Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Evaluate $\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$
Evaluate $$\lim_{n\to \infty} \sum_{r=1}^n \frac {1}{2^r}\tan \left(\frac {1}{2^r}\right)$$
I tried to create a telescoping sum but I couldn't. The last step I could reach was turning the limit into $$\lim_{n\to \infty} \sum_{r=... | Solution 1. By the sine double-angle formula,
$$ \sin x = 2^n \sin(2^{-n}x) \prod_{k=1}^{n} \cos(2^{-k}x). $$
Now log-differentiating both sides gives
$$ \cot x = \frac{1}{2^n}\cot\left(\frac{x}{2^n}\right) - \sum_{k=1}^{n} \frac{1}{2^k}\tan\left(\frac{x}{2^k}\right). $$
Taking $n\to\infty$ and simplifying gives
$$ \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find $a\in\mathbb R$, if it exists, so that $T(2,0,1)=(1,-\frac 52)$ Let $$T:\mathbb R^3\to \mathbb R^2\mid M_{B_1B_2}=\begin{pmatrix}a&1&2\\-1&0&\color{red}1\end{pmatrix}$$ a linear transformation and the basis $$B_1=\{(1,0,0),(0,-3,1),(0,0,-2)\},\qquad B_2=\{(2,0),(0,-1)\}.$$
Find $a\in\mathbb R$, if it exists, so th... | Note that $$[(2,0,1)]_{B_1}=(2,0,\frac{-1}{2})_{B_1}$$ because $$(2,0,1)=2(1,0,0)+0(0,-3,1)+\frac{-1}{2}(0,0,-2)$$, thus, $$T(2,0,1)=(2a-1,\frac{-5}{2})$$ ,take a=1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find the sum of the series -1^2-2^2+3^2+4^2-5^2… upto 4n terms? I tried by giving
$$ S = \sum_{k=0}^{n-1} \left((4k+3)^2+(4k+4)^2-(4k+1)^2-(4k+2)^2\right) $$
but I am stuck here. I have no idea what to do next. The answer in my book says 4n(n+1). How can I get it? I tried expanding (4k+1)^2, etc. and got $ \... | Expand
\begin{align}
&(4k+3)^2-(4k+2)^2+(4k+4)^2-(4k+1)^2\\
\qquad&=(4k+3-4k-2)(4k+3+4k+2)+(4k+4-4k-1)(4k+4+4k+1)\\
\qquad&=8k+5+3((8k+5)\\
\qquad&=4(8k+5)\\
\qquad&=32k+20
\end{align}
So your sum is
$$
\sum_{k=0}^{n-1}(32k+20)=
32\frac{n(n-1)}{2}+20n=16n^2-12n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $\dfrac{a+b+c}{b-a}$
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
Find the minimum value of $\dfrac{a+b+c}{b-a}$
Attempt:
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the express... | Rewrite the equation as $x^2+Bx+C$. Now, $B^2\le 4C$ So, the minimum value of $C=\dfrac{B^2}{4}$. Now, $\dfrac {a+b+c}{b−a} =\dfrac {1+B+(B^2/4)}{B−1}$. Find the minimum value of function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$? Is it true that $13|2^n5^{2n+1}+3^{2n+1}7^{n+1}$ for all $n$?
So what I did was basically simplify the terms on the right $\mod 13$.
$2^n5^{2n+1}+3^{2n+1}7^{n+1} \mod 13$
$= 5\cdot 2^n\cdot 5^n\cdot 5^n+3\cdot 7\cdot 3^{n}\cdot 3^n7^n \mod 13$
$= 5\cdot 50^n... | Yes it is correct and very nice. The only minor note, we can also conclude at that step of course $$...\equiv 13\cdot 11^n \mod 13$$
As an alternative you can try also by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum value of $\frac{pr}{q^2}$ in quadratic equation If $y^4-2y^2+4+3\cos(py^2+qy+r)=0$ has $2$ solutions and $p,q,r\in(2,5)$. Then minimum of $\displaystyle \frac{pr}{q^2}$ is.
solution I try
$$-3\cos(py^2+qy+r) =(y^2-1)^2+3\geq 3$$
$$\cos(py^2+qy+r)\leq -1\implies \cos(py^2+qy+r)=-1$$
$$py^2+qy+r=(4n+1)\pi\implies... | Rewrite the equation as $(y^2 - 1)^2 + 3 (1 + \cos (py^2 + qy + r)) = 0$. Both terms on the LHS are manifestly non-negative, so the equation is solved only when both terms are zero. The only possible solutions are $y = \pm 1$, and both are obtained only when $p+q+r$ and $p-q+r$ are both odd multiples of $\pi$. As $2 < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to prove that in a field with order $4$, $x^2 = x +1$ for $x$ different from $0$ and $1$? For a field $\mathbb{F}$ with order 4, prove that elements in $\mathbb{F}$ (except for $0$ and $1$) satisfies the equation:
$$
x^2 = x +1
$$
My thought is as followings:
Set $G = \{0,1,a,b\}$.
We can prove $Char \mathbb{F} = 2... | This is obviously not true for $x=0$ or $x=1$. But then the question looks a bit weird. Anyway,
Suppose that $x$ is neither $0$ nor $1$. We have $(x+1)(x^2 - x - 1) =(x+1)( x^2 - x + 1 ) = x^3 -1$. Since $F-\{0\}$ is cyclic of order $3$, $x^3 - 1 = 0$. Hence $x = -1 = 1$ or $x^2 = x+1$. So $x^2 = x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?
I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.
For instance:
$\sin(2x) = 2 \sin(x) \cos(x)$
| Since $\cos 2x=\pm\frac{7}{25}$, $\tan x=\frac{\sin 2x}{1+\cos 2x}\in\left\{\frac{24}{32},\,\frac{24}{18}\right\}=\left\{\frac{3}{4},\,\frac{4}{3}\right\}$ so $\sin x=\pm\frac{\tan x}{\sqrt{1+\tan^2 x}}\in\pm\left\{\frac{3}{5},\,\frac{4}{5}\right\}=\left\{-\frac{4}{5},\,-\frac{3}{5},\,\frac{3}{5},\,\frac{4}{5}\right\}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2786868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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About an interesting improper integral $\int_{0}^{\infty} \frac{x dx}{1+x^6 \sin^2 x}$ Is there a non-negative $C^\infty$ function $f(x)$ which is not bounded in $[a, \infty)$, but whose improper integral $\int_{a}^{\infty} f(x) dx$ exists?
Yes!
$\frac{x}{1+x^6 \sin^2 x}$ is not bounded in $[0, \infty)$ and not negativ... | Using the fact that
\begin{align}
\frac{2}{\pi}x \leq \sin x \ \ \text{ for } \ \ 0\leq x\leq \frac{\pi}{2}
\end{align}
and
\begin{align}
1-\frac{2}{\pi}\left(x-\frac{\pi}{2}\right) \leq \sin x \ \ \text{ for } \ \ \frac{\pi}{2}\leq x \leq \pi.
\end{align}
Hence we have
\begin{align}
\int^\pi_0 \frac{xdx}{1+(n\pi)^6\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$... | Note that, $(a^2-1)(a^{2m-2}+a^{2m-4}+\dots+a^2+1)\\=a^2(a^{2m-2}+a^{2m-4}+\dots+a^2+1)-(a^{2m-2}+a^{2m-4}+\dots+a^2+1)\\=a^{2m}+a^{2m-2}+a^{2m-4}+\dots+a^2-(a^{2m-2}+a^{2m-4}+\dots+a^2+1)\\=a^{2m}-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown of... | Continuing where you stopped:
$$
10b^2 = a^2 + 2ab
$$
Write this as
$$
b(10b-2a) = a^2
$$
Therefore, $b$ divides $a^2$. Since $\gcd(a,b)=1$, the only possibility is $b=1$. But then $\sqrt{11}-1=a$ is an integer, which implies $\sqrt{11}$ is an integer, which it clearly isn't because $3^2 < 11 < 4^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Finding the minimal polynomial of $2\sqrt[3]{3}+\sqrt[3]{4}$ in the most efficient way . How could one find the minimum polynomial of $2\sqrt[3]{3}+\sqrt[3]{4}$, withot using the method where we let $\alpha=2\sqrt[3]{3}+\sqrt[3]{4}$, and keep cubing until we get a polynomial for $\alpha.$
I know that method will work ... | Let $$x:= 2\sqrt[3]{3}+\sqrt[3]{4} = \sqrt[3]{24}+\sqrt[3]{4}$$
Then $$x^3 = 24+4+3(\sqrt[3]{24}\cdot\sqrt[3]{4})(\underbrace{\sqrt[3]{24}+\sqrt[3]{4}}_{=x})= 28+6x\sqrt[3]{12}$$
So $$216\cdot 12x^3 = (x^3-28)^3 =...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solving Recurrence Relation for Series Solution of an ODE I am trying to solve the below problem:
Assume $y = \sum_{n=0}^{\infty}a_nx^n$ is a solution to $(x-1)y''-(x-3)y'-y=0$. Find $a_n$.
I took both derivatives of $y$, plug them into the equation, modify the indices until each series has the same $x^n$, and take ter... | After Leucippus's answer.
Since Leucippus identified the proper sequences, if
$$a_{n+2} = \frac{(n+3) \, a_{n+1} - a_{n}}{n+2}$$ the coefficients can write
$$a_n=(2 a_0 -a_1) +e \,(a_1 -a_0 )\, \frac{\Gamma (n+1,1)}{n!}$$ where appears the incomplete gamma function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solving $4^m-3^n=p^2$ for natural $(m,n,p)$. I know that such exponential equations, like the one in question, $4^m-3^n=p^2$ (where $m,n,p$ are natural) are usually solved using numerical methods, so I tried the following.
My attempt
I’ve split the search into parts; at least intuitively (and WolframAlpha confirms), th... | Hint: Write like this
$$(2^m-p)(2^m+p) =3^n$$
Solution:
So $2^m-p = 3^a$ and $2^m+p = 3^b$ for some $a,b$ where $a+b=n$ and $b\geq a\geq 0$.
Now
$$ 2^{m+1}= 3^b+3^a = 3^a(3^{b-a}+1)\implies a=0 \;\;\;{\rm and}\;\;\; 2^{m+1}= 3^b+1$$
so
$$(-1)^{m+1}\equiv_3 1 \implies m+1=2k$$
so $$ (2^k-1)(2^k+1)=3^b$$ and we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Evaluate $\int_0^1x(\tan^{-1}x)^2~\textrm{d}x$
Evaluate $\int\limits_0^1x(\tan^{-1}x)^2~\textrm{d}x$
My Attempt
Let, $\tan^{-1}x=y\implies x=\tan y\implies dx=\sec^2y.dy=(1+\tan^2y)dy$
$$
\begin{align}
&\int\limits_0^1x(\tan^{-1}x)^2dx=\int\limits_0^{\pi/4}\tan y.y^2.(1+\tan^2y)dy\\
&=\int\limits_0^{\pi/4}\tan y.y^2d... | Let's try integration by parts:
\begin{align}
\int x(\arctan x)^2\,dx
&=
\frac{x^2}{2}(\arctan x)^2-
\int\frac{x^2}{2}2\arctan x\frac{1}{1+x^2}\,dx
\\
&=
\frac{x^2}{2}(\arctan x)^2-
\int\frac{1+x^2-1}{1+x^2}\arctan x\,dx
\\
&=
\frac{x^2}{2}(\arctan x)^2-
\int\arctan x\,dx+
\int\frac{1}{1+x^2}\arctan x\,dx
\end{align}
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$? I guess it's easy, but I still need help. The inequality is
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
If you set $t=2^x$, then it becomes
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
The set of solutions is
$$x \in (-\infty,-2) \cup \{1\}$$
which is not what I get.
... | Continue from your efforts, $2^x = t$ already implies $t \gt 0$.
$$\frac{1}{t+3}- \frac{1}{4t-1} \ge 0\\
\frac{3t-4}{(t+3)(4t-1)}\ge 0$$
Solving this inequality with the help of zero points of the three factors, gives $t\in (-3,1/4) \cup [4/3, \infty) $ but we also had $t\gt 0$. So final answer is $t \in (0,1/4) \cup... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Probability that $z$ is EVEN satisfying the equation $x + y + z = 10$ is Question
Three randomly chosen non negative integers $x, y \text{ and } z$ are found to satisfy the equation $x + y + z = 10$. Then the probability that $z$ is even, is"?
My Approach
Calculating Sample space -:
Number of possible solution for... | Presumably the intended problem is:
*
If $(x,y,z)$ is randomly chosen from the set of nonnegative integer triples $(x,y,z)$ satisfying $$x+y+z=10$$
what is the probability that $z$ is even?
Using that interpretation . . .
Suppose $x+y+z=10$, and $z$ is even.
Write $z=2c$.
Note that $x,y$ are either both even, o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
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Stuck with solving inequality to find the product of highest and lowest integer solutions The inequality in question is
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{6})^{2x}}\leq98$$
This time our job is to find the product of highest and lowest integer solutions.
My attempt
$$\sqrt{(5+2\sqrt{6})^{2x}}+\sqrt{(5-2\sqrt{... | We have
$$
(a+b)^2+(a-b)^2= 2a^2+2b^2 = 98
$$
with $a = 5$ and $b = 2\sqrt 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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You roll a die until you get a $5$, what is the expected value of the minimum value rolled? I am struggling to work out a simple way to answer this, and a rationale behind this approach using tail sum (as I do not understand):
$$E\left( x\right) =\sum ^{5}_{k=1}P\left( x\geq k\right) =\dfrac {1}{6}\sum ^{5}_{k=1}\left(... | I seem to get $\dfrac{137}{60}$, slightly more than $2$, which seems as plausible to me as your similar answer
More precisely: $$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15$$
I think you are saying that if only $6$s are thrown before the first $5$ then the minimum i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Sum of coefficients of even powers of x in $(1+x)^5 (1+x^2)^5$? the exact question is asking the sum of coefficients of even powers of x in the expansion of $(1+x+x^2+x^3)^5$ and in the solution this expression is simplified to $(1+x)^5(1+x^2)^5$. The solution to this question says this expression is equal to $(1+5x+10... | The second term $(1+x^2)^5$ will have only even powers.
Now you have to calculate the sum of coefficients,
$\Rightarrow$ $2^5=32$ as ${5 \choose 0}+{5 \choose 1}+{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}=2^5$
And calculate the sum of all even powers in the first term i.e. $(1+x)^5$ as sum of 2 even number... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to ... | Cauchy Schwarz Inequality
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}=\frac{\sin^4 x}{\cos x\sin x}+\frac{\cos^4 x}{\sin x\cos x}\geq \frac{(\sin^2 x+\cos^2 x)^2}{2\sin x\cos x}$$
$$\frac{\sin^3 x}{\cos x}+\frac{\cos^3 x}{\sin x}\geq \frac{1}{\sin 2x}\geq 1$$
equality hold when $$\frac{\sin^2 x}{\sin x\cos x}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Converting parametric $x = \sec \theta + \tan \theta$, $y = \csc\theta + \cot\theta$ to Cartesian form This question comes from Solomon C4 Paper K, Question 7b.
Consider the parametric equations:
$$\begin{align}
x &= \sec(\theta) + \tan(\theta) \\
y &= \csc(\theta) + \cot(\theta)
\end{align}$$
I would like to express ... | When in doubt, here are some heuristics that can sometimes help:
*
*Rewrite everything in terms of $\sin$ and $\cos$.
*Try putting things over a common denominator.
*Try computing $x+y$, $x^2$, $y^2$ and $xy$ and see if you can find relationships between them.
In this case, we have
$$x = \frac{1}{\cos\theta} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find a fraction $\frac{m}{n}$ which satisfies the given condition Find a fraction such that all of $\frac{m}{n}$, $\frac{m+1}{n+1}$, $\frac{m+2}{n+2}$, $\frac{m+3}{n+3}$, $\frac{m+4}{n+4}$, $\frac{m+5}{n+5}$ are reducible by cancellation. Condition: $m≠n$.
What I tried was... I wrote $$\frac{m}{n}=k$$ Then, I replaced ... | Simple computer search gives:
\begin{array}{|c|c|} \hline
m & n \\ \hline
212 & 2 \\
213 & 3 \\
214 & 4 \\
215 & 5 \\
300 & 90 \\
301 & 91 \\
324 & 114 \\
325 & 115 \\ \hline
\end{array}
Further observation:
\begin{array}{|c|c|} \hline
n \equiv m \pmod{\lambda} & \lambda \\ \hline
2,3,4,5 & 210 \\... | {
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"url": "https://math.stackexchange.com/questions/2803295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$ I have become stuck while solving a trig identity. It is:
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$
I have simplified the left side as far as I can:
\begin{align}
\frac{1-2\sin(x)}{\sec(x)}
&=\frac{1-2\s... | The identity is equivalent to
$$
\cos3x=\cos x(1-4\sin^2x)
$$
(except for the values where the denominators vanish). The right-hand side can be rewritten as
$$
\cos x(\cos^2x+\sin^2x-4\sin^2x)=\cos^3x-3\cos x\sin^2x
$$
which is known to be the same as $\cos3x$: by De Moivre
\begin{align}
\cos 3x+i\sin3x
&=(\cos x+i\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Factoring a given quartic form Description
Below is our polynomial:
$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4$$
I've gone to almost end of it but,it seems there's a small -or conversely huge- fault in my solving which made $\mathbf{A}\:and\: \mathbf{B}$ imaginary numbers as I didn't find any two numbers which Could have the su... | Assume by symmetry
$$(ax^2+bxy+cy^2)(cx^2+bxy+ay^2)=\\=acx^4+(ab+bc)x^3y+(a^2+b^2+c^2)x^2y^2+(ab+bc)xy^3+(ac)y^4$$
then we need
*
*$ac=3 \implies a=3\, c=1$
*$ab+bc=-8 \implies3b+b=-8\implies b=-2$
*$a^2+b^2+c^2=14 \implies 9+4+1=14$
thus
$$3x^4-8x^3y+14x^2y^2-8xy^3+3y^4=(3x^2-2xy+y^2)(x^2-2xy+3y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding all functions $ f : \mathbb R \to \mathbb R $ satisfying $ f ( x ) f ( y ) + f ( x y ) + f ( x ) + f ( y ) = f ( x + y ) + 2 x y $
Find all functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ f ( x ) f ( y ) + f ( x y ) + f ( x ) + f ( y ) = f ( x + y ) + 2 x y $$
for all $ x , y \in \mathbb R $.
I tried ... | "Quick" solution: let $g(x)=x-f(x)$. Then in terms of $g$, the functional equation can be rearranged to $$g(x+y)+g(x)g(y)=(1+x)g(y)+(1+y)g(x)+g(xy).$$ By my answer to this question, the solutions to this functional equation are $g(x)=0$, $g(x)=3x$, and $g(x)=x(x+1)$. We conclude that the solutions for $f$ are $f(x)=x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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Integral $\int_0^{\infty}\frac{1}{(1+x^2)(3-\cos x)}dx$ Greetings I tried to evaluate $$I=\int_0^{\infty}\frac{1}{(1+x^2)(3-\cos x)}dx$$ Here is my try, it is abit longer, but I wrote it all so that I wont have a silly mistake. My main ideea was to expand into fourier series $$g(t)=\frac{1}{3-\cos t}$$ so I took(I am n... | By following the same approach given by Random Variable in Prove $\int\limits_{0}^{\pi/2}\frac{dx}{1+\sin^2{(\tan{x})}}=\frac{\pi}{2\sqrt{2}}\bigl(\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\bigr)$ , we have that if $1<|a|<e$ then
$$\begin{align} \frac{1}{2a}\int_{0}^{\infty} \frac{1}{(1+x^{2})(\frac{1+a^2}{2a}-\cos x )} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$. Could someone please explain a step in the following proof?
Show that $\lim_{x\to 1}\frac{1}{x+1}=\frac{1}{2}$.
The above limit exists if for every $\varepsilon > 0$, there exists a real number $\delta > 0$ such that if $0<|x-1|<\delta$, then $\left |\frac{1}{x+1}-\... | So far we have $$\left |\frac{1}{x+1}-\frac{1}{2} \right |=\left | \frac{2-x-1}{2(x+1)} \right |=\left | \frac{1-x}{2(x+1)} \right |=\frac{|1-x|}{2|x+1|}=\frac{|x-1|}{2|x+1|}<\varepsilon$$ In order to achieve the last inequality, we have to control the denominator as well as the numerator.
By assuming $|x-1|<1$ we ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Find the number of ways to select $1,2,3,4,5,6,7,8,9,10$ fruits from a pile (using generating functions) Find the number of ways to select $1,2,3,4,5,6,7,8,9,10$ fruits from a pile of $3$ apples, $5$ oranges and $2$ bananas. (Use generating functions.)
Any tips?
| We encode zero up to
*
*three apples as: $\ \quad1+x+x^2+x^3=\frac{1-x^4}{1-x}$
*five oranges as: $\ \quad1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x}$
*two bananas as: $\quad 1+x+x^2=\frac{1-x^3}{1-x}$
Denoting with $[x^k]$ the coefficient of $x^k$ of a series, we are looking for
\begin{align*}
\color{blue}{[x^k]\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Special properties of the number $146$ I'm a math teacher. Next week I'll give a special lecture about number theory curiosities. It will treat special properties of numbers — the famous story with Ramanujan, taxicab numbers, later numbers divisible by all their digits, etc.
I was given class number $146$ for the lectu... | The sum of the first $n$ cubes, starting from $1^3$, is given by the square of the $n$th Triangular number, $T_n$:
\begin{align}
1^3 &= & 1^2 &= & T_{1}^2\\
1^3+2^3 &= & 3^2 &= & T_{2}^2 \\
1^3+2^3+3^3 &= & 6^2 &= & T_{3}^2 \\
1^3+2^3+3^3+4^3 &= & 10^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 19,
"answer_id": 18
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For any prime number $n >5$, prove the final digit of $n^4$ is $1$ So I am struggling a bit with this question
$n$ is prime
we can ignore $2$ and $5$ as $n>5$
now if $n$ is prime
for the digits: $\{0,1,2,3,4,5,6,7,8,9\}$
$\{0,2,4,6,8\}$ can be discounted as $n$ cannot be even that
$5$ can be discounted as $n$ is not a ... | Hint
$$n^4-1=(n^2-1)(n^2+1)=(n^2-1)(n^2-4+5)\\
=(n^2-1)(n^2-4)+5(n-1)(n+1)\\
=(n-2)(n-1)(n+1)(n+2)+5(n-1)(n+1)$$
Show that
$$10| (n-2)(n-1)(n+1)(n+2) \,, \mbox{ and }\\
10|5(n-1)(n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Without relating the problem to polynomials, how can we solve this inequality? $a,b,c$ are real;
$a<b<c$ ;
$a+b+c=6$ ;
$ab+bc+ca = 9$.
Prove that $0<a<1<b<3<c<4$.
I solved this by treating $a,b,c$ as roots of $x^3 - 6x^2 + 9x + d$ for some $d$, but I have not been able to solve it in a more direct way.
| Partial solution:
Let $\{x,y,z\} =\{a,b,c\}$. Since $z=6-x-y$ we get a quadratic equation:
$$ x^2+x(y-6)+(y-3)^2 =0$$
and since this one has to have a real solution we have a discriminant nonnegative, so
$$ -3y(y-4)\geq 0 \implies y\in [0,4]$$
and the same we can say for $x$ and $z$.
Now can any of $x,y,z\in\{0,4\}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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olympic mathematics For each integer $a_0 > 1$, define the sequence $a_0, a_1, a_2,
\ldots$ by:
$$a_{n+1} = \begin{cases} \sqrt{a_n}, & \mbox{if } \sqrt{a_n}\mbox{ is an integer,} \\a_n+3, &\mbox{otherwise,}\end{cases}$$
Determine all values of $a_0$ for which there is a number $A$ such that $a_n = A$ for infinitely m... | There is no integer $a$ so that $a^2 \equiv 2 \mod 3$ (as $0^2 \equiv 0 \mod 3$, $1^2 \equiv 1 \mod 3$ and $2^2 \equiv 1 \mod 3$ so if $a_0\equiv 2 \mod 3$ then by induction $a_{i+1} = a_i + 3 \equiv a_0 \equiv 2 \mod 3$ and the sequence never repeats.
There are infinitely many squares that are multiples of $3$. The s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1... | Since $(\forall x\in\mathbb{R}):4(x^3+2x^2+x)^3(3x^2+4x+1)=4x^3(3x+1)(x+1)^7$, both answers are correct. But your answer is more natural.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Finding the inverse of $x^3+x^2+1$ in $\Bbb F_2[x]/(x^4+x^2)$ with the Euclidean algorithm My first thought was successful: $x^4+x^2=x^2(x^2+1)$ and $x^3+x^2+1=x^2(x+1)+1$ so it is its own inverse because $(x^2(x+1)+1)^2\equiv x^4(x+1)^2+1\equiv x^4(x^2+1)+1\equiv1.$
The given solution claims to use the Euclidean algor... | For integers, if you're trying to find an inverse of $a$ mod $n$ using Euclidean algorithm, you would divide $n$ by $a$, find the remainder $r_1$, then divide $a$ by $r_1$, find the remainder $r_2$; divide $r_1$ by $r_2$, etc., until the remainder becomes 1. Here you're looking for the inverse of $x^3+x^2+1$ mod $x^4+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2824411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Show that the Huber-loss based optimization is equivalent to $\ell_1$ norm based. Dear optimization experts,
My apologies for asking probably the well-known relation between the Huber-loss based optimization and $\ell_1$ based optimization. However, I am stuck with a 'first-principles' based proof (without using Moreau... | The idea is much simpler. Use the fact that
$$\min_{\mathbf{x}, \mathbf{z}} f(\mathbf{x}, \mathbf{z}) = \min_{\mathbf{x}} \left\{ \min_{\mathbf{z}} f(\mathbf{x}, \mathbf{z}) \right\}.$$
In your case, the solution of the inner minimization problem is exactly the Huber function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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What's the u-v equation for u and v defined as follows? I can not solve the u-v equation for $u=\frac{1-r^2}{1-2r \cos \theta + r^2}$ for $0 < r < 1$ and $v=\frac{2r\sin\theta}{1-2r \cos \theta + r^2}$ for $0 < r < 1.$ I assume that it is a circle because it is the polar form of a bilinear function $ w=\frac{i(1-z)}{1... | Your expressions should read
\begin{align}
w &= \frac{i(1-z)}{1+z} \\
u+vi &= \frac{2r\sin \theta+i(1-r^2)}{1+2r\cos \theta+r^2} \\
\color{red}{u} &= \frac{2r\sin \theta}{1\color{red}{+}2r\cos \theta+r^2} \\
\color{red}{v} &= \frac{1-r^2}{1\color{red}{+}2r\cos \theta+r^2}
\end{align}
*
*$\theta=0 \implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the value of $\int^{1}_{0}(1+x)^{m}(1-x)^{n} \,\mathrm d x$
Find the value of $$\displaystyle \int^{1}_{0} (1+x)^{m}(1-x)^{n} \,\mathrm d x$$ where $m, n \geq 1$ and $m, n \in \mathbb{N}$.
Let
$$\displaystyle I_{m,n} = \int^{1}_{0}(1+x)^m(1-x)^n \,\mathrm d x = \int^{1}_{0}(2-x)^m\cdot x^n \,\mathrm d x$$
Pu... | By parts,
$$I_{m,n}:=\int_0^1(1+x)^m(1-x)^ndx\\
=\left.\frac{(1+x)^{m+1}(1-x)^n}{m+1}\right|_0^1+\frac n{m+1}\int_0^1(1+x)^{m+1}(1-x)^{n-1}dx.$$
This gives you the recurrence relation
$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}I_{m+1,n-1}.$$
From this,
$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}\left(-\frac1{m+2}+\frac{n-1}{m+2}I_{m+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le... | $f(x)=\frac{2x-7}{2x+5}$ is a concave function on $[5,8]$, hence by the Hermite-Hadamard inequality the wanted integral is bounded between
$$ \frac{1}{2}f(5)+f(6)+f(7)+\frac{1}{2}f(8) = \frac{11043}{11305} > \frac{42}{43}$$
and
$$ f\left(\tfrac{11}{2}\right)+f\left(\tfrac{13}{2}\right)+f\left(\tfrac{15}{2}\right) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
example of an inhomogenenous markov chain I have found a very interesting exercice about inhomogeneous markov chains, which I think you are also interested in and hope someone can solve it.
Let $X_{1},X_{2},...$ be a inhomogeneous markov chain with state space $\{0,1\}$ with transition probabilities
$$
\mathbb{P}(X_{n+... | So the markov matrices is,
$$
P_n = \begin{pmatrix}
1-\frac{1}{n^2} & \frac{1}{n^2} \\
\frac{1}{n^2} & 1-\frac{1}{n^2}
\end{pmatrix} =
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\frac{1}{n^2}
$$ Hence
$$
P_n = I + A\frac{1}{n^2}
$$ Now this means that all matrices $P_n$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Wave problem $u_{tt}=u_{xx}-u_t, 0\le x\le 2\pi$ with $u(x,0)=\phi(x)$, $u_t(x,0)=\psi(x)$ and $u(0,t)=u(2\pi,t)=0$ Solve the wave problem $$\begin{cases}u_{tt}=u_{xx}-u_t, 0\le x\le 2\pi\\ u(0,t)=u(2\pi,t)=0\\ u(x,0)=\phi(x)\\ u_t(x,0)=\psi(x)\end{cases}$$ Write the eigenvalues and eigenfunctions explicitly.
Are they ... | The boundary conditions give
$$ X(x) = \sin\left(\frac{n}{2}x\right) $$
As usual, we drop the remaining constant for now, since it'll absorbed into the constants of $T(t)$
This means $\lambda_n = X''/X = -n^2/4 = -\beta^2$ (you have a sign error). The other equation becomes
$$ T'' + T' + \frac{n^2}{4}T = 0 $$
The two r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2829941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Trying to find the mean of a Random Variable but I cannot do the integration. Please consider the problem below and my partial solution to it. Is it right so far? I do not know how to perform the integration and I am hoping somebody can point me in the right direction.
Thanks,
Bob
Problem:
The joint pdf of a bivariate ... | Don't go polar. Complete the square instead: $x^2 - xy + y^2 = \frac{3}{4}x^2 + (y - \frac{x}{2})^2$. The integrand $$\frac{x}{\sqrt{3 \pi } } e^{ -\frac{2}{3} ( x^2 - xy + y^2 ) }$$ becomes
$$\frac{x}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2} e^{-\frac{2}{3}(y-\frac{x}{2})^2}$$
Integrate it first over $dy$. Can you continue?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ If $M_A(x)=x^3-x$ So $M_{A^2}(x)=x^2-x$ Where $M$ is the minimal polynomial
$x^3-x=x(x^2-1)=x(x-i)(x+i)$ so $A$ is diagonalizable
$A=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \\
\end{pmatrix}P$
$A^2=P^{-1}\begin{pmatrix}
0 & 0 & 0 \\
0 & i & 0 \\
0 & 0 & -i \... | From $M_A(x) = x(x-1)(x+1)$ we conclude that $A$ is diagonalizable and $\sigma(A) = \{0, -1, 1\}$.
Hence $A^2$ is also diagonalizable and $\sigma(A)^2 = \sigma(A)^2 = \{0,1\}$. Therefore, the minimal polynomial $M_{A^2}$ has only linear factors and has zeroes $0, 1$.
The only option is $M_{A^2}(x) = x(x-1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Then general values of $\theta$ in inverse Trigo sum
If $\displaystyle \theta = \tan^{-1}\bigg(2\tan^2 \theta\bigg)-\frac{1}{2}\sin^{-1}\bigg(\frac{3\sin 2 \theta}{5+4\cos 2 \theta}\bigg).$ Then general values of $\theta$ is
Try: Let $\alpha =\tan^{-1}\bigg(2\tan^2 \theta\bigg)\Rightarrow \tan \alpha =2 \tan^2 \thet... | Hint:
multiply the original equation by $2$, take the sine of both sides and remember that $$\cot = \frac 1\tan\\1+\tan^2 = \sec^2\\1 + \cot^2 = \csc^2$$
Added:
$$\theta = \tan^{-1}u - \frac 12 \sin^{-1}v\\2\theta = 2\tan^{-1}u -\sin^{-1}v\\\sin 2\theta = \sin\left(2\tan^{-1}u -\sin^{-1}v\right)\\\sin 2\theta = \sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Equation for product of sequence involving 2 Supposing for all even numbers less that or equal to n say 16,
for example the set:
2, 4, 6, 8, 10, 12, 14, 16
2.4.6.8.10.12.14.16 = 10,321,920 = $2^{15}. 3^{2}. 5^{1}. 7^1$
I would like to find an equation that gives me the products of 2's only not the other factors in the... | Hint
$$2 \cdot 4 \cdot \cdots \cdot 2n = (2 \cdot 1) \cdot (2 \cdot 2) \cdot \cdots \cdot (2 \cdot n) = (2 \cdot 2 \cdot \cdots \cdot 2) \cdot (1 \cdot 2 \cdot \cdots \cdot n) .$$
Now, see this answer, which gives a formula for the highest power of a prime dividing $n!$.
| {
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"url": "https://math.stackexchange.com/questions/2835127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to prove this question by Ramanujan? click here for photo
$$1+2\sum_{k=1}^\infty \frac{1}{(4k)^3-(4k)}= \frac{3}{2}\ln(2)\,.$$
well i have attatched a photo which has been asked to prove without using calculus,but how to solve this using calculus ?
| I have no clue how to solve this without calculus. At least, I do not know how to define the natural logarithm without calculus. I am borrowing some part of this solution from mechanodroid's deleted solution.
First, write
$$\frac{1}{(4k)^3-(4k)}=-\frac{1}{4k}+\frac{1}{2(4k-1)}+\frac{1}{2(4k+1)}$$
for every positive... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Meaning of the sign of a complex number What is the meaning and the practical uses of the sign (signum function) of a complex number $z$ defined as $\frac{z}{|z|}$? Does it also extend to quaternions?
| You can think of a complex number $z=a+ib$ as a vector in the complex plane, and as such it has a direction and a length. The sign of $z$ gives the normalized direction of $z$ as a unit vector lying on the unit circle. We normalise $z$ by dividing by its modulus:
$$\operatorname{sgn}(z)=\frac{z}{|z|}=\frac{a+bi}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Integral of product of error function difference In the course of my research I came across the following integral:
$$\int_{-\infty}^{\infty}\,\left(\operatorname{erf}\left(ax-b\right) -\operatorname{erf}\left(\frac{a}{\gamma} x-b-
\dfrac{ar}{\gamma}\right) \right)
*
\left(\operatorname{erf}\left(cx-d\right)-\operatorn... | Let us take $a \ge 0$, $b \ge 0$, $c \ge 0$, $d \ge 0$, $\gamma \ge 0$ and $r \ge 0$ and let us define:
\begin{eqnarray}
{\mathfrak I}^{(a,b)}_{c,d}(\gamma,r):=
\int_{-\infty}^{\infty}\,\left(\operatorname{erf}\left(ax-b\right) -\operatorname{erf}\left(\frac{a}{\gamma} x-b-
\dfrac{ar}{\gamma}\right) \right)
*
\left(\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate a supremum over the unit sphere of $\mathbb{C}^2$
I want to calculate
$$K=\sup\left\{\left|y\overline{x}+|y|^2\right|^2+1;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\right\}.$$
I try to solve the problem as follows:
let $x=r_1e^{i\theta_1}$ and $y=r_2e^{i\theta_2}$ then
$$ |x|^2+ |y|^2=1 \quad \Rightarrow \... | We have
$$
f(x,y) = (y\bar x+y\bar y)(\bar y x+\bar y y)+1 = |y|^2\left(|x|^2+|y|^2+y\bar x+\bar y x\right)+1 = |y|^2\left(y\bar x+\bar y x+1\right)+1
$$
now calling
$$
x = \rho_x e^{i\phi}\\
y = \rho_y e^{i\psi}
$$
$$
f(\rho_x e^{i\phi},\rho_y e^{i\psi}) = \rho_y^2\left(2\rho_2\rho_y\cos(\phi-\psi)+1\right)+1
$$
so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is
Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a... | Let $\boldsymbol{x}=(a_1,a_2,a_3,a_4) \in \mathbb{R}^4$,
$$E=\boldsymbol{x}
\begin{pmatrix}
2 & -1 & 0 & -1 \\
-1 & 2 & -1 & 0 \\
0 & -1 & 2 & -1 \\
-1 & 0 &-1 & 2
\end{pmatrix}
\boldsymbol{x}^T$$
*
*Eigenvalues: $$\lambda_1=4, \, \lambda_2=\lambda_3=2, \, \lambda_4=0$$
*Unit eigenvectors:
\begin{align}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but ... | It is not too difficult to see that all three roots are real.
The roots are the same as roots of $$ x^3 -2x^2 +1/4 =0$$
Zeros could be approximated by Newton's method as $$ -0.32772....\\ 0.39462....\\ 1.9331...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Trying to evaluate $\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{dx}{1+x^3}$ I would like to work this out:
$$I=\large\int_{0}^{\infty}\frac{\ln(1+x^3)}{1+x^3}\frac{\mathrm dx}{1+x^3}$$
Making a sub: $u=x^3$, $dx=\frac{du}{3x^2}$
$$I=\frac{1}{3}\int_{0}^{\infty}\frac{\ln(1+u)}{u^{2/3}(1+u)^2}\mathrm du$$
Making a s... | In this solution we will not use the beta function or the digamma function.
Put
\begin{equation*}
f(s)=\int_{0}^{\infty}\dfrac{\ln(1+s^3x^3)}{(1+x^3)^2}\, \mathrm{d}x.
\end{equation*}
We want to calculate $f(1)$. However,
\begin{equation*}
f(1) = f(1)-f(0) = \int_{0}^{1}f'(s)\, \mathrm{d}s\tag{1}
\end{equation*}
and
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Factorization Manipulation Let $x$ and $y$ be real numbers. Consider $t=x^2+10y^2-6xy-4y+13$. So what is the $t$ as smallest number? The solution is $9$.
My trying:
$$t=x^2+10y^2-6xy-4y+13$$
$$=x^2+9+10y^2-6xy-4y+4$$
$$=(x+3)^2-3x+2y(5y-2)-6xy+4$$.
So let $x=-3$ and $y=0$. From this I got $t=13$. My answer is false... | Notice that
$$t=x^2+10y^2-6xy-4y+13=x^2-6y \cdot x+10y^2-4y+13,$$which could be regarded as a quadratic function with respect to the variable $x$.
Hence, $$t \geq \dfrac{4\cdot 1 \cdot (10y^2-4y+13)-(-6y)^2}{4 \cdot 1}=y^2-4y+13=(y-2)^2+9\geq 9,$$ with the equality holding if and only if $x=-\dfrac{-6y}{2 \cdot 1}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Equivalence in positive definiteness of $AB$ and $BA$ Let $A,B$ any two real matrices, with $AB$ and $BA$ not necessarily symmetric. I know that $AB$ and $BA$ have the same eigenvalues. Is it true that $AB$ is positive definite iff $BA$ is positive definite? The definition I use for positive definiteness of a general r... | The answer is no.
Consider $A = \begin{bmatrix}2 & 1 \\ 0 & 1\end{bmatrix}$ and $B = \begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix}$.
Then $AB = \begin{bmatrix}1 & 1 \\ -1 & 1\end{bmatrix}$ is positive definite:
$$\begin{bmatrix}x & y\end{bmatrix} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}\begin{bmatrix}x \\ y \end{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | My answer is a little roundabout but without calculus and without pictures or symmetry:
Arithmetic-geometric inequality:
$$ a+b \geq 2\sqrt{ab} $$
Harmonic-geometric inequality and some rearrangement:
$$ \sqrt{ab} \geq \frac{2}{\frac{1}{a}+\frac{1}{b}}$$
$$ \frac{1}{a} + \frac{1}{b}\geq \frac{2}{\sqrt{ab}}$$
Add bot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 0
} |
How to explain powers of $(x+1)^{2^n}$ appearing in the Babylonian approximation of $\sqrt x$? I'm working with this iteration used for approximating square roots and trying to see what I can draw out from it, and in doing so I found something very strange that I can't logically explain. I'm looking for any insight in... | Possible fixed points of the iteration are $\sqrt x$ and $-\sqrt x$,
this suggests to look at
$$
\frac{\rho_{n+1} - \sqrt x}{\rho_{n+1} + \sqrt x} =
\frac{\rho_n^2 - 2 \sqrt x \rho_n + x}{\rho_n^2 + 2 \sqrt x \rho_n + x} =
\left( \frac{\rho_n - \sqrt x}{\rho_n + \sqrt x} \right)^2 \, .
$$
It follows that
$$
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2851719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got a... | Pulling a small rabbit from a hat, consider
$$f(a,b)=3+{b+1\over a+b-2}={3a+4b-5\over a+b-2}$$
It's clear that $a^2+b^2=1$ implies $a+b-2\lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2\le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5\le0$ if $a^2+b^2=1$. Thus $f(a,b)\ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
} |
prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve ... | $$\color{brown}{\textbf{Final edition (13.08.18)}}$$
$\color{green}{\textbf{Task transformations.}}$
The problem is homogeneous with respect to unknowns $x,y,z.$
Let WLOG
$$x+y+z=6,\tag1$$
Then the equivalent inequality is
$$x^2y^2(6-x-y)^2\ge64(3-x)(3-y)(x+y-3),\tag2$$
where
$$(x,y,6-x-y)\in(0,6),$$
or
$$((3-y)+(x+y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving the equality case in triangle inequality Background
When plotted on a real number line, it may be deduced that if
$$a,b,c \in \mathbb{R} $$
$$a < b < c$$
then
$$\left| {a - c} \right| = \left| {a - b} \right| + \left| {b - c} \right|$$
Problem
But the problem is with the proof. How can the above statement be ... | Since both sides of the equality are nonnegative, we can square it:
\begin{align}
|a-c| = |a-b| + |b-c| &\iff |a-c|^2 = (|a-b| + |b-c|)^2\\
&\iff a^2-2ac+c^2 = a^2-2ab+b^2+b^2-2bc+c^2 + 2|a-b||b-c|\\
&\iff 0 = b^2 - ab - bc + ac + |a-b||b-c|\\
&\iff 0 = -(a-b)(b-c) + |a-b||b-c|\\
&\iff (a-b)(b-c) = |a-b||b-c|\\
&\iff ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
prove this inequality $\sum\frac{a}{b}\ge \sum a^2$ Let $a,b,c>0$ such $a+b+c=3$, show that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge a^2+b^2+c^2$$
I have show this not stronger inequality
$$\sum\dfrac{a}{b}\ge \sum a$$
| Let $\sum $ denote cyclic sum, then observe:
$$\sum a(a-b)^2(b-2c)^2 \geqslant 0$$
$$\implies \sum ab^4 + \sum a^3b^2 + 2\sum a^2b^3+4abc\sum ab -8abc\sum a^2 \geqslant 0 $$
$$\implies 2\left( \sum a \right)^2\sum ab^2 + abc\left(\sum a\right)^2 \geqslant 21abc\sum a^2$$
$$\implies 6 \sum \frac{a}b + 3 \geqslant 7\sum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof involving polynomial roots From USAMO 1977:
"If $a$ and $b$ are two roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$."
The solutions I've seen for this problem all involve manipulating Vieta's equations relating polynomial roots and coefficients. In particular, the solutions feel a bit l... | Considering
\begin{align}
x^4+x^3-1 &= (x-a)(x-b)(x-c)(x-d) \\
S_1 &= a+b+c+d \\
&= -1 \\
S_2 &= ab+ac+ad+bc+bd+cd \\
&= 0 \\
S_3 &= bcd+acd+abd+abc \\
&= 0 \\
S_4 &= abcd \\
&= -1 \end{align}
The sextic has a nice symmetry so that all the roots are $ab$, $ac$, $ad$, $bc$, $bd$ and $cd$.
Now, using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
inequality proving for intersection AB and CD are line segments, $AB=CD=1$, intersecting in point O, $\enspace$ $AB\cap CD =O$, $\angle AOC=60^{\circ}$. Prove that $AC+BD\geq1$. $\enspace$
What I tried:
$AO+BO=1$, $\enspace$ $CO+DO=1$
$AO=x$, $\enspace$ $CO=y$
$BO=1-x$, $\enspace$ $DO=1-y$
Law of cosines: $AC^2=x^2+y^... | Let $ABDK$ be a parallelogram.
Thus, $AK=DB$, $AD=AB=DC=1$ and $\measuredangle ADC=\measuredangle AOC=60^{\circ}.$
Hence, $\Delta KDC$ is an equilateral triangle, which says $KC=1$ and by the triangle inequality
$$AC+BD=KA+AC\geq KC=1.$$
Also, you can end your work by the triangle inequality again:
$$AC+BD=\sqrt{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate max/min of a 3 variable function, restricted to g(x,y,z)=0 Calculate extrema of $f(x,y,z)=xe^{yz}$ on boundary $3x^2 +y^2 +z^2 =27$
*
*I did Lagrange multiplier (4 equations 4 variable) but I can't figure out how to solve that system.
$f_x + \lambda g_x$,$f_y + \lambda g_y,f_z + \lambda g_z,g(x,y,z)=0$
$... | Hint.
Make $\mu = \frac{\lambda}{e^{yz}}$ and then solve
$$
1+\mu 6 x = 0\\
x z + \mu 2 y = 0\\
xy + \mu 2z = 0\\
3x^2+y^2+z^2-27=0
$$
giving
$$
\left[
\begin{array}{ccccc}
x & y & z & \mu & f \\\
-3 & 0 & 0 & \frac{1}{18} & 0 \\
3 & 0 & 0 & -\frac{1}{18} & 0 \\
-\frac{1}{\sqrt{3}} & -\sqrt{13} & -\sqrt{13} & \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{1}{x}+\frac{1}{y}= \frac{1}{2007}$ The number of positive integral pairs $(x<y)$ such that $\frac{1}{x}+\frac{1}{y}= \frac{1}{2007}$
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
| $$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4\cdot 223^2=(x-2007)(y-2007)$$
\begin{align}3^4\cdot 223^2 &=(3^0) \cdot (3^4\cdot 223^2)\\
&=(3^1) \cdot (3^3\cdot 223^2)
\\
&=(3^2) \cdot (3^2\cdot 223^2)
\\
&=(3^3) \cdot (3^1\cdot 223^2)\\
&=(3^4) \cdot (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limits for sequences: Prove that $\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$
Prove that $$\lim_{n \rightarrow \infty} \frac{2n^2 - 3n + 5}{9 - n^2} = -2$$
Note, this is a sequence.
I know that a sequence $(c_n)_{n=1}^{\infty}$ converges to a finite value L if for every $\epsilon > 0$, there exis... | If we take $n > 3+\frac6\varepsilon$ we have $ \frac{3}{n-3} < \frac{\varepsilon}2$, and if we take $n > \sqrt{\frac{28}\varepsilon + 9}$ we have $ \frac{14}{n^2-9} < \frac{\varepsilon}2$.
Therefore for $n \ge \max\left\{4, 3+\frac6\varepsilon, \sqrt{\frac{28}\varepsilon + 9}\right\}$ we have
\begin{align}
\left|\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to find the partial sum of $n/2^n$? I'm trying to find a formula for the partial sum of $n/2^n$.
I've tried this so far...
$$ S_n = \frac12 + \frac24 + \frac38 + \cdots + \frac{n}{2^n} $$
Then I tried to find a way to eliminate most of the terms by multiplying the whole sequence by $\frac{2^n}n\cdot\frac{n+1}{2^{n+... | Observe
$$2S_n = 1+ \frac{1+1}2 + \frac{2+1}4 + \cdots + \frac{(n-1)+1}{2^{n-1}}
\\=S_{n-1}+1+\frac12+\frac14\cdots+\frac1{2^{n-1}}
\\=S_n-\frac n{2^n}+2-\frac1{2^{n-1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the positive value of $x$ satisfying the given equation $${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
Tried removing roots. Got a degree $6$ equation which I didn't no how to solve. Also tried substituting $x = \sec(y)$ but couldn't even come close to the solution.
| HINT
We can try with
$${\sqrt {x^2- 1\over x}} + {\sqrt{x-1\over x}} = x$$
$$\sqrt {x+1}{\sqrt {x- 1\over x}} + {\sqrt{x-1\over x}} = x$$
$$(\sqrt {x+1}+1){\sqrt {x- 1\over x}} = x$$
$${\sqrt {x- 1\over x}} = \frac{x}{\sqrt {x+1}+1}\frac{\sqrt {x+1}-1}{\sqrt {x+1}-1}=\sqrt {x+1}-1$$
$${\sqrt {x- 1}} =\sqrt {x^2+x}-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Maximum $n$ such that ${n \choose k} \,2^{1 - {k \choose 2}} < 1$ (where $k$ is a constant) Maximum value of $n$ such that the expression given below does not exceed 1. ($k$ is a constant)
$${n \choose k} 2^{1 - {k \choose 2}} < 1$$
Any hints on how to approach this problem.
Thanks.
Edit:
I tried doing the problem thi... | I shall prove that, for a positive integer $k$, the maximum natural number $n$, denoted by $n_k$, such that
$$\binom{n}{k}\leq 2^{\binom{k}{2}-1}$$
satisfies
$$\lim_{k\to\infty}\,\frac{n_k}{k\,2^{\frac{k}{2}}}=\frac{1}{\text{e}\sqrt{2}}\,.\tag{*}$$
I also provide some bounds for the value of $n_k$.
First, we note that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_{-1}^{1} \cot^{-1} \left(\frac{1}{\sqrt{1-x^2}}\right) \cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-(x^2)^{|x|}}}\right)$
$$\int_{-1}^{1} \left(\cot^{-1} \dfrac{1}{\sqrt{1-x^2}}\right) \left(\cot^{-1}\dfrac{x}{\sqrt{1-(x^2)^{|x|}}}\right)= \dfrac{\pi^2(\sqrt a-\sqrt b )}{\sqrt c}$$
, where a,b, c are na... | Using the property $$\int_a^b f(x)=\int_a^b f(a+b-x)$$
The integral changes to $$I=\pi\int_0^1 \arctan \sqrt{1-x^2} dx$$
You might know the property that $$\int_a^b f(x)+\int_{f(a)}^{f(b)} f^{-1}(x) dx=-af(a)+bf(b)$$
Let $$J=\int_0^1 \arctan \sqrt{1-x^2} dx+\int_{\frac {\pi}{4}}^0 \sqrt {1-\tan ^2x } dx$$ Using above... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Laurent Series $e^{\frac1{1-z}}$, $|z|>1$. I tried to find Laurent expansion for:
$e^{\frac1{1-z}}$, $|z|>1$.
I tried next: $\frac1{1-z}=-\sum_{n=1}^{\infty}\frac1{z^n}$,
then using $e^{\frac1{1-z}}=1+\frac1{1-z}+\frac{1}{{2!(1-z)}^2}+\frac{1}{{3!(1-z)}^3}+...$
Then I have $e^{\frac1{1-z}}$ = $1$ + $(-\frac1{z}-\frac... | With $z=\dfrac{1}{w}$ we have $|w|<1$ then by $e^z$ expansion
\begin{align}
e^{\frac{1}{1-z}}
&= e^{\frac{-w}{1-w}} \\
&= \sum_{n\geq0}\dfrac{1}{n!}\left(\frac{-w}{1-w}\right)^n \\
&= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1-w\right)^{-n} \\
&= \sum_{n\geq0}\dfrac{(-w)^n}{n!}\left(1+nw+\dfrac{n(n+1)}{2}w^2+\cdots\right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification) I have to solve the following integral
$$\int{\frac{dx}{x^2(1-x^2)}}$$
What I've got:
\begin{split}
\int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\
&=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\
&=\int{\frac{dx}{x^2}}+\in... | $x = \tanh(u)$ then $dx = \operatorname{sech}^2(u) du$
$$
\begin{align}
\int \frac{dx}{x^2(1-x^2)}
&= \int \frac{\operatorname{sech}^2(u) du}{\tanh^2(u)(1-\tanh^2(u))}\\
&= \int \frac{\operatorname{sech}^2(u) du}{\tanh^2(u)(\operatorname{sech}^2(u))}\\
&= \int \frac{du}{\tanh^2(u)}\\
&= \int \frac{\cosh^2(u)}{\sinh^2(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2869976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Finding the area enclosed by the graphs of: $y=x\left(x-4\right)^2$ ,$y=4x-x^2$.
Find the area enclosed by the graphs of:
$y=x\left(x-4\right)^2$
, $y=4x-x^2$.
My answer was $\frac{7}{12}+\frac{45}{4}$, but apparently this is wrong: the right answer is $\frac{37}{2}$ according to the textbook.
Intersections occ... | You are correct. That is a typo. Furthermore it propagates all the way to the end. Your answer is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding integer solutions to $ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $ I was browsing through facebook and came across this image:
I was wondering if we can find more examples where this happens?
I guess this reduces to finding integer solutions for the equation
$$ \frac{a^3+b^3}{a^3+c^3} = \frac{a+b}{a+c} $$ for... | $$\frac{a^3+b^3}{a^3+c^3}=\frac{(a+b)(a^2-ab+b^2)}{(a+c)(a^2-ac+c^2)}=\frac{a+b}{a+c}$$
If $a+c \neq 0$ and $a+b \neq 0,$ then $$a^2-ab+b^2=a^2-ac+c^2,$$namely $$(b+c-a)(b-c)=0.$$
If $b=c$, the case is trivial. If $b \neq c$, then $$b+c=a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that: $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$ Given three positive numbers a,b,c satisfying $a+b+c=3$. Show that $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$
Things I have done so far:
$$a+b+c=3\Rightarrow b+c=3-a;0<a<3$$
$$\Rightarrow \frac{1}{(b+c)^2+a^2}=\frac{1}{(3-a)^2+a^2}=\fra... | Hint: Use the Tangent Line method.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{(3-a)^2+a^2}\leq\frac{3}{5}$$ or
$$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or
$$\sum_{cyc}\left(\frac{a^2-3a+2}{2a^2-6a+9}+\frac{1}{5}(a-1)\right)\geq0$$ or
$$\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| $\bmod n^2\!+1\!:\,\ \color{#c00}{n^2\equiv -1}\,\Rightarrow\, n^6\equiv (\color{#c00}{n^2})^3\equiv (\color{#c00}{-1})^3\equiv -1$ so $\,n^6-7\equiv -8$ is the remainder,
hence the quotient is $\ \dfrac{(n^6-7)-8}{n^2+1} = \dfrac{n^6+1}{n^2+1} = n^4 - n^2 + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2... | A slightly different approach: you have $a^2+b^2=7$ and $a^2 b^2=1$ with $a,b\in\mathbb{R}^+$, such that
$$ \operatorname*{Res}_{x=ia}\frac{1}{(x^2+a^2)(x^2+b^2)} = \lim_{x\to ia}\frac{1}{(x+ia)(x^2+b^2)} = \frac{1}{2i}\cdot \frac{1}{a(b^2-a^2)}$$
and similarly
$$ \operatorname*{Res}_{x=ib}\frac{1}{(x^2+a^2)(x^2+b^2)} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating... | Yes, the solution is correct.
You are using:
$$\arctan a\pm \arctan b=\arctan\frac{a\pm b}{1\mp ab}$$
to simplify:
$$x\tan^{-1}(x+2)-x\tan^{-1}x=x\tan^{-1}\frac{(x+2)-x}{1+(x+2)x}=x\tan^{-1}\frac{2}{1+2x+x^2}.$$
However, to evaluate the limit you can avoid using the L'Hospital's rule:
$$\lim_{x \to \infty} x \tan^{-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method Solving $(x^2 + 1) dy + 4 xy dx = x dx$ using separable variable method and integrating factor method
By integrating factor method -
I put it into the form of $ \frac{dy}{dx} + \frac{4x}{x^2 + 1} \cdot y = \frac{x}{x... | Yes, you have.
In separable form ,
$$\displaystyle \int \dfrac{1}{1-4y}~dy=\int\dfrac{x}{x^{2}+1}~dx$$
$\dfrac{-1}{4}\log(1-4y)=\dfrac{1}{2}\log(x^{2}+1)$ You have missed minus sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2881354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Non-PSD matrix with two different PSD submatrixes Let $A$ be an $n \times n$. Define $A_{-i}$ to be the matrix $A$ without the $i$-th column and row. For instance
$$
A=
\begin{pmatrix}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{pmatrix}
\implies
A_{-2}=
\begin{pmatrix}
1 & 3\\
7 & 9
\end{pmatrix}
$$
Is it possible to find ... | $$
\left(
\begin{array}{ccc}
5 & -3 & -3 \\
-3 & 5 & -3 \\
-3 & -3 & 5 \\
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc}
7 & -4 & -5 \\
-4 & 8 & -6 \\
-5 & -6 & 9 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
First four numbers of a sequence are $2,0,1,8,$. Each next is the last digit of the sum of the preceding four numbers. Will $2,0,1,8$ show up again? Let $a_n$ is a sequence of numbers. The firs four numbers are $2,0,1,8$ and each following number is the last digit of the sum of the preceding four numbers. The first ten... | Notice that given any four sequential numbers in the sequence, say $a_n, a_{n+1}, a_{n+2}, a_{n+3}$ with $n>1$, $a_{n-1}$ is determined uniquely; in fact, $a_{n-1} = a_{n+3} - a_{n+2} - a_{n+1} - a_{n} \bmod 10$.
Since there are only finitely many four-tuples of the digits $0-9$, the sequence of four-tuples must eventu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2884707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Taking square roots modulo $2^N$ I was trying to solve $y^2 - y \equiv 16 \pmod{512}$ by completing the square.
Here is my solution.
\begin{align}
y^2 - y &\equiv 16 \pmod{512} \\
4y^2 - 4y + 1 &\equiv 65 \pmod{512} \\
(2y-1)^2 &\equiv 65 \pmod{512} \\
2y - 1 &\equiv \pm 33 \pmod{512} &\text{Found by pointwi... | This isn't a general method, but here's a neat trick if $2^k|a-1$ for some large enough $k$.
Claim. If $x^2\equiv a\bmod 2^n$ and $a\equiv 1\bmod 2^k$, with $n\leq 2k$, then
$$x\equiv \pm \frac{a+1}{2}\bmod 2^{n-1}.$$
Proof. As there are only two residue classes $\bmod 2^{n-1}$ that will satisfy this, it suffices to sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Simple limit with asymptotic approach. Where's the error? Simply calculus question about a limit.
I don't understand why I'm wrong, I have to calculate
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\sin x}{1 - \cos\sqrt{x^3}}
$$
Using asymptotics, limits and De l'Hospital rule I would write these passages...
$$... | The mistake lies at the beginning :
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}(x-\frac{x^3}{6})}{1 - \cos\sqrt{x^3}} =\frac56$$
$$ \lim_{x \rightarrow 0} \frac{2x - \sqrt[3]{8 - x^2}\:x}{1 - \cos\sqrt{x^3}} =\frac16$$
At denominator $1-\cos(x^{3/2})$ is equivalent to $\frac12 x^3$. Thus one cannot neglect ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Product of binary matrices with binary eigenvalues Consider two binary matrices with obvious patterns:
$$
C=
\begin{bmatrix}
1 &0 &0 &0 &0 &0 &0\\
1 &0 &0 &0 &0 &0 &0\\
0 &1 &0 &0 &0 &0 &0\\
0 &1 &0 &0 &0 &0 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &1 &0 &0 &0 &0\\
0 &0 &0 &1 &0 &0 &0
\end{bmatrix}
$$
and
$$
T=
\begin{bmatrix}
... | Some thoughts:
Note that $T = I + N$, where $I$ is the identity matrix and
$$
N = \pmatrix{0&1\\&0&1\\&&0&1\\&&&0&1\\&&&&0&1\\&&&&&0&1\\&&&&&&0}
$$
Notably, $N^7 = 0$. Because $NI = IN$, we can compute $T^n = (I + N)^n$ by binomial expansion. That is, we have
$$
T^n = \binom n0 I + \binom n1 N + \cdots + \binom n6 N^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Parametrizing the intersection of a cylinder and a sphere I need to parametrize the intersection between the cylinder $
x^2 + y^2= \frac{1}{4}$ and the sphere $(x+ \frac{1}{2})^2 + y^2 +z^2 = 1$.
I tried parametrizing the first equation which gives $r(t) = (\frac{cos(t)}{2}, \frac{sin(t)}{2})$ since the radius is 1/2.... | $x^2+y^2=\frac{1}{4}$
$\left( x+\frac{1}{2} \right)^2 + y^2 + z^2 = 1$
Expand
$x^2 + x + \frac{1}{4} + y^2 + z^2 = 1$
Collect
$\left(x^2+y^2\right)+x+\frac{1}{4}+z^2=1$
Substitute first equation
$\frac{1}{4}+x+\frac{1}{4}+z^2=1$
The key equations now being$\ldots$
$x+z^2=\frac{1}{2}$
$x^2+y^2=\frac{1}{4}$
We pick $z=t$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding closed form for $\sum_{k=1}^n k2^{k-1}$ I am trying to use the perturbation method to find a closed form for:
$$
S_n = \sum_{k=1}^n k2^{k-1}
$$
This is what I’ve tried so far:
$$
S_n + (n+1)2^n = 1 + \sum_{k=2}^{n+1} k2^{k-1}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{k=1}^{n} (k+1)2^{k}
$$
$$
S_n + (n+1)2^n = 1 + \sum_{... | You have
$$S_n=\sum_{k=1}^n k 2^{k-1}=1\cdot 2^0 +2\cdot 2^1+3\cdot 2^2+\cdots+n2^{n-1}=$$
$$=\big(2^0+2^1+2^2+\cdots+2^{n-1}\big)+\big(1\cdot 2^1+2\cdot 2^2+\cdots+(n-1)2^{n-1}\big)=\quad (*)$$
$$=\sum_{k=0}^{n-1}2^k+2\big(1\cdot 2^0+2\cdot 2^1+\cdots+(n-1)2^{n-2}\big)=\quad (**)$$
$$=(2^n-1)+2 S_{n-1}.$$
So,
$$S_n=(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the integers that double if the first and last digits are swapped Find the integers that double if their first and last digits are swapped.
In other words find the digits: $a$, $b$ and $n_1$ ... $n_m$ in such a way that:
[$a$ $n_1$ $n_2$ ... $n_m$ $b$] $\times$ $2$ = [$b$ $n_1$ $n_2$ ... $n_m$ $a$]
For example, us... | I'll give an algebraic proof. Your answer is good, though.
Given a number $l$, greater than $10$ since this is obviously not true for single digit numbers, write it as $l = 10^nx + 10y + z$, where $x$ and $z$ are single digit numbers i.e. $10^n \leq l < 10^{n+1}$. Essentially, we are isolating the first and last digits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many solutions to each of the equations $2^x+3^y=5^z$, $2^x+5^y=3^z$, $3^x+5^y=2^z?$ Let $x,y,z\in\Bbb{N}$ How many total solutions are there to each of the three distinct equations below?
$$2^x+3^y=5^z \tag 1$$
$$2^x+5^y=3^z \tag 2$$
$$3^x+5^y=2^z \tag 3$$
I found 3 solutions to equation (1), 4 solutions to equati... | Mahler proved this equation has only finite solutions.$a^x+b^y=c^z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
A divisibility problem without using induction or exhaustion
Problem: Determine the values of $X$ such that $24\mid(18+37X-6X^2-X^3)$ without using induction or exhaustion.
This question was motivated by the solution set $$[a\quad b\quad c\quad d]=\left[\frac34\quad\frac{37}{24}\quad-\frac14\quad-\frac1{24}\right]$$... | It is necessary and sufficient that $18+37X-6X^2-X^3$ is divisble by $8$ and by $3$. Reducing mod $3$ yields
$$18+37X-6X^2-X^3\equiv X+2X^3\pmod{3},$$
where $X+2X^3\equiv-X(X+1)(X+2)\pmod{3}$, so this is satisfied for all $X$.
Reducing mod $8$ yields
$$18+37X-6X^2-X^3\equiv2+5X+2X^2+7X^3\pmod{8},$$
which is zero precis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of $\lim\limits_{x \rightarrow \infty}{(x-\sqrt \frac{x^3+x}{x+1})}$ - calculation correct? I just want to know if this way of getting the solution is correct.
We calculate $\lim\limits_{x \rightarrow \infty} (x-\sqrt \frac{x^3+x}{x+1}) = \frac {1}{2}$.
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \c... | I think, there is a little mistake in your computations.
It should be
\begin{align}
& \left(x-\sqrt \frac{x^3+x}{x+1}\,\right) \cdot \frac{x+\sqrt \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{x^2 - \frac{x^3+x}{x+1}}{x+\sqrt \frac{x^3+x}{x+1}} = \frac{\frac{x^2(x+1) - x^3-x}{x+1} }{x+\sqrt \frac{x^3+x}{x+1}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find lines of invariant points? Every time I try a question on this topic I get it wrong.
My textbook says:
Invariant points satisfy $B\begin{pmatrix}u\\ v\end{pmatrix}=\begin{pmatrix}u\\ v\end{pmatrix}$
Re-write this as a system of equations.
Check whether both equations are in fact the same.
If so, they give ... | Let
$$\mathbf{B} = \left [ \begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix} \right ], \quad \mathbf{p} = \left [ \begin{matrix} u \\ v \end{matrix} \right ]$$
If $\mathbf{p}$ is an invariant point with respect to $\mathbf{B}$, then
$$\mathbf{B} \mathbf{p} = \mathbf{p} \tag{1}\label{NA1}$$
which is equival... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $10^n \gt 6n^2+n$ Show for all $n \in \mathbb{N}$ $(n\geq1):$ $10^n \gt 6n^2+n$
My solution:
Base case: For $n=1$
$10^1 \gt 6 \cdot 1^2+1$
Inductive hypothesis:
$10^n \gt 6n^2+n \Rightarrow 10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
Inductive step:
$10^{n+1} \gt 6\cdot(n+1)^2+(n+1)$
$\Rightarrow$ $10^{n+1} \gt 6(n^2+... | You almost finished. Now you can transfer all to the left and group:
$$60n^2+10n \gt 6n^2+13n+7 \iff \\
54n^2-3n-7>0 \iff \\
44n^2+3n^2-3n+7n^2-7>0 \iff \\
44n^2+3n(n-1)+7(n^2-1)>0,$$
which is true because $n>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Let $ABCD$ be a parallelogram. Show that $\angle BQD = 90^ \circ$.
Let $ABCD$ be a parallelogram. There is point $P$ on the $BD$ such that $AP=BD$. $Q$ is the midpoint of the $CP$. Show that $\angle BQD = 90^ \circ$.
| Alternatively, draw the second diagonal and denote the center with $O$.
$\hspace{4cm}$
Using the formulas of medians:
$$\begin{align}DQ^2&=\frac{2CD^2+2PD^2-PC^2}{4};\\
BQ^2&=\frac{2BC^2+2BP^2-PC^2}{4};\\
OP^2&=\frac{2AP^2+2PC^2-AC^2}{4} \Rightarrow \color{red}{AC^2}=2AP^2+2PC^2-4OP^2=\color{red}{2BD^2+2PC^2-4OP^2}.\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Induction. Am I missing something or is there a mistake in the question? $$\sum_{k=1}^n k*3^k=\frac {3(3^n(2n-1)+1)} 4 $$
So let f(n)= $\sum_{k=1}^n k*3^k $
and g(n)=$\frac {3(3^n(2n-1)+1)} 4$
By induction hypothesis, $f(n+1) = f(n) + (n+1)3^{n+1} \overset{\text{i.h.}}{=} g(n) + (n+1) 3^{n+1} = g(n+1).$
$$\frac{3(3^{n... | We have
$$\sum_{k=1}^{n+1} k\cdot 3^k=(n+1)\cdot 3^{n+1}+\sum_{k=1}^n k\cdot 3^k\stackrel{Ind. Hyp.}=(n+1)\cdot 3^{n+1}+\frac {3(3^n(2n-1)+1)} 4=$$
$$=\frac {4(n+1)\cdot 3^{n+1}+3^{n+1}(2n-1)+3} 4=\frac {3^{n+1}(6n+3)+3} 4=$$
$$=\frac {3(3^{n+1}(2(n+1)-1)+1)} 4=f(n+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Evaluating an Integral with a $\sqrt{1-x^2}$ in it Below is a problem I did but my answer does not match the book's table of
integrals. I would like to know where I went wrong.
Thanks
Bob
Problem:
Evaluate the following integral:
\begin{eqnarray*}
\int x^2 \sqrt{ 1 - x^2 } \,\, dx \\
\end{eqnarray*}
Answer
\begin{eqna... | Let's try a different method:
$$
x^2\sqrt{1-x^2}=\frac{x^2-x^4}{\sqrt{1-x^2}}=(x^3-x)\frac{-x}{\sqrt{1-x^2}}
$$
Let's do integration by parts:
\begin{align}
I&=\int x^2\sqrt{1-x^2}\,dx\\[6px]
&=\int(x^3-x)\frac{-x}{\sqrt{1-x^2}}\,dx\\[6px]
&=(x^3-x)\sqrt{1-x^2}-\int(3x^2-1)\sqrt{1-x^2}\,dx\\[6px]
&=(x^3-x)\sqrt{1-x^2}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I determine general formula of this sequence? I am trying to find general formula of the sequence $(x_n)$ defined by
$$x_1=1, \quad x_{n+1}=\dfrac{7x_n + 5}{x_n + 3}, \quad \forall n>1.$$
I tried
put $y_n = x_n + 3$, then $y_1=4$ and
$$\quad y_{n+1}=\dfrac{7(y_n-3) + 5}{y_n }=7 - \dfrac{16}{y_n}, \quad \forall... | Here is a method that I read from a book. Yet I did not think deeply why it works in general.
If there exists real numbers $\alpha$, $\beta$ and $r$ such that
$$\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=r\cdot\frac{a_n-\beta}{a_n-\alpha}$$
for all $n\in\mathbb{N}$, then the sequence $\{b_n\}$, where $b_n=\frac{a_n-\beta}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the probability that the white ball labelled $1$ is drawn before all the black balls.
Suppose in an urn there are $20$ black balls labelled $1,2, \ldots , 20$ and $10$ white balls labelled $1,2, \ldots ,10$. Balls are drawn one by one without replacement. Find the probability that the white ball labelled $1$ is d... | To evaluate $\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}$:
\begin{align}
\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\frac{(30-k)!}{20!(10-k)!} \\
&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\binom{30-k}{20} \\
&=\frac{9!\cdot 20!}{30!}\left[\binom{20}{20}+\binom{21}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Is this proof correct (Rationality of a number)? Is $\sqrt[3] {3}+\sqrt[3]{9} $ a rational number? My answer is no, and there is my proof. I would like to know if this is correct:
Suppose this is rational. So there are positive integers $m,n$ such that $$\sqrt[3]{3}+\sqrt[3]{9}=\sqrt[3]{3}(1+\sqrt[3]{3})=\frac{m}{n}$$... | Your proof is correct.
Let $\alpha = \sqrt[3]{3} + \sqrt[3]{9} = \sqrt[3]{3}(1+\sqrt[3]{3})$. We have
$$\alpha^3 = 3(1+\sqrt[3]{3})^3 = 3(3 + 3\sqrt[3]{3} + 3\sqrt[3]{9} + 3) = 12 + 9(\sqrt[3]{3} + \sqrt[3]{9}) = 12 + 9\alpha$$
Hence $\alpha^3 - 9\alpha -12 = 0$.
However, the polynomial $x^3-9x-12$ is irreducible over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of the power series $\sum_{n=1}^\infty n*(n+1)*x^n$ $\sum_{n=1}^\infty n*(n+1)*x^n$
Hello everyone, I need help in solving the question above.
I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don... | Just to make the problem more general.
Consider
$$\sum_{n=1}^\infty (an^2+bn+c)\,x^n$$ rewrite $n^2=n(n-1)+n$ which makes
$$an^2+bn+c=a n(n-1)+a n+bn+c=an(n-1)+(a+b)n+c$$
$$\sum_{n=1}^\infty (an^2+bn+c)\,x^n=a \sum_{n=1}^\infty n(n-1)x^n+(a+b)\sum_{n=1}^\infty nx^n+c\sum_{n=1}^\infty x^n$$
$$\sum_{n=1}^\infty (an^2+bn+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Aperiodicity of $\sin[2x+\cos(\sqrt{2}x)]$
Is the function $f(x)=\sin[2x+\cos(\sqrt{2} \cdot x)]$ periodic?
I don't know how to prove that this function is not periodic. Can you help me out?
| First, some intuition. Here's a graph of $f(x)=\sin(2x+\cos(\sqrt{2} \cdot x))$:
We have to prove that the function is aperiodic. From the graph it appears that something stronger is true: even if we focus on any specific $y$-value $y_0$, the values of $x$ where $f(x) = y_0$ will not be evenly spaced. So let's pick $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
positive integer solutions to $x^3+y^3=3^z$ I am seeking all positive integer solutions to the equation $x^3+y^3=3^z$.
After doing number crunching, I think there are no solutions. But I am unable to prove it.
Attempt
If $x$ and $y$ have common divisor $d$, we have $d^3(m^3+n^3)=3^z$. So $d$ must be a power of $3$, and... | Solution by LTE, let $d=(x,y)>1$, $x=dx_1$, $y=dy_1$, $(x_1,y_q)=1$ then $$d^3\mid x^3+y^3=3^z$$ then $d\mid 3^a$ for a positive integer $a<z/3$. Dividing by $d^3$ we obtain $x^3_1+y^3_1=3^b$, for a positive integer $b=z-a$. This is a equation similar to the original, then we can assume $(x,y)=1$. If $3\nmid x+y$ then ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.