Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Is $\sin^2x$ uniformly continuous on$x\in [0,\infty]$ I have the question that is $sin^2x$ uniformly continuous on $x \in [0,\infty]$ ?
My approach:
Let $\left|x-y\right|<\delta$ we have:-
$$\left|sin^2x-sin^2y\right|=\left|(\sin x+\sin y)(sin x-sin y)\right|$$
$$=4\left|\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-... | Carrying your idea further, we have for $|x-y| < \delta(\epsilon) = \epsilon/2$
$$|\sin^2 x - \sin^2 y|= |\sin x + \sin y||\sin x - \sin y| \\ \leqslant 2 |\sin x - \sin y|\\ = 4\left|\sin\left(\frac{x-y}{2}\right)\right|\left|\cos\left(\frac{x+y}{2}\right)\right| \\ \leqslant 4\left|\sin\left(\frac{x-y}{2}\right)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find the center of circle given two tangent lines (the lines are parallel) and a point. How to find the center of a circle if the circle is passing through $(-1,6)$ and tangent to the lines $x-2y+8=0$ and $2x+y+6=0$?
| Let $(h, k)$ be the center of the circle then the distance of the center $(h, k)$ & $(-1, 6)$ will be equal to the radius ($r$) of circle $$r=\sqrt{(h+1)^2+(k-6)^2}\tag 1$$
Now, the perpendicular distance of the center $(h, k)$ from the line: $x-2y+8=0$
$$r=\left|\frac{h-2k+8}{\sqrt{1^2+(-2)^2}}\right|=\left|\frac{h-2k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Given complex $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$ : when and what is $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
If $z_{1},z_{2},z_{3}$ are three complex number Such that $|z_{1}| = 2\;\;,|z_{2}| = 3\;\;, |z_{3}| = 4\;\;$
Then $\max$ of $|z_{1}-z_{2}|^2+|z_{2}-z_{3}|^2+|z_{3}-z_{1}|^2$
... | The maximum is $87$.
First we note that by a simple compactness argument, the maximum must be attained for some choice of $z_1$, $z_2$, $z_3$.
Assume first that $z_1$ and $z_2$ are fixed and we let $z_3$ vary. Let $y_3$ be the midpoint of the segment between $z_1$ and $z_2$. The loci on which $Q = |z_3 - z_2|^2 + |z_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}... | $$\left|\frac{x}{x+4}\right|<4$$
Squaring, you get
$$\frac{x^2}{16(x+4)^2}<1$$
Rearranging we get
$$15x^2 + 128x+256 >0$$
Using quadratic roots
$$(x+\frac{16}{5}) (x+\frac{16}{3}) >0$$
Hence the solution is $$x \in (-\infty,-\frac{16}{3})\cup(-\frac{16}{5},\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Explicit Laplace approximation for tail of gaussian distribution I'm studying some lecture notes by S. R. Srinivasa Varadhan about Large Deviations Theory and I have some trouble understanding a simple equation right on page 2 where it says
$$\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp{\left(-\frac{nx^2}{2}\right)}\mathrm... | Regarding the left hand side,
\begin{align*}
\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp\left\{-\frac{nx^2}{2}\right\}\,dx
&=\frac{\sqrt{n}}{2\pi} \int_l^\infty \exp\left\{-\frac{1}{2}\left(\frac{x}{1/\sqrt n}\right)^2\right\}\,dx\\
&=\frac{\sqrt{n}}{2\pi}\cdot\sqrt{2\pi}(1/\sqrt n) \int_l^\infty \frac{1}{\sqrt{2\pi}(1/\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(... | We can prove that if a function $f:\mathbb R\to\mathbb R$ satisfies
$$f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)\tag0\label0$$
for every real numbers $x$ and $y$, then there is a constant real number $k$ such that $f(x)=kx$ for every real number $x$.
First, let $y=x$ in \eqref{0} and get $f(0)=0$. Next, let $x=1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| Another way is by induction:
$$
2^1+1 = 3 = 3 \cdot 1
$$
Then, if $2^k+1 = 3j, j \in \mathbb{N}$, then
\begin{align}
2^{k+2}+1 & = 4\cdot2^k+1 \\
& = 4(2^k+1)-3 \\
& = 4(3j)-3 \qquad \leftarrow \text{uses induction hypothesis} \\
& = 3(4j-1)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 5
} |
Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n... | For $n=2$ and $n=3,4$ we can give infinite families using a Pell equation and elliptic curves, respectively,
$n=2$:
$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3 = (2pq)^4$$
where $p,q$ solve $p^2-d^3q^2=1\tag1$.
$n=3$:
$$(a y)^2 + (ma)^3 + a^4 = a^5$$
and the elliptic curve solvable for an appropriate constant $m$,
$$a^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
What does the homogeneous system of equations represent under certain conditions?
Consider the following linear equations
$ax+by+cz=0,bx+cy+az=0,cx+ay+bz=0$
1) $a+b+c \neq o$ and $a^2+b^2+c^2=ab+bc+ca$
2) $a+b+c \neq o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
3) $a+b+c = o$ and $a^2+b^2+c^2 \neq ab+bc+ca$
4) $a+b+c = o$ and $... | hint
Note that $a^2+b^2+c^2=ab+bc+ca$ is another way of saying $a=b=c$ (just multiply both sides by $2$ and complete the squares).
So if (4) holds, then $a=b=c=0$, in which case the system is actually $0=0$, hence the solution space is all of $\mathbb{R}^3$.
Try to proceed and see if you can complete now. Otherwise I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$
Show that $a_n=\frac{n+1}{2n}a_{n-1}+1$ given that:
$a_n=1/{{n}\choose{0}}+1/{{n}\choose{1}}+...+1/{{n}\choose{n}}$
The hint says to consider when $n$ is even and odd. When $n=2k$ I get:
$$a_{n}=1/{{2k}\choose{0}}+1/{{2k}\choose{1}}+...+1/{{2k}\choose{2k}}$$
$$=1+1/{{2k}\choose... | Let $$f(x) =\sum_{n=1}^{\infty} a_n x^n =\sum_{n=1}^{\infty} \frac{n+1}{2n} a_{n-1} x^n +\sum_{n=1}^{\infty} x^n =1+\sum_{n=1}^{\infty}\frac{n+2}{2(n+1)} a_n x^{n+1} +\frac{x}{1-x}=\frac{1}{1-x} +\frac{1}{2} \int\sum_{n=1}^{\infty}(n+2) a_n x^n dx =\frac{1}{1-x} +\frac{1}{2}\int \frac{1}{x} \left(\sum_{n=1}^{\infty} a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $|\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$ I need to show that $ \forall x,y \in \mathbb R, |\log(1 + x^2) - \log(1 + y^2)| \le |x-y|$
I tried using concavity of log function: $\log(1 + x^2) - \log(1 + y^2)=\log(\frac{1 + x^2}{1 + y^2})=\log(\frac{x^2y^2}{(1 + y^2)y^2}+\frac{1}{1 + y^2}) \ge \frac{2(\log(x)... | The derivative of $f(x)=\log(1+x^2)$ is
$$
f'(x)=\frac{2x}{1+x^2}
$$
and
$$
\left|\frac{2x}{1+x^2}\right|\le1
$$
because
$$
2|x|\le 1+|x|^2
$$
since
$$
(1-|x|)^2\ge0
$$
Thus, by the mean value theorem,
$$
\frac{\log(1+x^2)-\log(1+y^2)}{x-y}=f'(z)
$$
for $z$ between $x$ and $y$ (assuming $x\ne y$ or the inequality is ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Equation with radicals and reciprocals Find all $x\in\mathbb{R}$ satisfying $$x=\sqrt{x-\frac{1}{x}}+\sqrt{1-\frac{1}{x}}.$$
Multiply both sides by $x^{1/2}$ to get $$x^{3/2} = \sqrt{x^2-1} + \sqrt{x-1}.$$Making the substitution $a = \sqrt{x^2 - 1}$, $b=\sqrt{x-1}$, we have $a+b = x^{3/2}$ and $a^2 - b^2 = x^2 - x$, s... | Here is a slicker solution I found.
Use David's substitution $y=\sqrt{x-\frac{1}{x}}, z=\sqrt{1-\frac{1}{x}}$, and note that $x=y^2-z^2+1$. Then multiply $x=y+z\;(\star)$ by $y-z$ to get $$y-z=\frac{y^2-z^2}{x}=z^2.$$ Summing this with $(\star)$ we get $$x-2y+z^2=0\iff(y-1)^2=0.$$ Thus $y=1$, and $x\geq1$ is the root o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $4 \over {{x^4} + {y^4} + {z^4}}$ from $x + y + z = 1$ and other conditions more? How to calculate $$4 \over {{x^4} + {y^4} + {z^4}}$$
from $$ x + y + z = 1, $$
$$ x^2 + y^2 + z^2 = 9, $$
$$ x^3 + y^3 + z^3 = 1. $$
Alternative answers: A) $1 \over {33}$, B) $2 \over {33}$, C) $4 \over {33}$, D) $16 \o... | \begin{align}
x^2+y^2+z^2&=(x+y+z)^2-2xy-2yz-2zx\\
9&=1-2(xy+yz+zx)\\
xy+yz+zx&=-4\\
(xy+yz+zx)^2&=16\\
x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)&=16\\
x^2y^2+y^2z^2+z^2x^2+2xyz&=16\qquad\text{since }x+y+z=1\\
\frac{(x^2+y^2+z^2)^2-(x^4+y^4+z^4)}{2}+2xyz&=16\tag{1}
\end{align}
On the other hand, from the identity
$$x^3+y^3+z^3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\sin x+\sin^{2} x=1$ , Find $\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
If $\sin x+\sin^{2} x=1$, then the value of
$\cos^{12} x+3\cos^{10} x+3\cos^{8} x+\cos^{6} x+2\cos^{4} x+\cos^{2} x-2 $
is equal to
$a.)\ 0 \\
b.)\ 1 \\
c.)\ 2 \\
\color{green}{d.)\ \sin^{2} x} $
$\boxed{... | Observe that $\sin x = 1-\sin^2 x =\cos^2 x$. Moreover let $\cos^4 x =t$ then the expression given is
\begin{align*}
& = \color{red}{t^3+3t^2\sqrt{t}+3t(\sqrt{t})^2+(\sqrt{t})^3}+2t+\sqrt{t}-2\\
&=\color{red}{(t+\sqrt{t})^3}+2t+\sqrt{t}-2\\
&=(\cos^4x+\cos^2x)^3+2\cos^4x+\cos^2x-2\\
\text{using the fact that $\sin x =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Probability of almost correct order Let's say I'm a teacher handing tests back to seven students.
If I do it with my eyes closed, what's the probability I hand exactly 5 of the tests back to the correct students?
There are many possible ways this could happen, and if my understanding of probability is correct, the su... | There are $$\binom{7}{2}$$ ways to choose the $2$ students who receive the wrong tests. Obviously, the only way for this to happen is if their tests are switched.
There are $7!$ total ways to hand back the tests, so the probability is
$$\frac{\binom{7}{2}}{7!} = \frac{1}{2 \cdot 5!} = \boxed{\frac{1}{240}}$$
A related... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Induction Proof for $F_{2n} = F^2_{n+1} - F^2_{n-1}$ As stated in the tag, I'm trying to prove by induction the claim $F_{2n} = F^2_{n+1} - F^2_{n-1}$, where $F_{n}$ is the $n^{th}$ Fibonacci number. I've spent hours on the inductive step without substantial progress, and am hoping someone can provide a path to the des... | As this is a difficult result to prove, I shall prove another result and then derive the said result through it.
Proposition: $F_{n+m} = F_n F_{m-1} + F_{n+1} F_m$
Base cases: $n = 1$:
$\begin{align}
F_{m+1} &= F_m + F_{m-1} \\
&= F_2 F_m + F_1 F_{m-1}
\end{align}$
Case $n = 2$:
$\begin{align}
F_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do I evaluate this series? How do I evaluate this series:
\begin{equation}
\sum_{n=2}^\infty \frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!} = \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} +\ldots
\end{equation}
I wanted to use the Comparison test to show convergence, but I didn't know what to compare it to s... | Defining$$a_n=\frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!}$$ you can get rid of the numerator if you notice that $$\prod_{k=1}^{n-1} (2k-1)= \frac{2^{n-1} \Gamma \left(n-\frac{1}{2}\right)}{\sqrt{\pi }}$$ which makes $$a_n=\frac{\Gamma \left(n-\frac{1}{2}\right)}{2 \sqrt{\pi } n!}$$ which makes $$\frac{a_{n+1}}{a_n}=1-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the straight line from an equation system. I have this problem:
the matrix below is an equation system:
$$
\begin{matrix}
1 & 1 & -1 & 19 \\
5 & 4 & -6 & 43 \\
7 & -1 & 7 & 80 \\
\end{matrix}
$$
It simplifies to
$$
\begin{matrix}
1 & 0 & 0 & 0 \\
... | The book is wrong, and as one of the commenters said. The solution set is the single point $\begin{bmatrix} 0 \\ 35.5 \\ 16.5 \end{bmatrix}$.
The general rule when row reducing $[A |\mathbf{b}]\rightarrow [RREF(A)|\mathbf{P}]$ where $\mathbf{P}\in\mathbb{R}^n$.
*
*$0$ free variables implies the solution set is only ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$? I have problems to solve this limit
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$
I tried with Taylor:
$$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac... | Another idea:
$$
\lim_{x\to\infty}\left(x\sqrt{x^2-x}-x^2\cos\left(\frac1{\sqrt{x}}\right)\right) =
\lim_{x\to\infty}\frac{
\left(x\sqrt{x^2-x}-x^2\cos\left(\frac1{\sqrt{x}}\right)\right)\left(x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)\right)}
{x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} =
$$
$$
\lim_{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Real Analysis : Value of $U(f,P)-L(f,P)$ in Darboux Integral? Given $f:[0,4]\rightarrow\mathbb{R}$ with the definition :
$f(x)=\begin{cases}
2x+3, & 0 \leq x<1 \\
3, & x=1 \\
-x+1, & 1<x\leq 3 \\
2, & 3<x\leq 4
\end{cases}$
If we're given the partition $P=\{0,1-h,1+h,3-h,3+h,4\}\subset[0,4]$... | If $\mathcal P = \{x_0,\ldots,x_n\}$ is a partition then the upper and lower sums are
\begin{align}
U_f(\mathcal P) &= \sum_{j=0}^{n-1}\sup_{x\ \in\ [x_j,x_{j+1}]}f(x)(x_{j+1}-x_j)\\
L_f(\mathcal P) &= \sum_{j=0}^{n-1}\inf_{x\ \in\ [x_j,x_{j+1}]}f(x)(x_{j+1}-x_j).
\end{align}
So we compute
\begin{align}
U_f(\mathcal P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Use the partial fraction to evaluate $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$. $\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$
My try：
$\int \frac{\left(3y^2+3y+2\right)}{\left(y^2-1\right)\left(y+1\right)}dy$ ＝
$\int \frac{\left(3y^2+3y+2\right)}{\left(y-1\r... | Try this format $$\frac{3y^2+3y+2}{\left(y-1\right)\left(y+1\right)^2}=\dfrac{A}{y-1}+\dfrac{B}{y+1}+\dfrac{C}{(y+1)^2}.$$
Actually this happens because of:
$$\begin{align}
\frac{3y^2+3y+2}{\left(y-1\right)\left(y+1\right)^2}
& =\dfrac{A}{y-1}+\dfrac{By+C}{(y+1)^2}\\
& =\dfrac{A}{y-1}+\dfrac{B(y+1)+(C-B)}{(y+1)^2}\\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating the integral $\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}}$ I successfully evaluated the following integral using partial fraction expansion, but am unsure of a few steps.
$$
\int{\frac{x^2-2x-1}{(x-1)^2(x^2+1)}} = \int\left( \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{Cx+D}{x^2+1} \right) \\\ \\
\\
x^2-2x-1 = A(x-1)... | Notice:
*
*Your way of using partial fractions is right, and it works in general.
*Use this to find the partial fractions.
*The integral of $\frac{1}{x}$ is equal to $\ln|x|+\text{C}$.
*For $i$ notice that $i^2=-1$
*When $a,b\in\mathbb{R}$:
$$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|\left(\cos(\arg(a+bi))+\sin(\arg(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how many $7$ digit numbers can be formed using $1,2,3,4,5,6,7,8,9,0$ How many seven digit numbers can be made if
$(a)$ they must be odd and repetition is not allowed
$(b)$ they must be even and repetition is not allowed
$0532129$ is not a seven digit number
So the question is asking, how many 7 digit numbers can be ma... | You have 10 numerals. $0,1,2,3,4,5,6,7,8,9$ and want to find the number of orderings of 7 of them so that $0$ is not first, and either $S_O=\{1,3,5,7,9\}$ or $S_E=\{0,2,4,6,8\}$ comes last.
Say we want 0 to come last. We can choose any $6$ of the 9 remaining digits and form a permutation of them, then stick 0 at the en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluation of all positive integer ordered pair $(n,r)$ for which $\binom{n}{r} = 2016$
$(1)$ Evaluation of all positive integer ordered pair $(n,r)$ for which $\displaystyle \binom{n}{r} = 120$
$(2)$ Evaluation of all positive integer ordered pair $(n,r)$ for which $\displaystyle \binom{n}{r} = 2016$
$\bf{My\; Try::... | I wrote a simple program
to search all
$\binom{n}{k}$
for
$n < 2016$.
The only solutions were the ones you found.
For a more sophisticated search,
I could use the factorization
$2016=2^53^27$,
to limit the possible values
of $n$ and $k$,
but not now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $3x(1-x^2)y^2\frac{dy}{dx}+(2x^2-1)y^3=ax^3$ I am solving this linear Differential equation which can be easily solve by using the formulas for the Bernoulli's Equations
I have solved till
$$\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)y^2}=\frac{ax^3}{3x(1-x^2)y^2}$$
$$y^2\frac{dy}{dx}+\frac{(2x^2-1)y^3}{3x(1-x^2)... | Notice, multiply the integration factor both the sides of the D.E. as follows $$\frac 13\frac{dt}{dx}\frac{1}{x\sqrt{1-x^2}}+\frac{(2x^2-1)t}{3x(1-x^2)}\frac{1}{x\sqrt{1-x^2}}=\frac{ax^3}{3x(1-x^2)}\frac{1}{x\sqrt{1-x^2}}$$
$$\frac{d}{dt}\left(\frac{t}{x\sqrt{1-x^2}}\right)=\frac{ax}{(1-x^2)^{3/2}}$$
$$\int d \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve $\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$ by algebraic methods I was trying to solve this equation without using calculus.
Is it possible to be solved by elementary algebraic methods?
$$\sqrt[3]{7x+19}+\sqrt[3]{7x-19}=\sqrt[3]{2}$$
| powering by $3$ we get
$$7x+19+3\sqrt[3]{7x+19}^2\sqrt[3]{7x-19}+3\sqrt{7x+19}(\sqrt[3]{7x-19})^2+7x-19=2$$
and we get
$$14x+3\sqrt[3]{7x+19}\sqrt[3]{7x-19}(\sqrt[3]{7x+19}+\sqrt[3]{7x-19})=2$$
$$14x+3\sqrt[3]{7x+19}\sqrt[3]{7x-19}\sqrt[3]{2}=2$$
can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
derivative of arctan(u) Im trying to find the derivative of $\arctan(x-\sqrt{x^2+1})$ here are my steps if someone could point out where I went wrong.
$$\begin{align}
\frac{\mathrm d~\arctan(u)}{\mathrm d~x} \;& =\; {1\over{1+u^2}}\cdot \frac{\mathrm d~u}{\mathrm d~x}
\\[1ex]
& =\; {1-{x\over{\sqrt{x^2+1}}}\over{1+(x-\... | It seems about right.
$$\begin{align}
\frac{\mathrm d~\arctan(u)}{\mathrm d~x} \;& =\; {1\over{1+u^2}}\cdot \frac{\mathrm d~u}{\mathrm d~x}
\\[1ex]
& =\; {1-{x\over{\sqrt{x^2+1}}}\over{1+(x-\sqrt{x^2+1})^2}}
\\[1ex]
& =\; \frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}\;(1+x^2-2x\sqrt{x^2+1}+x^2+1)}
\\[1ex]
& =\; \frac{\sqrt{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$\sin2(x) - \tan(x) = 0$ , solve for $-180\le x\le 180$ I have been unable to solve the following question,
If $$\sin(2x) - \tan(x) = 0$$
Find $x$ , $-\pi\le x\le \pi$
So far my workings have been
Use following identity:
$$\sin(2x) = 2\sin(x)\cos(x)\\2\sin(x)\cos(x) - \tan(x) = 0\\2\sin(x)\cos(x) - \frac{\sin(x)}{\co... | Once you got to
$$2\sin(x)\cos(x) - \frac{\sin(x)}{\cos(x)} = 0$$
you can pull out a factor of $\sin (x)$ to get
$$\sin(x)\left[2\cos(x)- \frac{1}{\cos(x)}\right]$$
Now, either $\sin(x) = 0$ or $2\cos(x)- \frac{1}{\cos(x)} = 0$. For $\sin(x) = 0$, we have $-180$, $0$, and $180$. For $2\cos(x)- \frac{1}{\cos(x)} = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Solve a linear system of equation involving some recursion $$
\begin{align*}
x_{1} &= 1 + x_{2}\\
x_{2} &= 1 + \frac{1}{2} x_{3} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{i} &= 1 + \frac{1}{2} x_{i+1} + \frac{1}{2} x_{1}\\
&\vdots\\
x_{n-2} &= 1 + \frac{1}{2} x_{n-1} + \frac{1}{2} x_{1}\\
x_{n-1} &= 1 + \frac{1}{2} x_{n} + \f... | I claim that $x_{n-k}=\left(1-\frac{1}{2^k}\right)\left(2+x_1\right)$. To see this, compute the first few terms of the "backwards" sequence $x_{n},x_{n-1},x_{n-2},...$
\begin{align}
x_n&=0 \\
x_{n-1}&=1+\frac{1}{2}\left(1+\frac{1}{2}x_1\right)+\frac{1}{2}x_1 \\
&=\left(1+\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + ... | How to start a proof by induction?
What a straight line!
You start at the beginning.
...
Then you show that the middle flows, and conclude that the end is inevitable.
But seriously... You start with $n = 2$.
Prove that $(1 - 1/4) = (2+1)/2*2$.
That's the initial or base step.
Then you do the induction step. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i... | You can for exemple complete the square $$x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2-2x^2 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1).$$
Now you can use the traditionnal method for each factor.
An other possibility is the used the polar form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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why does matlab give me a negative number? I have the following problem
A steel company has four different types of scrap metal (called Typ-1 to Typ-4) with
the following compositions per unit of volume
They need to determine the volumes to blend from each type of scrap metal so that
the resulting mixture has the fo... | Goal
The matrix $\mathbf{A}$ is a map between the $m=4$ material types and $n=4$ metals.
The fraction of the $m=4$ types (I-IV) is $f$. The relative abundance or parts of the $n=4$ elements is $p$.
Blend $m=4$ type to get a material with composition $a$.
Data
$$
%
\begin{align}
%
\mathbf{A} f &= p \\
%
\left(
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show $p(x)$ is a primitive polynomial First the definition:
Polynomial $q(x) \in \mathbb{Z}_p[x]$ of degree $n$ is called primitive, iff:
*
*$q(x) \mid x^{p^n-1}-1$
*$\forall k : 1 \leq k \leq p^{n}-1$ : $q(x) \nmid x^k - 1$
Now the polynomial from my exam, where I should show that it is primitive:
$p... | First we show that $p$ is irreducible. If it were not, it would have a root in $F_2$, $F_4$ or $F_8$.
If $a$ were a root in $F_4$, we'd have $a^3 = 1$, hence $0 = a^6 + a^5 + a^2 + a + 1 = a$, which is impossible.
If $a$ were a root in $F_8$, we'd have $a^7 = 1$, hence $0 = a^8 + a^7 + a^4 + a^3 + a^2 = a^4 + a^3 + a^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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integrate $\int\frac{\sin x}{1+\sin^{2}x}dx$
$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}$$
$$\int\frac{\sin x}{1+\sin^{2}x}\mathrm {dx}=\int\frac{\sin x}{2-\cos^{2}x}\mathrm {dx}$$
$u=\cos x$
$du=-\sin x dx$
$$-\int\frac{\mathrm {du}}{2-u^{2}}$$
How could I continue?
| Notice, factorize & use partial fractions as follows $$-\int \frac{du}{2-u^2}=-\int \frac{du}{(\sqrt2-u)(\sqrt2+u)}$$
$$=-\frac{1}{2\sqrt 2}\int \left(\frac{1}{\sqrt 2-u}+\frac{1}{\sqrt 2+u}\right)\ du$$
$$=-\frac{1}{2\sqrt 2} \left(-\ln|\sqrt 2-u|+\ln|\sqrt2+u|\right)+C$$
$$=-\frac{1}{2\sqrt 2}\ln\left|\frac{\sqrt 2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Probability of an even number of sixes
We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.
I tried doing this problem with induction, but I have problem with induction so I was wondering if my... | Your method is correct but the induction hypothesis should be
$$\text{Probabaility of an even number of sixes when you roll } k \text{ dice} = \frac{1}{2}\left[1+\left(\frac{2}{3}\right)^k\right]$$
(you have $n$ instead of $k$) and the thing you want to prove is
$$\text{Probabaility of an even number of sixes when you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $ \frac{4cos^{2}(2x)-4cos^2(x)+3sin^2(x)}{4cos^2(\frac{5\pi}{2} - x) -sin^22(x-\pi)} = \frac{8cos(2x)+1}{2(cos(2x)-1)}$
Question: Prove
$$ \frac{4cos^{2}(2x)-4cos^2(x)+3sin^2(x)}{4cos^2(\frac{5\pi}{2} - x) -sin^22(x-\pi)} = \frac{8cos(2x)+1}{2(cos(2x)-1)}$$
My attempt starting with the bottom line on the LHS... | In the numerator,
$$ 4\cos^2(2x) - 7 \cos^2(x) + 3 $$
As suggested in the comments, use,
$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$
Thus
$$ 4\cos^2(2x) - 7 \cos^2(x) + 3= 4\cos^2(2x)-\frac72-\frac72\cos(2x)+3=\frac12(8\cos^2(2x)-7\cos(2x)-1)=\frac12(8\cos(2x)+1)(\cos(2x)-1)$$
Dividing this with the expression for the denominat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the equation of tangent at origin to the curve $y^2=x^2(1+x+x^2)$ How do I find the equation of tangent at $(0,0)$ to the curve $y^2=x^2(1+x+x^2)$ ?
Differentiating and putting the value of $x$ and $y$ gives an indeterminate form.
Can we trace the curve and geometrically make tangents and find their equation ?
| $$y^2=x^2(1+x+x^2)\implies 2yy'=2x(1+x+x^2)+x^2(1+2x)=4x^3+3x^2+2x\implies$$
$$y'(0)=\lim_{x\to 0}\frac{4x^3+3x^2+2x}{2y}=\pm\lim_{x\to0}\frac{4x^2+3x+2}{2\sqrt{1+x+x^2}}=\pm1$$
and thus the tangent doesn't exist since $\;y=|x|\sqrt{1+x+x^2}\;$ and $\;\frac x{|x|}=\pm1\;$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this indefinite integral using integral substitution? So while working on some physics problem for differential equations, I landed at this weird integral $$ \int \frac 1 {\sqrt{1-\left(\frac 2x\right)}}\,dx $$
So since there is a square root, I thought I could use trig substitution, but I couldn't find a... | Some hints
First of all, simple, thing, just rewrite the integrand this way
$$\frac{1}{\sqrt{\frac{x-2}{x}}}$$
Then substitute
$$y = \frac{x-2}{x} ~~~~~~~ \text{d}y = \frac{1}{x} - \frac{x-2}{x^2}\ \text{d}x$$
You're now with
$$2\int \frac{1}{(1 + y^2)\sqrt{y}}\ \text{d}y$$
And for this one, use
$$z = \sqrt{y} ~~~~~~~ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Coordinates of a matrix So on the textbook, it gives an example:
If the basis of B matrix is{$\begin{bmatrix}1&0\\ 0&0\end{bmatrix}, \begin{bmatrix}0&1\\ 0&0\end{bmatrix},\begin{bmatrix}0&0\\ 1&1\end{bmatrix}, \begin{bmatrix}1&0\\ 1&0\end{bmatrix}$}
Then the B-coordinates of a matrix $\begin{bmatrix}a&b\\ c&d\end{bma... | The coordinates of a vector with respect to an ordered basis are unique so there is only one correct answer. In your case,
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} = (a + d - c) \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + d \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real Analysis proof, show there is a limit and find it. Assume$${ x }_{ 1 }=\sqrt { 3 } ,{ x }_{ 2 }=\sqrt { 3+\sqrt { 3 } } ,..,{ x }_{ n }=\sqrt { 3+\sqrt { 3+..+\sqrt { 3 } } } $$this sequence has a limit and find the limit.
I know I need to show that is is bounded and then show $x_{ n }<{ x }_{ n+1 }$ before fin... | We have the recurrence relationship
$$x_{n+1}=\sqrt{3+x_n}$$
with $x_1=\sqrt{3}$.
We propose that $\lim_{n\to \infty}x_n=\frac{1+\sqrt{13}}{2}$.
First note that $x_n>0$.
Second, we observe that if $x_{n}<\frac{1+\sqrt{13}}{2}$, then $x_{n+1}<\frac{1+\sqrt{13}}{2}$ also.
Therefore, for $x_1=\sqrt3<\frac{1+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$ Find $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
Let $I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx$
I took $\sqrt{2-x-x^2}$ as first function and $\frac{1}{x^2}$ as the second function and integrated it by parts,
$I=\int\frac{\sqrt{2-x-x^2}}{x^2}dx=\sqrt{2-x-x^2}\int\frac{1}{x^2}dx-\int\frac{-1-2x... | Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int \sqrt{2-x-x^2}\cdot \frac{1}{x^2}dx\;, $$ Now Using Integration by parts
$$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\int\frac{1+2x}{2\sqrt{2-x-x^2}}\cdot \frac{1}{x}dx $$
So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\underbrace{\int\frac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The sum of the least common multiple and greatest common divisors of the product Find number of solutions in positive integers the equation
$$[a^2,b^2]+[b^2,c^2]+[c^2,a^2]=(a^2,b^2)(b^2,c^2)(c^2,a^2),$$
where $[m,n] -$ least common multiple, $(m,n) -$ greatest common divisors of positive integers $m$ and $n$.
My work s... | The first step isn't possible: $p$ must be the gcd, and you cannot then require that $x$, $y$, and $z$ be coprime, e.g., if $a=6$, $b=10$, $c=15$.
If you write $a$ as $dABC$, where $d$ is $gcd(a,b,c)$, $A$ is $gcd(a/d,b/d)$, and $B$ is $gcd(a/d,c/d)$, and $b$ and $c$ in similar fashion, you should be able to cancel thi... | {
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"url": "https://math.stackexchange.com/questions/1659971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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integration of $\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$
$$\int \frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}$$
$$\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}*\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{{1+x}+2\sqrt{({1+x}... | Since $x=sin(t)$ then $cos(t)= \sqrt{1-x^2}$
To deal with the $tan(t/2)$ term use the half angle formulae to put this in terms of $cos(t)$ and $sin(t)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$
$$\int\frac{\sin(2x)}{2+\cos x} dx $$
Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 ... | Nothing's wrong ! Put $u=\cos x +2$
$$-2 (u - 2\ln u) + C=4\ln u -2u +\mathrm C =4 \ln(2+ \cos x) - 2\cos x - 4 + \mathrm C.$$
So $$-4+\textrm C = \rm constant $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Relationships between points lines and planes Develop the Cartesian equation of a plane with $x$-intercept $a$, $y$-intercept $b$ and $z$-intercept $c$.
Show that the distance $d$ from the origin to this plane is given by $$\frac{1}{d^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$
In the picture below I have included... | Another attempt, since the OP finds my first answer confusing ...
Let $A = (a,0,0)$, $B = (0,b,0)$, $C = (0,0,c)$. The normal to the plane is in the direction $N = (B-A)\times(C-A) = (bc, ca, ab)$. A unit vector in this direction is
$$
U = \frac{N}{\|N\|} = \frac{(bc,ca,ab)}{\sqrt{b^2c^2 + c^2a^2 + a^2b^2}}
$$
The dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$ Does this integral converge to any particular value?
$$\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$$
If the answer is yes, how should I calculate its value?
I tried to use convergence tests but I failed due to the complexity of the integral itself.
| Sure it does.
Collect $e^x$ in the denominator and you will get
$$\int_0^{+\infty}\frac{x}{e^{x}(1 + e^{-x})}\ \text{d}x$$
Since the integral range is from $0$ to infinity, you can see the fraction in this way:
$$\int_0^{+\infty}x e^{-x}\frac{1}{1 + e^{-x}}\ \text{d}x$$
and you can make use of the geometric series for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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Convergence of Infinite Sum Using Mathematica, I have been able to make the following statement based on numerical evidence:
$$\sum_{i=0}^\infty \frac{2^i}{x^i}=\frac{x}{x-2}$$
for any $x≥3$. How can this be proven?
| This can solved from scratch by taking one term out of the summation, changing the index range and refactoring to obtain
$$
\begin{align}
S=
\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^i
&=
1+\sum_{i=1}^{\infty} \left(\frac{2}{x}\right)^i \\
&=
1+\sum_{i=0}^{\infty} \left(\frac{2}{x}\right)^{i+1} \\
&=
1+\frac{2}{x}\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the smallest $a\in\mathbb{N}$, so that $a\equiv3\cdot 5^4\cdot 11\cdot 13^3 \pmod 7$ The problem I am trying to solve is:
Find the smallest $a\in\mathbb{N}$, so that $a\equiv3\cdot 5^4\cdot 11\cdot 13^3 \pmod 7$.
I noticed that $3,5,11$ and $13$ are primes, but I've no idea how that is supposed to help me.
The s... | Note $13^{3} \equiv (-1)^{3} \equiv -1 \pmod 7$
Also, note that $5^{4} \equiv 25^{2} \equiv 4^{2} \equiv 2 \pmod 7$.
Thus $3 \times 5^{4} \times 11 \times 13^{3} \equiv 3 \times 2 \times (-3) \times (-1) \equiv 18 \equiv 4 \pmod 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\righ... | Clearly $$(x + 1)(x + 2)(x + 3)(x + 4)(x + 5) = x^{5} + 15x^{4} + \cdots$$ and as we will show later we don't need to bother about the coefficients of $x^{3}, x^{2}, x, x^{0}$ in order to solve this problem.
Let $P(x)$ be a monic polynomial of degree $n$ and let the coefficient of $x^{n - 1}$ in $P(x)$ be $a$ so that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 5
} |
L'Hopital's Rule Values for an equation I have the following task:
For what values of $a$ and $b$ is the following equation true?
$$\lim\limits_{x \to 0} \left(\frac{\sin(2x)}{x^3} + a + \frac{b}{x^2}\right) = 0 $$
I want to know the steps I should follow in order to find the solution.
| We have
\begin{align}
L
&= \lim_{x \to 0} \frac{\sin(2x)}{x^3} + a + \frac{b}{x^2} \\
&= \lim_{x \to 0} \frac{\sin(2x) + a x^3 + b x }{x^3}\\
\end{align}
which is a limit of type $0 / 0$, so we try L'Hôpital's rule:
\begin{align}
L &= \lim_{x \to 0} \frac{2\cos(2x) + 3 a x^2 + b}{3x^2}
\end{align}
The denominator aga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Confused about infinite sum $\sum\limits_{a,b,c}\frac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}$ $$ \displaystyle \sum^{\infty}_{a=0} \displaystyle \sum^{\infty}_{b=0} \displaystyle \sum^{\infty}_{c=0}\dfrac{a+b+c+abc} {2^a(2^{a+b}+2^{b+c}+2^{a+c})}= \ ? \ $$
I calculated its value as $\frac{32}{3}$ but I'm not sure w... | Seems OK.
You've used
$$\frac{1}{xy+yz+zx} \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)=
\frac{1}{xyz}$$
and
$$\sum_{a=0}^{\infty} \frac{a}{2^{a}}=
\lim_{n\to \infty} \sum_{k=0}^{n} \frac{k}{2^{k}}=
\lim_{n\to \infty} \left(2-\frac{n}{2^{n}}-\frac{1}{2^{n-1}} \right)=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is this expression strictly positive? $n \left(\cos\frac{k\pi}{n}\right)\left(1-\cos\frac{k\pi}{n}\right)-\sin\frac{k\pi}{n}$ Let us define a function $f(k,n)$ by
\begin{equation}
f(k,n)=n \left (\cos\frac{k\pi}{n}\right) \left(1-\cos\frac{k\pi}{n}\right) - \sin \frac{k\pi}{n}
\end{equation}
where $\frac{k}{n}$ is irr... | Proof: Since $\frac{k}{n}$ is irreducible and $k\le \lfloor \frac{n}{2}\rfloor$, we have $k \le \frac{n-1}{2}$.
Let $u = \tan \frac{k\pi}{2n}$. Since $2\le k \le \frac{n-1}{2}$ and $n\ge 5$, we have
$$0 < \frac{\pi}{n} \le \tan \frac{\pi}{n} \le u \le \tan \Big(\frac{\pi}{4} - \frac{\pi}{4n}\Big)
= \frac{1 - \tan \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?
My Attempt
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0), (-2,0) are solution of the given ... | Any $x$ that satisfies the equation must be even. Let $x=2z$. Substituting and dividing by $2$ we get
$$2z^2+2y^2-2zy-2z-2y-4=0.$$
We can rewrite this as
$$(z-1)^2+(y-1)^2+(z-y)^2=6.$$
The only way the sum of three squares is $6$ is if the squares are, in some order, $1$, $1$, and $2$. Now it is a matter of examining ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
The following series with a polynomial as numerator and $e^n$ as denominator converge? The series is the following:
$$\sum_{n=1}^\infty \frac {8n^2-7}{e^n(n+1)^2} $$
| You can also use the ratio test $$u_n=\frac {8n^2-7}{e^n(n+1)^2}$$ $$\frac{u_{n+1}}{u_n}=\frac{(n+1)^2 \left(8 (n+1)^2-7\right)}{e (n+2)^2 \left(8 n^2-7\right)}=\frac 1e \, \frac{8 n^4+32 n^3+41 n^2+18 n+1}{8 n^4+32 n^3+25 n^2-28 n-28}$$ the limit of which being $\frac 1e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Algebraic identity involving powers of twin primes Yesterday, I verified that, if $a$,$b$ and $c$ are real numbers such that $a+b+c=0$, then
$$\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2}$$
and
$$\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}$$
Then I asked myself: will i... | $\newcommand{\weird}[4]{\dfrac{{#1}^{#4}+{#2}^{#4}+{#3}^{#4}}{#4}}$
I'd say your suggestion is false, because
$$\weird {(-y-z)}yz{13}-\weird {(-y-z)}yz{11}\cdot\weird {(-y-z)}yz2=\\=-y^3z^3(y+z)^3(y^2+yz+z^2)^2\ne0$$
It's always worth checking the first hard case with a calculator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
arithmetic progession question if $\sqrt{a-x}, \sqrt x, \sqrt{a+x}$ are in AP provided $a>x$ and $a,x$ are positive integers then what is the least possible value of $x$?
| $$\sqrt{a-x},\sqrt{x},\sqrt{a+x}$$ are in an AP. By the definition of an arithmetic progression,
$$\sqrt{a+x}-\sqrt{x}=\sqrt{x}-\sqrt{a-x}$$
Thus
$$2\sqrt{x}=\sqrt{a+x}+\sqrt{a-x}$$
Squaring both sides,
$$4x=a+x+a-x+2\sqrt{a^2-x^2}$$
$$2x-a=\sqrt{a^2-x^2}$$
Squaring both sides,
$$4x^2+a^2-4ax=a^2-x^2$$
$$5x^2-4ax=0$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-... | First establish a formula for the square root:
$$(a+ib)^2=a^2-b^2+i2ab=x+iy$$ is equivalent to the system
$$a^2-b^2=x,\\2ab=y.$$
If we notice that $$(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2,$$ we have
$$a^2-b^2=x,\\
a^2+b^2=\pm\sqrt{x^2+y^2}.$$
The real solutions are
$$a=\pm\sqrt{\frac{\sqrt{x^2+y^2}+x}2},\\b=\pm\sqrt{\frac{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to solve the $\min_x \int_c^d (u-x)^4 e^{-u^2/2} du$ How to solve the following equation
\begin{align}
\min_x \int_c^d (u-x)^4 e^{-u^2/2} du
\end{align}
Observer, that by taking the derivative we get
\begin{align}
\int_c^d -4(u-x)^3 e^{-u^2/2} du
\end{align}
So, if $c=-d$ then the solution is given by
\begin... | Just as Svetoslav answered, you first need to expand $$(u-x)^4=u^4-4 u^3 x+6 u^2 x^2-4 u x^3+x^4$$ and your are first left with the antiderivatives $$I_k=\int u^k e^{-\frac{u^2}{2}} \,du \qquad (k=0,1,2,3,4)$$ The calculations do not present major difficulties (integrations by parts) $$I_0=\sqrt{\frac{\pi }{2}} \text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Methods for Integrating $\int \frac{\cos(x)}{\sin^2(x) +\sin(x)}dx$ So I've found that there's the Weierstrass Substitution that can be used on this problem but I just want to check I can use a normal substitution method to solve the equation:
$$\int \frac{\cos(x)}{\sin^2(x) +\sin (x)}dx$$
Let $u = \sin(x)$
$du = \cos(... | Manipulating the Weierstrass result, we begin with
$$\ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right| +C$$
With $\tan\left(\frac x2\right)={\sin x\over 1+\cos x}$, we then get
$$\ln \left|{\sin x\over 1+\cos x}\right|-2\ln \left|{\sin x\over 1+\cos x}+1\right| +C\\
=\ln \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrating $\int^b_a(x-a)^3(b-x)^4 \,dx$ I came across a question today...
The value of $\int^b_a(x-a)^3(b-x)^4 \,dx$ is
First I tried the property $\int^b_af(x)=\int^b_af(a+b-x)$. I got $\int^b_a(x-a)^4(b-x)^3 \,dx$, which can be simplified to: $\dfrac{b-a}{2}\int^b_a(x-a)^3(b-x)^3 \,dx$. Well now what? Do I have t... | How about the following way? (though I'm not sure if you like it)
$$\begin{align}\\&\int_{a}^{b}(\color{red}{x-a})^3(b-x)^4dx\\&=\int_{a}^{b}(\color{red}{x-b+b-a})^3(x-b)^4dx\\&=\int_a^b\left((x-b)^3+3(b-a)(x-b)^2+3(b-a)^2(x-b)+(b-a)^3\right)(x-b)^4dx\\&=\int_a^b\left((x-b)^7+3(b-a)(x-b)^6+3(b-a)^2(x-b)^5+(b-a)^3(x-b)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt... | Reference that $$(a + b)^n = {n \choose 0}a^nb^0 + {n \choose 1}a^{n-1}b^1 + \cdots + {n \choose n-1}ab^{n-1} + {n \choose n}a^0b^n.$$
So we want $a = 2$ and $b = {x \over 3}$. So we are considering the terms $\displaystyle {n \choose 7}a^{n - 7}b^7$ and $\displaystyle {n\choose 8}a^{n-8}b^8.$ So, $${n \choose 7}a^{n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Solve fo a step by step $$\frac{81^{a}+9^{a}+1}{9^{a}+3^{a}+1}=\frac{7}{9} \Rightarrow a = ? $$
| Hint:
\begin{align}
\frac79&=\frac{81^a+9^a+1}{9^a+3^a+1}=\\
&=\frac{{(3^4)}^a+{(3^2)}^a+1}{{(3^2)}^a+3^a+1}=\\
&=\frac{{(3^a)}^4+{(3^a)}^2+1}{{(3^a)}^2+3^a+1};
\end{align}
Let $u=3^a$:
\begin{align}
\frac79&=\frac{u^4+u^2+1}{u^2+u+1}=\\
&=\frac{u^4+2u^2-2u^2+u^2+1}{u^2+u+1}=\\
&=\frac{{(u^2)}^2+2u^2+1+u^2-2u^2}{u^2+u+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How can I simplify this rational expression? I am trying to simplify $$ \frac{3(5^{k+1} - 1) + 4(3 \;\cdot\;5^{k+1})}{4}$$ to get to $$\frac{3(5^{k+2}-1)}{4} $$
I start with $$ \frac{3(5^{k+1} - 1) + 12\;\cdot\;5^{k+1}}{4}$$
$$= \frac{3\;\cdot\;5^{k+1} - 3 + 12\;\cdot\;5^{k+1}}{4} $$
then I am stuck.
| $$3(5^{k+1} - 1) + 4(3 \cdot 5^{k+1}) = \\ 3 \cdot 5^{k+1} - 3 + 12 \cdot 5^{k+1} = \\ 15 \cdot 5^{k+1} - 3 = \\ 3 \cdot 5 \cdot 5^{k+1} - 3 = \\ 3 (5 \cdot 5^{k+1} - 1) = \\ 3 (5^{k+2} - 1).$$
In going from the first line to the second line, I distribute the $3$ through on the first term in the sum: $3(5^{k+1} - 1) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Exponential of a non terminating matric So I understand how to calculate the exponential of matrices that eventually terminate; however, how to approach the cases in which the matrix does not seem to truncate? For example with the matrix $M=$$\quad
\begin{pmatrix}
2 & 1 \\
0 & 2
\end{pmatrix}$
I have calculated the f... | The basic observation that can be used to compute the exponent explicitly is that if $X,Y$ are matrices that commute ($XY = YX$) then $\exp(X + Y) = \exp(X) \exp(Y)$.
Your matrix $M$ can be written as $M = D + N$ where
$$ D = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \,\,\, N= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove by induction - can not prove the step case I have to prove by induction the following:
$$\sum^{n-1}_{k=2} k\log_{10} k \le \frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2$$
I have proven the base case, but I am stuck at the step case. I have tried different ways but always have stuck somewhere and could not finish it.
A... | What you want to show is that
$\sum^{n-1}_{k=2} k\log_{10} k
\le \frac{1}{2}n^2\log_{10} n-\frac{1}{8}n^2
$
implies
$\sum^{n}_{k=2} k\log_{10} k
\le \frac{1}{2}(n+1)^2\log_{10} (n+1)-\frac{1}{8}(n+1)^2
$.
From your assumption,
$\begin{array}\\
\sum^{n}_{k=2} k\log_{10} k
&=\sum^{n-1}_{k=2} k\log_{10} k+n\log_{10} n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An algorithm for creating a circle on a discrete plane and a limit for $\pi$ I know there is a well known algorithm which uses the circle equation to approximate it with pixels.
However, I wanted to approach this problem from the most basic principles.
So we start with a square with the side $a_0$ (all lengths are meas... | You state that
$\pi=2+4 \lim_{p \to \infty} \frac{1}{p} \sum_{n=1}^k \sqrt{1-\frac{2n}{p}-\frac{n^2}{p^2}}
$
This is a standard Riemann sum.
$\lim_{p \to \infty} \frac{1}{p} \sum_{n=1}^k \sqrt{1-\frac{2n}{p}-\frac{n^2}{p^2}}
\to \int_0^{\sqrt{2}-1} \sqrt{1-2x-x^2} dx
$
Since
$ \int \sqrt{1-2 x-x^2} dx
= \frac12 \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that sum of squares of all the values of $(\sqrt{3}+i)^{3/7}$ is $0$. Show that sum of squares of all the values of $(\sqrt{3}+i)^{3/7}$ is $0$.
Attempt:
Let $z=(\sqrt{3}+i)$ then simplifying I get
$z^{3/7}=2^{3/7}\Big\{\cos{\frac{13k+1}{2\times 7}\pi}+i\sin{{\frac{13k+1}{2\times 7}\pi}}\Big\}$, $k=0,1,2\cdots 6$ ... | Hint: Values of $(\sqrt3+i)^{3/7}$ are roots of $x^7-(\sqrt3+i)^3=0$. Now think Vieta and use $\sum x_i^2 = (\sum x_i)^2-2\sum x_i x_j$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
On equations $m^2+1=5^n$ I am looking for integer solutions of Diophantine equation $m^2+1=5^n$. I found that $m=0,n=0$ and $m=2,n=1$. I could not find any other solutions. I try to prove this but I could not. Could anyone help me to solve this equation?
| HINT.-If you are interested in another way of indicated in the link given by Yiyuan Lee (a good example of that elementary could be intricate enough and not easy), you could pay attention to the following:
$m^2+1=5^n\Rightarrow n$ is odd ($m^2+1=x^2$ is impossible since the difference between $x^2$ and the square of an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}... | Let $\left( 1+\dfrac1{\sin x}\right) \left( 1+\dfrac1{\cos x}\right)=1+y$
$\iff\sec x+\csc x=y-\sec x\csc x$
Squaring both sides, $$\sec^2x+\csc^2x+2\sec x\csc x=y^2-2y\sec x\csc x+\sec^2x\csc^2x$$
But $\sec^2x+\csc^2x=\dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}=\sec^2x\csc^2x$
$$\implies y^2-(2y-1)\sec x\csc x=0\iff0=\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
show this inequality with $a+b+c+d=1$ Let $a,b,c,d\ge 0$,and such $a+b+c+d=1$, show that
$$3(a^2+b^2+c^2+d^2)+64abcd\ge 1$$
use AM-GM
$$a^2+b^2+c^2+d^2\ge 4\sqrt{abcd}$$
it suffices to
$$4\sqrt{abcd}+64abcd\ge 1$$
| Here is a way without Schur. Let $p = a+b, u = ab$ and $q = c+d, v = cd$. Then we have $p+q = 1$ and wish to show $$F = 3(p^2+q^2-2u-2v)+64uv \ge 1$$
For any fixed $p, q$, the variables $u, v$ can range continuously in $u \in [0, p^2/4]$ and $v \in [0, q^2/4]$. Since $F$, viewed as a function of one variable (either... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | Set $\sin(2x) = u$
$7\cos(2x) - \sin(2x) + 5 = 0$
$\implies 7\sqrt{1-u^2} = u-5$
$\implies 49 - 49u^2 = u^2 - 10u + 25$
$\implies 50u^2 - 10u -24 = 0$
$\implies u_1 = 0.8, u_2 = -0.6$
$ \sin(2x) = 0.8$ or $\sin(2x) = -0.6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 5
} |
If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$. If $\csc \frac{\pi}{32}+\csc \frac{\pi}{16}+\csc \frac{\pi}{8}+\csc \frac{\pi}{4}+\csc \frac{\pi}{2}=\cot \frac{\pi}{k}$.Find $k$
Let $\frac{\pi}{32}=\theta$
$\csc \frac{\pi}{32}+\csc \frac{\pi}{16... | Hint: try to work out some smaller cases first, for example, $\csc\frac{\pi}{2}=\cot\frac{\pi}{k}$ or $\csc\frac{\pi}{4}+\csc\frac{\pi}{2}=\cot\frac{\pi}{k}$. Then try to generalize using induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit similar to $\lim_{n \to \infty} \left(1-\frac{1}{n} \right)^n = \text{e}^{-1}$ I want to show that
$$
\lim_{n \to \infty} \left(1-\frac{n}{n^2} \right) \left(1-\frac{n}{n^2-1} \right) \cdot \ldots \cdot \left(1-\frac{n}{n^2-n+1} \right)
= \lim_{n \to \infty} \prod_{k=0}^{n-1}
\left(1-\frac{n}{n^2 - k} \right)
=... | May be another way to see the limit and how it is approached.
Let $$u_i=1-\frac 1{n^2-i}\implies P_n=\prod_{i=0}^{n-1} u_i$$ Take logarithms $$\log(P_n)=\sum_{i=0}^{n-1} \log(u_i)$$ Now, use Taylor series for large values of $n$ $$\log(u_i)=-\frac{1}{n}-\frac{1}{2
n^2}-\frac{i+\frac{1}{3}}{n^3}-\frac{i+\frac{1}{4}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
If $a=b^c$, $b=c^a$, and $c=a^b$, prove that If $$a=b^c, \quad b=c^a, \quad \text{and} \quad c=a^b$$ prove that $$abc=1.$$
My Attempt;
Given,
$$a=b^c$$
$$b=c^a$$
$$c=a^b$$
Now,
$$
\begin{align}
\text{L.H.S.} & =abc \\
& = b^c\cdot b\cdot a^b \\
& =b^c\cdot b\cdot b^{bc} \\
& =b^{c+1+bc} \\
& \,\,\,\vdots
\end{align}
$$... | In $a=b^c$, substitute $b=c^a$, and you get $a=(c^a)^c=c^{ac}$. Next substitute $c=a^b$ and get $a=(a^b)^{ac}=a^{abc}$. So $a^{1}=a^{abc}$. Either $abc=1$ or $a=1$.
If $a=1$ then $c=a^b=1$ and $b=c^a=1$, so $abc=1$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solving $\int{\frac{2x \ dx}{\sqrt{9x^2-6x-8}}}$ I solved the problem but my solution seems too simple to be the right one.
$$\int{\frac{2x \ dx}{\sqrt{9x^2-6x-8}}}$$
$$\frac{1}{3}\int{\frac{2x \ dx}{\sqrt{x^2-2x-\frac{8}{9}}}}$$
$u = x^2 - 2x - \frac{8}{9}$
$du = 2x -2 \ dx$
$2x \ dx = du + 2 = 3 du$
$$\frac{1}{3}\int... | Wait, how did you go from $$du = (2x-2) \, dx$$ to $$2x \, dx = 3 \, du?$$ That is not right. You can't separate $(2x-2) \, dx$ into $2x \, dx - 2$ any more than you can write $(a+b)c = ac + b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1709552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| $$2(\sin^6x +\cos^6x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^2x +\cos^2x)(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=2(\sin^4x-\sin^2x \cos^2 x+\cos^4x) - 3(\sin^4x+\cos^4x)+1=$$
$$=-2\sin^2x \cos^2 x -\sin^4x-\cos^4x+1=$$
$$-(\sin^2x +\cos^2x)^2+1=-1+1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Limit of a sequence ratio Let's assume that $a_{n}$ and $b_{n}$ fulfill the equation
$$
a_{n} + b_{n}\sqrt{3} = (2+\sqrt{3})^n, \hspace{0.5cm} n=1,2,\dots
$$
Compute the limit
$$\lim_{n\to \infty}\frac{a_{n}}{b_{n}}.$$
Being honest, I do not know where to start and would appreciate any help. Thanks in advance!
| A start: Although this is not mentioned explicitly in the post, we assume that the $a_n$, $b_n$ are obtained by taking the natural expansion of $(2+\sqrt{3})^n$, and collecting terms that do not involve $\sqrt{3}$ together, and terms that involve $\sqrt{3}$ together. So for example if $n=2$, then $a_n=7$ and $b_n=4$.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to list all possible dimension of $\ker{T},\ker{T^2},...,\ker{T^{k-1}}$ and the corresponding canonical forms? Let $V$ be $5$-dimension vector space, and $T:\ V\rightarrow V$ a nilpotent linear transformation of order (index) $k$ where $1\le k\le 5$. How to list all possible dimension of $\ker{T},\ker{T^2},...,\ker... | You really should specify what you mean by "canonical form". I will assume that you meant rational canonical form since you did not specify that you were dealing with a complex vector space. In this particular problem, because of the nilpotency, these are also the Jordan forms, but that is just luck, the distinction of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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The Diophantine Equation: $x^3-3=k(x-3)$ I wish to know how to resolve the diophantine equation: $x^3-3=k(x-3)$ ? The problem is:
Find all integers $x\ne3$ such that $x-3\mid x^3-3$.
- From 250 Problem's in Elementary Number Theory, by W. Sierpinski
I have found the solutions to this problem by using rules in divisib... | Set $x-3=y\iff x=?$
$$\dfrac{x^3-3}{x-3}=\dfrac{(y+3)^3-3}y=y^2+9y^2+27y+\dfrac{3^3-3}y$$
So, $y$ must divide $3^3-3=24$ to keep $k$ an integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Question Regarding Integration Within Summation
It expresses an integral within a summation procedure.
| Compute everything from the inside out (one step at a time), don't look at the sum until you've considered the integral.
Observe that
$$
\int_{\frac{1}{k+b}}^{\frac{1}{k+a}}\frac{dx}{1+x}=\ln(1+x)|_{(k+b)^{-1}}^{(k+a)^{-1}}=\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right).
$$
By the properties of the log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to factorize a 4th degree polynomial?
I need help to factorise the following polynomial:
$x^4 - 2x^3 + 8x^2 - 14x + 7$
The solution I need to reach is $(x-1)(x^3 - x^2 + 7x - 7)$. I need to factorize to this exactly as it is for a limit question where I cancel out the $(x-1)$ in the numerator and denominator. How... | One may recognize the factorization of the polynomial(#): $$y^2+8y+7=(y+1)(y+7)$$
Also(^)
$$-2x^3-14x=-2x(x^2+7)$$
From (#), when $y=x^2$, we obtain(*)
$$x^4+8x^2+7=(x^2+1)(x^2+7)$$
Combine(*) and (^), we have
\begin{align}
x^4-2x^3+8x^2-14x+7&=(x^2+1)(x^2+7)-2x(x^2+7)\\&=(x^2-2x+1)(x^2+7)\\&=(x-1)^2(x^2+7)
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | You can use the odd-factorial
$$\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)} = \frac{(n!)^2 2^n}{(2n)!}
$$
Taking logarithm,
$$ \log\frac{(n!)^2 2^n}{(2n)!} = 2\log n! + n \log 2 - \log (2n!)
$$
so...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Maximum value of $f(x) = \cos x \left( \sin x + \sqrt {\sin^2x +\sin^2a}\right)$ Can we find maximum value of $$f(x) = \cos x \left( \sin x + \sqrt
{\sin^2x +\sin^2a}\right)$$
where '$a$' is a given constant.
Using derivatives makes calculation too complicated.
| Let $$y=\cos x\left[\sin x+\sqrt{\sin^2 x+\sin^2 a}\right] = \sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
We get $$(\sin^2 x+\cos ^2 x)\cdot \left[\cos^2 x+\sin^2 x+\sin^2 a\right]\geq \left(\sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}\right)^2$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Finding the matrix of projection Find the matrix A of the orthogonal projection onto the line L in R2 that consists of all scalar multiples of the vector $\begin{pmatrix} 2 \\ 3 \ \end{pmatrix}$. How do I begin to solve this? Any help would be appreciated.
| Use the orthonormal basis for ${\sf R}^2$
$$\gamma=\{u_1,u_2\}=\left\{\frac{1}{\sqrt{13}}\begin{pmatrix}2\\3\end{pmatrix},\frac{1}{\sqrt{13}}\begin{pmatrix}3\\-2\end{pmatrix}\right\},$$
we see that
${\sf T}(u_1)=u_1$, ${\sf T}(u_2)=0$, and $[{\sf T}]_\gamma=\begin{pmatrix}1&0\\0&0\end{pmatrix}$. Let $\beta$ be the stan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ x+iy = \sqrt{\frac{a+ib}{c+id} } ,$Show that$ (x^2+y^2)^2 = \frac {a^2+b^2}{c^2+d^2} $ $ x+iy = \sqrt{\frac{a+ib}{c+id} } , $Show that $ ({x^2+y^2})^2 = \frac {a^2+b^2}{c^2+d^2} $
How do i do this ?I tried squaring both sides but x+iy expansion becomes difficult when squaring the next time .I also tried conjugatin... |
Notice, when $q\space\wedge\space s\space\wedge\space z\in\mathbb{C}$, so we set $q=x+yi,s=a+bi,z=c+di$:
*
*$$\left(x^2+y^2\right)^2=\left(\Re^2[q]+\Im^2[q]\right)^2=|q|^4$$
*$$a^2+b^2=\Re^2[s]+\Im^2[s]=|s|^2$$
*$$c^2+d^2=\Re^2[z]+\Im^2[z]=|z|^2$$
So:
$$x+yi=\sqrt{\frac{a+bi}{c+di}}\Longleftrightarrow q=\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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3D Surface Area Integral find SA of the cone $$z=2\sqrt{(x^2+y^2)}$$ bounded by $$y=x$$ and $$y=x^2$$ in the first quadrant.
This is my integral setup for the surface area of that portion of the cone, what did I do wrong?
$\int_0^1 \int_0^{x^2} \sqrt{\frac{\ 2y+2x}{\sqrt{x^2+y^2}}+1} \, dydx $
Thanks for the help
| Mostly you forgot to square components before adding them under the square root. Along the surface, $\vec r=\langle x,y,2\sqrt{x^2+y^2}\rangle$, so
$$d\vec r=\langle1,0,\frac{2x}{\sqrt{x^2+y^2}}\rangle dx+\langle0,1,\frac{2y}{\sqrt{x^2+y^2}}\rangle dy$$
So
$$\begin{align}d^2\vec A & =\langle1,0,\frac{2x}{\sqrt{x^2+y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the volume of ice cream cone using cylindrical/spherical coordinates I'm stuck on what the boundaries are for the volume bounded by the cone $z=-\sqrt{(x^2+y^2)}$ and the surface $z=-\sqrt{(9-x^2-y^2)}$ $\,\,$-essentially an upside down ice cream cone
Remember that $r^2=x^2+y^2$
I assumed for cylindrical coordina... | The surfaces intersect when
$$-r=-\sqrt{9-r^2}$$
which is $r^2=\frac92$, not $r^2=9$ as you have in your cylindrical attempt.
For spherical coordinates $\phi$ should be from $\frac34\pi$ to $\pi$. I believe the rest is correct, as long as you are writing $r^2=x^2+y^2+z^2$ in the spherical case. (It is common to call ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
Show that $x^2 + y^2 + z^2 = x^3 + y^3 + z^3$ has infinitely many integer solutions.
I am not able to find an idea on how to proceed with the above questions. I have found only the obvious solution $(1,1,1)$.
Could you please provide... |
Let $z=-x$. Then
$$2x^2+y^2=y^3$$
$$2x^2=(y-1)y^2$$
If $\frac{y-1}2=k^2 - $ perfect square, then
$$y=2k^2+1, x=k(2k^2+1)$$
Answer: $$x=k(2k^2+1), y=2k^2+1, z=-k(2k^2+1)$$
Second method:
Let $y=1+a, z=1-a$. Then
$$x^3+2(1+3a^2)=2(1+a^2)+x^2$$
$$x^2-x^3=4a^2.$$
Let $1-x=4p^2$, then
$$x^2(1-x)=(4p^2-1)^24p^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1732049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Roots of quadratic equation are given by $b \pm \sqrt{b^2 - c}$ I was reading slides about the cancellation error in quadratic equations and it's written:
The roots of the quadratic equation:
$$x^2 - 2bx + c = 0$$
with $b^2 > c$ are given by $b \pm \sqrt{b^2 - c}$.
Fact that let me perplexed, since I always thought t... | If we write the quadratic like this $$ax^2+bx+c=0,$$ then the quadratic formula is right as you have written it; the roots are: $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But the quadratic $$x^2-2bx+c=0$$ has replaced $b$ with $-2b$ and $a$ with $1$. This makes the quadratic $$\frac{2b\pm\sqrt{(-2b)^2-4(1)c}}{2(1)}=\frac{2b\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is In a triangle $ABC,$if $(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{8a^2b^2c^2}{a^2+b^2+c^2}$ then the triangle is
$(A)$isosceles
$(B)$right angled
$(C)$equilateral
$(D)$ obtuse angled
$(a+b+c)(a+b-c)(a-b+c)(b+c-a)=\frac{... | You have
$\sin^2A+\sin^2B+\sin^2C=2
$.
What stands out to me
is that this is true
if the triangle is right:
$\sin C = 1$
and
$\sin^2A+\sin^2B
=\sin^2A+\sin^2(\pi/2-A)
=\sin^2A+\cos^2(A)
=1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Why does $\frac{1}{4}x^{-3/4} = \frac{1}{4x^{3/4}} = \frac{1}{4\sqrt[4]{x^3}}$? This is taken from Khan Academy, I don't understand how these equate:
$$\frac{1}{4}x^{-3/4}
= \frac{1}{4x^{3/4}}
= \frac{1}{4\sqrt[4]{x^3}}$$
How come the minus was remove from the original exponent?
| These are properties of exponentiation. In particular,
$$a^{-b} = \frac{1}{a^b}$$
combined with
$$a^{\frac{m}{n}} = \sqrt[\leftroot{-2}\uproot{2}n]{a^m}.$$
In your case,
$$\frac{1}{4}x^\frac{-3}{4} = \frac{1}{4x^\frac{3}{4}} = \frac{1}{4\sqrt[\leftroot{-2}\uproot{2}4]{x^3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1736981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-... | $
=2[4+6+12+14+20+22+\cdots+(8n-4)+(8n-2)]
\\ = 4[(2+3)+(6+7)+(10+11)+\cdots+(8n-3)]
\\ = 4[5+13+21+\cdots+(8n-3)]
\\ = 4 \sum_{k=1}^n (8k-3)
\\ = 4 (4n^2+n)
\\ = 16n^2+4n
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
How to solve this question in more time efficient way?
Q) if$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}$$ then find $4z^2(x^2+y^2)$a)$(x^2+y^2)^{3}$b)$(x^2-y^2)^3$c)$(x^2-y^2)^2$d)$(x^2+y^2)^2$
Ans:c
i solved this in a very long way:
$$x\sin a=y\cos a=\frac{2z\tan a}{1-\tan^2 a}=z\tan 2a$$$$\implies x= \frac{z\tan ... | How about this?
Since $\tan a=y/x$, we have
$$y\cos a=\frac{2z(y/x)}{1-(y/x)^2}\quad\Rightarrow\quad \cos a=\frac{2xz}{x^2-y^2}$$
So, from $1+\tan^2a=1/(\cos^2a)$, we have
$$1+\left(\frac yx\right)^2=\left(\frac{x^2-y^2}{2xz}\right)^2,$$
i.e.
$$4z^2(x^2+y^2)=(x^2-y^2)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to prove that my polynomial has distinct roots? I want to prove that the polynomial
$$
f_p(x) = x^{2p+2} - abx^{2p} - 2x^{p+1} +1
$$
has distinct roots. Here $a$, $b$ are positive real numbers and $p>0$ is an odd integer. How can I prove that this polynomial has distinct roots for any arbitrary $a$,$b$ and $p$.
Th... | Let $c$ denote $ab$. Note that
\begin{equation*}
f(x) = (x^{p+1}-1)^2 - cx^{2p}
\end{equation*}
and
\begin{equation*}
f'(x) = 2(p+1)x^p(x^{p+1}-1) - 2pcx^{2p-1}
\end{equation*}
Then, $f(x) = 0 \iff$
\begin{equation*}
c = \dfrac{(x^{p+1}-1)^2}{x^{2p}} = \varphi(x)\ (\text{say})
\end{equation*}
and $f'(x) = 0 \iff$
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Check: Radius of Convergence of the Sum of these Complex Taylor Series I just found the following Taylor series expansions around $z=0$ for the following functions:
*
*$\displaystyle \frac{1}{z^{2}-5z+6} = \frac{1}{(z-2)(z-3)} = \frac{-1}{(z-2)} + \frac{1}{(z-3)} = \sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}} - \sum_{n=... | You can use Hadamard's rule:
$$\frac 1R=\limsup_n a_n^{1/n}=\limsup_n \biggl(\frac1{2^{n+1}}-\frac1{3^{n+1}}\biggr)^{\!1/n}\!.$$
Now rewrite this expression as
$$\frac1{2^{(n+1)/n}}\biggl(1-\frac{2^{n+1}}{3^{n+1}}\biggr)^{\!1/n}=\frac1{2^{1+1/n}}\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{1/n}\!.$$
The first factor tend... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Study the convergence of the following series $\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $ I have to study the convergence of the following series:
$\sum\limits_{n=1}^\infty \frac{n}{n^2+3} \sin(\frac{1}{\sqrt{n+2}}) $
Is a positive series, so I should divide for $\frac{1}{\sqrt{n+2}}$
and th... | By the Taylor series expansion, as $u \to 0$, one has
$$
\sin u=u+o(u^2)
$$ giving, as $n \to \infty$,
$$
\sin\left(\frac{1}{\sqrt{n+2}}\right)\times \frac{n}{(n^2+3)}=\frac{1}{\sqrt{n+2}}\times \frac{n}{(n^2+3)}+O\left( \frac{1}{n+2}\times \frac{n}{(n^2+3)}\right)
$$ or
$$
\sin\left(\frac{1}{\sqrt{n+2}}\right)\times \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proof using deductive reasoning I need to deductively prove that the sum of cubes of $3$ consecutive natural numbers is divisible by $9$. I can prove deductively that they are divisible by $3$ but so far any combination I choose fails to prove the divisibility by $9$. As far as I can see. This is a high school question... | $(x-1)^3+x^3+(x+1)^3=3x^3+6x=3(x^3+2x)=3x(x^2+2)$. Now we just have to prove $3|x$ or $3|x^2+2$.
Case $1: x=3k$, then $3|x$.
Case $2: x=3k+1$, then $x^2+2=9k^2+6k+1+2=3(3k^2+2k+1)$.
Case $3: x=3k+2$, then $x^2+2=9k^2+12+4+2=3(3k^2+4k+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the product of $f(x)$ and $g(x)$ given one of them I'm given the final answer which is $$(g \cdot f)(x) = \frac{1}{x^2+4}\;.$$
Also, i'm given $f(x) = x^2+1$.
I've solved this using the composition, however the second part of the question asks me to find the $g(x)$ which would make this multiplication true. How wo... | So you are told that $(g \times f)(x) = \dfrac{1}{x^{2} + 4}$, and also told that $f(x) = x^{2} + 1$.
$(g \times f)(x)$ is just the name we give for the product of the two functions, i.e., $(g \times f)(x)$ really means $g(x)f(x)$.
So we know what this product is. It is $g(x)f(x) = \dfrac{1}{x^{2} + 4}$. We also know... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove inequality $n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$
Let $m,n\in\mathbb N$, $n>m$. Prove inequality
$$n\sqrt[n]{n!}-m\sqrt[m]{m!}\le\frac{(n−m)(n+m+1)}2.$$
My work so far:
$$\sqrt[n]{n!}=\sqrt[n]{1\cdot2\cdot...\cdot n}\le\frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}2.$$
Then $n\sqrt[n]{n!}\le\... | The result follows immediately by writing
\begin{align}
n\sqrt[n]{n!}
&=n\sqrt[n]{m!\cdot(m+1)(m+2)\cdots n}\\
&=n\sqrt[n]{(\sqrt[m]{m!})^m\cdot(m+1)(m+2)\cdots n}\\
&\le m\sqrt[m]{m!}+(m+1)+(m+2)+\cdots+n\\
&=m\sqrt[m]{m!}+\frac{(n-m)(n+m+1)}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Expand the function $f(z)=\frac{1}{(z-a)(z-b)}$ where $0 < |a| < |b|$ in a Laurent series in different annuli I have to expand the function $f(z) = \frac{1}{(z-a)(z-b)}$ where $a, b \in \mathbb{C}$, $0 < |a| < |b|$ in the following annuli:
(a) $0<|z|<|a|$
(b) $|a|<|z|<|b|$
(c) $|b|<|z|$
I made bonafide attempts at $(b)... | For $|z|<|a|<|b|$, there are no singularities of $f(z)=\frac{1}{(z-a)(z-b)}$. Thus, we write
$$\begin{align}
f(z)&=\frac{1}{a-b}\left(\frac{1}{z-a}-\frac{1}{z-b}\right)\\\\
&=\frac{1}{a-b}\left(\frac{1/b}{1-z/b}-\frac{1/a}{1-z/a}\right)\\\\
&=\frac{1}{a-b}\left(\frac{1}{b}\sum_{n=0}^\infty \left(\frac{z}{b}\right)^n-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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