Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Question involving some roots of the equation $31z^{15} - z^{10} + 32 = 0 $ Consider the equation $$31z^{15} - z^{10} + 32 = 0.$$ What would be the sum of all those roots of the equation whose real part is positive? Only trivially trying to solve the equation I find not helpful. Even factorizing we get , by putting $z^... | The equation $31x^3-x^2+32=0$, has the roots,
$$
-1\qquad \frac{4}{31(4 + i\sqrt{46})}\qquad \frac{4}{31(4 - i\sqrt{46})}
$$
so take all them to the $1/5$-th power and sum there real parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/227324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Formula for A119016 : Numerators of "Farey fraction" approximations to $\sqrt{2}$ Can someone please verify and maybe simplify the closed formula for A119016 :
$$ a(n)=\frac{1}{4} \left(1+\sqrt{2}\right)^{n/2} \left((-1)^n+1\right)+\frac{1}{2}
\left(\sqrt{2}-1\right)^{n/2} \cos \left(\frac{\pi n}{2}\right)
+\fra... | One observation is that
$$\cos\left(\frac{\pi n}2\right)=\left\{\begin{array}{cc}(-1)^{n/2}& n\equiv0,2(\mod4)\\0&n\equiv1,3(\mod4) \end{array}\right.$$
and
$$\sin\left(\frac{\pi n}2\right)=\left\{\begin{array}{cc}(-1)^{n/2}& n\equiv1,3(\mod4)\\0&n\equiv0,2(\mod4) \end{array}\right.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/228857",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"reverse" diophantine equation Suppose we define $A= (8+\sqrt{x})^{1/3} + (8-\sqrt{x})^{1/3}$. How can we find, algebraically, all values of x for which $A$ is an integer?
I was not able this problem save for with Mathematica. How can we solve this using the tool of our brains?
| $A^3 = 16+3A\sqrt[3]{64-x}$, or $A^3 = 16+3Ay$, where $y=\sqrt[3]{64-x}$.
For each $A \ne 0$ we have $y=\frac{A^3-16}{3A}$, and $x=64-y^3$.
If $A\in (0,4]$, then $x\ge 0$.
If $A\in (4,\infty)$, then $x<0$, so $\sqrt{x}$ will be imaginary.
(if $A<0$, then we have no solutions, because RHS of your formula has $Re > 0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/233928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the roots of $x^4 +1$ I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ is a positive term as the equation above.
Thanks
| Using this, $x^4=-1=e^{i\pi}=e^{(2n+1)\pi i}$ as $e^{2m\pi i}=1$ where $m,n$ are integers.
So, $x=e^{\frac{(2n+1)\pi i}4}=\cos\frac{(2n+1)\pi}4 +i\sin \frac{(2n+1)\pi}4 $ where $n$ has any $4$ in-congruent values $\pmod 4$, the most simple set of values of $n$ can be $\{0,1,2,3\}$.
If $x_m=e^{\frac{(2m+1)\pi i}4},x_{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluation of $\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$ I need an approach to analytically evaluating this limit:
$$\lim_{n\rightarrow\infty} \int_{-\pi}^\pi x^2 \frac{\sin(2nx)}{\sin x} dx$$
Numerically, I see that the answer is $-\pi^3$. Similarly, if I replace $x^2$ with $x^4$, I g... | Here is an (admittedly somewhat horrible) solution using both the Riemann-Lebesgue lemma and the residue theorem. Let
$$ f(x) = \frac{x^2}{\sin x} \quad \text{ and } \quad g(x) = \frac{\pi^2}{2} \frac{1-\cos x}{\sin x} $$
on $(-\pi,\pi)$. Then
$$ h(x) = f(x)-g(x) = \frac{x^2 - \frac{\pi^2}2(1-\cos x)}{\sin x}$$
is act... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/234338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to calculate $3^{45357} \mod 5$? I wrote some code, here is what it gives:
\begin{align*}
3^0 \mod 5 = 1 \\
3^1 \mod 5 = 3 \\
3^2 \mod 5 = 4 \\
3^3 \mod 5 = 2 \\\\
3^4 \mod 5 = 1 \\
3^5 \mod 5 = 3 \\
3^6 \mod 5 = 4 \\
3^7 \mod 5 = 2 \\\\
3^8 \mod 5 = 1 \\
3^9 \mod 5 = 3 \\
3^{10} \mod 5 = 4 \\
3^{11} \mod 5 = 2 \... | Since you say you are unfamiliar with modular arithmetic, here is an alternative proof. Your exponent $45357$ has form $\rm\:4n+1\:$ and your small table of values all satisfy $\rm\: mod\ 5\!:\ 3^{4n+1}\! \equiv 3,\:$ i.e. $\rm\:5\mid 3^{4n+1}\!-3 = 3(3^{4n}-1).\:$ This is true since $\rm\:5\cdot 16 = 3^4-1\:|\:3^{4n}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/234698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq... | (Note: The notation $\sum$ refers to a cyclic sum here.)
For the first one:
We can assume WLOG that $a\ge b\ge c$. We will prove a lemma:
Let $x, y$ be positive reals with $xy\ge 1$. Then: $$\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge \frac{2}{1+xy}$$
Proof: Upon direct expanding, the inequality is equivalent to $(x-y)^2(xy-1)\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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For any odd integer $x,y$, $(x^2+2) \nmid (y^2+4)$
Possible Duplicate:
Quadratics and divisibility
Prove that for any odd integers $x$ and $y$, we have $(x^2+2) \nmid (y^2+4)$
| If $x$ is odd, then $x^2+2\equiv 3\pmod{4}$, so $x^2+2$ must have a prime divisor $p$ of the form $4k+3$.
If $x^2+2$ divides $y^2+4$, then $y^2+4\equiv 0 \pmod{p}$, or equivalently $y^2\equiv -4\pmod{p}$. Let $w$ be the inverse of $2$ modulo $p$. Then $w^2y^2\equiv -4w^2\equiv -1\pmod{p}$, and therefore
$$(wy)^2\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/236251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove via induction $\frac{(n+1)^2}{2^n}\le\frac{9}{4}$ How do you prove in induction that:
$$\frac{(n+1)^2}{2^n}\le\frac{9}{4}$$
This is what I keep getting:
Checking for $n=1$ we get $2\le\frac{9}{4}$.
Assuming it's true for $n$ and checking for $n+1$ I get this:
$$\frac{(n+2)^2}{2^{n+1}}=\frac{2(n+1)^2-n^2+2}{2\time... | For another proof, you can see that $\displaystyle \frac{u_{n+1}}{u_n}= \frac{1}{2} \left( 1+ \frac{1}{n+1} \right)^2 \leq 1$ with $n\geq 2$ and $\displaystyle u_n= \frac{(n+1)^2}{2^n}$. So $(u_n)$ decreases for $n\geq 2$ whence $\displaystyle u_n \leq u_2=\frac{9}{4}$ for $n \geq 2$. Finally, you only have to check th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding a value of $a$ to satisfy an expression of the form $a*(1-\frac{1}{b})^{(a-1)} = r$ Consider the following expression:
$a*(1-\frac{1}{b})^{(a-1)} = r$
Provided some real number value for $b$, I need to find a positive real number $0 < a \leq b$ to satisfy the above equation, where $0 < r < 1$.
Must we appeal to... | We have
$$
\begin{align*}
a\left(1-\frac{1}{b}\right)^{a-1} &= r \\
a\left(1-\frac{1}{b}\right)^a &= \left(1-\frac{1}{b}\right)r \\
a e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r \\
a \log\left(1-\frac{1}{b}\right) e^{a \log\left(1-\frac{1}{b}\right)} &= \left(1-\frac{1}{b}\right)r\log\left(1-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/236902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that such sequence is monotone. I have trouble figuring out how to show that
$\frac{\sqrt{n+1}}{(n+1)^{2}+1}<\frac{\sqrt{n}}{n^{2}+1},\quad\forall n\geq1$.
I have solved several inequalities before, but I can't prove this one.
Thanks!
| This is equivalent to
$\sqrt{1+1/n} < ((n+1)^2+1)/(n^2+1)$.
Since $\sqrt{1+1/n} < 1+1/(2n)$
and $((n+1)^2+1)/(n^2+1) = 1+(2n+1)/(n^2+1)$,
this is true if
$1/(2n) < (2n+1)/(n^2+1)$
or $4n^2+2n > n^2+1$,
which is true for $n \ge 1$.
Note that more than this is true: Let's try to find the largest value of $c > 1$
such tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/236987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Unusual pair of dice Suppose we roll two ordinary, 6-sided dice, $f_i=s_i=\{1,2,3,4,5,6\} $.
All $36$ possible rolls with a pair of dice (first dice - $f_i$, second dice $s_i$):
$$
\begin{array}{ccccccc}
f_i/s_i & 1 & 2 & 3 & 4 & 5 & 6 \\
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
2 & 3 & 4 & 5 & 6 & 7 & 8 \\
3 & 4 & 5 & 6... | Yes.
If we factor the generating function $(x+x^2+x^3+x^4+x^5+x^6)^2$, we obtain
$$(x+x^2+x^3+x^4+x^5+x^6)^2= x^2(x+1)^2(x^2-x+1)^2(x^2+x+1)^2$$
and have the option of distributing the factors unevenly to the dice. But (apart from the standard dice) the only possibility to have nonnegative coefficients, no constant te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determining the number $N$ Let $1 = d_1 < d_2 <\cdots< d_k = N$ be all the divisors of $N$ arranged in increasing order. Given that $N=d_1^2+d_2^2+d_3^2+d_4^2$, determine $N$. The divisors include $N$. It seems that $130$ is an answer. Is there another possible answer for $N$?
| The equation $$N = 1 + d_1^2 + d_2^2 + d_3^2$$ implies $N$ even by reducing $\mod 2$. Suppose $2^2|N$ then there are three possible cases:
*
*$N = 1 + 2^2 + 4^2 + 8^2$ - but this is impossible by arithmetic.
*$N = 1 + 2^2 + 4^2 + p^2$ (with $p>3$) - but reducing $\mod p$ we find $0 \equiv 13 \pmod p$ so $p = 13$ bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/238677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Variance of binomial distribution Why for $X\sim B(n,p)$ is $Var(X)=np(1-p)$?
$Var(X)=\sum x_i^2 p_i -(\sum x_i p_i)^2=\sum_{r=0}^n r^2 \binom{n}{r}p^r(1-p)^{n-r}+( \sum_{r=0}^n r \binom{n}{r}p^r(1-p)^{n-r} )^2$
In my short-sightedness, I don't see any viable ways to derive the variance from this.
| An easier way is to recognize that $X = Y_1 + Y_2 + \cdots Y_n$ where $Y_k$ are independent Bernoulli random variables with parameter $p$. For a Bernoulli random variable $Y_k$, we have $$\text{Var}(Y_k) = p(1-p)$$ Since $Y_k$ are independent, we have that
$$\text{Var}(X) = \text{Var}(Y_1) + \text{Var}(Y_2) + \cdots +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/240070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find the area of largest rectangle that can be inscribed in an ellipse The actual problem reads:
Find the area of the largest rectangle that can be inscribed in the ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$$
I got as far as coming up with the equation for the area to be $A=4xy$ but then when trying to fin... | let L and H be the length and breadth of the required rectangle respectively
$$\frac{(L/2)^2}{a^2}+\frac{(H/2)^2}{b^2}=1$$
$$\frac{(L)^2}{4a^2}+\frac{(H)^2}{4b^2}=1$$
$$H=\frac{b}{a}\sqrt{4a^2-L^2}$$
Area=L*H
$$A= L*\frac{b}{a}\sqrt{4a^2-L^2}$$
$$\frac{dA}{dL} =\frac{b}{a}\sqrt{4a^2-L^2}-\frac{L^2b}{a\sqrt{4a^2-L^2}}=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/240192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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"answer_id": 1
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Integral $\int \sqrt{4+4x^2+1/x^2} dx$ I'm totally lost on how I would integrate the following function:
$f(x) = \sqrt{4+4x^2+1/x^2}$
If anyone could even just point me to the method of integration that would be grand.
| $$4 + 4x^2 + \dfrac1{x^2} = \left(2x + \dfrac1x \right)^2$$
Hence,
\begin{align}
\sqrt{4 + 4x^2 + \dfrac1{x^2}} & = \begin{cases} 2x + \dfrac1x & x > 0\\ -\left( 2x + \dfrac1x\right) & x < 0\end{cases}\\
& = 2 \vert x \vert + \dfrac1{\vert x \vert}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/240999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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(Wolframalpha) this shouldnt give x=50? This is my problem:
I need x or y for the triangle area that forms between the vertical axis(y) and the function y=100+2x where the area is equal to 2500.
so I used for condition to the linear function:
knowing that the triangle area in this case should be like:
x*y/2=area, so:
x... | You start with the equation $\frac{5100}{x} = 100 + 2x.$ If the left side and the right side are equal, then I can do the same to both sides and they'll still be equal. Let's multiply both sides by $x$. I get $5100 = 100x + 2x^2.$ Bringing all of the terms over to one side, we get $2x^2 + 100x - 5100 = 0.$ Next, notice... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the equation have infinite number of solutions in integers? Does either of the equations ${a^2} - 10{b^2} = \pm 1$ have infinite number of solutions in integers?
If the answers is yes, a hint about how to reduce this problem to the problem of Pythagorean triples in the Gaussian integers $\mathbb{Z}\left[ i \right... | Clearly $$3^2 - 10 \cdot 1^2$$ is a solution.
Now observe that:
*
*$(a + b \sqrt{10})(a - b \sqrt{10}) = a^2 - 10 b^2$ for any integers $a,b$.
*$(a + b \sqrt{10})(c + d \sqrt{10}) = (ac+10bd) + (ad + bc)\sqrt{10}$
therefore every power $r>1$, $(3+\sqrt{10})^r$ will give a new solution of the Diophantine equation ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Infinitely many odd and even integers in a sequence. Consider the sequence $\{a_n\}_{n\in\Bbb N}$ where $a_n = \lfloor n \sqrt2 \rfloor + \lfloor n \sqrt3 \rfloor $. Show that there exists infinitely many odd and even integers in this sequence.
| $1 \lt (n+1) \sqrt2 - n\sqrt2 \lt 1.5 $ so $1 \le \lfloor (n+1) \sqrt2 \rfloor- \lfloor n \sqrt2 \rfloor \le 2$
$1.5 \lt (n+1) \sqrt3 - n\sqrt3 \lt 2 $ so $1 \le \lfloor (n+1) \sqrt3 \rfloor- \lfloor n\sqrt3 \rfloor \le 2$
So $a_{n+1}-a_{n}$ must be $2$, $3$ or $4$.
Similarly $2 \lt (n+2) \sqrt2 - n \sqrt2 \lt 3 $ so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding Where A Parametric Curve Intersects Itself The problem I am working on is to find the where the curve intersects itself, using the parametric equations.
These are: $x=t^2-t$ and $y=t^3-3t-1$
For the graph to intersect itself, there must be two distinct t-values, $a$ and $b$, that when plugged into the parametri... | Well, $\,xy=\alpha\beta\,$ doesn't necessarily means $\,x=\alpha\,\,\vee\,\,x= \beta\,$ , of course. Form a quadratic in $\,a\,$ and solve:
$$a^2-a=b^2-b\Longrightarrow a^2-a+b-b^2=0\Longrightarrow \Delta=1-4(b-b^2)=(2b-1)^2\Longrightarrow$$
$$a_{1,2}=\frac{1\pm\sqrt\Delta}{2}=\begin{cases}b\\{}\\1-b\end{cases}$$
If $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Defining/constructing an ellipse Years ago I was confronted with a (self imposed) problem, which unexpectedly resurfaced just recently... I don't know whether it makes sense to explain the background or not, so I'll be brief.
If I was given two points in a plane, and two lines passing through them would I be able to co... | OK, if it's of any interest, here it goes...
Assuming the equation of the ellipse $ b^2x^2+a^2y^2=a^2b^2 $, we get:
$$ y=\pm\frac{b}{a}\sqrt{a^2-x^2}$$
$$y´=-\frac{b^2x}{a^2y}=\mp\frac{b}{a}\frac{x}{\sqrt{a^2-x^2}} $$
For two points $ i=1,2 $ equation of tangent is:
$$ t_i ... y=y´(x-x_i)+y_i=\pm\frac{b}{a}\frac{a^2-x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252368",
"timestamp": "2023-03-29T00:00:00",
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Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I wa... | By C-S $$\sum_{cyc}(a^2+bc)=\prod_{cyc}\sqrt{(a^2+bc)(ac+b^2)}\geq\prod_{cyc}\left(\sqrt{a^3c}+\sqrt{b^3c}\right)=abc\prod_{cyc}(a+b).$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/253015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $5/2 < e < 3$?
Prove that $$\dfrac{5}{2} < e < 3$$
By the definition of $\log$ and $\exp$,
$$1 = \log(e) = \int_1^e \dfrac{1}{t} dt$$
where $e = \exp(1)$.
So given that $e$ is unknown, how could I prove this problem? I think I'm missing some important facts that could somehow help me rewrite $e$ in s... | Let $$e_n=\left(1+\frac{1}{n}\right)^n,\quad s_n=\sum_{k=0}^{n}\frac{1}{k!}\ (0!=1),$$
then $e_1=2=s_1,e_2=\frac{9}{4}<\frac{5}{2}=s_2$, when $n>2$,
\begin{align*}
\left(1+\frac{1}{n}\right)^n
&=1+1+\sum_{k=2}^{n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\\
&<1+1+\sum_{k=2}^{n}\frac{1}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 8
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Proof of $\arctan(x) = \arcsin(x/\sqrt{1+x^2})$ I've tried following this way, but I haven't succeeded.
Thank you!
| Let $\displaystyle\arctan x= y$
$\implies(i) \tan y =x$
and $(ii)\displaystyle-\frac\pi2\le y\le\frac\pi2$ (using the definition of principal value)
$\implies \cos y\ge0$
We have $$\frac{\sin y}x=\frac {\cos y }1=\pm\sqrt{\frac{\sin^2y+\cos^2y}{x^2+1^2}}=\pm\frac1{\sqrt{x^2+1}}$$
$\displaystyle\implies \cos y=+\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/254561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
Determine that improper integral converge or diverge
Possible Duplicate:
Convergence/divergence of $\int_0^{\infty}\frac{x-\sin x}{x^{7/2}}\ dx$
Determine the improper integral $\int_0^{\infty} \frac{x-\sin x}{x^{7/2}}dx$ converge or diverge. Prove that please.
| Write for $0< \varepsilon<1$
$$\int\limits_{0}^{+\infty}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}=\int\limits_{0}^{\varepsilon}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}+\int\limits_{\varepsilon}^{1}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}+\int\limits_{1}^{+\infty}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}.$$
Since for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/257292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| A solution and some generalization:
First of all, we can replace $b$ by $-b$, since $\gcd(a,b)=\gcd(a,-b)$. So we need to show $\gcd(a^2+ab+b^2,a-b)$.
If $p$ is a prime dividing the gcd, then it also divides $a-b$ which means $a \equiv b \mod p$, hence:
$$0 = a^2+ab+b^2=a^2+a^2+a^2=3a^2 \mod p$$
So either $p|3$ or $p|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 3
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Prove that $\int_0^{+\infty} \frac{\ln x}{a^2+x^2} dx = \frac{\pi\ln a}{2a}$ Is that true $$\int_0^{+\infty} \cfrac{\ln x}{a^2+x^2} dx = \cfrac{\pi\ln a}{2a},$$ where $a>0$ ? And how to compute it?
| Another approach :
Consider
$$
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx.\tag1
$$
Rewrite $(1)$ as
\begin{align}
\int_0^\infty\frac{x^{b-1}}{a^2+x^2}\ dx&=\frac1{a^2}\int_0^\infty\frac{x^{b-1}}{1+\left(\frac{x}{a}\right)^2}\ dx.\tag2
\end{align}
Putting $x=ay\;\color{blue}{\Rightarrow}\;dx=a\ dy$ yields
\begin{align}
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
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Plotting $(x^2 + y^2 - 1)^3 - x^2 y^3 = 0$ I have no idea how this equation:
\begin{equation}
(x^2 + y^2 - 1)^3 - x^2 y^3 = 0
\end{equation}
Produces this picture:
Can someone provide a general explanation of plotting this function?
| It's easier to be understood in polar coordinates.
(x^2 + y^2 - 1)^3 - x^2*y^3 == 0 /. {x -> r*Cos[\[Theta]],
y -> r*Sin[\[Theta]]} // FullSimplify
% // ApplySides[#/r^5 &, #] &
$$
\left(r^2-1\right)^3=r^5 \sin ^3(\theta ) \cos ^2(\theta ) \\
\frac{\left(r^2-1\right)^3}{r^5}=\sin ^3(\theta ) \cos ^2(\theta )
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/263737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $m,n\in\mathbb{N}$ and $f(m) \mid f(n)$ for the $f$ given below, must $m=n$? If for two natural number $m$ and $n$, $(2^{2m+1}-1)2^{4m-2}(2^{2m+1}+2^{m+1}+1)\mid(2^{2n+1}-1)2^{4n-2}(2^{2n+1}+2^{n+1}+1)$, then $m=n$?
| Another answer. Let $p(z)=(2z^2-1)(2z^2+2z+1)$. Then you want $m<n$ and $p(2^m) \mid p(2^n)$.
Given $m$, let $D_m=p(2^m)$. Since $D_m$ is odd, if $n\equiv m \pmod {\phi(D_m)}$, then $2^n \equiv 2^m \pmod {D_m}$ and thus $D_n=p(2^n)\equiv P(2^m)=D_m \equiv 0\pmod {D_m}$
So for any $m$, there are infinitely many $n$.
Ac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help me solve this olympiad challenge? Given:
$$p(x) = x^4 - 5773x^3 - 46464x^2 - 5773x + 46$$
What is the sum of all arctan of all the roots of $p(x)$?
| Nothing additional, but for the OP a more elementary approach.
$(x-a_1)(x-a_2)(x-a_3)(x-a_4) = x^4 - c_1x^3 + c_2x^2 - c_3x + a_1a_2a_3a_4.$ Where
$c_1 = \Sigma a_i$, $c_2 = \Sigma_{i<j}a_ia_j$ and $c_3 = \Sigma_{i <j < k}a_ia_ja_k$. Let $t_i=\tan^{-1}a_i.$ You want to compute $\Sigma t_i.$ Here, the trick is knowi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We the... | This is a very fascinating problem, but in the $P_{(4)}$ case, we have
$$\tag{1}
M_{P_{(4)}}^{-1}=\begin{pmatrix}
1 & -1/2 & 1/6 & 0 & -1/30\\
& 1/2 & -1/2 & 1/4 & 0\\
& & 1/3 & -1/2 & 1/3\\
& & & 1/4 & -1/2\\
& & & & 1/5
\end{pmatrix}
$$
This factors out as
$$
M_{P_{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
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Smallest inradius in a triangle inradius http://sphotos-a.ak.fbcdn.net/hphotos-ak-ash4/603253_4699189150138_1902686882_n.jpg
Inside triangle ABC there are three circles with radius $r_1$, $r_2$, and $r_3$ each of which is tangent to two sides of the triangle and to its incircle with radius r. All of $r$, $r_1$, $r_2$, ... | For typographic convenience (and reduced visual clutter) in the following, I write "$A_2$", "$B_2$", "$C_2$" for the half-angles "$A/2$", "$B/2$", "$C/2$".
Let $P$ be the incenter of the triangle with radius $r =: s^2$; and let $P_1$, $P_2$, $P_3$ be the centers of the circles with respective radii $r_1 =: s_1^2$, $r_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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If $\sum\limits_{n=1}^{\infty}a_{n}$ diverges, does $\sum\limits_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge? Suppose $\displaystyle\sum_{n=1}^{\infty}a_{n}$ diverges. Does $\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{1+na_{n}}$ diverge?
| Eric has proven that the divergence of
$$
\sum_{n=1}^\infty a_n\tag{1}
$$
does not imply the divergence of
$$
\sum\limits_{n=1}^\infty\frac{a_n}{1+na_n}\tag{2}
$$
even if $a_n\ge0$.
However, if $a_n$ decreases monotonically to $0$, then $(2)$ also diverges.
First, note that $\displaystyle\frac{a_n}{1+na_n}$ also decre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/269583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{k=1}^{\infty} \large\frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$ Prove that
$$\sum_{k=1}^{\infty} \frac{k}{\text{e}^{2\pi k}-1}=\frac{1}{24}-\frac{1}{8\pi}$$
| Rewrite
$$\frac{1}{e^{2\pi k} -1} = \sum_{n=1}^\infty e^{-2\pi k n}.$$
So we need to evaluate
$$\sum_{n,k=1}^\infty k e^{-2\pi k n}.$$
Summing first over $k$, we have
$$ \sum_{k=1}^\infty k e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} \sum_{k=1}^\infty e^{-2\pi k n} = \frac1{2\pi}\frac{\partial}{\partial n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/272909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 3,
"answer_id": 2
} |
Cauchy comparison test - if the series limit tends to 1 from bottom I've a following task
Check convergence of:
$$\sum_{n=2}^\infty (-1)^{\lceil 1+\sin^{2} n^{5} \rceil}\left(\frac{n^{2}+3n+10}{n^{2}+5n+17}\right)^{n^{2}(\sqrt{n+1}-\sqrt{n-1})}$$
My solution is:
$\lceil 1+\sin^{2} n^{5} \rceil$ is always 2, because $\... | According to your calculation, there is no indication of whether the series converges or diverges. ($|a_n^{1/n}|$ converges or diverges as $n \rightarrow \infty$ according to the limit being > or < 1, and is inconclusive if = 1.)
You made a mistake above:
$$\frac{n^2+3 n+10}{n^2+5 n+17} \approx 1 - \frac{2}{n} $$
so t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/273481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What do cycle decompositions mean? I am reading about cycle decompositions where a cycle $(a_1 a_2 a_3\cdots a_m)$ is defined as the permutation which sends $a_i$ to $a_{i+1}$, $1\leq i\leq m-1$ and sends $a_m$ to $a_1$.
But I am unable to understand this particular notation:
$(1 2 3)\circ ( 1 2)(3 4)$. What is this su... | You want to think of your permutations as functions. In your example, let $f:\{1,2,3,4\}\rightarrow\{1,2,3,4\}$ be the function represented by $(1,2,3)$. This means that
\begin{align*}
f(1)=2,\quad f(2)=3,\quad f(3)=1,\quad \text{and} \quad f(4)=4.
\end{align*}
In a similar way, let $g:\{1,2,3,4\}\rightarrow\{1,2,3,4\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/274649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$? I was thinking about the problem:
What is the minimum value of $|z-w|$ where $z,w \in \mathbb C$ such that $|z|=11$, and $|w+4+3i|=5$?
My attempts: I notice that $|z-w| \geq |z|-|w|=11-|w|$. Also if i take $w=u+iv$ then $|w+4+3i|=5... | let $z=11(\cos A+i\sin A), w=r(\cos B+i\sin B)$ so $|w+4+3i|=\sqrt{(r\cos B+4)^2+(r\sin B+3)^2}=\sqrt{r^2+25+2r(4\cos B+3\sin B)}$
So, $r^2+25+2r(4\cos B+3\sin B)=25\implies r=-2(4\cos B+3\sin B)$ as $r\ne0$
$|z-w|^2=(11\cos A-r\cos B)^2+(11\sin A-r\sin B)^2=11^2+r^2-22r\cos(A-B)\ge (11-r)^2$
Now, let $R\sin C=4,R\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/275301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Proof of tangent half identity
Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$
I was unable to find any proofs of the above formula online. Thanks!
| Hints:
*
*$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$
*$1 - \cos x = 2 \sin^2 \frac{x}{2}$
*$1 + \cos x = 2 \cos^2 \frac{x}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/277106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Tips on proving this convergence. We have an inductively defined sequence $x_n=x_{n-1}+2y_{n-1}$ and $y_n=x_{n-1}+y_{n-1}$ where $x_n^2-2y_n^2=\pm 1$, where $x_0=1$ and $y_0=0$.
I need to prove that the sequence $\left(\frac{x_n}{y_n}\right)_{n=1}^\infty$ converges to $\sqrt2$.
Now I can see that $\left(\frac{x_n^2}{y_... | The recursion can be written as
$$
\begin{bmatrix}x_n\\y_n\end{bmatrix}=\begin{bmatrix}1&2\\1&1\end{bmatrix}\begin{bmatrix}x_{n-1}\\y_{n-1}\end{bmatrix}
$$
The matrix has two eigenvalues: the roots of $\lambda^2-2\lambda-1=0$; that is, $\lambda=1\pm\sqrt{2}$.
The eigenvector with the eigenvalue of $1-\sqrt{2}$ is
$$
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Help figuring out all the alternative solutions to the integrals of sine and cosine I always worry a lot when doing integrals with trigonometric functions because there's always many ways to write the final answer. I am trying to figure out the general pattern for the various different solutions.
The integral
$$ \int ... | Since
$$\sin^2x\cos^2x=\frac{1}{4}\left(2\sin x\cos x\right)^2=\frac{1}{4}\sin 2x$$
we get
$$\int\sin^2x\cos^2x=\frac{1}{4}\int\sin 2x\,dx=-\frac{1}{8}\cos 2x\;\;\;(**)$$
and then, integrating by parts:
$$u=\sin x\;\;,\;\;u'=\cos x\\v'=\sin^2x\cos^2 x\;;\,\;\;v=-\frac{1}{8}\cos 2x\,\Longrightarrow$$
$$\int \sin^3x\cos^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/280449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the maximum and minimum values of $A=\frac{y}{x}-\frac{x}{y}+\frac{z}{y}-\frac{y}{z}+\frac{x}{z}-\frac{z}{x}$ Find the maximum and minimum values of
$$A=\frac{y}{x}-\frac{x}{y}+\frac{z}{y}-\frac{y}{z}+\frac{x}{z}-\frac{z}{x}$$
$x,y,z$ are possitive real numbers satisfying $M\le4m$
where $M=max{(x,y,z)}$, $m=min{(x... | From $M \leq 4m$, you get $1/4 \leq a,b,c \leq 4$. And $abc=1$ implies atleast one of $a-1$, $b-1$, $c-1$ is non-positive and atleast one is non-negative. The third one can be either positive or negative (or zero).
For minimum value, without loss of generality, assume $a-1$ is negative and $b-1$, $c-1$ are positive. Fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/281369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx$ Question:
Solve the integral
$$
\int\frac{1+x+\sqrt{1+x^2}}{\sqrt{x+1}+\sqrt{x}}\,dx
$$
My solution:
Multiply both the numerator and denominator by $\sqrt{x+1}-\sqrt{x}$. This changes the integral to
$$
\begin{align}
\int&\left(1+x+\sqrt{1+x^2}\right)\... | One can be brought to the form $\int u^a (1-u)^b \mathrm{d}u$ which is discussed in this answer of mine:
$$\begin{eqnarray}
J &=& \int \sqrt{x} \sqrt{1+x^2} \mathrm{d}x \stackrel{u=x^2}{=} \frac{1}{2} \int u^{-1/4} (1+u)^{1/2} \mathrm{d} u \\ &=& \frac{2}{3} u^{3/4} \cdot {}_2F_1\left(-\frac{1}{2}, \frac{3}{4}; \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/282006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
Let be $a,b,c \geq 0$ such that: $a^2+b^2+c^2=3$.
Prove that:
$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27.$$
I try to apply $GM \leq AM$ for $x=a^3+a+1$, $y=b^3+b+1,z=c^3+c+1$ and
$$\displaystyle \sqrt[3]{xyz} \leq \frac{x+y+z}{3}$$ but still nothing.
Thanks :-)
| Final solution :
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=527572&p=2997801#p2997801
$$ (a-1)^4\geq0\Rightarrow a^3+a+1\leq\frac{(a^2+1)(a^2+5)}{4} $$
$$ \prod(a^3+a+1)\leq\frac{1}{64}\prod(a^2+1)\prod(a^2+5)$$
$$ \leq\frac{1}{64}\frac{(a^2+b^2+c^2+3)^3}{27}\frac{(a^2+b^2+c^2+15)^3}{27}=27 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/283895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "43",
"answer_count": 8,
"answer_id": 6
} |
Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as:
$$\frac{x^4}{x(y^2+z^2)}+\fr... | The function $f(w)=w^{3/2}/(3-w)$ is convex on $(0,3)$ so
$${1\over 3}\left[f(x^2)+f(y^2)+f(z^2)\right]\geq f\left([(x^2+y^2+z^2)/3]\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/290844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Relative error and roots of significant digits
a.) The maximal relative error of the volume of a ball is allowed
within $1\%$. What is the maximal relative error measuring the
radius of the ball?
b.) Given the equation $x^2-40x+1=0$, find its roots to five
significant digits using $\sqrt{399} = 19.975$, cor... | Solve for $x$ over the real numbers:
$x^2-40 x+1 = 0$
Solve the quadratic equation by completing the square.
Subtract $1$ from both sides:
$x^2-40 x = -1$
Take one half of the coefficient of $x$ and square it, then add it to both sides.
Add $400$ to both sides:
$x^2-40 x+400 = 399$
Factor the left hand side.
Write the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Problem on improper integrals I'm having some trouble evaluating the following integrals:
$$
I=\int_{-\infty}^0\frac{dx}{\sqrt{1-x^3}},\quad
J=\int_0^1\frac{dx}{\sqrt{1-x^3}}.
$$
Any help will be appreciated.
| The $J$ integral is reduced to the Euler integral of the first kind using $u$-substitution $u=x^3$:
$$
J = \frac{1}{3} \int_0^1 \left(1-u\right)^{1/2-1} u^{1/3-1} \mathrm{d}u = \frac{1}{3} \operatorname{Beta}\left(\frac{1}{3}, \frac{1}{2} \right)
$$
The $I$ integral can be reduced to the beta integral using $x^3 = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/291410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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even and odd perfect numbers existence Thank for my previous post. Also, thank you so much for this site (m.s.e)
1) If odd perfect numbers there, those numbers can be expressible $12k + 1$ or $324k + 81$
or $468k + 117$. If yes, please discuss, how far I am correct.
2) If $K$ = $(4^n - 2^n)$/2 is perfect, when $k = ... | For part 2):
The formula
$$\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}{4}$$
is well-known and leads to
$$ 1^3+3^3+\ldots+(2m-1)^3=\sum_{k=1}^{2m}k^3-\sum_{k=1}^{m}(2k)^3\\=\frac{(2m)^2(2m+1)^2}{4}-8\cdot\frac{m^2(m+1)^2}{4}\\=m^2(2m^2-1).$$
Since it is well-known that even perfect numbers $N$ are of the form $N=2^{p-1}(2^p-1)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/293470",
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"question_score": "2",
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Find $x$ such that $12+13^x$ be a perfect square Find $x \in N$ such that $12+13^x$ be a perfect square
I am going to limit $k < 12 + 13^x < k+i$ so that I can have $t<x<t+u$, I don't know how to do it, if $x=2k$, it pretty easy but x can also equal $2k +1$ too. So... Stuck here
Update 2:
I can prove that $x$ can't be... | I think algebraic approach is appropriate to this problem. With some calculation, we get
$$(y+2\sqrt{3})(y-2\sqrt{3})=(4+\sqrt{3})^x (4-\sqrt{3})^x.$$
and $4\pm\sqrt{3}$ is prime on $\mathbb{Z}[\sqrt{3}]$. And
$$\frac{5-2\sqrt{3}}{4+\sqrt{3}}=2-\sqrt{3}$$
$$\frac{47+2\sqrt{3}}{(4+\sqrt{3})^3}=2-\sqrt{3}$$
So I conjectu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How do I write a vector as a linear combination of other vectors. Write $\begin{pmatrix} 5 \\ 3 \\15 \end{pmatrix}$ as a linerar combination of the following vectors: $u=\begin{pmatrix} 1 \\ 2 \\5 \end{pmatrix}$, $v=\begin{pmatrix} 3 \\ -4 \\-1 \end{pmatrix}$, $w=\begin{pmatrix} -1 \\ 1 \\1 \end{pmatrix}$.
My atte... | You can set this problem up as linear system of 3 equations
$$ \begin{pmatrix} 5 \\ 3 \\ 15 \end{pmatrix}
= \begin{bmatrix} 1 & 3 &-1 \\ 2 & -4 & 1 \\ 5 & -1 & 1 \end{bmatrix}
\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix} $$
You do this by arranging the basis vectors as columns to the coefficient matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/299959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defi... | Let $x_0 = \sqrt{1}$, $x_1 = \sqrt{1 + \sqrt{2}}$, $x_3 = \sqrt{1 + \sqrt{2 + \sqrt{4}}}$, and so on. Then we have:
$$\sqrt{1 + \sqrt{2 + \sqrt{4 + \cdots}}} = \lim_{n \to \infty} x_n$$
Clearly, this sequence is monotonically increasing. It also converges, since we can see each new term appears under $n$ square roots, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"answer_id": 0
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Another limit from a math contest $\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$ Let $(x_{n})_{n\ge1}$, $(y_{n})_{n\ge1}$ be real number sequences and both converge to $0$.
Evaluate $$\lim_{n\to\infty}\frac{x_n^2y_n}{3x_n^2-2x_ny_n+y_n^2}$$
| I don't know how many of your alters (see http://en.wikipedia.org/wiki/United_States_of_Tara ) know Lagrange multipliers, so I will give a bit extra. $3 x^2 - 2 x y + y^2$ is a positive definite quadratic form. Setting it equal to a positive constant gives and ellipse, a bounded figure. As a result, the function $x^2$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .
$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
I tried solving it, but I got stuck near the end . . .
a. Basis Step:
$(1)(1!) = (1+1)!-1$
$1 = (2\cdot1)-1$
$1 = 1 \checkmark$
b. Inducti... | There is a mistake in the [LHS], it should look like this:
$$
\underbrace{1(1!) + \ldots + k(k!)}_{=(k+1)! - 1} + (k+1)(k+1)! = \ldots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/301615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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Find the value of "a," so that the lines intersect, Given: $$(l_1) =(x,y,z)=(2,1,2)+t(a,1,0)$$ $$(l_2) =(x,y,z)=(5,2,3)+t(2,3,1)$$
Find the value of "$a$", if any, so that $(l_1)$ and $(l_2)$ intersect
| For the two lines to intersect at a point we need all of their coordinates to be the same. For that reason we want to solve the following:
$$\begin{pmatrix}2\\1\\2\end{pmatrix} + t_1 \begin{pmatrix}a\\1\\0\end{pmatrix} = \begin{pmatrix}5\\2\\3\end{pmatrix} + t_2 \begin{pmatrix}2\\3\\1\end{pmatrix}$$
Which, if we split ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$ How to find the remainder of $1^2+3^2+5^2+7^2+\cdots+1013^2$ divided by $8$
| Hint $\rm\ mod\ 8\!:\,\ odd^2 = \{\pm 1,\, \pm3\}^2 \equiv \color{#C00}1,\:$ so $\rm\,1^2\!+3^2\!+\,\cdots + (2n\!-\!1)^2\equiv \color{#C00}1+\,\cdots\, + \color{#C00}1\equiv\, n\, $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Proving $4^{47}\equiv 4\pmod{12}$ I know this is a simple exercise, but I was wondering if I can make the following logical jump in my proof:
We see that $4\equiv 4\pmod{12}$ and $4^2\equiv 4\pmod{12}$. Then we can recursively multiply by $4$ to get $4^{47}\equiv 4\pmod{12}$.
| $$4^{47}\equiv 4\pmod{12}$$
$$4^{47}-4=4 \cdot \left(4^{46}-1\right)=4 \cdot \left(4^{23}-1\right)\cdot \left(4^{23}+1\right)=$$
$$=4 \cdot \left(4-1\right)(4^{22}+4^{21}\cdot1+\ldots+1)\cdot \left(4^{23}+1\right)=4 \cdot3 \cdot \left(4^{22}+4^{21}\cdot1+\ldots+1\right)\cdot \left(4^{23}+1\right)=$$ $$=12 \cdot \left(4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/304224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})$. what is the value of X? $$X=(1 + \tan 1^{\circ})(1 + \tan 2^{\circ})(1 + \tan 3^{\circ})\ldots(1 + \tan {45}^{\circ})$$
$$\tan(90-\theta)=\cot\theta=\frac{1}{\tan\theta}$$
| Let $\theta = 1^\circ = \frac{\pi}{180} \operatorname{rad}$. Then$$
X = \prod_{k=1}^{45} \left(1 + \tan(k \theta)\right) = \prod_{k=1}^{45} \frac{\sin(k \theta) + \cos(k \theta)}{\cos(k \theta)} = \prod_{k=1}^{45} \sqrt{2} \frac{\sin((k+45) \theta)}{\cos(k \theta)}
$$
Furthermore
$$
\prod_{k=1}^{45} \sqrt{2} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/304461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Real Analysis Inequality Questions I am working on a few questions, and I was hoping for some feedback regarding whether or not my answers make sense.
(a) Let $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^d$. If $||\mathbf{x}||<2$, $||\mathbf{y}||<3$ and $||\mathbf{z}||<4$, show that
$$| \mathbf{x} \cdot \mat... | a) follows from $|x \cdot y - x \cdot z| = |x \cdot (y -z)| \leq \|x\| \|y-z\| \leq \|x\| (\|y\|+\|z\|) = 2 (3+4)$.
The first inequality in your answer for b) is incorrect. First do the triangle inequality, then upper bound using the fact that $|x \cdot y| < 1$ when $x,y \in B$.
| {
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"source": "stackexchange",
"question_score": "4",
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Showing that $ |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$ For every nonnegative integer $n$ and every real number $ x$ prove the inequality:
$$\sum_{k=0}^n|\cos(2^kx)|= |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$$
| Actually, the statement is true even when one restrict the sum to start at $k=1$.
i.e. For all $x \in \mathbb{R}$,
$$\sum_{k=1}^n |\cos(2^k x)| \ge \frac{n}{2\sqrt{2}} \Leftrightarrow \varphi_n(2x) \ge 0$$
where $\varphi_n(x) = \sum_{k=0}^{n-1}|\cos(2^kx)| - \frac{n}{2\sqrt{2}}$.
Notice:
$$\begin{align}
\varphi_4(x) &=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate this rational limit Evaluate:
$$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6}$$
| $$\lim_{x\to0^+}\frac{(-3\sqrt{x}+x^3+\sin(x^6))(\text{e}^{4\sqrt{x}}-1)}{4\ln (1+x)+x^4-x^6} \\= \lim_{x\to0^+} \frac{-3\sqrt{x}+x^3+\sin(x^6)}{\sqrt x} \times \frac{e^{4\sqrt{x}}-1)}{\sqrt x} \times \frac{1}{4\frac{\ln (1+x)}{x} + x^3 - x^5 }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$ How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?
My attempt:
$$\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2} = \frac{1}{2}\int_{0}^{2\pi} \frac{d\theta}{(2+\cos\theta)^2}$$
To find the singularity, I solve:
$ (2+\cos\theta... | I will start from the point where you left off. The integral may be written as, upon substitution of $z=e^{i \theta}$:
$$\begin{align}\frac{1}{2} \int_0^{2 \pi} \frac{d\theta}{(2+\cos{\theta})^2} &= -i 2 \oint_{|z|=1} dz \frac{z}{(z^2+4 z+1)^2}\\ &= -i 2 \oint_{|z|=1} dz \frac{z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2}\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/308693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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How can we prove the Inequality : $ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$ How can we prove the following inequality? For every odd positive integer $n$,
$$ \frac {n!}{ 2^{n-1}((\frac {n-1}{2})!)^2} \leq \sqrt{n}$$
Thank You.
| For completeness, here is the proof by induction:
Base Case: The case $ n = 1$ is easy to check.
Inductive Step: Assume that $$\frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2} \leq \sqrt{k}.$$ We aim to show that this implies $$\frac{(k+2)!}{2^{k+1} \left( \frac{k+1}{2} \right)!^2} \leq \sqrt{k+2}.$$
We have
\begi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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limit of $\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$ using geometric mean How to find this question using Geometric Mean?
$$\sqrt[n]{\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots (2n)}}$$
Thanks!
| Your product is itself a geometric mean.
\[ \sqrt[n]{\frac{1\cdot 3\cdot \cdot (2n-1)}{2\cdot 4\cdot \cdot (2n)}}
= \left(\frac{1}{2}\right)^{1/n} \cdot \left(\frac{3}{4}\right)^{1/n} \dots \left(\frac{2n-1}{2n}\right)^{1/n}\]
The factor is approaching the same number:
\[ \frac{2n-1}{2n} = 1 - \frac{1}{2n} \to 1\]
If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/310216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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$3^{2n+3}+2^{n+3}$ is divisible with 7 Prove that $$3^{2n+3}+2^{n+3}$$ is divisible with $7$ for every $n \in \Bbb N$
$$3^{2n}\cdot27+2^{n}\cdot8=3^{2n}(28-1)+2^{n}(7+1)=\left(3^{2n}\cdot28+2^{n}\cdot7\right)-\left(9^{n}-2^{n}\right)=$$
$$\left(3^{2n}\cdot28+2^{n}\cdot7\right)-(9-2)\left(9^{n-1}+\ldots+2^{n-1}\right)$$... | Hint $\rm\ mod\ 7\!:\,\ 3^{2n+3}\!+2^{n+3}\equiv\, 27\cdot 9^n\! + 8\cdot 2^n \equiv\, 35\cdot 2^n\equiv 0\ \ $ by $\rm\ 9\equiv 2;\ 35\equiv 0$
Remark $\ $ Per your comment, for a proof without mod, subtract and add $\rm\,27\cdot 2^n\,$ above, i.e.
$\rm\quad 27\cdot 9^n\! + 8\cdot 2^n =\, 27\, (9^n\!-2^n)\! + (\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/312411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculus Midterm Question. I'm studying for my calc midterm right now, and I was finding the limit of the question:
$$\lim_{x \rightarrow 2}{\frac{\sqrt{x+2} - \sqrt{2x}}{x-2}}\tag{1}$$
When I was trying to solve it I came up with this:
$$\lim_{x \rightarrow 2}{\frac{\sqrt{x+2}^{2} - \sqrt{2x}^{2}}{x-2(\sqrt{x-2}+ \sqr... | Your error is your move from line $(3)$ to line $(4)$. Those two expressions are not equal.
Multiplying the numerator and denominator by the conjugate of $\;\displaystyle \frac{\sqrt{x+2} - \sqrt{2x}}{ x-2}\;$ gives us:
$$
\begin{align}
\frac{\sqrt{x+2} - \sqrt{2x}}{x-2}\cdot \frac{\sqrt{x+2} + \sqrt{2x}}{\sqrt{x+2} +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that
$$
a+b+c \geq ab+bc+ca
$$
I was able to prove that
$$
\begin{align}
a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\
&\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\... | $$9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 3(ab+bc+ca)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Inequality: $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ Prove that $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ for any $x,y,z \in \mathbb{R}$ and any $t \in [-1,2].$
One try:
for $t=-1$: $x^2+y^2+z^2-xy-yz-zx \geq 0$ is true .
for $t=2$: $x^2+y^2+z^2+2(xy+yz+zx) \geq 0$ is true
also for $t=0$.
But how can prove for $t \in (-1,0)$ and $t \in... | Note that
\begin{eqnarray}
x^2+y^2+z^2+t(xy+yz+zx)&=&\frac{2-t}{3}(x^2+y^2+z^2-xy-yz-zx)\\
&&+\frac{1+t}{3}(x^2+y^2+z^2+2xy+2yz+2yz) \, .
\end{eqnarray}
Your previous work shows that both terms on the right-hand side are always non-negative for $t \in [-1, 2]$; thus the left-hand side is as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving the inequality $\arctan\frac{\pi}{2}\ge1$ Do you see any nice way to prove that
$$\arctan\frac{\pi}{2}\ge1 ?$$
Thanks!
Sis.
| By Taylor's theorem with remainder in the integral form for the function $f=\arctan$ at the point $1$, we have:
\begin{equation}\arctan x=\arctan 1+(x-1)f'(1)+\int_1^x\frac{f''(t)}{2!}(x-t)dt \end{equation}
then we substitute $\displaystyle f'(1)=\frac{1}{2}$, $\displaystyle f''(t)=\frac{-2t}{(1+t^2)^2}$ and $\display... | {
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How to solve this irrational equation? How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$
| Write:
$$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$
As:
$$(\sqrt{8x + 1} - \sqrt{6x - 2}) = (2x^2 - 8x + 7).$$
Squaring both sides yields:
$$14 x-2 \sqrt{6 x-2}\sqrt{8 x+1}-1 = 4x^4-32x^3+92x^2-112x+49$$
Simplifying:
$$-2 \sqrt{6 x-2}\sqrt{8 x+1} = 4x^4-32x^3+92x^2-112x+49$$
Squaring both sides again yields... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Writing a matrix as multiplication of orthogonal and diagonal matrices Given is the matrix A = \begin{pmatrix}
5 & 3 \\
0 & -4 \\
\end{pmatrix}
I would like to write this matrix as the product A = $U S V^T$
whereas U and V both are orthogonal matrices and S is a diagonal matrix. What is the approach to find suit... | Given:
$$A = \begin{pmatrix}
5 & 3 \\
0 & -4 \\
\end{pmatrix}$$
We want to find the Singular Value Decomposition (SVD) where $U$ and $V$ both are orthogonal matrices and $\Sigma$ is a diagonal matrix, such that:
$$A = U \cdot \Sigma \cdot V^T$$
We arrive at:
$$A = U \cdot \Sigma \cdot V^T = \begin{pmatrix} 5 & 3... | {
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Critical point of a function? Can anyone show me how to simplify get the critical point of this function.
$f(x)=x^2\sqrt[3]{2+x}$
I did the product rule and got
$$2x\sqrt[3]{2+x}+\frac{x^2}{3\sqrt[3]{(2+x)^2}}$$
but I am having touble simplifying such this how would I simplify it can anyone show me how this would be do... | You need only find the points at which $f'(x) = 0$, and where $f'(x)$ is undefined.
To simplify, find a common denominator and add terms, then set equal to zero: The common denominator is $3\sqrt[\large 3]{(2 + x)^2}$. Do you recall how to bring all terms over this common denominator?
We have $$2x\sqrt[\large3]{2+x}+\... | {
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"timestamp": "2023-03-29T00:00:00",
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How to show $f(x,y)=|xy|^\alpha\log(x^2+y^2)$ is differentiable at $(0,0)$? Show that if $\alpha > 1/2$, then
$$f(x, y)=\begin{cases}
|xy|^\alpha\log(x^2+y^2), ~(x, y) \ne (0,0)\\\\
~~~0, ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\text{otherwise}\;.
\end{cases}$$
is differentiable at $(0,0)$.
| Note that $|x| = \sqrt{x^2} \leq \sqrt{x^2+y^2}$ and similarly $|y| = \sqrt{y^2} \leq \sqrt{x^2+y^2}$.
Thus $|xy|^\alpha \leq (x^2+y^2)^\alpha$.
Then
$$
\left|\frac{f(x,y)}{\sqrt{x^2+y^2}}\right| \leq (x^2+y^2)^{\alpha-\frac12} \log(x^2+y^2) = (r^2)^{\alpha-\frac12} \log(r^2)
$$
Of course $(x,y) \to 0$ if and only if $... | {
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Find reduction formula using integration by parts
Let $\displaystyle I(n,\alpha)=\int_0^\pi \sin(n\theta)e^{\alpha \theta} d\theta$. Show that $\displaystyle I(n,\alpha)=\frac{n}{n^2+\alpha^2}(1-(-1)^ne^{\alpha \pi})$.
This is done by parts, but I can't get the expression to equal the RHS.
| Note: I adapted this from Arturo's answer at: Integration by parts: $\int e^{ax}\cos(bx)\,dx$
Also, please forgive me for not using your variable names as I found them confusing (easy to modify the entire result by just changing them to your names).
Note: we are going to save the integration limits to the end for ease.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof: Two non identical circles have at most 2 same points I'm struggeling with an analytic proof for the fact, that two different circles have at most 2 same points. (I try to solve it analytical, because geometrical I already prooved it).
I tried to start with the equations
$r_1^2=(x-a_1)^2+(y-b_1)^2$ and $r_2^2=(x-... | You have worked too hard. The first circle has equation of the shape $x^2+y^2+ax+by+c=0$. The second circle has equation of the shape $x^2+y^2+dx+ey+f=0$. The two equations are not identical, since the circles are not.
If $(x,y)$ lies on both circles, then both equations are satisfied, and therefore (subtract) we have
... | {
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"source": "stackexchange",
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$2\frac{d^2y}{dx^2} + 7\frac{dy}{dx} -y =0$ where y(0)=1 and y'(0)=3 I need help with this slightly tedious question
the got the solution of the auxiliary equation to be $s=\frac{-7+\sqrt{57} }{2}$ and $t=\frac{-7-\sqrt{57} }{2}$ thus the general solution $y=Ae^{sx}+Be^{tx}$.
Given the initial conditions I obtained t... | As L.F. stated, you have an issue with your denominator from your auxiliary equation.
You should have, $2m^2 + 7m -1 = 0$, yielding:
$~m = \frac{1}{4} (-\sqrt{57}-7) $ and $m = \frac{1}{4} (\sqrt{57}-7)$
From this and the IC's, we arrive at:
$$\displaystyle \large y(x) = \frac{1}{6} e^{-\frac{1}{4} (7+\sqrt{57}) x} \le... | {
"language": "en",
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A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
| If we are allowed to use Coordinate Geometry,
WLOG we can assume the coordinates of $A,B,C$ to be $(x,y),(-a,0),(a,0)$ respectively and $D$ to be $(h,k).$
As $\triangle ABC$ is equilateral $(x-a)^2+(y-0)^2=(x+a)^2+(y-0)^2=(-a-a)^2+(0-0)^2$
From the first relation we find $4ax=0\implies x=0$ as $a\ne0$
From the second... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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How do I find the maximum and minimum values of $1−4\cos(2\theta)+3\sin(2\theta)$? To find the maximum of $$1 - 4\cos(2\theta) + 3\sin(2\theta) $$I tried:
$$1-4(1)+3(1)=0.$$
To find the minimum I tried to substitute with the minimum values of sin and cos:
$$1-4(-1)+3(-1)=2$$
I know I'm wrong, could someone explain why?... | Following is an algebraic way of finding the range of $
As $\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$
So, $1-4\cos2\theta+3\sin2\theta=\frac{(1+t^2)-4(1-t^2)+3(2t)}{1+t^2}=y$(say) putting $t=\tan\theta$
So, $t^2(y-5)-6t+y+3=0$
This is a Quadratic Equation i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Show $ \frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1 $ given $xyz = 1$ Please help me prove the equality: If $xyz=1$, prove that
$$
\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}=1
$$
| Hint: Rewrite the $1$ in the first denominator as $xyz,$ so that the left hand side becomes $$\begin{align}\frac1{y+1+yz}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1} &= \frac{1+y}{yz+y+1}+\frac{z}{xz+z+1}\\ &= 1-\frac{yz}{yz+y+1}+\frac{z}{xz+z+1}.\end{align}$$ Can you go from here?
| {
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"source": "stackexchange",
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Finding $u(x)$ using Green's Function Let $(Lu)(x)= -\frac{d}{dx}\big(\frac{1}{x} \frac{du}{dx} )$ where $u(x)$ is twice differentiable function defined on $[1,2]$.
A) I need to find Green's function $G(x,t)$ such that for any $h(x)$ defined on $[1,2]$ the solution to the boundary value problem
$Lu=h$ for $1\le x \le2... | Hint: The correct formula for the Green's function is
$$G(x,t) =
\begin{cases}
\frac{1}{2}(x^2-1), & x\le t \\
\frac{1}{2}(t^2-1), & t\le x.
\end{cases}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How to prove that $\frac{(5m)!(5n)!}{(m!)(n!)(3m+n)!(3n+m)!}$ is a natural number?
How to prove that $$\frac{(5m)! \cdot (5n)!}{m! \cdot n! \cdot (3m+n)! \cdot (3n+m)!}$$ is a natural number $\forall m,n\in\mathbb N$ , $m\geqslant 1$ and $n\geqslant 1$?
If $p$ is a prime, then the number of times $p$ divides $N!$ is... | Let $p$ be a prime number. You should show that
$$\left\lfloor \frac{5m}{p^k}\right\rfloor+\left\lfloor \frac{5n}{p^k}\right\rfloor \geq \left\lfloor \frac{m}{p^k}\right\rfloor+\left\lfloor \frac{n}{p^k}\right\rfloor+\left\lfloor \frac{3m+n}{p^k}\right\rfloor+\left\lfloor \frac{3n+m}{p^k}\right\rfloor$$
for $k \geq 1.... | {
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"source": "stackexchange",
"question_score": "24",
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Calculate: $\int_{2}^{7}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx $ How to calculate following integration?
$$\int_{2}^{7}\frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}}dx $$
| Replace $x$ by $9-x$. We then get
$$I = \underbrace{\int_2^7 \dfrac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} dx = \int_2^7 \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt{9-x}} dx}_{x \to 9-x}$$
Adding both we get that
$$2I = \int_2^7 dx = 5 \implies I = \dfrac52$$
In general, if we have
$$I = \int_a^b \dfrac{f(a+b-x)}{f(x) + f(a+b-x)} d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337576",
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"source": "stackexchange",
"question_score": "2",
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Finding polynomial given the remainders Question: Find a polynomial $f(x) \in \mathbb{Q} (x)$ of minimal degree that has both the following properties:
When $f(x)$ is divided by $(x-1)^2$, the remainder is $2x$; and when $f(x)$ is divided by $(x-2)^3$, the remainder is $3x$.
Answer provided: $f(x)=(x-2)^3 \cdot (4x-3)+... | Rather than attempting to match two solutions, construct the solution using one remainder, and then solve using the other one, like this:
$$
f(x) = (x-2)^3g(x)+3x
$$
Therefore, mod $(x-1)^2$,
$$
f(x) = (3x-4) g(x) + 3x = 2x
$$
Therefore, $(3x-4)g(x) = -x$ under mod $(x-1)^2$.
Now, we just need to find the polynomial th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Twice a triangle is triangle The question is to prove that there are infinitely many triangular numbers $T_n$ where $2 \times T_n$ is also a triangular number, and give the first few as an example.
My attempt:
$$2 \cdot {x(x+1) \over 2} = {y(y+1) \over 2} \\
\Leftrightarrow {x(x+1)} = {y(y+1) \over 2} \\
\Leftrightarro... | Note that
$$(a^2-2b^2)(c^2-2d^2)=(ac+2bd)^2 -2(ad+bc)^2.$$
This is a special case of the Brahmagupta Identity. It can be readily proved by expanding the two sides.
A nice way of looking at things is to let $(a,b)$ be a particular solution of your Pellian $x^2-2y^2=1$, say $(1,1)$. Then you can generate infinitely man... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $
where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$
My try:: I have expand the expression
$\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4... | Suppose we seek to evaluate
$$\sum_{k=0}^r (-1)^k (k+2)(k+1) {n\choose r-k}.$$
Start from
$${n\choose r-k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{r-k+1}} (1+z)^n \; dz.$$
This yields the following expression for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\sum_{k=0}^r (-1)^k (k+2)(k+1)
\frac{1}{z^{r-k+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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The quadratic diophantine $ k^2 - 1 = 5(m^2 - 1)$ Here's the problem.
Find the solutions of the following equation:
$$ k^2 - 1 = 5(m^2 - 1).$$
Here's my idea:
The original equation can be written as:
$$ k^2 = 5m^2 - 4 \Longleftrightarrow k^2 - 5m^2 = -4$$
I know this is Quadratic Diophantine Equation and i've done som... | $$ A =
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right) ,
$$ and
$$ A^{-1} =
\left( \begin{array}{cc}
9 & -20 \\
-4 & 9
\end{array}
\right).
$$
$$
\left( \begin{array}{cc}
9 & 20 \\
4 & 9
\end{array}
\right)
\left( \begin{array}{c}
1 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/340142",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Factorize $3m^4-6m^3+14m^2-6m+11$ I have this expression:
$3m^4-6m^3+14m^2-6m+11=0$ and I want to factorize it in $(m^2+1)(3m^2-6m+11)$.
How can I do it? Thanks for any help!
| You already factorized the $3m^4 −6m^3 +14m^2 −6m+11$ with $(m^2 +1)(3m^2 −6m+11)$.
First factor equals to $(m^2 +1)$ and second $(3m^2 −6m+11)$.
$(m^2 +1)(3m^2 −6m+11) = 3m^4-6m^3+11m^2+3m^2-6m+11 = 3m^4-6m^3+14m^2+11$
Note:
But if you need solve $3m^4 −6m^3 +14m^2 −6m+11 = 0$ using factorization $(m^2 +1)(3m^2 −6m+... | {
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"source": "stackexchange",
"question_score": "6",
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$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$ Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$
Thanks in advance .
| We already now that $\frac{a^p+b^p}{a+b}=a^{p-1}-a^{p-2}b+\ldots+b^{p-1}$
Using polynomial division, we find that
$$a+b\,|\,a^{p-1}-a^{p-2}b+\ldots+b^{p-1}=(a^{p-2}-2a^{p-2}b+3a^{p-3}b^2-\ldots-pb^{p-2})+pb^{p-1}$$
Hence $\gcd(\frac{a^p+b^p}{a+b},a+b)=\gcd(pb^{p-1},a+b)=\gcd(p,a+b)=1\text{ or } p$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$ I solved $\sum_{k=1}^n(k-1)(n-k)$ algebraically
\begin{eqnarray*}
\sum_{k=1}^n(k-1)(n-k)&=&\sum_{k=1}^n(nk-n-k^2+k)\\
&=&\sum_{k=1}^nnk-\sum_{k=1}^nn-\sum_{k=1}^nk^2+\sum_{k=1}^nk\\
&=&\frac{n(n^2+n)}{2}-n^2-\frac{n(2n^2+3n+1)}{6}+\fr... | The left hand side represents picking a "middle" element in a set of $3.\ $ Then you have $k-1$ choices for picking the smallest element and $n-k$ choices for picking the largest element.
For example if $n=5$ and $k=3$, then you have $2 \times 2$ ways of $3$ being the middle element:
$$(1 2) 3 (4 5)$$
If $n = 7$ and $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove ${a^2+ac-c^2=b^2+bd-d^2}$ and $a > b > c > d \implies ab + cd$ is not prime Let $a>b>c>d$ be positive integers and suppose that
$${a^2+ac-c^2=b^2+bd-d^2}$$
Prove that $ab+cd$ is not prime? I don't know if this problem is true.
I found that this same problem has also been posted on AOPS.
But I can't prove this pro... | Rewrite as:
$$a^2-b^2+ac-bc=bd-bc+c^2-d^2$$
$$(a-b)(a+b+c)=(c-d)(c+d-b)$$
Since $a>b>c>d$, each of $a-b, a+b+c, c-d, c+d-b$ is positive. By factoring lemma (excerpted below) there exists $w, x, y, z \in \mathbb{Z}^+$ s.t.
$$a-b=wx, a+b+c=yz, c-d=wy, c+d-b=xz$$
Solving for $a, b, c, d$, we get:
\begin{align}
5a=3wx+2yz-... | {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$
Prove that $\gcd(5^{98} + 3, \; 5^{99} + 1) = 14$.
I know that for proving the $\gcd(a,b) = c$ you need to prove
*
*$c|a$ and $c|b$
*$c$ is the greatest number that divides $a$ and $b$
Number 2 is what I'm struggling with. Does anybody have any ideas?
| By euclid's algorithm $$\begin{align}gcd(5^{98}+3,5^{99}+1) &= gcd(5^{98}+3,5^{99}-1-5*(5^{98}-3))
\\ & =gcd(5^{98}+3,14)\end{align}$$Now we can see that 2 divides $5^{98}+3$. Also $5^3$ is -$1 (mod 7)$. Therefore $5^{98}=5^2*5^{96}=4(mod 7)$. So $5^{98}+3=0(mod 7)$. Hence 14 divides $5^{98}+3$. So $gcd(5^{98}+3,14)=14... | {
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"timestamp": "2023-03-29T00:00:00",
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Showing that $T(n)=2T([n/2]+17)+n$ has a solution in $O(n \log n)$ How can we prove that $T(n)=2T([n/2]+17)+n$ has a solution in $O(n \log n)$? What is the resulting equation I get after the substitution?
$$ T(n) = 2c \cdot \frac n2 \cdot \log \frac n2 + 17 + n $$
or
$$ T(n) = 2c \cdot \left(\frac n2 + 17\right) \cdo... | After the inductive step we have:
$T(n) \le 2c(\lfloor \frac{n}{2} \rfloor + 17) \log (\lfloor \frac{n}{2} \rfloor + 17) + n$
Note that $2 \lfloor \frac{n}{2} \rfloor \le n$ and we can get
$T(n) \le c(n + 34) \log (\lfloor \frac{n}{2} \rfloor + 17) + n$
Expand to
$ T(n) \le cn\log (\lfloor \frac{n}{2} \rfloor + 17) + ... | {
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"timestamp": "2023-03-29T00:00:00",
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What is the remainder when $25^{889}$ is divided by 99? What is the remainder when $25^{889}$ is divided by 99 ?
$25^3$ divided by $99$ gives $26$ as a remainder.
$25*(25^3)$ divided by $99$ gives (remainder when $25*26$ is divided by $99$) as a remainder.
i.e. $25*(25^3)$ divided by $99$ gives $56$ as a remainder.
$(2... | Using Carmichael Function,
$\lambda(99)=$lcm $(\lambda(9),\lambda(11))=$lcm$(3(3-1),10)=30$
So, $5^{30}\equiv1\pmod {99}$
Now, $25^{889}=(5^2)^{889}=5^{1788}$
Also, $1778\equiv 8\pmod {30}\implies 25^{889}=5^{1780}\equiv5^8\pmod {99}$
$5^2=25,5^3=125\equiv26\pmod{99},5^4\equiv26\cdot5\equiv31\pmod{99},$
$5^8=(5^4)^2\e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity?
$$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3}... | Let $w = \cos \left ( \frac{2\pi}{7} \right ) + i\sin \left ( \frac{2\pi}{7} \right )$ so that $w^7 = 1$. Thus
$$\begin{align*}
w^7 - 1 &= 0\\
(w-1)(w^6 + w^5 + w^4 + w^3 + w^2 + w + 1) &= 0\\
w^6 + w^5 + w^4 + w^3 + w^2 + w + 1 &= 0 &&\text{since } w \ne 1\\
\left ( w^3 + w^{-3} \right ) + \left ( w^2 + w^{-2} \right ... | {
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"url": "https://math.stackexchange.com/questions/347112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Proving a system of n linear equations has only one solution I have been given the system:
$-x_1 + 2x_2 + ... + 2x_{n-1} + 2x_n = 1$
$2x_1 - x_2 + ... 2x_{n-1} + 2x_n = 2$
$.$
$.$
$.$
$2x_1 + 2x_2 + ... + 2x_{n-1} - x_n = n$
And the assignment to prove that it has only one solution. I am aware of the existence of discr... | Your equation is of the form $Ax = b$ where $$A = \begin{bmatrix}-1 & 2 & 2 & 2 & 2 & \cdots & 2 & 2\\ 2 & - 1 & 2 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & -1 & 2 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & -1 & 2 &\cdots & 2 & 2\\ 2 & 2 & 2 & 2 & -1 &\cdots & 2 & 2\\ \vdots & \vdots & \vdots & \vdots & \vdots &\ddots & \vdots & \vdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Equation $(8\cos^3x+1)^3=162\cos x-27$
Solve equation $$(8\cos^3x+1)^3=162\cos x-27$$
I saw this equation before 5 month, and I couldn't solve it. This isn't homework, etc. (I don't do stuff like this anymore). I am just curious.
| This approach is based off knowing that the answers are $ \frac {2 \pi}{9} , \frac {4 \pi }{9} , \frac { 8 \pi }{9} $. Another equation whose solutions are exactly those values, is $ 2 \cos 3 x + 1 = 0$, so we'd look to factorize that out.
We know that $ 2 \cos 3x + 1 = 8\cos^3 - 6 \cos x + 1$, so $ (8 \cos^3 x + 1) ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove $(n^5-n)$ is divisible by 5 by induction. So I started with a base case $n = 1$. This yields $5|0$, which is true since zero is divisible by any non zero number. I let $n = k >= 1$ and let $5|A = (k^5-k)$. Now I want to show $5|B = [(k+1)^5-(k+1)]$ is true....
After that I get lost.
I was given a supplement that... | Your induction hypothesis is that $5\mid k^5-k$, which means that $k^5-k=5n$ for some integer $n$. Now
$$\begin{align*}
(k+1)^5-(k+1)&=\left(k^5+5k^4+10k^3+10k^2+5k+1\right)-(k+1)\\
&=k^5+5k^4+10k^3+10k^2+5k-k\\
&=(k^5-k)+5k^4+10k^3+10k^2+5k
\end{align*}$$
can you see why that must be a multiple of $5$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of $\sqrt{2^{2^k}} = 2^{2^{k-1}}$? It's quite easy to observe that for $k \ge 0$:
$$
\begin{align}
2^{2^k} &= 4, 16, 256, 65536, \dots\\
\sqrt{2^{2^k}} &= 2, 4, 16, 256,\dots
\end{align}
$$
More in general:
$$
\sqrt{2^{2^k}} = 2^{2^{k-1}}
$$
How can I prove this identity?
| $$\begin{align}
\left(2^{2^{k-1}}\right)^2 & = \left(2^{2^{k-1}}\right)\left(2^{2^{k-1}}\right)\tag{1}\\
& = \left(2^{2^{k-1}+2^{k-1}}\right)\tag{2}\\
& = \left(2^{2(2^{k-1})}\right)\tag{3}\\
& = 2^{2^{k}}\tag{4}\\
\end{align}$$
To go from $(1)$ to $(2)$, we apply the rule that $a^ba^c=a^{b+c}$.
Going from $(2)$ to $(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Sin inverse of a complex number Is it possible to calculate the value of $\delta$ from the relation
$\delta=\sin^{-1}(5.4i)$ ? where $i=\sqrt{-1}$
| Let $\delta=x+iy$
So, $\sin(x+iy)=5.4i$
Now, $\sin(x+iy)=\sin x\cos(iy)+\cos x\sin(iy)=\sin x\cosh y+i\cos x\sinh y$
Equating the real parts, $\sin x\cosh y=0\implies \sin x=0$ as $\cosh y\ge 1$ for real $y$
So, $\cos x=\pm1$
If $\cos x=1, x=2m\pi$ where $m$ is any integer
and $\sinh y=5.4\implies \frac{e^y-e^{-y}}2=... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$? Integration by parts is of no success. What else to try?
$$\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$$
| If $a$ is real-valued and $|a| >1$, $$\sum_{n=1}^{\infty} \frac{\cos (n \theta)}{n} \left(\frac{1}{a} \right)^{n} = - \frac{1}{2} \log \left(1- \frac{2}{a} \cos \theta+ \frac{1}{a^{2}} \right) = \log|a| - \frac{1}{2} \log \left(1-2 a \cos \theta +a^{2} \right). $$
This identity can be derived from the Maclaurin series... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 3
} |
Finding all orthogonal matrices of a given form I am given this problem. I need to find all $a,b,c,d \in \mathbb{R}$ for which the matrix $A$ is orthogonal.
$$A:=\frac{1}{3} \begin{pmatrix} -1 & 2 & a \\ 2 & 2 & b \\ 2 & c & d \end{pmatrix}$$
I know that $A$ is orthogonal if $A^T=A^{-1}$. But I am stuck how to find a... | Let $u,v,w$ be the first, second and third collumns of $3A$, respectively.
You wish to have $u\cdot v=0\wedge u\cdot w=0 \wedge v\cdot w=0$ (among other things).
This yields $-2+4+2c=0 \wedge -a+2b+2d= 0\wedge 2a+2b+cd=0$. Note that this is a linear system in (a very poor) disguise. So:
$$\begin{align} \begin{cases} ... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Strategies to prove inequalities with interval notation How to prove a inequalities with interval notation, for example:
Find minimum of $a^3+b^3+c^3$ with $a,b,c \in [-1;\infty), a^2+b^2+c^2=9$
| By Holder: if $a,b,c>0$
$$(a^3+b^3+c^3)(a^3+b^3+c^3)(1+1+1)\ge (a^2+b^2+c^2)^3$$
then
$$a^3+b^3+c^3\ge 9\sqrt{3}$$
and when $a=b=c=\sqrt{3}$
case 2:let $a\ge b\ge 0>c\ge -1$
then $a^2+b^2>8$,then $a>2sqrt{2}$,so $a^3+b^3+c^3>a^3+c^3>a^3-1>\sqrt{512}-1>9\sqrt{3}$
case 3:let $a>0>b>c\ge -1$,and we have $a^2>7$,so $a>\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.