Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{... | One of the other roots is $1-i,$ since complex roots of real polynomials come in complex conjugate pairs.
The sum of the roots is $-1$ times the coefficient of $x^2.$ So, if $r$ is the third root, $(1+i)+(1-i)+r=1,$ or $r=-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Proving contour integral equal to zero Let $G$ be the path traversed once as shown:
Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i} \right)$
The singular p... | The shortest proof is as follows:
Let $\displaystyle f(z)=\frac{1}{z^4-1}$. Let $\displaystyle I=\int \limits _Cf(z)\,dz$.
Note that both the contour $C$ and the function $f$ are invariant under the rotation operation $z\mapsto iz$. Therefore, $I=iI$. Then $I=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $x\tan^2 x-\tan x+x-1=0$ One of my homework questions is to find where the gradient of $\frac{x}{1+\tan x}$ is zero. Using the quotient rule got me to $\frac{1+\tan x-x\sec^2 x}{(1+\tan x)^2}$. I know I can ignore the denominator when the equation is equal to $0$, thus resulting in $1+\tan x-x\sec^2 x$, wh... | If you remember that, up to few weeks ago, $x=\cos(x)$ did not show explicit solution even with special functions, there is no hope to find the zero of function
$$f(x)=x\tan^2 (x)-\tan (x)+x-1$$and numerical methods would be required.
However, you can have quite good approximations.
First, inspection
$$f(0)=-1 \qquad f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $(1+\frac{a^2+b^2+c^2}{ab+bc+ca})^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$
Assuming $a,b,c>0$, show that
$$\Big(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Big)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq \Big(1+\frac{a}{b}\Big)\Big(1+\frac{b}{c}\Big)\Big(1+\frac{c}{a}\Big).$$
I know... | Let $x = \frac{a^2+b^2+c^2}{ab + bc + ca}$. Then $x \ge 1$.
Using Bernoulli inequality, we have
$$\mathrm{LHS} = (1 + x)^{1 + 2/x} = (1 + x)\Big((1 + x)^{1/x}\Big)^2
\le (1 + x)\left(1 + x\cdot \frac{1}{x}\right)^2 = 4 + 4x.$$
It suffices to prove that
$$4 + \frac{4(a^2+b^2+c^2)}{ab + bc + ca} \le \frac{(a + b)(b + c)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proof $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $ I want to prove that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $, I started proving but I got stuck.
Let $ x, y \in \mathbb{R} $. Therefore:
$ \lfloor{x}\rfloor \le x $
$ \lfloor{y}\rfloor \le y $
$ \Rightarrow $
$ \lfloor... | Is this proof OK, using both (1) integers $n = \left\lfloor n \right\rfloor$, and (2) if $a\le b$ then $\left\lfloor a\right\rfloor \le \left\lfloor b \right\rfloor$?
$$\left\lfloor x\right\rfloor + \left\lfloor y \right\rfloor
= \big\lfloor \left\lfloor x\right\rfloor + \left\lfloor y \right\rfloor \big\rfloor
\le \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that $\lim_{n\to\infty} \sum_{k=1}^{n}$ $(\frac {k}{n^2})^{{k\over n^2}+1} = {1\over 2}$. To simplify the question I imagined what if $k\over n^2$= $x$ then it will look like $x^{x+1}$.Else I couldn't think on how to proceed ahead.
In one of my previously asked question viz. Prove that $\lim_{n\to\infty} n^2 \in... | We will show that
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}=\frac{1}{2}$$
Your desired limit will then follow from $\lim_{n\to\infty}\left(1+1/n\right)=1$ and the product rule for limits.
$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $0
Show that $0<e-\sum\limits_{k=0}^n\frac{1}{k!}<\frac{1}{n!n}$, where $n>0$.
Hint:
Show that $y_m:=\sum\limits_{k=n+1}^{m+n}\frac{1}{k!}$ has the limit $\lim\limits_{m\to\infty}y_m=e-\sum\limits_{k=0}^n\frac{1}{k!}$ and use that $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$.
If I use the hint then th... | We have
$$\begin{align}y_m &= \frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots + \frac{1}{(n+1)(n+2)\cdots(n+m)}\right] \\&<\frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots + \frac{1}{(n+1)^m}\right]\end{align}$$
Evaluating the finite geometric sum on the RHS it follows that
$$y_m < \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all positive integer solutions for $3^x-2^y=1$.
Find all positive integer solutions for $3^x-2^y=1$.
Quickly we can find the solutions $(1,1)$ and $(2,3)$. Now the claim is that for $x \ge 2$ and $y \ge 3$ there are no solutions.
The equation can be expressed as $3^x=2^y+1$ from where we can get to $3^x-3 = 2^y-... | We are going to assume there is a larger solution than $9-8=1$ and get a contradiction.
We have
$$ 3^u = 2^v + 1. $$ Subtract $9$ from both sides, we have
$$ 3^u - 9 = 2^v - 8. $$
Taking $u = x + 2$ and $v = 3 + y$ gives $9 \cdot 3^x - 9 = 8 \cdot 2^y - 8,$ or
$$ 9 (3^x - 1 ) = 8 (2^y - 1). $$ We think this is only... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $ab+bc+ca$ Given that $a,b,c \in \mathbb{R^+}$ and $(a+b)(b+c)(c+a)=1$
Find the Minimum value of $ab+bc+ca$
My try: Letting $x=a+b, y=b+c, z=c+a$ we get
$$a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2},c=\frac{y+z-x}{2}$$
$$xyz=1$$
Now the problem is to minimize $$ab+bc+ca=\frac{xy+yz+zx}{2}-\frac{x^2+y^2+z^2}... | This doesn't have a well-defined minimum. Solve
$$(a+b)(b+c)(c+a)=1$$
for $c$ to get
$$
c=\frac12\left(-(a+b)\pm\sqrt{(a+b)^2-4\left(ab-{1\over a+b}\right)}\right)
$$
Let $a=b=1/N$ where $N$ is some
"big enough" number, then $a+b\approx0$ and $c\approx\sqrt N$ and $ab+bc+ca\approx {1/\sqrt N}\approx 0$, up to factors o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Compute expected value from two dimensional normal distribution
Random vector (X, Y) has the two-dimensional normal distribution
with the density $ f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{
-\frac{1}{8} \left[ 4x^2 +8x(y+3)+6(y+3)^2 \right] \right\} $
Compute $\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ]$
Find the co... | The term in the exponent of $f(x,y)$ can be written as
$$\begin{split}-\frac 18\left[4x^2+8x(y+3)+6(y+3)^2\right] &= -\frac 12 \left[x^2+2x(y+3)+\frac 32(y+3)^2\right]\\
&=-\frac 12\begin{pmatrix}x\\
y+3\end{pmatrix}^T\begin{pmatrix}1&1\\1&3/2\end{pmatrix}\begin{pmatrix}x\\y+3\end{pmatrix}\end{split}$$
Thus $\Sigma^{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?
I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$... | This answer extends the comment of zwim.
First, see
Stirling approximation For factorials.
I know that as $n$ goes to $\infty$, that the geometric mean of
$\{1,2,\cdots,n\}$ approaches $~\displaystyle \frac{n}{e}~$ from above. Further, as $n$ increases, the ratio between $n$ and the geometric mean of $\{1,2,\cdots,n\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is wrong with my proof that $\int 2x dx= 2x^2$ by writing $2x=\underbrace{2+2+\cdots+2}_{x\;\text{times}}$? I know $\int 2x \,dx = x^2 + C$ (by the power rule) but why does the following proof not give the same answer?
\begin{align*}
\int 2x \,dx &= \int \underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}} \, dx... | The first integral is wrong because $x\in\mathbb{R}$ need not be a natural number. So the multiplication cannot be split as you did. Similar reasoning applies to the second case, that is to say, since $x$ need not be a natural number, the interpretation for $2^{x}$ is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I c... | $x^2+y^2-z^2=1$, $-\sqrt{3}\le z\le\sqrt{3}$
Convert to cylindrical coordinates. Fitting given the symmetry, it's a hyperboloid of 1 sheet.
$z^2=r^2-1\implies r=\sqrt{z^2+1}\implies dr/dz=\frac{z}{\sqrt{z^2+1}}$
Lateral surface area of a cylinder is $2\pi rz$. To generalize, we keep the $2\pi r$ and the $z$ becomes the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge4$
Let $a,b,c>0$. Prove that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{ab+bc+ca}{a^2+b^2+c^2}\ge4$$
I know $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge 3$ but $\dfrac{ab+bc+ca}{a^2+b^2+c^2}\le1$. And then I try $\dfrac{a}{b}+\dfrac{b}{c}+\d... | Using Cauchy-Schwarz inequality,
\begin{aligned}
\frac ab+\frac b c+\frac ca+\frac{ab+bc+ca}{a^2+b^2+c^2}&\ge \frac{(a+b+c)^2}{ab+bc+ca}+\frac{ab+bc+ca}{a^2+b^2+c^2}\\
&=2+\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{ab+bc+ca}{a^2+b^2+c^2}\\
&\ge 4
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Given matrix $B$, find all possible matrices $A$ satisfying $ A(A-2B) = -(A-2B)A $
Let $A, B \in \Bbb R^{3 \times 3}$ such that $A(A-2B) = -(A-2B)A$. Given $$ B = \begin{pmatrix} 2 & -2 & 1 \\ -1 & 3 & -1 \\ 2 & -4 & 3 \end{pmatrix} $$ find all possible matrices $A$ satisfying the equation above.
I hope there is som... | Following the hint of @P.Quinton, we can rewrite the problem as $(A-B)^2=B^2$, and that reduces the problem to finding all $C$ such that $C^2=B^2$. The difficulty here is that $B$ has a repeated eigenvalue, which means that finding all the square roots of $B^2$ is not quite straight forward. After all, any reflection... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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How can I show that $x$ and $y $ must have the same length where $ x+y $ and $x-y $ are non-zero vectors and perpendicular?
Problem: Let $x$ and $y$ be non-zero vectors in $\mathbb{R}^n$.
(a) Suppose that $\|x+y\|=\|x−y\|$. Show that $x$ and $y$ must be perpendicular.
(b) Suppose that $x+y$ and $x−y$ are non-zero and ... | It's unclear from where you got the starting relations. That $x+y$ and $x-y$ are perpendicular means that
$$(x+y)\cdot(x-y) = 0 \tag 1$$
and you can start reasoning from there:
$$
0 \stackrel{(1)}= (x+y)\cdot(x-y) = x^2 +yx - xy -y^2 = \|x\|^2 - \|y\|^2
$$
Thus $\|x\|^2 = \|y\|^2$ and taking square root yields $\|x\| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Show that $\lim_{n \to \infty} \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \frac{1}{2}$.
Show that
$$ \lim_{n \to \infty} \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \frac{1}{2}. $$
Progress:
I rewrote the inside as
$$ S_n = \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \sum_{k = 0}^\infty (-1)^k \left( 1 - \frac{k... | If you are willing to accept a solution that doesn't use generating functions, then here are such ones:
1st Solution. Fusing each consecutive even and odd terms into a single term,
\begin{align*}
S_n := \sum_{k=0}^{\infty} (-1)^k \frac{n}{n + k}
&= \sum_{k=0}^{\infty} \left[ \frac{n}{n + 2k} - \frac{n}{n + 2k+1} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Why is $(3, 1+\sqrt{-26})^3=( 1+\sqrt{-26})$ in $\mathbb Z[\sqrt{-26}]$? $a = 3, b = 1+\sqrt{-26}$ then $(a,b)^3=(a^3,b^3,a^2b,ab^2)$
each generator except $a^3$ has a $b$ factor and $\bar b b=27$, so $"\subseteq"$. Now the question is how to obtain $b$ using these generators. Is there a general formula, which states i... | To be explicit, you could use linear algebra:
$$\begin{align}
a^3&=27&&\mapsto\begin{bmatrix}27\\0\end{bmatrix}\\
b^3&=-77-23\sqrt{-26}&&\mapsto\begin{bmatrix}-77\\-23\end{bmatrix}\\
a^2b&=9+9\sqrt{-26}&&\mapsto\begin{bmatrix}9\\9\end{bmatrix}\\
ab^2&=-75+6\sqrt{-26}&&\mapsto\begin{bmatrix}-75\\6\end{bmatrix}\\
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving a $n\times n$ determinant. $$
\left|
\begin{array}{cccccc}
3&2&0&0&0&\ldots\\
1&3&1&0&0&\ldots\\
0&2&3&2&0&\ldots\\
0&0&1&3&1&\ldots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{array}
\right|
$$
I solved the determinant by realising it's matrix is tridiagonal and using the continuant, and have gotten a rec... | If you develop along the first column, you get
$$\begin{align*} D_n &= 3\begin{vmatrix}3 & 1 & 0 & 0 & \cdots \\
2 & 3 & 2 & 0 & \cdots \\
0 & 1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots \end{vmatrix} -
\begin{vmatrix} 2 & 0 & 0 & 0 & \cdots \\
2 & 3 & 2 & 0 & \cdots \\
0 & 1 & 3 & 1 & \dots \\
\v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Standard limit proof Everyone knows the standard result $$\boxed{\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=e}$$ My friend gave a proof for this result using binomial expansion of $$\left(1+\dfrac{1}{x}\right)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\dots$$ which is valid only when $|1/x|<1$ and $n\... | This is correct for each $\vert x \vert > 1$:
$$ \left( 1+\frac{1}{x} \right) ^ x =
1+1+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots$$
It is simply Newton's Generalised Binomial Theorem with $x$ replaced by $1$, $y$ replaced by $\frac{1}{x}$ and $r$ replaced by $x.$
However, It is not obviou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Computing the integral of trig function under square root How can we solve this integral
$$\int \sqrt{\csc^2x -2}
\mathrm{d}x$$
My idea was substituting $\csc^2x=2\csc^2\theta$. Then the integral became $$\sqrt{2}\int \frac{\csc^2\theta-1}{\sqrt{2\csc^2\theta-1}} \mathrm{d}\theta$$ after a few simplifications. I don'... | Letting $u=\sqrt{\csc ^{2} x-2}$ transform the integral into
$$\begin{aligned}
I&=-\frac{1}{2} \int \frac{u^{2} d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}\\& = \frac{1}{2} \underbrace{\int \frac{d u}{\sqrt{u^{2}+1}}}_{\sinh ^{-1} u+C_1} + \underbrace{ \int \frac{d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}}_{J}\end{aligne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How to prove that $\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$? $$\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$$
Equality seems to be when x = 1
I have managed to show that the derivative is 0 at x = 1, and that this is a minimum (by the second der... | First, we can take the first derivative of this function
$$\frac{d}{dx}\sqrt{\frac{x^2+1}{x+1}}\;+\;\frac{2}{\sqrt{x}+1}$$
we get
$$\frac{x^2+2x-1}{2\sqrt{x^2+1}\left(x+1\right)^{\frac{3}{2}}}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}$$
We then want to find the critical number by letting the first derivative be $0$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the inequality $3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}$ I tried to group the summands so that I could decompose them into multipliers, but nothing worked...
$$3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}\Leftrightarrow 3^{x^2+6x+9}+\frac{1}{9}\leq 3^{x^2-2}+3^{6x+9}$$
$$3^{x^2+6x+7}+1\leq 3^{x^2-4}+3^... | Multiplying both sides by $9=3^2$,
\begin{align*}
3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}&\Longleftrightarrow 3^{(x+3)^2\color{red}{+2}}+1\leq 3^{x^2-2\color{red}{+2}}+3^{3(2x+3)\color{red}{+2}}\\
&\Longleftrightarrow 3^{x^2+6x+11}+1\leq 3^{x^2}+3^{6x+11}\\
&\Longleftrightarrow \left(3^{x^2}-1\right)\left(3^{6x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Cholesky-like decomposition that works on singular matrices? Is there a variant of Cholesky-like decomposition that works for singular matrices?
My problem is that all implementations of Cholesky I found fail when the matrix is singular in machine precision.
One idea for such decomposition is to consider duality betwee... | An approach using $QR$ decomposition: the following results in a decomposition of the form $A = MM^T$ with $M$ of full rank of size $n \times r$ (with $r$ equal to the rank of $A$), but $M$ is not necessarily in lower-triangular form. If it is desired, $M$ can be put into lower-triangular form with a further $QR$ decom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is... | The polynomial $x^{k+2}+x^k$ is divisible by $x^3+x$ for $k\ge 1.$ Hence $[x^{13}+x^{14}+\ldots +x^{20}]$ and $x^{10}+x^{12}$ are divisible by $x^3+x.$ We are down to $x^{11}$ (thanks @Cathedral ) and $$\displaylines{x^{11}=x^{11}+x^9-(x^9+x^7)+(x^7+x^5)\\ -(x^5+x^3)+(x^3+x)-x}$$
The remainder is equal $-x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
Regular heptagon area formula Be a regular heptagon. Can anyone demonstrate this formula?
I know:
Let apothem = ap
side = l
$ S=\frac{p.ap}{2}\implies S = \frac{7l.ap}{2}\\
\triangle OBC: OC^2 = a^2+l^2\\
\triangle PAB: b^2+l^2 = PB^2\\
OC^2-a^2=PB^2-b^2$
but i can't go on
| Let $d$ be the diameter of the circumcircle of $ABCDEFG$. Note that $a = BC \sin \frac{2\pi}{7}=d\sin \frac{\pi}{7}\sin \frac{2\pi}{7}$ and $b=AD \sin \frac{2\pi}{7} = d \sin \frac{3\pi}{7} \sin \frac{2\pi}{7}$. On the other hand, the area of $ABCDEFG$ equals $7 \cdot \frac 12 \left(\frac{d}{2}\right) \sin \frac{2\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to ... | Choosing a positive number $\delta < \frac{\epsilon}{1 + \epsilon}$ will make $\frac{\delta}{1 - \delta} < \epsilon$; if $0 < |x| < \delta$, then by the reverse triangle inequality $$\left|\frac{x}{x+1}\right| \le \frac{|x|}{1 - |x|} < \frac{\delta}{1 - \delta} < \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Is there a general method to find the asymptotic order for this sequence? Given $$a_{n+1}=a_n+\frac{n}{a_1+\dots+a_n},\qquad a_1>0$$ The answer is $$\lim_{n\to\infty} a_n\sim\sqrt{3}\cdot\sqrt{n}-\frac{\sqrt{3}}{4}\cdot\frac{1}{\sqrt{n}}$$
It is easy to show this sequence is increasing, and is divergent, because if ass... |
$$a_{n+1}=a_n+\frac{n}{\sum_{k=1}^n a_k}$$
This equation can be written as:
$$\sum_{k=1}^n a_k=\frac{n}{a_{n+1}-a_n}$$
Further, we can find $a_n$ by taking subtraction:
$$\begin{align}
a_n&=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k\\
\\
a_n&=\frac{n}{a_{n+1}-a_n}-\frac{n-1}{a_{n}-a_{n-1}}\\
\\
a_n(a_{n+1}-a_n)(a_{n}-a_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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How to show $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$, with $n$=odd? In previous post, I got the answer that $\gcd \left(\frac{p^2+1}{2}, \frac{p^5-1}{2} \right)=1$, where $p$ is prime number.
I am looking for more general case, that is for $p$ prime,
When is $\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2}... | For primes of the form $6k-1$, and choosing $m$ odd and $n$ even, you will obtain $p^m=6a-1$ and $p^n=6b+1$. Hence $p^m+1=6a$ and $p^n-1=6b$ with $\gcd(6a,6b) \ge 6$. Your final conjecture $\gcd \left( \frac{p^m+1}{2}, \frac{p^n-1}{2}\right)=1$ is never true in the circumstances considered here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Solving the system $x^3+y= 3x+4$, $2y^3+z = 6y+6$, $3z^3+x=9z+8$
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x^3+y= 3x+4 \\
2y^3+z = 6y+6 \\
3z^3+x=9z+8
\end{split}
\end{equation}$$
By the theorem of triviality, I assumed $x=y=z=k$ and then solved to fortunately get a solution. Since, it ... | Taking help from the previous answer, we can write the equations as
$$x^3-3x-2=(x+1)^2(x-2)=2-y$$
$$2(y^3-3y-2)=2(y+1)^2(y-2)=2-z$$
$$3(z^3-3y-2)=3(z+1)^2(z-2)=2-x$$
Multiplying the three, we get
$$6(x+1)^2(y+1)^2(z+1)^2(x-2)(y-2)(z-2) = -(x-2)(y-2)(z-2)$$
If none of $x,y,z$ is equal to 2 we can cancel $(x-2)(y-2)(z-2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2... | $$x + \frac{1}{x} = \sqrt2 $$
squaring both sides,
$$x^2 + \frac{1}{x^2} = 0 $$
then,
$$x^{2022} + \frac{1}{x^{2022}} = (x^2 + \frac{1}{x^2})(x^{2020} + \frac{1}{x^{2020}}) - (x^{2018}+ \frac{1}{x^{2018}}) = - (x^{2018}+ \frac{1}{x^{2018}})$$
repeat the above process,
$$x^{2022} + \frac{1}{x^{2022}} = x^2 + \frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2} \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) ... | I think you are correct, although your answer for $(c)$ is unnecessarily long-winded. You would benefit from using the formula for a finite sum of geometric series, rather than using the infinite sum everywhere. It is:
$$ \sum_{k=1}^{n} ar^k = a\left( \frac{1-r^{n+1}}{1-r} \right).$$
where $a$ is the first term and $r$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Number of paths in a square lattice with restrictions It is known that the number of paths with only rightward and upward moves in a square lattice from the lower left corner to the upper right corner is $\binom{a+b}{b}$, where $a$ is the height of the lattice and $b$ is the width. What is the number of paths in the la... | The problem is equivalent to finding the number of permutations of $a$ letters $U$ and $b$ letters $R$ such that there aren't four or more consecutive $U$s in the permutation. right?
If first, we put $R$s in a row, then there will be $b + 1$ places between the letters:
\begin{align*}
-\ R\ -\ R\ -\ R\ - \cdots -\ R\ -\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Other approaches to simplify $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$ I want to simplify the trigonometric expression $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$.
My approach,
Here I used the abbreviation $s,c,t$ for $\sin x$ and $\cos x$ and $\tan x$ respectively,
Numerator is, $$\frac{s^2}{c^2}-s^2=\frac{s^2-s^2c^2}{c... | There are many ways to do it.
For axample, let $x=\tan^{-1}(t)$ and, after simplifications
$$\frac{\tan ^2(x)-\sin ^2(x)}{\tan (2 x)-2 \tan (x)}=\frac{1}{2} t \cos \left(2 \tan ^{-1}(t)\right)=\frac{t \left(1-t^2\right)}{2 \left(1+t^2\right)}=\frac 1 2 \times t\times \frac{ \left(1-t^2\right)}{ \left(1+t^2\right)}$$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why this transformation matrix $A$ has $\begin{pmatrix}0 \\ 1\end{pmatrix}$ as Eigenvector? I have the following transformation matrix:
$$
A=\begin{pmatrix}
1 & 0 \\
-1 & 4
\end{pmatrix}
$$
If I resolve to find the eigenvalues I get:
$$
\begin{vmatrix}
A-\lambda I
\end{vmatrix} = 0
$$
which leads to:
$$
\lambda_1 = 1;... | $\begin{pmatrix} 3\\1\end{pmatrix}$ is not an eigenvector for eigenvalue $\lambda_2=4$; it is an eigenvector for $\lambda_1=1$.
$\begin{pmatrix} 0\\1\end{pmatrix}$ is the eigenvector for eigenvalue $\lambda_2=4$, given that plugging in $x_1=3$, $x_2=1$ gives the zero vector in that calculation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the range of $\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}$ Here is the expression:
\begin{equation}
\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}
\end{equation}
I have tried to maximize the denominator, so I got the maximum value of denominator equal to 7, but I am lost what to do next
I did t... | If $x\in[-\pi/2,\pi/2]$ then $|\cos x|=\cos x$ so
\begin{equation}
\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}
={5\cos x+5/6\over6\cos x+1}+{-2\sin^2 x+ 4\sin x -3\frac56\over6\cos x+1}\\
=5/6+{-2\sin^2 x+ 4\sin x -3\frac56\over6\cos x+1}\\
\end{equation}
The $-2\sin^2 x$ term is always negative, so let $x=-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
When solving $\sin^2 x = \frac14(2+ \sqrt 3)$, is it better to substitute $\sin^2x=1-\cos^2x$ or $\sin^2x=\frac12(1-\cos 2 x)?$ I am trying to evaluate
$$\sin^2 x = \frac{2+ \sqrt 3}{4}$$ where $x$ is $0$ to $2 \pi$.
For $ \sin^2 x$, I can substitute it to either:
*
*$ 1 - \cos ^2 x$ (Pythagorean identity)
*$ \frac... | $$\frac{2+\sqrt 3}{4} = \sin^2x = \frac{1-\cos 2x}{2},$$
$$\cos 2x = 1-\frac{2+\sqrt 3}{2} = -\frac{\sqrt 3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| We know that: $x=\frac{-7\pm\sqrt D}2$
For natural solutions: $$x=\frac{-7+\sqrt D}2$$
Also, $D=n^2$ where $n$ is an natural number. And, $n$ has to be odd and $n\geq7$.
$$\Rightarrow 7^2-4(4-2^y)=n^2$$
$$\Rightarrow 7^2-16-2^{y+2}=n^2$$
$$\Rightarrow n^2-33=2^{y+2}$$
We put $n=7$ (our first choice), and get $y=2$ (and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Solving $\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$ I have started self-studying differential equations and I have come across the following initial value problem
$$\frac{dx}{dt}=\frac{xt}{x^2+t^2}, \quad x(0)=1$$
Now, since $f(t,x)=\frac{xt}{x^2+t^2}$ is such that $f(rt,rx)=f(t,x)$ for every $r\in\mathbb{R}\setminus\{... | With homogeneous equations, you have two choices for the substitution.
Here we have
$$(x^2+t^2)\,dx=xt\,dt $$
You made the substitution $x=yt$, but you also had the choice of letting $t=ux$. As a general rule, the substitution which makes the algebraic simplification easier also makes the integration easier. So lets i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find possible values of $|z|$ if both the real and imaginary parts of $(\bar z)^2+\frac{1}{z^2}$ are integers Let $\bar z$ denote the complex conjugate of a complex number $z$. If $z$ is a non-zero complex number for which both real and imaginary parts of $$(\bar z)^2+\frac{1}{z^2}$$ are integers, then which of the fol... | Let $r = |z|$ be a possible value of the modulus. We claim that the above condition is equivalent to
$$
\frac{(|z|^4 +1)^2}{|z|^4} = a^2 + b^2; \ \ a,b \in \mathbb{Z}
$$
It is clear that the condition is necesary, because this the left side is just $|f(z)|^2$. To prove the other implication, consider the map $f(z) = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as fol... | Let $z=x+iy$
$$z^2 =\bar z\cdot\dfrac{1-i}{1+i}=\cdots=-i\cdot\bar z$$
$$\implies x^2-y^2+i(2xy)=-i(x-iy)=-y-ix$$
Equating the imaginary parts, $-x=2xy\iff x(2y+1)=0\ \ \ \ (1)$
Equating the real parts, $x^2-y^2=-y\ \ \ \ (2)$
From $(1),$
either $x=0, $ using $(2), 0^2-y^2=-y\implies y=?$
or $2y+1=0, $ using $(2), x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $xdx-ydy=y^2(x^2-y^2)dy$ Question:
$$xdx-ydy=y^2(x^2-y^2)dy$$
I'm having trouble matching the solution in the book, which is $\frac{1}{2}\ln(x^2-y^2)=\frac{1}{3}y^3+C$.
I'm getting an integral that requires the incomplete gamma function.
My attempt:
Rewrite the equation:
$x + (-(x^2 - y^2) y^2 - y)\frac{dy}{dx} =... | Looks easier if you try to substitute $u=x^2-y^2 \implies u'=2xx'-2y$
$$xdx-ydy=y^2(x^2-y^2)dy$$
$$xx'-y=y^2(x^2-y^2)$$
$$u'=2y^2u$$
The DE is separable
More simply:
$$xdx-ydy=y^2(x^2-y^2)dy$$
$$dx^2-dy^2=2y^2(x^2-y^2)dy$$
The DE is separable.
$$\dfrac {d(x^2-y^2)}{x^2-y^2}=2y^2dy$$
Integrate.
$$\ln (x^2-y^2)=\dfrac 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1... | $\sin(\alpha)-\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}-\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(-\sin(\alpha)\cos\left(\frac{3\pi}{4}\right)-\cos(\alpha)\sin\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)$
$\sin(\alpha)+\cos(\alpha)=\sqrt{2}\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
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Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-... | For $x=y=z=1$ we obtain on the L.H.S. the value two. Which is far away from the R.H.S. $9/4$. What about showing a stronger inequality under the given constraints?
(Of course, we are slightly changing the "weakest" term...)
$$
\boxed{\qquad
xy + yz + zx - \color{blue}{\frac 34} xyz \ \le\ \frac 94
\qquad}
$$
It is a go... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Doubts in solving $\int_0^1 x\sqrt{x+2} dx$ I was solving the following integral and got stuck in finding the new limits after the substitution. $$\int_0^1 x\sqrt{x+2} dx$$
Here's my work so far:
Putting $(x+2) = t^2$ so that $dx = 2t\ dt$ and $x = t^2 - 2$
Thus the original integral changes to,
$$\int_{?}^{?} (t^2 - ... | Given that : $$=\int _0^1\:\left(x\sqrt{x+2}\right)dx$$
Let : $$u=x+2$$ $$\frac{du}{dx}=1$$ $$du=dx$$
Also : $$u=x+2$$ $$x=u-2$$
Substitute into the equation : $$=\int \left(\left(u-2\right)\sqrt{u}\right)du$$
Then : $$=\int \left(u^{1.5}\:-2u^{0.5}\right)du$$
Integrate : $$=\frac{u^{2.5}}{2.5}-\frac{2u^{1.5}}{1.5}$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | I think your current description is not satisfactory.
Take Case 3 for example. You said,
This means, $f(x) \ge \frac {3a^2}{4}-5a+4$. Thus, for $f(x) > 0$, it
is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a \ge
0\wedge 4-\frac a2\ge 0$.
This description does not emphasize the sufficient and necess... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 5
} |
Rotation matrix exponential form derivation (only acts in 2 dimension) I am trying to derive the rotation matrix in expoonential form. I start by considering a rotation in three dimensions about an arbritrary axis, $\hat{\textbf{n}}$ by an infinitesimally small angle $\delta\theta$ as depicted above. The coordinate tra... | Does the following help ?
\begin{align}
&\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}+
\sum_{n=1}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\\
&=\begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}+\sum_{n=0}^\infty \frac{(-1)^n\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all real solutions of the equation $x^{10} - x^8 + 8x^6 - 24x^4 + 32x^2 - 48 = 0$ I have been able to factorize the polynomial as follows: $$(x^2 - 2)(x^8 + x^6 + 10x^4 - 4x^2 + 24)$$ from which $\sqrt2$ and $-\sqrt2$ are obvious solutions. My guess is that $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ does not have any rea... | $x^8 + x^6 + 10x^4 - 4x^2 + 24\geq 10x^4 - 4x^2>0$ if $x^{2}>\frac 2 5$ and $x^8 + x^6 + 10x^4 - 4x^2 + 24\geq x^8 + x^6 + 10x^4 - \frac 8 5 + 24\geq 24 -\frac 8 5 >0$ if $x^{2}<\frac 2 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$ Without any software and approximations prove that $$\sec(52^{\circ})-\cos(52^{\circ})>1$$
We can use some known trig values like $18^{\circ}$,$54^{\circ}$,etc
My try:
I considered the function:
$$f(x)=\sec(x)-\cos(x)-1,\: x\in \le... | Thanks to "user" for giving me thought to complete the proof.
Here is the proof:
We are aiming to prove $\sec(52^{\circ})-\cos(52^{\circ})>1$. Let $\phi$ be the Golden ratio and $\psi$ be its reciprocal..
Consider $$f(x)=\sec x-\cos x,\:\:0<x<\frac{\pi}{2}$$
Its evident that $f$ is Monotone increasing.
Also $$\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
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Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
From a question I asked before this, I have trouble actually with the numbers manipulating part.
Using trigo identity, $\sin^2 \frac{\pi}{12} ... | Another way would be to use the double angle formula
$$\sin 2\theta=2\sin\theta\cos\theta$$
With $\theta=\frac {\pi}{12}$, then
$$\sin 2\theta=\sin\frac {\pi}6=\frac 12$$
Thus
\begin{align*}
\cos\frac {\pi}{12} & =\frac {\sin 2\theta}{2\sin\theta}\\
& =\frac 1{4\sin\theta}\\
& =\frac 1{\sqrt 2(\sqrt 3-1)}
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Proving that $\frac{xy^2}{x^2+y^4}$ is bounded I am trying to prove that there exists $M>0$ such that, for all $(x,y) \in \mathbb{R}^2$,
$|\frac{xy^2}{x^2+y^4}|\leq M$. Is the following proof correct?
I claim that there exists such an $M$, and that $M=1$.
Suppose, first, that $|x^2|\geq |xy^2|$. Then $|\frac{xy^2}{x^2+... | Your proof is a good attempt, but it is not complete. The "mistake" in the proof is that the inequality in red below is not justified:
Suppose, second, that $x^2<xy^2$. Then $|\frac{xy^2}{x^2+y^4}|\color{red}{\leq} |\frac{y^4}{x^2+y^4}|\leq |\frac{y^4}{y^4}|=1$.
You do not explain why this is true. Your assumption is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what positive integers $A$ is the sequence eventually constant?
For each integer $n \ge 0$ let $S(n)=n-m^2$, where $m$ is the greatest integer such that $m^2\le n$. Define $(a_k)_{k=0}^\infty$ as $a_0=A$ and $a_{k+1}=a_k+S(a_k)$ for $k \ge0$. For what positive integers $A$ is the sequence eventually constant?
I'm... | You're misunderstanding the $A = 4 $ case. The $a_1$ part will be $4 + (4 - 4)$.
This sequence will be constant for all perfect squares because if $a_i = k^2$, then $S(a_{i}) = k^2 - k^2 = 0 $, since $k^2$ is the largest perfect square $\leq k^2$. This implies that $a_{i + 1} = a_i = k^2$. Notice that the sequence will... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the locus of points |z-1|= -Im(z). If I wish to find the locus of complex points satisfying $ |z-1|= -\text{Im}(z)$, then would I be right in supposing it represents the half-circle
$(x-1)^2 + (y-1/2)^2 = 1/4, y \leq 0$?
My work follows:
*
*First, notice $|z-1| \geq 0 \Rightarrow \text{Im}(z) \leq 0.$
*Second,
... | HINT
You are on the right track, but you forgot to square $\text{Im}(z)$. If we let $z = x + yi$, it results that
\begin{align*}
|z - 1| = -\text{Im}(z) = -y & \Longleftrightarrow
\begin{cases}
|z - 1|^{2} = y^{2}\\\\
y\leq 0
\end{cases}\\\\ & \Longleftrightarrow
\begin{cases}
|z|^{2} - 2\text{Re}(z) + 1 = y^{2}\\\\
y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Partial Fraction of $\frac{1-x^{11}}{(1-x)^4} $ for Generating Function The original question involves using generating functions to solve for the number of integer solutions to the equation $c_1+c_2+c_3+c_4 = 20$ when $-3 \leq c_1, -3 \leq c_2, -5 \leq c_3 \leq 5, 0 \leq c_4$.
Using generating functions I was able to ... | $$\begin{align}
\frac{1-x^{11}}{(1-x)^4}&=(1-x^{11})\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n\\
&=\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n-\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^{n+11}\\
&=\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n-\sum_{n\ge11}\frac{(n-10)(n-9)(n-8)}{6}x^{n}\\
&=\sum_{n\ge0}\frac{1}{6}\bigg((n+1)(n+2)(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
2022 Gr11 Fermat math contest question #20 Question: A sequence of numbers $t_1, t_2, t_3$,... has its terms defined by $t_n = \frac{1}{n} - \frac{1}{n+2}$ for every integer $n \geq 1.$ For example, $t_4 = \frac{1}{4} - \frac{1}{6}$. What is the largest positive integer k for which the sum of the first k terms (that is... | $S_k = \sum_{n=1}^k t_n = \sum_{n=1}^k\frac{1}{n} - \frac{1}{n+2} = A_k - B_k< 1.499$. Now, since $B_k = A_{k+2} - \frac{1}{1} - \frac{1}{2}$,
$$
S_k = A_k - A_{k+2} + \frac{3}{2} = -\frac{1}{k+1} - \frac{1}{k+2} + 1.5 < 1.499.
$$
This gives
$\frac{1}{k+1} + \frac{1}{k+2} > 1.5 - 1.499 = 0.001 = \frac{1}{2000} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Determine the cartesian equation of the tangent plane to M in p. Given the torus and given the point p $\in$ M corresponding to the parameters $s=\frac{\pi }{4}$ and $t=\frac{\pi }{3}$.
Determine the cartesian equation of the tangent plane to M in p.
$\begin{cases} x=\left(3+\sqrt{2}cos\left(s\right)\right)cos\left(t\r... | Since you have the parametric equation of the surface, which is
$ \mathbf{P}(t, s) = ( x(t,s), y(t,s), z(t,s) ) $
Then find $\dfrac{\partial \mathbf{P}}{\partial t} $ and $\dfrac{\partial \mathbf{P}}{\partial s } $ as follows
$ \dfrac{\partial \mathbf{P}}{\partial t} =( - (3 + \sqrt{2} \cos(s) ) \sin(t) , (3 + \sqrt{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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The convergence speed of $ \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x $? I have already known how to prove
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \sqrt{\frac{\pi}{2n}}
\end{equation*}
with Wallis's formula
\begin{equation*}
\quad \frac{... | If you are familiar with the Gaussian hypergeometric function, if $0 \leq x \leq \frac \pi 2$
$$\int \sin^n(x)\,dx=-\cos (x) \, _2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos^2(x)\right)$$ and the definite integral just becomes
$$I_n=\int_0^{\frac \pi 2}\sin^n(x)\,dx=\frac{\sqrt{\pi }} 2\,\,\frac{ \Gamma \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Which numbers result in a chain other than $2 \to 0$? Let $n$ be a natural number and $k$ the number of its divisors. Calculate $n-k$, then repeat this procedure by taking $n-k$ as the starting value. If you do this repeatedly, which numbers results in chains other than $2 \to 0$?
I found this problem in the exam prep ... | Let us denote the iteration function with $f(n) = n - d(n)$, where $d(n)$ is the number of divisors of $n$.
For $n \ge 3$ is $1 \le f(n) \le n-2$, so that the iteration $n, f(n), f(f(n)), \ldots$ is strictly decreasing until it reaches either $1$ or $2$. Let $t(n)$ be this “terminal” non-zero number in the chain start... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $
\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
=
\begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
-1
... | It's easy to see that $0$ is a multiple eigenvalue with multiplicity two as we can exhibit two linearly independent vector of the kernel (since all columns are obviously a multiple of the first one). Then the last eigenvalue $\lambda$ can be found using the Trace as the sum of the diagonal elements is also equal to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus
*
*Verify the following limits
(iii) $\lim\limits_{n\to \infty} \left
[\sqrt[8]{n^2... | The solution considers $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[8]{n^2}\right) \tag{1}$$ and $$\lim_{n \to \infty} \left( \sqrt[4]{n} - \sqrt[4]{n+1}\right), \tag{2}$$ yet you are asking about $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[\color{red}{4}]{n^2}\right), \tag{3}$$ which is neither of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4552533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Creating a complex number so that its norm equals to 1 I would like to create a complex number c so that its norm is equal to some number a (for the purpose of this question let's assume a = 1) if I already have either its real or imaginary part. I know that:
$$\lVert \mathbf{c} \rVert = \sqrt{\sum_{i=1} ^{n} c_i \over... | The Problem
You can't just use any number for $a$ and $b$! If $a$ and $b$ are real numbers, which they are, their square is always positive or zero, which can only be the case if $|a| \leq 1$ and/or $|b| \leq 1$ are less than or equal to 1!
$$
\begin{align*}
a^{2} + b^{2} &= 1 \quad\mid\quad -\left(b^{2}\right)\\
a^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the closed formula for the following recusive sequence. Find the closed formula for the recursive sequence defined by $a_0 = 4$, $a_1 = 12$, $a_n = 6a_{n-1}-a_{n-2}$ for $n>1$. This question is stumping me. I don't know any methods besides guess and check.
Current progress:
The first five terms are $4, 12, 68, 396... | Solution by generating function:
The difference equation can be seen in the form
$$ a_{n+2} = 6 \, a_{n+1} - a_{n} \, \hspace{10mm} a_{0} = 4, a_{1} = 12$$
leads to
\begin{align}
\sum_{n=0}^{\infty} a_{n+2} \, t^{n+2} &= 6 \, t \, \sum_{n=0}^{\infty} a_{n+1} \, t^{n+1} - t^2 \, \sum_{n=0}^{\infty} a_{n} \, t^n \\
\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Confusion with solution to UKMT question There was a question I was trying to solve from the UKMT Senior Maths Challenge. It goes like this:
Four positive integers a, b, c, and d are such that
abcd + abc + bcd + cda + dab + ab + bc + cd + da + ac + bd + a + b + c + d = 2009
What is the value of a + b + c + d?
I have... | $$abcd + abc + bcd + cda + dab + ab + bc + cd + da + ac + bd + a + b + c + d$$
You can gather together all the terms containing for example $a$, and take it out to get this:
$$a(bcd + bc + cd + bd + b + c + d + 1) + bcd + bc + cd + bd + b + c + d$$
Clearly all the later terms are almost the same as the terms inside the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Summation of a rational function Calculate $\sum_{r=2}^{n} \frac{3r^2-1}{(r^3-r)^2}$.
My approach till now: $T_r=\frac{3r^2 -1}{r^2(r+1)^2(r-1)^2}$. Now let, $$3r^2-1= a(r^2)+ b(r+1)^2 + c(r-1)^2$$
we find, $a=4$, $b=c=-\frac{1}{2}$. now we break it up, $$T_r = \frac{4}{(r+1)^2(r-1)^2}-\frac{1}{2(r)^2(r+1)^2}-\frac{1}... | You can do standard partial fraction decomposition. That is, let $$\frac{3r^2-1}{(r^3-r)^2} = \frac{a}{r}+\frac{b}{r^2}+\frac{c}{r-1}+\frac{d}{(r-1)^2}+\frac{e}{r+1}+\frac{f}{(r+1)^2}.$$ Then you will get $$-\frac{1}{r^2}+\frac{1}{2(r-1)^2}+\frac{1}{2(r+1)^2}.$$ Can you do the telescoping from here and conclude?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Can we make the inequality $A-B \leq \frac{A^2}{4}$ strict? I have an interesting problem:
Given that $$
A=\frac{1}{2022}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)$$
$$B=\frac{1}{2023}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2023}\right)
$$
I got an upper bound on $A-B$.The Lower bound is zero,... | Your proof for the $A-B \le A^2/4$ is correct, but for the strict inequality, the irrationality of $\sqrt{1-B}$ is needed, which is not obvious (at least not to me).
Here is a different approach which gives the strict inequality without the need to argue about irrational numbers. It works for an arbitrary number of ter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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$\int\sqrt{1+t^2}\, dt$ using the substitution $t = \frac{1}{2}(e^u-e^{-u})$ This is what I have done so far:
$$\frac{dt}{du}=\frac{d}{du}\left(\frac{1}{2}(e^{u}-e^{-u})\right)=\frac{1}{2}(e^{u}+e^{-u}) \implies dt=\frac{1}{2}(e^{u}+e^{-u})\,du$$
$$L=\frac{1}{2}\int_{0}^{2}\sqrt{1+t^2}\, dt=\frac{1}{2}\int_{u(0)}^{u(2)... | Starting with
$$L=\frac{1}{2}\int_{a}^{b}\sqrt{1+t^2}\, dt=\frac{1}{4} \, \int_{u(a)}^{u(b)} (e^{u}+e^{-u}) \, \sqrt{1+\left(\frac{1}{2}(e^{u}-e^{-u})\right)^2} \, du$$
and then using
$$ 1 + \frac{(e^{u} - e^{-u})^2}{4} = \frac{e^{2 u} + 2 + e^{- 2 u}}{4} = \left(\frac{e^{u} + e^{-u}}{2}\right)^2 $$
the integral $L$ be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Multiple solutions for trig function with period Find all angles with limits $−\pi \leq \theta \leq \pi$ which satisfy $\sin 4 \theta = 1$
The working out is given as:
If $\sin 4 \theta = 1$ then $4 \theta = \pi/2 + 2 k \pi$, so $\theta = \pi/8 + k \pi/2$
For $−\pi \leq \theta \leq \pi$ we have $\theta = \pi/8, 5 \pi/8... | When you reach at: $θ=\displaystyle \frac {\pi} 8 +\frac{k \pi} 2 ,k\in\mathbb{Z}\text{ } (1) $
We are given that: $θ\in[-\pi, \pi] \text{ } (2)$
So: $(1),(2)\implies -\pi<\displaystyle \frac {\pi} 8 +\frac{k \pi} 2<\pi\iff -\frac {\pi} 8-\pi<\frac{k \pi} 2< -\frac {\pi} 8+\pi\iff $
$$ \displaystyle -\frac {9\pi} 8<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all the real roots of $P(Q(x))=0$ Let $$P(x)=x^2+\frac{x}{2}+b$$ and $$Q(x)=x^2+cx+d$$ be two polynomials with real coefficients such that $\displaystyle P(x)\cdot Q(x)=Q(P(x))$ for all real $x$. Find all the real roots of $P(Q(x))=0$.
I found out that
\begin{align}
x^4+x^3c+x^2d+\frac{x^3}{2}+\frac{x^2c}{2}+\fra... | So the equation you've found gives
$$x^4+x^3(c+1/2)+x^2(d+c/2+b)+x(d/2+bc)+bd=x^4+x^3+(2b+1/4+c)x^2+(b+c/2)x+(b^2+cb+d)$$
Comparing coefficients, we get
*
*$c+1/2=1\implies c=1/2$
*$d+c/2+b=2b+1/4+c\implies d+1/4+b=2b+1/4+1/2\implies d=b+1/2$
*$d/2+bc=b+c/2\implies d/2+b/2=b+1/4\implies d=b+1/2$
*$bd=b^2+bc+d\impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$.
Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$.
Firstl... | My solution:
Let us consider two positive reals $\alpha,\beta$ defined as$$\alpha=\sqrt{4+x} \quad \beta=\sqrt{4-x} \Rightarrow \alpha^2+\beta^2=8 \Rightarrow (\alpha-\beta)^2+2\alpha\beta=8$$
Now the equation becomes
$$\frac{2+\alpha}{\alpha}=\frac{2-\beta}{\beta} \Rightarrow \frac{\alpha}{\alpha}+\frac{\beta}{\beta}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Prove that $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=2\pi$. Probably equation $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=\int_0^{2\pi} \frac{ab}{a^2\sin^2t+b^2\cos^2t}\mathrm dt$ is useful. Double integration is also a tool, but I don't know next step.
| I thought it might be instructive to present an approach that uses the identities, $\cos^2(t)=\frac{1+\cos(2t)}{2}$ and $\sin^2(t)=\frac{1-\cos(2t)}{2}$ to streamline the analysis. Proceeding, we can write
$$\begin{align}
\int_0^{2\pi}\frac{ab}{a^2\cos^2(t)+b^2\sin^2(t)}\,dt&=\int_0^{2\pi}\frac{2ab}{(a^2+b^2)+(a^2-b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
A right triangle and a point inside that divides it into three equal areas $ABC$ is a right-angled triangle ($\measuredangle ACB=90^\circ$). Point $O$ is inside the triangle such that $S_{ABO}=S_{BOC}=S_{AOC}$. If $AO^2+BO^2=k^2,k>0$, find $CO$.
The most intuitive thing is to note that $AO^2+BO^2=k^2$ is part of the c... | As @DavidQuinn noted, point $O$ is the centroid of the right triangle.
If we place $C$ at $(0,0)$ and $B$ at $(a, 0)$ and $A$ at $(0,b)$, then
$O = \dfrac{1}{3} (a, b ) $
Therefore
$OA^2 + OB^2 + OC^2 = \dfrac{1}{9} \bigg( a^2 + b^2 + 4 a^2 + b^2 + a^2 + 4 b^2 \bigg) = \dfrac{1}{9} ( 6 ) (a^2 + b^2 ) = \dfrac{2}{3} (a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Tangent plane at a point to surface $\mathbf{F}=\mathbf{0}$ Let $\mathbf{F}:\mathbb{R}^5\to\mathbb{R}^3, \mathbf{F}\begin{pmatrix}x_1\\x_2\\y_1\\y_2\\y_3\end{pmatrix}=\begin{bmatrix}2x_1+x_2+y_1+y_3-1\\x_1x_2^3+x_1y_1+x_2^2y_2^3-y_2y_3\\x_2y_1y_3+x_1y_1^2+y_2y_3^2\end{bmatrix}$ and $\mathbf{a}=\begin{bmatrix}0\\1\\-1\\... | The result is correct. Another way to check this is that the tangent plane at the point can be obtained by translating the subspace $\mathbf{N}([D\mathbf{F(a)}])$ so that it passes through $\mathbf{a}$.
A basis of $N(D\mathbf{F}(\mathbf{a}))$ is $\left\{
\begin{bmatrix}4\\-1\\-7\\2\\0\end{bmatrix}
, \begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4570100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove that $a_{n} := \frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2}$ converges to $2$ when $n\to\infty$ by the definition? Just wondering if this is the correct way to approach and $\epsilon$ n proof for convergence.
I am starting with the sequence
$\frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2}$
First step,
\begin{al... | HINT
As a start, I would recommend you to split the argument into two parts. The first part corresponds to:
\begin{align*}
\frac{n^{2}}{(n + 1)^{2}} - 1 = -\frac{2n + 1}{(n + 1)^{2}} = \frac{1}{(n + 1)^{2}} - \frac{2}{n + 1}
\end{align*}
Similarly, the second part corresponds to:
\begin{align*}
\frac{(n + 1)^{2}}{(n + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the coefficient without a calculator? I wanted to solve this question without using a calculator.
Question: The number of non-negative integer solutions to
$$3x+y+z=24$$
By creating generating functions you have to find the coefficient of $x^{24}$ in the expression: $$\left(\frac{1}{1-x}\right)^{2}\left(\fr... | Pen, paper, and effort is what is required outside of a computer to calculate higher terms in expansions. In this case
\begin{align}
\frac{1}{(1-x)^2} &= \sum_{n=0}^{\infty} (n+1) \, x^n \\
\frac{1}{1-x^3} &= \sum_{n=0}^{\infty} x^{3 n}
\end{align}
for which the product of these series gives
$$ P = \frac{1}{(1-x)^2 (1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[... | You can obtain the antiderivative. Let $x^n=t$ to face
$$I=\frac 1 n \int \frac{\log (t+1)}{t+1}\,t^{\frac{1}{n}-1}\,dt $$
$$I=n (t+1)^{\frac{1}{n}} \,
_3F_2\left(-\frac{1}{n},-\frac{1}{n},-\frac{1}{n};1-\frac{1}{n},1
-\frac{1}{n};\frac{1}{t+1}\right)+t^{\frac{1}{n}} \log (t+1)-$$
$$\frac{n t^{\frac{1}{n}+1} }{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s... | Too long for a comment.
It's equivalent to evaluate a certain digamma series, perhaps a more attainable goal:
$$\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\psi\left(\frac{2m+1}{n}\right)$$
$\newcommand{\d}{\,\mathrm{d}}$We have, for $s>0$: $$\begin{align}J_n&:=\frac{1}{n}\int_0^1\frac{(1-t)^{s-1}}{1+t^{2/n}}t^{1/n-1}\d t\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
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How many ways to get a sum of 29 by adding 5 & 2? ex 5+5+5+5+5+2+2 = 29, is one way. Ex: $2+2+3$, $2+3+2$, $3+2+2$ these are three ways to get a sum of $7$ with $3$ and $2$. But my example is with $5$ and $2$ and a sum of $29$.
I believe there are three ways to get a sum of $29$ by using fives and nines. But is there a... | Here's an observation: $(x^2+x^3)^k$ is a polynomial whose $x^n$ coefficient is the number of ways to get $n$ by using $k$ total $2$'s and $3$'s. So, if we don't care how many total we use, the $x^n$ coefficient of the geometric series
$$
\sum_{k=0}^\infty (x^2+x^3)^k = \frac{1}{1-x^2-x^3}
$$
is the number of ways of g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On The Question of A Squeeze-Derivative Definition Question: Find $g'(0)$ where $$g(x)=\left(x+1\right)^2 \left(1+\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\right)+\sqrt{x^4+4x+1}\left(1-\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\right).$$
The above question was asked by my professor as an exam question. When I was in the ex... | Let $\;g_1(x)=x^2+2x+1+\sqrt{x^4+4x+1}\;.$
Let $\;g_2(x)=\left[x^2\!+\!2x\!+\!1\!-\!\sqrt{x^4\!+\!4x\!+\!1}\right]\cos\left(\!\sqrt{|\tan(x+1)|}\right).$
It results that
$\dfrac{g(x)-g(0)}x=\dfrac{g_1(x)-g(0)}x+\dfrac{g_2(x)}x\;.$
$\dfrac{g_1(x)-g(0)}x=\dfrac{x^2+2x+1+\sqrt{x^4+4x+1}-2}x=$
$=\dfrac{x^2+2x}x+\dfrac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proof that $\frac{(1+\sqrt 5)^n+(1-\sqrt 5)^n}{2^n}$ is an integer I want to prove, by induction, that
$$\frac{(1+\sqrt 5)^n+(1-\sqrt 5)^n}{2^n}\quad\text{is an integer}.$$
Instinctively, this sort of thing looks like the solution of a second order recurrence relation (like the explicit formula for the Fibonacci sequen... | You have\begin{multline}\frac{\left(1+\sqrt5\right)^{n+1}+\left(1-\sqrt5\right)^{n+1}}{2^{n+1}}-\frac{\left(1+\sqrt5\right)^n+\left(1-\sqrt5\right)^n}{2^n}=\\=\frac{\left(-1+\sqrt5\right)\left(1+\sqrt5\right)^n+\left(-1-\sqrt5\right)\left(1-\sqrt5\right)^n}{2^{n+1}}=\\=\frac{\left(1+\sqrt5\right)^{n-1}+\left(1-\sqrt5\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Source: https://www.hkage.edu.hk/uploads/file/202207/6cda89c718b674f6ac3aa2c19049abe5.pdf
I tried substituting $y = 2x+3$ and ended up with $\frac{y^2}{2}-\frac{7}{2}y^{\frac 3 2} ... | You were correct in recognizing that $6x+9 = 3(2x+3)$ but we don't have to put everything in terms of $2x+3$
If we let $y=2x+3$ we get $2x^2 -7x\sqrt y +3y$ and we might recognize this as a quadratic. If we replace $w =\sqrt y = \sqrt{2x+3}$ we get
$2x^2 - 7xw + 3w^2=0$ and maybe we can factor $2x^2 - 7xw + 3w^2=0$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In $\triangle ABC$, if the angles are in an A.P and $b:c=\sqrt{3}:\sqrt{2}$, compute $\angle A$. As title suggests, we have a triangle $\triangle ABC$ with angles in an arithmetic progression, and sides $b$ and $c$ in a ratio of $\sqrt{3}:\sqrt{2}$. This is a problem I saw from a mathematics textbook in India. While I ... | Since angles are in an arithmetic progression, $(\angle B - \alpha) + \angle B + (\angle B +\alpha) =180^\circ$ then $\angle B= 60^\circ$
Let's say $b=\sqrt3, c=\sqrt2$.
Draw $AH \perp BC$ then $\angle BAH=30^\circ$and $AH=\frac{\sqrt6}{2}$.
Then in $\triangle AHC$, $HC=\frac{\sqrt6}{2}$ (Pytagorean theorem) and $AH=HC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How might I have anticipated that $\frac14(\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}})$ simplifies to a single surd (namely, $\frac14\sqrt{25+10\sqrt{5}}$)? This is perhaps a silly question related to calculating with surds. I was working out the area of a regular pentagon ABCDE of side length 1 today and I ended up with the... | Yes. There is a way to formalize this particular type of sum of square roots, similar to the way a determinant is developed for quadratic equations.
We can write the generic form of the expression in the first place as follows.
$\sqrt{a+b\sqrt{s}}+\sqrt{c+d\sqrt{s}}=\sqrt{x+y\sqrt{s}}$
Note that
*
*$a, b, c, d, s$ ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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School Outing Combinatorial Design Problem So this is an actual organization problem I am dealing with right now as a high school teacher. There is a school outing, with $8$ groups of students. At the venue there are $7$ stations, where $2$ groups can compete against each other. Is there a way to organize the groupings... | In terms of graph theory, you have a complete graph $K_{2n}$ on $2n$ nodes (one node per group) and want to decompose it into $2n-1$ perfect matchings (one matching per time interval). See Perfect 1-factorization of $K_{2n}$
But you also want its "transpose" to form such a decomposition. For $n=4$, here is a solution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding the laurent expansion of $\frac{z}{(z-1)(z-2)}$ I want to find the Laurent series for $\frac{z}{(z-1)(z-2)}$ in the region $1 < |z| < 2$. This implies that $\frac{1}{|z|} < 1$, so noticing that $(z-1) = z(1 - \frac 1 z)$ I can rewrite the desired function as
$$\frac{z}{z(1- \frac 1 z)(z-2)} = \frac{1}{z-2} \cd... | For $1 < |z| < 2$,
\begin{align*}
\frac{1}{(z-1)(z-2)} &=-\frac{1}{z-1} + \frac{1}{z-2}\\
&=-\frac{1}{z}\left[\frac{1}{1-(1/z)}\right]-\frac{1}{2}\left[\frac{1}{1-(z/2)}\right]\\
&=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{z^n}-\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\\
\end{align*}
\begin{align*}
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $x_1x_2+x_1x_3+x_2x_4+x_3x_4$ without explicitly finding the roots of $x^4-2x^3-3x^2+4x-1=0$
The equation $x^4-2x^3-3x^2+4x-1=0$ has $4$ distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1\lt x_2\lt x_3\lt x_4$ and product of $2$ roots is unity, then find the value of $x_1x_2+x_1x_3+x_2x_4+x_3x_4$
This quest... | I offer to substitute $-3x^2$ as $\left(-4x^2+x^2\right)$, therefore, we will have
$$
x^4-2x^3-3x^2+4x-1=0,
\\
\left(x^4-2x^3+x^2\right)-\left(4x^2-4x+1\right)=0,
\\
x^2\left(x^2-2x+1\right)-\left(2x-1\right)^{2}=0,
\\
x^{2}\left(x-1\right)^{2}-\left(2x-1\right)^{2}=0,
\\
\left(x\left(x-1\right)\right)^{2}-\left(2x-1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Sum binomial coefficients $$\sum_{k=0}^{n} {\frac{k^2+k}{3^{k+2}} {n \choose k}}=?$$
What I've tried:
$$(k^2+k){n \choose k}=k(k+1){\frac{n!}{k!(n-k)!}}$$
$$k^2+k = k^2-k+2k=k(k-1)+2k$$ =>
$$
\begin{align}
(k^2+k){n \choose k} &= k(k-1){\frac{n!}{k!(n-k)!}}+2k{\frac{n!}{k!(n-k)!}}\\&={\frac{n!}{(k-2)!(n-k)!}}+2{\frac{n... | Starting where you stopped:
$$\begin{align}\sum_{k=0}^n(k^2+k)\binom nkx^k&=n(n-1)x^2\sum_{k=2}^n\binom{n-2}{k-2}x^{k-2}+2nx\sum_{k=1}^n\binom{n-1}{k-1}x^{k-1}\\
&=n(n-1)x^2(1+x)^{n-2}+2nx(1+x)^{n-1}\\
&=nx(1+x)^{n-2}((n-1)x+2(1+x))\\
&=nx(1+x)^{n-2}((n+1)x+2)\end{align}$$
hence
$$\begin{align}\sum_{k=0}^n\frac{k^2+k}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[... | It is correct so far. Note that\begin{align}-10+5i&=5\sqrt5\left(-\frac2{\sqrt5}+\frac i{\sqrt 5}\right)\\&=5\sqrt5\exp\left(\left(\pi-\arctan\left(\frac12\right)\right)i\right).\end{align}Therefore, if $\alpha=\pi-\arctan\left(\frac12\right)$, then $z^3=-10+5i$ if and only if $z$ is one of the numbers$$\sqrt5\exp\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$?
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ?
The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$
How to proceed?
| Another way to realize the solution (through inspection):
\begin{align*}
20x^{2} + 13x - 15 & = (20x^{2} - 12x) + (25x - 15)\\\\
& = 4x(5x - 3) + 5(5x - 3)\\\\
& = (4x + 5)(5x - 3)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$
My attempt:
$$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$
$$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1... | Less painful:
$$\begin{align}
\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}&=
\frac{x-2}{x-2-\frac{x(x-2)}{x(x-2)-x+1}}\\
&=\frac{(x-2)\bigl(x(x-2)-x+1\bigr)}{(x-2)\bigl(x(x-2)-x+1\bigr)-x(x-2)}.
\end{align}
$$
Now cancel by $x-2$ and multiply out to get
$$\frac{x^2-3x+1}{x^2-4x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \i... | Thanks to @David G. Stork who gave an alternative method as below:
$$
\begin{aligned}
I&= \int_0^{2 \pi} \frac{d x}{1+a \cos x} d x\\& =\int_0^{2 \pi} \frac{1-a \cos x}{1-a^2 \cos ^2 x} d x\\&=\int_0^{2 \pi} \frac{d x}{1-a^2 \cos ^2 x}-a \int_0^{2 \pi} \frac{\cos x}{1-a^2 \cos ^2 x} d x \\
& =4 \int_0^{\frac{\pi}{2}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Sum of two subspaces: representing it with equations I found the following excercise:
Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\}$.
Give a base, a dimension and an eq... | Note that $(1,0,−1,−1,0,1)\in W_1$. Therefore, if$$W_3=\operatorname{span}\bigl\{(1,−1,1,−1,1,−1),(1,0,2,1,0,0),(2,1,0,0,0,0)\bigr\},$$then $W_1+W_2=W_1+W_3$.
Now, let us see what $W_1\cap W_3$ is. Asserting that$$\overbrace{\alpha(1,−1,1,−1,1,−1)+\beta(1,0,2,1,0,0)+\gamma(2,1,0,0,0,0)}^{\text{arbitrary element of }W_3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $|z+2|-|z-2|=2$ and $\dfrac{z-a}{z+a}=ik$, where $k$ is a real parameter, $a\in \mathbb R$ has exactly two solution. Then interval of $a$ is?
If $|z+2|-|z-2|=2$ and $\dfrac{z-a}{z+a}=ik$, where $k$ is a real parameter, $a\in \mathbb R$ has exactly two solution. Then interval of $a$ is?
My Approach:
First Curve i... | Well, it is not hard to show that:
$$\frac{\text{z}-\text{a}}{\text{z}+\text{a}}=i\text{k}\space\Longrightarrow\space\begin{cases}
\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)-\text{a}^2=0\\
\\
\text{k}=\frac{2\text{a}\Im\left(\text{z}\right)}{\Im^2\left(\text{z}\right)+\left(\Re\left(\text{z}\right)+\text{a}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there other method to evaluate $\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x, \textrm{ where }n\in N?$ Letting $x\mapsto \frac{1}{x}$ transforms the integral into
$\displaystyle I=\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x=-\int_0^1 \frac{x^{2 n-1} \ln x}{x^n+1} d x \tag*{} $
Splittin... | Making it more general $$I=\int \frac{\log( x)}{x^{m}\left(1+x^n\right)}\, d x$$
$$x=t^{\frac{1}{n}} \quad \implies \quad I=\frac 1{n^2}\int \frac {\log(t)}{t^a\,(1+t)}\,dt \qquad \text{with} \qquad a=\frac{m+n-1}{n}$$ The antiderivative express in terms of two hypergeometric functions.
$$J=\int_1^\infty \frac{\log( x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4609287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $3^{3x-1}+27^x=2^{2x+1}+7\cdot4^x$ Solve the equation $$3^{3x-1}+27^x=2^{2x+1}+7\cdot4^x$$
We can rewrite the equation as $$\dfrac{3^{3x}}{3}+3^{3x}=2\cdot2^{2x}+7\cdot2^{2x}\\4\cdot3^{3x}=27\cdot2^{2x}\\\dfrac{3^{3x}}{2^{2x}}=\dfrac{27}{4}$$ I don't know how to approach it from here, as the LHS isn'... | $3^{3x-1} + 27^{x} = 2^{2x+1} + 7\cdot4^{x}
\\ 3^{3x}\left(\frac{1}{3} + 1\right) = 2^{2x}(2+7)
\\ 3^{3x}\left(\frac{4}{3}\right) = 2^{2x}\cdot9
\\ \frac{3^{3x}}{2^{2x}} = \frac{27}{4}
\\ \left(\frac{27}{4}\right)^{x} = \frac{27}{4} \\ \textsf{clearly it's x=1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of $\sqrt{\cos x+3}+\sqrt{2\sin x+7}$ without derivative How do we find the minimum of
$$f(x)=\sqrt{\cos x+3}+\sqrt{2\sin x+7}$$
without using derivatives?
This problem is probably related to circles of Apollonius.
I have tried AM-GM and Cauchy-Schwarz inequality but I can't work it out.
Anyway, I have... | Here's a rough idea for a solution (since there are some details I skim over). I'm pretty sure the algebra can be simplified using some AM-GM variation, but I couldn't make it work. Although I'm hopeful that someone else can give a more optimal solution.
Note that for all $t \in \mathbb{R}$
\begin{align*}
&0 \le(t+2)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}$
I'm struggling with the integral, $$\int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx.$$
I tried it as follows:
Substituting $x-2 = \frac1t \implies dx = \frac{-dt}{t^2}.$
$$\therefore \int\frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} = \int \frac{- dt}{\frac{t^2}{t^4} \sqrt{(\f... | I'm going to continue from the last known position above. In the end, you'll get a polynomial in secants, which just results in using standard tricks from calculus II (along with integration by parts).
Let's start with the standard secant substitution $t + \frac{5}{18} = \sqrt{\frac{7}{18}} \sec y,$ so $dt = \sqrt{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$ Solve the equation $$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$$
We have $$D_x:\begin{cases}1-2x>0\\6x^2-5x+1>0\\1-3x>0\\1-3x\ne1\\4x^2-4x+1>0\iff(2x-1)^2>0\iff x\ne\dfrac12\end{cases}\iff x\in(-\infty;0)\cup(0;\dfrac{1}{3})$$
Also the quadratic... | You're almost there, but you may use a simplified method such as this:
$$\log_{1-2x}(2x-1)(3x-1) - 2\log_{1-3x}(1-2x)=2$$
$$\log_{1-2x}(1-2x)(1-3x) - 2\log_{1-3x}(1-2x)=2$$
$$\log_{1-2x}(1-2x)+\log_{1-2x}(1-3x) - 2\log_{1-3x}(1-2x)=2$$
$$\log_{1-2x}(1-3x) - 2\log_{1-3x}(1-2x)=1$$
$$\frac{1}{\log_{1-3x}(1-2x)} - 2\log_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Rearranging cylinder equation Given this equation $(x-y)^2+(y-z)^2+(z-x)^2=r^2$, which when plotted for some radius $r$ results in a "rotated" cylinder:
Plotted equation with $r=1$
Is it possible to rearrange this equation to a form where to rotation is clearer? Perhaps to something around the lines of $(\overrightarro... | We have
$ \begin{bmatrix} x - y \\ y - z \\ z - x \end{bmatrix} = \begin{bmatrix} 1 && - 1 && 0 \\ 0 && 1 && - 1 \\ -1 && 0 && 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$
Therefore,
$(x - y)^2 + (y - z)^2 + (z - x)^2 = \begin{bmatrix} x && y && z \end{bmatrix} \begin{bmatrix} 1 && 0 && -1 \\ -1 && 1 && ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Convergence of series $\sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^4}}$ In the study of the following series
$$
\sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^p}}
$$
it is not hard to prove that it diverges for $p \leq 3$, since the sequence itself does not converge to 0. You can also conclude that the series conv... | We do not need to apply the explicit formula for $1^2+2^2+\ldots +n^2.$ Instead observe that
$$n^3-(n-1)^3=3n^2-3n+1\ge 3n^2$$
Hence $$1^2+2^2+\ldots +n^2\ge {1\over 3}\left ([1^3-0^3]+[2^3-1^3]+\ldots +[n^3-(n-1)^3]\right ]={1\over 3}n^3$$
Hence the $n$th term of the series for $p=4$ is greater or equal $\displaystyle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How fast does $a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}$ decay as $n \to \infty$? Let
\begin{equation}
a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}
\end{equation}
for $n \in \mathbb{N}$.
Then, $a_n$ is clearly a monotone decreasing sequence of positive numbers... | We can also try to find the asymptotics of the integral directly.
Making the substitution $x=\frac{t}{\sqrt2}$
$$a_n=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}=2^n\int_1^\sqrt2\frac{dt}{(1+t^2)^{n+1/2}}=2^n\int_1^\sqrt2\frac{e^{-n\ln(1+t^2)}}{\sqrt{1+t^2}}dt$$
Making one more substitution $t=1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
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