Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How can $S_n=n$ for the series $S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$ How can $S_n=n$ for the series $$S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots$$
My try:
We can Re-write $S_n$ as
$$S_n=\frac{0+1}{2}+\frac{1+2}{2^2}+\frac{3+2^2}{2^3}+\frac{7+2^3}{2^4}+\cdots \frac{2^{n-1}-1+2^... | Given:
$$S_n=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\cdots+a_n,$$
as pointed out by John Omielan, you are missing the first (or last) term:
$$S_n=\frac{\color{red}0+\color{blue}1}{2}+\frac{\color{red}1+\color{blue}2}{2^2}+\frac{\color{red}3+\color{blue}{2^2}}{2^3}+\frac{\color{red}7+\color{blue}{2^3}}{2^4}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3331620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Deriving the vector field from a curl If $B= (x/r^3,y/r^3,z/r^3)$ find a vector $A$ such that $curl A = B$.
My attempt:
$∂A_3∂Y−∂A_2∂Z=x/r^3$
$∂A_1∂Z−∂A_2∂y=y/r^3$
$∂A_2∂x−∂A_1∂y=z/r^3$
I assumed $A_3=0$, ending up having to solve:
$A_2=\int xr^{-3}dz$
Here, I got stuck. Is there an easier way of doing this?
| I assume your book asks for any solution even though physically meaningless. Take for instance $A_3 = 0$. Then
\begin{equation}
B_1 = \frac{\partial A_3}{\partial y}-\frac{\partial A_2}{\partial z} = \frac{x}{(\sqrt{x^2+y^2+z^2})^3}
\end{equation}
Thus,
\begin{equation}
A_2 = -\int \frac{x}{(\sqrt{x^2+y^2+z^2})^3} \mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do I find the real or imaginary part of this complex equation? I am trying to solve the below equation for surface plasmon polariton(SPP) wave
$ \sqrt{\dfrac{1}{\beta^2-(\frac{\omega^2}{c^2})}} + \sqrt{\dfrac{\epsilon_r}{\beta^2-(\frac{\omega^2}{c^2}\epsilon_r)}} = -j \dfrac{\sigma(w)}{\omega \epsilon_0}$
Because ... | You replaced $\epsilon_r$ with $1$ in some but by no means all places, but more to the point, you committed a fallacy of the form $a+b=c\implies a^2+b^2=c^2$. Squaring should give $$2\sqrt{\frac{\epsilon_r}{(\beta^2-\omega^2/c^2)(\beta^2-\epsilon_r\omega^2/c^2)}}=-\frac{\sigma^2}{\omega^2\epsilon_0^2}-\frac{1}{\beta^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ Find value for $k$ such that $(x^2-k)$ is a factor for, $$f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$$
My Try
Since $x^2-k=0$ when we substitute $x=\pm k$ to $f(x)$ it should be equal to $0.$
But ... | I would begin by replacing $x^2$ with $k$ in the polynomial. Where there is an odd power of $x$ you leave the remaining power of $x$ as is. Thus
$2x^4=2k^2$
$(3k-4)x^3=(3k^2-4k)x$
$(2k^2-5k-5)x^2=2k^3-5k^2-5k$
$(2k^3-2k^2-3k-6)x=(2k^3-2k^2-3k-6)x$
$6=6$
Add the right sides together and collect all $x$ terms to get
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How to factorize $\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$? How to factorize and simplify the following?
$$\frac{4x^3+4x^2-7x+2}{4x^4-17x^2+4}$$
I've tried everything I know. Trying to factorize the numerator first then denominator, but I get no where. Usual identities like $(x+y)^2=x^2+2xy+y^2$ don't work either, and ne... | The polynomial in the denominator can be rewritten as $4t^2 - 17t + 4$ where $t = x^2$
Use the quadratic formula to find that this polynomial has roots $t_1 = 4, t_2 = \frac{1}{4}$
Since $t = x^2$, we can factor the bottom polynomial as $(x-2)(x+2)(x-\frac{1}{2})(x + \frac{1}{2})$. The roots are solutions for $x^2 = t_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to equalize correctly? If i have this number:
$2 \sqrt{2-\sqrt{3}}$ and i want to find some $x,y$ nonzero real numbers such that $2\sqrt{2-\sqrt{3}} = \sqrt{x} + \sqrt{y}$
And for that, i do this:
$(2 \sqrt{2-\sqrt{3}})^2 = x + 2\sqrt{xy} + y$
$4(2-\sqrt{3})=(x+y)+2\sqrt{xy}$
$(8)+(-4\sqrt{3})=(x+y)+(2\sqrt{xy})$
... | You should assume a binomial of the form whose root you desire. In this case, you should have supposed the root is $$\sqrt x -\sqrt y$$ instead.
To be specific, the problem in the above calculation is in your step ii, where you set $$-\sqrt{12}=\sqrt{xy}.$$ But this is impossible if you're dealing only with real numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sqrt{{x^2+y^2+z^2}}\, dx \,dy \,dz$ by converting to spherical coordinates
I would like to know how to evaluate the following triple integral with the help of spherical coordinates
$$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sqrt{{x^2+y^2+z^2}} \,dx \,dy\, dz$$
The re... | Substitute:
$$z=\sqrt{x^2+y^2} t$$
$$I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1/\sqrt{x^2+y^2}} (x^2+y^2) \sqrt{{1+t^2}} dt dy dx$$
Integrating w.r.t. $t$:
$$I=\frac{1}{2} \int_{0}^{1} \int_{0}^{1} (x^2+y^2) \left(\frac{1}{\sqrt{x^2+y^2}}\sqrt{1+\frac{1}{x^2+y^2}}+\sinh^{-1} \frac{1}{\sqrt{x^2+y^2}} \right) dy dx$$
Using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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For which $a$, $b$, $c$ does this linear system have exactly one solution? $x+ay+a^2z=0$, $x+by+b^2z=0$, $x+cy+c^2z=0$
For which $a$, $b$, $c$ does this linear system have exactly one solution?
$$x + ay + a^2z = 0$$
$$x + by + b^2z = 0$$
$$x + cy + c^2z = 0$$
I started this problem by recognizing that if the RREF of ... | HINT
According to Cramer's rule, this system admits a unique solution iff the determinant of the coefficient matrix is different from zero. Precisely,
\begin{align*}
\begin{vmatrix}
1 & a & a^{2}\\
1 & b & b^{2}\\
1 & c & c^2\\
\end{vmatrix}\neq 0
\end{align*}
Such matrix is widely known as the Vandermonte's matrix,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The units digit of a power tower of consecutive numbers, from 2019 to 1
Is it possible to find the units digit of
$2019^{2018^{2017^{.^{.^{.^{3^{2^{1}}}}}}}}$?
Where the expression contains all natural numbers $[1,2018]$ as powers and $2019$ as the main base.
Any help would be appreciated. THANKS!
| HINT : Assume $a_n=n^{n-1^{n-2^{.^{.^{.^{3^{2^{1}}}}}}}} \mod 10$, then the sequence will be :
$$
1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0,1,2,9,4,5,6,9,8,1,0...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3338112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Knowing that $m$ and $n$ are positive integers satisfying $mn \mid m^2 + n^2 + m$, prove that $m$ is a square number.
Knowing that $m$ and $n$ are positive integers satisfying $$\large mn \mid m^2 + n^2 + m$$, prove that $m$ is a square number.
We have that $mn \mid m^2 + n^2 + m \implies mn \mid (m^2 + n^2 + m)(n + ... | Let $p$ be prime and let $k\ge0 $ be maximal with $p^k\mid m$. Similarly, let $l\ge 0$ be maximal with $p^l\mid n$. So $m=p^kr$, $n=p^ls$ for some positive integers $r,s$.
So $$p^{k+l}rs\mid p^{2k}r^2+p^{2l}s^2+p^kr. $$
Conclude that $2l\ge k$ and
$$p^{l}rs\mid p^{k}r^2+p^{2l-k}s^2+r. $$
of $k>0$ then also $l>0$ and s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $ If $ a + b + c = 90^{\circ}$, prove $ \tan(a) \cdot \tan(b) + \tan(b) \cdot \tan(c) + \tan(c) \cdot \tan(a) = 1 $
Attempt:
Notice that $$ \tan(a+b+c) = \frac{\tan(a+b) + \tan(c)}{1 - \tan(a+b)\tan(c)} $$... | It's wrong for $a=90^{\circ}$ for example.
But on the domain of our tangents we obtain:
$$1-\sum_{cyc}\tan{a}\tan{b}=1-\tan{a}\tan{b}-\tan{c}(\tan{a}+\tan{b})=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\frac{\tan{c}(\tan{a}+\tan{b})}{1-\tan{a}\tan{b}}\right)=$$
$$=\left(1-\tan{a}\tan{b}\right)\left(1-\tan{c}\tan(a+b)\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Show that $\operatorname{Aff}(3)$ is isomorphic to $S_3$, the symmetric group of all permutations of 3 objects. Show that $\operatorname{Aff}(3)$ is isomorphic to $S_3$, the symmetric group of all permutations of 3 objects.
Where
$$\operatorname{Aff}(3):={\{( \begin{array}{cc}
a & b \\
0 & 1
\end{array}): a,b\in\math... | If $a\mapsto \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$ and $b\mapsto \begin{pmatrix} 2 & 0 \\ 0 & 1\end{pmatrix}$, then $\operatorname{Aff}(3)$ has as a presentation $$\langle a, b\mid a^3, b^2, bab=a^{-1}\rangle,$$ which is, in turn, a presentation for $S_3$. Hence $\operatorname{Aff}(3)\cong S_3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Expression of $x^n+\frac1{x^n}$ by $x+\frac1{x}$ where $n$ is a positive odd number. There was a problem in a book:
Denote that $y=x+\dfrac{1}{x}$, express $x^7+\dfrac{1}{x^7}$ using $y$.
It's not a hard question, but I find a special sequence:
$x+\dfrac{1}{x}=y\\x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\lef... | Let $x=e^{iz}$. We want to express $x^n+\frac{1}{x^n}=2\cos(nz)$ in terms of $x+\frac{1}{x}=2\cos(z)$, which can be done through Chebyshev polynomials of the first kind:
$$ x^n+\frac{1}{x^n}=2\cos(nz) = 2\,T_n(\cos z) = 2\, T_n\left(\frac{x+\frac{1}{x}}{2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Continuity of $f(x,y)=\dfrac{(xy+\sqrt{x^2+y^2})\sin^2 (x+y)}{x^2+y^2}$
$f(x,y)=\dfrac{(xy+\sqrt{x^2+y^2})\sin^2 (x+y)}{x^2+y^2}$ if $xy\neq0$ and $f(x,y)=0$ if $xy=0$, check whether $f(x,y)$ is continous at origin.
For it to be continuous $$\lim_{(x,y)\rightarrow (0,0)} f(x,y)=0$$
using epsilon-delta definition,$$\b... | $\sin^{2}(x+y)\leq (x+y)^{2} \leq 2(x^{2}+y^{2})$. Hence $|f(x,y)|\leq 2|xy|+2\sqrt {x^{2}+y^{2}} \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit without l'Hopital rule I have the following limit:
$$
\lim_{x\to 7}\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}
$$
I could easily calculate the limit = 12 using the l'Hopital rule.
Could you please suggest any other ways to solve this limit without using the l'Hopital rule?
Thank you
| Just another way to do it.
Consider
$$A=\dfrac{x^2-4x-21}{x-4-\sqrt{x+2}}$$ and let $x=y+7$ to work around $y=0$. So,
$$A=\frac{y (y+10)}{y+3-\sqrt{y+9}}$$ Now, using the binomial theorem or Taylor series
$$\sqrt{y+9}=3+\frac{y}{6}-\frac{y^2}{216}+O\left(y^3\right)$$ making
$$A=\frac{y (y+10)}{\frac{5 y}{6}+\frac{y^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a$, $b$, $c$ are three positive integers such that $a^3+b^3=c^3$ then one of the integer is divisible by $7$ Let on contrary that none of the $a$, $b$, $c$ is divisible by $7$. Then either $a^3\equiv b^3\pmod{7}$ or $b^3\equiv c^3\pmod{7}$ or $c^3\equiv a^3\pmod{7}$.
Now how to go further?
| For any integer $x$ we have $x^3 \equiv -1 \text{ or } 0 \text{ or }1\pmod{7}$.
$a^3+b^3=c^3$
$a^3+b^3+(-c)^3=0 \equiv 0\pmod{7}$ [Take modulo 7 on the whole equaton]
This sum can be zero iff one of them is $-1 \bmod7$, one is $1 \bmod 7$ and one is $0 \bmod 7$. Or if each one of them is $0 \bmod 7$.
The result follow... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions to $\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2$? I am looking for all non-negative real solutions to
$$
\left( \frac{1+3x}{1+2x} \right)^{\frac{1+x}{x}}=2.
$$
Numerically it seems that there is a unique solution $x \approx 0.4256$. Any ideas how to prove/find it?
| This is probably nowhere near a full answer, but steps could potentially be useful.
Recognize that $\frac{1+x}{x}=\frac1{x}+1$. Then, the equation can be rewritten as
\begin{align}
\bigg(\frac{1+3x}{1+2x}\bigg)^{\frac{1+x}{x}}=\bigg(\frac{1+3x}{1+2x}\bigg)^{\frac1{x}+1}&=2\\
\bigg(\frac{\frac1{x}+3}{\frac1{x}+2}\bigg)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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What is $\alpha^{4} + \beta^{4} + \gamma^{4}$? If
$$ \alpha + \beta + \gamma = 14 $$
$$ \alpha^{2} + \beta^{2} + \gamma^{2} = 84 $$
$$ \alpha^{3} + \beta^{3} + \gamma^{3} = 584 $$
What is $\alpha^{4} + \beta^{4} + \gamma^{4}$?
Attempt:
Notice that
$$ \alpha + \beta + \gamma = 14 \implies \alpha^{2} + \beta^{2} + \ga... | Well, its visible clearly that numbers are 8, 4 and 2.
From first two equations, $$y = 1/2 (-\sqrt{-3 x^2 + 28 x - 28} - x + 14)$$
$$z = 1/2 (\sqrt
{-3x^2 + 28 x - 28} - x + 14)$$
Putting them into third equation, we get a cubic equation
$$3x^3 - 42 x^2 + 168 x + 392=584$$ solving which by cardano or any method or wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Contradiction when solving a linear system with Gauss-Jordan elimination Consider a linear system with an unknown constant $a$:
$$
\left\{
\begin{aligned}
x+(a-1)y+az&=1 \\
ax+ay+az&=1 \\
a^2x+y+z&=a
\end{aligned}
\right.
$$
This gives us an augmented matrix:
$$A= \left[
\begin{array}{ccc|c}
1&a-1&a&1\\
a&a&a&1\\... | If you want to rref and avoid any division at all you'll get
$$\begin{pmatrix}
a(a-2)(a^2-1)& 0 & 0 & (a-2)(a^2-1)\\
0 & -a(a-2)(a+1) & 0 & (a-2)(a^2-1)\\
0 & 0 & a(a^2-1) & (a+1)(a-1)^2
\end{pmatrix}.$$
From here it's eazy to consider all different cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given concurrent cevians $AD$, $BE$, $CF$ of $\triangle ABC$, show $\text{area }\triangle DEF\leq \tfrac14\text{}\triangle ABC$
Let $M$ be a point in $\triangle ABC$. Lines $AM$, $BM$, $CM$ respectively intersect $BC$, $CA$, $AB$ at $D$, $E$, $F$. Prove that
$$S_{DEF}\leq \tfrac{1}{4}S_{ABC}$$
where $S_{XYZ}$ ind... | Let the segment ratios be
$$x= \frac {CD}{DB}, \>\>\> y= \frac {AE}{EC}, \>\>\> z= \frac {BF}{FA}$$
According to the Ceva's theorem,
$$xyx=1\tag{1}$$
Evaluate the area ratio
$$\frac{S_{AEF}}{S_{ABC}} = \frac{ \frac 12 AE\cdot AF\sin A}{\frac 12 AB\cdot AC\sin A}
=\frac{1}{(1+z)(1+\frac 1y)}=\frac{y}{(1+z)(1+y)}$$
Simi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Angle between a vector and cross product of two vectors
The angle between the vectors $\overrightarrow {a}$ and $\overrightarrow {b}$ is $\pi/3$, $\overrightarrow {b}$ and $\overrightarrow {c}$ is $\pi/4$, $\overrightarrow {c}$ and $\overrightarrow {a}$ is $\pi/6$.
Find the angle between $\overrightarrow {a}$ and $\... | I just got another solution.
Consider $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ to be unit vectors.
$$\left| \overrightarrow {a}\cdot \left( \overrightarrow {b}\times \overrightarrow {c}\right) \right| ^{2}=\begin{vmatrix} \overrightarrow {a}\cdot \overrightarrow {a} & \overrightarrow {a}\cdot \overrig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will b... | Once you’ve rewritten your sum as $\frac14\sum_{k=1}^n k^4+2k^3+k^2$ it’s not terribly difficult to compute this using generating functions if you use some key tools for manipulating them. To wit, if $g(x)$ is the ordinary generating function for the sequence $\{a_n\}_{n=0}^\infty$, then $g(x)/(1-x)$ is the o.g.f. for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Differentiation using l´Hopital I need to use L´Hopital's rule with this functions:
$$\lim_{x\rightarrow\frac{\pi}{2}} (1-\sin(x))^{\cos(x)}$$
$$\lim_{x\rightarrow\frac{\pi}{4}} (\tan(x))^{\tan(2x)}$$
I take the exponent down using the properties of logarithms and then make the denominator like: $\lim_{x\rightarrow\fra... | For the first one we have
$$(1-\sin x)^{\cos x}= (1-\sin x)^{\cos x}\frac{(1+\sin x)^{\cos x}}{(1+\sin x)^{\cos x}}=\frac{(\cos^2 x)^{\cos x}}{(1+\sin x)^{\cos x}}\to \frac 11=1$$
indeed by standard limits
*
*$\lim_{x\rightarrow\frac{\pi}{2}} (\cos^2 x)^{\cos x}=\lim_{t\to 0}(t^2)^t=1$
*$\lim_{x\rightarrow\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3374957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find a $k$ such that $3^k \equiv -6 \pmod{43}$ I have been trying to find this $k$, but I am stuck.
The only information I could extract was from the Fermat's Little Theorem:
Since $43$ doesn't divide 3 and it is a prime, it follows $3^{42} \equiv 1 \pmod{43}$
However, I have no idea how to proceed from now.
All help i... | Notice $3$ and $43$ are relatively prime. Thos if we hae $3m \equiv 3n \pmod{43}$ we can safely conclude that $m \equiv n\pmod{43}$
So if $3^k \equiv -6 \pmod {43}$ then
$3^{k-1} \equiv -2\equiv -45\pmod {43}$
$3^{k-2} \equiv -15$
$3^{k-3}\equiv -5\equiv -48$
$3^{k-4} \equiv -16\equiv 27$
$3^{k-7} \equiv 1\pmod {43}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Roots in equation In the equation
$\sqrt{x-2} - \sqrt{6x-11} + \sqrt{x+3} =0$
I got the roots of $x$ being $6$ and $7\sqrt{3}$.
Considering the graph shows only $6$ as being a valid solution, how should I go as figuring this out in the equation itself?
| Square both sides
$$\sqrt{x-2} + \sqrt{x+3} = \sqrt{6x-11} $$
$$2x+1 + 2\sqrt{(x-2)(x+3)}=6x-11$$
$$\sqrt{(x-2)(x+3)}=2x-6$$
Square again
$$x^2+x-6 = 4x^2 - 24x+36$$
$$ 3x^2 - 25x+42=0$$
$$ (x-6)(3x-7)=0$$
$$x=6,\>\>\>x=\frac 73$$
Plug into the original equation and check validity. Only $x=6$ is the true solution. (Edi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$ is an integer and find its value I had proceed this question by taking
$$x =(5^{\frac 12} + 2)^{\frac 13}- (5^{\frac 12} - 2)^{\frac 13}$$
Then
$$x + (5^{\frac 12} -2)^{\frac 13} =(5^{\frac 12} + 2)^{\frac 13}$$
And then cubing both sides and the... | Let $$x=\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2},$$
raise it to the third power:
$$\begin{align}
x^3&=\sqrt5+2-3\sqrt[3]{\sqrt5+2}^2\sqrt[3]{\sqrt5-2}
+3\sqrt[3]{\sqrt5+2}\sqrt[3]{\sqrt5-2}^2-(\sqrt5-2)\\
&=4-3\cdot\sqrt[3]{\sqrt5+2}\sqrt[3]{\sqrt5-2}\cdot\bigl(\sqrt[3]{\sqrt5+2}-\sqrt[3]{\sqrt5-2}\bigr)\\
&=4-3\cdot\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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In an integral domain prove that $a=b$
Let R be an integral domain and $a,b\in R$ are elements satisfying $a^7=b^7$ and $a^{12} = b^{12}$. Prove that $a=b$.
My attempt:-
Since $R$ is an integral domain, R is commutative. That is all I could conclude...could you please help me out?
| If $b=0$ then $a^6=0$ but $R$ is an integral domain, so $a=0=b$
We suppose that $b\neq 0$. Then
$a^{12}-b^{12}=(a^6-b^6)(a^6+b^6)=0$
but $R$ is an integral domain, so
$a^6-b^6=0$ or $a^6+b^6=0$
If $a^6=b^6$ then
$b^7=a^7=aa^6=ab^6$ that implies
$b^6(b-a)=0$
but $b\neq 0$ then $b=a$
Conversely, if $a^6=-b^6$, then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Find an approximate sum of the series $\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}$ $$
S=\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}\ \ \ (*)
$$
I have to find $S_n: |S-S_n|<\epsilon=10^{-4}$
Also, I cannot use expansion series for any functions.
This is what I came up with so far:
$$
a_k=\frac{1}{2^{2k+1}(2k+1)}\leq... | $S
=\sum_{k=0}^{\infty}\frac{1}{2^{2k+1}(2k+1)}\ \ \ (*)
$
Let
$t_n
=\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}(2k+1)}
$.
Then
$\begin{array}\\
t_n
&\lt\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}(2n+1)}\\
&=\frac1{2n+1}\sum_{k=n}^{\infty}\frac{1}{2^{2k+1}}\\
&=\frac1{2n+1}\sum_{k=0}^{\infty}\frac{1}{2^{2k+2n+1}}\\
&=\frac1{(2n+1)2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$ Question :
Solve for natural number the equation :
$5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
My try :
Let : $X=5^{x}$ and $Y=2^{y}$ so above equation
equivalent :
$2X^{2}+(Y-4)X-6Y^{2}-Y+2... | Now substitute back: you get that $5^x=3\cdot 2^{y-1}+1$ or $5^x=1-2^{y+1}$. The second one is impossible cause the RHS is $\leq 0$. For the first one to hold, looking mod 3 you see that $x$ needs to be even, so write $x=2x'$. Then $(5^{x'}-1)(5^{x'}+1)=3\cdot 2^{y-1}$. So the only chance is that $5^{x'}-1=2^{y_1}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find all $n$ natural numbers such that $10\mid n^{10}+1$ Determine all natural numbers $n$ such that :
$10$ divisor of $n^{10}+1$
My attempt :
Let $n=r(\mod{10})$ so $n^{10}+1=(r^{10}+1)(\mod{10})$
This mean : $r^{10}+1=0(\mod{10})$
Now $r\in {0,1,2,3,4,5,6,7,8,9}$ after try I get $r=3,7$
So : $n=10k+3,10k+7$
Is... | By Fermat's theorem, $n^5 \equiv n \bmod 5$. Therefore, $n^{10} \equiv n^2 \bmod 5$ and so $n^{10} +1 \equiv n^2 +1 \bmod 5$. Thus, if $10\mid n^{10}+1$, then $n^2 \equiv 4 \bmod 5$ and so $n \equiv \pm 2 \equiv 2,3 \bmod 5$. Thus, $n \equiv 2,3,7,8 \bmod 10$. Since $n$ is odd, we're left with $n \equiv 3,7 \bmod 10$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Find the explicit form of $ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1} $. Find the explicit form of
$$
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}.
$$
Let $S(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n(n+2)}x^{n-1}$. It has radius of convergence $1$.
Let $S_1(x)=xS(x)$. Then $S_1'(x)=\sum_{n=1}^{\infty}\... | @ Tao X : Your method is not the simplest. But what you did is correct :
$$S_1'(x)=\frac{1}{2x}-\frac{1}{x^2}+\frac{\ln (x+1)}{x^3}$$
Then integrate :
$$S_1(x)=\int S_1'(x)dx=\frac{1}{2x}-\frac{\ln (x+1)}{2x^2}+\frac12\ln(x+1)+C$$
We know that $S_1(0)=0$. This allows to find $C$. Expand $\ln(x+1)$ to series around $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3386371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Divergence theorem, Gauss's theorem $$\mathbf F(x,y,z)=x\mathbf i + y\mathbf j+z\mathbf k$$ and
$$ D=\left\{x,y,z\in \mathbf R^3 : 0\le z \le1-x^2-y^2\right\}.$$
I want to calculate both
$$\iiint_D\nabla\cdot\mathbf F\,dV = \oint_{\delta D}\mathbf F \cdot\mathbf N\,dS.$$
So far:
$$\nabla f=1+1+1=3$$
$$\oint_{\delta D}f... | As I said in the comments, $D$ is not a ball. It's bounded below by the unit disk in the $xy$-plane, and above by the paraboloid $z=1-x^2-y^2$. (You may be thinking of $z = \sqrt{1-x^2-y^2}$; that surface is the upper unit hemisphere.) In cylindrical coordinates,
$$
D = \left\{(r,\theta,z) : 0 \leq r \leq 1,\ 0 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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To prove that $a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$ To prove: $$a_n=[x^n]\frac {x-2x^2+x^3}{1-2x+x^2-x^4}$$
Given that, $a_n$is the number of compositions of $n$ in which the number of parts is odd with the first part being equal to the number of parts.
| For $k\geq 1$ and $n\geq 3$, the number of positive integer solutions of
$$(2k+1)+x_2+x_3+\dots+x_{2k+1}=n$$
that is of
$$x_2+x_3+\dots+x_{2k+1}=n-(2k+1)$$
which is equal to
$$\binom{n-(2k+1)-1}{2k-1}.$$
Hence
$$a_n=\sum_{1\leq k\leq \lfloor (n-1)/4\rfloor}\binom{n-(2k-1)-1}{2k-1}.$$
Now by partial fraction decompositi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3389238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Maximum value of $a+b+c$ in an inequality Given that $a$, $b$ and $c$ are real positive numbers, find the maximum possible value of $a+b+c$, if
$$a^2+b^2+c^2+ab+ac+bc\le1.$$
From the AM-GM theorem, I have
$$a^2+b^2+c^2+ab+ac+bc\geq 6\sqrt[6]{a^4b^4c^4} = 6\sqrt[3]{a^2b^2c^2} \\
6\sqrt[3]{a^2b^2c^2} \le1 \\
a^2b^2c^2 \l... | Notice that
$$ab+bc+ca\leq\frac{(a+b+c)^2}{3}.$$
Hence
\begin{align}
a^2+b^2+c^2+ab+bc+ca&=(a+b+c)^2-(ab+bc+ca)\\&\geq (a+b+c)^2-\frac{(a+b+c)^2}{3}\\&=\frac{2}{3}(a+b+c)^2.
\end{align}
That is
$$\frac{2}{3}(a+b+c)^2\leq 1,$$
which we get
$$a+b+c\leq\sqrt{\frac{3}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Rational with minimal denominator between two rationals My question from an easy problem.
$p,q$ are positive integers such that $$
\frac{5}{9}<\frac{p}{q}<\frac{4}{7}
$$ find $p,q$ such that $q$ is the smallest number that satisfies this inequality.
Draw the line of $ y<\frac{9}{5}x$ and $y>\frac{7}{4}x$ , we can "ob... | What do we get with continued fractions?
$\dfrac{5}{9}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{4}}}$
$\dfrac{4}{7}=\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}$
The fractions are identical until we get to the last layer where one has a $4$ and the other has a $3$. Were there an integer between $3$ and $4$ we could replace the las... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The sum of the series
$$\frac{1}{\log_24}+\frac{1}{\log_44} + \frac{1}{\log_84}.....\frac{1}{\log_{2^n}4}$$
MY SOLUTION
We can write it as
$$\frac{\log2}{\log4} + \frac{\log4}{\log4} + ....\frac{\log 2^n}{\log4}$$
$$=\frac{1}{\log4}\left[\log 2 + \log 4....\log 2^n\right]$$
$$=\frac{1}{\log 4}\left [\log(2.4.8....2^... | $\dfrac{1}{2\log 2}[\log 2(1+2+3+....n)]=$
$\dfrac{n(n+1)}{2\cdot 2}$;
Used:
1)$\log 4= \log 2^2=2\log/2$;
2)$\log 2+\log 2^2+.......\log 2^n=$
$\log 2 +2\log 2 +....+n\log 2=$
$\log 2(1+2+3+.....n)=$
$\log 2\dfrac{n(n+1)}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How do I solve a problem with term $a^{n} + b^{n}$?
Given two non-zero numbers $x$ and $y$ such that $x^{2} + xy + y^{2} = 0$.
Find the value of
$$\left(\frac{x}{x + y}\right)^{2013} + \left(\frac{y}{x + y}\right)^{2013}$$.
I found out that $(x + y)^2 = xy$ and I'm stuck at $\frac{x^{2013} + y^{2013}}{(x + y)^{2013}}... | We have
$${y}^{2}=-{x}^{2}-xy\\
{y}^{3}=-{x}^{2}y-xy^2
$$
Then after substition and elimination $y, y^2,y^3$ we get
$$
(x+y)^3={x}^{3}+3\,{x}^{2}y+3\,x{y}^{2}+{y}^{3}=-x^3.
$$
Thus
$$
\left(\dfrac x{x+y}\right)^{2013}=\left(\left(\dfrac x{x+y}\right)^{3}\right)^{671}=\left(\dfrac{ x^3}{(x+y)^3}\right)^{671}=\left(\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3404062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Following Fibonacci Prove that $ F_ {n + 2} = \sqrt{\frac{F_n {F_ {n + 1} ^ 2}(3 {F_n} +4 {F_ {n + 1}}) + 1 }{{F_n} ^ 2 + {F_ {n + 1}} ^ 2}} $?
I discovered this property from an attempt to solve the following problem:
Defining $ A_n = \sqrt{{F_n} ^ 2 + {F_{n + 2}} ^ 2}, $ the numbers $ A_n, $ $ A_ {n + 1} $ and $ A_{... | The hypothesis is equivalent to $g(n)=0$ with$$g(n):=F_nF_{n+1}^2(3F_n+4F_{n+1})+1-F_n^2F_{n+2}^2-F_{n+1}^2F_{n+2}^2\\=-F_n^4-2F_n^3F_{n+1}+F_n^2F_{n+1}^2+2F_nF_{n+1}^3-F_{n+1}^4+1.$$Since $F_0=0,\,F_1=1$, we have the base case $g(0)=0$. For the inductive step use the abbreviations $a:=F_n,\,b:=F_{n+1}$, so$$g(n+1)-g(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3407394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Limiting value of the coefficients of the power series $(1-x)^{-\frac{1}{2}}$ Suppose we have the power series $(1-x)^{-\frac{1}{2}}$. We know that this power series converges iff $|x|<1$. Suppose $a_n$ denote the coefficieis nt of this power series. So $a_0=1, a_1=\frac{1}{2}, a_2= \frac{3}{8}, a_3=\frac{5}{16} \ldots... | Determine a general formula for the coefficients
How do we compute these coefficients? The way to expand this is:
$$(1-x)^{-1/2}=\frac1{0!}(-x)^0+\frac1{1!}(-x)^1\left(-\frac12\right)+\frac1{2!}(-x)^2\left(-\frac12\right)\left(-\frac32\right)+\frac1{3!}(-x)^3\left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.
I started by using Bernoulli's inequality:
$(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$
I can say that:
$(1+\frac{2}{\sqrt{n}}) = (1+\frac... | $(1+x)^n \ge (n^2/4) x^2 $, $n \ge 1$, $x \ge 0$.
Proof:
$(1+x)^n=$
$1+nx + (n(n-1)/2!)x^2+...\gt (n(n-1)/2)x^2 \ge (n^2/4)x^2,$
since we have
$(n)((n-1)/2) \ge n( n/4)$, for $n\ge 2.$
Let $x=2/√n$, in
$(1+x)^n > (n^2/4)x^2$;
$1+2/√n \gt \sqrt[n]{n}$, $n \ge 1$.
Finally
$(1+2/√n)^2 >(1+2/√n) \gt \sqrt[n]{n}$, $n \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
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$\int \frac{1}{(x^2-4)^2}dx$ Calculate:
$$\int \frac{1}{(x^2-4)^2}dx.$$
I tried Partial Fractions method first I write:
$$\frac{1}{(x^2-4)^2}=\frac{A}{X-2}+\frac{Bx+C}{(x-2)^2}+\frac{D}{x+2}+\frac{Ex+F}{(x+2)^2}.$$
We have:
$$A(x-2)(x+2)^2+(Bx+C)(x+2)^2+D(x+2)(x-2)^2+(Ex+F)(x-2)^2=1.$$
$$(A+B+D+E)x^3+(4A-2A+4B+C-4D+2D... | It is $$\frac{1}{(x^2-4)^2}=1/16\, \left( x-2 \right) ^{-2}-1/32\, \left( x-2 \right) ^{-1}+1/16\,
\left( x+2 \right) ^{-2}+1/32\, \left( x+2 \right) ^{-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Find the minimum of :$P=a+b+c-ab-bc-ca$ soluLet $a,b,c$ be positive real numbers and $a+b+c+abc=4$.
We can rewrite the first equation as $a+b+c=4-abc.$ Then,
\begin{align*}
P&=a+b+c-ab-bc-ca\\&=(4-abc)-ab-bc-ca\\&=4-abc-ab-bc-ca-(a+b+c+1)+(a+b+c+1)\\&=4-(abc+ab+bc+ca+a+b+c+1)+a+b+c+1\\&=4-(a+1)(b+1)(c+1)+a+b+c+1\\&=5+a... | We will show that $a+b+c \geq ab+bc+ca$, from which it follows that the minimum is 0. This is attained when $a=b=c=1$ or permutations of $ (2,2,0)$.
The equality conditions suggest that Shur's inequality is involved, and we likely want to use
$$ a^3 + b^3 + c^3 + 3abc \geq \sum ab(a+b)$$.
Let's rewrite this in terms of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Calculate the approximate value I am supposed to calculate the approximate value of $$\cos 151^\circ$$ My idea was that I can divide it in the form: $$\cos 90^\circ+ 61^\circ= \cos \frac{\pi}{2}+ \left (\frac{\pi}{3} +\frac{\pi}{180} \right )$$ Then I use the derivation for cosx:
$$-\sin \frac{\pi}{2}\left ( \frac{\pi}... | \begin{equation}
\cos\left(150^\circ\right)=\cos\left(90^\circ-\left(-60^\circ\right)\right)=\sin\left(-60^\circ\right)=-\sin\left(60^\circ\right)=-\frac{\sqrt3}2\\
\sin\left(150^\circ\right)=\sin\left(90^\circ-\left(-60^\circ\right)\right)=\cos\left(-60^\circ\right)=\cos\left(60^\circ\right)=\frac12
\end{equation}
Usi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3410735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Prove the equality (Taylor series).
Prove the equality:
$$
\frac{1}{3}\left(e^x+2e^{-x/2}\cos\frac{x\sqrt{3}}{2}\right)=
\sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!},\ \ -\infty<x<+\infty
$$
I tried to apply Euler's formula ($e^{ix}=\cos x+i\sin x$) to this problem but it went rather unsuccessful. Here is what I did:
$$... | Hint: (A followup to Lord Shark the Unknown's observation) You're already half-way there. You've established that
\begin{align}
\frac{1}{3}\left(e^x + 2e^\frac{-x}{2}\cos \frac{x\sqrt{3}}{2}\right) &=\frac{1}{3}\left(e^x + e^\frac{x\left(-1+i\sqrt{3}\right)}{2}+ e^\frac{x\left(-1-i\sqrt{3}\right)}{2}\right)\\
&= \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find eigenvalues of $x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi = 0$ with boundary conditions $\phi(1) = \phi(b) = 0$
Since this is an equidimensional equation, determine all positive eigenvalues.
\begin{align*}
x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\
\phi(1) &= 0 \... | You have to take the change of variable $y=\ln x$ and the equation becomes
$$
\frac{d^2\phi}{dy^2}+\lambda\phi=0
$$
that is a standard form having as a solution
$$
\phi(y)=A\cos\sqrt{\lambda} y+B\sin\sqrt{\lambda} y.
$$
I think now you can go on from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3414750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Square-root equation
Solve square-root equation: $\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10$
$\left (\sqrt{5+2\sqrt{6}} \right )^x+\left (\sqrt{5-2\sqrt{6}} \right )^x=10\\
\left (\sqrt{\left (\sqrt{3}+\sqrt{2} \right )^2} \right )^x+\left (\sqrt{\left (\sqrt{3}-\sqrt{2} \righ... | As an alternative,
quadratic with roots $\alpha=\sqrt{3}+ 2$ and $\beta=\sqrt{3}-2$ is $x^2-2\sqrt{3}x+1=0$
Sum of roots is $\alpha+\beta=2\sqrt{3}$
$\alpha^2+\beta^2=(\alpha+\beta)^2 - 2\alpha\beta = 12-2(1) = 10$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Writing polynomial using powers of (x-a) I have a question related to writing a polynomial using powers of binomial of form $(x-a).$
I found an example: polynomial $P(x) = x^4 + 2x^3-3x^2-4x+1$ can be written as
$ (x+1)^4-2(x+1)^3-3(x+1)^2+4(x+1)+1$ using powers of $(x+1)$ and Horner's Method. How do we obtain this r... | You want an expression $a_4(x+1)^4+a_3(x+1)^3+a_2(x+1)^2+a_1(x+1)+a_0$. Repeatedly apply the method for $x=-1$. The last number in each row will give you the next coefficient starting from the lowest power.
\begin{array}{|c|c|c|c|c|c|}
\hline
& 1 & 2 & -3 & -4 & 1 \\ \hline
-1 & 1& 1& -4 & 0 & 1 \rightarrow a_0\\ \hlin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3420690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find integers $x$ and $y$ such that $8^x-9^y=431$ Find integers $x$ and $y$ such that $8^x-9^y=431$
My working:
By taking mod 9 and 16, I got $x$ odd and $y$ even.
Also $8^x>431\implies x\ge 3$
For $x=3$ I got $y=2$
| SECOND ANSWER: We can stick with smaller primes in the other direction
We suspect that $512-81$ is the largest solution. Proof by contradiction:
Giving new names to $x,y,$ we say
$$ 512(8^x - 1) = 81 (9^y - 1) $$
We ASSUME both $x \geq 1, y \geq 1.$
First, $9^y \equiv 1 \pmod {512}.$ A calculation shows that $y$ mus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3421712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determine the value of $k$ such that the line determined by the points $(3,2)$ and $(1,-2)$ is tangent to the graph of $y=\frac{k}{x+1}$. I have done the following work but I am stuck on solving for k. Am I doing this right? If so what do I do next?
| You are given a function $$y=\frac{k}{x+1}.$$
The line defined by the two points has a slope of $$m=\frac{\Delta y}{\Delta x}=\frac{2-(-2)}{3-1}=2$$
and therefore has an equation $$y-2=2(x-3)\Rightarrow2x-y-4=0$$
We want the original function and the tangent line to have exactly one point of intersection (by the defini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3423556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving the differential equation $y''=\frac{1}{y}$ I tried to solve the differential equation $\frac{d^2y}{dx^2}=\frac{1}{y}$ by assuming $y(0)=1$ and $y'(0)=0$ and finding a taylor expansion for $y$ at $x=0$. By differentiating $\frac{d^2y}{dx^2}$, I got:
$\frac{d^3y}{dx^3}=-\frac{1}{y^2}\frac{dy}{dx}$
$\frac{d^4y}{d... | Multiply both sides of $y''= 1/y $ by $ y'$ to get $$y''y' = \frac {y'}{y}$$
Integrate to get $$(y')^2 = 2\ln y $$
which gives us $$y'=\sqrt {2\ln y+K}$$
Separate variables to get $$\int \frac {dy}{\sqrt {2\ln y+K}} =x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3427545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the units digit of $572^{42}$ The idea of this exercise is that you use the modulus to get the right answer.
What I did was:
$$572\equiv 2\pmod {10} \\
572^2 \equiv 2^2 \equiv 4\pmod{10} \\
572^3 \equiv 2^3 \equiv 8\pmod{10} \\
572^4 \equiv 2^4 \equiv 6\pmod{10} \\
572^5 \equiv 2^5 \equiv 2\pmod{10} \\
572^6 \... | Notice the pattern in the indices, the indices that get mapped to each of $2,4,8$ or $6.$
Note that the indices $1,5,9,\cdots, 1+4n$ get mapped to $2.$ The indices $2,6,10,\cdots,2+4n$ to $4,$ the indices $3,7,11,\cdots 3+4n$ get mapped to $8,$ and finally all multiples of $4,$ of the form $4n$ get mapped to $6,$ where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3428727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g... | Like you have suggested, consider the function $$g(x)=(x+1)(x+2)(x+3)(x+4).$$
$g'(x)$ potentially has minimum values at its critical numbers, which we can find by setting the derivative equal to $0$.
\begin{align}
g(x)&=(x+1)(x+2)(x+3)(x+4)\\
\log g(x)&=\log\big((x+1)(x+2)(x+3)(x+4)\big)\\
\log g(x)&=\log(x+1)+\log(x+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need ... | $$ a_n = \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n} $$
is bounded between $$ \frac{n}{\left(n+\frac{1}{2}\right)+n}\quad\text{and}\quad\frac{n}{n+n} $$
so $\frac{1}{2}-a_n$ is bounded between
$$ 0\quad\text{and}\quad\frac{1}{2(4n+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding the minimal polynomial of an $n \times n$ matrix How would one find the minimal polynomial of
$$ A= \begin{pmatrix}0&-1&-2&\cdots&1-n\\1&0&-1&\cdots&2-n\\2&1&0&\cdots&3-n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n-1&n-2&n-3&\cdots&0\end{pmatrix}$$
Where $A$ is an $n \times n$ matrix with $n\ge 3$?
| Let
$$
U=\begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 2 & -1 \\ \vdots & \vdots \\ n-1 & -1
\end{pmatrix}
\;\;\;\text{and}
\;\;\;
V=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & n-1
\end{pmatrix}
$$
Then $A=UV.$ Therefore, $\mathrm{rank}(A)=2,$ and $A$ has only $2$ non-zero eigenvalues. $iA$ is hermitian. This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\f... | Since for moderately large values of $n$ we have
$$ H_n \approx \log(n)+\gamma+\frac{1}{2n}-\frac{1}{12n^2} $$
we also have
$$ H_{n(n+1)/2}-H_{n(n-1)/2} \approx \frac{2}{n}-\frac{4}{3n^3}$$
and
$$ \sum_{n\geq 1}\left[H_{n(n+1)/2}-H_{n(n-1)/2}\right]^2 \approx 1+\sum_{n\geq 2}\left(\frac{2}{n}-\frac{4}{3n^3}\right)^2=1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Solve for positive real solutions of cyclic equations $x+y^2+2xy=9$, $y+z^2+2yz=47$, $z+x^2+2xz=16$
Solve over positive reals $$x+y^2+2xy=9$$
$$y+z^2+2yz=47$$
$$z+x^2+2xz=16$$
With standard manipulation we get that $x+y+z=8$. Thus $x=8-y-z$ and we have two equations,
$$(8-y-z)^2+y^2+2(8-y-z)y=9$$
$$y+z^2+2yz=47$$
Whi... | Continue by substituting $z=\frac{-y^2+15y-1}{2y+1}$ into $y+z^2+2yz=47$,
$$y+\left(\frac{-y^2+15y-1}{2y+1}\right)^2+2y\left(\frac{-y^2+15y-1}{2y+1}\right)=47$$
Simplify,
$$-3y^4+32y^3+69y^2-219y-46=0$$
and factorize
$$(y-2)(3y^3-26y^2-121y-23)=0$$
which yields the positive real solutions $x=1,\>y=2,\>z=5$. The other t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $\sin(x+28^\circ)=\cos(3x-78^\circ)$, then find x $$\sin(x+28^\circ)=\sin (90^\circ-3x+78^\circ)$$
$$x+28^\circ=168^\circ-3x$$
$$x=35^\circ$$
Pretty straightforward question, but then answer is $35^\circ$ and $8^\circ$. How is $8^\circ$ the answer?
| Hint: Use $$\sin(x)-\cos(y)=-2 \sin \left(-\frac{x}{2}-\frac{y}{2}+\frac{\pi
}{4}\right) \sin
\left(-\frac{x}{2}+\frac{y}{2}+\frac{\pi }{4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this exercise with limits? I have some difficulties solving this exercise:
\begin{align}
a_n & = \sum_{k=1}^n \frac 1 {\sqrt {n^2+k}} \\[6pt]
b_n & = n\\[6pt]
\lim_{n→∞} (a_n)^{b_n} & = \text{?}
\end{align}
I believe that I have to find the limit of $a_n$ first but then I have a limit of something... | For $1 \le k \le n$, we have that $n < \sqrt{n^2+k} < \sqrt{n^2+n+\frac{1}{4}} = n+\frac{1}{2}$. Therefore, $$\frac{n}{n+1/2} = \sum_{k=1}^n \frac{1}{n+1/2} < \sum_{k=1}^n \frac{1}{\sqrt{n^2+k}} < \sum_{k=1}^n \frac{1}{n} = 1.$$
Thus, by the Squeeze Theorem, we have that $\lim_{n\to \infty} a_n = 1$. However, this on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1... | Let me propose an alternative pathway.
Suppose there exists $y$ such that
$$
\frac{2x}{x^2-1} = \tan(y).
$$
This seems resonable as $\tan$ range is the whole $\mathbb{R}$.
Now observe
$$
\frac{1}{\cos^2(y)} = \frac{\sin^2(y) + \cos^2(y)}{\cos^2(y)} = 1+\tan^2(y)
= 1 + \frac{4x^2}{(x^2-1)^2} = \frac{x^4 - 2x^2 + 1+4x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$ Let $x,$ $a,$ $b,$ $c$ be nonnegative real numbers. Prove that
$$(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le \left( x + \frac{a + b + c}{3} \right)^3.$$
I know that this problem is a RHS-AM-GM-HM problem, but I am unsu... | $$F=(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc~~~~(1)$$ By AM-GM
$$F \ge x^3 + 3(abc)^{1/3} x^2+3(abc)^{2/3} x+ abc= (x+(abc)^{1/3})~~~~(2)$$
Using $$\frac{(a+b+c)^2}{3} \ge (ab+bc+ca)$$ and AM-GM in (1), we get
$$F \le x^3+3 \frac{(a+b+c)}{3} x^2 +3\frac{(a+b+c)^2}{9}x+ \left(\frac{a+b+c}{3}\right)^3 = \left(x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do we prove this inequality? Suppose $a,b,c > 0$. Prove that
$$\frac{a^2}{b^2} +\frac{b^2}{c^2} + \frac{c^2}{a^2} \geq \frac ab + \frac bc + \frac ca.$$
I've tried multiplying everything by the denominator and then I tried to use the rearrangement inequality, but it didn't yield the result I was looking for. I coul... | Because by AM-GM we obtain: $$\sum_{cyc}\frac{a^2}{b^2}=\frac{1}{6}\sum_{cyc}\left(\frac{4a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\geq\sum_{cyc}\sqrt[6]{\left(\frac{a^2}{b^2}\right)^4\cdot\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=\sum_{cyc}\frac{a}{b}.$$
Just another way:
Let $x^3=\frac{a}{b}$, $y^3=\frac{b}{c}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Compute $\int\frac{du}{\sqrt{a^2-u^2}}$ Please, can you explain why isn't the $$\int\frac{du}{\sqrt{a^2-u^2}} = \int\frac{du}{a\sqrt{1-(u/a)^2}}=\frac1a\sin^{-1}\bigg(\frac{u}{a}\bigg)+c$$ if $a > 0$ is a positive constant.
| We have $$\int\frac{du}{\sqrt{a^2-u^2}}.$$
Let $u=a\sin\theta\implies\theta=\arcsin\bigg(\dfrac{u}{a}\bigg)$. Then $du=a\cos\theta\ d\theta$ and we have
$$\int\frac{du}{\sqrt{a^2-u^2}}=\int\frac{a\cos\theta\ d\theta}{\sqrt{a^2-a^2\sin^2\theta}}=\int\frac{a\cos\theta\ d\theta}{a\sqrt{1-\sin^2\theta}}=\int\frac{\cos\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$ $\underset{n}\lim{\Big(\sqrt[3\;]{\frac{\sin n}{n}+n^3-n^2}-\sqrt[3\;]{n^3+n}\Big)}$
I thought of using the squeeze theorem for$\;\frac{\sin n}{n},$ but I think it might incorrect here.
Another attempt was:
$$\underset{n}\lim{\frac{\fr... | Starting from Paramanand Singh's first equation
$$A=n\left(\sqrt[3]{1-\frac{1}{n}+\frac{\sin (n)} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)$$ since $\sin(n)$ is bounded and very small when divided by $n^4$, consider, for the time being
$$B=n\left(\sqrt[3]{1-\frac{1}{n}+\frac{a} {n^4}}-\sqrt[3]{1+\frac{1}{n^2}}\right)\qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve for $y'\cos{y}=\sin(x+y)$ Solve for $y'\cos{y}=\sin(x+y)$
My attempt
$$y'\cos{y}=\sin x \cos y + \cos x \sin y$$
Divide both side by $\cos y$
$$\frac{dy}{dx}=\sin x + \cos x \tan y$$
Integrate both sides and I got
$$y = -\cos x + \int \cos x \tan y \,dx$$
As Maximilian Janisch said that $\int\cos() \tan(())d≠\sin... | As said in comments, power series solution seems to be the only way.
Writing
$$y=\sum_{n=0}^\infty \frac{a_n}{n!} x^n$$ the very first terms would be
$$\left(
\begin{array}{cc}
1 & t \\
2 & t^3+t+1 \\
3 & 3 t^5+4 t^3+t^2+1 \\
4 & 15 t^7+27 t^5+7 t^4+9 t^3+8 t^2-3 t \\
5 & 105 t^9+240 t^7+63 t^6+144 t^5+103 t^4+6 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3447588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Transforming coordinates system to obtain an ellipse in the standard form Let $E$ be the ellipse $x^2+xy +y^2 = 1$. I would like to obtain the area of $E$ with the formula $ab\pi$. I am not able to transform the equation in the form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$. The textbook I am following uses a transformat... | Let $X=\dfrac x2$ and $Y=\dfrac x2 + y$.
Then $\dfrac{X^2}{\left(\dfrac{1}{\sqrt{3}}\right)^2}+\dfrac{Y^2}{1^2}=3X^2+Y^2=\dfrac34x^2+\dfrac14x^2+xy+y^2=1$.
Addendum:
The above explains the intermediate steps that led to the transformation, as the original post requested, but it should be noted, as alluded to in the co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1$ Hello ladies and gentlemen, here I have another inequality that I am struggling with:
Let $a,b,c>0$ Then
$$\sum_\text{cyc} \frac{ab}{ab+b^2+ca}\le 1.$$
I try to show $$\frac{ab}{ab+b^2+ca}\le\frac{a}{a+b+c}$$ but this is doesn't work because $$\frac{ab}{ab+b^2+ca}... | Let $a=x^3$, $b=y^3$ and $c=z^3$.
Thus, by Muirhead $$\sum_{cyc}\frac{ab}{ab+b^2+ca}=\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^3+(zx)^3}\leq$$
$$\leq\sum_{cyc}\frac{x^3y^3}{x^3y^3+\left(y^2\right)^2\cdot(zx)+(zx)^2\cdot(y^2)}=\sum_{cyc}\frac{x^2y}{x^2y+y^2z+z^2x}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Logical statements: $A+B+C=A+B+C+AB+BC+AC$
Let $A,B, C$ be logical statements.
Then:
$A+B+C=A+B+C+AB+BC+AC$
Prove it without a table.
My attempt:
$$A=A\cdot(1+B+C)$$
$$B=B\cdot(1+A+C)$$
$$C=C\cdot(1+A+B)$$
$$\implies A+B+C= A+AB+AC+B+AB+BC+C+BC+AC$$
$$\implies A+B+C=A+B+C+AB+AB+BC+BC+AC+AC$$
$$\implies A+B+C=A+... | It's correct, if you are using Monotone laws of Boolean algebra or Logical equivalence, we can write$:$
\begin{align}
&\hspace{3ex}A+B+C\\
&=A+AB+B+BC+C+CA\tag*{Absorption law}\\
&=A+B+C+AB+BC+AC\tag*{Commutative law}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3452920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
$$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{... | Use that $$m_a=\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine the minimum of:$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d},\text{ if }a,b,c,d>0$ Determine the minimum of:$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d}$, if $a,b,c,d>0$
I tried partial derivatives of the unction and also its natural log aI came up with system of equation that is not easy to so... | Let $a=\frac{x}{\sqrt3},$ $b=\frac{y}{\sqrt3},$ $c=\frac{z}{\sqrt3}$ and $d=\frac{t}{\sqrt3}$.
Thus, by C-S we obtain:
$$\frac{\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}}{a+b+c+d}=\frac{\sqrt{(x^2+3)(y^2+3)(z^2+3)(t^2+3)}}{9(a+b+c+d)}\geq$$
$$\geq\frac{\sqrt{2((x+y)^2+4)\cdot2(4+(z+t)^2)}}{9(a+b+c+d)}\geq\frac{2(2(x+y)+2(y+t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Use DeMoivre's Theorem to prove $ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$ I need to prove the following equalities using DMT:
$ \cos 5x = 16 \cos^5x$$ - 20\cos^3x$$+5\cos x$
and
$ \sin 5x = 16 \sin^5x$$ - 20\sin^3x$$+5\sin x$
Can someone help me with this question?
(Attempt: $\cos5x+i\sin5x=(\cos x+i\sin x)^5$)
| Use the short-hands $c= \cos x$, $s=\sin x$ and continue with
$$\cos5x+i\sin5x=(c+is)^5
=c^5-10s^2c^3+5s^4c+i(s^5-10s^3c^2+5sc^4)$$
$$=c^5-10(1-c^2)c^3+5(1-c^2)^2c+i(s^5-10s^3(1-s^2)+5s(1-s^2)^2)$$
$$=16c^5-20c^3+5c+i(16s^5-20s^3+5s)$$
where $c^2+s^2=1$ is used. Thus,
$$ \cos 5x = 16 \cos^5x - 20\cos^3x+5\cos x$$
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Convergence of $\int_0^{\infty}\frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx $
Find out if the following integral diverges or converges:
$$
\int_0^\infty \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx
$$
First I split the integral as $\displaystyle \int_0^1 \frac{\ln (1 + x^2)}{\sqrt {2x^5 + x^6}}\,dx + \int_1^\infty... | For small $x$, $$\frac{\log(1+x^2)}{x^2\sqrt{2x+x^2}}=\frac1{\sqrt{2x}}+o(x^{-1/2}).$$
For large $x$, $$\frac{\log(1+x^2)}{x^2\sqrt{2x+x^2}}=\frac{\log(x)}{x^3}+o\left(\frac{\log(x)}{x^3}\right)=o\left(x^{-2}\right).$$
In both cases, the integrals of the asymptotic expressions do converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
combinatorics - arranging consecutive numbers One is arranging the $10$ numbers $0$ to $9$ in a row.
what is the probability of not having a consecutive of $7$ numbers or more?
for example, $2034567891$ is forbidden as it has $7$ consecutive numbers.
Edit :
What I have tried:
for $0-6,$ I treat it as one block along wi... | Might not be the adaptable way to solve and certainly wouldn't be practical for just about any numbers but certainly the easiest is:
There is $1$ way to have all ten consecutive.
There are$2$ ways to have exactly nine ways consecutive: ($0$ to $8$ after a $9$ or $1$ to $9$ before a $0$.
exactly eight. There are $3$ ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Closed expression for sum $\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2}$ Generalizing a recent post Closed expression for sum $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\left\lfloor \sqrt{k}\right\rfloor}{k}$ where convergence was assured by alternating the sign here's a similar problem in which con... | My solution attempt
I have not found a closed form expression but the following integral representation
$$s_{2} =\sum_{k = 1}^{\infty} \frac{\left\lfloor \sqrt{k} \right \rfloor}{k^2} = \int_0^{\infty } \frac{t \left(\vartheta _3\left(0,e^{-t}\right)-1\right)}{2 \left(1-e^{-t}\right)} \, dt\tag{1}$$
Here
$$\vartheta _... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3455835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\be... | The determinant in your problem is equal to $p(-\lambda),$ where $p$ is the characteristic polynomial of $A = vv^T,$ where $ v = [a, \ b, \ c]^T.$ The $0$-eigenspace (i.e, the kernel of $A$) is the two dimensional hyperplane given by $v^Tx = 0,$ so $0$ is an eigenvalue with algebraic multiplicity at least two and the f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Two Primitive Stochastic Matrices with Eventually Equal Sequence of Powers Let $A$ and $B$ be two $n\times n$ primitive row-stochastic matrices. That is, all of their entries are non-negative, all the rows sum up to $1$, and there is an integer $p\geq 1$ such that all the entries of $A^p$ and $B^p$ are strictly positiv... | The specified conditions are not sufficient to force $A=B$.
Here's an example . . .
Choose $a,b\in (0,\frac{1}{2})$ with $a\ne b$, and let $A,B$ be given by
$\\[5.5pt]$
$$
A =
\pmatrix
{
a &\frac{1}{2}-a&\frac{1}{2}\cr
a &\frac{1}{2}-a&\frac{1}{2}\cr
\frac{1}{2}-a&a&\frac{1}{2}
}
\\
$$
$$
B =
\pmatrix
{
b &\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} dx$ How can we evaluate
$$\int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} dx$$
Any kind of help is appreciated.
| I found the answer:
$$\small \int_0^1 \frac{\log ^2(x+1) \log \left(x^2+1\right)}{x^2+1} \, dx=\frac{\pi ^2 C}{48}-\frac{15}{4} C \log ^2(2)+20 \Im(\text{Li}_4(1+i))+2 \log (2) \Im(\text{Li}_3(1+i))+\frac{35 \pi \zeta (3)}{64}-\frac{5}{24} \pi \log ^3(2)-\frac{21}{64} \pi ^3 \log (2)-\frac{3}{256} \left(\psi ^{(3)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to compute $\int \frac{16 x^3 - 42 x^2+2x}{\sqrt{-16x^8+112x^7-204x^6+28x^5-x^4+1}}\,\mathrm dx.$ I want to compute the following integral:
$$\int \frac{16 x^3 - 42 x^2+2x}{\sqrt{-16x^8+112x^7-204x^6+28x^5-x^4+1}}\,\mathrm dx.$$
First I tried substituting $y=\text{denominator}$ but it gets very messy.
Also, I trie... | $-16x^8+112x^7-204x^6+28x^5-x^4+1=-(4x^4-14x^3+x^2+1)(4x^4-14x^3+x^2-1)$
Set $a=4x^4-14x^3+x^2.$ The integral rewrites
$$\int \frac{\mathrm da}{\sqrt{-a^2+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3464809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$ The following problem is from CHKMO 2018 Problem 1:
If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$
I tried to use Cauchy–Schwarz inequality, by try multiplying different... | AM-GM leads to
\begin{align*}
\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}&\geqslant 2\sqrt{\frac{a^2b^2}{c^2}\cdot \frac{b^2c^2}{a^2}}=2b^2
\end{align*}
Similarly
\begin{align*}
\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\geqslant 2b^2\qquad \frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}&\geqslant 2c^2\qquad \frac{c^2a^2}{b^2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3465571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Can any number of squares sum to a square? Suppose
$$a^2 = \sum_{i=1}^k b_i^2$$
where $a, b_i \in \mathbb{Z}$, $a>0, b_i > 0$ (and $b_i$ are not necessarily distinct).
Can any positive integer be the value of $k$?
The reason I am interested in this: in a irreptile tiling where the smallest piece has area $A$, we hav... | Yes.
For $k = 2$:
$3^2 + 4^2 = 5^2$
For $k > 2$:
Start with a solution for $k-1$
Multiply both sides by $5^2$
Replace one $(5a)^2$ on the left with $(3a)^2$ + $(4a)^2$.
$3^2 + 4^2 = 5^2$
$15^2 + 20^2 = 25^2$
$9^2 + 12^2 + 20^2 = 25^2$ (k = 3)
$45^2 + 60^2 + 100^2 = 125^2$
$27^2 + 36^2 + 60^2 + 100^2 = 125^2$ (k = 4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 8,
"answer_id": 2
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In how many ways could six of the colored Easter eggs been crushed? Problem:
Mother was carrying a basket with 11 red, 8 blue, 5 green and 5 yellow decorated Easter eggs (eggs
of the same color are identical).
When entering the room of her son Thomas, she tripped
over door step and crushed six eggs.
Thomas paused in ... | It doesn't matter which specific red, blue, green, or yellow were crushed. Eggs the same color are considered equivalent.
Let $R = $ then number of red eggs. $R$ can be $0$ or it can be as high as $6$.
Let $B = $ be the number of blue eggs. $B$ can be as low as $0$ or as high as $6-R$
Let $G = $ the number of green ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3468403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$? Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$?
$340=2\cdot2\cdot5\cdot17$
I just noticed that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ can it be useful to prove first equation?
Also I tried to construct numbers $a,b,c$ such that $a+b$ would be even, $b... | Your observation that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ i.e. even, could help. Since $a,b,c\in \mathbb{N}$, so the least value of each of the factors in LHS is $2$. So, there are four possibilities $340=2\times 2\times 85=2\times 5\times 34=2\times 10\times 17=4\times 5\times 17$. The first three are not acceptable as their... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+... | By the Carl Schildkraut's hint we have: $$(1+x^2)(1+y^2)\leq\left(1+\left(\frac{x+y}{2}\right)^2\right)^2$$ it's
$$(x-y)^2(x^2+y^2+6xy-8)\geq0,$$ which is true by AM-GM:
$$x^2+y^2+6xy-8\geq2xy+6xy-8\geq0.$$
Thus, it's enough to prove that
$$\prod_{cyc}\left(1+\left(\frac{x+y}{2}\right)^2\right)\leq\left(1+\left(\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $\lim\limits_{x \to \infty} \left( \frac{\sqrt{x^2+2x-3}}{x+2} \right)^{3-2x}$ How can I find this limit?
$$\lim\limits_{x \to \infty} \bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$$
Firstly I thought I can use the limit:
$$\lim\limits_{x \to \infty} \bigg ( 1 + \dfrac{1}{x} \bigg )^x=e$$
by adding $1$ and s... | Let $y =\bigg ( \dfrac{\sqrt{x^2+2x-3}}{x+2} \bigg )^{3-2x}$. Then,
$$\ln y = (3-2x)\ln \sqrt{\dfrac{x^2+2x-3}{(x+2)^2}} $$
$$= \frac{3-2x}2 \ln \left(1-\frac2{x+2}+O(\frac1{(x+2)^2})\right)
= \frac{\ln\left( 1-\frac2{x+2}\right)+O(\frac1{(x+2)^2})}{\frac1{-(x+2)}+O(\frac1{(x+2)^2})}$$
Therefore,
$$\lim\limits_{x \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For real positive $a,b$ given $a^2b^2(a^2b^2+4)=2(a^6+b^6)$, show that at least one of the numbers is irrational I think I am close to the answer, but still not sure how to finish it.
$$a^2b^2(a^2b^2+4)=2(a^6+b^6)$$
I know I can rewrite that equation as
$$(a^2-b\sqrt2)(a^2+b\sqrt2)(a\sqrt2-b^2)(a\sqrt2+b^2)=0$$
Thus $$... | Lets do it the long way. Distribute everything and move it to one side
$$a^4b^4+4a^2b^2-2a^6-2b^6=0$$
Lets substitute
$$x=ab$$
Note x is always positive by the positive definition of $1$ and $b$. Then,
$$a=\frac{x}{b}$$
$$b=\frac{x}{a}$$
Which are also positive definite relations. Now substitute to get
$$x^4 + 4x^2 -2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3472944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n... | The sequence $2,1,2,1,2,1,...$ alternates between $2$ and $1$, being $2$ for even-numbered terms and $1$ for odd-numbered terms. The generating function is thus $$\sum\limits_{n \ge0}(2)x^{2n}+\sum\limits_{n\ge0}(1)x^{2n+1}=\sum\limits_{k\ge0}(1)x^k+\sum\limits_{k\ge0}(1)x^{2k}=\dfrac{1}{1-x}+\dfrac{1}{1-x^2}=\dfrac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do you evaluate $\int_{0}^{1} \frac{(3x^3-x^2+2x-4)}{\sqrt{x^2-3x+2}} \, dx$? Saw this problem on a FaceBook meme that said the pin code to his ATM debit card is the solution to the following problem:
$$\int_{0}^{1} \frac{(3x^3-x^2+2x-4)}{\sqrt{x^2-3x+2}} \, dx$$
I was trying to see how we could break this up into ... | As an alternative to the substitution described in the comments, the anti-derivative of expressions of the form $P(x)/\sqrt{ax^2+bx+c}$, $(a\ne 0)$, where $P(x)$ is a non-constant polynomial is:
$$\int \frac{P(x)}{\sqrt{ax^2+bx+c}}\mathrm{d}x=Q(x)\sqrt{ax^2+bx+c}+\lambda\int\frac{1}{\sqrt{ax^2+bx+c}}\mathrm{d}x $$
wher... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find area of region bounded by $xy = a^2$, $xy = b^2$, $x = py$, $x = qy$ I need to obtain the area of the region bounded by the curves
$xy = a^2$, $xy = b^2$, $x = py$, $x = qy$, $0 < a < b$, $0 < p < q.$
I guess I should use Jacobian of the transformation. I tried $u = \sqrt{xy}$ and $v = \frac{x}{y}$, but it didn't... | Here is to integrate in polar coordinates. The boundaries are
$a^2\le r^2\sin\theta\cos\theta\le b^2$ and $\frac1q\le \tan\theta \le \frac1p$, or, the corresponding integral limits,
$$\theta_1= \tan^{-1}\frac1q,\>\>\>\>\> \theta_2= \tan^{-1}\frac1p,
\>\>\>\>\>r_1^2(\theta)=\frac{a^2}{\sin\theta\cos\theta},
\>\>\>\>\>r_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the angle between the asymptotes of the hyperbola $5x^2-2\sqrt 7 xy-y^2-2x+1=0$?
What is the angle between the asymptotes of this hyperbola?
$$5x^2-2\sqrt 7 xy-y^2-2x+1=0$$
I used $S+\lambda=0$ and used straight line condition to find combined equation to asymptotes. Then how to find angle between them?
| Solve the quadratic equation for $y$:
$$y=-x\sqrt{7}-\sqrt{12x^2-2x+1},y=-x\sqrt{7}+\sqrt{12x^2-2x+1}$$
Let the assymptotes be $y=a_1x+b_1$ and $y=a_2x+b_2$. Then:
$$\lim_{x\to\infty} \frac{-x\sqrt{7}-\sqrt{12x^2-2x+1}}{a_1x+b_1}\stackrel{LR}{=}\lim_{x\to\infty} \frac{-\sqrt{7}-\frac{12x-1}{\sqrt{12x^2-2x+1}}}{a_1}= \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it r... | Yes, I think it's right. Of course, you might have to justify that your approximations still preserve the limit.
Another way would be to write
$$
\left(1+ \frac{3}{n^2+n^4}\right)^n=\exp\left( n\log\left(1+ \frac{3}{n^2+n^4}\right)\right)\sim\exp\left( \frac{3n}{n^2+n^4}\right)\to e^0=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Simplify $\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$ Simplify $$\frac{d}{dx}\frac{x^2}{1+\sqrt{x^2+1}}$$
I have manage to arrive to $$\frac{x^3+2x+2x\sqrt{x^2+1}}{(x^2+2)\sqrt{x^2+1}+2x^2+2}$$
But Wolfram manage to simplify to $$\frac{x}{\sqrt{x^2+1}}$$
| Hint: Expand the fraction with $1-\sqrt{x^2+1}$:
$$\frac{x^2}{1+\sqrt{x^2+1}} =\frac{x^2(1-\sqrt{x^2+1})}{1-(x^2+1)} = \sqrt{x^2+1}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to solve $x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$ Question:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
My attempts to solve:
$x^{\log(x)-1 } = \frac{1}{\sqrt[4]{10}}$
$x^{\log(x)-\log(10) } = 10^\frac{-1}{4}$
$x^{\log(\frac{x}{10}) } = 10^\frac{-1}{4}$
$-\frac{1}{4}\log{_x 10} = \log(\frac{x}{10})$
$-\frac{1}{4}\l... | $$ x^{\log_{10} x-1}=10^{-1/4} \implies 10^{\log_{10} (x^{\log_{10} x-1})}=10^{1/4} $$ $$\implies
10^{\log^2_{10} x -\log_{10} x-1/4}=10^0$$ $$\implies (\log_{10} x-1/2)^2=0 \implies \log_{10}x=1/2 \implies x=10^{1/2}=\sqrt{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3479328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $\int \sqrt{3-2x-x^2}dx$ $\int \sqrt{3-2x-x^2}\,dx$
First I did:
$$\begin{align}\int \sqrt{3-2x-x^2}\,dx &= \int \sqrt{-((x+1)^2-4)}\,dx \\
&=\int \sqrt{4-(x+1)^2}\,dx \\
\end{align}$$
Then I set $(x+1)=2\sin(t)$, $dx = 2\cos(t)\,dt$
$$\begin{align}
\int \sqrt{4-(2\sin(t))^2}2\cos(t)\,dt &= \int\sqrt{4-4\sin^2... | HINT:
Note that $\sin(2x)=2\sin(x)\cos(x)$, $\sin(\arcsin(x))=x$, and $\cos(\arcsin(x))=\sqrt{1-x^2}$.
Can you reconcile now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3480370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int_0^{2\pi}{dx/(1+\epsilon \cos{x})}$ Please give hints or help to solve this integral, $\epsilon^2<1$
$$\int_0^{2\pi}\frac{dx}{1+\epsilon\cos x}$$
| Let $a^2<1$, then
$$I=\int_{9}^{2\pi} \frac{dx}{1+a \cos x}= 2 \int_{0}^{\pi} \frac{dx}{1+a\cos x}~~~~(1)$$
$$I=2\int_{0}^{\pi}\frac{dx}{1-a \cos x}~~~~(2)$$
Adding (1) and (2), we get
$$2I=4\int_{0}^{\pi}\frac{dx}{1-a^2\cos^2 x} =8 \int_{9}^{\pi/2} \frac{sec^2 xdx}{sec^2 x-a^2}=8\int_{0}^{\pi/2} \frac{sec^2 x dx}{\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3481958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Determine all pairs $(a, b)$ for which $s (an + b) - s (n)$ assumes finitely many values For every positive integer n we define $s (n)$ as the sum of the digits of $n. $ Determine all pairs $(a, b)$ of
positive integers for which$$s (an + b) - s (n)$$assumes a finite number of values by varying $n$ in positive integers... | If $a \ne 10^k$, then choose $L$ greater than both $a$ and $b$ and let $n=10^L+10^{2L}+... +10^{mL}$. Then $S(an+b)-S(n)=m(S(a)-1)+S(b)$ takes infinitely many values.
If $a =10^k$ and $b\ge a$, then choose $m$ greater than $k$ and let $n=10^m-1$. Then $S(an+b)-S(n)=1+S(b-10^k)-9m$ takes infinitely many (negative) value... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}... | Using $\cos x=1-x^2/2+O(x^4)$ gives
$\cos 2x=1-2x^2+O(x^4)$ and
$\cos 3x=1-9x^2/2+O(x^4)$. Therefore
$$\cos x\cos 2x\cos 3x=1-7x^2+O(x^4)$$
and so
$$\lim_{x\to0}\frac{1-\cos x\cos 2x\cos 3x}{x^2}=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have ... | $xy = 6 \implies y= \dfrac6x$ . So ,
$$x^2-y^2 = 5 \implies x^2 - \dfrac{36}{x^2} =5$$
Taking $x^2 = a$ , we get :
$$a^2-36=5a\implies (a-9)(a+4) = 0$$
We have $a=9\implies x=\pm3$ and $y = \pm2$
And we have $a=-4 \implies x = \pm 2i$ and $y=\pm 3i$
So the solutions are $(x,y) = (3,2)$ , $(x,y) = (-3,-2)$ ,
$(2i,-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Justification for Rudin exercise 3.2
Rudin 3.4. Calculate $\lim\limits_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$.
The solution to this seems obvious: after enough arithmetic, let $n \to \infty$ and get $\frac{1}{2}$. I am not sure how to rigorously justify the final step, though. Here is what I have:
Leveragin... | Here's how I solved it back when I self-studied my way through Baby Rudin. This has never been seen by anyone other than me before, so apologies if there are any errors in it.
As a lemma, we will show that $\lim_{n\rightarrow\infty}\sqrt{n+\alpha}-\sqrt{n}=0$
for any $\alpha\in\mathbb{P}$. To that end, let $\epsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does $x^2 \equiv 2 \pmod{5}$ have no solution? Can someone explain why $x^2 \equiv 2 \pmod{5}$ has no solution other than trial and error?
So I've shown that modulo 5, we have $0^2 \equiv 0, 1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 4, 4^2 \equiv 1$ and I'm seeing that it doesn't look like there is a solution but wha... | You have already shown what you need. Every integer is of the form $0, 1, 2, 3$, or $4$ modulo $5$, and $0^2$ mod $5 = 0$, $1^2$ mod $5 = 1$, $2^2$ mod $5 = 4$, $3^2$ mod $5 = 4$, and $4^2$ mod $5 = 1$, so we know that there is no integer $x$ for which $x^2 \equiv 2(mod\;5)$ i.e. $x^2(mod\;5) \neq 2$ for all integers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Jordan canonical form of a matrix $A$ Can you help me to find solution to this problem?
I have matrix $ A \in M_n(\mathbb{C}): A^2 + A^4 = 0$ and I know, that maximal number of linear independent eigenvectors is $4$. Also known, that minimal polynomial not equal to characteristic polynomial.
What we can say about Jor... | Here is my second answer that I believe is a sufficient guide to a full classification of all Jordan Canonical forms for arbitrary dimension $n$.
So first we need to find the possible minimal polynomials, which are:
\begin{align*}
x, \ x+i, \ x−i, \ x^2, \ x^2+1, \ x(x+i), \ x(x−i), & \\
x(x^2+1), \ x^2(x−i), \ x^2(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$. If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.
Here's what I have got,
By Vieta's rule
$p+q+r=1\text{. ...........}(1)$
$pq+qr+pr=1\text{. ...........}(2)$
$p... | $$p³+q³+r³ = (p+q+r)³-3(p+q)(q+r)(r+p)$$
$$ = (1)³- 3(1-r)(1-p)(1-q)$$
$$ = 1-3(1-1+1-2) =4$$
Here, 1-1+1-2= the sum of coefficients of the given equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.