Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Product of generating functions Let $f(x) = \sum_{i=0}^\infty a_ix^i$ and $g(x) = \sum_{i=0}^\infty b_ix^i$ where $a_n = 1$ and $b_n = 2^n$ for all natural numbers $n$
What are the first three terms of the sequence generated by $f(x)g(x)$
?
So I know that $f(x)$ will generate $\{1,1,1,1,1,.... \}$ and $g(x)$ will g... | In general:
$\begin{align}
\left( \sum_{n \ge 0} a_n z^n \right)
\cdot \left( \sum_{n \ge 0} b_n z^n \right)
= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} a_k b_{n - k} \right) z^n
\end{align}$
In your particular case, $a_n = 1$, switching the product around:
$\begin{align}
\left( \sum_{n \ge 0} b_n z^n \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can two dozen identical robots be assigned to four assembly lines?
In how many ways can two dozen identical robots be assigned to four
assembly lines with
(a) at least three robots assigned to each line?
(b) at least three, but no more than nine assigned to each line ?
Two dozen = 24,
Now we use ... | Hint: We have, by the usual formula for the sum of a finite geometric progression, that if $x\ne 0$ then
$$1+x+x^2+\cdots+x^6=\frac{1-x^7}{1-x}.$$
It follows that
$$(1+x+x^2+\cdots+x^6)^4=(1-x^7)^4(1-x)^{-4}.$$
Note that we only need two easy terms of the expansion of $(1-x^7)^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is the sum of the series $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$? $x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n}$
I'm not sure if this can even be summed. Any help is appreciated.
| \begin{align}
& x^{n-1} + 2x^{n-2} + 3x^{n-3} + 4x^{n-4} + \cdots + nx^{n-n} \\[10pt]
& x^{n-1}\left( 1 + \frac 2 x + \frac 3 {x^2} + \frac 4 {x^3} + \cdots + \frac n {x^{n-1}} \right) \\[10pt]
= {} & x^{n-1}\left( 1 + 2y + 3y^2 + 4y^3 + \cdots + ny^{n-1} \right) & (\text{where }y = 1/x) \\[10pt]
= {} & x^{n-1} \frac d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Computing $\sum{\frac{1}{m^2+n^2}}$ I want to prove that $\sum_{1\leq m^2+n^2\leq R^2}{\frac{1}{m^2+n^2}}=2\pi\log R+O(1)$ as $R\rightarrow\infty$.
For this, I'm trying to approximate the sum by using the integral $\int_{1\leq r\leq R}{\frac{1}{x^2+y^2}}dxdy=\int_{0}^{2\pi}\int_{1}^{R}{\frac{1}{r}}drd\theta=2\pi\log R$... | Just an idea:
Let's use
$$x-1<[x]\leq x$$
$$\sum \frac{1}{m^2+n^2}=\int \frac{1}{[x]^2+[y]^2}$$
Then
$$\int \frac{1}{x^2+y^2}\leq\int \frac{1}{[x]^2+[y]^2}<\int \frac{1}{(x-1)^2+(y-1)^2}$$
And hopefully we can show
$$\lim_{R\to\infty}\int \frac{1}{x^2+y^2}=\lim_{R\to\infty}\int \frac{1}{(x-1)^2+(y-1)^2}$$
?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$, If $z_1$ and $z_2$ are two complex numbers,and if $z_1^3-3z_1^2z_2=2,3z_1z_2^2-z_2^3=11$,then find the value of $|z_1^2+z_2^2|$.
$z_1^3-3z_1^2z_2=2$
$3z_1z_2^2-z_2^3=11$
Adding them,we get
$z_1^3-3z_1^2z_2+3z_1z_2^2-z_2^3=13$
$(z... | $$z_{1}(z^2_{1}-3z^2_{2}) = 2\Rightarrow z^3_{1}-3z_{1}z^2_{2} = 2\cdots \cdots (1)$$
Similarly $$z_{2}(3z^2_{1}-z_{2}^2) = 11\Rightarrow 3z^2_{1}z_{2}-z^3_{2} = 11\cdots (2)\times i$$
Now Add and Subtract these two equation, we get
$$(z_{1}+iz_{2})^3 = 2+11i$$
$$(z_{1}-iz_{2})^3 = 2-11i$$
So $$(z^2_{1}+z^2_{2})^3 = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Determine the general solution of the inhomogenous system (Difference&Differential Equations) Determine the general solution of the system
$y(n+1)=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} y(n) + 4^n \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix},\quad n \in \mathbf{N}$
The homogenous part of finding general solution ... | $\begin{bmatrix} a_{1}\\ b_{1} \\ \end{bmatrix}
=\begin{bmatrix} 5 & 1\\ -1 & 3 \\ \end{bmatrix} \begin{bmatrix} a_0\\ b_0 \\ \end{bmatrix} + \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=
\begin{bmatrix} 5a_0 + b_0+1\\ -a_0+3b_0 -1 \end{bmatrix}
$
$\begin{bmatrix} a_{2}\\ b_{2} \\ \end{bmatrix}
=\begin{bmatrix} 5 & 1\\ -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number of solutions to Pell-type equation $k(x^2-2y^2) = d$ Given equation $$k(x^2-2y^2) = d$$
Where d is a constant. k,x,y are variables. All are positive integers.
Is there some characterization for the values of d for which there are unique solutions, multiple solutions or no solutions? If this is unsolved, then any... | If there is a solution then there are infinitely many solutions. If $a$ and $b$ are so that $a^2-2b^2=1$ (of which there are infinitely many), and $x,y,k$ are solutions to $k(x^2-2y^2)=d$, then $n=ax+2by$ and $m=ay+bx$ will also satisfy $k(n^2-2m^2)=d$.
This is because $1=a^2-2b^2=(a+\sqrt{2}b)(a-\sqrt{2}b)$ and $(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
Use Complex Integrals/ Residue to evaluate $\int_0^\infty \frac{dx}{(x+1)^3 + 1}$
I'm not sure how to do this integration. It looks like partial fractions but I'm unsure.
| Generally, we can evaluate integrals of the form
$$\int_0^{\infty} dx \, f(x)$$
using a contour integral of the form
$$\oint_c dz \, f(z) \log{z} $$
As long as $f$ is sufficiently well-behaved at the origin and at infinity and in between, we have
$$-i 2 \pi \int_0^{\infty} dx \, f(x) = i 2 \pi \sum_k \operatorname*{Re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Using $y=\sum_{n = 0}^\infty a_nx^n$ in order to find recurrence relation of $a_n$'s for $y'=x-y^2$ I have to solve the differential equation $y'=x-y^2 => y''' = -2y'y' - 2yy''$.
I plugged in the series $\sum_{n = 0}^\infty a_nx^n$ to get:
$$\sum_{n = 0}^\infty a_{n+3}(n+3)(n+2)(n+1)x^n + \sum_{n = 0}^\infty \sum_{n =... | Note that $$ \left ( \sum_{n=0}^\infty a_n x^n \right )^2 = \sum_{n=0}^\infty\left ( \sum_{k=0}^na_ka_{n-k} \right ) x^n$$Thus $$y' = \sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty(n+1)a_{n+1}x^n$$ And you'll get$$y'+y^2 = \sum_{n=0}^\infty(n+1)a_{n+1}x^n + \sum_{n=0}^\infty\left ( \sum_{k=0}^na_ka_{n-k} \right ) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If an integer $n$ is chosen at random from $1$ to $96$ inclusive ,what is the probability that $n(n+1)(n+2)$ is divisible by 8? In this one If I consider that n is even then probability that the number $n(n+1)(n+2)$ will be divisible by 8 will be 1/2 ,Now if n is odd then for n(n+1)(n+2) to be divisible by 8 ,n+1 shoul... | HINT:
*
*$n\equiv0\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv0\cdot(0+1)\cdot(0+2)\equiv\color\green0\pmod8$
*$n\equiv1\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv1\cdot(1+1)\cdot(1+2)\equiv\color\red 6\pmod8$
*$n\equiv2\pmod8 \implies n\cdot(n+1)\cdot(n+2)\equiv2\cdot(2+1)\cdot(2+2)\equiv\color\green0\pmod8$
*$n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Is my integration of $\sqrt{4z^2 - 4z + 2}$ correct? I'm trying to
$$
\int \sqrt{4z^2 - 4z + 2}\ dz
$$
the integrand I first rewrite to (completing the square)
$$
4\left(z^2 - z +\frac{1}{4} - \frac{1}{4}\right) + 2 = 4\left(z - \frac{1}{2}\right)^2 + 1
$$
I apply $\int \sqrt{a^2 + x^2} \ dx = \frac{x}{2} \sqrt{a^2+x^2... | You have made some little mistakes. I give you a hint to solve it more easier:
$$\int\sqrt{4z^2-4z+2}\space\text{d}z=$$
$$\int\sqrt{(2z-1)^2+1}\space\text{d}z=$$
Subsitute $u=2z-1$ and $\text{d}u=2\space\text{d}z$:
$$\frac{1}{2}\int\sqrt{u^2+1}\space\text{d}u$$
Subsitute $s=\arctan(u)$ and $\text{d}u=\sec^2(s)\space... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int_0^2 x(8-x^3)^{\frac{1}{3}}\ dx$ What substitution would you use to get from
$$\int\limits_0^2 x(8-x^3)^{\frac{1}{3}}\ dx$$
to
$$\int\limits_0^1 (1-t)^{a-1}t^{-a}\ dt, \ a\in(0,1)\ ?$$
I know how to evaluate the second integral and I thought that if I substitute $t={x^3\over8}$ I would reduce this to th... | $$\frac { 8 }{ 3 } \int _{ 0 }^{ 1 }{ \left( { \left( 1-y \right) }^{ \frac { 1 }{ 3 } }{ y }^{ \frac { -1 }{ 3 } } \right) dx } \\ =B\left( \frac { 4 }{ 3 } ,\frac { 2 }{ 3 } \right) \\ =\frac { \Gamma \left( \frac { 4 }{ 3 } \right) \Gamma \left( \frac { 2 }{ 3 } \right) }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | You can define $\frac{0}{0}$ to be anything you want; it can be $0$, or $1$, or $\pi$. But here's the catch: we want the choice of value for $\frac{0}{0}$ to be compatible with the usual laws of arithmetic.
From this standpoint, what you've really proved is that if $\frac{0}{0}$ is anything except $0$, then the law $2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 6
} |
Definite integral $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\ln(\cos x + \sin x) dx$ Evaluate the following integral:
$$\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\ln(\cos x + \sin x) dx$$
| $$I=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} log(sinx+cosx)^2dx=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}log(1+sin2x)dx \tag{1}$$ using $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we have
$$I=\frac{1}{2} \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}log(1-sin2x)dx \tag{2}$$
Adding $(1)$ and $(2)$ we get
$$2I=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which is more likely with perfect six-side dice? Find the probability of the following
$(i)$ to roll a $1$,then a $2$ and then a $3$ on consecutive roll of a single dice,
$(ii)$to roll three identical dice and have the outcome be a $1$,a $2$ and a $3$.
Which is more likely with perfect six-side dice?
My Attempt:
Prob... | Let's rephrase the question:
Which is more likely with perfect six-sided dice?
*
*Roll $1$ die $3$ times and get 1, then 2, then 3
*Roll $3$ dice $1$ time and get 1, 2, 3
Obviously, without even computing the probabilities, the second option is more likely.
This is because in the first option the order matters an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How many solutions to the (general) equation? I've been trying to hash this one out for the last few days.
Please determine the number of solutions to the equation $x_1 + x_2 + x_3 + x_4 = n$, where $x_i \in N$, $x_1$ is even, $ 0 \leq x_2 \leq 2$, 3 divides $x_3$, and $x_4 \in \left\{0,1\right\}$.
My thoughts, so far:... | Here is a generating function solution.
First your constraints can be encoded by power series which include all powers of $x$ that satisfy your constraint:
$\bullet$ $x_1$ is even$\rightarrow 1 + x^2 + x^4 + \cdots$
$\bullet$ $0 \leq x_2 \leq2 \rightarrow 1+x+x^2$
$\bullet$ $3|x_3 \rightarrow 1+x^3+x^6+ \cdots$
$\bulle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Equation with different bases (exponential) I seem to be stuck with the following equation right here:
$$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$
| \begin{align}
2^x+2^{x+1}&=3^{x+2}+3^{x+3}\\
2^x+2\cdot2^x&=3^2\cdot3^x+3^3\cdot3^x\\
(1+2)2^x&=(3^2+3^3)3^x\\
3\cdot2^x&=36\cdot3^x\\
2^x&=12\cdot3^x\\
\left(\frac23\right)^x&=12\\
\end{align}
Now use logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Finding the remainder with Modular Arthmetic I have this math question that I'm kind of stuck on.
What is the remainder when $1^5+2^5+3^5+\cdots +99^5+100^5$ is divided
by $4$?
I'm supposed to use modular arithmetic and equivalences. I know that when dividing by $4$ the only possible remainders are $0, 1, 2, 3$ an... | The even numbers are not of interest here. If $a$ and $b$ are odd integers from $1$ to $99$, call $a$ and $b$ friends if $a+b=100$.
If $a$ and $b$ are friends, then $a+b\equiv 0\pmod{4}$, so $b\equiv -a\pmod{4}$. It follows that $b^5\equiv -a^5\pmod{4}$, and therefore $a^5+b^5\equiv 0\pmod{4}$.
The sum of the fifth po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$A$ be a real symmetric matrix of size $n$ ; is $I_n+A$ always non-singular ? Is $I_n - A$ always singular ? Let $A$ be a real symmetric matrix of size $n$ ; is $I_n+A$ always non-singular ? Is $I_n - A$ always singular ?
| Take
$$
I= \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
\quad
A= \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
$$ We have
$$
I+A= \begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}
$$
which is clearly singular.
Take
$$
I= \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Derivative of $10^x\cdot\log_{10}(x)$
Derive $10^x\cdot\log_{10}(x)$
$$10^x\cdot \ln(10)\cdot \log_{10}(x)+\frac{1}{x\cdot \ln(10)}\cdot 10^x$$
But WolframAlpha gives another solution. Where am I wrong?
| \begin{align}
\text{your answer} & = 10^x\cdot \ln10\cdot \log_{10} x+\frac{1}{x\cdot \ln10}\cdot 10^x \\[10pt]
& = 10^x \ln x + \frac{10^x}{x\ln 10} \\[10pt]
& = 10^x \left( \ln x + \frac 1 {x\ln 10} \right) = \text{Wolfram's answer}.
\end{align}
Note that where Wolfram writes $\log x$ or $\log 10$ with no base specif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 4
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Residue of $\frac{g(z)}{\cos^{2}z}$ I would like to show that the residue of the function $$\text{Res}\left(\frac{g(z)}{\cos^{2}z}\right) = g'(z_n)$$ at $z_{n}=(n+\frac{1}{2})\pi$, where $g$ is analytic.
I tried the Limit formula for higher order poles and it fails. What other method can I use?
| Using the addition theorem for the cosine and the fact that
$\sin(z_n) = \pm 1$, $\cos(z_n) = 0$, you can compute the initial part of the
Taylor series for $\cos^2(z_n + h)$ at $h = 0$:
$$
\cos^2(z_n + h) = \sin^2(h) = \bigl( h + \frac{h^3}{3!} + O(h^5) \bigr)^2 \\
= h^2 \bigl( 1 + \frac{h^2}{6} + O(h^4) \bigr)^2 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$ $$ \lim_{x\to 0} \frac{1-\sqrt{3x+1}}{2-\sqrt{5x+4}}$$
How do I solve this without using derivatives or integrals.
| Both the numerator and the denominator are of the form
$$
a-\sqrt{bx+a^2}
$$
(with $a>0$). In the numerator we have $a=1$, $b=3$; in the denominator we have $a=2$, $b=5$.
Let's try
\begin{align}
\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x}
&=\lim_{x\to0}\frac{a-\sqrt{bx+a^2}}{x}
\frac{a+\sqrt{bx+a^2}}{a+\sqrt{bx+a^2}}\\
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
chinese remainder theorem problem in one of the steps I need to calculate the following:
$$x=8 \pmod{9}$$
$$x=9 \pmod{10}$$
$$x=0 \pmod{11}$$
I am using the chinese remainder theorem as follows:
Step 1:
$$m=9\cdot10\cdot11 = 990$$
Step 2:
$$M_1 = \frac{m}{9} = 110$$
$$M_2 = \frac{m}{10} = 99$$
$$M_3 = \frac{m}{11} = 90... | $x \equiv -1 \pmod 9$ and $x \equiv -1 \pmod {10}.$ So $x \equiv -1 \pmod {90}$ and $x = 90 n - 1.$ But $90 = 88 + 2,$ so $90 \equiv 2 \pmod {11}.$
$$ x = 90 n - 1 \equiv 2n - 1 \pmod {11}. $$
$$ 2n \equiv 1 \pmod {11}, $$
$$ n \equiv 6 \pmod {11}. $$
Start with $n=6,$ so $x = 540 - 1 = 539.$
$$ \bigcirc \bigcirc \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Finding $\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$ $$\int_{-\infty}^\infty \frac{x^2}{x^4+1}\;dx$$
I'm trying to understand trigonometric substitution better, because I never could get a good handle on it. All I know is that this integral is supposed to reduce to the integral of some power of cosine. I tried $x^2=... | Note\begin{align*}
\int_{-\infty}^{\infty} \frac{x^2}{x^4+1} dx
&\overset{x\to \frac1x}=\int_{-\infty}^{\infty} \frac{1}{x^4+1}dx
=\int_{0}^{\infty} \frac{x^2+1}{x^4+1}dx
=\int_{0}^{\infty} \frac{1+\frac1{x^2}}{x^2+\frac1{x^2}}dx
\\
&= \int_{0}^{\infty} \frac{d(x-\frac{1}{x} )}{\left(x-\frac{1}{x}\right)^2+2} =\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 13,
"answer_id": 9
} |
Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm ... | Suppose $\sum_{i=1}^\infty \frac{1}{4i + 1}$ converges. Then as for all $i \geq 1$,
$$\frac{1}{4i + 1} > \frac{1}{4i + 2} > \frac{1}{4i + 3} > \frac{1}{4i + 4} $$
it must be the case each of these series also converge:
$$\sum_{i=1}^\infty \frac{1}{4i + 2}, \quad \sum_{i=1}^\infty \frac{1}{4i + 3}, \quad \sum_{i=1}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Determine the form of set containing the real solutions
From the following inequality:
$\dfrac{\log_{2^{x^2+2x+1}-1}(\log_{2x^2 + 2x + 3}(x^2 - 2x)}{\log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geq 0 $, the set of all real solutions to this inequality is of the form:
$$(a) \ (a,b) \cup (b,c) \\ (b) \ (-\infty,a) \cup (c... | Do you know about the change of base formula? If $a,b,x > 0$, $a \not= 1$, and $x \not= 1$ then
$$ \log_a b = \frac{\log_x b}{\log_x a}.$$ Your original inequality reads
$$ \log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0$$ provided that $2^{x^2 + 2x + 1} - 1 > 0$, $2^{x^2 + 2x + 1} - 1 \not= 1$, $x^2 + 6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving For $X$ In A 4th Degree Polynomial The examples they use on my book are basically from basic arithmetic like 1, 2 and 3 to calculus and calculating derivatives, which is really annoying because I can't build the fundamental skill required to do harder questions. Anyhow, I have to solve an non-factorable inequal... | After manipulating a little bit and rearranging terms when the whole thing is $\leq 0$:
$$x^4 + 2x^3 - 4x^2 - 6x + 3 \\
x^4+2x^3-6x-4x^2+3 \\
x^4 + 2x(x^2-3)-4x^2+3 \\
x^4 + 2x(x^2-3)-3x^2-x^2+3 \\
x^4 + 2x(x^2-3)-3x^2-(x^2-3) \\
x^4 -3x^2 + 2x(x^2-3)-(x^2-3) \\
x^4 -3x^2 + (x^2-3)(2x-1) \\
x^2(x^2-3) + (x^2-3)(2x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1568881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For $n>1$, estimate the product $(n+1)^{n+1}(n/1)^n\dots (1/n)$ from above If $ n $ be a positive integer $>1$, prove that
$$2^{n(n+1)}\gt(n+1)^{n+1}\biggl(\frac{n}{1}\biggr)^{n}\biggl(\frac{n-1}{2}\biggr)^{n-1}...\biggl(\frac{2}{n-1}\biggr)^{2}\biggl(\frac{1}{n}\biggr)$$
| Consider $(n+1)$ positive numbers $\binom{n}{0}, \binom{n}{1}, ...\binom{n}{n}$. $$ A.M.= \frac{\binom{n}{0}+ \binom{n}{1}+ ...+\binom{n}{n}}{n+1}=\frac{(1+1)^n}{n+1}=\frac{2^n}{n+1}$$
$$G.M.=\sqrt[(n+1)]{\binom{n}{0} \binom{n}{1}. ...\binom{n}{n}}=\sqrt[(n+1)]{\biggl(\frac{n}{1}\biggr)^n \biggl(\frac{n-1}{2}\biggr)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find formula for recursive sequence sum? I have the following sequence:
$$a(1) = 1$$
$$a(n) = a(n-1) + n$$
For example:
$$a(1) = 1$$
$$a(2) =3$$
$$a(3) =6$$
$$a(4) =10$$
$$a(5) =15$$
$$a(6) = 21$$
Which approach should I use in order to find the formula for the sum of elements $a(1)$ through $a(n)$?
Thanks :)
| The equation you write is equivalent to
$$a(n)-a(n-1)=n$$
Sum both sides from $n=1$ to $x$ to find that the left hand side telescopes.
$$\sum_{n=1}^{x} (a(n)-a(n-1))= \sum_{n=1}^{x} n$$
$$a(x)-a(0)=\frac{x(x+1)}{2}$$
But $a(1)-a(0)=1-a(0)=1$ so $a(0)=0$.
Now that we have:
$$a(x)=\frac{x(x+1)}{2}$$
Then you can go from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What rule is used for this simplification? $$ \frac{8}{(s+1)^2 + 2^2} \times \frac{1}{s} = \frac{8}{5} - \frac{1}{s} + \frac{16}{10}\times \frac{s+1}{(s+1)^2 + 2^2} + \frac{8}{10}\times \frac{2}{(s+1)^2 + 2^2}
$$
| Do some partial fraction magic?
$$\frac{8}{s(s^2+2s+5)} = \frac{As+B}{s^2+2s+5} + \frac{C}{s}$$
This gives $C=8/5, A=-8/5, B=-16/5$ I think, so
$$\frac{8}{s(s^2+2s+5)} = \left(\frac{8}{5}\right)\left[\frac{1}{s}-\frac{s+2}{s^2+2s+5}\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\det$ in terms of $k$ Consider the following matrix:
\begin{bmatrix}
1 & 2 & 3 \\
2 & k-3 & 4 \\
3 & 4 & k-4 \\
\end{bmatrix}
I have the following problems:
*
*How to find $\det(B)$ in terms of $k$?
*For what value(s) of $k$ are the column vectors of $B$ linearly dependent?
| You can find the determinant by applying row transformations:
\begin{align}
B=\begin{bmatrix}
1 & 2 & 3\\
2 & k-3 & 4\\
3 & 4 & k-4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 2 & 3\\
0 & k-7 & -2\\
0 & -2 & k-13
\end{bmatrix}
&&\begin{aligned}R_2&\gets R_2-2R_1\\R_3&\gets R_3-3R_1\end{aligned}
\\&\to
\begin{bmatrix}
1 & 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Generalized "square root" of a real positive definite symmetric matrix with respect to another Related post: Square root of Positive Definite Matrix, which is a special case of this question ($A = I$).
Let $A$ and $B$ be real, symmetric, positive definite matrices of the same size $n \times n$. The question is: Does ... | Define $S$ by
$$ S = A^{-\frac{1}{2}} \left( A^{\frac{1}{2}} B A^{\frac{1}{2}}\right)^{\frac{1}{2}} A^{-\frac{1}{2}}. $$
Then $S$ is symmetric and so, by Theorem 2 of Wigner's "On Weakly Positive Matrices", $S$ is positive definite and
$$ SAS = A^{-\frac{1}{2}} \left( A^{\frac{1}{2}} B A^{\frac{1}{2}}\right)^{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $7^n-3^n$ is divisible by $n>1$, then $n$ must be even. I tried using factorization of $a^n-b^n$ for odd $n$ in an attempt to work through to a situation where the factors are such that they cannot have n as a factor. But I reached nowhere. Here's how I proceeded -
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^... | If $n\mid 7^n-3^n$ and $n>1$, then let $p$ be the least prime divisor of $n$.
Clearly $\gcd(21,p)=1$, so $\left(7\cdot 3^{-1}\right)^n\equiv 1\pmod{p}$, i.e. $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid n$.
By Fermat's Little theorem $\text{ord}_p\left(7\cdot 3^{-1}\right)\mid p-1$. Therefore $\text{ord}_p\left(7\cdot 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Finding the logarithm of a matrix? Find $B$ if $A=e^B$ and
$A=\begin{bmatrix}
2&1&0\\
0&2&0\\
0&0&4\\
\end{bmatrix}$.
Besides, I would be very happy if give some general remark(Best approach). I have seen the wiki article on log of a matrix but it was too complicated(for me).
| You can make use of the block structure of $A$:
$$
A =
\begin{pmatrix}
C & 0 \\
0 & 4
\end{pmatrix}
= e^B
= \sum_{k=0}^\infty \frac{1}{k} B^k
\Rightarrow
B =
\begin{bmatrix}
D & 0 \\
0 & x
\end{bmatrix}
$$
so we can assume $4 = e^x \Rightarrow x = \ln(4)$.
For the block matrices we get
$$
C =
\begin{pmatrix}
2 & 1 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1578647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac{1}{a(a-b)(a-c)} +\frac{1}{b(b-c)(b-a)} +\frac{1}{c(c-a)(c-b)} =\frac{1}{abc}$ for all sets of distinct nonzero numbers $a,b,c$.
Prove that $$\cfrac{1}{a(a-b)(a-c)} +\cfrac{1}{b(b-c)(b-a)}
+\cfrac{1}{c(c-a)(c-b)} =\cfrac{1}{abc}$$ for all sets of distinct nonzero numbers $a,b,c $.
Now my question ... | For a proof that might satisfy a thirst for more symmetry, but which uses a very similar technique, consider the equivalent identity $$\cfrac{bc}{(a-b)(a-c)} +\cfrac{ac}{(b-c)(b-a)} +\cfrac{ab}{(c-a)(c-b)} =1.$$ Let $f(x)$ be the polynomial function defined by $$f(x) = \cfrac{(x-b)(x-c)}{(a-b)(a-c)} +\cfrac{(x-c)(x-a)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
explicit solution to heat equation without an integral sign Consider the 1-dimensional heat equation:
$$\left\{ \begin{align}
& {{u}_{t}}\left( x,t \right)={{u}_{xx}}\left( x,t \right),\text{ }x\in R,\text{ }t>0 \\
& u\left( x,0 \right)={{e}^{a{{x}^{2}}}},\text{ }x\in R \\
\end{align} \right.$$
Find ... | Inspired by Heat equation with initial value answered by Mercy King.
To simplify, let $a=1$.
Setting
$$\xi=\sqrt{\frac{1-4t}{4t}}\left(y-\frac{x}{1-4t}\right),$$
we have:
\begin{eqnarray}
\frac{(x-y)^2}{4t}-y^2&=&\frac{(1-4t)y^2-2xy+x^2}{4t}\\
&=&\frac{1-4t}{4t}\left[y^2-\frac{2x}{1-4t}y+\frac{x^2}{1-4t}\right]\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using substitution in complex equation $x^{10}-2x^5+\frac{4}{3}=0$
I substituted
$t=x^5$
Then I got with the quadratic formula the result
$t_1=1+\frac{\sqrt\frac{4}{3}}{2}i$
$t_2=1-\frac{\sqrt\frac{4}{3}}{2}i$
How do I calculate the roots now?
When I now resubstitute with
$t=x^5$
and form the polar coordinates do... | You now need to solve $$x^5 = \left(1 \pm \frac{i}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}}\exp\left({i\left(\frac{\pi}{6} + 2\pi k\right)}\right).$$
This should net you ten solutions, $5$ from each - which is intuitively right since you were originally solving a polynomial of degree $10$.
So $$x = \left(\frac{2}{\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integrate: $\int \frac{2x^2-3x+8}{x^3+4x} \, dx$ $$\int \frac{2x^2-3x+8}{x^3+4x}\,dx$$
My main problem is calculating the $B$ and $C$. This is the algebra part. Thus, what is a technique I can use that is in line with what I did to calculate $A$?
| Since you're going to set $x=2t$ anyway, because of the factor $x^2+4$, do it at the start: the integral becomes
$$
\int\frac{8t^2-6t+8}{4t(t^2+1)}\,2dt
=
\int\frac{4t^2-3t+4}{t(t^2+1)}\,dt
=
\int\left(\frac{4}{t}-\frac{3}{t^2+1}\right)\,dt=4\log|t|-3\arctan t+c
$$
So your original integral is
$$
\int\frac{2x^2-3x+8}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqr... | HINT:
Let's try to show that the RHS satisfies the same equation as the LHS.
Consider
$$z = \cos \frac{2\pi}{19} + i \sin \frac{2\pi}{19}$$
Then $z^{19}-1 = 0$ and $z \ne 1$ so
$$\sum_{k=0}^{18} z^k = 0$$
On the right had side you have
$$ z^2 + z^{-2} + z^3 + z^{-3} + z^5 + z^{-5} = z^2 + z^{17} + z^3 + z^{16} + z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Trigonometry identity proof sum I am normally good at solving trigonometry sums. However I stumbled upon this and could not figure out how to do it. I solved a few lines and then I got stuck. If anyone can teach me how to prove the equation, I'll be highly grateful.
If $\cos^2 A- \sin^2 A = \tan^2 B$, prove that $\co... | From $\cos^2A-\sin^2A=\tan^2B$ you get
$$
\cos^2A=\frac{1+\tan^2B}{2}=\frac{1}{2\cos^2B}
$$
and
$$
\sin^2A=\frac{1-\tan^2B}{2}=\frac{\cos^2B-\sin^2B}{2\cos^2B}
$$
Thus
$$
\tan^2A=\frac{\cos^2B-\sin^2B}{2\cos^2B}\,2\cos^2B
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Prove that $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+ac}{1+c} \geq 3.$
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that $$\dfrac{1+ab}{1+a}+\dfrac{1+bc}{1+b}+\dfrac{1+ac}{1+c} \geq 3.$$
This looks symmetric, so should I prove it for just $a \leq b \leq c$ and then the other cases follow?
| Another way.
Let $a=\frac{y}{x},$ $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by AM-GM we obtain: $$\sum_{cyc}\frac{1+ab}{1+a}=\sum_{cyc}\frac{1+\frac{z}{x}}{1+\frac{y}{x}}=\sum_{cyc}\frac{x+z}{x+y}\geq3\sqrt[3]{\prod_{cyc}\frac{x+z}{x+y}}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$ Find the value of the integral $\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$\int_{0}^{\frac{\pi}{12}}\frac{\tan^2x-3}{3\tan^2x-1}dx$
$=\int_{0}^{\frac{\pi}{12}}\frac{1}{3}\frac{\tan^2x-\frac{1}{3}+\frac{1}{3}-3}{\tan... | HINT: convert your integrand into $\frac{4\cos(x)^2-1}{4\cos(x)^2-3}$ and use the tan-half angle formulas
the result should be $$2/3\,\sqrt {3}{\rm arctanh} \left(\cot \left( {\frac {5\,\pi }{12}}
\right) \sqrt {3}\right)+\pi /12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Replacing in equation introduces more solutions Let's say I have an equation $y=2-x^2-y^2$.
now, since I know that $y$ is exactly the same as $2-x^2-y^2$ I can create the following, equation by replacing $y$ with $2-x^2-y^2$.
$y=2-x^2-(2-x^2-y^2)^2$
doing this replacement introduces new solutions such as $(-1, 0)$. Rep... | We know that every $(x,y)$ such that
$$y=2-x^2-y^2\quad\text{i.e.}\quad x^2+y^2+y-2=0$$
satisfies
$$y=2-x^2-(2-x^2-y^2)^2\quad \text{i.e.}\quad (2-x^2-y^2)^2+x^2+y-2=0\tag1$$
This implies that $(2-x^2-y^2)^2+x^2+y-2$ is divisible by $x^2+y^2+y-2$ :
$$(2-x^2-y^2)^2+x^2+y-2=(x^2+y^2+y-2)(x^2+y^2-y-1)$$
Hence,
$$(1)\iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Product of two series to get a series decomposition of zeta in the critical strip $\def\sfrac#1#2{%
\small#1%
\kern-.05em\lower0.1ex/\kern-.025em%
\lower0.4ex\small#2}$I've been working on gaining an intuitive understanding of the analytic continuation of the zeta function, but I've gotten stuck at this par... | the series for $\eta(s)$ is absolutely convergent for $\Re(s) > 0$ if you group the terms by two :
$\eta(s) = \displaystyle\sum_{n=1}^\infty (2n-1)^{-s} - (2n)^{-s} = \sum_{n=1}^\infty \mathcal{O}(s (2n)^{-s-1})$ ( from the Taylor expansion of order 1 of $(1-x)^{-s}$ when $x \to 0$)
$\eta(s) = (1-2^{1-s}) \ \zeta(s)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1... | For fun, I created a trigonograph:
$$\frac{1 + \cos\theta + \sin\theta}{1 + \sin\theta - \cos\theta} = \frac{1 + \cos\theta}{\sin\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polyloga... | We will outline of a way forward leaving some of the work to the reader.
Denote the integral of interest by $I$ where
$$I=\int \arcsin(x) \log^3(x)\,dx \tag 1$$
Integrating $(1)$ by parts by letting $u=\arcsin(x)$ and $v=x\left(\log^3(x)-3\log^2(x)+6\log(x)-6\right)$, we find that
$$\begin{align}
I&=x\arcsin(x)\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Which quantity is greater, $\frac{x^2}{y+\frac1y}$ or $\frac{y^2}{x+\frac1x}$? $x \gt y$, $ xy \neq 0$
A= $ x^2\over {y+{1\over y}}$
B= $ y^2\over {x+{1\over x}}$
Options:
1) Quantity A is greater.
2) Quantity B is greater.
3) The two quantities are equal.
4) The relationship cannot be determined from the given informa... | A nice approach is as follows: note first of all that $x + 1/x$ is positive if $x$ is positive and negative if $x$ is negative. Now, if $x$ and $y$ have the same sign, then
$$
A > B \iff\\
\frac{x^2}{y+1/y} > \frac{y^2}{x+1/x} \iff \\
x^2(x + 1/x) > y^2(y + 1/y) \iff \\
x^3 + x > y^3 + y
$$
Note, however, that $f(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to compute $\lim_{x \to 0} (\frac{x^5 e^{-1/x^2}+x/2 - \sin(x/2))}{x^3})$? I have a problem with this limit. I have no idea where is the problem.
Can you correct my mistake? Thanks
$$\lim\limits_{x \to 0} \left(\frac{x^5 e^\frac{-1}{x^2}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}\right)$$
I used the developments of McLa... | Setting $y=x^{-1}$, it is equivalent to
$$
y^3 \cdot \left(\frac{1}{e^{y^2}y^5}+\frac{1}{2y}-\sin\left(\frac{1}{2y}\right)\right) \approx y^3 \cdot \left(\frac{1}{2y}-\left(\frac{1}{2y}-\frac{1}{48y^3}\right)\right)
$$
as $y\to \infty$. Conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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} |
Limit problem with an exponential function without L'Hopital How can I calculate this limit without L'Hospital rule and Taylor series?
$${\lim_{x \to 1} \big(4^x - 3^x\big)^{\frac{1}{x - 1}}}$$
| In THIS ANSWER, I showed using standard, non-calculus based analysis that
$$\frac{x}{x+1}\le \log(1+x)\le x$$
for $x\ge -1$ and
$$1+x\le e^x \le \frac{1}{1-x}$$
for $x<1$.
Now, using $(1)$ we can write
$$\frac{4^x-3^x-1}{(4^x-3^x)(x-1)}\le \frac{\log(4^x-3^x)}{x-1}\le \frac{4^x-3^x-1}{x-1}$$
Next, using $(2)$ we see ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\sin(1/x)$ - why there is no limit? $$ \lim_{x\to 0+} \sin\left(\frac{1}{x}\right)$$
I know that there is no limit.
but, why there is no limit?
I tried $x=0.4$, $x=0.3$, $x=0.1$, it looks like the limit is $0$.
And how can I show that there is no limit? I tried to calculate it like all the other functions, an... | Why there is no limit?
The graphic can help you understand why and suggest you some approach for the proof:
Remark: You have to be careful with tables of values because they can be misleading:
\begin{array}{ c | c c c c }
x & \frac{1}{2\pi} & \frac{1}{3\pi} & \frac{1}{4\pi} &\frac{1}{5\pi} \\ \hline
\sin\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| $$\text {We have } \tan A+\tan B=-p\; \text { and }\; \tan A \tan B=q.$$ Let $C=A+B.$ Let $F=\sin^2 C + p\sin C \cos C + q\cos^2 C-q.\quad$ CASE 1. $\cos C=0.$ Then $$F=1-q,\; \text {and}$$ $$0=\cos C=\cos (A+B)\implies \tan B=\cot A\implies q=\tan A \tan B=1.$$ CASE 2. $\cos C\ne 0.$ Then $\;\tan C\;$ is defined : ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If $\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{2^n-1}{6}$ then find n. Problem :
If $\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{2^n-1}{6}$ then find n.
My approach :
$\sum^n_{k=1} (\frac{k+2}{k+1})^nC_k = \frac{3}{2}^nC_1 +\frac{4}{3}^nC_2 +\frac{5}{4}^nC_3 +\cdots \frac{n+2}{n+1}$
Now how to proceed further pl... | We have
$$\sum_{k=1}^n \left(\dfrac{k+2}{k+1}\right) \dbinom{n}k = \sum_{k=1}^n \dbinom{n}k + \sum_{k=1}^n \dfrac1{k+1} \dbinom{n}k$$
We have
$$\sum_{k=0}^n \dbinom{n}k x^k = (1+x)^n \,\,\,\,\, (\spadesuit)$$
Taking $x=1$, we obtain
$$\sum_{k=1}^n \dbinom{n}k x^k = 2^n-1$$
Integrating $(\spadesuit)$, we obtain
$$\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.
| You can make use of Horner's method. By further applying the formula $(a-b)(a+b) = a^2-b^2$ while sometimes factoring out common factors (-1 and 3 in this case), calculating the result becomes astonishingly nice:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|cc}
& 1 & 2 & -2 & -2 & 1 & 0 & 0 & 0 & 3 & 6 & 1\\
& + & + & + & + &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How to analyze the time complexity $\Theta$ of this recurrence I am trying to understand how to show that
$$T(n) = T(n/2) + T(n/4) + n^2$$
is $\Theta(n^2)$ by using a recursion tree. I tried substitution at first but it got real messy real fast.
This is self-study. Problem is from this pdf, problem 1-12.
| Suppose we start by solving the following recurrence:
$$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + n^2$$
where $T(1) = 1$ and $T(0) = 0.$
Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
We unroll the recursion to obtain an exact formula for $n\ge 1$
$$T(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $a$ such that $4\sin^2x+4\sin x+a^2-3<0$ for atleast one negative $x$ Find the value of $a$ such that $4\sin^2x+4\sin x+a^2-3<0$ for atleast one negative $x$.
$4\sin^2x+4\sin x+a^2-3<0$
$4\sin^2x+4\sin x<-a^2+3$
or $-a^2+3>4\sin^2x+4\sin x$
I found the minimum and maximum values of $4\sin^2x+4\sin x$... | First of all, previous considerations:
a) For such $a$, if there is at least $x<0$ satisfiyng the inequality, then there is an $x^\prime\in[0,2\pi]$ that satisfies it (cause periodicity of sine function), and the inverse is also true. Thus, we can change the problem to find values of $a$ for which inequality is satisfi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve irrational inequality $\sqrt{x^2-8x+15} + \sqrt{x^2+2x-15} > \sqrt{4x^2-8x+18}$ Here's what I tried
$$\sqrt{x^2-8x+15} + \sqrt{x^2+2x-15} > \sqrt{4x^2-8x+18}$$
$$\sqrt{(x-3)(x-5)} + \sqrt{(x-3)(x+5)} > \sqrt{4x^2-8x+18}$$
Now,
$$(x-3)(x-5) > 0 \Leftrightarrow x \in (- \infty, 3) \cup (5, \infty)$$
$$(x-3)(x+5) > ... | You are correct so far. for $|x|\geq 5$ both sides of your last equation are positive, so you can square the equation again and solve the remaining equation (it is cubic at most).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to evaluate $\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{((2x+1)^2+\ln x)x!}$$
| $$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(1+\frac{\ln x}{(2x+1)^2})(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!+4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)!}{(2x+1)^2(x)!} + \lim\limits_{x \to +\infty} \frac{4^x}{(2x+1)^2(x)!} = $$
$$\lim\limits_{x \to +\infty} \frac{(x+2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converge? Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converges?
Well, I'm trying the approach using Cauchy's test for convergence. If $a_n >= 0$ and $\lim \sqrt[n]a_n = L $ exists. then, I need to check whether it is bi... | Observe this variant of rational test.
$$\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}=\sum_{n=1}^\infty 2^n \frac{n!}{n^n}=\sum_{n=1}^\infty 2^n \prod\limits_{k=1}^{n} \frac{k}{n}$$
Now $$\sum_{n=1}^\infty 2^n \cdot b_{n}$$ could be understood as sort of a binary expansion and we are sure that it co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$ Evaluating
$$\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!}{(m+n+2)!}$$
involving binomial coefficients.
My attempt: $$\frac{1}{(m+1)(n+1)}\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{(m+1)!(n+1)!}{(m+n+2)!}=\frac{1}{(m+1)(n+1)} \sum_{m=0}^\infty \... | One may observe that,
$$
\frac{m!}{(m+n+2)!}=\frac{m!}{(n+1)(m+n+1)!}-\frac{(m+1)!}{(n+1)(m+n+2)!}
$$ giving, by telescoping terms,
$$
\sum_{m=0}^N\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}-\frac{(N+1)!}{(n+1)(N+n+2)!}
$$ thus, as $N \to \infty$,
$$
\sum_{m=0}^\infty\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}.
$$ Then the ini... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Explanation of formula for integer sequence with integers being repeated according to polynom I have two questions concerning OEIS sequence A056556:
$m$ is repeated $\frac{1}{2}(m+1)(m+2)$ times:
$$0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, ... | Let's study the cubic $\;P_n(x):=x^3+3x^2+2x-6n\,$
Its discriminant is given by $\;\Delta=4\,(1-243\,n^2)\,$ and will thus be negative for $n>0$ :
$P_n(x)$ will always have two conjugate complex solutions with negative real part and one real positive solution (thus the largest).
Since $\;P_n(x):=x(x+1)(x+2)-6\,n\ \ $ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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When is the limit of a sum equal to the sum of limits? I was trying to solve a problem and got stuck at the following step:
Suppose ${n \to \infty}$ .
$$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$
Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.
Now we have
$$\lim \limits_{... | Because the number of terms goes up exactly as the size of each term goes down.
Specifically $$\lim \limits_{n \to \infty} \Big(\underbrace{\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}}_{n\text{ times}}\Big) = \lim \limits_{n \to \infty} \sum_{i=1}^n \frac 1n$$
Does that help?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
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Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose ... | Write out the determinant. Alongside write the first 3 columns"Downward diagonal multiplication" gives ( O + O + 2OO + O ). "Upward diagonal multiplication" gives -(O + O + O + 18O) . det = 2O.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 5
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\... | HINT:
$$\dfrac2{x^4+x^2+1}=\dfrac{x^2+1-(x^2-1)}{x^4+x^2+1} =\dfrac{1+\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}-\dfrac{1-\dfrac1{x^2}}{x^2+1+\dfrac1{x^2}}$$
Now as $\int\left(1+\dfrac1{x^2}\right)dx=x-\dfrac1x,$
write the denominator as $\left(x-\dfrac1x\right)^2+3$
Similarly for the second integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to use parametric equation/trigonometric identity to show an ellipse? I have the equation $16x^2+25y^2=400$, and the parametric equation $(x,y)=(5\cos t, 4\sin t)$.
If I plug in the parametric equation into the first equation, I end up with the trigonometric identity $\cos^2 t+ \sin^2 t= 1$. How does this identity... | The pair of parametric equations $x = 5\cos t, y = 4\sin t$ in fact corresponds to the ellipse described by the Cartesian equation $16x^2 + 25y^2 = 400$. Substituting the first equation into the second would naturally give you an identity (like you got). However, to properly prove this, it's better to manipulate one fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine convergence when AST does not work Consider this series --
$$\frac{1}{2^3}-\frac{1}{3^2}+\frac{1}{4^3}-\frac{1}{5^2}+\frac{1}{6^3}-\frac{1}{7^2}+\frac{1}{8^3}-\frac{1}{9^2}+\dots$$
Apparently, alternating series test (AST) cannot be applied. What other methods should I consider in order to determine if the se... | Generically, if a series converges absolutely, then it converges (i.e. without absolute values around each term). So you might consider the absolute version,
$$ \frac{1}{2^3} + \frac{1}{3^2} + \frac{1}{4^3} + \frac{1}{5^2} + \cdots$$
This looks very similar to some very common series. It's clear that the $n$th term is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all $n$ such that $3^{2n+1}+2^{n+2}$ is divisible by $7$ Find all $n$ such that
$3^{2n+1}+2^{n+2}$ is divisible by $7$
Prove that your answer is correct
So I am not allowed to use mods, as is a calculus question, I have tried by induction but can't get to prove that it works for $k+1$, by multiplying the equati... | *
*Setting $n=1$, we get $$3^{2\cdot 1+1}+2^{1+2}=35$$ above number $35$ is divisible by $7$ hence it holds for $n=1$
*Assume the number $3^{2n+1}+2^{n+2}$ is divisible by $7$ for $n=k$ then $$3^{2k+1}+2^{k+2}=7\lambda \tag 1$$
*Setting $n=k+1$, we get $$3^{2k+2+1}+2^{k+1+2}$$
$$=9\cdot 3^{2k+1}+2\cdot 2^{k+1}$$
$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 5
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A simple inequality, does it always hold? I am looking at the following inequality:
$$\Gamma(-\frac{x+1}{x})\lt x , \forall x\gt \frac{7}{2}$$
It seems that the LHS and RHS eventually diverge for large enough $x$, but I have failed in a proof of that inequality. Any help would be greatly appreciated. Thanks.
| We shall show this using the helpful observation made by Claude Leibovici.
First, we shall show that $0 < z < 2/7$,
$$
f(z) \equiv \log\Gamma(1 - z) - z + \frac{ z^2 } { 2} < 0.
\qquad (1)
$$
From the Bohr-Mollerup theorem, $\log\Gamma(1-z)$ is convex, so is $-z + z^2/2$.
So $f(z)$ is convex, and for $z < a = 2/7$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Pythagorean triples with the same c value $a^2 + b^2 = c^2$
There are, Primitive Pythagorean Triples, that share the same c value. For example,
$63^2 + 16^2 = 65^2$ and $33 ^2 + 56^2 = 65^2$.
I have been trying to figure out why the following theorem for finding such triples works.
Take any set of primes. Ex: $5,13,1... | The system of equations:
$$\left\{\begin{aligned}&x^2+y^2=z^2\\&q^2+t^2=z^2\end{aligned}\right.$$
Formulas you can write a lot, but will be limited to this. Will make a replacement.
$$a=p^2+s^2-k^2$$
$$b=p^2+s^2+k^2-2pk-2ks$$
$$c=p^2+k^2-s^2+2ps-2kp$$
$$r=s^2+k^2-p^2+2ps-2ks$$
The solution then is.
$$x=2ab$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Closed form of $\sum\frac{1}{k}$ where $k$ has only factors of $2,3$
Consider the set containing $A$ all positive integers with no prime
factor larger than $3$, and define $B$ as
$$
B= \sum_{k\in A} \frac{1}{k}
$$
Thus, the first few terms of the sum are:
$$
\frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4... | In general, you use the same logic used for the Euler product. Notice that your form becomes $$(1+\sum\limits_{k=1}^{\infty}\frac{1}{2^k})(1+\sum\limits_{k=1}^{\infty}\frac{1}{3^k})$$ since you have all powers of 2 and all powers of 3 and when you multiply them you have all possible combinations without repetition, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Find the minimum roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$ given certain conditions on $f(x)$. Problem:
Let $f(x)$ be a thrice differentiable function satisfying:
$$|f(x) - f(4-x)| + |f(4-x)-f(4+x)| = 0, \forall x \in R$$
If $f'(1)=0$, then find the minimum number of roots of $f'(x)\cdot f'''(x)+(f''(x))^2 =0$, on $x... | Since $f(x) = f(4-x) = f(4+x)$ we have $f'(x) = -f'(4-x) = f'(4+x)$ for all $x \in \mathbb{R}$.
Since $f'(1) = 0$, using the above identity for $x = 1$ gives us $f'(1) = f'(3) = f'(5) = 0$.
Also, using the above identity for $x = 2$ yields $f'(2) = -f'(2) = f'(6)$, so $f'(2) = f'(6) = 0$.
Similarly, for $x = 0$, we get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\frac{x}{2}}$ Compute the following limit: $$L=\lim_{x\to 0}\frac{\sin(\ln(x+1))-\ln(\sin(x)+1)}{\sin^4\left(\frac{x}{2}\right)}$$
| $$\begin{align}\sin{\log{(1+x)}} &= \sin{\left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )} \\ &= \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right ) - \frac16 \left (x - \frac12 x^2 + \frac13 x^3 - \cdots \right )^3 + \cdots \\ &= x - \frac12 x^2 + \frac16 x^3 + \left ( \frac32 \cdot \frac16 - \frac14 \right )... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove inequality $ a^4+b^4 \leq 2 a^2 b^2 +1. $ Let $a,b$ be such that $|a|\leq 1$, $|b|\leq 1$. Prove that then
$$
a^4+b^4 \leq 2 a^2 b^2 +1.
$$
| Staring at this, I have an uncontrollable urge to complete the square.
$$a^4 + b^4 \le^{\text{?}} 2a^2 b^2 + 1
$$
$$a^4 - 2a^2 b^2 + b^4 \le^{\text{?}} 1
$$
$$(a^2 - b^2)^2 \le^{\text{?}} 1
$$
Now it's easy. If $|a|, |b| \le 1$, it follows that $0 \le a^2, b^2 \le 1$, and so $|a^2 - b^2| \le 1$ and so $(a^2 - b^2)^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the minimum of $\sum\limits_{cyc} {\sqrt{\frac a{2(b+c)}}}$ with $a,b,c \gt 0$ As said in the title, I have to find the minimum of the following:
$$\sum_{cyc} {\sqrt{\frac a{2(b+c)}}} $$ with $a+b+c>0$
In my very last attempt, I tried to work it out using AM-GM:
Since $$\sqrt{\frac a{2(b+c)}}= \frac 1{\sqrt 2}\f... | By AM-GM $\sum\limits_{cyc}\sqrt{\frac{a}{2(b+c)}}=\sum\limits_{cyc}\frac{\sqrt2a}{2\sqrt{a(b+c)}}\geq\sum\limits_{cyc}\frac{\sqrt2a}{a+b+c}=\sqrt2$,
which is infimum because we can use $c\rightarrow0^+$ and $b=c$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second d... | Throughout this solution, I will use the Leibniz notation $\frac{dy}{dx}$ to represent the derivative of $y$ with respect to $x.$ Notice that the second derivative is notated $\frac{d^{2}y}{dx^{2}}.$ We begin as you did - find the first derivative.
Given: $x^{3} + y^{3} = 1$
$3x^{2} + 3y^{2} \cdot \frac{dy}{dx} = 0$ (*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6,\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$,then find $\frac{a}{b}+\frac{c}{d}$ $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=6,\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=8$,then find $\frac{a}{b}+\frac{c}{d}$
I have tried multiplying the two given equa... | Hint:
Let $$\alpha=\frac ab+\frac cd,\beta=\frac bc +\frac da$$
Consider
$$\alpha\beta=(\frac ab+\frac cd)(\frac bc +\frac da)=\frac ac+\frac bd+\frac db+\frac ca=8$$
And note that
$$\alpha+\beta=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the probability of a region inside a pyramid I have a region in a 3-D space with a density of
$$
\
f_{x,y,z}(X,Y,Z) = \begin{cases}
1 & \text{if $ (x,y,z)\in W$}; \\
0 & \text{if $(x,y,z)\notin W$};\\
\end{cases}
\ $$
Being $W$ the set of points inside the pyramid $(0,0,0)$, $(2,0,0)$,$(0,3,0)$ and $(0,0,1)$
I... | Really, you've done just about all of the work. The only missing part is the easy one. You know the volume of the pyramid is one, so that the probability is the fraction of the volume between $z \in [1/3,2/3]$, or
$$3 \int_{1/3}^{2/3} dz \, (1-z)^2 = \left (\frac23 \right )^3 - \left (\frac13 \right )^3 = \frac{7}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $$\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$$
I am having a hard time relating the square roots in the inequality to the given condition. I was think... | Another simple solution.
We show that
$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^4 \ge 16(xy+yz+zx) \quad\forall x,y,z \ge 0$$
or
$$(a+b+c)^4 \ge 16(a^2b^2+b^2c^2+c^2a^2) \quad\forall a,b,c \ge 0.$$
WLOG, assume that $a=\max(a,b,c)$, then we have $$a^2b^2+b^2c^2+c^2a^2 = a^2(b+c)^2 + bc(bc-2a^2) \le a^2(b+c)^2.$$
Thus it reduces ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I get a irreducible polynomial of degree 8 over $Z_2[X]$? I have got one of degree 5: $x^5+x^2+1$, but I need one of degree 8.
| The only way I know is trial and error. A polynomial of degree $8$ is irreducible/prime over $\mathbb Z_2$ if it divides:
$$\frac{x^{2^8}-x}{x^{2^4}-x}=\frac{x^{255}-1}{x^{15}-1} = 1+x^{15}+x^{30}+\cdots x^{225}+x^{240}$$
This polynomial is exactly the product of all the prime polynomials of degree $8$, once each.
It's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \... | You can use $\displaystyle4\csc^2 x+9\sin^2x=\big(\frac{2}{\sin x}-3\sin x\big)^2+12$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is ... | The conjugate of $\sqrt{a} + \sqrt{b}$ is $\sqrt{a} - \sqrt{b}$. To rationalize an expression of the form $$\frac{1}{\sqrt{a} + \sqrt{b}}$$ we multiply the numerator and denominator by $\sqrt{a} - \sqrt{b}$. Since the conjugate of $2\sqrt{2} + \sqrt{3}$ is $2\sqrt{2} - \sqrt{3}$, we obtain
\begin{align*}
\frac{1}{2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the coefficient of $x^4$ in the expansion of $\sqrt[3]{1+x}$ Here's what I tried:
$$\sum_{n \ge0} {\frac{1}{3} \choose n} x^n= \sum_{n \ge0} = \frac{\frac{1}{3}!}{n!(n-\frac{1}{3})!}x^n=\sum_{n \ge0} \frac{(\frac{1}{3}-1)(\frac{1}{3}-2)\cdot ...\cdot(\frac{1}{3}-(n-1)) }{n!}x^n$$
What to do more, or is this all... | The fourth term should be:
$\dfrac{\left(\dfrac{1}{3}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{5}{3}\right)\left(-\dfrac{8}{3}\right)}{4!}x^4 $
Which should come out to:
$-\dfrac{10}{243}x^4 $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a set that spans the set of all 3 by 3 semimagic matrices and prove it Call a 3 × 3 matrix a semi-magic square if all its rows and columns (but not necessarily its diagonals) have the same sum.
The questions asks me to find a set that spans the set of all 3 by 3 semimagic matrices and prove it
| In order to answer the question we have to cope with two issues
*
*Semimagic matrices: We need a proper characterisation of these specific matrices.
*Linear span: Once we have a candidate for the set of all semimagic matrices, we have to show this set is a linear span, which means a subspace in the vector spac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x$ How to evaluate$$\int_{0}^{1}\frac{x^{4}\arctan x}{\sqrt{1+x^{2}}}\mathrm{d}x$$
I tried to define:$$I\left ( \alpha \right )=\int_{0}^{1}\frac{x^{4}\arctan \left ( \alpha x \right )}{\sqrt{1+x^{2}}}\mathrm{d}x$$
and
$$I'\left ( \alpha \rig... | Starting from $$I'\left ( a \right )=\int_{0}^{1}\frac{x^{5}}{\sqrt{1+x^{2}}\left ( 1+a ^{2}x^{2} \right )}\mathrm{d}x$$ change variable $x=\sqrt t$ and get $$I'\left (a \right )=\int_{0}^{1}\frac{t^2}{2 \sqrt{t+1} \left(a^2 t+1\right)}\,dt$$ Now $$\frac{t^2}{2 \sqrt{t+1} \left(a^2 t+1\right)}=\frac{\sqrt{t+1}}{2 a^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I simplify and evaluate the limit of $(\sqrt x - 1)/(\sqrt[3] x - 1)$ as $x\to 1$? Consider this limit:
$$ \lim_{x \to 1} \frac{\sqrt x - 1}{ \sqrt[3] x - 1}
$$
The answer is given to be 2 in the textbook.
Our math professor skipped this question telling us it is not in our syllabus, but how can it be solve... | Note the following identities,
$$ y^2-1 = (y-1)(y+1) $$
$$ y^3-1 = (y-1)(y^2+y+1) $$
we can use these to rewrite the numerator and the denomiantor by substituting $\sqrt{x}$ and $\sqrt[3]{x}$ for $y$ respectively,
$$ x - 1 = (\sqrt{x}-1)(\sqrt{x}+1) \Rightarrow \sqrt{x}-1 = \frac{x-1}{\sqrt{x}+1} $$
$$ x - 1 = (\sqrt[3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Calculate this integral $\int _{\frac{1}{4}}^2\:\frac{\ln\left(2x\right)}{x\ln\left(4x\right)}dx$ I have to calculate this integral .
$$\int _{\frac{1}{4}}^2\:\frac{\ln\left(2x\right)}{x\ln\left(4x\right)}\,dx$$
I have no idea how to start , help someone ?
Thanks.
| $$\int _{ \frac { 1 }{ 4 } }^{ 2 } \: \frac { ln\left( 2x \right) }{ xln\left( 4x \right) } dx=\int _{ \frac { 1 }{ 4 } }^{ 2 }{ \frac { \ln { 2 } +\ln { \left( x \right) } }{ \ln { 4 } +\ln { \left( x \right) } } d\left( \ln { \left( x \right) } \right) } =\int _{ \frac { 1 }{ 4 } }^{ 2 }{ \frac { \ln { 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How to evaluate $\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+sinx}-3}$? $$\lim _{x\to \:0^+}\frac{\left(e^x+e^{2x}\right)^2-4}{\sqrt{9+\sin x}-3}$$
I have tried with Taylor:
$$\lim _{x\to \:0^+}\frac{\left(1+x+\frac{x^2}{2}+1+2x+2x^2\right)^2-4}{\sqrt{9+x-\frac{x^3}{6}+\frac{x^5}{120}}-3}=\lim _{x\to \... | To simplify, let us first factor $(e^x+e^{2x})^2-4=(e^x+e^{2x}-2)(e^x+e^{2x}+2)$ and get rid of the square root with $(\sqrt{9+\sin x}-3)(\sqrt{9+\sin x}+3)=\sin x$.
Then
$$\frac{e^x+e^{2x}-2}{\sin x}=\frac{1+x+\frac12x^2\cdots+1+2x+2x^2\cdots-2}{x-\frac16x^3\cdots}=\frac{3x+\frac52x^2\cdots}{x-\frac16x^3\cdots}$$tends... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}$ converges. Show that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges.
I recognized that that since the integrand is even then
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
Then, I started with t... | $$
\begin{align}
\int_{-\infty}^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}
&=2\int_0^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag1\\
&=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_1^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag2\\
&=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag3\\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Find $\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$ I don't know how to find the sum of $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$. After rationalization we have
$\left(\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{2... | $$\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k+2}+\sqrt{2k}}\right)$$
$$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\sum_{n =1}^\infty \frac 1 {5^{n+1}-5^n+1}$ $$\sum_{n = 1}^\infty \dfrac 1 {5^{n+1}-5^n+1}$$
I can factorize denominator to $4\times5^n+1$ to confirm the series does not diverge,
But how do I calculate its actual sum?
The series is not a telescoping series nor I can partial factorise.
I get confused ... | The series doesn't have a closed form (except for a very complicated one involving Q-Polygamma function, as was said in a comment), however, we can transform it to get much better convergence.
$$\frac{1}{4 \cdot 5^n+1}=\frac{1}{4 \cdot 5^n} \left(1-\frac{1}{4 \cdot 5^n}+\frac{1}{4^2 \cdot 5^{2n}}-\frac{1}{4^3 \cdot 5^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
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Calculate $\lim\limits_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right)$ I was trying to calculate the limit of the following function:
$$ \lim_{x\to0} \frac{1}{\sin x} \cdot \ln \left(\frac{e^x -1}{x}\right) $$
My first thought was using L'Hopital's rule since $\Large \frac{e^x -1}{x}$ goes to 1 so the... | Use Taylor expansion: $e^x-1=x+x^2/2+o(x^2)$. Thus
\begin{align*}
\frac{1}{\sin x}\log\left(\frac{e^x-1}{x}\right)
&=\frac{x}{\sin x}\frac1x\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)\\
&=\frac{x}{\sin x}\frac1x\frac{\log\left(1+\frac{x^2/2+o(x^2)}{x}\right)}{(x^2/2+o(x^2))/x}\frac{x^2/2+o(x^2)}{x}\\
&=\underbrace{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 4
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$ x \ge 0\text{ and } y \ge 0 \implies \frac{x+y}{2} \ge \sqrt{xy} $ The above applies $\forall x,y \in \mathbb{R}$
I've tried: $x + y \ge 0$
$$x + y \ge x$$
$$ (x + y)^2 \ge 2xy$$
$$\frac{(x + y)^2}{2} \ge xy$$
But the closest I get is $\dfrac{x+y}{\sqrt{2}} \ge \sqrt{xy}$
Any ideas?
| $$(x-y)^2 \ge 0$$
$$x^2 - 2xy + y^2 \ge 0 $$
$$x^2 + y^2 \ge 2xy $$
$$x^2 + 2xy + y^2 \ge 4xy $$
$$(x+y)^2 \ge 4xy $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
| *
*$7-5\sqrt 2 = 1-3\sqrt{2} + 6 - 2 \sqrt{2} = 1^3+\dbinom{3}{1}(-\sqrt{2})+\dbinom{3}{2}*(-\sqrt{2})^2+(-\sqrt{2})^3=(1-\sqrt{2})^3$ (By binomial theorem:https://en.wikipedia.org/wiki/Binomial_theorem)
*$7+5\sqrt 2 = 1+3\sqrt{2} + 6 + 2 \sqrt{2} = (1+\sqrt{2})^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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how to prove this inequality $(ab+bc+ac)^2 ≥ 3abc(a+b+c)$
Prove that if $a,b,c$ are non-negative real numbers, then
$(ab + bc + ca)^2 \geq 3abc(a+b+c)$.
I tried to compute from $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$.
|
$$a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\ge3a^2bc+3ab^2c+3abc^2$$
$$a^2b^2+a^2c^2+b^2c^2-a^2bc-ab^2c-abc^2\ge0$$
$$2a^2b^2+2a^2c^2+2b^2c^2-2a^2bc-2ab^2c-2abc^2\ge0$$
$$(a^2b^2-2a^2bc+a^2c^2)+(a^2c^2-2abc^2+b^2c^2)+(b^2c^2-2ab^2c+a^2b^2)\ge0$$
$$(ab-ac)^2+(ac-bc)^2+(ab-bc)^2\ge0$$
the upper operation is re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatri... | You have that there is some $\theta$ for which you can write
$$a = \cos\theta \qquad\text{and}\qquad c = \sin\theta$$
Now, note that the vanishing elements of the matrix product give the equation
$$a b + c d = 0 \tag{1}$$
and that the assumed relation $\det A = 1$ gives
$$a d - b c = 1 \tag{2}$$
Simply solve $(1)$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Finding a linear transformation with a given null space The problem statement is,
Find a linear transformation $T: \mathbb R^3 \to \mathbb R^3$ such that the set of all vectors satisfying $4x_1-3x_2+x_3=0$ is the (i) null space of $T$ (ii) range of $T$.
For (i),
I found out the basis of the null space for the system,
$... | For (i), consider the matrix
$$ A = \begin{pmatrix} 4 & -3 & 1 \\ 4 & -3 & 1 \\ 4 & -3 & 1 \end{pmatrix}. $$
Denoting by $T_A \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ the corresponding linear map $T_A(v) = Av$, we have
$$ T_A \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 4x_1 - 3x_2 + x_3 \\ 4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How do I finish this trig integral $\int_0^{\pi/4}\frac{\sin^2 \theta}{\cos \theta}d\theta$? I got up to the part where it's $$\frac{9}{125}\int_0^{\large \frac{\pi}{4}}\frac{\sin^2\theta}{\cos\theta}\,\,d\theta$$
but I can't figure out how to finish it off.
By the way the original problem was:
$$\int_0^{0.6}\frac{x^2... | Your substitution is a bit off. Begin with
$$\begin{align*}
\int_0^{0.6}{\frac{x^2}{\sqrt{9-25x^2}}}\, dx
&= \int_0^{0.6}{\frac{x^2}{\sqrt{25\left(\frac{9}{25}-x^2\right)}}}\, dx \\
&= \frac{1}{5}\int_0^{0.6}{\frac{x^2}{\sqrt{\frac{9}{25}-x^2}}}\, dx.
\end{align*}
$$
Now let
$$\begin{align*}
x&=\frac{3}{5}\sin \theta ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists ... | Consider that $$\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \dots \text{ (for } -1 < x < 1).$$
Then since $\frac{\pi}{5}$ is in $(-1,1)$, we have:
$$\underbrace{\frac{1}{1 - \frac{\pi}{5}}}_{\dfrac{5}{5-\pi}} = 1 + \underbrace{\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots}_{\text{infinite sum of ir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "113",
"answer_count": 23,
"answer_id": 12
} |
Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$
$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarro... | For $x\gt0$, repeatedly integrating from $0$ to $x$ gives
$$
\cos(x)\le1\implies\sin(x)\le x\implies1-\cos(x)\le\frac{x^2}2\implies x-\sin(x)\le\frac{x^3}6
$$
Noting that both sides are odd, we get
$$
\left|x-\sin(x)\right|\le\frac{\left|x^3\right|}6
$$
Since $\frac{n}{n^2+k^2}\le\frac1n$,
$$
\begin{align}
\sum_{k=1}^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) wi... | $$\sin x+\cos x=\frac{7}{5}$$
Let $\sin x=t$
$$\implies t+\sqrt{1-t^2}=\frac{7}{5}$$
Shifting, squaring and simplifying, we get
$$25t^2-35t+12=0$$
$$\implies t=\frac{35 \pm 5}{50}$$
Hence, $$\sin x= \frac{4}{5},\ \cos x=\frac{3}{5} \ \text{or} \ \cos x= \frac{4}{5}, \ \sin x=\frac{3}{5}$$
But as we need to find $$\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 5
} |
Probability of a random Permutation Pick up a random permutation in S5(assuming all elements have the equal chance to be picked). Find the probability that the sum of the first three entries of σ is less than or equal to sum of last two.
My try: I mean there will be 5! different combination possible, do I have to look ... | Hint: the permutations you want are just :
$ (...,5,4)$, $(...,5,3)$, $(...,4,5)$, $(...,3,5)$, $(...,2,5)$ and $(...,2,5)$ and you don't care about the order of the other elements where I put the dots.
So the probability is:
$$\frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{4} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.