Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{... | Hint.
Making
$$
\cases{
u = x+y\\
v = x-y
}
$$
we have
$$
\sqrt{u^2-v^2}=2\frac uv
$$
so
$$
u^2 = \frac{v^4}{v^2-4}
$$
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Apporoaches to solve the given algebraic expression If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displa... | Let me try it
we have putting $t=\tan\Big(\frac{x_1}{2}\Big)$
$$\cos(x_1)=\frac{1-t^2}{1+t^2}$$
It's the Weierstrass substitution
Putting $y=t^2$ we get :
$$\cos(x_1)=\frac{1-y}{1+y}$$
Now we put $x=y$ to get :
$$\Big(-\cos(x_1)\Big)^{\frac{1}{3}}+\Big(-\frac{1}{\cos(x_1)}\Big)^{\frac{1}{3}}=?$$
After that I have tried... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,... | Multiply both sides of $x+y=5xy$, $y+z=6yz$ and $z+x=7zx$ by $z,x$ and $y$ respectively and add them together to obtain:
$xz+zy+xy+xz+zy+xy=2(xz+zy+xy)=18xyz$ so $xz+zy+xy=xy+z(x+y)=xy+5xyz=9xyz$.
So $xy=4xyz$ and hence $z=\frac{1}{4}$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=... | We write inequality we have
$$4(a^3+b^3+c^3-3abc) \geqslant 9 |(a-b)(b-c)(c-a)|,$$
or
$$2(a+b+c)[(a-b)^2+ (b-c)^2+ (c-a)^2] \geqslant 9 |(a-b)(b-c)(c-a)|.$$
It's easy to check $a + b \geqslant |a-b|,$ now using the AM-GM inequality, we have
$$2(a+b+c)[(a-b)^2+ (b-c)^2+ (c-a)^2] $$
$$\geqslant \left(|a-b|+|b-c|+|c-a|\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\tan(x)\tan(x+\frac{\pi}{3})+\tan(x)\tan(\frac{\pi}{3}-x)+\tan(x+\frac{\pi}{3})\tan(x-\frac{\pi}{3}) = -3$ Let's assume that $\tan(x) = y$.
So, $\tan\Big(x+\dfrac{\pi}{3}\Big) = \dfrac{\tan(x) + \tan\Big(\dfrac{\pi}{3}\Big)}{1-\tan(x)\tan\Big(\dfrac{\pi}{3}\Big)} = \dfrac{y+\sqrt{3}}{1-\sqrt{3}y}$
Similarly... | As written, this statement is false, as per @quasi 's comment. I suspect the identity should have used $\tan(x-\pi/3)$ rather than $\tan(\pi/3-x)$. In this case, by the angle-sum formula for tangent,
$$\tan(x)\tan\left(x+\frac\pi3\right)=\tan(x)\left(\frac{\tan x+\sqrt 3}{1-\sqrt 3\tan x}\right)=\tan(x)\left(\frac{4\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving the recurrence relation $T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$ I have been trying to solve the following recurrence relation
$T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$
$T(0) = 0$
I've tried to use substitution which wasn't very useful as I couldn't figure out a way to simplify the resu... | Multiply the recurrence relation by $n$ to get $$nT_n = T_0+T_1+\cdots+T_{n-1}+5n^2.$$ Replacing $n$ with $n+1$ gives $$(n+1)T_{n+1} = T_0+T_1+\cdots+T_{n-1}+T_n+5(n+1)^2.$$ Subtracting the first equation from the second gives $$(n+1)T_{n+1}-nT_n = T_n+5(n+1)^2-5n^2,$$ which can be simplified to $$T_{n+1}-T_n = 10-\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating the volume under the surface $z = x^2 + y^2$ and above $D$ Given:
$$D = \left\{(x,y) | 1 \leq x^2 +y^2 \leq 100, ~~ \frac{x \sqrt{3}}{3} \leq y \leq x \sqrt{3} \right\}$$
Calculate the Volume under the surface $$z = x^2 + y^2$$
And above $D$.
My try:
$$I = \iint_D x^2 + y^2 \,dx\,dy$$
We can write $D$ as:
$... | In polar coordinates, the region $D$ is equivalent to $$D = \{(r, \theta) : (1 \le r \le 10) \cap (\tfrac{\pi}{6} \le \theta \le \tfrac{\pi}{3})\}.$$ The proof is straightforward after noting that $r^2 = x^2 + y^2$ and $\tan \theta = \frac{y}{x}$. Consequently, the desired volume is easily expressed as an integral in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n = a_0 + {a_1}x + {a_2} x^2 + .... +{a_{4n}}x^{4n}$ , then prove that $a_r=a_{4n-r}$ I tried solving this question by attempting to prove $(a_r)=(a_{4n-r})$
Now, $a_r$ is the coefficient of $x^r$ which we can obtain by adding up the coefficients of $x^r$ from the 3 sep... | Let $$P(x)=(1+x)^{4n} +(1+x+x^2)^{2n} +(1+x+x^2+x^3+x^4)^n$$ Then it's reciprocal polynomial is $$\hat{P}(x)=x^{4n}P(1/x)=x^{4n}\left(\left(1+\frac{1}{x}\right)^{4n} +\left(1+\frac{1}{x}+\frac{1}{x^2}\right)^{2n} +\left(1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}\right)^n\right)=P(x)$$ Hence $P(x)$ is a pal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\lim _{x\to \infty }\left(x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex\right)$ with Taylor expansion $L=\lim _{x\to \infty }\left(x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex\right)$
If I do $u=\frac{1}{x} \Rightarrow x=\frac{1}{u}$
Then $L=... | Composing Taylor series is a piece of cake if you are patient, going from inside to outside.
$$A=x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex$$ As you properly did, let $x=\frac 1u$ to get
$$A=\frac{\left(1+u (u+1)^{\frac{1}{u}}\right)^u-e u^2-1}{u^3}$$ Now, let us work the pieces
$$a=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all pairs of integers $(x, y)$ such that $x^3+y^3=(x+y)^2.$
Find all pairs of integers $(x, y)$ such that $$x^3+y^3=(x+y)^2.$$
Since $x^3+y^3 = (x+y)(x^2-xy+y^2)$ we get that $$x^2-xy+y^2=x+y$$
this can be expressed as $$x^2-(y-1)x+y^2-y=0.$$
Since we want integers we should probably look at when the discriminan... | The solutions when $x=0,$ $y=0,$ $x=y$ and $x+y=0$ have already
been presented. I give a new argument for the remaining case.
Let $y=-x+t,$ where both $x$ and $t$ are nonzero. With this
substitution, the original equation becomes
$$3x^2-3tx+t^2-t=0\tag1$$
As a polynomial in $x,$ the discriminant for (1) is
$$ D=12t-3t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I c... | $$
\begin{aligned}
\int \frac{1}{\sin ^4 x+\cos ^4 x} d x = & \int \frac{1}{1-\frac{\sin ^2 2 x}{2}} d x \\
= & 2 \int \frac{1}{2-\sin ^2 2 x} d x \\
= & \int \frac{\sec ^2 2 x}{2 \sec ^2 2 x-\tan ^2 2 x} d(2 x) \\
= & \int \frac{d\left(\tan ^2 x\right)}{2\left(1+\tan ^2 2 x\right)-\tan ^2 2 x}\\=& \int \frac{d(\tan 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 8
} |
Sides $\frac{|b - c|}{\sqrt{(b^2 + 1)(c^2 + 1)}}, \frac{|c - a|}{\sqrt{(c^2 + 1)(a^2 + 1)}}, \frac{|a - b|}{\sqrt{(a^2 + 1)(b^2 + 1)}}$ of a triangle.
Prove that for all pairwise distinct $a, b, c \in \mathbb R$, $$\frac{|b - c|}{\sqrt{b^2 + 1}\sqrt{c^2 + 1}}, \frac{|c - a|}{\sqrt{c^2 + 1}\sqrt{a^2 + 1}}, \frac{|a - b... | For complex numbers $z, w \in \Bbb C$ is
$$
d(z, w) = \frac{|z - w|}{\sqrt{|z|^2 + 1}\sqrt{|w|^2 + 1}}
$$
(apart from a constant factor) the “spherical distance” of $z$ and $w$, that is the euclidean distance of the stereographic projections of $z, w$ onto a sphere. See for example A metric in $\mathbb{C}^{\infty}$ or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Does the inequality $ \sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2}$ have a name? This is somewhat embarrassing but if $a,b$ are nonnegative real numbers the following seems to hold
$$
\sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}.
$$
Does this inequality have a name?
| Squaring both sides we obtain
$$
\sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2} \iff a+b\ge \frac{a+b+2\sqrt{ab}}2\iff \frac {a+b}2\ge \sqrt{ab}
$$
which is true by AM-GM and squaring again we obtain
$$\iff \frac{a^2+2ab+b^2}{4}\ge ab \iff a^2-2ab+b^2=(a-b)^2 \ge 0$$
which is a more foundamental result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n... | First, let's note: $$(\forall n\in\mathbb{N}) \ \ \ 1+2+3+...+n= \frac{(n+1)n}{2} $$ so $$\eqalign{\sum_{1 \le n \le \infty } \frac{1+2+...+n}{n!}&=\frac{1}{2} \sum_{1 \le n \le \infty } \frac{n^2+n}{n!}=\cr &= \frac{1}{2} \sum_{1 \le n \le \infty } \frac{n^2}{n!}+\frac{1}{2} \sum_{1 \le n \le \infty } \frac{n}{n!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
Find a determinant of a large matrix Evaluate $$\det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a \\
1 & 1& \dots &\ldots&a&1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&1\\
a & 1& \dots &\ldots&1&1 \\
\end{vmatrix}$$
I tried cofactor expansion and it leads ... | Add all the columns to the last one : you get
$$\det \begin{vmatrix}
1 & 1& \dots &\ldots&1&a \\
1 & 1& \dots &\ldots&a&1 \\
\vdots & \vdots& \ddots &&\vdots&\vdots \\
\vdots & \vdots&&\ddots&\vdots&\vdots \\
1 & a& \dots &\ldots&1&1\\
a & 1& \dots &\ldots&1&1 \\
\end{vmatrix} = \det \begin{vmatrix}
1 & 1& \dots &\ldot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3811716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
| Easiest proof is actually the proof of Schur. Alternatively, you can even prove it using simple calculus.
Assume $a\geq b\geq c.$ Consider:
$$f(x) = x^3-x^2(b+c) - x(b^2+c^2) +3xbc+b^3+c^3-bc(b+c),\,\, x\geq b$$
Then, $f'(x) = 3x^2-2x(b+c) - b^2-c^2+3bc$ and $f''(x) = 6x - 2b-2c\geq 0.$
So $f'$ is increasing on $[b,\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Finding $\int_S z\,dS$ where $S=\left\{ \big(x,y,z\big):x^2+y^2+z^2=a^2,z\ge 0, a>0\right\}$ $$\int_Sz\,dS$$
$$S=\big\{ \big(x,y,z\big):x^2+y^2+z^2=a^2,z\ge 0, a>0\big\}$$
I've already calculate this surface integral " by hand " :
$$z=\sqrt{a^2-x^2-y^2}\text{, thus,}$$
$$\int_SzdS=\int\int_{x^2+y^2\le a^2}\sqrt{a^2-x^... | With the same trick, you can use Stokes' theorem in one of two ways. First is directly:
$$\nabla \times H = (0,0,a) \to H = (0,ax,0)$$
which gives us the line integral
$$\int\limits_{x^2+y^2=a^2\:\cap\:z=0}(0,ax,0)\cdot dr = a^3\int_0^{2\pi}\cos^2 t\:dt = \pi a^3$$
Or we can use a corollary of Stokes' theorem, which sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $x_{n}$ be the positive real root of $(x^{n-1}+2^n)^{n+1} = (x^{n} + 2^{n+1})^{n}$, how to prove that $x_{n} > x_{n + 1}$? Let $x_{n}$ be the positive real root of equation
$$(x^{n-1}+2^n)^{n+1} = (x^{n} + 2^{n+1})^{n}$$
How to prove that $x_{n} > x_{n + 1}$?
Actually, $x_{n} > 2$ and I get that $x_{1} = 5, x_{2} \... | replace $x$ by $2x$, we have
\begin{equation}
1+\frac{x^n}{2} = \Big(1+\frac{x^{n-1}}{2}\Big)^{1+\frac{1}{n}} > 1 + \big(1+\frac{1}{n}\big)\frac{x^{n-1}}{2}
\end{equation}
hence $x> 1+1/n$. Let
\begin{equation}
f(y) = 1+\frac{y^{n+1}}{2} - \Big(1+\frac{y^n}{2}\Big)^{1+\frac{1}{n+1}}
\end{equation}
then
\begin{equation}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Probability that a quadratic equation has real roots Problem
The premise is almost the same as in this question. I'll restate for convenience.
Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(-1,+1)$. What is the probability that the polynomial $Ax^2+Bx+C$ has real roots?
Note: The di... | We know from the quadratic formula that the polynomial $Ax^2 + Bx + C$ has real roots if $B^2 - 4AC \geq 0$. We can think of this problem in terms of volumes. To do so, it's easier if we rename the coefficients as $x \equiv A$, $y \equiv C$, and $z \equiv B$. Hence, in order to have real roots we require that $z^2 \geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "57",
"answer_count": 3,
"answer_id": 2
} |
How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl... | $$
\begin{align}
&\sum_{r=1}^n\frac{r}{(2r-1)(2r+1)(2r+3)}\\
&=\sum_{r=1}^n\left(\frac1{16(2r-1)}+\frac1{8(2r+1)}-\frac3{16(2r+3)}\right)\tag1\\
&=\frac1{16}\sum_{r=1}^n\left(\frac1{2r-1}-\frac1{2r+1}\right)+\frac3{16}\sum_{r=1}^n\left(\frac1{2r+1}-\frac1{2r+3}\right)\tag2\\
&=\frac1{16}\left(1-\frac1{2n+1}\right)+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How does this division work? $\frac{\;\frac{6^6}{1}\;}{2^{-3}}\cdot2^{-10}$ I came across this division and can't wrap my head around how it is solved:
$$\frac{\;\;\frac{6^6}{1}\;\;}{2^{-3}}\cdot2^{-10}$$
They subtract the exponent of $2^{-10}$ from the denominator's exponent $2^{-3}$:
$$2^{-3-(-10)}$$
Which gives us:
... | $b^nb^m = b^{n+m}$ and $b^{-n} = \frac 1b$ and $\frac {b^n}{b^m} = b^{n-m} = \frac 1{b^{m-n}}$. That is all that is going on.
$\frac {BLAH}{\frac 1{2^{-3}}\cdot 2^{-10}}=$
$\frac {BLAH}{\frac 1{2^{-3}}\frac 1{2^{-(-10)}}}=\frac {BLAH}{\frac 1{2^{-3 - (-10)}}}$
Frankly seems like a convoluted way to make things as hard... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Solving for $x$ when $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x$ Suppose $x$ is a real number such that $\sqrt{\sqrt{3} - \sqrt{\sqrt{3} + x}} = x.$ Find $x.$
I first squared to get rid of the first square root, which gave me $\sqrt{3} - \sqrt{\sqrt{3} + x} = x^2.$ However, I'm not sure how to move on from there. Can ... | From where you left off:
$$\begin{align}
\sqrt{3}-\sqrt{\sqrt3+x}
&=x^2\\
\sqrt{3}-x^2
&=\sqrt{\sqrt3+x}\\
3-2\sqrt{3}x^2+x^4
&=\sqrt{3}+x\\
x^4-2\sqrt{3}x^2-x+\left(3-\sqrt3\right)&=0\\
\left(x^2+x+(1-\sqrt{3})\right)\left(x^2-x-\sqrt{3}\right)&=0
\end{align}$$
So if there is a solution to the original equation, it is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving that $\left[ \mathbb{Q} \left( \sqrt[3]{4+\sqrt{5}} \right ) : \mathbb{Q} \right] = 6$ Let $\alpha = \sqrt[3]{4+\sqrt{5}}$. I would like to prove that $\left[ \mathbb{Q} \left( \alpha \right ) : \mathbb{Q} \right] = 6$. We have $\alpha^3 = 4 + \sqrt{5}$, and so $(\alpha^3 - 4)^2 = 5$, hence $\alpha$ is a root o... | As requested by OP I am rewriting my comment as an answer. We will show that $[\mathbb{Q}(\sqrt[3]{4+\sqrt{5}}) : \mathbb{Q}(\sqrt{5})] = 3$ by showing that $f(x) = x^3 - (4+\sqrt{5})$ has no solution in $\mathbb{Q}(\sqrt{5})$.
Rather than the approach in the question we notice that $\operatorname{Nm}_{\mathbb{Q}(\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Induction proof of a known harmonic sum I want to prove that $$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1$$ only by induction!
I check for the first one, $\frac12 \leq 1 $ correct.
Then I assume for $n=k$ : $$\frac12 + \frac14 + \dots + \frac{1}{2^k} \leq 1$$
And Try and prove for $n=k+1$
$$\frac12 + \fr... | We can instead try to prove the following
$$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n} \leq 1-\frac1{2^n}\le 1$$
then the induction step becomes
$$\frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^n}+ \frac{1}{2^{n+1}} \le 1-\frac1{2^n}+ \frac{1}{2^{n+1}}= 1-\frac1{2^n}\left(1- \frac{1}{2}\right)=1-\frac1{2^{n+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Maximizing $\frac{y+1}{x+2}$ when $(x-3)^2 + (y-3)^2 = 6$ Suppose $x$ and $y$ are real numbers such that $(x-3)^2 + (y-3)^2 = 6.$ Than, maximize $\frac{y+1}{x+2}.$
I do in fact realize that this is a double post, but it's a 5 year old question and I don't feel as if it is appropriate to bump it. I did as the original ... | The function and the constraint are
$$
f(x,y) = \frac{y+1}{x+2}\\
\quad g(x,y) = (x-3)^2 + (y-3)^2 - 6 = 0
$$
using Lagrange multipliers
$$
\nabla f = (-\frac{y+1}{(x+2)^2}, \frac{1}{x+2}), \quad \nabla g = (2x,2y)
$$
We get the equations
$$
\nabla f - \lambda \nabla g = 0\\
g(x,y) = 0
$$
in $\lambda, x$, and $y$. The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to shrink explicit function definitions to 1 line? Assume maximum domain for all following examples.
We know that $f(x)=|x-1|$ can be expanded(split) into branches like this: $f(x)=\left\{\begin{gather}-x+1, x\le 1\\x-1, x \gt 1\end{gather}\right\}$
or $f(x)=min(x^{2}-2, 2)$ can be written as $f(x)=\left\{\begin{ga... | We can use the indicator function to write for instance
\begin{align*}
f(x)&=|x-1|\\
&=(1-x)\mathbf{1}_{x\le 1}+(x-1)\mathbf{1}_{x> 1}\\
\\
&\qquad\qquad\text{or}\\
\\
g(x)&=
\begin{cases}
-2x+7&\qquad x<-10\\
x^{3}-2x^{2}+8&\qquad -10 \le x \le 3 \\
\ln (x+8)&\qquad x > 3
\end{cases}\\
\\
&=(-2x+7)\mathbf{1}_{x<-10}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Where did I go wrong in solving this equation?
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$
Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \lefta... | Your problem lies in $(1)$
Since the equation is
$$f(x)=ax^2+bx+1$$
And the value when $x=2$ is $1$, equation one should be
$$f(2)=a2^2+b2+1\\
1=4a+2b+1\\
2a=-b\\
a=\frac{-b}{2}$$
Substituting in (2)
$$f(3)=9a+3b+1\\
4=9\left(\frac{-b}{2}\right)+3b+1\\
6=-9b+6b\\
b=-2\\
\therefore a=1$$
So when $x=4$
$$f(4)=4^2-2\cdot4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$ then evaluate $|a|$, $|b|$ and $|c|$
Solve the following:
If $|ax^2 + bx + c| \le 1$ for all $x$ in $[0,1]$, then
i) $|a| \le 8$
ii) $|b| \ge 8$
iii) $|c| \le 1$
iv) $|a| + |b| + |c| \le 17$
Solution from my textbook:
Put $x = 0, 1 \text{ and } \frac{1} {2}$ to get:
... | Let $f(x)=ax^2+bx+c$.
Thus, $$a+b+c=f(1),$$
$$\frac{1}{4}a+\frac{1}{2}b+c=f\left(\frac{1}{2}\right)$$ and $$c=f(0),$$ which gives
$$b=4f\left(\frac{1}{2}\right)-f(1)-3f(0),$$ $$a=2f(1)+2f(0)-4f\left(\frac{1}{2}\right)$$ and by the triangle inequality we obtain: $$|a|+|b|+|c|\leq2+2+4+4+1+3+1=17.$$
It's interesting that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3839626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that the diophantine equation $(xz+1)(yz+1)=az^{k}+1$ has infinitely many solutions in positive integers. Given two positive integers $a$ and $k>3$ : From experimental data, it appears the diophantine equation
$(xz+1)(yz+1)=az^{k}+1$
has infinitely many solutions in positive integers $x,y, z$.
To motivate the que... | This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$
Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Why doesn't trig substitution work for definite integrals? In the following example, I am attempting to find the area of a semicircle using calculus, which is obviously $\frac{\pi{r}^2}{2}$. Effectively, I am trying to find
$$\int_{-r}^r\sqrt{r^2-x^2}dx$$
Here goes:
Let $x=r\sin\theta$:
$$\frac{dx}{d\theta}=r\cos\theta... | Ideally, as others have pointed out, you will set up the bounds of your trigonometric substitution so that each value of $x$ you need is produced once and only once, in a continuous stream.
For $-\frac\pi2 \leq \theta \leq \frac\pi2,$
if $x = r\sin\theta$ then $x$ increases monotonically from $-r$ to $r$
as $\theta$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}$ for $a,b,c\in\mathbb{R}^+$ with $abc=1$ Suppose that $a,b,c$ are positive reals such that $abc=1$. Prove that $$\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}\geq \frac{3}{2}.$$
Hint: Use Titu's lemma.
My approach: I am trying to use Titu's lemma direct... | Also, we can use AM-GM and SOS:
$$\sum_{cyc}\frac{a^3}{b+c}=\frac{a^2+b^2+c^2}{3}+\sum_{cyc}\left(\frac{a^3}{b+c}-\frac{a^2}{2}\right)\geq$$
$$\geq\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{a^2(2a-b-c)}{b+c}=\frac{3}{2}+\frac{1}{2}\sum_{cyc}\frac{a^2(a-b-(c-a))}{b+c}=$$
$$=\frac{3}{2}+\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Fibonacci sequence and convergence of $t_n=\cfrac{x_{n+1}}{x_n}$ Fibonacci sequence define as: $x_1=x_2=1 , x_n=x_{n-1}+x_{n-2}$ for $n \ge 1$ prove the sequence $t_n=\cfrac{x_{n+1}}{x_n}$ is convergent.
First of all I start by calculating some of the first terms of the $t_n$ sequence:
$\begin{array}{rcc}
n:&1&2&3&4&5&... | $$X_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]=\frac{a^n-b^n}{\sqrt{5}}$$
So $\lim_{n \to \infty} \frac{X_{n+1}}{X_n}=a=\frac{1|+\sqrt{5}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results th... | $$\cos 3x =4\cos^3x -3\cos x$$
So, $$\frac{1}{12}\color{green}{\cos 3x} - \frac{3}{4}\cos x=\frac{1}{12}(\color{green}{4\cos^3x -3\cos x})-\frac{3}{4}\cos x$$
$$=\frac{1}{3}\cos^3x-\cos x$$
Hence both the answers are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$
Evaluate $$\lim_{x \to 3\pi/4}\frac{4\sin^2x \cos x-\cos x+\sin x}{\sin x+\cos x}$$
Of course L-hospitals rule kills it.The question is how to do without L-hospitals rule.The substituition $x-3\pi/4=h$ seems promising but only complica... | We try to simplify the numerator and factor out $(\sin x+\cos x)$ to avoid using L'Hospital.
$4\sin^2x \cos x-\cos x+\sin x\\
=\sin x + 3\sin^2 x\cos x+(-\cos x+\sin^2 x\cos x)\\
=\sin x + 3\sin^2 x\cos x +\cos x(\sin^2 x-1)\\
=\sin x + 3\sin^2 x\cos x -\cos^3 x+(\sin^3x-\sin^3x)\\
= \sin x (1-\sin^2x)+ 3\sin^2 x\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$\sqrt{a^2+5b^2}+\sqrt{b^2+5c^2}+\sqrt{c^2+5a^2}\geq\sqrt{10(a^2+b^2+c^2)+8(ab+ac+bc)}$ for any real numbers. I think that this inequality is strong, though I do not have knowledge of many techniques. There goes my work:
Positive variables only make the inequality stronger, hence suppose $a,b,c\geqslant0$
$$
\sqrt{a^2+... | Probably not the proof you are looking for, but a proof nonetheless.
The inequality is really sharp, and I don't think that a manual solution exists. Concretely, I don't think that one can find a lower bound on the LHS, such that we can algebraically confirm that it upper bounds the RHS. However, it is easy to numerica... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3850734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
A binomial inequality raising from an exercise on the Rademacher Process Consider the inequality below
\begin{gather*}
\frac{1}{2^{n+1}}\left[\binom{n+1}{k-1} + \binom{n+1}{k+1}\right] > \frac{1}{2^n}\binom{n}{k}
\end{gather*}
It can be reduced through the following passages
\begin{gather*}
\frac{1}{2}\left[\binom{n+1}... | You used $\dbinom n k \le \dbinom n{k+1}$ for $k\le \left\lfloor\dfrac {n-1}2\right\rfloor$.
Similarly for $k \ge \left\lfloor\dfrac {n+1}2\right\rfloor$ we have $\dbinom n k > \dbinom n{k+1}$.
That is, for $k \ge \left\lfloor\dfrac {n+1}2\right\rfloor + 1$:
$$\binom {n+1}{k-1} = \binom {n}{k-2} + \binom {n}{k-1} \ge \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration - Evaluate $\int \frac{x^7+2}{(x^2+x+1)^2} \ dx$
Evaluate $$\int \frac{x^7+2}{(x^2+x+1)^2} \ dx$$
This problem is from G N Berman, no. 2056 (integrate using ostrogradsky's method). I referred to this question as well as this article but I could neither understand the method nor come up with a solution for... | Standard long division is not really that bad. I got $$\int (x^3-2x^2+x+2-\frac{4x^3+6x^2+5x}{(x^2+x+1)^2})dx$$
Integrate each terms. You can integrate the first four terms to get $\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+2x$ use Ostrogradsky's method on the last term.
Ostrogradsky's method
$Q(x)=(x^2+x+1)^2$, $Q_1(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Multiple proofs of $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$ Here is my question:
Let $a,b,c\in\mathbb{R^+}$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}2$$
Here is my solution:
From C-S inequality, we get $$\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq3^2,$$
which is equivalent to ... | Because
$$\frac{a}{b+c} - \frac{8a-b-c}{4(a+b+c)} = \frac{(2a-b-c)^2}{4(b+c)(a+b+c)} \geqslant 0,$$
so
$$\frac{a}{b+c} \geqslant \frac{8a-b-c}{4(a+b+c)}.$$
Therefore
$$\sum \frac{a}{b+c} \geqslant \sum \frac{8a-b-c}{4(a+b+c)} = \frac 32.$$
Note. In addition, you can see 45th-proof-Nesbitt.pdf (Vietnamese)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Conjugating simultaneously two matrices to an integer matrix I have the matrices $$A=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1 \end{pmatrix},\quad B=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&\cos\frac{2\pi}{3}&-\sin\frac{2\pi}{3}\\0&0&\sin\frac{2\pi}{3}&\cos\frac{2\pi}{3} \end{pmatrix}.$$
Problem: I'm trying to ... | No such $P$ exists. Suppose the contrary. Let
$$
X=P^{-1}AP=\pmatrix{1&0&0&0\\ 0&-1&0&0\\ 0&0&1&1\\ 0&0&0&-1},
\ Y=P^{-1}BP=\pmatrix{a&b&c&d\\ e&f&g&h\\ i&j&k&l\\ m&n&o&p}.
$$
Since $AB=BA$, we have $XY=YX$, i.e.
$$
\pmatrix{a&b&c&d\\ -e&-f&-g&-h\\ i+m&j+n&k+o&l+p\\ -m&-n&-o&-p}
=\pmatrix{a&-b&c&c-d\\ e&-f&g&g-h\\ i&-j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3863503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find a closed formula (not including $\sum$) for the expression $\sum_{k=0}^{n-1}\binom{2n}{2k+1}$ Find a closed formula (not including $\sum$) for the expression
$$\sum_{k=0}^{n-1}\binom{2n}{2k+1}$$
I started by using the fact that
$$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$
to get that
$$\sum_{k=0}^{n-1}\binom{2... | Note that
$$\sum_{k\ge0} a_{2k+1} = \sum_{k\ge0} \frac{1-(-1)^k}{2}a_k.$$
Taking $$a_k=\binom{2n}{k}$$ yields
\begin{align}
\sum_{k\ge0} \binom{2n}{2k+1}
&= \sum_{k\ge0} \frac{1-(-1)^k}{2} \binom{2n}{k} \\
&= \frac{1}{2}\sum_{k\ge0} \binom{2n}{k} - \frac{1}{2}\sum_{k\ge0} (-1)^k\binom{2n}{k} \\
&= \frac{1}{2}(1+1)^{2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.... | Let $x=b+c$, $y=a+c$ and $z=a+b$.
Thus, $a$, $b$ and $c$ are positives, $a+b+c=1$ and
$$xy+xz+yz-xyz=\sum_{cyc}(a^2+3ab)\sum_{cyc}a-\prod_{cyc}(a+b)=$$
$$=\sum_{cyc}(a^3+a^2b+a^2c+3a^2b+3a^2c+3abc)-\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)>$$
$$>\sum_{cyc}(a^3+3a^2b+3a^2c+2abc)=(a+b+c)^3=1.$$
For $a=b\rightarrow\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}... | Here's another way . . .
On the domain $[-1,1]$, let $f(x)=\sqrt{1-x^2}$ and let $g=f^2$.
If $f$ was differentiable at $x=1$ then by the power rule, the equation
$$
g'=2ff'
$$
would hold at $x=1$, but at $x=1$ the $\text{LHS}$ evaluates to $-2$ (since $g'(x)=-2x$) while the $\text{RHS}$ evaluates to $0$ (since $f(1)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Antiderivative to $\int\frac{1}{(\cos x+\sin x)^2} \ dx$ I have tried the following:
$\int\frac{1}{(\cos x+\sin x)^2} \ dx \ = \int \frac{\sec^2x}{(\tan x+1)^2}\ dx \ $.
After using the substitution $t=\tan x$, I got the solution: $- \frac{1}{\tan x+1} + C$.
Wolfram alpha gives the solution: $\frac{\sin x}{\sin x+\cos ... | $$
\begin{aligned}
\int \frac{1}{(\cos x+\sin x)^{2}} d x &=\int \frac{\sec ^{2} x}{(1+\tan x)^{2}} d x \\
&=\int \frac{d(\tan x)}{(1+\tan x)^{2}} \\
&=-\frac{1}{1+\tan x}+C \\
&=-\frac{\cos x}{\cos x+\sin x}+C\\
&=-\frac{\cos x+\sin x-\sin x}{\cos x+\sin x}+C\\
&=-1+\frac{\sin x}{\cos x+\sin x}+C\\
&=\frac{\sin x}{\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
By application of calculus of residues, can you please solve this problem? By application of calculus of residues, prove that
$$
\int_{0}^{2\pi}
\frac{\cos^{3}\left(3\theta\right)}
{1 - 2p\cos\left(2\theta\right) + p^{2}}
\,\mathrm{d}\theta = \frac{\pi\left(1 - p + p^{2}\right)}{1 - p}
$$
I have attempted the above que... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that... | Any number that is a square mod $100$ is necessarily a square both mod $4$ and mod $5$, which is to say $0$ or $1$ mod $4$ and $0$, $1$, or $4$ mod $5$. The only number in the $70$s that satisfies both criteria is $76$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Hard limit $\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$ Prove that :
$$\lim_{x\to\infty}\Bigg((x(x+1)(x+2))^{\frac{1}{3}}-\Big(2(x+1)-\Big(x^x(x+1)^{x+1}(x+2)^{x+2}\Big)^{\frac{1}{3x+3}}\Big)\Bigg)=0$$
I can prove that :
$$\lim_{x\to\... | Expanding at $\infty$ with Taylor formula, we get
$$x^{\frac{x}{3 x+3}} (x+1)^{\frac{x+1}{3 x+3}} (x+2)^{\frac{x+2}{3 x+3}}\sim 1+x;\;x\to\infty$$
and
$$\sqrt[3]{x (x+1) (x+2)}\sim 1+x;\;x\to\infty$$
Therefore as $x\to\infty$ limit becomes
$$1+x-(2(x+1)-(1+x))=0$$
$$.....................................$$
To expand at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find $a$ and $b$ such that probability is the biggest Basketball player A hit the basket with probability $\frac{1}{2}$, and B with $\frac{1}{10}$. Player A can shoot $a$ times, and B $b$ times, $a+b=20$. Find $a$ and $b$ such that probability to make at least one point each is the biggest.
I just begging my course in ... | Denote the possible events of no hits/hits by (no hits, hits). We get the following sequence for A:
$E_A=(0,1),(1,1),(2,1),....,(a-1,1)$ and similarly for b: $E_B=(0,1),(1,1),(2,1),....,(b-1,1)$. The probablities are:
$$
P(E_A)=(\frac{1}{2})^0(\frac{1}{2})^1+(\frac{1}{2})^1(\frac{1}{2})^1+(\frac{1}{2})^2(\frac{1}{2})^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ Prove that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \leq 3\sqrt{n+1} - 3$ for every natural $n$.
I've already tried to write it like this:
$$\frac{\sq... | Have you tried using induction?
Let's assume the following proposition is true for $n$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{k}} \leq 3\sqrt{n+1} -3$$
By induction we must prove it works for the case $n=1$ and for all the cases $n+1$.
$P(1)$:
The base case is trivial, since
$$\frac{1}{\sqrt{1}} = 1 \leq 3\sqrt{2} -3 = 3(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b... | Your application of rearrangement is correct, in either case, $(a, b, c)$ and $(a^2+a, b^2+b, c^2+c)$ are similarly ordered, so
$$\sum_{cyc} \frac{a}{a^2+2a} \leqslant \sum_{cyc} \frac{a}{b^2+2b}$$
For another way, which generalises, consider
$$f(x) = \sum_{cyc} \frac{a+x}{b+x}, \quad f'(x) = \sum_{cyc} \frac{b-a}{(b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$? Why I cannot get the same answer if I do substitution with $x=a \cos\theta$ for $\int \sqrt{a^2-x^2}dx$ compared with substitution $x=a \sin\theta$?
$\int \sqrt{a^2-x^2}dx... | $I(x)$ integral done by two different methods yields $I_1(x), I_2(x)$. These two differ only by a constant independent of $x$: $I_1(x)-I_2(x)=$Cosnstant, eventually.
In @lab Bhattacharjee 's answer above
$$I_2(x)=\frac{x}{2}\sqrt{a^2-x^2}-\frac{a^2}{2} \cos^{-1}(x/a)$$
and$$I_2(x)=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .
What I Tried: Here is the diagram :-
You can see I ma... | $DC=x$ and the altitude of $\Delta DEC$ from $E$ is $\frac{x}{2}$. This gives us
$$[\Delta DEC] = \frac{x^2}{4}$$
And the result
$$\frac{[\Delta ADE]}{[\Delta DEC]}=\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that... | The sequence is
$$1=1 \ , \ 3+5 = 8 \ , \ 7+9+11 = 27 \ , \ \dots$$
So you can see that
$$a_n = \sum_{k=1}^{1+2+...+n} (2k-1) - \sum_{k=1}^{n-1} a_k =$$
$$\sum_{k=0}^{n(n+1)/2} (2k-1) - a_{n-1} =\frac{1}{4}n^2(n+1)^2 - \sum_{k=1}^{n-1} a_k$$
because you can view the $n$-th term as the sum of $1+2+\dots+n$ odd numbers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
On the number of prime divisors of a product of 4 consecutive integers In PEN (Problems in Elementary Number Theory) Project, there is a problem A111 as follows
Find $n$ if $n(n+1)(n+2)(n+3)$ has exactly three distinct prime divisors.
It's easy to see that at least 2, 3 should be two of the prime divisors of $n(n+1)(... | First not ethat $n=1$ is only almost a solution as $n(n+1)(n+2)(n+3)$ has only two prime divisors. Hence we ma assume $n>1$.
Let $p$ denote the third prime.
If $n$ is a multiple of $3$, then so is $n+3$. Then $n+1, n+2$ are divisible only by $2$ and $p$, and non of them is divisible by both. In other words, we have eit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges How would you show that $\sum\limits_{n=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^n}{n}(-1)^n$ converges?
My approach
At first glance I would go for the alternate series criterion and show that $\frac{\left(1+\frac{1}{n}\ri... | Since
$$
2\le\left(1+\frac1n\right)^n\le3
$$
and
$$
\sum_{n\ge1}\frac{(-1)^n}n=\log2
$$
Calling $S$ the desired limit (that is, your sum), you can immediately get that
$$
2\log 2\le S\le3\log 2\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Allowed values of $x$ in $x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$ when $y=0$? Can someone please check the given explanation? I have this two-variable equation
$$x \sin 6 y-\left(3 x^2-5\right) \sin ^23 y=0$$
for $0\leqslant x,y<\infty$, and I want to check which values of $x$ satisfy this equation when $y=0$.
Fr... | When you "rewrite" an equation by multiplying or dividing by some factor, the new equation is equivalent to the old when that factor is not $0$. When it is $0$, the two are generally not equivalent.
So, your first equation
$$ x \sin 6 y - (3 x^2 - 5) \sin^2 3 y $$
is indeed always true when $y = 0$, or more generally ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a +... | Your method might work, but I couldn't really see how to proceed. A substitution that I am sure would work is as follows:
First expand your polynomial: $x^4-(m^2+2m+2)x^2+m^2+2m+1$.
Next, you can use your $a=x^2$ substitution: $a^2-(m^2+2m+2)a+m^2+2m+1$. Then, you use essentially the same $b$ substitution, but expanded... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3904731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Problem of minimum in physics.
It is necessary to go from a point $A(0,0)$ to a point $B(a,b)$ walking from A to $P(x,0)$ with a speed $v_1$ and then until the point B with a speed $v_2$.
Find where is the point in which it is necessary to abandon x axis in order to have the minimum time to complete the path.
I've cal... | Your calculations are mostly correct, you just have to interpret the results correctly.
You have a continuous function $t : \Bbb{R} \to \Bbb{R}$ and hence it must attain its minimum on the compact set $[0,a]$ in some point $x_{0} \in [0,a]$.
Moreover, $t$ is differentiable on $\langle 0,a\rangle$ and if it happens that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating matrix exponential
Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?
I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as
$$P(\lambda)=\lambda^2+89$$
Is there an easier w... | Your matrix $M$ is diagonalizable with eigenvalues $\pm i\sqrt{89}$. This means that $e^M = p(M)$ where $p \in \Bbb{C}[x]$ is the unique polynomial of degree less than $2$ such that
$$p(i\sqrt{89}) = e^{i\sqrt{89}}, \quad p(-i\sqrt{89}) = e^{-i\sqrt{89}}.$$
Using Lagrange interpolation formula, we see that
$$p(x) = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Solving equation $a_n=1+\frac12a_{n-1}+\frac12a_{n+1}$ Let's consider sequence following for $n=0...,T$ for which $a_0=a_T=0$. I want to solve equation $$a_n=1+\frac12a_{n+1}+\frac12a_{n-1}$$
My work so far
$$a_1=1+\frac12a_2 \Leftrightarrow a_2=2a_1-2$$
$$a_2=1+\frac12a_3+\frac12a_1 \Leftrightarrow2a_1-2=1+\frac12a_3 ... | Consider the generating function $$f(x)=\sum_{n\ge0}a_nx^n,\qquad|x|<1.$$
We may write
$$2a_{n+1}x^{n+2}=2x^{n+2}+a_{n+2}x^{n+2}+a_{n}x^{n+2},$$
and sum over $n\ge 0$:
$$2x\sum_{n\ge0}a_{n+1}x^{n+1}=2x^2\sum_{n\ge0}x^n+\sum_{n\ge0}a_{n+2}x^{n+2}+x^2\sum_{n\ge0}a_nx^n,$$
which is
$$2x(f(x)-a_0)=\frac{2x^2}{1-x}+f(x)-a_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Showing $\frac{f(a+2b)+f(a-2b)}{2f(a)}\leq\sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$ for $f(x)=\sqrt{x(1-x)}$ with some constraints My question is from a inequality that is not proved (it is just implicitly mentioned I guess) in a book.
Specifically, let $a \in (0,1)$ and $b \in (0,1)$ with $a - 2b \geq 0$ and $a + 2b ... | I will just bash it. Let's work on the square of the LHS first. You will actually find that:
$$f^2(a+2b) + f^2(a-2b) = 2a(1-a) - 8b^2 = 2f^2(a) - 8b^2$$
and
$$f(a+2b)f(a-2b) = \sqrt{(a^2-4b^2)((1-a)^2 - 4b^2)} = \sqrt{(f^2(a)+4b^2)^2 - 4b^2}$$
Therefore, the squared version of your inequality is:
$$LHS^2 = \dfrac{f^2(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$ We have to prove the convergence and find the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$. I tried to find $a_{n+1}$ in terms of $a_{n}$ and got $a_{n+1}=\frac{n+... | OK, I see what needs to be done here. Denote the sum in brackets by $S_n$
$$S_n = \left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$$
First we prove by induction (for $n \ge 4$) the following:
$$S_n \gt \frac{ 2^{n+1} (n+2)}{(n+1)n} \tag{1}$$
Then using (1) it is easy to prove by induction that:
(2) the sequen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is the solution of $x^3+x=1$? According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:
$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$
$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is ... | $$ \left( \frac{1}{2} \left( 1 + \sqrt{\frac{31}{27}} \right) \right)^{1/3} + \left( \frac{1}{2} \left( 1 - \sqrt{\frac{31}{27}} \right) \right)^{1/3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is |
The centroid $G$ belongs to the line
$\mathcal{L}:$ $y=x+1$.
Since the circle $\mathcal{C}$ must pass through
$G(2,3)$ and $D(1,1)$, it must also pass through
the point $K(4,4)$,
which is a reflection of $D$ wrt the perpendicular through $G$.
Hence, the sought circle is the circumcircle of $\triangle GDK$
with the sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3917805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove $ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$ I want to prove that
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = 1$
My approach is the most logical one, rewrite as follows:
$ \sum_{n=1}^{\infty} \frac{2}{4n^2-1} = \sum_{n=1}^{\infty} \frac{1}{2n-1} - \sum_{n=1}^{\infty} \frac{1}{2n+1} $
But the remaining series are both... | Rewriting the term,
$\color{blue}{\left(\sum_{n=1}^{\infty} \frac{2}{4 n^{2} - 1}\right)}=\color{black}{\left(\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + \frac{1}{2}\right)} + \frac{1}{2 \left(n - \frac{1}{2}\right)}\right)\right)}$
This is the telescoping series:
$\sum_{n=1}^{\infty} \left(- \frac{1}{2 \left(n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrating factor for the ODE $(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$ I was asked to find an integrating factor $\mu = \mu(x+y^2)$ for the ODE
$$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0.$$
So the natural approach was define $P = 3x+2y+y^2$ and $Q = x+4xy+5y^2.$ Then,
$$\partial_yP = 2+2y,~~~\partial_xQ = 1+4y$$
and hence $\par... | $$(3x+2y+y^2)dx + (x+4xy+5y^2)dy = 0$$
That you can rewrite as:
$$(x+y^2)dx+2(x+y)dx + (x+y^2)dy+4y(x+y)dy = 0$$
$$(x+y^2)d(x+y)+2(x+y)(dx+2ydy) = 0$$
$$(x+y^2)d(x+y)+2(x+y)d(x+y^2) = 0$$
$$\dfrac {d(x+y)}{x+y}+2\dfrac {d(x+y^2)}{x+y^2} = 0$$
Integrate.
$$(x+y)(x+y^2)^2 = C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the relationship between $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ and $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$? Is $\frac{1}{3}x_1+\frac{1}{3}x_2+\frac{1}{3}x_3$ > or < or = $x_1^{\frac{1}{3}}x_2^{\frac{1}{3}}x_3^{\frac{1}{3}}$, given $x_1$, $x_2$ and $x_3$ are all positive?
I know there exis... | For what it's worth, you can prove this inequality without knowing the more general result: write $x_i=y_i^3$ so$$\frac{y_1^3+y_2^3+y_3^3}{3}-y_1y_2y_3=\tfrac16(y_1+y_2+y_3)((y_1-y_2)^2+(y_2-y_3)^2+(y_3-y_1)^2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Combination of a piecewise defined function and a square function Let's say I have two functions, $f(x)=x^2$, and:
$$g(x)=
\begin{cases}
- 4 & x \leq 0\\
|x- 4| & x > 0
\end{cases}$$
Now I have to write combinations $f\circ g$ and $g\circ f$.
I think $f\circ g$ should be:
$$f\circ g(x)=
\begin{cases}
16 & x \leq 0\\
(x... | $f(x) = x^2$
$$
g(x) = \begin{cases}
-4 & \text{if \(x \leq 0\)}\\
|x - 4| & \text{if \(x > 0\)}
\end{cases}
$$
You correctly found that
\begin{align*}
(f \circ g)(x) & = f(g(x))\\
& = \begin{cases}
f(-4) & \text{if \(x \leq 0\)}\\
f(|x - 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$? Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
I know the following elementary facts. We have
\begin{equation}
\math... | The number field $K=\Bbb Q(\sqrt{5})$ has class number one because its Minkowski bound satisfies $B_K<2$ . Hence its ring of integers $\mathcal{O}_K=\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is even a PID and hence a UFD.
On the other hand, it is enough to see that $\mathcal{O}_K/(4+\sqrt{5})$ is a field, so that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$? I need to prove something like that:
For $a,b,c>0$ prove: $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$.
I know that $3abc≤(a^3+b^3+c^3)$, but then I derived $3(ab+bc+ca) ≤ (a+b+c)^2$, I can't move on.
Can anyone help me?
| Hint:
$$-(a+b)(a+c)(b+c)(a^2-ab-ac+b^2-bc+c^2)=abc(a+b+c)^2-(a^3+b^3+c^3)(ab+bc+ca)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3926463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the asymptotic expansion of $x_n$ where $x_{n+1} = x_n+1/x_n$? Let
$x_{n+1} = x_n+1/x_n,
x_0 = a \gt 0$
and
$y_n = x_n^2$.
What is the asymptotic expansion
of $x_n$ ($y_n$ will do)?
I can show that
$y_n
=2n+\dfrac12 \ln(n) + O(1)
$.
Is there an explicit form
for the constant
implied by the $O(1)$?
What is the a... | Not a complete answer. Let me first reproduce what I imagine is your argument. We have $y_{n+1} = y_n + \frac{1}{y_n} + 2$. In particular $y_{n+1} \ge y_n + 2$ which gives $y_n \ge 2n + a^2$ and hence $\frac{1}{y_n} \le \frac{1}{2n + a^2}$. This gives
$$y_{n+1} \le y_n + 2 + \frac{1}{2n + a^2}$$
which gives
$$y_n \le 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1) $28x+17y\equiv 18 \pmod{41}$
(2) $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular a... | $$-13x+17y\equiv 18 \pmod{41}\tag1$$
$$-10x+11y\equiv -6 \pmod{41}\tag2$$
$(1)\times 10-(2)\times 13$
$$
27y\equiv 12 \pmod{41} \implies y \equiv \frac{12}{27} \equiv \frac{4}{9} \equiv \frac{20}{45} \equiv \frac{20}{4} \equiv 5 \pmod{41}
$$
$$
x \equiv \frac{11y+6}{10}\equiv \frac{61}{10} \equiv \frac{20}{10} \equiv 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
number of positive integer solutions of $x+y+z+w=24$ Determine the number of positive interger solutions of $x+y+z+w=24$ such that $x\leq 6, y \leq 7, z\leq 8, w\leq 9$
My try:
I used generating polynomial as
$$f(x)=(x+x^2+\cdots+x^6)(x+x^2+\cdots+x^7)(x+x^2+\cdots+x^8)(x+x^2+\cdots+x^9$$
$\implies$
$$f(x)=x^4(1-x^6)(1... | Your answer is right here is an easier way to go about this which may bring less calculation errors
set $x=6-u.y=7-v,z=8-t,w=9-p$
then with $0\le u\le 5,0\le v\le 6,0\le t\le 7,0\le p\le 8 $ we have to find the number of solutions of $$u+v+t+p=6$$ but $u\neq 6$ hence number of solutions is $\binom{6+4-1}{4-1}-1=83$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3936560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
critical points of $f(x,y) = \frac{\sin(x)}{1+y^2}$ so I determined: $f'(x,y) = \left(\begin{array}{c} \frac{\cos\left(x\right)}{y^2+1}\\ -\frac{2\,y\,\sin\left(x\right)}{{\left(y^2+1\right)}^2} \end{array}\right)$
and: hess-f = $\left(\begin{array}{cc} -\frac{\sin\left(x\right)}{y^2+1} & -\frac{2\,y\,\cos\left(x\right... | Note that for:
$$f'(x,y) = (0,0) \iff \frac{cos(x)}{y^2+1} = 0 \quad \wedge \quad \frac{y\operatorname{sin}(x)}{(y^2+1)^2} = 0 $$
The first condition can only be achieved if $x\in \frac{\pi}{2}+\pi\mathbb{Z}$, and since cos and sin are linearly independent, then the only way they can both be 0 at the same time is for y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3939071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$ Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).
Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \... | Alternatively, $\alpha$ is a root of $x^3 - 3 x^2 - 60 x - 176$ and so its inverse is $\frac{1}{176}(\alpha^2-3\alpha-60)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative? Someone gives a solution here:
\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{... | $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ = $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+k\cos\alpha}$
then use $cos\alpha$ = ($cos$ ^2)$\alpha/2$ - $(sin$^2)$\alpha/2$.
divide numerator and denom by ($cos$ ^2)$\alpha/2$
then take tan$\alpha/2$ = t
then u can get a quadratic in denominator... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$. Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.
Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sq... | Notice that
$$\frac{1}{\sqrt{an+b}} \cdot \frac{2n+1}{2n+2} \le \frac{1}{\sqrt{a(n+1)+b}} \\
\iff (a(n+1)+b)(2n+1)^2 \le (2n+2)^2 (an+b) \\
\iff an+a-4bn-3b \le 0$$
Hence if $a=3$, then $b=1$ would work. Of course, you need to prove the initial case ($n$=1).
BTW: how amazing it is that the first two answers got $e$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Simplification of multiplying fractions Simplify $(1+\frac{1}{20\times22})(2+\frac{2}{21\times23})(2+\frac{2}{22\times24})(2+\frac{2}{23\times25})(13+\frac{13}{24\times26})$
I noticed the denominator of the five fractions are all in the form of:$$\frac{...}{(n-1)(n+1)}$$
Which can be simplified to:$$\frac{...}{n^2-1}$$... | $$m+\frac{m}{(n-1)(n+1)}=m\times(1+\frac{1}{(n-1)(n+1)})=m\times (\frac{n^2-1+1}{(n-1)(n+1)})=m\times\frac{n^2}{(n-1)(n+1)}$$
$$(1+\frac{1}{20\times22})(2+\frac{2}{21\times23})(2+\frac{2}{22\times24})(2+\frac{2}{23\times25})(13+\frac{13}{24\times26})=\frac{21^2}{20\times 22}\times 2\times \frac{22^2}{21\times 23}\times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to deduce the Vieta's substitution $y = z-\frac {p}{3z}$ for cubic equation $y^3+py +q=0$ In deriving the formula of cubic equations, Vieta substituted the following
$$y = z-\frac {p}{3z}$$
for the depressed cubic equation
$$y^3+py +q=0$$
and transformed it into a quadratic one.
My question: How did he get that sub... | This is a sligntly different way of looking at your problem and getting the same results.
You have an equation of the form $$t^3 + pt + q=0 \tag{A.}$$
Let $t=z-w$. Then
$\begin{align}
t^3 &= z^3 - 3z^2V + 3zw^2 - w^3 \\
t^3 &= 3zw(z - w) + (z^3 - w^3) \\
t^3 &= 3zwt + (z^3 - w^3)
\end{align}$
Hence
$$t^3 - 3zw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3956419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show factorization $ x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1) $ I'm interested in how to show that
$$
x^{2n}-1=(x^2-1) \prod_{k=1}^{n-1}(x^2-2x \cos \frac{\pi k}{n} + 1)
$$
I've seen this equality too often, but have no idea how to derive it. I've tried the following:
$$
x^{2n}-1=(x^n-1)(x^n+... | Note
\begin{align}
x^{2n}-1
&=\prod_{k=0}^{2n-1} (x-e^{i\frac{\pi k}n})
=\prod_{k=0}^{n-1} (x-e^{i\frac{\pi k}n})\cdot
\underset{k=2n-j}{ \prod_{k=n}^{2n-1} (x-e^{i\frac{\pi k}n})}\\
&=(x-1)\prod_{k=1}^{n-1} (x-e^{i\frac{\pi k}n})
\cdot \prod_{j=1}^{n-1} (x-e^{-i\frac{\pi j}n}) (x-e^{-i \pi})\\
&=(x-1)(x+1)\prod_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3... | You need to show the sequence
$$
x_n = \sqrt{12 + \sqrt{12 + \cdots \text{n times} }}
$$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps.
It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$.
As answered by @user2661923 and @QC_QAOA the sequence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
How to integrate $\int_0^1\frac{dx}{1+x+x^2+\cdots+x^n}$ I am interested in finding a solution to the integral
$$I_n=\int_0^1\frac{dx}{\sum_{k=0}^nx^k}$$
Since the denominator is a geometric series with $a=1$ and $r=x$ and it is within the radius of convergence, we should be able to say
$$\sum_{k=0}^nx^k=\frac{1-x^{n+1... | $$I_n=\int_{0}^{1}\frac{dx}{1+x+x^2+\ldots+x^n}=\underbrace{\int_{0}^{1}\frac{1-x}{1-x^{n+1}}dx}_{x\rightarrow y^{\frac{1}{n+1}}}=\frac{1}{n+1}\int_{0}^{1}\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{y(1-y)}dy$$
$$=\frac{1}{n+1}\int_{0}^{1}\left(\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}{y}+\frac{y^\frac{1}{n+1}-y^\frac{2}{n+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
How to solve the equation $x^2-2y=z^2$? Consider the following equation
$$x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can h... |
Not a 'real' answer, but it was too big for a comment. I think that you're looking for a solution without using a calculator or PC but maybe this gives some insight. I did only a quick search with the following bounds: $-10\le x\le10$, $-10\le y\le10$, $-10\le z\le10$, and $x$, $y$, $z$ are all integers with $xy\ne0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Cannot find limit using epsilon delta definition Prove
$$\lim\limits_{x\to 2} x^3 = 8$$
using epsilon delta definition.
I try as below.
Let $\varepsilon>0$. We choose $\delta>0$.
Consider that
\begin{align}
\vert x^3-8\vert &= \vert (x-2) (x^2+2x+4)\vert\\
&=\vert (x-2) \vert\vert(x^2+2x+4) \vert \\
&=\vert (x-2) \vert... | Let $\epsilon > 0$ be arbitrary and start off with $\delta_{1} := 1$ so that we have $|x-2| < \delta_{1} = 1$.
Then $x>1$ and $x < 3$ so that $(x-2)^{2} < 1$ and $6x < 18$.
Then $|(x-2)^{2} + 6x|<19$ and set $\delta _{2}$ so that $|x-2| < \delta_{2}$ where $\delta_{2} := \frac{\epsilon}{19}.$
Hence, if $\delta := \min\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how ... | $$45x^2-30x+1=(-9x^2+6x+7)^2$$
$$-5(-9x^2+6x+7)+36=(-9x^2+6x+7)^2$$
Substitute $u=-9x^2+6x+7$ and solve the quadratic equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a ( n ) = b( n + 2)$
let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that $a(n) = b(n+2)$. for $n=1,2,3...$
My approach:
$a(1) = 1$ (only... | For compositions into ones and twos we have the OGF
$$A(z) = \sum_{q\ge 1} (z+z^2)^q =
\frac{z+z^2}{1-z-z^2}
= -1 + \frac{1}{1-z-z^2}.$$
Compositions into parts at least two have OGF
$$B(z) = \sum_{q\ge 1} (z^2+z^3+\cdots)^q
= \frac{z^2/(1-z)}{1-z^2/(1-z)}
= \frac{z^2}{1-z-z^2}.$$
We then have for $n\ge 1$
$$[z^n] A(z)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \in... | With difference of squares and difference of cubes, you get that the limit simplifies to
$$\lim_{x\to\infty} \frac{(x+1)-(x-1)}{(x+2)-(x-2)}\cdot \frac{f(x)}{g(x)} = \frac{1}{2}$$
because $f$ and $g$ both grow at the rate $\sim 6 x$ for large $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Research question - prove inequality for $n \log n$. The following inequality has come up in some research and appears true when plotted, but I'm not sure how to to prove it. If $g(z) = z \log z$, then for $x>3$ ($x$ real),
$$
g(x) < g(x-1)\cdot \left(1 + \frac{g(x-1) - g(x-2)}{g(x-1)-1} \right).
$$
| Alternative solution:
Fact 1: $\frac{2u}{2+u} \le \ln (1 + u) \le \frac{2u + u^2}{2 + 2u}$ for all $u \ge 0$.
(The proof is easy. Hint: Take derivative.)
Now, first, by Fact 1, we have
$$g(x) - x\ln (x-1) = x \ln(1 + \tfrac{1}{x-1})
\le x \cdot \frac{2x-1}{2x(x-1)} = \frac{2x-1}{2(x-1)}$$
which results in
$$g(x) \le x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3964226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to do integral $\int\frac1{x^3+x+1}dx$? I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$
| $x^3+x+1$ has one real and a pair of complex roots. The real root is $r= -0.6823$, given analytically by the Cardano’s formula. Decompose the integrand in terms of $r$
\begin{align}
\frac1{x^3+x+1}&= \frac1{(x-r)(x^2+rx-1/r)}
=\frac r{2r +3}\left(-\frac1{x-r}+\frac{x+2r}{x^2+rx-1/r}\right)
\end{align}
and then integrat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x + x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 - x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$ (I think this is where I have gone wrong.)
The actual... | From your expansions of $e^x$ and $e^{-x}$, we have $e^x+e^{-x}=2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots$,
so sech$( x)=\dfrac2{e^x+e^{-x}}=\dfrac{2}{2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots}=\dfrac1{1+\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots}$
$=1-\left(\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots\right)+\left(\dfrac{x^2}{2}+\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Indefinite integral $\int \frac {\mathrm d x} {p^2 + q^2 \cosh^2 a x}$ Spiegel's "Mathematical Handbook of Formulas and Tables" (Schaum, 1968), item $14.584$ gives:
$$\int \dfrac {\mathrm d x} {p^2 + q^2 \cosh^2 a x} = \begin{cases} \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \left({\dfrac {p \tanh a x + \sqrt {p^2 + q^2} ... | Note
$$ \frac12\ln {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }
= \frac12\ln {\frac {\frac {p \tanh a x }{\sqrt {p^2 + q^2} } + 1}{\frac{p \tanh a x }{\sqrt {p^2 + q^2} }-1}}
=\tanh^{-1}\frac {p \tanh a x }{\sqrt {p^2 + q^2} }\\
$$
So, there is a typo in the second expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$ Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.
That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4... | Using a result from Some Logarithmic Inequalities, E.R.Love :
$$
\forall x>0 \quad ; \quad \ln (1+x) > \frac{2x}{2+x}
$$
Making the transformation $x \to \frac 1x$ gives us :
$$
\forall x > 0 \quad ; \quad \ln\left(1+\frac 1x\right) > \frac{\frac 2x}{2+\frac 1x} = \frac 2{2x+1}
$$
Thus if $g(x) = x^2 + \frac {2}{2x+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find all positive integer solutions for the following equation: Find all positive integer solutions for the following equation:
$(x^2+1)(y^2+1)+2(x-y)(1-xy)=4(1+xy)$
I've tried simplifying the equation and then refactoring but I can't find any solutions.
| Using wolfram alpha, all the solutions are given by the expression : $$\begin{cases}y=\frac{x-1}{x+1}, x\neq -1 \\y = \frac {x+3}{x+1}, x\neq -1 \end{cases}$$
For example, the solutions for $x=1$ are $y=0$ and $y=2$.
You could try to find these solutions by hand solving the quadratic equation in respect to $y$. If you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3968900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
Given that $a = \sqrt[3]4 + \sqrt[3]2 + 1$, find $\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}$.
What I Tried: I only figured out that:- $$\rightarrow a = 2^\frac{2}{3} + 2^\frac{1}{3} + 2^\frac{0}{3}$$ Yet this does not he... | Note that
\begin{eqnarray*}
\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{(a+1)^3-a^3}{a^3}.
\end{eqnarray*}
Note also that $a+1=\sqrt[3]{2}a$ so ...
\begin{eqnarray*}
\frac{3}{a} + \frac{3}{a^2} + \frac{1}{a^3}= \frac{2a^3-a^3}{a^3}=1.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art o... | $\sqrt{3}f(x) = \frac{1}{\sqrt{3^{2x-1}}+1}$ and $\frac{1}{\sqrt{3^{a}}+1}+\frac{1}{\sqrt{3^{-a}}+1}=1$.
You can pair like $(-11,11),(-9,9),(-7,7)\cdots(-1,1)$ then answer is $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Sums of integer powers similar to Prouhet–Tarry–Escott problem Recall Prouhet–Tarry–Escott problem. Its solution shows that certain sums of powers of integers can be made to vanish simultaneously if their signs are chosen to follow the Thue–Morse sequence, e.g.
$$\sum_{k=0}^{2^4-1}(-1)^{\sigma_2(k)}\,k^n=0\quad\text{fo... | The exponential generating function for $S_{n,m}=\displaystyle\sum_{k=0}^{2^n-1}(-1)^{\sigma_2(k)}k^m$ (w.r.t. $m$) is $$\sum_{m=0}^\infty S_{n,m}\frac{x^m}{m!}=\sum_{k=0}^{2^n-1}(-1)^{\sigma_2(k)}e^{kx}=\prod_{k=0}^{n-1}(1-e^{2^k x})=2^{n(n-1)/2}(-x)^n\exp\sum_{k=0}^{n-1}f(2^k x),$$ where $$f(x)=\log\frac{e^x-1}{x}=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{2}^{7} \frac{x}{1-\sqrt{2+x}} d x$ We have the following integral:
$$
\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\, dx
$$
And this is my solution, which seems to be wrong, and I am failing to see where exactly I failed at:
We have $u=1-\sqrt{2+x}, x=u^2-2u-1, dx=-2\sqrt{2+x}\, du$, and we know that $x\geq -2$ a... | exactly, when you have $u=1-\sqrt{2+x}\ => x=u^2-2u-1$
Then $du=2(u-1)dx$
Plugging it into your formula we get
$=2\int\ \frac{(u^2-2u-1)(u-1)}{u}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
For what $a$ and $b$ are there explicit expressions for $I(a, b) =\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b} $? For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?
This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-f... | The following is trivial due to expansion of digamma (formula 14 of this page) $$\sum _{n=0}^{\infty } \frac{1}{(a+n) (b+n)}=\frac{\psi ^{(0)}(a)-\psi ^{(0)}(b)}{a-b}$$ Thus OP's integral equals to $$I(a,b)=\sum _{n=0}^{\infty } \frac{a b}{(a b n+b) (a b n+a)}=\frac{\psi ^{(0)}\left(\frac{1}{b}\right)-\psi ^{(0)}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$ Find $\displaystyle \lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$
My attempt:
\begin{align*}
A&=\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
\ln A&=\lim_{k\to 2}\ln\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
&=\lim_{k\to 2}(\ln(2^{2^k}-2^{k^2})-\ln(2^k-k^2))
\end{align*}
I st... | Applying L'Hopital's rule you get
$$
\lim_{x\to 2}\dfrac{2^{x+2^x} \log ^2 2-2^{x^2+1} x \log (2)}{2^x \log 2-2 x} = 16 \log 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Singular Value Decomposition gives a row permutated matrix I want to find the SVD of the following matrix.
$$A = \begin{bmatrix}1&2\\2&2\\2&1\end{bmatrix}$$
The singular values of the above matrix are $\sqrt{17}$,$1$. The following results are posted here in order to make this question shorter.
$$AA^T=\begin{bmatrix}5&... | With the help of Lutz Lehmann's comment above I figured out the way to obtain the SVD without worrying about the problem I previously faced. I am posting my answer here so that it would be a help to anyone who has the same problem.
In the SVD, we express the matrix A as a product of 3 matrices which I am denoting as U,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Number of solutions of $ax+by+cz=n$ in nonnegative integers Let $a,b,c,n\in\mathbb{Z}_0^+$. The number of solutions of
$$ax+by+cz=n$$
(i.e. the number of ordered triplets $(x,y,z)$ satisfying the equation) in $\mathbb{Z}_0^+$ as a function of $n$ is
$$\frac{1}{n!}\lim_{w\to 0}\frac{d^n}{dw^n}\frac{1}{(1-w^a)(1-w^b)(1-w... | The topic is standardly known as the Frobenius coin problem, and
it is quite complicated in general.
In the 2D case $ax+by=n$ there is an interesting theorem, the Popovicius' theorem
which tells that the number of non-negative solutions $ p_{\left\{ {a,b} \right\}} (n)$ is given by
$$
\eqalign{
& p_{\left\{ {a,b} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.