Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove : $\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$ with $a+b+c=1$ Claim :
Let $a\geq b\geq c\geq 0$ and $a+b+c=1$ then we have :
$$\sqrt{\frac{a^2+b^2+1}{a^2+c^2+1}}+\sqrt{\frac{a^2+c^2+1}{b^2+c^2+1}}+\sqrt{\frac{b^2+c^2+1}{a^2+b^2+1}}\geq 3$$
To... | Hint:clearly this is of form $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge 3$$ which is true by am-gm
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Polynomial with root $α = \sqrt{2}+\sqrt{5}$ and using it to simplify $α^6$ Find a polynomial $\space P(X) \in \mathbb{Q}[X]\space$ of degree 4 such that
$$\alpha = \sqrt{2} + \sqrt{5}$$
Is a root of $P$.
Using this polynomial, find numbers $\space a, b, c, d \space$ such that
$$\alpha^{6} = a + b\alpha + c\alpha^{2} ... | Hint
$a^2=(\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}=7+2\sqrt{10}\Rightarrow a^2-7=2\sqrt{10} \Rightarrow (a^2-7)^2=40$
Edit
$(a^2-7)^2=40 \Rightarrow a^4-14a^2+49=40 \Rightarrow a^4-14a^2+9=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating $\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ I was trying to evaluate
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$
I have tried taking natural logarithm first:
$\lim_{x\to0}\frac{\ln(\sin x)-\ln x}{1-\cos x}=\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}\quad\qua... | Let me try an approach without Taylor series, starting with your step
$$\lim_{x \to 0} \frac{\frac{\cos x}{\sin x}-\frac{1}{x}}{\sin x}=\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x}$$
We continue from here:
$$
\begin{align}
\lim_{x \to 0}\frac{x\cos x - \sin x}{x \sin^2 x} &\overset{\mathrm{H}}{=} \lim_{x \to 0} \ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate the arc length of a difficult radical function I have been struggling with an arc length question, and I want to make sure I get this right. I have the function of:
\begin{align}
f(x) = \sqrt{7.2 (x-\frac {1}{7}}) - 2.023, [0.213, 0.127].
\end{align}
I have found the derivative of the function and set ... | I amssuming you work prior to the $u$-substitution is correct.
Instead of using the substitution $u=C\tan^2(\nu)$, consider the more simple one $u=\nu^2$, which reduces your integral to
$$I=\frac{2\sqrt{A}}7\int_{\sqrt{a}}^{\sqrt{b}}\sqrt{1+\left(\frac{\nu}{\sqrt{A}}\right)^2}d\nu.$$
There are various approaches of eva... | {
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Solve the inequality $x^4-3x^2+5\ge0$
Solve $$\sqrt{x^4-3x^2+5}+\sqrt{x^4-3x^2+12}=7.$$
$D_x:\begin{cases}x^4-3x^2+5\ge0 \\x^4-3x^2+12\ge0\end{cases}.$ We can see that $x^4-3x^2+12=(x^4-3x^2+5)+7,$ so if $x^4-3x^2+5$ is non-negative, $x^4-3x^2+12$ is also non-negative (even positive). So I am trying to solve $$x^4-3x... | Denote:
$x^4-3x^2+5 = a - 7/2$
Then:
$x^4-3x^2+12 = a + 7/2$
Now you want to solve this one:
$\sqrt{a-7/2} + \sqrt{a + 7/2} = 7$
This is quite symmetrical and nice to work with.
Raise it to power $2$ and proceed, should be trivial from there.
At the end do a direct check to see if the values you found for $a$ give rise... | {
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Verification to solution an inequality and proving another. I need solution verification an inequality, that I have solved because it seems too good to be true.
But first, I attempted this but couldn't complete:
Let $a$, $b$ and $c$ be the sides of a triangle with perimeter $3$. Prove that
$$
\sum_{cyc}{\frac{a^2}{a +... | The first inequality.
Since $$a+2\sqrt{b}-1=\frac{1}{3}(3a+6\sqrt{b}-a-b-c)=$$
$$=\frac{1}{3}\left(2a+2\sqrt{3b(a+b+c)}-b-c\right)>\frac{1}{3}(a+b-c)>0,$$ by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{a+2\sqrt{b}-1}\geq\sum_{cyc}\frac{a^2}{a+b+1-1}=\sum_{cyc}\frac{a^2}{a+b}$$ and it's enough to prove that:
$$\sum_{cyc}\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadrilateral with given angles
We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have f... | A proof by Sine Law goes as follows:
Considering triangles $AED, BED, ABD, ABE$:
\begin{align}
\frac {\sin y}{AD} &= \frac {\sin 48^\circ}{ED}\\
\frac {\sin x}{EB} &= \frac {\sin 38^\circ}{ED}\\
\frac {\sin 46^\circ}{AD} &= \frac {\sin 64^\circ}{BA}\\
\frac {\sin 22^\circ}{EB} &= \frac {\sin 74^\circ}{BA}\\
\end{align}... | {
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Algebra - Piecewise Function Let $p(x)$ be defined on $2 \le x \le 10$ such that$$p(x) = \begin{cases} x + 1 &\quad \lfloor x \rfloor\text{ is prime} \\ p(y) + (x + 1 - \lfloor x \rfloor) &\quad \text{otherwise} \end{cases}$$where $y$ is the greatest prime factor of $\lfloor x\rfloor.$ Express the range of $p$ in inter... | If $\lfloor x \rfloor \in \mathbb{P}$, then
$$ p(x) = x + 1 $$
So we cover the interval $[\lfloor x \rfloor + 1, \lfloor x \rfloor + 2)$
If $\lfloor x \rfloor \not\in \mathbb{P}$, then
$$ p(x) = p(y) + \{x\} + 1 = y + 2 + \{x\}$$
So we cover the interval $[y+2, y+3)$
The prime values of $\lfloor x \rfloor$ we take on ... | {
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"timestamp": "2023-03-29T00:00:00",
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Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab... | Not quite: in general, for $a,b,c\in\mathbb{Z}$, $a|b+c$ does not imply $a|b$ and $a|c$.
Instead, simply suppose $ab|a^2+2ab+b^2$. That is, $abk=a^2+2ab+b^2$ for some $k\in\mathbb{Z}$. The rest sort of follows like the end of your idea, if you subtract $4ab$ from both sides of the equation, then $abk-4ab=a^2-2ab+b^2$, ... | {
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Prove inequality $|a-b+c-d| \leqslant \frac{1}{16}$ Let $a,b,c,d$ be positive real numbers that fulfill two conditions:
$$a+b+c+d \leqslant 2$$$$ab+bc+cd+ad \geqslant 1$$
Prove that $|a-b+c-d|\leqslant \frac{1}{16}$
Let:
$a+c=x$ and $b+d=y$
Both $x$ and $y$ are positive.
$$x+y \leqslant 2$$$$xy \geqslant 1$$
$$-4xy \le... | Your conclusion is correct. One can shorten the argument as follows:
$$
(a-b+c-d)^2 = (a+b+c+d)^2 - 4(a+c)(b+d) \\
= (a+b+c+d)^2 - 4(ab+bc+cd+da)\le 4 - 4 = 0
$$
which implies $a-b+c-d = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
If you replace $\sec^2$ with $\tan^2 + 1$, it should be ... | The correct statement is $\sec^2(y)=1+\tan^2(y)$.
Now let $y=\tan^{-1}\left(\dfrac x2\right)$.
I hope this helps you see why.
| {
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"url": "https://math.stackexchange.com/questions/3884630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$ via $\epsilon$-$\delta$
Show that $\lim_{x\rightarrow0}\frac{1}{x^4}=\infty$.
Consider our preliminary work:
$$\begin{align}
|\frac{1}{x^4}-0|<\epsilon &\implies|x^4-0|<\frac{1}{\epsilon} \tag1\\
&\implies|x^2-0||x^2+0|<\frac{1}{\epsilon} \tag2\\
&\implies|x-0||x+0|... | Let $M$ be given. Then, take $\delta = \dfrac{1}{\sqrt[4]{M}}$. Thus,
$$0< |x| < \delta \implies |x| < \dfrac{1}{\sqrt[4]{M}} \implies |x^4| < \dfrac{1}{M} \implies |\dfrac{1}{x^4}| > M $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = ... | Here are more details for @Phicar's suggested approach.
Let $A(z)=\sum_{n \ge 0} a_n z^n$ be the ordinary generating function of $(a_n)$.
Then $z A'(z)=\sum_{n \ge 0} n a_n z^n$, and the recurrence relation implies that
\begin{align}
z A'(z)
&= \sum_{n \ge 5} \left((n-4)a_{n-1} + 12n H_n\right) z^n \\
&= z \sum_{n \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration of $\int_{0}^{1} \frac{x^k}{1-x} dx$ Solving this using integration by parts and by letting $u = x^k$ and $dv = (1-x)^{-1}$, it follows:
$$\int_{0}^{1}\frac{x^k}{1-x} dx = -x^{k}\ln(1-x) + k\int_{0}^{1} \ln(1-x)x^{k-1} dx$$
It is known that
$$\ln(1-x) = (1-x) - (1-x)\ln(1-x)$$
But doesn't this make the inte... | Let $$ \mathcal{I} = \int_0^1 \frac{x^k}{1-x} \, \mathrm{d}x $$ Now using integration by parts we get
\begin{align*}
\mathcal{I} &= -x^k\, \ln(1-x) \bigg{|}_0^1 + \int_0^1 k \, x^{k-1} \, \ln(1-x) \, \mathrm{d}x \\
&= 1 + \int_0^1 k \, x^{k-1} \left( \sum_{n=1}^\infty - \frac{x^n}{n} \right) \, \mathrm{d}x \\
&= 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3891855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2... | $(a^2+ac+c^2)-(a^2+ab+b^2)=a(c-b)+c^2-b^2=(c-b)(a+b+c)\\(b^2+bc+c^2)-(a^2+ac+c^2)=b^2-a^2+(b-a)c=(b-a)(a+b+c)$
Are they equal?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$. I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that $$x-... | Hint: Simplify that LHS to $\displaystyle -\frac{2x}{1+\sqrt x}< 0$, which is obvious for $x>0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate a function with the use of a development point and geometric series Assuming I have this function :
$$ f(x) = \frac{1 + x^3}{2-x} $$
The question was to calculate this with the use of the geometric series and the development point $x_{0} = 0$.
Well setting first the function to a geometric series is quite c... | We are looking for a representation
\begin{align*}
f(x)=\sum_{n=0}^\infty a_n(x-x_0)^n=\sum_{n=0}^\infty a_nx^n\tag{1}
\end{align*}
where the series expansion is at the point $x_0=0$.
We obtain
\begin{align*}
\color{blue}{f(x)}&\color{blue}{=\frac{1+x^3}{2-x}}
=\frac{1}{2}\cdot\frac{1+x^3}{1-\frac{1}{2}x}\\
&=\frac{1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$
Using the $\epsilon,\delta$ definition of uniform continuity, prove that $f(x)=\frac{1}{1+x}$ is uniformly continuous on $(-\infty,-\frac{3}{2})$.
Attempt:
We must show that for any $\epsilon>0$ we can find a $\delta > 0$, such that i... | Since $x$ and $a$ are both assumed to satisfy the inequalities $x,a < -\frac{3}{2}$ then sure, you may use those inequalities in your proof.
The proof looks fine. Your scratchwork has a mistaken inequality and a mistaken implication sign, although that seems not to have affected the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given X+Y+Z = 0, find $\frac{(X^3+Y^3+Z^3)}{XYZ}$ The result can be found using the equation :
$(X^3+Y^3+Z^3) - 3XYZ = (X+Y+Z)(X^2 - XY +Y^2 - YZ +Z^2 - XZ)$
Since X+Y+Z = 0, the right side of the equation is equal to 0. Therefore $X^3+Y^3+Z^3 = 3XYZ$ and the answer to the problem is 3.
However what if we calculate $X^... | In your second set of calculations, where you're subtracting terms from $(X + Y + Z)^3$, you're missing a $-6XYZ$ term. Also, going from $3X + 3Y + 3Z = 0$ gives $-3X - 3Y = 3Z$, not $-3Z$, with this sign error occurring for the other $2$ terms as well. When you include the $-6XYZ$ term, you'll get an additional $\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Product Series Formula through Induction For the series:
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$ $(1+\frac{1}{n})^n$ I have the formula $\frac{(n+1)^n}{n!}$ for n$\in$ $\Bbb N$
I used induction to try and solve but I'm stuck at trying to prove it for n+1 since
$(1+\frac{1}{1})^1$$(1+\frac{1}{2})^2$$.....$$(1+\fr... | Hint: We have
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n = \frac{(n+1)^{n}}{n!}
\end{eqnarray*}
So
\begin{eqnarray*}
\left( 1+\frac{1}{1} \right) \left( 1+\frac{1}{2} \right)^2 \cdots \left( 1+\frac{1}{n} \right)^n \color{red}{\left( 1+\frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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solving a system of cubic equations with real roots Consider the following system of equations:
\begin{align}
\frac{1}{8} (\alpha +2 x)^2 \left((\alpha +2 x)^2-12\right)+\\ +\frac{4}{9} y^2 \left((\alpha
+2 x)^2+2\right)+\frac{32 y^4}{81}&=0\\
\alpha ^3-6 \alpha +8 x^3-4 \alpha x^2-2 \left(\alpha ^2+2\right) x&=0
\... | The second equation write
$$\alpha^3-2x\alpha^2-2(3+2x^2)\alpha+4x(2 x-1)=0$$ The discriminant is
$$\Delta=32 \left(128 x^4+18 x^2+27\right) > 0 \qquad \forall x$$
Using the trigonometric method for three real roots leads to
$$\alpha_k=\frac{2}{3} \left(x+ \sqrt{2(8 x^2+9)} \cos \left(\frac{1}{3} \left(2 \pi k-\cos
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many students in the school?
If they put the rows of 13, there are 8 students left; if they put the rows of 15, there are 3 students left and if they put the rows of 17, there are 9 students left. How many students are there given that the total students are < 5000?
The following congruences are: $$x\equiv 8 \pmo... | Let $N$ be the number of students. Let $M = 13\times 15 \times 17 = 3315$
$\qquad\ M_1 = 15 \times 17 = \color{#c00}{255},\,\ \ y_1 = 255^{-1} \equiv \ \ 8^{-1}\equiv \ \ \color{#c00}{5}\ \pmod{13} $
$\qquad\ M_2 = 13 \times 17 = \color{#0a0}{221},\,\ \ y_2 = 221^{-1} \equiv 11^{-1} \equiv \color{#0a0}{11}\pmod{15} $
$... | {
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"url": "https://math.stackexchange.com/questions/3909216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of all possible valuse of $\frac{a}{b}+\frac{c}{d}$? If a,b,c,d are real numbers and $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$ and $\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$, then find the sum of all possible valuse of $\frac{a}{b}$+$\frac{c}{d}$ ?
I tried this problem for a while but made no p... | \begin{align}
17 &= \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \\
&= \frac{a}{b}+\frac{c}{d}+\frac{b}{c}+\frac{d}{a} \\
&= \color{red}{\frac{ad+bc}{bd}}+\color{blue}{\frac{ab+cd}{ac}} \\
&= \color{red}{x}+\color{blue}{y} \\
20 &= \frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \\
&= \frac{a}{c}+\frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911548",
"timestamp": "2023-03-29T00:00:00",
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Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$ I found the following exercise reading my calculus notes:
If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$
I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.
| By A.M.$\geq$G.M.$$(x^2+1)(y^2+1)(z^2+1)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\geq8\times \sqrt[8]{(x^2y^2z^2)\times(x^2y^2)\times(y^2z^2)\times(z^2x^2)\times(x^2)\times(y^2)\times(z^2)\times1}=8xyz$$
| {
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How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral? So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)
I was hoping someone ou... | Suppose $6 | 12^{n}-4^{n}-3^{n}+1$ for some $n$.
Then $6| 12(12^{n}-4^{n}-3^{n}+1) = 12^{n+1}-12\cdot4^{n}-12\cdot3^{n}+12=$
$12^{n+1}-3\cdot4^{n+1}-4\cdot3^{n+1}+12.$
Since $3^{m} \equiv 3 \pmod 6$, and $4^{m} \equiv 4 \pmod 6$ for every $m$, we have that
$2\cdot4^{n+1} + 3\cdot3^{n+1}-11 \equiv2+3+1\equiv0\pmod 6.$
H... | {
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Limit with parameter $p$ Find $ p $ that will make limit convergent
$$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$
I know how to solve most of the basic limits. In this problem I don't know if... | $$ \lim _ { x \to \infty } \left( \left( n + \left( \frac { x ^ 2 + 2 } { x ^ 2 - x + 1 } \right) ^ { 2 x + 1 } \right) \cdot \sin \left( \frac { x \pi } 2 \right) \right) $$
since as mentioned the SIN is diverging because it keeps oscillating. Therefore, the first part has to converge to 0.
We have that the first part... | {
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How does $f^{(3)} (0) \geq 3\ $?
Let $f(x)$ be a real polynomial of degree $4.$ Suppose $f(-1) = 0,f(0) = 0, f(1) = 1$ and $f^{(1)} (0) = 0,$ where $f^{(k)} (a)$ is the value of $k^{\text {th}}$ derivative of $f(x)$ at $x = a.$ Which of the following statements are true?
$(1)$ There exists $a \in (-1,1)$ such that $... | The conditions $f(0)=0$ and $f'(0)=0$ tell us that $f(x)=mx^4+ax^3+bx^2$ for some $a,b\in\Bbb R$. Furthermore, $f(-1)=0$ implies $b-a=-m$ and $f(1)=1$ implies $a+b=1-m$. Thus the polynomial is $$f(x)=mx^4+\frac12x^3+\left(\frac12-m\right)x^2.$$
| {
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Is this $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$? Is this $x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$ irreducible in $\mathbb Q[x]$? Can you help please?
| Let $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$. Define the new polynomial $f(x+1)=x^6+7x^5+21x^4+35x^2+21x+7$, using the binomial theorem expand $f(x+1)$. This polynomial is irreducible over $\mathbb{Q}$ by Eisenstein criterion, for $p=7$
If $f(x)$ were reducible over the rationals then let $p(x), q(x)\in \mathbb{Q}[x]$ such that ... | {
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The orthogonal projection of the vector b is not equal to $AA^+b$ Given matrix $A:= \begin{pmatrix}1&2\\3&6\end{pmatrix}$
I am trying to solve this task:
... compute $P:=AA^+$. Finally, evaluate $Pb$ with
$b:=\begin{pmatrix}2\\4\end{pmatrix}$, and show that $Pb$ is indeed
the orthogonal projection of the vector b onto... | Your solution for $P$ and $Pb$ is correct.
You can simply verify that $Pb$ is in the column space of $A$ and that $b-Pb$ is orthogonal to this column space, which is now the span of $\pmatrix{1\\3}$.
The other method works only if $u_1,u_2$ are orthogonal, and in that case it produces the projection to the plane spanne... | {
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How to solve the homogeneous system of a differential equation I want to solve the system $\dot{x}=Ax$ where $ A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right).$ I found the eigenvalues of $A$ as $\lambda_1=4,\lambda_2=-1$ with correspongind eigenvectors $v_1=(2,3)^\intercal,v_2 =(1,-1)^\intercal $ respect... | $$x(t)=\left( \begin{matrix} \frac{3}{5}e^{-t} + \frac{2}{5}e^{4t} & -\frac{2}{5}e^{-t} + \frac{2}{5}e^{4t} \\ \frac{3}{5}e^{4t} + -\frac{3}{5}e^{-t} & \frac{3}{5}e^{4t} + \frac{2}{5}e^{-t} \end{matrix} \right) x_0$$
$$x(t)=\left( \begin{matrix} \frac{1}{5}e^{-t}(3c_1-2c_2) + \frac{2}{5}e^{4t} (c_1+c_2) \\ - \frac{1}{... | {
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find recurrence relation for strings of length n over {A,B,C}(none of the strings should have consecutive As) Assume that an is number of strings of length n over A B C and none of strings have consecutive As
I have no idea how to find a relation for this
| For a recurrence relation, define $u_n$ to be the number of strings of length $n$ with letters $A,B,C$ such that there are no consecutive $A$s, and additionally, the string must end in $A$. Define $v_n$ to be the number of strings of length $n$ with letters $A,B,C$ such that there are no consecutive $A$s, and additiona... | {
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Proof that $\int_0^\infty\frac{\ln x}{x^3 - 1} \, dx = \frac{4 \pi^2}{27}$ I realise this question was asked here, but I'm not able to work with any of the answers. The hint given by my professor is
Integrate around the boundary of an indented sector of aperture $\frac{2 \pi}{3}$
but when I try that I can't figure ou... | Here we discuss two ways of tackling the integral using contour integral. I added Solution 1 to help you implement the hint suggested by the professor. However, I personally recommend you to jump directly to Solution 2.
Solution 1. Consider
$$ f(z) = \frac{\log z}{z^3 - 1}, $$
where $\log(\cdot)$ is the principal comp... | {
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Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$... | Spoiler alert: this is the solution.
Let's take $ a = \sqrt[5]{16+\sqrt x} $ and $ b = \sqrt[5]{16-\sqrt x} $ . Note that because $ \sqrt x \ge 0 $, $ a $ is always positive (technically greater than or equal to $ \sqrt[5]{16} $), but $ b $ can be positive, zero, or negative. We have:
$$ a + b = 2 $$
$$ (a + b)^5 = 32 ... | {
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Double root in Trigonometric equation?
Consider the function $ f( x) = \sin^2 x + \sin x$ in the domain $ [0, 2 \pi]$, find the subset of the domain where $f \geq 0 $
Factoring by treating as quadratic in $ \sin$:
$$ f = ( \sin x + 1 ) ( \sin x)$$
Hence, the roots are $ \sin x = \{ 0, -1 \}$, hence the roots in $x$ ... | Your factorization $ \ (\sin x)·(1 + \sin x) \ $ will also explain how the sign of $ \ f(x) \ $ behaves. Since $ \ (1 + \sin x) \ $ is non-negative for all values of $ \ x \ \ , $ the sign of $ \ f(x) \ $ is entirely determined by the sign of the factor $ \ (\sin x) \ \ . $ As this is negative in quadrants III and I... | {
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What is the $n$th term of the sequence $1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1 ...$ So there exist a lot of similar questions which ask the $n$th term of
$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider e... | The number of terms in $x$ rows is $1+2+...+x=x(x+1)/2$. The row number $m$ of $n^{\text{th}}$ term is thus the smallest positive integer solution of $x(x+1)/2\ge n$. The roots of $x^2+x-2n=0$ are $\frac{-1\pm\sqrt{8n+1}}2$. First, we reject the negative root. Next, we take the ceiling value of the positive root as it ... | {
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Basic Geometric Series Question-Stuck I'm studying Calc 2 and I have a basic series question.
A geometric series $\sum _{n=0}^{\infty }\:Ar^n$ is convergent if |r|<1 and the sum equals $\frac{a}{1-r}$ if the series is convergent.
Question: $\sum _{n=1}^{\infty }\:\frac{5}{\pi ^n}=-\frac{5}{\pi }+\sum _{n=0}^{\infty \:}... | You did almost everything right. You took a sum from 1 to $\infty$ and added a $n=0$ term, and then subtracted it (your $\frac{-5}{\pi}$ term).
But you wrote it wrong, because
$$
\frac{5}{\pi^0} = \frac{5}{1} = 5.
$$
Nice work otherwise!
There's another way to deal with a series starting at $1$. You write
\begin{alig... | {
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Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ $\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}Show that for $a,b>0$ and $n \in \mathbb N^+$ and $a \ne b$ the following does hold:
$$\left[a-\left(n+1\right)\left(a-b\right)\right]a^{n}<b^{n+1}$$
I tried to simplify the inequality to $$na^n(b-a)<b(b^n-... | We want to prove that, for any $n\in\mathbb{N}^+$ and $a,b>0;\;a\ne b$ the following inequality holds
$$a^n (a-(n+1) (a-b))-b^{n+1}<0$$
Proof by induction. It is true for $n=1$
$$a^1 (a-(1+1) (a-b))-b^{1+1}=-(a-b)^2<0$$
Now suppose it is true for $n$ that is
$$P(n)=-n a^{n+1}+b a^n+b n a^n-b^{n+1}<0\tag{1}$$
and let's ... | {
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Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$ Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable
Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equati... | Another Way to look at it
$$\frac{d^2y}{d^2x} = \frac{1}{y^3} - y => \frac{d^2y}{d^2x} = \frac{1-y^4}{y^3} =>$$
$$\frac{y^3}{1-y^4} d^2y = d^2x $$
Now this can be double integrated using the integral of a division formula
| {
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How do this question? A discrete random variable $X$ can only take the value of 1, 2, 3 and 4, with the probabilities $P(X=1)=P(X=3)$ and $P(X=2)=P(X=4)=2P(X=1)$.
A random sample of size 3 is chosen from $X$ with replacement.
(a) given that $X_1,\ X_2,\text{ and }X_3$ are the three independent random values of $X$, fin... | Your calculations for (a) is correct. You choosing three events from
event
probability
$\{X \le 3 \}$
$2/3$
$\{X > 3 \}$
$1/3$
You need $2!$ in the denominator as events with two '$\le3$' like $\{X_1\le3, X_2\le3, X_3>3\}$ are double-counted by $3!$.
Your error in (b) comes from $\bar{X} \sim \mathcal{N... | {
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Evaluate $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$ for $k=3$ $\sum _{n=1}^{k}\frac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\frac{i}{i+1}$
How can I think this for k=3 ?
Here is my idea: $\prod _{i=n}^{6}\frac{i}{i+1}=\frac{1}{2}.\frac{3}{4}.\frac{4}{5}.\fra... | That's a telescoping product so
$\begin{array}\\
\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\prod _{i=n}^{2k}\dfrac{i}{i+1}
&=\sum _{n=1}^{k}\dfrac{{\left(-4\right)}^{n+2}}{(n-2{)}^{2}}\dfrac{n}{2k+1}\\
\end{array}
$
but then there is a problem at $n=2$
when there is a division by zero.
I'll change that... | {
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Why $\omega\left[\begin{pmatrix} 0 &Q\\ R &0 \end{pmatrix}\right]\leq\omega\left[\begin{pmatrix} P &Q\\ R &S \end{pmatrix}\right]$? Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;, \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all operators on $E$. The numerical... | Note that $\omega(T) = \omega(-T)$. Thus it suffices to find a unitary operator such that satisfies
$$U^*TU=\begin{pmatrix}
P & -Q\\
-R &S
\end{pmatrix}.$$
But for this we can take
$$U=\begin{pmatrix}
I & 0\\
0 & -I
\end{pmatrix}.$$
| {
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Numbers of the form $\sum_{n=0}^{N} 2^{a_n} 3^n$ Just out of curiosity.
Does the numbers of the form
\begin{align}
\sum_{n=0}^{N} 2^{a_n} 3^n \text{,}
\end{align}
with $a_n \ge 0$ and $N \ge 0$, cover all integers not divisible by 3?
These numbers can never be divisible by three.
It is also easy to discover integers w... | If $x \le 8$, then all numbers not divisible by $3$ have the desired form as one can check by hand (e.g. $1=1$, $2=2$, $4=4$, $5=2+3$, $7=1+2 \cdot 3$, $8 = 8$).
If $x > 8$, and $x \equiv 1,2,4,5,7,8 \bmod 9$, then let $m = 4,8,1,2,4,2$ respectively, so $m$ is a power of $2$ and $x \equiv m \bmod 3$ but $x \not\equiv m... | {
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Jacobian linearization of trigonometric functions If I have to linearize a nonlinear trigonometric system around the origin $(0,0)$:
$$\dot{x_1} = x_2$$
$$\dot{x_2} = \cos(x_1)$$
I can apply the small angle approximation to find the matrix A:
$$ A = \pmatrix{0&1\\1&0} $$
However, if I apply Jacobian linearization and t... | Your small angle approximation is wrong and your Jacobian linearization is incomplete.
If $x_1 \approx 0$ then $\cos(x_1) \approx 1$ and so the system
$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} &= \cos(x_1) \\
\end{align}
$$
behaves close to $(0,0)$ approximately like
$$
\begin{align}
\dot{x_1} &= x_2 \\
\dot{x_2} ... | {
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Derive half angle formula $ \cos\frac x 2= \sqrt{\frac {1 + \cos x} {2}}$
How can the cosine of half angle formula
$$ \cos(\frac x 2) = \sqrt{\frac {1 + \cos x} {2}}$$
can be derived using the bisector angle theorem $ OA:OB = AM:MB$ for the $ \triangle AOM $?
Can anybody explain this to me?
|
Beside $\frac{AM}{BM} = \frac{OA}{OB}$ per the bisector angle theorem, $MN= \frac{BM}{OM}AM$ holds from similar triangles. Substitute them below
\begin{align}
\cos^2\frac x2= \frac{ON}{OA} \frac{OB}{OM}
&=\left( \frac{OM}{OA} + \frac{MN}{OA} \right) \frac{OB}{OM}
= \frac{OB}{OA} + \frac{\frac{AM}{BM}}{\frac{OA}{... | {
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Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$ I found the following exercise in a problem book (with no solutions):
Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$
I tried AM-GM for the fraction on the LHS but got stuck ... | let $p=a+b+c,q=ab+bc+ca,r=abc$
Now it is easy to prove the following as $r=1$
*
*$pq\ge 9r=9$
*$p^2\ge 3q$
Now we have to prove $$4(3+2p+q)\le (p+3)(2+p+q)$$ or $$p^2+pq\ge 6+3p+q$$ $$p^2+9\ge 6+3p+q$$ using $q\le p^2/3$ $$\frac{2p^2}{3}-3p+3\ge 0$$ which is true as $p\ge 3$
| {
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Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, u... | Hint: every term but the last is divisible by $a$, so your remainder is congruent to $(2x)^n$. But that's not quite enough either, because that's still too 'large' if $n\gt 1$. You need to figure out how to divide $(2x)^n$ by $x^2-x+1$. The easiest way (IMHO) to do that is this: modulo $x^2-x+1$, we can just say that $... | {
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Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.
Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.
What I Tried: I am not sure about any specific approach to this problem. Of course as the number is... | Because it must be a multiple of $30 = 2 \cdot 3 \cdot 5$, then the number must be of the form $$N = 2^{\alpha+1} \cdot 3^{\beta+1} \cdot 5^{\gamma+1} \cdot M $$ where $M$ is $1$ or a product of primes that are greater than $5$ and $\alpha, \beta$, and $\gamma$ are non negative integers.
Let $d(n)$ represent the number... | {
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"answer_id": 4
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How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| $f(x)=\frac{1-\cos x}{x^2}$ is an even function, so we only look at $x \in [0,1]$. It's easy to prove $f(x)$ is continuous and differentiable at $x=0$, and
$$f'(x) = \frac{x \sin x + 2 \cos x - 2}{x^3}$$
Now $$x \sin x + 2 \cos x - 2 \le 2 \tan \frac x2 \cdot \sin x - 4 \sin^2 \frac x2\\= 2 \tan \frac x2 \cdot 2 \sin \... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$ Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.
We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin... | After your 3rd line, you are actually done, without realizing it; you simply went down the wrong path.
The RHS
$$= \frac{\sin(a) + \cos(a)}{\sin(a) - \cos(a)}.$$
Edit
See the comments following this answer.
A case can be made that my analyis is flawed, since I didn't bother to prove that the denominator above is alway... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Finding the maximizer Let be $ A\in \mathbb{R}^{2,2} $ a symmetrical and positive definite matrix with different eigenvalues $ \lambda_1>\lambda_2>0 $. Further we have the rayleigh quotient $$ \max_{x\neq 0}\frac{\overline{x}^T\cdot A\cdot x}{\overline{x}^T\cdot x}=\max_{\|x\|_2=1}\overline{x}^T\cdot A\cdot x $$.
For h... | Note that the set $\|x\|=1$ is compact, so a $\max$ exists.
Solve $\max_{ \|x\|^2 =1 } {1 \over 2}x^T A x$ using Lagrange multipliers to get
$Ax = \lambda x$, so we see that the Lagrange multiplier must be an eigenvalue.
Hence $\max_{ \|x\|^2 =1 } x^T A x = \lambda_\max$ and in the given example,
since the eigenvalues ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Least residue power modulo May someone confirm that the least residue of $44^8$ modulo $7$ is $4$ please?
And that $2^2 \equiv 4\, (\!\!\!\mod 7)$?
Thanks in advance.
| It is correct: $44\equiv 2\mod 7$, so $44^8\equiv 2^8\mod 7$. Now $2^3\equiv 1\mod 7$, so $2^8\equiv 2^{8\bmod 3}=2^2\mod 7=4$, since $0\le 4<7$.
| {
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"url": "https://math.stackexchange.com/questions/3976854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Misunderstanding on Taylor series expansion Sometimes, I try to find the series expansion of some random functions but I am usually wrong, especially when using already known formulas such as $$\frac{1}{1-X} = 1 + X + X^2 + X^3 + ... \quad (1)$$
Now imagine I want to find the series expansion of $\frac{1}{2+x^2}$, here... | The expansion $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ... $ only works if $|x|<1$ as you pointed out yourself. Because $2+x^2>1$ the substitution you make is guaranteed to give you a divergent sum.
Now, by factoring out $\frac{1}{2}$ you solve that problem, since now it works whenever $|x|<\sqrt{2}$. So the substitution i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $x^3 + \frac{1}{x^3}$ Question
$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 + \frac{1}{x^3}$ and choose the correct answer from the following options:
*
*198
*216
*200
*186
What I have did yet
I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$... | Hint $(x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})$ also $(x+\frac{1}{x})=6$ not $36$ you seem to have typed an incorrect calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve $e^{2z} - 2e^z + 2= 0$ So I've started by looking at
\begin{align} e^{z} = x:\\
x^2 - 2x + 2 = 0
\end{align}
Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}
Then I did the following:
\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2... | Yes.....
Once you get $e^z = 1 +i, 1-i$ you can take the following as a formula (assuming $a,b$ are real):
$a + bi= \sqrt{a^2 + b^2}e^{\arctan \frac ab i}= e^{\ln(a^2 + b^2)}e^{\arctan \frac ab i}= e^{\ln(a^2+b^2) + \arctan \frac ab i}$
[I suppose another way putting this is the principal natural log $\operatorname{Ln}... | {
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"source": "stackexchange",
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"answer_id": 1
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How to find a upper bound of the set $A= \{ (1+\frac{1}{n})^n : n \in \mathbb{N}^{*} \} $ i have the set $A= \{(1+\frac{1}{n})^{n} : n \in \mathbb{N}^* \}$ and the exercise ask me to find at least $2$ upper bound least equal to $\frac{14}{5}$.
My first question is, how i know $\frac{14}{5}$ is a upper bound of $A$?
My ... | If you are not allowed to use the definition of $e$, then by observing that the sequence terms are bound by (refer to this)
$$
\begin{align}\left(1+\frac 1n\right)^n&=\sum_{k=0}^{n}\binom{n}{k}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{n(n-1)\cdots(n-k+1)}{k!}\left(\frac 1n\right)^k\\&=\sum_{k=0}^{n}\frac{1}{k!}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3989503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$ How to get a closed form for the following sum:
$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$
Where $F_k$ is the $k$th Fibonacci number.
I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\le... | $F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1,ab=-1, a^2=a+1, b^2=b+1$
We will make multiple use of there relations below:
$$S=\frac{1}{5}\sum_{k=0}^{n} (-1)^k[a^{2k}+b^{2k}-2(ab)^k]~~~~(1)$$ $$S=\frac{1}{5}\sum_{k=0}^{n}(-1)^k\left[a^{2k}+b^{2k}-2(n+1)]\right]~~~~(2)$$
$$S=\frac{1}{5}\left[\frac{(-1)^{n+1}a^{2(n+1)}-1}{-a^2-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991692",
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"source": "stackexchange",
"question_score": "1",
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Minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$? Given: $\begin{cases}x;y;z \in\Bbb R\\x+y+z=3\end{cases}$
Then what is the minimum value of $A = (x^2 + 1)(y^2 +1)(z^2 +1)$ ?
I start with $AM-GM$
So $A \geq 8xyz$
Then I need to prove $xyz\geq 1$ right?
How to do that and if it's a wrong way, please help me to solve it... | Let $A(x,y,z)=(1+x^2)(1+y^2)(1+z^2)$, defined on $\{(x,y,z)\in\mathbb{R}^3{\,\mid\,}x+y+z=3\}$.
Since $\max(|x|,|y|,|z|)\ge 3$ would yield $A > 8$, it follows $A$ has a global minimum value, $a$ say.
Since $(x,y,z)=1$ yields $A=8$, it follows that $a\le 8$.
Suppose $A(x,y,z)=a$.
Without loss of generality, assume $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Area bounded by $2 \leq|x+3 y|+|x-y| \leq 4$
Find the area of the region bounded by $$2 \leq|x+3 y|+|x-y| \leq 4$$
I tried taking four cases which are:
$$x+3y \geq 0, x-y \geq 0$$
$$x+3y \geq 0, x-y \leq 0$$
$$x+3y \leq 0, x-y \geq 0$$
$$x+3y \leq 0, x-y \leq 0$$
But it becomes so confusing to draw the region. Any be... | Well, if you're looking for a plot:
Now, if we want to find the area of the part where $y\ge0$ and $x\ge0$, we can take a look at the following picture:
We can find the shaded area by finding:
$$3\cdot\left(\int_0^\frac{1}{2}\left(1-x\right)\space\text{d}x-\frac{1}{2}\cdot\frac{1}{2}\right)+\frac{1}{2}\cdot\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Finding standard deviation given mean and probability of a sample (algebraically). I am trying to solve for the standard deviation of a normal distribution given the probability of a sample and mean:
$P(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp(-0.5(\frac{x-\mu}{\sigma})^2)$
So given $P(x)=a, x=b, \mu=c$ is it possible to fin... | $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Indeed, this case is suitable for solution in terms of the Lambert $\W$ function as follows:
\begin{align}
\frac1{\sqrt{2\pi}\sigma}\exp\left(-\frac{(b-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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what value of K does the system have a unique solution $\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$
I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$
So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$
I'... | If $k\neq -\frac12$
$$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
2 & -1 & k & | & 5 \\
1 & 10 & -6 & | & 1 \\
\end{matrix}
\right) \xrightarrow[\text{$R_3=R_3-R_1$}]{\text{$R_2=R_2-2R_1$}}$$ $$\left(
\begin{matrix}
1 & k & -1 & | & 2\\
0 & -1-2k & k+2 & | & 1 \\
0 & 10-k & -5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$
The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:
\begin{align}
\int \... | As substitution by parts goes nowhere It simply do what you did but skip the step of multiplying out the polynomial $ \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}=\frac {(x(x+1) + 3)(x^3+7)}{x+1}= x(x^3 +7) + \frac 3{x+1}(x^3 + 7)=$
$x(x^3 +7) +\frac 3{x+1}(x^2(x+1) -x(x+1) + x+1 + 6)=x(x^3 +7) + 3(x^2-x+1) + \fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution? In this answer, I give an elementary solution of the Diophantine equation $$y^2=3x^4+3x^2+1.$$
In this post, the related equation $$y^2=3x^4-3x^2+1 \tag{$\star$}$$ is solved (in two different answers!) using elliptic curve theory. Is there an elementary m... | An elementary (and general) method
This is a little-known method which will solve many equations of this type. For this example it will show that the only positive rational solution is $p=1$. We start by letting $p=\frac{y}{x}$, where $x$ and $y$ are coprime integers. Thus we seek positive integer solutions of $$ x^4-3... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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The value of $\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$ What is the value of this expression :
$$\sin\left(\frac{1}{2}\cot^{-1}\left(\frac{-3}{4}\right)\right)$$
Using calculator and wolfram alpha, the answer is, $-\frac{1}{\sqrt{5}}$
But, by solving it myself the result comes out to be different.... | You got the expression $\cot \theta= \frac{-3}{4}$
As you may know, the value of $\cot \theta$ is postive in the first and third quadrants, which means it is negative in the second and fourth quadrants.
When $\cot \theta=\frac{-3}{4}$, $\cos \theta$ can either equal $\frac{-3}{5}$ or $\frac{3}{5}$
$\cos\theta=\frac{3}{... | {
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"url": "https://math.stackexchange.com/questions/4004465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Sum of square roots inequality
For all $a, b, c, d > 0$, prove that
$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$
The idea would be to use AM-GM, but $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$ is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a t... | Alternative solution using Cauchy-Schwarz, which finishes the problem off immediately. By C-S, we have:
\begin{align}
& (a+b+c+d)(1+1+1+1) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\
& \Rightarrow 4(a+b+c+d) \geq (\sqrt{a}+ \sqrt{b}+ \sqrt{c}+ \sqrt{d} )^2 \\
& \Rightarrow 2\sqrt{a+b+c+d} \geq \sqrt{a}+ \sqrt{... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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$ \sum_{i=1}^{10} x_i =5 $, and $ \sum_{i=1}^{10} x_i^2 =6.1 $ ; Find the largest value of these ten numbers
Ten non-negative numbers are such that $ \sum_{i=1}^{10} x_i =5 $,
and $ \sum_{i=1}^{10} x_i^2 =6.1 $. What is the greatest value of the
largest of these numbers?
It presents a lot of confusion to me, but I t... | A General Problem
Suppose
$$
\frac{\sum_{i = 1}^n x_i}{n} = a
$$
and
$$
\frac{\sum_{i = 1}^n x_i^2}{n} = b^2
$$
Set $y_i = x_i - a$. Then (as you can check)
$$
\frac{\sum_{i = 1}^n y_i}{n} = 0
$$
So
$$
b^2 = \frac{\sum_{i = 1}^n (y_i + a)^2}{n} = \frac{\sum_{i = 1}^n y_i^2 + 2ay_i + a^2}{n} = \frac{\sum_{i = 1}^n y_i^... | {
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"url": "https://math.stackexchange.com/questions/4006963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Numerically evaluate a series involving Stirling numbers of the second kind How can I numerically evaluate the following for $n$ above one million:
$$ \sum_{k>=n} \sum_{i=0}^{n-1} (-1)^i {{n-1}\choose{i}} \left( \frac{n-1-i}{n} \right)^{k-1} k (k+1)$$?
I have tried changing the order of summation and the getting closed... | Recall that $${n \brace k}=\frac{1}{k!}\sum _{i=0}^k(-1)^i\binom{k}{i}(k-i)^n,$$
so your expression is equal to
$$\sum _{k\geq n}\frac{(n-1)!k(k+1)}{n^{k-1}}{k-1\brace n-1}=2(n-1)!\sum _{k\geq n}\frac{1}{n^{k-1}}\binom{k+1}{2}{k-1\brace n-1},$$
if $n$ is large, then $k$ is large, so ${k-1\brace n-1}\sim \frac{(n-1)^{k-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^3+2x+1=2^n$ over positive integers? I just have proved that $x^3+2x+1$ can not be divided by $2^n$ ($n>3$), for $x=8k+t$ and $t=0,1,2,3,4,6,7$, but for $x=8k+5$, I have no way to deal with this case, may someone give me some hints for this question? It seems that $(x,n)=(1,2)$ is the unique solution. Also what... | Use the Quadratic Reciprocity - https://en.wikipedia.org/wiki/Quadratic_reciprocity (The supplementary laws using Legendre symbols):
\begin{align}
\left(\frac{-2}{p}\right) = (-1)^{\frac{p^2+4p-5}{8}}=\begin{cases}
1 & p\equiv 1,3~\text{(mod 8)}\\
-1 & p\equiv 5,7~\text{(mod 8)}
\end{cases}
\end{align}
\begin{align}
{... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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maximum and minimum value of $(a+b)(b+c)(c+d)(d+e)(e+a).$
Let $a,b,c,d,e\geq -1$ and $a+b+c+d+e=5.$ Find the maximum and minimum value of $S=(a+b)(b+c)(c+d)(d+e)(e+a).$
I couldn't proceed much, however I think I got the minimum and maximum case.
For minimum, we get $-512$ with equality on $(-1,-1,-1,-1,9).$
For maxi... | We can solve the problem by substitution.
Step 1: Let
\begin{align}
&A = a+b \\
&B = b+c \\
&C = c+d \\
&D = d+e \\
&E = e+a \\
\end{align}
Then the problem is equivalent to find the maximum and minimum of $ABCDE$, where $A,B,C,D,E\ge -2$ and $A+B+C+D+E = 10$.
\begin{align}
{}
\end{align}
Step 2: Consider ea... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How can I find the domain of this sqrt log function?
Can someone help me to find the domain of this function step by steps,
I know how to find the domain for a log by making it >0 but I am unsure about the process doing it with 2 different logarithm and a square root.
$$f(x)=\sqrt{\log_{0.5}(5-x)+\log_2(2x-4)-1}$$
| Let's first define both the $\log$s, input of $\log$ should be a positive quantity.
$5 - x > 0 \implies x \in (-\infty, 5)$
$2x - 4 > 0 \implies x \in (2, \infty)$
Now we know that $\log_{a}{b} = -\log_{\frac{1}{a}}{b} = \log_{\frac{1}{a}}{\frac{1}{b}}$
$\implies\log_{0.5}(5-x) = \log_{2}{\frac{1}{5-x}}$
Now, we know... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4012619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$ Prove that
$$x = \frac 21 \times \frac 43 \times \frac 65 \times \frac 87 \times \cdots \times \frac{9998}{9997} \times \frac {10000}{9999} > 115$$
saw some similar problems like
show $\fra... | Looking at your method, you've actually proven $\frac{3}{4}x^2 > 10000$ as well – just, instead of writing $\frac{2}{1} \geqslant \frac{3}{2}$, you can incorporate that $\frac{3}{4}$ here, and you'll have exactly $\frac{3}{2} = \frac{2}{1} \cdot \frac{3}{4}$. Now we can deduce that
$$ x > \sqrt{\frac{40000}{3}} \approx... | {
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"url": "https://math.stackexchange.com/questions/4012920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Holomorphic Function $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$ Let $g(z)=\sum_{n=1}^\infty (2-3i)\cdot \dfrac{1}{n! z^n}$
Prove that this function is holomorphic in $\mathbb{C} \setminus \{0\}$ and calculate $$\int_{|z|=2} g(z)\,\mathrm{d}z$$
Edit:
What I have done:
\begin{align}
\begin{split}
g(z) & = \s... | It was discussed in the comments that $g(z)=(2-3i)(e^{1/z}-1)$ is holomorphic in the complex plane excluding the origin. The function $e^{1/z}$ is a composition of holomorphic functions, thus holomorphic.
The residue is $2-3i$ so the integral evaluates to $2 \pi i \cdot(2-3i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If$x^2+y^2+z^2+t^2=x(y+z+t)$prove $x=y=z=t=0$
If
$$x^2+y^2+z^2+t^2=x(y+z+t)$$ Prove $x=y=z=t=0$
I added $x^2$ to both side of the equation:
$$x^2+x^2+y^2+z^2+t^2=x(x+y+z+t)$$
Then rewrite it as:
$$x^2+(x+y+z+t)^2-2(xy+xz+xt+yz+yt+zt)=x(x+y+z+t)$$
$$x^2+(x+y+z+t)(y+z+t)=2(xy+xz+xt+yz+yt+zt)$$
But it doesn't seem usefu... | We have$$x^2+y^2+z^2+t^2=x(y+z+t)\iff x^2-\frac34x^2+(y-\frac x2)^2+(z-\frac x2)^2+(t-\frac x2)^2=0$$ i.e.$$\left(\frac x2\right)^2+\left(y-\frac x2\right)^2+\left(z-\frac x2\right)^2+\left(t-\frac x2\right)^2=0$$
We are done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4014634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Divisibility by 7 Proof by Induction Prove by Induction that
$$7|4^{2^{n}}+2^{2^{n}}+1,\; \forall n\in \mathbb{N}$$
Base case:
$$
\begin{aligned}
7&|4^{2^{1}}+2^{2^{1}}+1,\\
7&|7\cdot 3
\end{aligned}$$ Which is true.
Now, having $n=k$, we assume that:
$$7|4^{2^{k}}+2^{2^{k}}+1,\;\; \forall k\in \mathbb{N}$$
We have to ... | $7|8\cdot 2^{2^{k}}+2\cdot 2^{2^{k}}-2\cdot 2^{2^{k}}-2^{2^{k}}$
$7| 2^{2^{k}}\cdot (8+2-2-1)$
$7| 2^{2^{k}}\cdot (7)$
I think I proved it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 4
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Maximum value of $(x−1)^2+ (y−1)^2+ (z−1)^2$ with constraint $x^2+y^2+z^2 ≤2 , z≤1$ So the problem is that I have $D(f)=\{(x,y,z), x^2+y^2+z^2 ≤2 , z≤1\}$ and I have to determine the maximum value for the function $(x−1)^2+ (y−1)^2+ (z−1)^2$ in $D$.
I'm just confused as I don't actually know if $z\le1$ counts as a con... | From the inequality: $(a+b)^2 \le 2(a^2+b^2) $, we have $$(-(x+y))^2 \le 2(x^2+y^2) \implies -(x+y) \le\sqrt{2(x^2+y^2)}$$
And by applying the constraint $x^2+y^2+z^2 ≤2$,
\begin{align}
(x−1)^2+ (y−1)^2 &= (x^2+y^2) -2(x+y)+2 \\
& \le (x^2+y^2) +2 \sqrt{2(x^2+y^2)}+2\\
& \le (2-z^2) +2 \sqrt{2(2-z^2)}+2\\
\end{align}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4017526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Find a matrix $A$ with no zero entries such that $A^3=A$ I took a standard $2 × 2$ matrix with entries $a, b, c, d$ and multiplied it out three times and tried to algebraically make it work, but that quickly turned into a algebraic mess. Is there an easier method to solve this?
| If $ad-bc=1,$ with $a,b,c,d$ non-zero, you can use:
$$\begin{pmatrix}-bc&-bd\\ac&ad\end{pmatrix}$$
For example, $a=2,b=1,c=1,d=1$ gives:
$$\begin{pmatrix}-1&-1\\2&2\end{pmatrix}$$
I got this by taking:
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}\begin{pmatrix}0&0\\0&1\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 3
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$3^n+4^n<5^n$ for all $n>2$ I'm doing the following induction proof and wanted to know if this was valid. I think it is, but I'm seeing more complicated solutions than what I did. What I did seems much easier.
Prove that $3^n+4^n<5^n$ for all $n>2$.
When $n=3$ we get $91<125$. No problem, now assume the result is true ... | Here's a method I like
for showing
$f(n) > g(n)$
for $n \ge n_0$.
*
*Show that
$f(n_0) > g(n_0)$.
*Show that,
if $n \ge n_0$
and $f(n) > g(n)$ then
$f(n+1)-f(n)
\ge g(n+1)-g(n)
$.
Then $f(n) > g(n)$
for $n \ge n-0$.
In this case,
$f(n) = 5^n,
g(n) = 3^n+4^n$.
Choose $n_0 = 3$.
$f(n_0) = 5^3=125,
g(n_0) = 3^3+4^3 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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At least how many different numbers are there from $a_1,a_2,\cdots,a_{1394}$?
Consider positive integers $a_1,a_2,\ldots,a_{1394}$ so that neither
of two numbers of
$\dfrac{a_1}{a_2},\dfrac{a_2}{a_3},\cdots,\dfrac{a_{1393}}{a_{1394}}$
be equal to each other.
at least how many different numbers are there
from $a_1,a_2,... | You are on the right track.
What you need is to realize that essentially you are looking for the smallest $n$ such that $2{n\choose 2}\geq 1394$.
See, the first sequence $2,3,2$ has length of pairs $2=2{2\choose 2}$.
The second sequence $2,3,2,5,3,5,2$ has length of pairs $6=2{3\choose 2}$
The third sequence $2,3,2,5,3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to use telescoping series to find: $\sum_{r=1}^{n}\frac{1}{r+2}$ I am a bit confused in this one, how do i do the required modification in this case?
| As stated in the comments, $\sum_{r = 1}^{n} \frac{1}{r + 2}$ does not telescope. As for an example of a sum that does telescope, consider $\sum_{r = 1}^{n} \frac{1}{r(r + 1)}$. We can rewrite this summation as
$$\sum_{r = 1}^{n} \frac{1}{r(r + 1)} = \sum_{r = 1}^{n} \left(\frac{1}{r} - \frac{1}{r + 1}\right) = \left(\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proof by induction: Inductive step struggles Using induction to prove that:
$$
1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} +...+\left( \frac{1}{2}\right)^{n} = \frac{2^{n+1}+(-1)^{n}}{3\times2^{n}}
$$
where $ n $ is a nonnegative integer.
Preforming the basis step where $ n $ is equal to 0
$$
1 = \frac{2^{1}+(-1)^{0}}{... | Multplying numerator and denominator of the first term by $2$ we get $$
\frac{2^{k+1}+(-1)^{k}}{3*2^{k}} + (- \frac{1}{2})^{k+1} =\frac{2^{k+2}+2(-1)^{k}}{3*2^{k+1}} + (- \frac{1}{2})^{k+1} =\frac {2^{k+2}+2(-1)^{k}+3(-1)^{k+1}} {3*2^{k+1}}
$$ Now use the fact that $2(-1)^{k}+3(-1)^{k+1}=(-1)^{k+1}$ (which can be chec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{n \to \infty}\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}$ Find the following limit $$\lim_{n \to \infty}\left(\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+ \dots +\frac{2n-1}{2^n}\right)$$
I don´t get catch a idea, I notice that
$$\left(\frac{3}{2}+\frac{5}{2^2}+\frac{7}{2^3}+\cdots + \frac{... | 1)First you need to proof that this sequence is a absolute convergent sequence.
Hence the convergence properties wont be affected by the finite basic algebraic manipulation ,such as multiplying by a real number ,or exchanging the position of elements of the S(n).
*let S=lim S(n) (n⇒+∞)
1/2 S=1/2 limS(n) (n⇒+∞)
S（n）-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Picking out unique solution of $x,y,z$ from two constraints
Let $x,y,z$ be positive real numbers in $\mathbb{R}$ satisfying the conditions:$x+y+z=12$ and $x^3 y^4z^5=(.1)600^3$. Then the value of $x^3 + y^3 +z^3=?$
This is a question from JEE mains, in it I have a doubt of how we are able to uniquely determine an $x,... | For positive reals, we can use AM-GM inequality. This inequality for $12$ terms applies as
$$\frac{3\cdot (x/3)+4\cdot (y/4)+5\cdot (z/5)}{3+4+5} \ge \left(\frac{x^3}{3^3}\cdot \frac{y^4}{4^4}\cdot \frac{z^5}{5^5} \right)^{1/(3+4+5)}$$
$$\Rightarrow x^3\cdot y^4 \cdot z^5 \le 3^3 \cdot 2^8 \cdot 5^5$$
$$\Rightarrow x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$ I hit a wall on this question. Below are my steps
$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$
Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$
$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)... | Mistake appeared in evaluating:
$$\int\frac{1}{2u\pm a^2}\,\mathrm{d}u.$$
To solve this you have to substitute $v=2u\pm a^2$, $\mathrm{d}u=1/2 \mathrm{d}v$. We get
$$\frac{1}{2}\int\frac{1}{v}\,\mathrm{d}v.$$
From now on everything should be fine.
There is also a much easier way to evaluate this integral.
Substitute $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What is $(7^{2005}-1)/6 \pmod {1000}$? What is $$\frac{7^{2005}-1}{6} \quad(\operatorname{mod} 1000)\:?$$
My approach:
Since $7^{\phi(1000)}=7^{400}=1 \bmod 1000, 7^{2000}$ also is $1 \bmod 1000$.
So, if you write $7^{2000}$ as $1000x+1$ for some integer $x$, then we are trying to $((1000x+1)\cdot(7^5)-1)/6 = (16807000... | By geometric sum formula we have
$$\frac{7^{2005}-1}{6} = \frac{7^{2005}-1}{7-1} = 1+7+7^2+\dots+7^{2004}$$
The sequence $1,7,7^2,\dots$ has period $20 $ modulo $1000$ (since $7^{20} \equiv 1 \pmod{1000}$).
$$1+7+7^2+\dots+7^{2004} \equiv 100(1+\dots+7^{19})+1+7+7^2+7^3+7^4 \equiv 100\cdot 0+801\equiv 801 \pmod{1000} $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4043349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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How to find the second time when a piezoelectric crystal vibrates given a cubic trigonometric equation? The problem is as follows:
In an electronics factory a quartz crystal is analyzed to get its
vibration so this frequency can be used to adjust their components.
The length that it expands from a certain charge is gi... | You made the same mistake a few times: you neglected the $t$ in $\tan\frac{\pi t}{24}$ in some of your equations, which may be the cause of your confusion. We have the equation
$$x^3+3x^2-3x-1=0$$ where $x=\tan\frac{\pi t}{24}$. Simply by inspection, after trying a few different values for $x$ we see that $x=1$ is a so... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Alternative proof of the Fibonacci property $\sum\limits_{j=n}^{n+9}F_j=11\cdot F_{n+6}.$ If we consider the standard Fibonacci sequence$$F_0=0;\quad F_1=1;\quad F_n=F_{n-1}+F_{n-2}\quad\forall n\geq 2,$$ it is easy to proof that, given any chunk of $10$ consecutive Fibonacci, its sum is equal to $11$ times the seventh... | After OP changes 10 to 9.
Use $F_j=\frac{a^j-b^j}{\sqrt{5}}, a+b=1, ab=-1, a^2=a+1, b^2=b+1$
$$S_n=\sum_{j=n}^{n+9} F_j= \sum \frac{a^j-b^{j}}{\sqrt{5}}$$
USE sum of finite GP, changing the index as $j-n=k$
$$S_n=\sum_{k=0}^{9} \frac{a^{n+k}-b^{n+k}}{\sqrt{5}}=\frac{1}{\sqrt{5}} \left(a^n \frac{a^{10}-1}{a-1}-b^n\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4049985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$ I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}$$
applying Taylor's series.
$$\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \tan 2x}=\lim_{x\to 0^{+}} \frac{\ln (1- \cos 2x)}{\ln \frac{\sin 2x}{\cos 2x}}= \lim_{x\to 0^{+}} \frac... | Now @AdamRubinson has identified the error, note the limit is $L+2$ with$$L:=\lim_{x\to0^+}\frac{\ln\frac{1-\cos2x}{\tan^22x}}{\ln\tan2x}=\lim_{x\to0^+}\frac{\ln\tfrac12}{\ln2x}=\tfrac{-\ln2}{-\infty}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4050600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Integrate $\frac{\log(x^2+4)}{(x^2+1)^2}$.
Using residue calculus show that
$$\int_0^{\infty}\frac{\log(x^2+4)}{(x^2+1)^2}dx=\frac{\pi}2\log 3-\frac{\pi}6.$$
I was thinking of using some keyhole or semi-circular contour here. But the problem is apart from poles at $x=-i$ and $x=i$, the logarithm has singularities whe... | Integrating
$$
\int_\gamma\frac{\log\left(z^2+4\right)}{\left(z^2+1\right)^2}\,\mathrm{d}z\tag1
$$
along the contour
gives $2\pi i$ times the residue at $z=i$.
The integral along the curved pieces vanish as the radius of the large semi-circle grows to $\infty$.
The integral along the downward line along the right side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4052162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. If $\gcd(a,b,c) = 1$ and $c = {ab\over a-b}$, then prove that $a-b$ is a square. $\\$
Well I tried expressing $a=p_1^{a_1}.p_2^{a_2} \cdots p_k^{a_k}$ and $b = q_1^{b_1}.q_2^{b_2}\cdots q_k^{b_k}$ and $c=r_1^{c_1}.r_2^{c_2}\cdots r_k^{c_k}... | Well $\frac {ab}{a-b}$ being an integer seems unlikely and specific.
In particular if $p$ is a prime divisor of $c$ then $p|\frac {ab}{a-b}$ so $p|ab$ so $p|a$ or $b|b$. But $p$ can't divide both as $\gcd(a,b,c) =1$.
But we have that if $p|c$ then either $p|a$ or $p|b$ but not both but then $p\not \mid a-b$ yet for an... | {
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"url": "https://math.stackexchange.com/questions/4052683",
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"source": "stackexchange",
"question_score": "1",
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Prove that $\forall x,y, \, \, x^2+y^2+1 \geq xy+x+y$ Prove that $\forall x,y\in \mathbb{R}$ the inequality $x^2+y^2+1 \geq xy+x+y$ holds.
Attempt
First attempt: I was trying see the geometric meaning, but I´m fall.
Second attempt: Consider the equivalent inequality given by $x^2+y^2\geq (x+1)(y+1)$
and then compare ... | Apply Cauchy-Schwarz or Buniakovsky inequality: $(xy+x+y)^2 = (x\cdot y + 1\cdot x + y\cdot 1)^2 \le (x^2+1^2+y^2)(y^2+x^2+1^2) = (x^2+y^2+1)^2\implies xy+x+y \le x^2+y^2+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
} |
generalized f-mean/power mean inequalities I'd like to prove the following inequality (which seems to be true by numerics)
$$
(p-1)\frac{a^2+b^2}{2} \leq \Big(\frac{a^p+b^p}{2}\Big)^{2/p}
$$
for all $a,b\in [0,1]$ and $p\in [1,2]$. I'd eventually like a generalization when I am summing over many numbers and not just $2... | By applying the Holder's inequality where $\frac{1}{m}+\frac{1}{n} = 1$
$$|x_1 y_1 +x_2y_2| \le (x_1^m +x_2^m)^{\frac{1}{m}} (y_1^m +y_2^n)^{\frac{1}{n}}$$
with $(m,n)=\left(\frac{p}{2},\frac{p}{p-2} \right)$, $(x_1,x_2) = (a^2,b^2)$ and $(y_1,y_2) = (1,1)$, we have
\begin{align}
a^2 +b^2 \le \left((a^2)^{\frac{p}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4054284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Expected value of the sum of two dice. Came across this question:
We roll two dice. Let $X$ be the sum of the two numbers appearing on the dice.
*
*Find the expected value of $X$.
*Find the variance of $X$.
I'm not sure how to do either, but this was my thinking for part 1:
$$E(X) = 2((1/6)^2) + \\
3(2(1/6)^2) + \\... | $$E(X)=E(X_1+X_2)=E(X_1)+E(X_2)=3.5+3.5=7$$
can you calculate $V(X)$?
$$V(X_1)=E(X_1^2)-E^2(X_1)\approx 2.917$$
thus
$$V(X)=2\times 2.917=5.8\overline{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4057700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{x-1389}{7}}+\sqrt{\frac{x-1390}{6}}$
How many solutions the following equation has in real numbers
?$$\sqrt{\frac{x-8}{1388}}+\sqrt{\frac{x-7}{1389}}+\sqrt{\frac{x-6}{1390}}=\sqrt{\frac{x-1388}{8}}+\sqrt{\frac{... | Write $x=1396+t$. Then
$$\sqrt{1+\frac t{1388}}\:+\sqrt{1+\frac t{1389}}\:+\sqrt{1+\frac t{1390}}$$
$$=\sqrt{1+\frac t8}\:+\sqrt{1+\frac t7}\:+\sqrt{1+\frac t6}.$$
Clearly $t=0$ is one solution. Now, $t$ cannot be positive, because that would make the terms on the LHS each smaller than the terms on the RHS. Similarly, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Triangle Properties application In a triangle ABC, AD is the altitude from A. Given b>c, $\angle C = {23^o}$,$AD = \frac{{abc}}{{{b^2} - {c^2}}}$. Then $\angle B = \_\_\_{\_^o}$
My approach is as follow
$\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R$
$AD = \frac{{8{R^3}sinA\sin B\sin C}}{{4{R^2}{{\... | We can use the extended law of sines and the area of a triangle in terms of sides and circumradius formulae.
$$\bigtriangleup=\frac{1}{2}a\cdot AD \implies AD= \frac{2\triangle}{a} = \frac{{abc}}{{{b^2} - {c^2}}}=\frac{4R \bigtriangleup}{{b^2} - {c^2}}$$
$$\implies 2Ra={b^2} - {c^2}$$
$$\implies 2R(2R\cdot \sin(A))=(2R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4059876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true? Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)... | This is just an idea, why don't you view it like this
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$
Where $$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4063411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Probabilities in a game John plays a game with a die. The game is as follows: He rolls a fair six-sided die. If the roll is a $1$ or $2$, he gets $0$ dollars. If he rolls a $3, 4$ or $5$, he gets $5$ dollars. If he rolls a $6$, he wins $X$ dollars where $X$ is a continuous random variable that is uniform on the interva... | In short $Y= 0\mathbf 1_{1\leq X\leq 2}+5\mathbf 1_{3\leq X\leq 5}+Z\mathbf 1_{X=6}$ where $X\sim\mathcal U\{1,2,3,4,5,6\}$ and independently $Z\sim\mathcal U(10..30)$. That means:-
$$Y=\begin{cases}0 &:& 1\leq X\leq 2\\5&:& 3\leq X\leq 5\\Z&:& X=6\\0&:&\textsf{else}\end{cases}$$
(i) The CDF will be a piecewise functi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4067662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$63^{63^{63}} \mod 100$ I need to find $63^{63^{63}} \bmod 100$.
This is what I've got so far:
Since $\gcd(63,100)=1$ we can use Euler's theorem. We have $\phi (100)=40$ so $63^{40} \equiv 1 \mod 100$
Again $\gcd(16,100)=1$ and $\phi (40)=16$, that is $63^{16} \equiv 1 \mod 40$
Using this I got that $63^{63} \equiv ... | $(63)^7 = (60+3)^7 = (60^7 + {7\choose 1} 60^6\cdot 3 + \cdots + {7\choose 6} 60\cdot 3^6 + 3^7 $
All the terms to the left of the last 2 are equivalent to 0 modulo 100, and can be dropped.
Leaving $7\cdot 60\cdot 3^6 + 3^7\pmod {100}$
We only need to worry about the one figure for the first term since we know it will ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to show that there is no positive rational number a such that $a^3$ = 2? I'm stuck on this one.
So far, I've tried this:
\begin{align}
a^3 &= 2 \\
a &= \left(\frac mn\right) \\
a^3 &= \left(\frac mn\right)^3 = 2 \\
m^3 &= 2n^3 \\
m &= 2p \\
m^3 &= (2p)^3
\end{align}
I'm really confused about what to do after this—... | Alternative route:
Let $\alpha^3=2$, and assume $\alpha\in\Bbb Q$, then we have $$\frac{p^3}{q^3}=2\\p^3=2q^3\\p^3=q^3+q^3$$
The last line contradicts Fermat's Last Theorem.
Let $x,y,z$ be non-zero integers. Then if $x^3+y^3=z^3$ we have $xyz=0$.
Proof: We can assume that $x,y,z$ are relatively prime, since if they ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4068683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be? If we let $x = 0$.
\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}
Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.
But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $... | If we let $x=0$ then we do indeed find that the expression equals $21-9y$. But this does not mean that $21-9y$ must equal $0$; it only means that $21-9y$ is a constant. If, as Raffaele has suggested, we let $x=1$, then we find that the expression equals $24-11y$. Hence, $21-9y$ and $24-11y$ must be equal to the same co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Can anyone tell me if my proof is right? Let $0<a<b,$ compute $\lim_{n\to \infty}{\frac{a^{n+1}{+}b^{n+1}}{a^n+b^n}}$
My proof are followings :
Assuming that $r=b-a>0,$ so $b=a+r.$ So the original formula equals to $\frac{a\cdot a^{n}{+}{(a+r)\cdot}b^{n}}{a^n+b^n}=a+\frac{r\cdot b^n}{a^n+b^n}=a+\frac{r}{1+(a/b)^n},$ so... | Your proof is correct. Here's another way to see why the limit is $b$:
Notice that
\begin{align*}
\frac{a^{n+1}+b^{n+1}}{a^n+b^n} &= a^n\cdot\frac{a}{a^n+b^n}+b^n\cdot\frac{b}{a^n+b^n}\\
&= \frac{1}{\frac{1}{a^n}}\cdot\frac{a}{a^n+b^n}+\frac{1}{\frac{1}{b^n}}\cdot\frac{b}{a^n+b^n}\\
&= \frac{a}{1+\left(\frac{b}{a}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Question about basis change/transformation Let
$B=\left[ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 0\\ 1 \end{pmatrix}\right]$
$B_1=\left[ \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \\ -2\end{pmatrix}, \begin{pmatrix} 2 \\ ... | Yes, you are correct. In general, given a linear transformation from $V$ to $W$ vector spaces $f:V \to W$, $\alpha=\{v_1, \cdots, v_n\}$ basis of $V$ and $\beta=\{w_1, \cdots, w_n\}$ basis of $W$ to construct $[f]_{\alpha,\beta}$ you proceed like this: The $i$-th column of the matrix $[f]_{\alpha,\beta}$ is given by th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.