Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to prove $e^x\ge \left(1+\frac xn\right)^n$ for any real numbers $x, n > 0$ Can someone provide a detailed proof? I saw a proof here
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right... | Note that we have
$$\begin{align}
\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}n\right)^n}&=\left(\frac{1+\frac{x}{n+1}}{1+\frac xn}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(\frac{\frac{n+1+x}{n+1}}{\frac {n+x}n}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(\frac{n(n+1+x)}{(n+1)(n+x)}\right)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Given the list of six positive integers 9, 13, $a$, 12, $b$, and 8, how many ordered pairs $(a,b)$ will result in a mean of 10 for the entire list? Given the list of six positive integers $9$, $13$, $a$, $12$, $b$, and $8$, how many ordered pairs $(a,b)$ will result in a mean of $10$ for the entire list?
I took the sum... | We want $\frac{8+9+12+13+a+b}{6} = 10$. We have $8+9+12+13 = 42$, so we are looking for two positive integers $a$ and $b$ that sum to $60 - 42 = 18$. Because we are looking at ordered pairs, $a = 1$, $b=17$ is a distinct solution from $a = 17$, $b= 1$. There are $17$ possible values of $a$ for which there is a correspo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Sum of a trigonometric series involving sin and tan I'm trying to prove that the following expression is true:
$$\sum_{k=1}^{N-1} \frac{\sin\left(\frac{\pi k n}{N}\right)}{\tan\left(\frac{\pi k}{2N}\right)} = N-n$$
I think that substituting $\tan\left(\frac{\pi k}{2N}\right)=\frac{\sin\left(\frac{\pi k}{N}\right)}{1+\c... | Use $e^{i\theta}=\cos(\theta)+i\sin(\theta)$, $\theta\in\textbf{R}$ to write
$$
\cos\left(\theta\right)=\frac{e^{i\theta}+e^{-i\theta}}{2}\textbf{, }\theta\in\textbf{R}.
$$
Hence
$$
\cos\left(\frac{\pi k}{N}\right)\cos\left(\frac{n\pi k}{N}\right)=
$$
$$
=\frac{1}{4}e^{-ik\pi/N-ikn\pi/N}+\frac{1}{4}e^{ik\pi/N+ikn\pi/N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculating $\int^1_0\frac{x \log x}{1 + x^2} dx$ Trying to reinvent the spirit of Calculus, I was trying to Understand the Leibniz Rule of Differentiation under Integration sign. To check my understanding, I was considering this integral:
$$ I(b) = \int^1_0 \frac{x \log(b + x)}{1 + x^2} dx $$
After some simplification... | This calculation may contains some convergence problem, but it'll be resolved easily.
Using integration by parts, we have
$$
I(0)=\int_{0}^{1}\frac{x\log x}{1+x^{2}}dx = \left[\frac{1}{2}\log(1+x^{2})\log x\right]_{0}^{1} - \int_{0}^{1} \frac{1}{2}\log(1+x^{2})\frac{dx}{x}
$$
Since
$$
\lim_{x\to 0} \log(1+x^{2})\log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many different positive integers can be obtained as a sum of some or all off the numbers: $1,3,5,10,25$? How many different positive integers can be obtained as a sum of some or all of the numbers:
$1,3,5,10,25$?
I have just started discrete mathematics and I am having problems with some tasks. This is one of them.... | A sum is of the form $a*1 + b*3 + c*5 + d*10 + e*25$ where $a,b,c,d,e$ are either $0$ or $1$. There are $2$ choices for each $a,b,c,d,e$ so there are $2^5 = 32$ possible sums. But $0+0+0+0+0 = 0$ is not acceptable so there are only $2^5 - 1 = 31$ possible sums.
But we need to worry about if there are two different wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we calculate $ i\sqrt { i\sqrt { i\sqrt { \cdots } } }$? It might be obvious that $2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { \cdots } } } } } } $ equals $4.$ So what about $i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } } \text{ ?} $ The answer might be $-1$, but I'm not sure as ... | \begin{eqnarray*}
x= a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { a\sqrt { \cdots } } } } } } \\
x=a^{ 1+1/2+1/4+1/8+\cdots} \\
x=a^2
\end{eqnarray*}
So it would seem that
\begin{eqnarray*}
i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { \cdots } } } } } }=\color{red}{-1}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
"answer_count": 7,
"answer_id": 0
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If $A+B+C=180^{\circ}$, then show that: $\cos B=\sin A\sin C-\cos A\cos C$ Here is the question :
If $A+B+C = 180^{\circ}$, then show that: $\cos B = \sin A \sin C - \cos A \cos C$.
EDIT : Here is my reviewed working :
$$
\cos B=-\cos (A+C)
$$
Since $$\space A+B+C = 180^\circ, \space B =180^\circ-(A+C)$$
And
$$\begi... | Your first step $\cos(B) = \cos(A+C)$ is incorrect. It should be $\cos(B) = -\cos(A+C)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing that $\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$
Prove that if $x>-1$ then
$$\frac 1 {(x+1)^2}=\frac 1 {(x+1)(x+2)}+\frac {1!} {(x+1)(x+2)(x+3)} + \frac {2!} {(x+1)(x+2)(x+3)(x+4)} + \cdots$$
Could I have a hint for this? I tried writing... | Hint: This is equivalent to
$$\frac{1}{x} = \frac{0!}{(x+1)}+\frac{1!}{(x+1)(x+2)}+\frac{2!}{(x+1)(x+2)(x+3)}+\cdots.$$
Let $$f_n(x)=\frac{n!}{x(x+1)\cdots(x+n)}.$$
We have that
$$f_n(x)=f_{n-1}(x)-f_{n-1}(x+1),$$
and we wish to prove that
$$\frac{1}{x}=\sum_{n=0}^{\infty} f_n(x+1).$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}$ is convergent by Cauchy Criterion. Prove that the series
$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{\sqrt{n}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\cdots$ is convergent by Cauchy Criterion
Let $\epsilon >0$. We w... | Note that the partial sums alternately overshoot and undershoot the limit.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove for any odd integer $n$, $n^2-1\equiv 1\pmod{8}$ Let, $\exists k \in \mathbb{Z}, n=2k+1\implies n^2-1= 4k(k+1) \pmod{8}$
Now, if $k$ is odd, $k+1$ is even, and vice-versa, so $4k(k+1) \pmod{8}=\forall k\, \exists k' \in \mathbb{Z},\,8k' \pmod{8}$
Unable to proceed further.
| The source of confusion stated the following:
$2\mid n \,\Rightarrow\, 8\mid \color{#c00}{n^3},\,$ else $\,(n,2)=1 \,\Rightarrow\, 8\mid \color{#0a0}{n^2-1}\ $ by $\,{\rm odd}^2\equiv \{\pm1,\pm3\}^2\equiv 1 \pmod{\!8}$
That is they claim that if $n$ is odd, then $8|n^2-1$.
$8|n^2-1$ is equivalent to $$n^2-1 \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Standard Deviation for sums of fair dice given the number of dice, and the number of sides on each die $n$ fair die are rolled, and each dice has $x$ sides, with the numbers on the sides going from $1$ to $x$, and with each side having a different number from the other sides. how do I figure out the standard deviation ... | I am assuming you mean the standard deviation of the sum of $n$ dice, each with $x$ sides.
Let the random variables $X_1,X_2,X_3,..,X_n$ denote the results on the first, second, third$,...,n^{th}$ dice. The $X_i$ are independent. Because of this, we can simply take
$$\begin{align*}
Var(X_1+X_2+X_3+...+X_n)
&=Var(X_1)+V... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Ind... | As an alternative for induction step assume $n^3+5n=6k$ is true and we need to prove that $(n+1)^3+5(n+1)=6h$.
Let observe that
$$(n+1)^3+5(n+1)=n^3+3n^2+3n+1+5n+5=3n^2+3n+6+(n^3+5n)=\\=3n^2+3n+6(k+1)$$
then we need to show that $3n^2+3n$ is divisibleby $6$ wich can be easily verified by plugging $n=2r$ and $n=2s+1$.
| {
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"url": "https://math.stackexchange.com/questions/2687992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find Laurent series about $z_0=0$ for $\frac{1}{z-1} -\frac{1}{z+1}$ in $2<|z|<\infty$ Find Laurent series about $z_0=0$ for $\frac{1}{z-1} -\frac{1}{z+1}$ in $2<|z|<\infty$.
I know how to do this for $|z|>1$ but I don't understand how I should set it up for $|z|>2$.
Here is how I would do it for $|z|>1$.
$\frac{1}{z-1... | HINTS:
$$\frac{1}{z+b}=\frac{1}{z-c+(b+c)}=\left(\frac{1}{b+c}\right)\,\left(\frac{1}{1+\left(\frac{z-c}{b+c}\right)}\right)$$
and
$$\frac{1}{z+b}=\frac{1}{z-c+(b+c)}=\left(\frac{1}{z-c}\right)\,\left(\frac{1}{1+\left(\frac{b+c}{z-c}\right)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving ${l\choose n}+nl-{d+n\choose d}<0$ for $n\geq 6$, $d\geq 3$, $lLet $n\geq 6$ and $d\geq 3$. Prove that if $l<d+n-1$, then
$${l\choose n}+nl-{d+n\choose d}<0.$$
This is the necessary condition in my paper. Experiments show that it is true and $<0$. I tried by expanding but could not find any special for showing ... | Because$$
\binom{l}{n} < \binom{d + n - 1}{n}, \quad nl < n(d + n - 1),
$$
then it suffices to prove that$$
\binom{d + n}{d} \geqslant \binom{d + n - 1}{n} + n(d + n - 1).
$$
Since\begin{align*}
&\mathrel{\phantom{=}}{} \binom{d + n}{d} - \binom{d + n - 1}{n} = \binom{d + n}{d} - \binom{d + n - 1}{d - 1}\\
&= \binom{d ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why do I get that $0 = 1$ when trying to prove that $\forall n\in\mathbb{N}, \ 3 + 2\left(n + \sum_{i=1}^{2n-1}(i+1)\right) = (2n+1)^2\,?$ I was playing around with numbers and by looking at some similar observations, I found an interesting pattern. For a natural number $n$, it seemed like the following was always true... | The formula you have found is just simply a way of rewriting $(2n+1)^2$.
$$3+2\left(n+\sum_{i=1}^{2n-1}I+1\right)=3+2\left(n+2n-1+\frac{(2n-1)(2n)}{2}\right)$$
$$=3+2(3n-1+2n^2-n)=4n^2+4n+1=(2n+1)^2$$
Therefore
$$3+2\left(n+\sum_{i=1}^{2n-1}I+1\right)\equiv (2n+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to use integration by parts to solve an integral? I'm having trouble with solving this integral by parts. Here is what I have so far. Can anyone please help me out?
Solve the integral $\int x^3\sqrt{x^2-1}dx$ by parts, choosing u = $x^2$ and
$dv = x\sqrt{x^2-1} dx$
$du_1 = 2xdx$
$v_1$ = $\int x\sqrt{x^2-1}dx$ $ t =... | You don't have to use integration by parts for $\displaystyle \frac{1}{3}\int 2x (x^2-1)^\frac{3}{2}dx$.
$$\displaystyle \frac{1}{3}\int 2x (x^2-1)^\frac{3}{2}dx=\frac{1}{3}\int(x^2-1)^\frac{3}{2}d(x^2-1)$$
So the integral is
\begin{align*}
\frac{1}{3} (x^2-1)^\frac{3}{2}x^2 - \frac{1}{3}\int(x^2-1)^\frac{3}{2}d(x^2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by Induction: Summation of Factorial (n! * n) Prove by induction (weak or strong) that:
$$(1! \cdot 1) + (2! \cdot 2) + \cdots + (n! \cdot n) =\sum_{k=1}^nk!\cdot k= (n + 1)! - 1$$
My base case is:
$n = 1$, which is true.
And my Inductive Hypothesis is:
$(1! \cdot 1) + (2! \cdot 2) + \cdots + (k! \cdot k) =... | \begin{align*}
(1!\cdot 1) + (2!\cdot 2) + \cdots + (k!\cdot k) + ((k+1)!\cdot(k+1))
&= (k+1)!-1 + ((k+1)!\cdot(k+1)) \\
&= ((k+1)!\cdot(k+2))-1 \\
&= (k+1)!-1.
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$ C++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't ... | When you expand something squared, you multiply each term by each other term, so in this case you have $$\begin{aligned}(n^2+3n+1)^2&=n^2\cdot n^2 + n^2\cdot 3n+n^2\cdot 1\\&\;+3n\cdot n^2+3n\cdot3n+3n\cdot1\\&\;+1\cdot n^2+1\cdot3n+1\cdot1\\&=n^4+3n^3+n^2+3n^3+9n^2+3n+n^2+3n+1\\&=n^4+6n^3+11n^2+6n+1\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$ I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$
$\sum_{i=n}^{2n-1} 3+ \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ am I right?
$4 \sum_{i=0}^{2n-1-n} i+n $
$4... | Note $4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )=4 \left( \frac{n(n-1)}{2}+\color{red}{n^2}\right)$ since from $0$ to $n-1$ there are $n$ numbers.
Another comment is use braces
$4 \sum_{i=0}^{2n-1-n} i+n $ should be $4 \sum_{i=0}^{2n-1-n} (i+n) $
Alternative approach: Note that you are summing an arithmetic progres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Multivariable delta-epsilon limit proofs The Question asks: Find the delta epsilon limit if it exist
$$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$
What I've done so far:
$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$
Now, since $... | In fact
$$\left|\frac{x^2y^3}{2x^2+y^2}\right| \leq \frac{1}{2}|y^3|$$ which is equivalent to
$$2x^2\leq 2x^2+y^2.$$
| {
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"url": "https://math.stackexchange.com/questions/2706833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$ Solve
$$\left ( 1- \sqrt{2}\sin x \right )\left ( \cos 2x+ \sin 2x \right )= \frac{1}{2}$$
Now I did not understand how can i solve that.
I have tried substituting $\cos(2x)=\cos^2(x)−\sin^2(x)$ and$\,$ $\sin(2x)=2\sin(x)\cos(x)... | Multiplying by $1+\sqrt 2 \sin x$ on both sides and using $1-2 \sin^2 x = \cos 2x$ we get
$\cos 2x (\cos 2x + \sin 2x ) = \dfrac{1+\sqrt 2 \sin x}{2}$
or $1+\cos 4x + \sin 4x = 1+\sqrt 2 \sin x$
or $\dfrac{\sin 4 x + \cos 4x}{\sqrt 2} = \sin x$
or $\sin \left(4x+\dfrac{\pi}{4} \right) = \sin x$ which can be easily solv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let
\begin{eqnarray}
x= \sqrt[3]{a} \\
y= \sqrt[3]{b} \\
z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that
\begin{equation}
x^3+y^3+z^3 = (x+y+z)^3... | I try this:
$y=\sqrt[3]{2}$,$y^3=2$, $x= \sqrt[3]{\sqrt[3]{2}-1}$
Therefore, $x^3=y-1$
$y^3-1=(y-1)(y^2+y+1)=1$
Because $y^2+y+1=\frac{3y^2+3y+3}{3}=\frac{y^3+3y^2+3y+1}{3}=\frac{(y+1)^3}{3}$
and $y^3+1=3=(y+1)(y^2-y+1)$
Hence
$x^3=y-1=\frac{1}{y^2+y+1}=\frac{3}{(y+1)^3}=\frac{1}{9}(y^2-y+1)$
Therefore, based on the ... | {
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"url": "https://math.stackexchange.com/questions/2707393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
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Visualizing $3^3+4^3+5^3=6^3$ I was wondering how would the visualization of $3^3+4^3+5^3=6^3$ look? I tried some 3D sketches but when calculating none work. Any ideas would be highly appreciated.
| $$ 5^3 + 3^3 + 4^3 = 5^3 + 3 \cdot (3^2 + 4^2) + 4^2 = 5^3 + 3 \cdot 5^2 + 3 \cdot 5 + 1 = (5 + 1)^3 $$
So a cube of side $6$ units can be made by taking a $\color{green}{\text{cube of side 5 units} }$, adding $ \color{blue}{\text{3 5-by-5 squares}}$ at the top, right and left, then adding $\color{red}{\text{3 lin... | {
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"answer_id": 0
} |
$A289B$ is divisible by $90$, then what is $A+B$? It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?
My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
sin... | You can simply take the sum of digits $A+2+8+9+0 \equiv A + 1 \equiv 0 \pmod 9$, so as to know that $A \equiv 8 \pmod 9$ is the desired solution because $10^k \equiv 1 \pmod 9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Partial Derivative of arctan Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$
Find $f_x(x,y)$
My attempt,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
B... | Hint:
$$1+x^2+y^2+x^2y^2=(1+x^2)+y^2(1+x^2)=?$$
Alternatively, see Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
limit of a recursive sequence_1
I have this sequence
$$ \begin{cases}a_1= \alpha +2 & \alpha >0\\
a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} & n \ge1\end{cases}$$
and I want to know
$\lim_\limits{n\rightarrow \infty} a_n$
I can prove by induction that the sequence is a monotone increasing.
$$
a_1= \a... | This series diverges towards infinity as you noted. In this case, instead of a good old "$\varepsilon - \delta$" proof, we need a $K-\delta$ proof.
Note first that:
$$
a_n >\frac{n+1}{n} \Rightarrow a_{n+1}>2+\frac{1}{n+1}
$$
Because:
$$
a_{n+1} = \left(a_n-\frac{1}{n} \right)^n +\frac{n+2}{n+1}> \left(\frac{n+1}{n}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding an implicit solution of the ODE $y' (x) = \frac{y^2-xy}{2xy^3+xy+x^2}$ I am currently trying to find an implicit solution of the nonlinear ODE:$$y' (x) = \frac{y^2-xy}{2xy^3+xy+x^2}$$
But to be quite honest, I am a little bit lost. I am pretty new to this field and have not learned many methods of solving diffe... | $\dfrac{dy}{dx}=\dfrac{y^2-xy}{2xy^3+xy+x^2}$
$(y^2-xy)\dfrac{dx}{dy}=x^2+(2y^3+y)x$
$(y-x)\dfrac{dx}{dy}=\dfrac{x^2}{y}+(2y^2+1)x$
This belongs to an Abel equation of the second kind.
Let $u=y-x$ ,
Then $x=y-u$
$\dfrac{dx}{dy}=1-\dfrac{du}{dy}$
$\therefore u\left(1-\dfrac{du}{dy}\right)=\dfrac{(y-u)^2}{y}+(2y^2+1)(y-u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Sum of the first integer powers of $n$ up to k Pascal's triangle has a lot of interesting patterns in it; one of which is the triangular numbers and their extensions. Mathematically:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2... | The general result is called Faulhaber's Formula.
A hint in the general direction.
Let $f_i(x)=x(x+1)(x+2)\cdots(x+(i-1)),$ and $f_0(x)=1.$ We call $f_i$ the "rising factorial," and is sometimes written $x^{(i)}.$
Note the property that $f_{i+1}(x)-f_{i+1}(x-1)=(i+1)f_i(x).$
So, we can telescope the sum:
$$\sum_{n=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^... | As pointed out within the other answers we want to prove that
$$\mathfrak{I}=\int_0^{\pi/2}\frac{x^2}{\sin x}~dx=2\pi G-\frac72\zeta(3)$$
As the OP showed $\mathfrak{I}$ can be reduced to a linear combination of $x$ and the function $\log(\cot x)$
$$\mathfrak{I}=\int_0^{\pi/2}\frac{x^2}{\sin x}~dx=8\int_0^{\pi/4}x\lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 3
} |
Simplistic Odd Collatz formulas I've been experimenting with the Collatz for a couple of months.
I just want to give you some formulas and show tables that give the same results for odd inputs. I have also several new functions, but ill only provide a few of them here. I tweaked the formulas. It seems very strange tha... | There is nothing wrong with using an alternate form of the original Collatz ($3x+1$) rule, as long as you have a reason for the modification.
Without using a floor function, it is much easier to show that your method is equivalent to the $\frac{(3n+1)}{2}$ rule. The floor function truncates 0.5 after dividing an odd in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{... | $$2x^2-9x+9\tag1$$
Factorise
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)\tag2$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2+{9\over 2}-\left(9\over 4\right)^2\right]\tag3$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2-{9\over 16}\right]\tag4$$
$$2x^2-9x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Show the equivalence of norms on polynomial space Let $\|ax+b\|^2:=2a^2+3b^2+2ab$ be a norm on the space of all polynomials of degree one, where $a,b\in\mathbb{R}$.
Hence, $\|ax+b\|_{L^2} ^2:=\int_0^1(ax+b)^2\,\mathrm{d}x$ be another norm ($L^2$-norm).
I want to show equivalence, in the sense that, for all polynomials ... | \begin{align*}
\|ax+b\|_{L^2}^2 &= \int_0^1 (ax+b)^2\,\mathrm{d}x = a^2/3+ab+b^2\le a^2 + ab + b^2
\le
2(a^2+ab+b^2) \le 2a^2+2ab+3b^2=\|ax+b\|^2
\end{align*}
Therfore c is atleast $2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Basis for an eigenspace Let $A=\begin{pmatrix}
6&3&-8\\
0&-2&0\\
1&0&-3\\
\end{pmatrix}$. I found that the two distinct eigenvalues are $\lambda=5,-2$. Now, I am asked to find a basis for each eigenspace corresponding to each eigenvalue. I found the basis for the eigenspace corresponding to $\lambda=5$ to be $(8,0,1)$... | From here,
$$\begin{pmatrix}
8&3&-8\\
0&0&0\\
1&0&-1\\
\end{pmatrix}\vec{x}=\vec{0}$$
we see that $\vec{x}=(1,0,1)$ is the correct solution.
Indeed from here
$$\begin{pmatrix}
8&3&-8&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}\stackrel{R_1-8R_3}\to \begin{pmatrix}
0&3&0&0\\
0&0&0&0\\
1&0&-1&0\\
\end{pmatrix}
\implies x_2=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proving expression$\geq 2$, where variables are $x,y,z$
If $x,y,z$ are distinct real number. Then prove that $$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2\geq 2$$
Try : let $\displaystyle \frac{x}{y-z}=p\Rightarrow x=py-pz\Rightarrow x-py+pz=0$
And let $\displaystyle \frac{y}{... | $$\bigg(\frac{x}{y-z}\bigg)^2+\bigg(\frac{y}{z-x}\bigg)^2+\bigg(\frac{z}{x-y}\bigg)^2-2=\frac{\left(\sum\limits_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2}{\prod\limits_{cyc}(x-y)^2}\geq0.$$
Also, we can use the following way.
Since $$\sum_{cyc}\left(\frac{x}{y-z}\cdot\frac{y}{z-x}\right)=\sum_{cyc}\frac{xy(x-y)}{\prod\limits_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Get variable value from Logarithm? I have below equation.
$$ x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2 $$
I am not very familiar with Logarithm.
How can I grab value of $x$ from this equation?
| For $x>0$ write the equation
$$x = \left(\frac{1}{-2 \log_{10}\left(9 + \frac{46}{x}\right)}\right)^2$$
as
$$\dfrac{1}{-2\sqrt{x}}=\log_{10}\left(9 + \frac{46}{x}\right)=\dfrac{1}{\ln10}\ln\left(9 + \frac{46}{x}\right)$$
or
$$\dfrac{\ln10}{-2\sqrt{x}}=\ln\left(9 + \frac{46}{x}\right)=\ln9+\ln\left(1 + \frac{46}{9x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How has this result been obtained?? We have the quadratic equation
$$ 2y^2 -(1+x)y + x = 0 $$
The author derives the following result from the above equation -
$$ y' = \frac{y-1}{4y-1-x} = \frac{(x-3)y-x+1}{x^2-6x+1} $$
Now, I understand the first part. It has been obtained by simply differentiating w.r.t to $x$ and th... | There's a way to simplify the denominator without explicitly solving for $y$. Note that
\begin{align}
\big(4y-(1+x)\big)^2 &= 16y^2 - 8(1+x)y + (1+x)^2 \\
&= 8\big(2y^2 - (1+x)y\big) + (1+x)^2 \\
&= -8x + (1+x)^2 \\
&= x^2 - 6x + 1
\end{align}
Therefore
\begin{align}
y' = \frac{y-1}{4y-(1+x)} &= \frac{(y-1)\big(4y-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Recreational math problem: $f(x)=\frac{x^2}{2x+101}$ Here is a math problem (just for fun) for the residents of MSE to enjoy:
Let $f(z)$ be defined as
$$f(z)=\frac{z^2}{2z+101}$$
Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that
$$f(z_1)=z_2$$
$$f(z_2)=z_3$$
$$f... | HINT:
Let's do some reductions of $f$:
$$f(101 x) = \frac{(101 x)^2}{2\cdot 101 x + 101}= 101\cdot \frac{x^2}{2x+1}$$
so we have
$$f= s \circ g \circ s^{-1}$$ where $s(x) = 101 x$ and $g(x)=\frac{x^2}{2x+1}$. Further
$$g(x)=\frac{x^2}{2x+1}= \frac{x^2}{(x+1)^2 - x^2}= \frac{1}{(1+\frac{1}{x})^2 -1}= t^{-1}\circ h \cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Probability - Peculiar die
A peculiar die has the following properties: on any roll the probability of rolling either a $4$, a $6$, or a $1$ is $1/2$, just as it is with an ordinary die. Moreover, the probability of rolling either a $1$, a $3$, or a $2$ is again $1/2$. However, the probability of rolling a $1$ is $5/1... | I agree with your solution. I wrote a very similar solution below.
Let $E_j$ be the event that we roll a $j$. Then
$$
P(E_2 \cup E_3) = P(E_1 \cup E_2 \cup E_3) - P(E_1) = \frac12 - \frac{5}{16} = \frac{3}{16}.
$$
It follows that
\begin{align}
P(E_5) &= 1 - P(E_1 \cup E_2 \cup E_3 \cup E_4 \cup E_6) \\
&= 1 - P(E_1 \cu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $\lim_{x \to \infty} e^{-x^2} \int_x^{x + \ln(x)/x} e^{t^2} dt $
Evaluate
$$\lim \limits_{x \to \infty} \frac{1}{e^{x^2}}\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt.$$
I should probably use such inequality:
$$ \int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt > \frac{e^{x^2} \ln x}{2x} \text { for } x ... | Let $$I(x) =\int \limits_x^{x + \frac{\ln x}{x}} e^{t^2} dt \Rightarrow I(x) \geq \frac{\ln x}{x}e^{x^2} \stackrel{x\rightarrow\infty}{\longrightarrow}\infty$$
So, we have with $\frac{I(x)}{e^{x^2}}$ a L'Hospital case of $\frac{\infty}{\infty}$ for $x\rightarrow\infty$:
$$\frac{I(x)}{e^{x^2}} \sim \frac{e^{(x^2+2\ln x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2730599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If a,b, and c is real, and following the equation, solve for
If $a,b,c$ are real and aren't 0, that complete the following equations
$a^2 + a = b^2$
$b^2 + b = c^2$
$c^2 + c = a^2$
Solve for $(2a+c)(2b+a)(2c+b)$,
What I've done:
*
*Adding all of the equation of it and substact with $(a^2 + b^2 + c^2)$ of both si... | Hah!
$a + b + c = 0$
And $a = b^2 - a^2 = (b-a)(b+a) = (b-a)(-c) = c(a-b)$
$b = c^2 - b^2 = (c-b)(c+b) = (c-b)(-c) = a(b-c)$
$c = a^2 - c^2 = (a-c)(a+c) = (a-c)(-b) = b(c-a)$.
and $(2a + c) = (a +(a+c)) = (a-b) = \frac ac$
$(2b + a) = (b + (b + a)) = (b-c) = \frac ba$
$(2c + b) = (c + (c+b)) = (c -a) = \frac cb$
So
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Can this equation be solved by radicals? $x^6+x^5+2x^4-2x^3-x^2+1=0$ Can this equation be solved by radicals? If it can be solved, what is the method? At least, I will resolve it myself if you specify reference or source.
$$x^6+x^5+2x^4-2x^3-x^2+1=0$$
| Yes because$$x^6+x^5+2x^4-2x^3-x^2+1=0\iff (x^3-1)^2+x^2(x^3-1)+2x^4=0$$ It follows solving as a quadratic in $x^3-1$
$$2(x^3-1)=-x^2\pm\sqrt{-7}x\iff2x^3+x^2\mp\sqrt{-7}x^2-2=0$$ so you can apply Cardano's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2734487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+... | Or another way:
$$9^n + 4^{n+1} = (10-1)^n +(5-1)^{n+1}=10c+(-1)^n + 5d+(-1)^{n+1}=10c+5d = 5e$$
Edit:
Another similar way:
$$9^n + 4^{n+1}= 9^n+4\cdot(9-5)^n = 9^n+4(9^n+5c)= (1+4)9^n + 20c = 5e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Showing that $\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$ As the title suggests, I would like to show that: $$\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$ for every positive integer $n$.
How can I go about doing that? I understand the that there are alrea... | You can solve the integral by induction. First of all, you may notice that
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}(2\cos\theta)^2 d\theta&=\frac{1}{\pi}\Big[\theta+\cos\theta\sin\theta\Big]_0^{2\pi}=2=\frac{2!}{1!1!}
\end{align*}
Thus the case $n=1$ holds. Now, integrating by parts:
\begin{align*}
\frac{1}{2\pi}\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find value of floor function of a given number? I am considering numbers $n = 3m + 1$ and $p = 3q$, where $m$ and $q$ are some positive integers.
I want to find the value of $\lfloor \frac{3m + 1}{2} \rfloor$ and $\lfloor \frac{3q}{2} \rfloor$. I know somewhere the answer depends on the values of $m$ and $q$.
F... | Assume that $m=2k+1$ ($k$ is a positive integer), then
$$\lfloor \frac{3m + 1}{2} \rfloor=\lfloor \frac{3(2k+1)+1}{2} \rfloor=\lfloor \frac{6k+4}{2} \rfloor=\lfloor 3k+2 \rfloor=3k+2.$$
If $m$ is an even number or $m=2k$, then
$$\lfloor \frac{3m + 1}{2} \rfloor=\lfloor \frac{3(2k)+1}{2} \rfloor=\lfloor \frac{6k+1}{2} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integral $\int \frac {dx}{\sin^2 x + \tan^2x}$
I am trying to find the following integral:
$$\int \frac {dx}{\sin^2 x + \tan^2x}$$
I have tried the common thing to do when encountering rational functions that contains rational functions and converting everything in terms of $\tan \frac{x}{2}$ then substituting it... | $$\begin{align}\int\dfrac{dx}{\sin^2{(x)}+\tan^2{(x)}}\cdot\dfrac{\sec^2{x}}{\sec^2{x}}&=\int\dfrac{\sec^2{x}\ dx}{\tan^2{x}+\sec^2{x}\tan^2{x}}\\
&=\int\dfrac{\sec^2{x}\ dx}{\tan^2{x}+(\tan^2{x}+1)\tan^2{x}}\\
&=\int\dfrac{\sec^2{x}\ dx}{2\tan^2{x}+\tan^4{x}}\end{align}$$
Let $u=\tan{x}$ and $du=\sec^2{x}\ dx$
$$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many solution does $1/a+1/b+1/c+1/d=1$ have? From my friend, he gives me a competition question:
"How many solution $(a,b,c,d)$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$ have where $a,b,c,d$ are positive integers? (the size of $a,b,c,d$ doesn't matter, either one can be the biggest or smallest, and t... | In my opinion your approach is fine and I am not aware of a faster method.
However I got a different number of solutions.
Assuming that $a\leq b\leq c\leq d$ then
1) if $a=2$ and $b=3$ then $(c,d)$ can be
$$(7,42),\;(8,24),\;(9,18),\;(10,15),\;(12,12).$$
2) if $a=2$ and $b=4$ then $(c,d)$ can be
$$(5,20),\;(6,12),\;(... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 3,
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Taylor series $\sqrt{1-2x+x^2+o(x^3)}$ I need to solve this asymptotic expansion for $x\to0$
$$\sqrt{1-2x+x^2+o(x^3)}$$
This is the expansion to use:
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$
let:
$$t = -2x+x^2+o(x^3)$$
My question is, should I stop to the first order or keep going?
How can I ... | Since $1-2x+x^2 = (1-x)^2$, it is more practical to check which coefficient $K$ ensures
$$\left[1-x+K x^2+o(x^3)\right]^2 = 1-2x+x^2+o(x^3) $$
but the coefficient of $x^2$ in the LHS is $2K+1$, hence $K$ has to be $0$.
Just a different way for putting Lubin's argument.
| {
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Stuck calculating the derivative of $f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$. I have to calculate the derivative of this:
$$f(x)=\log_{10}{\frac{x}{1+\sqrt{5-x^2}}}$$
But I'm stuck. This is the point where I have arrived:
$$f'(x) = \frac{(1+\sqrt{5-x^2})(\sqrt{5-x^2})+x^2}{x(\ln 10)(1+\sqrt{5-x^2})(\sqrt{5-x^2})}$$
H... | Since the denominator is positive (when it's defined, that is, for $-\sqrt5\le x\le \sqrt5 $), the function is only defined for $0<x\le \sqrt5 $, so you can write it as
$$
f(x)=\log_{10}x-\log_{10}(1+\sqrt{5-x^2})
$$
which should simplify the computation very much. Recall that the derivative of $\log_{10}{x}$ is
$$
\fr... | {
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"source": "stackexchange",
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If $x$, $y$, and $z$ are real numbers such that $x+y+z=8$ and $x^2+y^2+z^2=32$, what is the largest possible value of $z$? I tried swapping $z$ from the first equation to the second, and got $$x^2 + x y - 8 x + y^2 - 8 y + 16=0$$ Not sure where to go from there, and if I'm on the right track at all.
| \begin{align*}
x^2+(8-x-z)^2+z^2&=32\\
x^2-(8-z)x+z^2-8z+16&=0
\end{align*}
As $x$ is real,
\begin{align*}
[-(8-z)]^2-4(1)(z^2-8z+16)&\ge0\\
3z^2-16z&\le 0\\
0\le z&\le\frac{16}{3}
\end{align*}
The largest value of $z$ is $\displaystyle \frac{16}{3}$.
| {
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"question_score": "4",
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Find all positive integers $n$ such that $n^2 + 1$ is divisible by $n+ 1$
Find all positive integers $n$ such that $n^2 + 1$ is divisible by $n+ 1$
I know a lot about number theory, for example conjectures, theories, equations and so on. But I just can't solve this type of problems. Which algorithm should i use in or... | Another proof:
$$gcd(n+1,n^2+1)=gcd(n+1, n^2+1-n(n+1))=gcd(n+1,1-n)=gcd(n+1,1-n+(1+n))=gcd(n+1,2).$$
If $n$ is even, then $n+1$ is odd. Then $gcd(n+1,2)=1$ (is this clear?), this is:
$gcd(n+1,n^2+1)=1$ so that $n+1,n^2+1$ are relative primes.
If $n$ is odd, then $n+1$ is even, so $gcd(n+1,2)=2$, thus $gcd(n+1,n^2+1)=2$... | {
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"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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To find a relation between $a,b,c$ given a quadratic and a cubic epression If $x^2+px+1$ is a factor of $ax^3+bx+c$ then which of the following is true:
a)$a^2-b^2=ac$
b)$a^2-c^2=ab$
c)$a^2+b^2=bc$
d)$a^2+c^2=ab$
I assumed the roots of $x^2+px+1$ to be $k$ and $\frac{1}{k}$ since product of roots is 1 . $ax^3+bx+c$ w... | $ax^3+bx+c$ is the product of $x^2+px+1$ and a linear polynomial. This polynomial must be $ax+c$.
$$ax^3+bx+c=(ax+c)(x^2+px+1)$$
So, $c+ap=0$ and $a+pc=b$.
$$0=c(c+ap)=c^2+acp=c^2+a(b-a)$$
| {
"language": "en",
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AC^2 Algebraic simplifications Suppose I know that:
$AB^2=DF^2+CF^2$;
$AD^2=BE^2+CE^2$;
$AC^2=AE^2+CE^2$;
$AC^2=AF^2+CF^2$
And also:
$DF=AF-AD$ and $BE=AE-AB$
How can I prove that $AB * AE + AD * AF = AC^2$?
I start like this, but then I get stucked:
$AB*AE + AD*AF = AB*(BE+AB) + AD*(DF+AD) =
AB*BE + AB^2 + AD*DF + A... | Let $AB=x$, $BE=y$, $AD=p$, $DF=q$, now from the first $4$ equations we have
\begin{eqnarray*}
p^2+(x+y)^2=y^2+CE^2+(x+y)^2=y^2+AC^2 \\=y^2+(p+q)^2+CF^2=y^2+p^2+2pq+x^2
\end{eqnarray*}
and so $xy=pq$.
Now
\begin{eqnarray*}
AB \times AE+ AD \times AF = x(x+y)+p(p+q)
\end{eqnarray*}
and
\begin{eqnarray*}
AC^2 = (x+y)^2+... | {
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Solve a trigonometric equation: $|1-2\sin^2 x|=|\cos x|$ I have difficulty in solving this equation:
$$|1-2\sin^2 x|=\lvert\cos x\rvert.$$
What are the indications to solve the exercise correctly?
| $1 - 2\sin^2 x = 2\cos^2 x - 1\\
2\cos^2 x - 1 = \pm \cos x\\
2\cos^2 x \pm\cos x - 1 = 0$
Apply the quadratic formula:
$\cos x = \pm \frac 14 \pm \frac 34\\
\cos x = \pm \frac 12, \pm 1 $
| {
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Find the polynomial if remainder is given If $f$ is a quintic polynomial which leaves remainder $1$ when divided by $(x-1)^3$, and $-1$ when divided by $(x+1)^3$ , then find the value of first derivative of $f$ at $x=2$.
My approach
Let $$ f = A(x-1)^5 + B(x-1)^4 + C(x-1)^3 +1 $$
Also
$$ f = A(x+1)^5 + D(x+1)^4 + E(x+... | Another approach is the extended Euclidean algorithm to find the quintic $f,$ although there is now the possibility of fractions:
$$ \left( x^{3} + 3 x^{2} + 3 x + 1 \right) \left( \frac{ 3 x^{2} - 9 x + 8 }{ 16 } \right) - \left( x^{3} - 3 x^{2} + 3 x - 1 \right) \left( \frac{ 3 x^{2} + 9 x + 8... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluating $\lim _{x\to \infty} (\sqrt[3]{(x+1)(x+2)(x+3)}-x)$
Evaluate: $\lim _{x\to \infty}( \sqrt[3]{(x+1)(x+2)(x+3)}-x)$
I set $y= x-3$ for simplification and then tried to solve $y\to \infty$.
I tried to use the tayor expansion of the cubic function f(y) in the cuberoot but that didn't help.
How do I approach... | Using general Binomial expansion, $((x+1)(x+2)(x+3))^\frac{1}{3}-x = ((x+2)^3-(x+2))^\frac{1}{3}=(x+2)-\frac{1}{3}(x+2)^{-2}(x+2)-\frac{1}{9}(x+2)^{-5}(x+2)^3+..-x=2-P(x+2)$,
where the power of the polynomial $P(x+2)$ is negative. Letting $x \to \infty$, we get the limit 2.
| {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$... | \begin{align}
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|&=
\left|
\begin{matrix}
a+b+c&a-b&a \\
a+b+c&b-c&b \\
a+b+c&c-a&c \\
\end{matrix}\right|\\
&= (a+b+c)\left|
\begin{matrix}
1&a-b&a \\
1&b-c&b \\
1&c-a&c \\
\end{matrix}\right|\\
&= (a+b+c)\left|
\begin{matrix}
1... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to find the sum $1+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{8}-\frac{1}{10}-\frac{1}{11}+\cdots =\ ?$ Let $\phi(x)=\begin{cases}0, & 0\lt x\lt 1\\ 1, & 1\lt x\lt3 \end{cases}$
We have that the Fourier cosine series is given by $$\phi(x)=\begin{cases}0, & 0\lt x\lt1\\ \frac{4}{3}+\displaystyle\sum_{m... | The key observation, which will be exploited below in computing the series
$$
S=\sum_{n=1}^\infty\frac{\sigma_n}{n},
$$
where $\sigma_n$ is repeating sequence of $(1,1,0,-1,-1,0)$, is that
$$
\sigma_n=\frac{2}{\sqrt3}\sin\frac{\pi n}{3}.
$$
It follows:
$$
S=\frac{2}{\sqrt3}S',\text{ with } S'=\sum_{n=1}^\infty\frac{\si... | {
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$\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$ using integral(lower-/upper- or Riemann-sums) how can I show that $$
\lim_{n\rightarrow \infty} \frac{1^4+2^4+...+n^4}{n^5}=\frac{1}{5}$$
using integral(lower-/upper- or Riemann-sums).
I tried the following:
$\lim_{n\rightarrow \infty} \frac{1^4+2^4+.... | Hint
$$\frac{\sum_{k=1}^n k^4}{n^5}=\frac{1}{n} \sum_{k=1}^n \left(\frac{k}{n} \right)^4=\frac{1}{n} \sum_{k=1}^n f\left(\frac{k}{n} \right)$$
with $f(x)=x^4$, so you can recognize a Riemann sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $?
What is the value of
$\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?
The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (1... | We have to calculate $$\prod^{45}_{k=1}\sin^2((2k-1)^\circ)$$
Because $\sin(180-\theta)=\sin(\theta)$
Now Let $$P=\prod^{45}_{k=1}\sin((2k-1)^\circ)$$
Then $$\prod^{45}_{k=1}\sin(2k^\circ)\cdot P=\prod^{45}_{k=1}\sin((2k-1)^\circ)\cdot \prod^{45}_{k=1}\sin(2k^\circ)$$
So $$\prod^{45}_{k=1}\sin(2k^\circ)\cdot P=\frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Find the parametrization of the curve resulting from intersection of a function and a curve I have the following function $f(x,y) = 2-x^2-4y^2$ and the surface $2x+4y+z-1 = 0.$
How do i go about finding the parametrization of the curve resulting from intersection of these surfaces? I see that $f(x,y)$ is the equation o... | You have
*
*$z = f(x,y)= 2- x^2-4y^2$
*$2x+4y+z-1 = 0 \Leftrightarrow z=1-2x-4y$
It follows
$$1-2x-4y = 2- x^2-4y^2 \Leftrightarrow x^2-2x+4y^2-4y = 1$$
Square completion gives
$$(x-1)^2-1 +(2y-1)^2-1 = 1 \Leftrightarrow (x-1)^2 +(2y-1)^2 = 3 \Leftrightarrow$$ $$ \left( \frac{x-1}{\sqrt{3}} \right)^2 + \left( \fr... | {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
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$\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}$ doesn't exist (?) Is what I did correct?
We have: $\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}$
We know that as $x\rightarrow 0$ $\frac{\ln(1+x^3)}{x^3}=1$ , can be proven using taylor series..
So we have: $\lim_{(x,y)\rightarrow0} \frac{\ln(1+x^3)}{... | Yes you derivation is correct, indeed since $x\to 0$
$$\frac{\ln(1+x^3)}{x^3}\to1$$
it is true and we can conclude that for the trajectories $y=Cx^2\to 0$
$$\lim_{(x,y)\rightarrow0} \frac{x\ln(1+x^3)}{y(x^2+y^2)}=\frac1C$$
then the limit doesn't exist.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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How to solve the integration How to find $$ \frac{\int_0^{\pi}x^3\log(\sin x)\,dx}{\int_0^{\pi} x^2 \log(\sqrt{2} \sin x)\,dx} $$
I couldn’t resolve it by using integration by parts.
| Using Fourier series, we see that
\begin{align}
\log(\sin x)= -\log 2-\sum^\infty_{k=1}\frac{\cos(2kx)}{k}
\end{align}
for $0\le x\le \pi$. Hence we have
\begin{align}
\int^\pi_0x^3\log(\sin x)\ dx =&\ -\log 2\int^\pi_0x^3\ dx -\sum^\infty_{k=1} \frac{1}{k}\int^\pi_0 x^3\cos(2kx)\ dx\\
=&\ -\frac{\pi^4}{4}\log 2 - \fra... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
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Finding Sum of $\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty$ Finding Sum of $$\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}+\frac{12!}{16!}+\cdots \cdots \cdots \infty\; \bf{terms}$$
Try: Writting it as $$\sum^{\infty}_{r=0}\frac{(4r)!}{(4r+4)!}=\sum^{\infty}_{r=0}\frac{1}{(4r+1)... | As Piyush Divyanakar remarks, the sum is
$$\frac16\sum_{r=0}^\infty\left(
\frac{1}{4r+1}-\frac{3}{4r+2}+\frac{3}{4r+3}-\frac{1}{4r+4}\right).$$
This equals
$$\frac16\sum_{r=0}^\infty\int_0^1(t^{4r}-3t^{4r+1}+3t^{4r+2}-t^{4r+3})\,dt
=\frac16\int_0^1\frac{(1-t)^3}{1-t^4}\,dt.$$
You just have to do this integral...
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$
i have to prove:
$$\begin{vmatrix}
-2a &a+b &a+c \\
b+a&-2b &b+c \\
c+a&c+b &-2c
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$
I have tried many calculations between the rows an... | Brute force: expand using Cramer's rule:
$$
\begin{split}
\det A = &-2a\left( (-2b)(-2c) - (b+c)^2 \right)\\
&-(b+a)\left( (-2a)(-2c) - (a+c)^2\right)\\
&+(c+a)\left( (a+b)(b+c)-(-2b)(a+c)\right)
\end{split}
$$
and simplify out.
Similarly multiply out the RHS. It is painful, but maybe less painful tha... | {
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"source": "stackexchange",
"question_score": "1",
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Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then ... | Lemma: If $x^2+mx+n\geq 0$ for all $x$ then discriminant $m^2-4n \leq 0$.
First note that $x^2+x+1>0$ for all $x$ (since discriminant =$-3$)
$$0\leq f(x)\leq 2 \implies 0\leq x^2+ax+1\leq 2x^2+2x+2$$
*
*From $0\leq x^2+ax+1$ we get $a^2-4\leq 0$ so $|a|\leq 2$ so $\boxed{-2\leq a\leq 2}$.
*From $x^2+ax+1\leq 2x^2+... | {
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"question_score": "2",
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Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just ... | Hint:
\begin{align}
\frac{(3-x)(4-x)}{(3-x)(4-x)-x}&=\frac{(3-x)(4-x)}{x^2-8x+12}=\frac{(3-x)(4-x)}{(x-2)(x-6)} \\
&=1+\frac x{(x-2)(x-6)} \qquad\text{(by Euclidean division)}\\
&=1+\frac A{x-2}+\frac B{x-6}\qquad\text{(partial fractions)}
\end{align}
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $a^3+b^3=10^5$ has no positive integer solution.
Prove that there do not exist two positive integers, $a$ and $b$, such that $$a^3+b^3=100\,000$$
I tried to use congruence modulo $7$ and some other modulo, but it does not seem to work. If possible, please prove it with modular congruences.
| The possible cubic residues modulo $19$ are:
$$1,7,8,11,12,18$$
Further, $100000\equiv 3\pmod{19}$
The only two residues whose sum is equivalent to $3$ would be $11$ and $11$.
This implies that both $a$ and $b$ are one of $5\pmod{19},16\pmod{19}$ and $17\pmod{19}$
Further suppose that $a\geq b$. This implies that $a^3... | {
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"question_score": "4",
"answer_count": 3,
"answer_id": 2
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What is the probability of the drawn ball from U2 being white? This question is already posted here ,but i want to know flaw in my approach.
Question
Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls, and $U_2$ contains only $1$ white ball.
A fair coin is tossed:
*
*If head appear... | Following your approach it should be:
Case 1 (Head Appears): $U_2=WW$ with probability $\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$ and $U_2=RW$ with probability $\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}$.
Case 2 (Tail Appears): $U_2=WWW$ with probability $\frac{1}{2}\cdot\frac{3}{10}=\frac{3}{20}$, $U_2=RWW$ with probab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Cyclotomic polynomials and products of cosines problem I've run into an inconsistency I can't figure out while trying to find products of the cosine of various roots of unity.
For example: $\cos(\frac{2\pi}{5})\cdot cos(\frac{4\pi}{5}) \cdot cos(\frac{6\pi}{5}) \cdot cos(\frac{10\pi}{5})$ Multiply everything by $\frac{... | $$z=\cos(2k\pi/5)+i\sin(2k\pi/5)$$
($k\in\{1,2,3,4\}$) are the zeros of $z^4+z^3+z^2+z+1$. Then
$w=z+1/z=2\cos(2k\pi/5)$ and
$$0=z^2+z+1+z^{-1}+z^{-2}=(z+z^{-1})^2+(z+z^{-1})-1=w^2+w-1.$$
Each $\cos(2k\pi/5)$ is a zero of
$$4x^2+2x-1$$
and so a zero of
$$(4x^2+2x-1)^2=16x^4+16x^3-4x^2-4x+1.$$
I suspect this is the equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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let $f(x) = (x-1)\ln x$, and given $0 < a < b$. If $f(a) = f(b)$, prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$ Let $f(x) = (x-1)\ln x$, and given $0 < a < b$.
If $f(a) = f(b)$, how to prove that $\frac{1}{\ln a}+\frac{1}{\ln b} < \frac{1}{2}$?
| Proof.
Let $f(x) = (x-1)\ln x$.
Clearly, $f(1) = 0$, $\lim_{x\to 0^{+}} f(x) = \infty$, and $\lim_{x\to \infty} = \infty$.
We have $f'(x) = \ln x + 1 - \frac{1}{x}$. Clearly, $f(x)$ is strictly decreasing on $(0, 1)$, and strictly increasing on $(1, \infty)$. Thus, $0 < a < 1$ and $1 < b$.
We have
$$
\frac{1}{\ln a} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$ where $b>a$
My Attempt
$$
\begin{align}
&\int \frac{1}{\sqrt{(x-a)(b-x)}}dx=\int\frac{dx}{\sqrt{-x^2+(a+b)x-ab}}\\
&=\int\frac{dx}{\sqrt{-(x^2-2.\frac{a+b}{2}x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4})}}\\
&=\int\frac{dx}{\sqrt{... | Use the Euler substition $$\sqrt{(x-a)(x-b)}=(x-a)t$$ Then is $$x=\frac{at^2+b}{1+t^2}$$ and then is $$dx=\frac{2(a-b)t}{(1+t^2)^2}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$ To find the general solution $(x,y)$ satisfying the system of equations
\begin{align} 5(\sin x + \sin y) &= 1 \\ 5(\sin 2x + \sin 2y) &= 1
\end{align}
I applied $\sin C + \sin D$ rule and then divided these two equations, then I am stuck at
$$\c... | Starting from Alex Francisco's answer
$$16 (u^2)^3 - \left( \frac{4}{25} + 17 \right) (u^2)^2 + \left( \frac{4}{25} + 4 \right) u^2 - \frac{1}{25} = 0$$ let $x=u^2$ to get the cubic
$$400 x^3-429 x^2+104 x-1=0$$ and use the trigonometric method for three real roots. This would give the nasty
$$x_k=\frac{143}{400}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
find range for dependent
There are three vector $a$, $b$, $c$ in three-dimensional real vector space, and the inner product between them $a\cdot a=b\cdot b = a\cdot c= 1, a\cdot b= 0, c\cdot c= 4$. When setting $x = b\cdot c$, answer the following question: when $a, b, c$ are linearly dependent, find all possible valu... | Linear dependence places the vectors in the same plane. This makes the problem easier because we can focus on the angle.
From $c\cdot c=4$ we get that $|c|=2$.
So $a\cdot c=2\cos\theta=1\implies\cos\theta=\frac{1}{2}\implies\theta=60^{\circ}$.
From $b\cdot b=1$ and $a\cdot b=0$ we get that $b$ is a unit vector perpendi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Finding the probability that the matrix has full rank
There are only two entries, $0$ and $1$, over $\mathbb{Z}_2$. Thus,
only $16$ possible $2\times 2$ matrices over $Z_2$, and $6$ of them
have full rank:
$\begin{pmatrix}0&1\\ 1&0\end{pmatrix} \quad \begin{pmatrix}1&1\\
> 1&0\end{pmatrix} \quad \begin{pmatr... | The second formula seems correct to me. The rows of the matrix must be linearly independent. The first row can be any vector but the zero vector, so there are $2^n-1$ choices. Suppose that $k$ linearly independent rows have been added. The next row can be any vector not in the linear span of the the first $k$ rows.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
For $0 < x < \pi/2 $, prove that $x + \frac{x^{3}}{3} <\tan x $ I proved as follows:
Let $f(x) = \tan x - x - \frac{x^{3}}{3}.$
Then $f '(x) = \sec^{2}x - 1 - x^{2}$ and $f''(x) = 2\sec^{2}x\tan x-x > \tan x - x,$ since $\\tan x > x,$ $f''(x) > 0$ so $f'(x)$ is increasing.
Since $f '(0) = 0,$ $f '(x) > 0$ for all $x ... | Differentiate the given inequation three times
$$1+x^2\stackrel?<\tan^2x+1,$$
$$x\stackrel?<\tan x(\tan^2x+1),$$
$$1\stackrel?<(\tan^2x+1)^2+2\tan^2x(\tan^2x+1).$$
The final one is obviously true and equality holds at $x=0$, so that the second holds, by integration. That one holds at $x=0$, so that the first holds. An... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3... | $$
x+\frac{1}{x} = b\Rightarrow \left(x+\frac{1}{x}\right)^2=b^2
$$
then
$$
x^2+\frac{1}{x^2}= b^2-2 = a\Rightarrow b = \pm\sqrt{a+2}
$$
$$
\left(x^2+\frac{1}{x^2}\right)^2=a^2\Rightarrow x^4+\frac{1}{x^4} = a^2-2
$$
$$
\left(x^2+\frac{1}{x^2}\right)^2\left(x+\frac{1}{x}\right) = x^3+\frac{1}{x^3}+x+\frac{1}{x}
$$
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Question on an identity the representation of the Riemann Zeta function using Bernoulli numbers I'm reading Serre's A Course in Arithmetic.
In Section 4.1, we are showing that the Riemann Zeta function has a representation using the Bernoulli numbers.
The trick was to take the logarithmic derivative of
$$ \sin(z) = ... | Note that and using some of the below, you should be ale to recover the steps in the last line above.
\begin{align}
1+2z^2\sum_{n=1}^{\infty} \frac{1}{z^2-n^2}&= 1-2z^2 \sum_{n=1}^{\infty}\frac{1}{n^2}\frac{1}{1-z^2/n^2} \\
&=1-2z^2\sum_{n=1}^{\infty}\frac{1}{n^2}\sum_{k=0}^{\infty}\left(\frac{z}{n}\right)^{2k} \\
&= 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is the Method of Frobenius Appropriate for this DE? If so, how to proceed? To not bore with motivation, I'll get straight to the point. I have struggled for quite some time finding analytical solutions to the differential equation
$$y'' + \frac{a}{x}y' - \frac{b}{x^2\left(1 - \frac{c}{x^{n-2}}\right)}y = 0,\tag{1}$$ wh... | $y''+\dfrac{a}{x}y'-\dfrac{b}{x^2\left(1-\dfrac{c}{x^{n-2}}\right)}y=0$
$x^2\dfrac{d^2y}{dx^2}+ax\dfrac{dy}{dx}-\dfrac{bx^{n-2}}{x^{n-2}-c}y=0$
With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0216.pdf:
Let $r=x^{n-2}$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=(n-2)x^{n-3}\dfrac{dy}{dr}$
$\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the value of $S$ in term of $k$ (telescoping sums) Let $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550}$.
Find the value of $S=\frac{1275}{1276}+\frac{1276}{1277}+\frac{1277}{1278}+\cdots+\frac{2548}{2549}$ in term of $k$.
I've tried to write the terms of $S$ in the follo... | $k=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+\cdots+\frac{1}{2549\times2550} = \frac{1}{1}-\frac{1}{2} + \frac{1}{3}-\frac{1}{4} + ... + \frac{1}{2549}-\frac{1}{2550}$
Collecting even and odd and with some manipulation, we get
$k = H_{2550} - H_{1275}$ where $H_n$ = harmonic number
$S = 1 - \frac{1}{1276... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
How do I calculate how many ways 14 non-attacking bishops can be placed on a chessboard? It's possible to place up to 14 non-attacking bishops on a chessboard.
How can I calculate the number of valid configurations, without writing a program to brute force it?
I've checked Non Attacking Chess Pieces, but it doesn't cov... | Let's convert the chessboard into points: (rank,file). The normal chessboard refers to file A through H. We will use 1 through 8. Consider the number rank minus file:
$$\begin{matrix} & & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline 8 & | & 7 & 6 & 5 & 4 & 3 & 2 & 1 & 0 \\ 7 & | & 6 & 5 & 4 & 3 & 2 & 1 & 0 & -1 \\ 6 & | & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way.
I ... | $$\sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}=\frac{2\tan\dfrac{\alpha}{2}}{\sec^2 \dfrac{\alpha}{2}}=\frac{2\tan\dfrac{\alpha}{2}}{1+\tan^2 \dfrac{\alpha}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
If $4|n(n-1)$ then $4|n$ or $4|(n-1)$ Since $4$ is not prime, we can't use Euclid's lemma. Assume $4$ does not divide $n$, we'd like to show that $4|(n-1)$.
Suppose $n=4a+b$ where $b\neq 0$. It suffices to prove that $b=1$, but I feel like this is a dead end. Is my setup wrong?
| Method 1: You can't use Euclid's Lemma with $4$ but you can use it with $2$. So either i) $2^2|n$ or ii) $2|n$ and $2|n-1$ or iii) $2^n|n-1$. And ii) $2|n$ and $2|n-1$ is impossible.
Method 2: $n$ and $n - 1$ are relatively prime. So If $2|n$ then $2\not \mid n-1$ so $2^2|n$. Likewise if $2|n-1$ then $2\not \mid ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Linear equation help $$\frac{1-x}{4} + \frac{5x+1}{2} = 3 - \frac{2(x+1)}{8}$$
I got x=1 but the book says x=4/5 and I don't understand how to get to that, I tried working backwards too but I just can't figure it out. Any help would be appreciated, thanks.
Here's my work
using 8 as the lcm
$$
\begin{align}
\frac 81(\f... | Get everything over as a fraction with a denominator of $8$ and equate, so:
$$\frac{2(1-x)}{8}+\frac{4(5x+1)}{8}=\frac{24}{8}-\frac{2(x+1)}{8}$$
Thus we have:
$$2-2x+20x+4=24-2x-2$$
See if you can go from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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When is the given matrix not diagonizable?
Given the following matrix $$A = \begin{bmatrix} 0 & 1\\ bk+b & -bk\end{bmatrix}$$ where $b \neq 0$ and $k \neq 0$, find when the given matrix is not diagonizable.
My initial thought would be when the matrix doesn't have $2$ linearly independent eigenvectors.
In this case i... | $$b\neq 0, k\neq 0, A=\begin{pmatrix}
0 & 1\\
bk+b & -bk
\end{pmatrix}\Rightarrow \det (A-xI)=\begin{vmatrix}
-x & 1\\
bk+b & -bk-x
\end{vmatrix}=0\Rightarrow$$
$$\Rightarrow x^2+bkx-(bk+b)=0\Rightarrow x=\dfrac{-bk\pm \sqrt{b^2k^2+4bk+4b}}{2}$$
$$x_1=x_2\Rightarrow b^2k^2+4bk+4b=0\Rightarrow b^2k^2=-4(bk+b)\Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2819992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Congruence 10 modulo 7 I'm reading some proofs for divisibility rules and see that $10 \equiv -1 \pmod{11}$, then $10^2 \equiv 1 \pmod{11}$ and etc., it's clear for me, but I can't understand the situation with modulo 7. If we have $10 \equiv 3 \pmod{7}$ how can we get the rest of the values? They are $10^2 \equiv 2 \p... | As a start, we use the rule that
$$a \equiv b \pmod{n} \Rightarrow a^k \equiv b^k \pmod{n},\forall k\in \mathbb{N} \tag{1}$$
Hint
$$10 \equiv 3 \pmod{7} \Rightarrow 10^2 \equiv 3^2 \pmod{7} \Rightarrow \\
10^2 \equiv 9\equiv 7+2\equiv 2 \pmod{7}$$
or
$$10^3 \equiv 3^3 \pmod{7} \Rightarrow \\
10^3 \equiv 27\equiv 28-1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find all real values of $a$ such that $\sum_{n=0}^{\infty} \sin(\pi\sqrt{a^2+n^2})$ diverges To find all $a$ that makes the series diverge, it is sufficient to make the limit of $\sin(\pi\sqrt{a^2+n^2})$ not exist or exists but not equal to $0$ as $n\to\infty.$
I suspect only $a=0$ would converge the series, and the l... | HINT:
Observe that we can write
$$\begin{align}
\sin(\pi\sqrt{a^2+n^2})&=\sin\left(n\pi\sqrt{1+\frac{a^2}{n^2}}\right)\\\\
&=\sin\left(n\pi+\frac{a^2\pi}{2n}+O\left(\frac1{n^3}\right)\right)\\\\
&=(-1)^n\sin\left(\frac{a^2\pi}{2n}+O\left(\frac1{n^3}\right)\right)
\end{align}$$
Can you finish now? Is there any real val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Remainder of trinomial What will be the remainder if I divide $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ by $(a+b+c)^3-a^3-b^3-c^3$.
I have tried trinomial expention. But its still too big for long division. Is there a shorter method for this problem?
| Since $$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(a+c)(b+c)$$ and
$$(a+b+c)^{333}-a^{333}-b^{333}-c^{333}=0$$ for $a=-b$, for $a=-c$, for $b=-c$
and $(a+b+c)^{333}-a^{333}-b^{333}-c^{333}$ is divided by $3$,
we obtain that the remainder is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Sequence $(f_n)$ equibounded, equicontinuous but with not uniformly convergent subsequence
Let $f_n:[0,\infty)\rightarrow R$, given by $f_n(x) = \sin(\sqrt{x+4\pi^2n^2})$. Prove that $(f_n)$ is equicontinuous and equibounded, but has no uniformly convergent subsequence.
I'm having trouble to prove that this sequence... | Let's prove that this sequence of function is equicontinuous. Indeed fix $\epsilon > 0$, then consider $\delta > 0$ for a $\delta$ to be chosen shortly. Now consider $|f_n(x) - f_n(y)|$ = $|\sin(\sqrt{x+4 \pi^2n^2}) - \sin(\sqrt{y+4 \pi^2n^2})|$. Indeed, now (assume $x < y$) by MVT we see there exists a $\xi \in (x,y)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that the substitution $x=X-1$ and $y=Y+3$ turns $\frac{dy}{dx}=...$ into a homogenous equation. $$\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$$
After substitution of the given variable I get $$\frac{dY}{dX} = \frac{4-\frac{Y}{X}}{2+\frac{Y}{X}}$$ which seems to give a homogenous equation. (There is no answer given for th... | Solving for $u,v$
$$
4x-y+7=v\\
2x+y-1=u
$$
and substituting into
$$
\frac{dy}{dx} = \frac{4x-y+7}{2x+y-1}\rightarrow \frac{dv}{du} = \frac{4u-v}{2u+v}\;\;\;\mbox{which is homogeneous}
$$
Now making $z = v = \lambda u$ and correspondingly $dz = \lambda du+ud\lambda$ we have the new DE
$$
\frac{dv}{du} = \frac{dz}{du} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lagrange multipliers exercise Given the function $$f(x,y,z)=xy+xz+yz$$ I have to find its extremes on the set $E=\{(x,y,z)\in\mathbb R|x^2+y^2+z^2=1\}$.
Given the fact that $E$ is the border of a sphere, I can conclude that it is a compact set and because $f$ is continuous then there will be some extremes.
I proceed wi... | On the intersection of the plane $x+y+z=0$ and the unit sphere,
$$xy+xz+yz=\frac12\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{0-1}2=-\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Floor function of a real number The floor of a real number $x$, denoted by $⌊x⌋$, is the largest integer that is less than or equal to $x$. The ceiling of a real number x, denoted by $⌈x⌉$, is the smallest integer that is greater than or equal to $x$. Let $x$ and $y$ be any real numbers. Let $n, m$ and $k$ be any integ... |
The following does not use the fact that $e$ is transcendental.
First , note that $$\tag 1\lfloor x+y\rfloor \le\lfloor x\rfloor +\lfloor y\rfloor +1$$
for all real $x,y$.
Take the exponential of both sides of teh target equation to arrive at the equivalent condition
$$ \lfloor x+y\rfloor =e^{\lfloor x\rfloor}e^{\lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Pedal form to polar form of an ellipse How do I convert the equation
$$\frac{a^2b^2}{p^2}=a^2+b^2-\frac{1}{u^2}$$
into the following equivalent form?
$$u^2=\frac{\sin^2 \theta}{b^2}+\frac{\cos^2 \theta}{a^2}$$
where $$\frac{1}{p^2}=u^2+\left( \frac{du}{d\theta} \right)^2$$
EDIT:
I have tried and found out
$$\frac{uab}{... | Assume $a>b>0$,
\begin{align}
t &= \sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta} \\
t^2 &= (a^2-b^2)\sin^2 \theta+b^2 \\
\sin^2 \theta &= \frac{t^2-b^2}{a^2-b^2} \\
t^2 &= a^2+(b^2-a^2)\cos^2 \theta \\
\cos^2 \theta &= \frac{a^2-t^2}{a^2-b^2} \\
dt &=
\frac{(a^2-b^2)\sin \theta \cos \theta}
{\sqrt{a^2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq 2x^2$ for $0
Find the smallest value of $c$ such that $(1-x)\ln(1-x)+(1+x)\ln(1+x)\leq c x^2$ holds for $0<x<1$.
I saw the curve and realized this is true for $c=2$. How can I prove it? What is the smallest $c$ that still makes the inequality valid?
I think we should use the Taylo... | The radius of convergence of the Maclaurin series of $(1-x)\log(1-x)+(1+x)\log(1+x)$ is one.
In explicit terms
$$\begin{eqnarray*} (1-x)\log(1-x)+(1+x)\log(1+x) &=& x^2+\frac{x^4}{6}+\frac{x^6}{15}+\frac{x^8}{28}+\ldots\\&=&x^2+\sum_{n\geq 2}\frac{x^{2n}}{n(2n-1)} \end{eqnarray*}$$
hence
$$ f(x)=\frac{(1-x)\log(1-x)+(1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
} |
Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 & 1& 4 \end{matrix}\right)^T$ Consider the linear map $T:\mathbb{R}^2->\mathbb{R}^3$ with
$$T\left(\begin{matrix} x \\ y \end{matrix}\right)=\left(\begin{matrix} x-2y \\ 2x+y \\ -5x+8y \end{matrix}\right)$$ ... | What you need to solve is the system of linear equations:
$$\left\{\begin{align} x-2y&=3, \\ 2x+y&=1, \\ -5x+8y&=4. \end{align}\right.$$
The point of the Gauss elimination method (I recommend you check out this Wikipedia page!) is that we "encode" a system of equations by its augmented matrix, and then we apply the so-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2835392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ I am just wondering how to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ (This one is taken from Wo... | You will probably have to use the identity:
$$\dfrac{1}{x-y} = \dfrac{(x+y)(x^2+y^2)}{x^4-y^4} = \dfrac{x^3+xy^2+x^2y+y^3}{x^4-y^4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$ Eliminate $\theta$ in following equations
$$\begin{align}
a \cos(\theta-\alpha) &= x \\
b \cos(\theta- \beta) &=y
\end{align}$$
I am trying to solve this problem but still I am unable to get the perfect answer
I added both the equations bu... | \begin{align}
\frac{x}{a}+\frac{y}{b} &= \cos (\theta-\alpha)+\cos (\theta-\beta) \\
&=2\cos \frac{\alpha-\beta}{2} \cos \frac{\alpha+\beta-2\theta}{2} \\
\frac{x}{a}-\frac{y}{b} &= \cos (\theta-\alpha)-\cos (\theta-\beta) \\
&=-2\sin \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta-2\theta}{2} \\
1 &=
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$
$x^2 -xy + y^2 \equiv 0 \pmod{43}$
We could transform this into
$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$
and thus
$x^3 + y^3 \equiv 0 \pmod{43}$
From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $... | The polynomial $p(T)=T^2-T+1$ is the sixth cyclotomic polynomial. Meaning that its zeros have order six.
Because $43-1=6\cdot7$ an element $a$ of the cyclic group $\Bbb{Z}_{43}^*$ is of order six if and only if it is a seventh power but is not a 14th or a 21st power.
Let's just list seventh powers and do a bit of scree... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find all $z$ for which $z^2+2z+2$ is real positive. Find all $z$ for which $z^2+2z+2$ is real positive.
My Attempt :
Let $z=x+iy, (x,y\in R)$:
$$z^2+2z+2= (x^2-y^2+2ixy) +2(x+iy)+2$$
$$=(x^2-y^2+2x+2)+i(2xy+2y)$$
| Let $p$ be a real number.
$$(z+1)^2+1=p^2\iff z=\pm\sqrt{p^2-1}-1.$$
If $|p|<1$, $z=\pm i\sqrt{1-p^2}-1$ runs from $-1-i$ to $-1+i$,
If $|p|\ge1$, $z=\pm\sqrt{p^2-1}-1$, runs from $-\infty$ to $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 6
} |
Find a close formula using generating functions I want to find the sum:
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}$$
I did it by using the binomial theorem and I got
$$\sum \limits_{k=0}^n {n\choose k}2^{k-n}
= 2^{-n} \sum \limits_{k=0}^n {n \choose k}2^k
= \left(\frac{1}{2}\right)^n3^n$$
I would like to know how to d... | Another look at the problem, using the fact that
$${\displaystyle {\binom {n}{k}}={\binom {n-1}{k-1}}+{\binom {n-1}{k}}}$$
we can construct the recurrence relation for the sequence:
$$a_n=\sum \limits_{k=0}^n {n\choose k}2^{k-n}=
{n\choose 0}2^{-n}+\sum \limits_{k=1}^{n-1} {n\choose k}2^{k-n}+{n\choose n}2^{0}=\\
{n\ch... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
Find the inverse function of $f(x) = \frac{x^3+3x}{2}$
My thought on this are as follows:
I've solved a similar problem where I had to find the inverse of:
$$
g(x) = \sqrt[3]{x+\sqrt{x^2-1}} + \sqrt[3]{x-\sqrt{x^2-1}}
$$
So let $a = \sqrt[3]{x+\sqrt{x^2-1}}$ and ... | Hint:
Let $$x:=\sqrt[3]u-\frac1{\sqrt[3]u}$$
Then
$$x^3+3x=u-3\sqrt[3]u+\frac3{\sqrt[3]u}-\frac1u+3\sqrt[3]u-\frac3{\sqrt[3]u}=u-\frac1u.$$
Now solve the equation
$$u-\frac1u=2y$$
for $u$ (it can be reduced to a quadratic one).
In the end,
$$x=\sqrt[3]{u(2y)}-\frac1{\sqrt[3]{u(2y)}}.$$
Notice that the product of the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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