infty}\frac{4x^2}{[(1+(1+x^2)^2]^2}dx=\frac{\pi}{15}$$
$u=\tan(z)$ $\rightarrow$ $du=\sec^2(z)$
$u$ $\rightarrow \infty$, $\tan(z)=\frac{\pi}{2}... | Hint. One may write
$$
\begin{align}
\int_{0}^{\infty}\frac{4x^2}{(x^4+2x^2+2)^2}dx&=4\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:\frac{dx}{x^2}
\\\\&=2\sqrt{2}\int_0^{\infty}\frac1{\left(\left(x-\frac{\sqrt{2}}x\right)^2+2+2\sqrt{2}\right)^2}\:dx \quad (x \to \sqrt{2}/x)
\\\\&=... | {
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"source": "stackexchange",
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How to verify $(1+\frac{1}{n})^2(1-\frac{1}{n^2})^{n-1}\geq \exp(\frac{1}{n})$ How to verify this inequality? Assuming that $n\in \mathbb{N}^+$, we have:
$$\left(1+\frac{1}{n}\right)^2\left(1-\frac{1}{n^2}\right)^{n-1}\geq \exp\left(\frac{1}{n}\right).$$
| Consider $$A_n=\left(1+\frac{1}{n}\right)^2\times\left(1-\frac{1}{n^2}\right)^{n-1}$$ Take logarithms $$\log(A_n)=2\log\left(1+\frac{1}{n}\right)+(n-1)\log\left(1-\frac{1}{n^2}\right)$$ Now, use the Taylor series $$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+O\left(x^6\right)$$ and replace $x$ b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $(a,b)+(c,d) = (a+c,b+d)$
Let $a<b$ and $c<d$ be real numbers. Show that $(a,b)+(c,d) = (a+c,b+d)$.
I don't understand the question. Since $(a,b)$ and $(c,d)$ are intervals, what does it mean to add them?
| In terms of sets and set notation:
$(a,b)$ = all the points of R that are between a and b exclusively =$\{x\in \mathbb R| a < x < y\}$
If $A$ and $B$ are sets, than we say $A + B =\{x+y|x \in A; y \in B \}$.
So the statement $(a,b)+(c,d) = (a+b,c+d)$ means that $(a,b) + (c,d)=\{x+y|a <x <b;c <y<d\}$ is the same as $(a+... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I'm stuck in a logarithm question: $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ If $4^{y+3x} = 64$ and $\log_x(x+12)- 3 \log_x4= -1$ so $x + 2y= ?$
I've tried this far, and I'm stuck
$$\begin{align}4^{y+3x}&= 64 \\
4^{y+3x} &= 4^3 \\
y+3x &= 3 \end{align}$$
$$\begin{align}\log_x (x+12)- 3 \log_x 4 &= -1 \\
\log_x ... | You're right up to $y+3x=3$.
Now consider the other statement $\log_x(x+12)-3\log_x 4=-1$
$\log_x{x+12 \over 64 }=-1$
${x+12 \over 64 }={1 \over x}$
| {
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"source": "stackexchange",
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Let $n \in \mathbb{N}$. Proving that $13$ divides $(4^{2n+1} + 3^{n+2})$ Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $
Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13 = \bigg( ( 4^{2n} \mod 13) \cdot ( 4... | By the binomial theorem,
$$
4^{2n+1} + 3^{n+2}
=4\cdot 16^n+9\cdot 3^n
=4\cdot (13+3)^n+9\cdot 3^n
=4(13a+3^n)+9\cdot 3^n
=13(4a+3^n)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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$\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}$ when $\alpha + \beta + \gamma = \pi$ Assume: $\alpha + \beta + \gamma = \pi$ (Say, angles of a triangle)
Prove: $\sin\alpha + \sin\beta + \sin\gamma = 4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}... | You may go the other way around:
$$
\cos\frac{\gamma}{2}=\cos\frac{\pi-\alpha-\beta}{2}=
\sin\frac{\alpha+\beta}{2}=
\sin\frac{\alpha}{2}\cos\frac{\beta}{2}+
\cos\frac{\alpha}{2}\sin\frac{\beta}{2}
$$
so the right hand side becomes
$$
4\cos\frac{\alpha}{2}\cos\frac{\beta}{2}
\sin\frac{\alpha}{2}\cos\frac{\beta}{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding all pairs of integers that satisfy a bilinear Diophantine equation The problem asks to "find all pairs of integers $(x,y)$ that satisfy the equation $xy - 2x + 7y = 49$.
So far, I've got
\begin{align}
xy - 2x + 7y &= 49 \\
x\left(y - 2\right) + 7 &= 49
\\
y &\leq 49
\end{align}
I can't get any further. Any hel... | hint: $xy+7y = 2x+49 \implies (x+7)y = 2x+49 \implies y = \dfrac{2x+49}{x+7}= 2 + \dfrac{35}{x+7}\implies (x+7) \mid 35\implies x+7 = \pm 1, \pm 5, \pm 7, \pm 35$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Condition on $a$ for $(x^2+x)^2+a(x^2+x)+4=0$ Find the set of values of $a$ if $$(x^2+x)^2+a(x^2+x)+4=0$$ has
$(i)$ All four real and distinct roots
$(ii)$ Four roots in which only two roots are real and distinct.
$(iii)$ All four imaginary roots
$(iv)$ Four real roots in which only two are equal.
Now if I set $x^2+x=t... | Alternatively, graphing it:
Let $a=y$, then $(x^2+x)^2+y(x^2+x)+4=0$
$\displaystyle y=\frac{1}{4}-\left( x+\frac{1}{2} \right)^{2}+
\frac{4}{\frac{1}{4}-\left( x+\frac{1}{2} \right)^{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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proof that diagonals of a quadrilateral are perpendicular if $AB^2+CD^2=BC^2+AD^2$ proof that diagonals of a quadrilateral are perpendicular if $AB^2+CD^2=BC^2+AD^2$.
My Attempt:we know that if diagonals of a quadrilateral are perpendicular then we have $AB^2+CD^2=BC^2+AD^2$.But have to proof opposite of it?
| Assume the intersection of the to diagonals is $O$. Let $|OA|=a,|OB|=b,|OC|=c,|OD|=d$. Assume $\angle AOB=\gamma$. Then
$$|AB|^2=a^2+b^2-2ab \cos\gamma,$$
$$|CD|^2=c^2+d^2-2cd \cos\gamma,$$
$$|BC|^2=b^2+c^2-2bc \cos(\pi-\gamma)=b^2+c^2+2bc \cos\gamma,$$
$$|AB|^2=a^2+d^2-2ad \cos(\pi-\gamma)=a^2+d^2+2ad \cos\gamma.$$
F... | {
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Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but that it has two different irreducible factors in $\mathbb{R}[X]$
Prove that $X^4+X^3+X^2+X+1$ is irreducible in $\mathbb{Q}[X]$, but it has two different irreducible factors in $\mathbb{R}[X]$.
I've tried to use the cyclotomic polynomial as:
$$X^5-1=... | let $$P(x)=x^4+x^3+x^2+x^1+1$$
We know if $x=\frac{a}{b}$ is root of $P(x)$ then $b|1\,$ , $\,a|1$. In the other words $a=\pm 1 $ and $b=\pm 1 $ but $P(1)=5$ and $P(-1)=1$, thus we let
$$P(x)=(x^2+ax+b)(x^2+cx+d)$$
as a result
\begin{align}
& bd=1 \\
& ad+bc=1 \\
& b+d+ac=1 \\
& a+c=1 \\
\end{align}
This sys... | {
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When a loop with inverse property is commutative Question
How to prove that a loop $L$ with inverse property and $x^3=e$ for all $x$ is commutative iff $(x y)^2=x^2 y^2$ for all $x,y$?
Definitions:
A loop is a quasigroup with identity $e$.
$L$ has the inverse property if every element has a two sided inverse and $x... | Note that $x^3 = e$ means that $x^2 = x^{-1}$. Similarly, this means that $(xy)^2 = (xy)^{-1}, y^2 = y^{-1}$. We now write our condition as
$$
(xy)^{-1} = x^{-1}y^{-1}
$$
This implies that $(x^{-1}y^{-1})^{-1} = xy$.
We now apply the inverse property repeatedly:
$$
x^{-1} = (x^{-1}y^{-1})y\\
(x^{-1}y^{-1})^{-1}x^{-1}... | {
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"answer_id": 0
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Prove that as $PP'$ varies,the circle generates the surface $(x^2+y^2+z^2)(\frac{x^2}{a^2}+\frac{y^2}{b^2})=x^2+y^2.$ $POP'$ is a variable diameter of the ellipse $z=0,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ and a circle is described in the plane $PP'ZZ'$ on $PP'$ as diameter.Prove that as $PP'$ varies,the circle generates... | The circle has two endpoints $(-X,-Y)$ and $(X,Y)$ with $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1$ and is perpendicular to the $z$-plane.
Let $(x,y,z)$ be a point on this circle.
Then we have:
$x^2+y^2+z^2=X^2+Y^2$ and $y/x=Y/X$.
So let $Y/y=X/x=c$.
We get: $c^2\left (\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} \right)=1$
as well as:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 ... | By the double angle formula we get $$\tan(2x)+\tan(x)=\frac{2\tan(x)}{1-\tan^2(x)}+\tan(x)=\frac{3-\tan^2(x)}{1-\tan^2(x)}\tan(x),$$ so that $\tan(x)=0$ is certainly a solution.
But $\tan(x)=\pm\sqrt3$ as well, so that the initial claim is false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Sum $1+(1+x)^2+(1+x+x^2)^2+\cdots+(1+x+x^2+\cdots+x^{N-2})^2 $ Is there a way to find the sum the following series:
$$1+(1+x)^2+(1+x+x^2)^2+\cdots+(1+x+x^2+\cdots+x^{N-2})^2 \text{ ?}$$
Any ideas ? Perhaps someone knows already the result..
Thank you in advance for your time.
| $$
1+(1+x)^2+(1+x+x^2)^2+\cdots+(1+x+x^2+...+x^{N-2})^2 = \sum_{i=0}^{N-2}(1+x+\cdots + x^i)^2$$
Let $x<1$, though same can be repeated for $x>1$ - we do not consider $x=1$ since the answer is straightforward in this case
$$\sum_{i=0}^{N-2}(1+x+\cdots + x^i)^2 = \sum_{i=0}^{N-2}\left(\frac{1-x^{i+1}}{1-x}\right)^2 = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{dy}{dx} = -\frac1{(1+x)^2}$ for given that $x\sqrt{1+y} + y\sqrt{1+x} = 0$ $$x\sqrt{1+y} + y\sqrt{1+x} = 0$$
Please tell me where I went wrong. Why I am not getting correct answer ?
| There is nothing wrong. Put the value of $y$ to get your result.
However, a simpler approach:
$$x\sqrt{1+y} + y\sqrt{1+x} = 0$$
$$x\sqrt{1+y} = - y\sqrt{1+x}$$
Squarring both sides, we get
$$x^2(1+y) = y^2(1+x)$$
$$x^2(1+y) - y^2(1+x)=0$$
$$(x-y)(x+y+xy)=0$$
So either $x-y=0$ or $x+y+xy=0$.
Now if, $x-y=0$, then we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the second derivative of $f(x) = \frac{4x}{x^2-4}$. What is the second derivative of
$$f(x) = \frac{4x}{x^2-4}?$$ I have tried to use the quotient rule but I can't seem to get the answer.
| Avoiding the quotient rule, just for an option:
$$\begin{align}
\ln(f(x))
&=\ln(4)+\ln(x)-\ln(x+2)-\ln(x-2)\\
\frac{f'(x)}{f(x)}
&=x^{-1}-(x+2)^{-1}-(x-2)^{-1}\\
f'(x)
&=f(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)\\
f''(x)
&=f(x)\left(-x^{-2}+(x+2)^{-2}+(x-2)^{-2}\right)+f'(x)\left(x^{-1}-(x+2)^{-1}-(x-2)^{-1}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x... | $$\begin{align}
S&=1+2q+3q^2+\qquad\cdots\qquad \qquad+nq^{n-1}\\
qS&=\qquad q+2q^2+3q^3+\cdots +\quad(n-1)q^{n-1}+nq^n \\
\text{Subtracting,}&\\
(1-q)S&=1+\;\ q \ +\ q^2 +\ q^3+\cdots \qquad \qquad +q^{n-1}-nq^n\\
&=\frac {\;\ 1-q^n}{1-q}-nq^n\\
S&=\frac{1-q^n-nq^n(1-q)}{(1-q)^2}\\
&=\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
What is the sum of the first $17$ terms of an arithmetic sequence if $a_9=35$? What is the sum of the first $17$ terms of an arithmetic sequence if $a_9=35$?
This is what I did: $a_9=a_1+8d=35$
$S_{17}=\frac{17}{2}(a_1+a_{17})=\frac{17}{2}(a_1+a_1+16d)=\frac{17}{2}(2a_1+16d)=\frac{17}{2}\cdot 70= 595$
This solution i... | Dang, @gt6989b is right:
\begin{align}
a_1 + a_{17} &= a_1 + (a_1 + 16 d) = 2 a_1 + 16 d = 2(a_1 + 8 d) = 2 a_9 \\
a_2 + a_{16} &= (a_1 + d) + (a_1 + 15 d) = 2 a_1 + 16 d = 2 a_9 \\
& \vdots \\
a_8 + a_{10} &= (a_1 + 7d) + (a_1 + 9 d) = 2 a_1 + 16 d = 2 a_9
\end{align}
so
$$
\sum_{i=1}^{17} a_i = 8 \cdot 2 a_9 + a_9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx $ $$\int_0^\infty \frac{ x^{1/3}}{(x+a)(x+b)} dx$$ where $a>b>0$
What shall I do?
I have diffucty when I meet multi value function.
| Let us assume $A^3=a, B^3=b$, for simplicity. Now make a substitution $x=t^3$ which will transform the integral like this $$I=\int_{0}^\infty \frac {3t^3dt}{(t^3+A^3)(t^3+B^3)}$$ Now break this into partial fractions like this
$$I=3\int_{0}^\infty [\frac {1}{(t^3+B^3)}-\frac {A^3}{B^3-A^3}(\frac {1}{t^3+A^3}-\frac {1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find stationary points of the function $f(x,y) = (y^2-x^4)(x^2+y^2-20)$ I have problem in finding some of the stationary points of the function above. I proceeded in this way: the gradient of the function is:
$$ \nabla f = \left( xy^2-3x^5-2x^3y^2+40x^3 ; x^2y+2y^3-x^4y-20y \right) $$
So in order to find the stationary... | WA gets
$$
\DeclareMathOperator{grad}{grad}
\grad((x^2+y^2-20) (y^2-x^4)) = (-6 x^5-4 x^3 (y^2-20)+2 x y^2, 2 y (-x^4+x^2+2 y^2-20))
$$
(link) and nine real critical points (link).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A sum of squared binomial coefficients I've been wondering how to work out the compact form of the following.
$$\sum^{50}_{k=1}\binom{101}{2k+1}^{2}$$
| $$\begin{align}\sum_{k=0}^m \binom {2m+1}{2k+1}^2
&=\sum_{k=0}^m \binom {2m+1}{2k+1}\binom {2m+1}{2m-2k}
\color{lightgrey}{=\sum_{j=0}^m\binom {2m+1}{2(m-j)+1}\binom {2m+1}{2j}\quad \scriptsize (j=m-k)}\\
&=\frac 12 \sum_{k=0}^m \binom {2m+1}{2k}\binom {2m+1}{2(m-k)+1}+\binom {2m+1}{2k+1}\binom{2m+1}{2m-2k}\\
&=\frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}$ if $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$? If $$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 1$$ then find the values of $$\frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b}.$$ How can I solve it? Please help me. Thank you in ad... | HINT:
$$\dfrac{a^2}{b+c}+a=\dfrac{a(a+b+c)}{b+c}$$
$$\sum_{\text{cyc}}\left(\dfrac{a^2}{b+c}+a\right)=(a+b+c)\sum_{\text{cyc}}\dfrac a{b+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Limit of the minimum value of an integral
Let $$f(a)=\frac{1}{2}\int_{0}^{1}|ax^n-1|dx+\frac{1}{2}$$ Here $n$ is a natural number. Let $b_n$ be the minimum value of $f(a)$ for $a>1$. Evaluate $$\lim_{m \to \infty}b_mb_{m+1}\ldots b_{2m}$$
Some starters please. Thanks.
| $$\begin{eqnarray*}f(a) = \frac{a}{2}\int_{0}^{1}\left| x^n-\frac{1}{a}\right|\,dx+\frac{1}{2}&=&\frac{1}{2}+\frac{a}{2}\int_{0}^{1}(x^n-1/a)\,dx+a\int_{0}^{\frac{1}{\sqrt[n]{a}}}\left(\frac{1}{a}-x^n\right)\,dx\\&=&\frac{1}{2}+\frac{a}{2n+2}-\frac{1}{2}+\frac{1}{\sqrt[n]{a}}-\frac{1}{(n+1)\sqrt[n]{a}}\\&=&\frac{a}{2n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | $$
\begin{aligned}
I & =\int_0^{\infty} \frac{\frac{1}{x^2}}{x^2+\frac{1}{x^2}+1} d x \\
& =\frac{1}{2} \int_0^{\infty} \frac{\left(1+\frac{1}{x^2}\right)-\left(1-\frac{1}{x^2}\right)}{x^2+\frac{1}{x^2}+1} d x \\
& =\frac{1}{2}\left[\int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+3}-\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 7
} |
How do I find the Integral of $\sqrt{r^2-x^2}$? How can I find the integral of the following function using polar coordinates ?
$$f(x)=\sqrt{r^2-x^2}$$
Thanks!
| $\displaystyle\int \sqrt{r^2-x^2}dx$
Let be $\;x=r.\sin\alpha$ or $\quad x=r.\cos\alpha$,
Let be $\;x=r.\sin\alpha$,
and $\quad dx=r.\cos \alpha \;d\alpha$
Integral be,
$\displaystyle\int \sqrt{r^2-x^2}dx=\displaystyle\int r.\sqrt{1-\sin^2\alpha}\;.r.\cos \alpha \;d\alpha=\displaystyle\int r^2.\cos^2\alpha\; d\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$5^{th}$ degree polynomial expression
$p(x)$ is a $5$ degree polynomial such that
$p(1)=1,p(2)=1,p(3)=2,p(4)=3,p(5)=5,p(6)=8,$ then $p(7)$
$\bf{My\; Try::}$ Here We can not write the given polynomial as $p(x)=x$
and $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ for a very complex system of equation,
plz hel me how can i solve that... | hint : write the polynomial in this form $$f(x)= a(x-1)(x-2)(x-3)(x-4)(x-5)+b(x-1)(x-2)(x-3)(x-4)(x-6) +c(x-1)(x-2)(x-3)(x-5)(x-6)+d(x-1)(x-2)(x-4)(x-5)(x-6)+e(x-1)(x-3)(x-4)(x-5)(x-6)+f(x-2)(x-3)(x-4)(x-5)(x-6)$$ now finding constants are easy
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
How to solve asymptotic expansion: $\sqrt{1-2x+x^2+o(x^3)}$ Determinate the best asymptotic expansion for $x \to 0$ for:
$$\sqrt{1-2x+x^2+o(x^3)}$$
How should I procede?
In other exercise I never had the $o(x^3)$ in the equation but was the maximum order to consider.
| You have the following asymptotic expansion :
$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+o(x^3)$$
So :
$$\sqrt{1-2x+x^2+o(x^3)}=1+\frac{-2x+x^2+o(x^3)}{2}-\frac{(-2x+x^2+o(x^3))^2}{8}+\frac{(-2x+x^2+o(x^3))^3}{16}+o((-2x+x^2+o(x^3))^3)\\=1+\frac{-2x+x^2}{2}-\frac{4x^2-4x^3+x^4}{8}+\frac{-8x^3+12x^4-6x^5+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Matrix with orthonormal base I have the two following given vectors:
$\vec{v_{1} }=\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$
$\vec{v_{2} }=\begin{pmatrix} 3 \\ 0 \\ -3 \end{pmatrix} $
I have to calculate matrix $B$ so that these vectors in $\mathbb{R}^{3}$ construct an orthonormal basis.
The solution is:
$$B=\begin... | Maybe these calculations would help you.
We need to find vertor $\vec{v}_3$ such that $\vec{v}_3\perp\vec{v}_1$ and $\vec{v}_3\perp \vec{v}_2$, i.e.
$$
\begin{cases}
(\vec{v}_1, \vec{v}_3) = 0, \\
(\vec{v}_2, \vec{v}_3) = 0.
\end{cases}
$$
Here $(\vec{x},\vec{y})$ is a scalar product of vectors $\vec{x}$ and $\vec{y}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $\sin(\pi \cos\theta) = \cos(\pi\sin\theta)$, then show ........ If $\sin(\pi\cos\theta) = \cos(\pi\sin\theta)$,
then show that $\sin2\theta = \pm 3/4$.
I can do it simply by equating $\pi - \pi\cos\theta$ to $\pi\sin\theta$,
but that would be technically wrong as those angles could be in different quadrants. So how... | We first rewrite
$$
\cos(\pi/2 - \pi \cos \theta) = \cos(\pi \sin \theta)
$$
(cofunction identity).
We notice that $\cos(x)$ is a periodic function with period $2\pi$, so we need a period offset term to be sure that we find all solutions. We also need to account for the fact that $\cos(x)$ is symmetric, so:
$$
\cos(\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$then.. If $f(x) = \frac{\cos x + 5\cos 3x + \cos 5x}{\cos 6x + 6\cos4x + 15\cos2x + 10}$ then find the value of $f(0) + f'(0) + f''(0)$.
I tried differentiating the given. But it is getting too long and complicated. So there must be a way... | we can simplify the fraction as $$\frac{2\cos3x\cos2x+5\cos3x}{2\cos^23x-1+6[2\cos3x\cos x]+9\cos2x+10}$$
$$=\frac{(2\cos2x+5)\cos3x}{2\cos^23x+12\cos3x\cos x+18\cos^2x}$$
$$=\frac{(2\cos2x+5)\cos3x}{2(\cos3x+3\cos x)^2}$$
$$=\frac{[2(2c^2-1)+5](4c^3-3c)}{2(4c^3)^2}$$
$$=\frac{(4c^2+3)(4c^2-3)}{32c^5}$$
$$=\frac 12\sec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Determining a basis for a space of polynomials. Let $V = \mathbb R[x]_{\le 3}$
I have the space of polynomials $U_2 = \{ p = a_0 + a_1x + a_2x^2 + a_3x^3 \in V \mid a_1 - a_2 + a_3 = 0, a_0 = a_1 \}$
I am asked to find a basis, so I proceed by noticing that in $U_2$:
$$a_0 + a_1x + a_2x^2 + a_3x^3 = a_0 (1+x-x^3) + a... | These are both bases. Your basis is $\{1+x-x^3, x^2+x^3\}$; the solution gives the basis $\{1+x+x^2, -1-x+x^3\}$. But each of these is expressible in terms of the other:
\begin{gather*}
1+x+x^2 = (1+x-x^3) + (x^2+x^3),\quad -1-x+x^3 = -(1+x-x^3) \\
1+x-x^3 = -(-1-x+x^3),\quad x^2+x^3 = (1+x+x^2) + (-1-x+x^3).
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I solve this inequality? Have a nice day, how can I solve this inequality?
$$a<b<-1$$
$$ |ax - b| \le |bx-a|$$
what is the solution set for this inequality
| There may be more efficient ways to do these but for clarity I like to break absolute values into cases:
Case 1: $ax -b \ge 0$ and $bx -a \ge 0$.
[This implies $ax \ge b\implies x \le b/a$ and likewise $x \le a/b$ so $x \le \min(a/b,b/a) = b/a < 1$. Let's keep in mind $b/a < 1 < a/b$]
Then $|ax - b| \le |bx -a| \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}... | Elaborating user @Dr. MV's answer, we have
\begin{equation}
\int_0^\infty\frac{1}{a^2x^4+bx^2+c^2}\ dx=\frac{c\pi}{2a\sqrt{b+2ac}}
\end{equation}
Putting $a=1$, $b=a$, and $c^2=b$, then
\begin{equation}
I(a,b)=\int_0^\infty\frac{1}{x^4+ax^2+b}\ dx=\frac{\pi}{2}\sqrt{\frac{b}{a+2\sqrt{b}}}
\end{equation}
Hence
\begin{eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
prove inequation a,b,c,d $\in \mathbb{R}$ $a,b,c,d \gt 0$ and $ c^2 +d^2=(a^2 +b^2)^3$ prove that $$ \frac{a^3}{c} + \frac{b^3}{d} \ge 1$$
If I rewrite the inequation like $ \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}$ and manage to simplfy it brings me nowhere. I try with
Cauchy-Schwarz Inequa... | Using Titu's Lemma, we have
$$ \dfrac{a^3}{c} + \dfrac{b^3}{d} \ge \dfrac{(a^2+b^2)^2}{ac+bd}$$
So, we are left to prove that
$$ (a^2+b^2)^2 \geq (ac+bd)\tag{1} $$
Using Cauchy-Schwarz inequality, we have
$$(a^2+b^2)(c^2+d^2) \geq (ac+bd)^2\tag{2}$$
Using the given proposition, $$c^2+d^2 =(a^2+b^2)^3$$ in $(2)$ and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $1... | Helping out with the problem.
I'm stuck at the basis step.
$p(n)$: $3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \dfrac{3(5^{n+1} -1)}{4}$. Where $n \in \{0, 1, 2, \dots \}$.
We can rewrite the predicate. $p(n)$: $\sum_0^n3\cdot5^n = \dfrac{3(5^{n+1} -1)}{4}$
Base case:
Here you need to start at 0 because we onl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ such that $Q(x)|P(x)$, find $a+b$
Let $P(x)=ax^{2014}-bx^{2015}+1$ and $Q(x)=x^2-2x+1$ be the polynomials where $a$ and $b$ are real numbers. If polynomial $P$ is divisible by $Q$, what is the value of $a+b$.
This is what I have tried so far: Since $Q(x)|P(x)$ we h... | Since $Q$ divides $P$, we have that $x-1$ divides $P$ twice. Hence $1$ is a root of $P$ and of its derivative $P'(x)=2014ax^{2013}$. So $a-b^{2015}+1=0$ and $2014a=0$. Now we can solve to get $a=0$, but in that case $Q$ cannot divide $P$, since $P$ has degree $0$.
If we actually have $P(x)=ax^{2015}-bx^{2014}+1$ then i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how t... | Because $$\sum_{cyc}\left(\frac{1}{a^2}-\frac{1}{ab}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{1}{a}-\frac{1}{b}\right)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?
| Multiply the inequality by $\;(x+2)^2>0\;$ ( obviously, $\;x\neq-2\;$) :
$$x(x+2)^2<x+2\iff x^3+4x^2+3x-2<0\iff$$
$$\iff (x+2)(x+1-\sqrt2)(x+1+\sqrt2)<0\iff \color{red}{x<-1-\sqrt 2}\;\;\text{or}\;\color{red}{-2<x<-1+\sqrt2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$ Prove that $$\sum_{n=1}^{\infty }\frac{B_{2n}}{(2n-1)!}=\frac{1}{2}-\frac{1}{(e-1)^2}$$
My idea is to find the Taylor series of $\frac{1}{(e^x-1)^2}$, but it seems not useful.
Any helps, thanks
| An alternative approach is to use the integral representation
$$ B_{2n} = (-1)^{n}4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} \, dx.$$
Specifically,
$$ \begin{align}\sum_{n=1}^{\infty} \frac{B_{2n}}{(2n-1)!} &= \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} (-1)^{n-1} 4n \int_{0}^{\infty} \frac{x^{2n-1}}{e^{2 \pi x}-1} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\le... | $\sin A=0\implies A=n\pi$
$\cos B=0\implies B=(2m+1)\pi/2$
$m,n$ are arbitrary integers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
If $x\in \left(0,\frac{\pi}{4}\right)$ then $\frac{\cos x}{(\sin^2 x)(\cos x-\sin x)}>8$
If $\displaystyle x\in \left(0,\frac{\pi}{4}\right)\;,$ Then prove that $\displaystyle \frac{\cos x}{\sin^2 x(\cos x-\sin x)}>8$
$\bf{My\; Try::}$ Let $$f(x) = \frac{\cos x}{\sin^2 x(\cos x-\sin x)}=\frac{\sec^2 x}{\tan^2 x(1-\ta... | We are to prove that $$\frac{\cos x}{\sin x (\cos x - \sin x)}> 8 \sin x $$
By Cauchy-Schwarz Inequality, since all quantities involved are positive
$$\bf{LHS = }\frac{1}{\cos x} + \frac{1}{\cos x-\sin x} \ge \frac{4}{\sin x + \cos x - \sin x} = \frac{4}{\sin x}$$
For the given range of x, we have $1> 2 \sin^2 x$
So $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Need a solution to this Integration problem How to evaluate:$\displaystyle\int_{0}^{r}\frac{x^4}{(x^2+y^2)^{\frac{3}{2}}}dx$
I have tried substituting $x =y\tan\ A$, but failed.
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to find $\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$ How to find ?$$\int \frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}} dx$$
I tried using the substitution $x^2=z$.But that did not help much.
| By setting $x^2=z$ we are left with:
$$ \frac{1}{2}\int\frac{z-1}{z^2\sqrt{2z^2-2z+1}}\,dz=C+\frac{\sqrt{2z^2-2z+1}}{2z}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Finding the coefficient of $x^{50}$ in $\frac{(x-3)}{(x^2-3x+2)}$ First, the given answer is: $$-2 + (\frac{1}{2})^{51}$$
I have tried solving the problem as such:
$$[x^{50}]\frac{(x-3)}{(x^2-3x+2)} = [x^{50}]\frac{2}{x-1} + [x^{50}]\frac{-1}{x-2}$$
$$ = 2[x^{50}](x-1)^{-1} - [x^{50}](x-2)^{-1}$$
$$=2\binom{-1}{50}-\... | The beginning looks good, but I do not see how you justify the last line. I would use the geometric series instead:
$$\begin{align*}\frac{x-3}{x^2-3x+2} &= \frac{2}{x-1} - \frac{1}{x-2}\\
&= -2\frac{1}{1-x} + \frac{1}{2}\frac{1}{1-\frac 12 x} \\
& = -2 \sum_{n=0}^\infty x^n + \frac{1}{2}\sum_{n=0}^\infty \frac{x^n}{2^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the range of $λ$? Suppose $a, b, c$ are the sides of a triangle and no two of them are equal.
Let $λ ∈ IR$. If the roots of the equation $x^
2 + 2(a + b + c)x + 3λ(ab +
bc + ca) = 0$ are real, then what is the range of $λ$?
I got that $$λ ≤\frac{
(a + b + c)^
2}
{3(ab + bc + ca)}$$
After that what to do?
| For a triangle with sides $a,b,c$ by triangle inequality, we have
$$|a-b|<c$$
Squaring both sides we get,
$$(a-b)^2<c^2\tag{1}$$
Similarly,
$$(b-c)^2<a^2\tag{2}$$
And
$$(c-a)^2<b^2\tag{3}$$
Adding $(1),(2)$ and $(3)$, we get
$$a^2+b^2+c^2 <2(ab+bc+ca) \Longleftrightarrow (a+b+c)^2 <4(ab+bc+ca)\tag{4}$$
From $(4)$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Series with digammas (Inspired by a comment in answer https://math.stackexchange.com/a/699264/442.)
corrected
Let $\Psi(x) = \Gamma'(x)/\Gamma(x)$ be the digamma function. Show
$$
\sum_{n=1}^\infty (-1)^n\left(\Psi\left(\frac{n+1}{2}\right)
-\Psi\left(\frac{n}{2}\right)\right) = -1
$$
As noted, it agrees to many de... | Using the integral representation $$\psi(s+1) = -\gamma +\int_{0}^{1} \frac{1-x^{s}}{1-x} \, dx ,$$ we get
$$ \begin{align} \sum_{n=1}^{\infty} (-1)^{n} \left(\psi \left(\frac{n}{2} \right)- \psi \left(\frac{n+1}{2}\right) \right) &= \sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{1} \frac{x^{(n+1)/2-1} - x^{n/2-1}}{1-x} \, dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to turn the reflection about $y=x$ into a rotation. If we reflect $(x,y)$ about $y=x$ then we get $(y,x)$. And because $x^2+y^2=y^2+x^2$ this can also be represented by a rotation.
Using this we get:
$$(x,y)•(y,x)=2xy=(x^2+y^2)\cos (\theta)$$
Hence
$\theta=\arccos (\frac{2xy}{x^2+y^2})$
So using complex numbers w... | If $\theta$ is the angle between the x-axis and the line from 0 to $(x,y)$, then $\theta = \arccos(\frac{y}{\sqrt{x^2+y^2}})$. We reflect about a line with angle $\frac{\pi}{4}$, so the angle between $(x,y)$ and the line $x=y$ is $\frac{\pi}{4} - \theta$. The angle between $(x,y)$ and the reflected point will be doub... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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A rectangle with perimeter of 100 has area at least of 500, within what bounds must the length of the rectangle lie? Problem
The problem states that there is a rectangle that has a perimeter of $100$ and an area of at least $500$ and it asks for the bounds of the length
which can be given in interval notation or in th... | You are correct that $x + y = 50$ and that $xy \geq 500$. We can solve the inequality by completing the square.
\begin{align*}
xy & \geq 500\\
x(50 - x) & \geq 500\\
50x - x^2 & \geq 500\\
0 & \geq x^2 - 50x + 500\\
0 & \geq (x^2 - 50x) + 500\\
0 & \geq (x^2 - 50x + 625) - 625 + 500\\
0 & \geq (x - 25)^2 - 125\\
125 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is $77!$ divisible by $77^7$?
Can $77!$ be divided by $77^7$?
Attempt:
Yes, because $77=11\times 7$ and $77^7=11^7\times 7^7$ so all I need is that the prime factorization of $77!$ contains $\color{green}{11^7}\times\color{blue} {7^7}$ and it does.
$$77!=77\times...\times66\times...\times55\times...\times44\times...\... | If $p$ is a prime number, the largest number $n$ such that $p^n \mid N!$ is
$\displaystyle
n = \sum_{i=1}^\infty \left \lfloor \dfrac{N}{p^i}\right \rfloor$.
Note that this is really a finite series since, from some point on, all of the $\left \lfloor \dfrac{N}{p^i}\right \rfloor$ are going to be $0$. There is also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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How to find the union&intersection of two lines by their equations? I will try to be as clear as possible concerning my confusion, and I will use some examples(several ones).
Case number 1.
Assume two equations(in cartesian form) of two planes.
$2x+2y-5z+2=0$ and $x-y+z=0$
Now,we need to find their vectors.
For the fir... | First of all: $Ax+By+Cz+D=0$ is plane equation.
Case 1:
Intersection of to planes is line. To find equation of that line you have to solve system of equations:
$$
2x+2y-5z+2=0\\
x-y+z=0 \Rightarrow x=y-z \\
$$
If we substitute second equation into first we got
$$
2(y-z)+2y-5z+2=0 \Rightarrow 4y-7z+2=0 \Rightarrow y=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Double integration over a general region $\iint x^2 +2y$ bound by $y=x$ $y=x^3$ $x \geq 0$
this is either a type I or type II since the bounds are already nicely given for a type I, I integrated it as a type I:
Finding the bounds:
$x^3=x \to x^3-x=0 \to x(x^{2}-1)= 0 \to x=0, x=\pm1$
Since $-1\lt 0$ my bounds for $x$... | Note that $x^3\lt x$ in the interval $(0,1)$. (A picture always helps in this kind of problem.)
So $y$ travels from $y=x^3$ to $y=x$.
One can see without checking details that the answer $-\frac{4}{21}$ cannot be right. Your integrand is positive in the region, so the answer must be positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $2^{2a+1}+2^a+1$ is not a perfect square given $a\ge5$ I am attempting to solve the following problem:
Prove that $2^{2a+1}+2^a+1$ is not a perfect square for every integer $a\ge5$.
I found that the expression is a perfect square for $a=0$ and $4$. But until now I cannot coherently prove that there are n... | I will assume that $a \ge 1$
and show that
the only solution to
$2^{2a+1}+2^a+1 = n^2$
is
$a=4, n=23$.
This is very non-elegant
but I think that
it is correct.
I just kept charging forward,
hoping that the cases
would terminate.
Fortunately,
it seems that they have.
If
$2^{2a+1}+2^a+1 = n^2$,
then
$2^{2a+1}+2^a = n^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Finding the basis and dimension of a subspace of the vector space of 2 by 2 matrices I am trying to find the dimension and basis for the subspace spanned by:
$$
\begin{bmatrix}
1&-5\\
-4&2
\end{bmatrix},
\begin{bmatrix}
1&1\\
-1&5
\end{bmatrix},
\begin{bmatrix}
2&-4\\
-5&7
\end{bmatrix},
\begin{bmatrix}
1&-7\\
-5&1
\en... | Inputs
$$
\alpha =
\left(
\begin{array}{rr}
1 & -5 \\ -4 & 2
\end{array}
\right),
\qquad
\beta =
\left(
\begin{array}{rr}
1 & 1 \\ -1 & 5
\end{array}
\right),
\qquad
\gamma =
\left(
\begin{array}{rr}
2 & -4 \\ -5 & 7
\end{array}
\right),
\qquad
\delta =
\left(
\begin{array}{rr}
1 & -7 \\ -5 & 1
\end{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many integers $\leq N$ are divisible by $2,3$ but not divisible by their powers? How many integers in the range $\leq N$ are divisible by both $2$ and $3$ but are not divisible by whole powers $>1$ of $2$ and $3$ i.e. not divisible by $2^2,3^2, 2^3,3^3, \ldots ?$
I hope by using the inclusion–exclusion principle ... | The rules permit all numbers divisible by $6$, but excluding those also divisble by $4$ or $9$.
This is given by:
$$\lfloor\frac{N}{6}\rfloor-\lfloor\frac{N}{12}\rfloor-\lfloor\frac{N}{18}\rfloor+\lfloor\frac{N}{36}\rfloor$$
Firstly - Start by enumerating number of numbers divisible by 6.
Next term: Remove numbers divi... | {
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"url": "https://math.stackexchange.com/questions/1859057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor$ means that $x$ is close to an integer
Suppose $x>30$ is a number satisfying $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor$. Prove that $\{x\}<\frac{1}{2700}$, where $\{x\}$ is the fractional part of $x$.
My heuristic is that $x$ needs ... | Let $\lfloor x \rfloor =y$ and $\{x\}=b$ Then
$\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor
=y\lfloor y^2+2by+b^2 \rfloor=
\lfloor y^3+3y^2b+3yb^2+b^3\rfloor$
One way this can happen is that $b$ is small enough that all the terms including $b$ are less than $1$, which makes both sides $y^3$. This... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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A closed form for $1^{2}-2^{2}+3^{2}-4^{2}+ \cdots + (-1)^{n-1}n^{2}$ Please look at this expression:
$$1^{2}-2^{2}+3^{2}-4^{2} + \cdots + (-1)^{n-1} n^{2}$$
I found this expression in a math book. It asks us to find a general formula for calculate it with $n$.
The formula that book suggests is this:
$$-\frac{1}{2}... | We wish to show that
$$
1^{2}-2^{2}+3^{2}-4^{2} + \dotsb + (-1)^{n-1} n^{2}=
(-1)^{n+1}\frac{n(n+1)}{2}\tag{1}
$$
To do so, induct on $n$.
The base case $n=1$ is simple to verify.
Now, suppose that $(1)$ holds. Then
\begin{align*}
1^{2}-2^{2}+3^{2}-4^{2} + \dotsb + (-1)^{n} (n+1)^{2}
&= (-1)^{n+1}\frac{n(n+1)}{2}+(-1)... | {
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"url": "https://math.stackexchange.com/questions/1859620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 5
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Prove That If $(a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd)$ Then $a=b=c=d$ If the following equation holds
$$(a + b)^2 + (b + c)^2 + (c + d)^2 = 4(ab + bc + cd)$$
Prove that $a$,$b$,$c$,$d$ are all the same.
What I did is I let $a$,$b$,$c$,$d$ all equal one number. Then I substituted and expanded. I'm sort of p... | Consider the following steps
$$\begin{align}
(a + b)^2 + (b + c)^2 + (c + d)^2 &= 4(ab + bc + cd) \\
\left[ (a + b)^2-4ab \right] + \left[ (b + c)^2-4bc \right] + \left[ (c + d)^2-4cd \right] &=0 \\
\left[ a^2+b^2+2ab-4ab \right] + \left[ b^2+c^2+2bc-4bc \right] + \left[ c^2+d^2+2cd-4cd \right] &=0 \\
\left[ a^2+b^2-2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$ I stumbled on the following inequality: For all $n\geq 1,$
$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.$$
However I cannot find the proof of this anywhere.
Any ideas how to proceed?
Edit: I posted a... | \begin{align*}
2\sqrt{n+1}-2\sqrt{n} &= 2\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})} \\
&= 2\frac{1}{(\sqrt{n+1}+\sqrt{n})} \\
&< \frac{2}{2\sqrt{n}} \text{ since } \sqrt{n+1} > \sqrt{n}\\
&=\frac{1}{\sqrt{n}}
\end{align*}
Similar proof for the other inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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Consider the function $f(x) = x^2 + 4/x^2$ a) Find$f ^\prime(x)$ b) Find the values of $x$ at which the tangent to the curve is horizontal. So far I have this...
a) $f^\prime(x) = 2x + (0)(x^2)-(4)\dfrac{2x}{(x^2)^2}$
$= 2x - \dfrac{8x}{x^4}$
$= \dfrac{2x^5 - 8x}{x^4}$
$= \dfrac{2(x^4 - 4)}{x^3}$
I believe I derived ... | Your derivative is correct. You could have saved yourself some work by using the power rule.
\begin{align*}
f(x) & = x^2 + \frac{4}{x^2}\\
& = x^2 + 4x^{-2}
\end{align*}
Using the power rule yields
\begin{align*}
f'(x) & = 2x^1 - 2 \cdot 4x^{-3}\\
& = 2x - 8x^{-3}\\
& = 2x - \frac{8}{x^3}
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the values of $b$ for which the equation $2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$ has only one solution Find the values of 'b' for which the equation
$$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$ has only one solution.
=$$-2/2\log_{5}(bx+28)=-\log_5(12-4x-x^2)$$
My try:
After removing the logarithmic... | You have
$$x^2+(4+b)x+16=0\tag1$$
This is correct.
However, note that when we solve
$$2\log_{\frac{1}{25}}(bx+28)=-\log_5(12-4x-x^2)$$
we have to have
$$bx+28\gt 0\quad\text{and}\quad 12-4x-x^2\gt 0,$$
i.e.
$$bx\gt -28\quad\text{and}\quad -6\lt x\lt 2\tag2$$
Now, from $(1)$, we have to have $(4+b)^2-4\cdot 16\geqslant ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What are the constraints on $\alpha$ so that $AX=B$ has a solution? I found the following problem and I'm a little confused.
Consider $$A= \left( \begin{array}{ccc}
3 & 2 & -1 & 5 \\
1 & -1 & 2 & 2\\
0 & 5 & 7 & \alpha \end{array} \right)$$ and $$B= \left( \begin{array}{ccc}
0 & 3 \\
0 & -1 \\
0 & 6 \end{array} \ri... | Ignoring the fourth column, notice that
$$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \\ 0 \end{pmatrix}$$
and
$$\begin{pmatrix} 3 & 2 & -1 \\ 1 & -1 & 2\\ 0 & 5 & 7 \end{pmatrix} \begin{pmatrix} \frac15 \\ \frac65 \\ 0 \end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int\sin^{7}x\cos^4{x}\,dx$
$$\int \sin^{7}x\cos^4{x}\,dx$$
\begin{align*}
\int \sin^{7}x\cos^4{x}\,dx&= \int(\sin^{2}x)^3 \cos^4{x}\sin x \,dx\\
&=\int(1-\cos^{2}x)^{3}\cos^4{x}\sin x\,dx,\quad u=\cos x, du=-\sin x\,dx\\
&=-\int(1-u^{2})^3u^4{x}\,du\\
&=-\int (1-3u^2+3u^4-u^6)u^4\,du\\
&=u^4-3u^6+3u^8-u^{1... |
is it correct?
No, it isn't. You have errors in the following part :
$$=-\int(1-u^{2})^3u^4{x}du=-\int (1-3u^2+3u^4-u^6)u^4du$$
$$=u^4-3u^6+3u^8-u^{10}=\frac{u^5}{5}-\frac{3u^7}{7}+\frac{3u^9}{9}-\frac{u^{11}}{11}+c$$
They should be
$$-\int(1-u^{2})^3u^4du=-\int (1-3u^2+3u^4-u^6)u^4du$$
$$=\int \left(-u^4+3u^6-3u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to solve this Sturm Liouville problem? $\dfrac{d^2\phi}{dx^2} + (\lambda - x^4)\phi = 0$
Would really appreciate a solution or a significant hint because I could find anything that's helpful in my textbook. Thanks!
| Hint:
Let $\phi=e^{ax^3}y$ ,
Then $\dfrac{d\phi}{dx}=e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}y$
$\dfrac{d^2\phi}{dx^2}=e^{ax^3}\dfrac{d^2y}{dx^2}+3ax^2e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y=e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y$
$\therefore e^{ax^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Squeeze fractions with $a^n+b^n=c^n+d^n$ Let $0<x<y$ be real numbers. For which positive integers $n$ do there always exist positive integers $a,b,c,d$ such that $$x<\frac ab<\frac cd<y$$ and $a^n+b^n=c^n+d^n$?
For $n=1$ this is true. Pick any $a,b$ such that $x<\frac ab<y$ -- this always exists by the density of the r... | Partial answer I: if $x < 1 < y$, then we can find $a,b$ with $\frac{a}{b}, \frac{b}{a}$ arbitrarily close to one, satisfying the requirements for any $n$. Then it can be seen that that it suffices to prove the result for $y<1$ or $1<x$, since we have symmetry about 1 by inversion: $$x < \frac{a}{b} < \frac{c}{d} < y <... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem ... | Using $t$-formula
Let $\displaystyle t=\tan \frac{x}{2}$, then $\displaystyle \cos x=\frac{1-t^2}{1+t^2}$ and $\displaystyle \tan x=\frac{2t}{1-t^2}$.
Now
\begin{align*}
\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2} &=4 \\
\frac{(1+t)^{2}}{1-t^2} &= 4 \\
\frac{1+t}{1-t} &= 4 \quad \quad (t\neq -1) \\
t &= \frac{3}{5} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 12,
"answer_id": 0
} |
How do I find the terms of an expansion using combinatorial reasoning? From my textbook:
The expansion of $(x + y)^3$ can be found using combinatorial reasoning instead of multiplying the three terms out. When $(x + y)^3 = (x + y)(x + y)(x + y)$ is expanded, all products of a term in the first sum, a term in the seco... | Expanding $(x+y)(x+y)(x+y)$ amounts to adding up all the ways you can pick three factors to multiply together. For example, you could pick an $x$ from the first $(x+y)$, a $y$ from the second $(x+y)$, and another $x$ from the third $(x+y)$ to get $xyx=x^2 y$.
You are right, the only possible products we can get are $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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.