Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Applications of the inequality $e^x\geqslant 1+x$ I am interested in proofs of famous theorems or inequalities which can be proved by the inequality
$$
e^x\geqslant 1+x.
$$
For example, the divergence of harmonic series can be proved by assigning $x=\frac{1}{k}$.
$e^{\pi}>\pi^e$ can be proved by assigning $x=\frac{\p... | Define $p_i$ as the i-th prime number. Taking $x=\frac{1}{p_i -1}$ and we have
$$e^{\frac{1}{p_i-1}}≧1+\frac{1}{p_i -1}=1+\frac{1}{p_i} +\frac{1}{p_i ^{2}}+\cdots.$$
The product from $i=1$ to $n$ is
$$e^{1+\sum_{i=2}^n \frac{1}{p_i-1}}≧\prod_{i=1}^n \left(1+ \frac{1}{p_i} +\frac{1}{p_i ^{2}}+\cdots \right).$$
And here... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2496387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Induction proof: $2^n + 3^n ≡ 5^n (mod 6)$ I'm trying to prove that $2^n + 3^n ≡ 5^n\ (mod\ 6)$ using induction.
$n=1$:
$2+3≡5\ (mod\ 6)$
$n=k$:
$2^k + 3^k ≡ 5^k\ (mod\ 6)$
$n=k+1$:
$2^{k+1} + 3^{k+1} ≡ 5^{k+1}\ (mod\ 6)$
$2*2^k + 3*3^k ≡ 5*5^k\ (mod\ 6)$
$6\ |\ 5*5^k - 2*2^k - 3*3^k$
$6\ |\ (2+3)*5^k - 2*2^k - 3*3^k$
... | Multiply both sides of $$a^k+b^k\equiv(a+b)^k\pmod{ab},$$ by $a+b$ to find $$(a+b)^{k+1}\equiv(a+b)(a^k+b^k)\equiv a^{k+1}+b^{k+1}+a\cdot b^k+b\cdot a^k\pmod{ab}$$
Use the face $a\cdot b^k,b\cdot a^k,$ are both divisible by $ab$ for $k\ge1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.
Plane is g... | Given the points
$$
\cases{
A=(a,0,0)\\
B=(0,b,0)\\
C=(0,0,c)\\
p = (x,y,z)
}
$$
the sphere center $p_0$ is determined by the intersection of plane $\Pi$ with the line $L$ such that
$$
\cases{
\Pi\to (p-A)\cdot\vec n=0\\
\vec n = (A-B)\times(B-C)\\
L\to \left((p-\frac{A+B}{2})\cdot(A-B)=0\right)\cap \left((p-\frac{A+C}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a syst... | Disclaimer: This answer is probably not appropriate considering (algebra-precalculus) tag, but I'll write it anyway since it might be useful to others stumbling upon this question.
The system is easier to solve when considering original equation $$(a+b\sqrt{13})^3=18+5\sqrt{13}.$$
If we look at norm $N(x+y\sqrt{13}) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Computing the determinant of a large matrix? How would I go about computing the determinant of large matrices, such as $6 \times 6$. I believe that I need to use multilinear maps, but I am not sure how I can go about computing the determinant in a nice and efficient way.
Can anyone show me how I can determine the dete... | First, note that row 6 is mostly '$0$'s. Next note the red values in row 2. The non-zero part of the determinant has to go through that $1$, so swap rows 2 and 5,
\begin{align}
\begin{vmatrix}
0&0&1&1&1&1 \\
1&\color{red}0&0&\color{red}0&0&1 \\
1&0&1&1&1&1 \\
0&1&1&1&0&1 \\
0&1&0&1&0&0 \\
0&0&\color{red}1&0&0&0
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\f... | $$\underbrace{1^4+2^4+3^4\cdots\ + n^4}_{\sum_{k = 1}^n k^4} = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$$
See the Faulhaber's formula and the Bernoulli number: $\text{B}_{j}(x)$
\begin{align}
\sum_{k=0}^{n} k^p = \frac{1}{p+1} \sum_{j=0}^p (-1)^j {p+1 \choose j} B_j n^{p+1-j}=\frac{\text{B}_{p+1}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
calculating $\int_0 ^3 x\sqrt{|1-x^2|}dx$ I tried to calculate $\int_0 ^3 x\sqrt{|1-x^2|}dx$ using $x=\sin t$, but couldn't solve it and got stuck. Is $x=\sin t$ correct or should I have used another value for $x$? don't know how to solve it.
hoping you can help me with that.
thank you very much for helping.
| From x = 0 to x = 1,
$$1 - {x^2} \ge 0$$
so
$$\left| {1 - {x^2}} \right| = 1 - {x^2}$$
From x = 1 to x = 3
$$\left| {1 - {x^2}} \right| \le 0$$
so
$$\left| {1 - {x^2}} \right| = {x^2} - 1$$
$$\int_0^3 {x\sqrt {\left| {1 - {x^2}} \right|} dx} = \int_0^1 {x\sqrt {1 - {x^2}} dx} + \int_1^3 {x\sqrt {{x^2} - 1} } dx$$
F... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Residue of $\frac{\cos (\pi z)}{z^2-1}$ I want to find the residue $\displaystyle\frac{\cos (\pi z)}{z^2-1}$ at $z=1$.
The function has two poles, at $-1,1$, both of order $1$. The series expansion for the function gives $\displaystyle\frac{1}{z^2-1}\left(1- \frac{z^2\pi^2}{2!}+O(z^4z\pi^4) \right)$.
(1) Is then Laure... | You have the function $f(z) = \frac{g(z)}{z^2-1}$, where $g(z)= \cos(\pi z)$ and therefore holomorf. The residue at $z=1$:
Using limits.
\begin{align}
\text{Res}_{z=1} f(z) =\lim_{z\to 1} (z-1)\frac{g(z)}{(z-1)(z+1)} = \frac{g(1)}{2} = \frac{-1}{2}
\end{align}
Using Laurent series. You must have powers of $z-1$ in your... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify solutions for $y'' + 4y = 2 \tan x$? The original equation is $$y'' + 4y = 2 \tan x$$
What I did so far:
$$\lambda^2+4 = 0$$
$$\lambda_1 = -2i \quad \lambda_2 = 2i$$
$$y(x) = C_1\cos2x+C_2\sin2x$$
Write down the system:
$$\begin{cases} C_1'\cos2x + C_2'\sin2x = 0 \\
2C'_2\cos2 x - 2C_1'\sin2x = 2\tan x... | Mutiply the first equation by $\cos(2x)$ and the second by $\sin(2x)$ and subtract them from each other to get:
$$C_1'\cos^2(2x) + C_2'\cos(2x)\sin(2x) - C_2'\cos(2x)\sin(2x) + C_1'\sin^2(2x) = - \tan(x)\sin(2x)$$
$$C_1' = - 2\frac{\sin x}{\cos x} \cos(x)\sin(x) = - 2\sin^2(x) = \cos(2x) - 1$$
This is easily integrable... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A deck of 40 cards, 4 suits from 1 to 10, 2 cards extracted, probability the sum is 9? You have a deck of $40$ cards, $4$ suits from $1$ to $10$, pick randomly $2$ cards, no reimmission, what is the probability you get $9$ as a sum?
How can I solve using the basic principle of counting?
The favorable outcome are $(8+1... | What is the probability that we select $1$ and $8$? There are $4$ $1$'s and $4$ $8$'s and we select $2$ cards from the $40$ so we have
$$\frac{{4\choose{1}} \cdot {4\choose{1}}}{40\choose{2}}$$
This is the same as the probability of selecting a $2$ and a $7$, a $3$ and a $6$, and a $4$ and $5$.
So the probability that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating limits using Taylor's expansion. Evaluate the following limit : $\lim_{x \to 0}(\frac{1}{\log(\cos(x))}+\frac{2}{\sin^2(x)})$
Generally speaking I have no idea at what order should I stop the Taylor's expansion.
I've tried this using two different orders and i've gotten both the results 1 and 0.
If anyone co... | $$\lim \limits_{x \to 0}\left(\frac{1}{\log(\cos(x))}+\frac{2}{\sin^2(x)}\right)=\lim \limits_{x \to 0}\frac{\sin^2x+2\ln\cos x}{\sin^2x\ln\cos x}$$
We can stop Taylor's expansion at the denominator to the first (different from zero) term, because it's a product and nothing is going to cancel. So if $x$ tends to $0$ we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that, if $a,b,c$ are positive real numbers $\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$
Prove that, if $a,b,c$ are positive real numbers:
$$\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$$
Can the proof be written without the need for high mathematics?
| We'll prove a stronger inequality:
$$\sum_{cyc}\frac{a}{2a+b+c}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{1}{4}-\frac{a}{2a+b+c}\right)\geq0$$ or
$$\sum_{cyc}\frac{b+c-2a}{2a+b+c}\geq0$$ or
$$\sum_{cyc}\frac{c-a-(a-b)}{2a+b+c}\geq0$$ or
$$\sum_{cyc}\left(\frac{c-a}{2a+b+c}-\frac{a-b}{2a+b+c}\right)\geq0$$ or
$$\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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For each $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix}1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^n$
If $a \in \mathbb{R}$, evaluate $$ \lim_{n \to \infty}\left(\begin{matrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{matrix}\right)^{n}$$
My attempt: Let $$A = \left(\begin{matrix... | The best approach is to diagonalize your matrix. Both $(1,i)$ and $(1,-i)$ are eigenvectors. So, let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix}.$$Then$$T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}.T=\begin{pmatrix}1+\frac ani&0\\0&1-\frac 1ni\end{pmatrix}.$$Therefore$$T^{-1}.\begin{pmatrix}1&\frac an\\-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
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Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$
Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$
1)$\dfrac{\cos(x)}{1+\sin(x)}$
2)$\dfrac{\cos(x)}{1-\sin(x)}$
3)$\dfrac{1-\sin(x)}{\cos(x)}$
4)$\dfrac{1+\sin(x)}{\cos(x)}$
Of course we can find the correct choice by testing eac... | Multiply numerator and denominator by $1-\sin x+\cos x$: you get
\begin{align}
\frac{1-\sin x+\cos x}{1+\sin x+\cos x}
&=\frac{(1-\sin x+\cos x)^2}{(1+\cos x)^2-\sin^2x} \\[6px]
&=\frac{1+\sin^2x+\cos^2x-2\sin x+2\cos x-2\sin x\cos x}{1+2\cos x+\cos^2x-(1-\cos^2x)}\\[6px]
&=\frac{2(1-\sin x+\cos x-\sin x\cos x)}{2\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Find $6^{273} + 8^{273}\pmod{49}$
The number $6^{273} + 8^{273}$ divided by $49$ has a remainder, what is its value?
I used the totient function to compute for modulo 49.
$6^{42}$ and $8^{42}$ are $-1$ and $1$ mod $49$ respectively, $273/49$ is equal to $5$ with a remainder of $21$.
We would then look for the remain... | Note that $6\cdot 8\equiv 48 \equiv -1 \bmod 49$, so $8^{-1}\equiv -6 \bmod 49$.
Then $6^{273}+8^{273} \equiv 6^{273}+(-6)^{-273} \equiv 6^{273} -6^{-273}\bmod 49$ (since $-1^{-1} \equiv -1$)
Since $\phi(49)=42$ Euler gives us $6^{42}\equiv 1\bmod 49$. Then note that $6^{273}\not\equiv 0 \bmod 49$ and $6^{273} (6^{273}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533754",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Derive an $O(h^4)$ five point formula to approximate $f'(x_0)$ that uses $f(x_0 - h), f(x_0), f(x_0 + h), f(x_0 + 2h), f(x_0 + 3h)$ This is a Numerical Analysis textbook question on Numerical Differentiation (like https://en.wikipedia.org/wiki/Numerical_differentiation):
Official Textbook Question
Q: Derive an $O(h^4)$... | Consider that your formula must give the correct values for $f (x) = 1$, $f (x) = x$, $f (x) = x^2$, $f (x) = x^3$ and $f (x) = x^4$. The hint is weird, since it assumes that you are not using f (x).
So you get the equations:
$A + B + C + D + E = 0$, $A(x-h) + B(x) + C(x+h) + D(x + 2h) + E(x + 3h) = 1$, $A(x-h)^2 + B... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of integer reciprocals The following problem results in a cubic, but is there a way to simplify it easier?:
The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of these integers?
I ended up with $47{ x }^{ 3 }-180{ x }^{ 2 }-47x+60 =0$. However, unlike other methods where I've learne... | This answer is based upon symmetry and a guess.
We try to exploit symmetry and use the approach
\begin{align*}
\color{blue}{\frac{47}{60}}&=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\\
&=\frac{n(n+1)+n^2-1+(n-1)n}{n(n^2-1)}\\
&\color{blue}{=\frac{3n^2-1}{n(n^2-1)}}
\end{align*}
Let's take a closer look at
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2537557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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if $\frac{m}{n},\frac{m+1}{n+2},\frac{m+2}{n+4},\frac{m+3}{n+6}$be positive integers, find the minimum of the m let $m$ be a positive integer,and $n$ be an positive integers,such $$\dfrac{m}{n},\dfrac{m+1}{n+2},\dfrac{m+2}{n+4},\dfrac{m+3}{n+6}$$be positive integers,find the minimum of the $m$.
It seem use CRT to solv... | If $\frac{m}{n}, \frac{m+1}{n+2}, \frac{m+2}{n+4}, \frac{m+3}{n+6}$ are all integers, so are $\frac{2m}{n}, \frac{2m+2}{n+2}, \frac{2m+4}{n+4}, \frac{2m+6}{n+6}$, and so are $\frac{2m-n}{n}, \frac{2m-n}{n+2}, \frac{2m-n}{n+4}, \frac{2m-n}{n+6}$. Hence $2m-n$ is a multiple of $\mbox{lcm}(n, n+2, n+4, n+6)$.
By Sridhar's... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\... | Consider the function and its first derivative$$f(x)=\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
$$f'(x)=\frac{1}{2} x \left(\frac{5-x^2}{\sqrt{-x^4+10 x^2-9}}+\sqrt{3}\right)$$ The first derivative cancels for $x=0$ which has to be excluded.
Now, let $x^2=y$ and solve
$$\frac{5-y}{\sqrt{-y^2+10 y-9}}+\sqr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve $h_n = 3h_{n-1} -4n$ using generating functions Solve $h_n = 3h_{n-1} -4n$, where $h_0 = 2$ using generating functions.
I am struggling to figure out how to solve this using generating functions. I know the answer should be $h_n = -3^n +2n + 3$.
Here is the method that both my textbook and professor used.
Let $g... | From where you left off, the first problem is to find the generating function for sequence $a_n = -4n$. This is not so hard if know the relation between operations on generating functions and operations on their corresponding sequences (see here): $\langle a_n \rangle = -4( \langle 1, 1, 1, \cdots \rangle * \langle 1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find value of $x^2+y^2$ using given expression
Consider two real numbers $x , y$ such that $$\left(x^2+1\right)\left(y^2+1\right)+9=6\left(x+y\right)$$ Hence find the value of $x^2+y^2$.
At first I tried to factorise the condition but the $(xy)^2$ created much problems. I also tried to create a complete square to get... | Continuing from Dylan's answer, $xy = 1$ and $x+y = 3$. Then:
$$x^2 + y^2 = (x+y)^2 - 2xy = (3)^2 - 2(1) = \boxed{7}.$$
Moreover, note that we can swap the values of $x$ and $y$ without changing the expression. Therefore from user16394's answer, $x = \frac{1}{2}(3 + \sqrt{5}), y = \frac{1}{2}(3 - \sqrt{5})$ or vice ver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Number of seven digit numbers without repetition of digits divisible by $3$ Find number of seven digit numbers divisible by $3$ with
$1.$ Repetition
$2.$ Without Repetition
For Part $1.$ The least seven digit number divisible by $3$ is $1000002$ and highest seven digit number is $9999999$
So total is $3000000$
For Part... | Let $N_7 = \text{what you are looking for}$
However, this consists of one set that doesn't use any zeros - $(NZ)_7$ - and another that does. The number in the set that uses zeros can be defined recursively as $6(NZ)_6$ because the zero can be placed in 6 different places.
Notice that $(NZ)_7 = \frac{\text{pair of two n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Division by $0$ and its restrictions Consider the following expression:
$$\frac{1}{2} \div \frac{4}{x}$$
Over here, one would state the restriction as $x \neq 0 $, as that would result in division by $0$.
But if we rearrange the expression, then:
$$\begin{align}
\frac12\div\frac4x &= \frac{1}{2} \times \frac{x}{4} \\... | When we make a simplification like that, we are implicitly assuming that $x\neq 0$. Really, one should write that
$$
\frac{1}{2} \div \frac{4}{x}=\begin{cases}
x/8&\text{if $x\neq 0$}\\
\text{undefined}&\text{if $x=0$}.
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2548196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "100",
"answer_count": 9,
"answer_id": 4
} |
Simplifying expression of algebra I am having trouble simplifying the following,
$$\frac{-1}{2(3+n\pi i)} \left( e^{-(3+n\pi i )}- e^{3+n\pi i }\right)$$
I am wondering how it reduces to the following, looking for someone to show me the steps,
$$\frac{1}{3+n\pi i} \left( \cos(n\pi) + i\sin (n\pi) \right )\frac{e^3-e^{-... | The mistake is in the first line of your attempt:
$$\frac{1}{2(3+n\pi i)} \left( -e^{-(3+n\pi i )}+ e^{3+n\pi i }\right)$$
$$\frac{1}{(3+n\pi i)} \left( -e^{-n\pi i }+ e^{n\pi i }\right) \frac{e^3 - e^{-3}}{2}$$
The two expressions are not the same (multiply the second one out and check for yourself). In particular... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2552841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit
I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
| The most elementary method is simply to observe patterns of the last digits ( there are only 10 possible digits to check so it is quite fast) by multiplying last digit by $a,a^2,a^3,a^4 $, for greater powers - the pattern is repeating ..
$0 \rightarrow 0,0,0,0,.. $
$1 \rightarrow 1,1,1,1,.. $
$2 \rightarrow 4,8,6,2,..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Operating with Exponential Generating Function I am having troubles understanding how to operate with generating functions to obtain a final formula. I have been looking to the answer to this question: Exponential generating function and number of balls .
$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color... | The term with $\frac{1}{16}$ arises as follows:
\begin{align*}
\frac{x^2}{2}\cdot\frac{1}{2}e^{2x}
&= \frac{1}{4}x^2\sum_{n=0}^\infty \frac{2^nx^n}{n!} \\
&= \frac{1}{4}\sum_{n=0}^\infty \frac{2^nx^{n+2}}{n!} \\
&= \frac{1}{16}\sum_{n=0}^\infty \frac{2^{n+2}x^{n+2}}{n!} \\
&= \frac{1}{16}\sum_{n=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solution of: Let $k,n \in \mathbb{N}$ and $\frac{n(n-1)}{4} = \frac{k(k-1)}{2} $ Let $k,n \in \mathbb{N}$, $k,n < 101 $ and $\frac{n(n-1)}{4} = \frac{k(k-1)}{2} $.
How can I solve this and find $n$ and $k$ in an analytic way without trying some solutions?
| solving the given equation $$n^2-n=2k^2-2k$$ for $n$ we get
$$n=\frac{1}{2}+\sqrt{\frac{1}{4}+2k^2-2k}$$
or
$$n=\frac{1}{2}-\sqrt{\frac{1}{4}+2k^2-2k}$$
now you Can solve the inequality
$$n=\frac{1}{2}+\sqrt{\frac{1}{4}+2k^2-2k}<101$$
writing the inequality in the form
$$\sqrt{\frac{1}{4}+2k^2-2k}<101-\frac{1}{2}$$
squ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$ Can someone please check my conclusion to the evaluation of the following limit?
$$\lim_{x\to \infty} x(16x^4 + x^2+1)^{1/4}-2x^2$$
I got that the limit is equal to infinity. If limit is equal to infinity does this mean that limit does not exist?
| $\text{HINT}$
$$x(16x^4 + x^2+1)^{1/4}-2x^2 $$ $$=x^2(16+ \frac{1}{x^2}+\frac{1}{x^4})^{1/4}-2x^2$$ $$=x^2(\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2)$$ $$=\frac{\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}-2}{\frac{1}{x^2}}$$
You can multiply the last fraction with $$\sqrt[4]{16 + \frac{1}{x^2}+\frac{1}{x^4}}+2$$ both... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Taylor series expansion of sin(x) I understand that Taylor series expansion for $\sin(x)$ is derived as follow:
$$
\sin(x) = x - \frac{x^3}{3!}+\frac{x^5}{5!}-...
$$
Now, what exactly is the first, second, and third term?
Is the first term just $\sin(x) = x$?
Is the second term $\sin(x) = x-\frac{x^3}{3!}$?
| In the Taylor expansion at $0$ of the function $\sin(x)$, the even powers of $x$, i.e. the "missing" terms, are zero because $\sin(x)$ is an odd function:
$$
\begin{align}\sin(x)&=\sum_{k=0}^{\infty}\frac{D^k(\sin(x))_{x=0}}{k!}\cdot x^k
\\&=\sin(0)+\cos(0)x+\frac{-\sin(0)}{2!}\cdot x^2+\frac{-\cos(0)}{3!}\cdot x^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ I saw this problem somewhere recently and I was having some difficulty getting started on it.
The problem is twofold. The first is to evaluate:
$$\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac... | Note
\begin{eqnarray}
&&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)\\
&=&\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2}\right) -\sum_{k=0}^\infty\left( \frac{1}{5k+3}-\frac{1}{5k+4} \right)\\
&=&\sum_{k=0}^\infty\frac{1}{(5k+1)(5k+2)}-\sum_{k=0}^\infty\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
} |
Taylor series for $\sqrt{1+x^2}$ I want to expand
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\f... | To give the general term expansion let $y=x-2$, then
\begin{align}
\sqrt{1+x^2}
&= \sqrt{1+ (y+2)^2}
= \sqrt{5 + 4y + y^2}
= \sqrt{5} \sqrt{1 + \frac{y}{5}(4+y)}\\
&= \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \left[\frac{y}{5}(4+y)\right]^n
= \sqrt{5}\sum_{n=0}^{\infty} \binom{1/2}{n} \frac{y^n}{5^n}\sum_{k=0}^n\binom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int ydx + zdy + xdz$ using Stokes' Theorem?
Evaluate $\int ydx + zdy + xdz$ where $C $ is intersection of $x+y=2$
and $x^2+y^2+z^2=2(x+y) $ traversed counterclockwise as viewed from
origin
I am using Stokes' theorem to solve this question so
We want $\int \int curl F.N \; dS$ where $N$ is the normal u... | Since $\nabla\times F\cdot \vec{N}=\sqrt{2}$, you only need to evaluate
$$\sqrt{2}\iint_S dS$$
which is $\sqrt{2}$ multiplied by the area of the disk. (Note that the intersection is a disk, and it passes through the center of the original ball.) The original ball can be written as
$$(x-1)^2+(y-1)^2+z^2=2$$
whose cente... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x)... | Observe that
$$\phi\left(x\right)=\exp\left(\int\frac{2}{x}{\rm d}x\right)=\exp\left(2\ln\left|x\right|+C\right)=Ax^{2}$$
with $A\equiv\exp C$. Thus the particular solution is
$$y_{\rm p}=Ax^{2}\int\frac{x^{3}}{Ax^{2}}{\rm d}x=\frac{x^{4}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| Alright, you want difference of squares have some difference of squares
$$(\frac{1+i \sqrt7}{2})^4 + (\frac{1-i \sqrt7}{2})^4=$$
Lets add $0$ cleverly
$$=(\frac{1+i \sqrt7}{2})^4 -2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2+ (\frac{1-i \sqrt7}{2})^4 + 2(\frac{1+i \sqrt7}{2})^2(\frac{1-i \sqrt7}{2})^2$$
$$=((\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2572494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Find the limits of find the limits :
$$\lim_{x\to 0}\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}$$
i know that :
$$\sin 5x=5\sin x-20\sin^3x+16\sin^5x$$
$$5\sin 3x=15\sin x-20\sin^3x$$
so we have :
$$\sin 5x-5\sin 3x+10\sin x=16\sin^5x$$
$$\lim_{x\to 0}\frac{16\sin^5x}{\sin (\sin x)+\tan x-2x}=\frac{16\sin... | Use Taylor's formula at order $5$:
\begin{align}
&\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}=\\[1ex]
&\frac{5x-\frac{125x^3}6+\frac{625x^5}{24}+o(x^5)-5\Bigl(3x-\frac{27x^3}6+\frac{81x^5}{40}+o(x^5)\Bigr)+10\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\Bigr)}{\sin\Bigl(x-\frac{x^3}6+\frac{x^5}{120}+o(x^5)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculate the limit: $\lim_{x\to+\infty}(\frac{x^2 -x +1}{x^2})^{\frac{-3x^3}{2x^2-1}}$ without de l'Hôpital rule I was wondering how can I calculate the limit:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}$$
without de l'Hôpital rule.
I tried to reconduct the limit at the well known one... | Since $e^x$ and $\ln$ are continuous functions, we obtain:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\lim_{x\to+\infty}\left(1+\frac{x^2 -x +1}{x^2}-1\right)^{\frac{1}{\frac{x^2 -x +1}{x^2}-1}\cdot\left(\frac{x^2 -x +1}{x^2}-1\right)\frac{-3x^3}{2x^2-1}}=$$
$$=
\lim_{x\to+\infty}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Argument principle in the evaluation of integral
Evaluate $$I=\int_{-\infty}^{\infty}\frac{x}{x^{2}+x+1}dx$$
Attempt: Note that $$\int_{-\infty}^{\infty}\frac{x}{x^{2}+x+1}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2x+1}{x^{2}+x+1}dx-\int_{-\infty}^{\infty}\frac{1}{x^{2}+x+1}dx$$
Denote $\Gamma $ as the circle with r... | Using, as you did,$$I=\int\frac{x}{x^{2}+x+1}\,dx=\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}\,dx-\int\frac{1}{x^{2}+x+1}\,dx$$
$$I=\frac{1}{2} \log \left(x^2+x+1\right)-\frac{1}{\sqrt{3}}\tan ^{-1}\left(\frac{2
x+1}{\sqrt{3}}\right)$$ making, after simplifications,
$$J=\int_a^a\frac{x}{x^{2}+x+1}\,dx=\frac{1}{\sqrt{3}}\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
I don't know how to solve this integral
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$
What I've done so far is:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$
So
$$x^2+x-4 = A(x^2-4)+Bx$$
And I did this so I could figure out the v... | The partial fractions decomposition is not what you think: it should be
$$\frac{x^2+x-4}{x(x^2+4)}=\frac Ax+\frac{Bx+C}{x^2+4},$$
because the condition on the numerators is not they're constants, but they're polynomials with degree less than the degree of the irreducible factor in the denominator. Here the second irr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Factorising $x^2-x+1$ I attempted to factorise $x^2-x+1$, which seemed fairly straightforward:
$$
\begin{align}
x^2-x+1 &= (x+1)^2-3x \\
&= (x+1-\sqrt{3x})(x+1+\sqrt{3x}) \\
&= (x-\sqrt{3x}+1)(x+\sqrt{3x}+1)
\end{align}$$
Technically neither $(x-\sqrt{3x}+1)$ nor $(x+\sqrt{3x}+1)$ are linear functions,... | What you propose is not a polynomial factorisation. The polynomial function $x^2-x+1$ is defined for all $x$, whereas your ‘factorisation’ requires $x\ge 0$.
Anyway, since the roots of $x^2+x+1$ are the non-real cubic roots of unity $j$ and $j^2$, the roots of $x^2-x+1$ are $-j$ and $-j^2$n, whence the factorisation
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$ Got these problems as separate sections of a question in a book's chapter on 'Divisibility & primes'.
*
*Show that if $x^2 + 5y^2 =1$, then $x = \pm 1$.
Can state it in terms of two factors as : $(x + y\sqrt{-5})(x-y\sqrt{-5}) =x^2 + 5y^... | For 3, the squares modulo $5$ are $-1,0,1$, so the primes are: $5,11,19,29,31,41$.
Question 2 doesn't make sense. Need provide context.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Extremal problem with positive integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}-b\sqrt{7}>0$. Find the minimum value of
$$
S=(a\sqrt{3}-b\sqrt{7})(a+b).
$$
Attempt I have tried and guess that the minimum value of $S$ is $(55+36)(55\sqrt{3}-36\sqrt{7})$, where $(55,36)$ is the integer solut... | We need to minimise
$$S = \sqrt3 \left(1+\frac{a}b\right)\cdot \color{blue}{b^2\left(\frac{a}b-\sqrt{\frac73} \right)}$$
There are infinitely many convergents from the regular continued fraction expansion of $\alpha = \sqrt{\frac73}$ which keep the terms in blue bounded, i.e. satisfy $b^2\left(\frac{a}b-\alpha \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Can I use the method of seperation variables to solve the following ODE Can I use the method of seperation variables to solve the following ODE
$$F'(x)=\frac{a}{(F(x))^2}+c$$
where $a,c>0$
If without $c>0$, I can solve it , but now how can I solve this equation?
| Rearrange to get $$\frac{F^2}{a+cF^2} dF = dx$$ which is a separable equation. Integrate both sides to get $$x = \int \frac{F^2}{a+cF^2} dF = \frac 1a \int \frac{F^2}{1+\frac caF^2} dF.$$ Let $\tan u = \sqrt{\frac ca} F$, then $\sec^2u \,du = \sqrt{\frac{c}{a}} \, dF$ so we get $$x = \frac 1a \int \frac{\frac ac \tan^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice f... | $$4^3\equiv 2^3\equiv 1\pmod{7} \tag{A}$$
$$ 4^{3x}\equiv 2^{3x}\equiv 1\pmod{7}\tag{B}$$
$$ 4^{3x+1}+2^{3x+1}+1\equiv 4+2+1 \equiv 0\pmod{7}.\tag{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How to prove that 2 is not a Gaussian prime? To prove an algebraic number is not a Gaussian prime, need find factors of it.
Let us assume, $(a + bi)(x +yi) = 2, \exists a,b,x,y \in \mathbb{Z}$; so $(ax -by) + i(ay + bx) =2$.
As imaginary part is null in $2$, so $ax - by =2, ay + bx =0$.
Now, how to reach from here, the... | Let $(a+ib)(c+id)=2$.
There is obviously no solution with real integers, so $b$ and $d$ are both nonzero (if only one is zero, the product is complex or 0, so it also fails). Also, since the product is real, their arguments are opposite: they are multiples of a conjugate pair of gaussian integers.
So it's really $k(a+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve the recurrence relation $y_{n+2}+2y_{n+1}+2y_n=n^3.$ First I solved the characteristic equation $r^2+2r+2=0$ with $r_1=-1+i$ and $r_2=-1-i.$ In polar form this becomes $r_{1,2}=\sqrt{2}e^{\pm i3\pi/4},$ thus the general solution to the homogenous equation is $$y_h=\sqrt{2}^n\left(A\cos{\frac{3\pi}{4}n}+B\sin{\fra... | You might do better to choose $n+1, n, n-1$ which simplifies the arithmetic and promises more cancellation. Also you might organise the computation so that you get $$(n-1)^3=a\left((n+1)^3+2n^3+2(n-1)^3\right)+b\left((n+1)^2+2n^2+2(n-1)^2\right)+c\left((n+1)+2n+2(n-1)\right)+5d=$$$$=5an^3+(-3a+5b)n^2+(8a-4b+5c)n+(-a+3b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\co... | $$\begin{align}\tan A+\cot A&=\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\\&=\dfrac{\sin^2A+\cos^2A}{\sin A\cos A}\\&=\dfrac1{\sin A\cos A}\\&=\dfrac2{2\sin A\cos A}\\&=\dfrac2{(\sin A+\cos A)^2-1}\\&=\dfrac2{(\sqrt2)^2-1}\\&=2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \f... | Let $y=-x:$
$F(y) = \dfrac{y^2 -\sqrt{y^4 +1}}{-y^3 - \sqrt{y^6+1}}=$
$(-1)\dfrac{y^2 - \sqrt{y^4+1}}{y^3 +\sqrt{y^6+1}}.$
Conjugates and rearrangement , and then considering $\lim_{y \rightarrow \infty}$ should get you there .
Or:
$0\le |F(y)| \le \dfrac{2\sqrt{y^4+1}}{2y^3} \lt \dfrac{4y^2}{2y^3}$.
And the limit is?.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
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Differences between using partial fractions or completing the square to solve an integral? I have this question
$$\int \frac{1}{4x^2-4x-3}\, dx$$
I tried to solve it by using completing square method, and I got
$$\frac{1}{4} \arctan\left(\frac{2x-1}{2}\right),$$
but when I saw the answers, I found that it should be s... | $$\int { \frac { dx }{ 4x^{ 2 }-4x-3 } } =\int { \frac { dx }{ \left( 2x+1 \right) \left( 2x-3 \right) } } =-\frac { 1 }{ 4 } \int { \left[ \frac { 1 }{ 2x+1 } -\frac { 1 }{ 2x-3 } \right] dx } \\ =-\frac { 1 }{ 4 } \left[ \int { \frac { dx }{ 2x+1 } } -\int { \frac { dx }{ 2x-1 } } \right] =-\frac { 1 }{ 8 } \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Integral by change of variable $\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $ What is $$\int_0^{\pi} \sin^4(x)\cos^6(x)\,dx $$
Putting $\sin(x) = t$, then $\cos(x) = \sqrt{1-t^2}$ and $\cos(x)\,dx = dt$
At $x = 0$, $\sin(x) = 0$, $\therefore\ t = 0$
At $x = \pi$, $\sin(x) = 0$, $\therefore\ t = 0$
Integral becomes
$$\int_0^0 t... | $$\sin^4x\cos^6x=\frac{1}{16}\cos^2x\sin^42x=\frac{1}{128}(1+\cos2x)(1-\cos4x)^2=$$
$$=\frac{1}{128}(1+\cos2x)\left(1-2\cos4x+\frac{1+\cos8x}{2}\right)=$$
$$=\frac{1}{256}(1+\cos2x)(3-4\cos4x+\cos8x),$$
which says that our integral it's
$$\int\limits_0^{\pi}\frac{3}{256}dx=\frac{3\pi}{256}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Geometry problem involving circles and triangles This is the question:
$E$ is a point on the side $AB$ of a rectangle $ABCD$ such that $DE=6$ cm, $DA= 8cm$ and $DC=6cm$. If $CE$ extended meets the circumcircle of the rectangle at $F$, then what is the value of $BF$ ? (use $\sqrt{2}$=$1.414$).
Here is how I tried to so... |
If $DA=8$ and $DE=6$, then $AE=2$.
Since $AC$ and $BD$ are diagonals of a $6 \times 8$ rectangle, $AC = BD = 10.$
$\triangle CDE$ is an isosceles right triangle. So $CE = 6 \sqrt 2.$
$\triangle CDE \sim \triangle AFE$. So $\triangle AFE$ is an isosceles right triangle. Since $AE=2$, then $EF = AF = \sqrt 2.$
$\triangl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Verify that for $k=3/2 ,$ $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$ This is a part of a proof I am studying on:
Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\}) $
For $k=3/2 ,$ it's shown that $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$
How is it conc... | Look at the Taylor expansion $$\sin t = t - \frac{t^3}{3!} +\frac{t^5}{5!} - \frac{t^7}{7!} + \cdots$$Now put $t = 2x$:$$\sin 2x = 2x - \frac{8x^3}{3!} +\frac{32x^5}{5!} - \frac{128x^7}{7!} + \cdots$$Divide by $2x$: $$\frac{\sin 2x}{2x} = 1 - \frac{4x^2}{3!} +\frac{16x^4}{5!} - \frac{64x^4}{7!} + \cdots$$Plugging $k=3/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder when $27^{40}$ is divided by $12$ What is the remainder when ${27}^{40}$ is divided by $12$ ?
The answer is supposed by $9$ but i’m getting it’s as $3$ . Please correct me in my approach to the problem.
This is how I did it :-
${27}^{40} = {(3^3)}^{40} = {3}^{120}$
$\frac{{3}^{120}}{12} = \frac{3^{119}}{4}$... | $27(4K) + 27$ is the numerator in your fraction $\frac{(3^{4})^{29}3^3}{4}$. Did you forget about the $4$ in the denominator? Here is another approach to solve the problem. The remainder of 27 in dividing by $12$ is $3.$ Thus the remainder of $27^{40}$ is the same as the remainder of $3^{40}$ which is $3(27^{13}).$ Rep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2}\ge 1$ for real numbers $x$, $y$, and $z$ with $xyz=1$. I've been thinking to prove each one individually then sum them up which seems too complex to because I haven't made any notable success proving $\frac{x^2}{(x-1)^2}> 1$.
I know that $(x-1)^... | If
$x=1+a, y=1+b, z=1+c$
then
$a, b, c \ge 0$
since all are positive integers
and
$\begin{array}\\
\dfrac{x^2}{(x-1)^2} + \dfrac{y^2}{(y-1)^2} + \dfrac{z^2}{(z-1)^2}
&=\dfrac{(1+a)^2}{a^2} + \dfrac{(1+b)^2}{b^2} + \dfrac{(1+c)^2}{c^2}\\
&=1+\dfrac{1+2a}{a^2} +1+ \dfrac{1+2b}{b^2} +1+ \dfrac{1+2c}{c^2}\\
&=3+\dfrac{1}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Dividing polynomial fractions with varying term quantities I'm working through an old algebra book as a refresher and I've come across what should be a simple polynomial division. The exercise prompts the reader to perform the following operation:
$$
\frac{a^2-9}{a^2+3a} \div \frac{a-3}{4}
$$
I started off by inverting... | Because $$\frac{a^2-9}{a^2+3a} \cdot \frac{4}{a-3}=\frac{4(a-3)(a+3)}{a(a+3)(a-3)}=\frac{4}{a}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all ordered pairs such that $x + y^2 = 2$ and $y + x^2 = 2$ How can I find all ordered pairs $(x,y)$ such that $x + y^2 = 2$ and $y + x^2 = 2$?
A solution/explanation would be greatly appreciated.
If it helps, the solutions given in the textbook are
x=-2, y=-2; x=1, y=1; x= (1+root 5)/2, y=(1-root 5)/2; ... | HINT
$$y+x^2= x+y^2\iff y-x-y^2+x^2=0\implies y-x-(y-x)(y+x)=0\implies (y-x)(1-x-y)=0$$
thus
$$y-x=0 \implies y=x \implies x^2+x-2=0$$
or
$$x+y=1\implies x^2-x-1=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right).$ This is what my prof did:
Rewrite as $$x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\frac{x}{e}\left(1-e\cdot e^{-x\ln{\left(1+\frac{1}{x}\right)}}\right).$$
Here he uses that
$$\ln{(1+t)}=t-\frac{t^2}{2}+O(t^3),$$
so
$$\... | $$\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\lim_{x\rightarrow0}\frac{\frac{1}{e}-\left(\frac{\frac{1}{x}}{\frac{1}{x}+1}\right)^{\frac{1}{x}}}{x}=\lim_{x\rightarrow0}\frac{(1+x)^{\frac{1}{x}}-e}{ex(1+x)^{\frac{1}{x}}}=$$
$$=\frac{1}{e^2}\lim_{x\rightarrow0}\left((1+x)^{\frac{1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the equation of the line through the point $(1,-1)$ Find the equation of the line through the point $(1,-1)$ which cuts off a chord of length $4\sqrt {3}$ from the circle $x^2+y^2-6x+4y-3=0$
My Attempt:
Let the equation of line be $y=mx+c$
It passes through $(1,-1)$ so
$$-1=m+c$$
$$c=-1-m$$
| $$(x-3)^2+(y+2)^2=3+9+4=4^2$$
Let the center be $O(3,-2)$
Let one of the intersections of the line with the circle be $P$
So, $|OP|=4$
Let the perpendicular from the center to the line intersect at $Q$
Using The perpendicular from the center of a circle to a chord bisects the chord (Proof)
$|PQ|=2\sqrt3$
$$OP^2=PQ^2+OQ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How to evaluate the integral $\int_0^{\pi/2}x^2(\sin x+\cos x)^3\sqrt{\sin x\cos x} \, dx$? How to evaluate the integral $$\int_0^{\pi/2}x^2(\sin x+\cos x)^3(\sin x\cos x)^{1/2} \, dx \text{ ?}$$
I tried to subsititution $x=\frac{\pi}{2}-t$, but it doesn't work. can someone
help me, any hint or trick are appreciated.
| $$\color{blue}{\int_0^{\pi/2} {{x^2}{{(\sin x + \cos x)}^3}\sqrt {\sin x\cos x} dx} = \frac{\pi}{384\sqrt{2}}\left(35 \pi ^2-84 \ln ^22+132 \ln 2-150 \right)}$$
Note that $$\int_0^{\pi/2} {{x^2}{{(\sin x + \cos x)}^3}\sqrt {\sin x\cos x} dx} = \int_0^{\pi/2} {{x^2}{{(\sin x + \cos x)}^3}\sin x\sqrt {\cot x} dx}$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$ QuestionProve that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$=
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$
My Approach R.H.S
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$=$\frac{x^{2^{k-1}\left(1-x^{2}\ri... | $x^{2^k}=x^{2\times2^{k-1}}=(x^{2^{k-1}})^2$
⇒$1-x^{2^k}=(1-x^{2^{k-1}})(1+x^{2^{k-1}})$
⇒ $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^k}}=\frac{1+x^{2^{k-1}}-1}{1-x^{2^k}}=\frac{x^{2^{k-1}}}{1-x^{2^k}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Solve $(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$ I wish to solve the equation
$$(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$$
I tried factorizing until I reached
$(\sqrt{2}+\sqrt{3})^x + (\sqrt{2}-\sqrt{3})^x = 10$
But from there I don't know what to do any help would be welc... | Since $\sqrt{3} - \sqrt{2} > 0$, your equation should simplify to
$$ (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10 $$
Also note that
$$ \sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3}+\sqrt{2}} $$
Let $t = (\sqrt{3}+\sqrt{2})^x$, then
$$ t + \frac{1}{t} = 10 $$
$$ t^2 - 10t + 1 = 0 $$
which gives $t = 5 \pm 2\sqrt{6}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator. Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $... | For an approximation, we might start (using a not generally adopted, but not unfamiliar notation) with$$H^{(-1/2)}_n=1 + \sqrt{2} + \ldots + \sqrt{n}=\frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + \zeta\left(-\frac12\right)+o(1),$$ mentioned in this question ($\zeta$ is the famous Riemann zeta function).
Then, $$\sqrt{2n}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Negafibonacci Identity Consider the sequence of the negafibonacci numbers: $$..., \overbrace{-8}^{F_{-6}}, \overbrace5^{F_{-5}}, \overbrace{-3}^{F_{-4}}, \overbrace2^{F_{-3}}, \overbrace{-1}^{F_{-2}}, \overbrace1^{F_{-1}}$$
I have discovered that the following identity holds for integers I've tried:
$$\sum_{n=1}^kF_{-... | Here is an answer based upon generating functions. But at first note that the identity
\begin{align*}
\sum_{k=1}^nF_{-k}=1-F_{1-n}\qquad\qquad n\geq 1\tag{1}
\end{align*}
gives for $n=1$: $F_{-1}=1-F_{0}$ and so we have to add the initial condition
\begin{align*}
F_{0}=0
\end{align*}
in order to fully specify the iden... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derive the conditions $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ and $xy>-1$ for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \text{, }xy<1\\
\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \text{, }xy>-1
$$
But, How do I reach the conditions $... | We prove $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$. The other inequality is likewise. Suppose not. Therefore $xy>1$. The case $xy=1$ would be discussed later. Also the case where at least one of them is zero is trivial, so we assume both to be nonzero. Then we have four cases to check:
Case 1: $x,y>0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
existence of solutions of $a^n+b^n+c^n=6^n$ $$2^2+4^2+4^2=6^2$$
$$3^3+4^3+5^3=6^3$$
There is no such thing for exponent 4. It's not hard to prove that the only integer solutions of $a^4+b^4+c^4=6^4$ are trivial ones.
For which $n$ there exist $a,b,c\in\mathbb{N}$ such that $a^n+b^n+c^n=6^n$?
(own question, a very prob... | There are only solutions for $n=2,3$ as you suspect.
I assume you're looking for non-trivial solutions, so $0\leq a,b,c \leq 5$. Assuming that $a,b,c$ satisfy your equation, then
$$6^{n}=a^{n}+b^{n}+c^{n}
\leq 5^{n}+5^{n}+5^{n} = 3\times 5^{n} \\
\Rightarrow n \leq \frac{\log(3)}{\log(6)-\log(5)}\approx 6.0257$$
So tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$ Let $a, b, c$ be real numbers such that $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$. Prove that $a^2+b^2+c^2\gt \frac {1}{2018}$.
It looks like an easy question, but I thought for a while a... | By C-S $$a^2+b^2+c^2=$$
$$=\frac{\sqrt{(3^2+28^2+35^2)(a^2+b^2+c^2)\cdot(20^2+23^2+33^2)(a^2+b^2+c^2)}}{2018}\geq$$
$$\geq\frac{(3a+28b+35c)(20a+23b+33c)}{2018}=\frac{1}{2018}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is this : $\sqrt{3+\sqrt{2+\sqrt{3+\sqrt{2+\sqrt{\cdots}}}}}$ irrational number? I'm confused how do i show if the bellow number could be irrational or no
$$\sqrt{3+\sqrt{2+\sqrt{3+\sqrt{2+\sqrt{\cdots}}}}}$$ ? can I use irrationality of $\sqrt{2}+\sqrt{3}$ ?
| As Masacroso has already pointed out in the comments, you have to find rational solutions to $c=\sqrt {3+ \sqrt {2+c}}$.
$c^2=3+\sqrt {2+c}$
$c^2-3=\sqrt {2+c}$
$(c^2-3)^2=2+c$
$c^4-6c^2+9=2+c$
$c^4-6c^2-c+7=0$
So the answer to your question is whether there are any rational solutions to $c^4-6c^2-c+7=0$ or not.
Actu... | {
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"url": "https://math.stackexchange.com/questions/2610818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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$\tan{A} \cdot \tan{B} \cdot \tan{C}=9$, find $\tan^2{A}+\tan^2{B}+ \tan^2{C}$ In $\triangle{ABC}$,
$$\tan{A}\cdot \tan{B}\cdot \tan{C}=9$$
$$\tan^2{A}+\tan^2{B}+ \tan^2{C}=\lambda$$
then,
$\lambda$ lies in the interval?
| As lab bhattacharjee noted in comments, for triangle angles we have
$$\tan A\cdot \tan B \cdot \tan C = \tan A + \tan B + \tan C.\tag{1}$$
Denote $\;\;x=\tan A,\quad y=\tan B, \quad z=\tan C$.
We have
$$x+y+z=9;\\
xyz=9;\tag{2}$$
and we try to estimate expression $$x^2+y^2+z^2.$$
Let use $z$ as parameter ($0<z<9$). Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How and where can I calculate $\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots\right)?$ We have
$$A=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\right)=\ln(2)$$
$$B=\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\cdots\right)$$
Here in $A$ $+-+-$, in $B$ $+... | $$B=\sum_{k\geq 0}\left[\frac{1}{4k+1}+\frac{1}{4k+2}-\frac{1}{4k+3}-\frac{1}{4k+4}\right]=\int_{0}^{1}\sum_{k\geq 0}x^{4k}(1+x-x^2-x^3)\,dx $$
$$ B = \int_{0}^{1}\frac{1+x-x^2-x^3}{1-x^4}\,dx = \int_{0}^{1}\frac{1+x}{1+x^2}\,dx = \color{red}{\frac{\pi}{4}+\frac{\ln 2}{2}}.$$
| {
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"url": "https://math.stackexchange.com/questions/2624829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5... | Just to avoid L' Hospital's rule, consider the following:
$$\frac{x^{1/5}-1}{x^{1/6}-1}=\frac{x^{6/30}-1}{x^{5/30}-1}=\frac{(x^{1/30})^6-1}{(x^{1/30})^5-1}=\frac{(x^{1/30}-1)(x^{5/30}+x^{4/30}+\dots+1)}{(x^{1/30}-1)(x^{4/30}+x^{3/30}+\dots+1)}$$
So:
$$\lim_{x\to1}\frac{x^{1/5}-1}{x^{1/6}-1}=\lim_{x\to1}\frac{(x^{1/30}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 7
} |
Finding minimum of $\max(\arg(a),\arg(b),\arg(c))$ in complex number
If $a,b,c$ be $3$ complex number (not all real ) such that $|a|=|b|=|c|=1$ and $2(a+b+c)-3abc$ is real , Then minimum of $\max(\arg(a),\arg(b),\arg(c))$ Given argument of $a,b,c$ all are positive.
Try: Assuming $z =2(a+b+c)-3abc$. Then $z=\bar{z}$
S... | We claim that the answer is $\dfrac{\pi}{6}$ and proceed to argue by contradiction.
Let $a = e^{ix}, b = e^{iy}, c = e^{iz}$ and assume to the contrary that there exists $x,y,z \in [0, \dfrac{\pi}{6})$, such that $$2(e^{ix}+e^{iy}+e^{iz})-3e^{i(x+y+z)} = 2(e^{-ix}+e^{-iy}+e^{-iz})-3e^{-i(x+y+z)}$$ or $$2(\sin x+\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Jordan's Canonical Form of a Matrix I'm a little tied up about Jordan's canonical form of a square matrix. How would Jordan's Canonical form of the following square matrix be obtained?
$$A = \begin{pmatrix}
2 & 1 & 1 \\
-2 & -1 & -2\\
1 & 1 & -2
\end{pmatrix}$$
| You have to find a Jordan basis for $A$. As $-1$ is a double eigenvalue, to determine whether $A$ is diagonalisable, we have toknow whether the eigenspace for this value has dimension $2$ or not, and in the latter case determine a basis for the generalised eigenspace.
Now $\ker(A+I)$ has dimension $1$ since
$$A+I= \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trig substitution for $\int \frac{x^2dx}{\sqrt{4 - x^2}}$ According to my textbook the answer is: $2\arcsin(\frac{1}{2}x) - \sin(2\arcsin(\frac{1}{2}x))$, but I'm getting something slightly different.
First I tried setting $x = 2 \sin \theta, dx = 2 \cos \theta, \theta = \arcsin(\frac{x}{2})$
$$\int \frac{4\sin^2 \thet... | You've apparently used an incorrect trigonometric identity,
$$\sin^2 t = \frac{1 + \sin 2t}{2}.$$
This is correct only if you erase all the sines and write cosines instead; it is false if $t = 0$, for example.
The correct identity for $\sin^2 t$ is (among several others, of course)
$$\sin^2 t = \frac{1 - \cos 2t}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is wrong with my solution to this equation : $346 x+ 1250 \equiv 49 \pmod{105}$? I'm not sure how to get $x$ but with my way $x = 39$. The solution for $x$ is 29 though. Could anyone possibly help me with the calculating method?
Mine:
$\gcd(346,105) = 1 \Rightarrow x = a^{-1} \cdot b \pmod m \Rightarrow 346^{-1}... | Note that
$$346x+ 1250 \equiv 49 \pmod {105} \iff 31x \equiv 59 \pmod {105}$$
thus since $gdc(31,105)=1$ we can find the inverse of $31 \pmod {105}$ by Euclidean's algorithm.
As an alternative by CRT we get
$$\begin{cases}31x \equiv 59 \pmod {3}\iff x \equiv 2 \pmod {3} \\31x \equiv 59 \pmod {5} \iff x \equiv 4 \pmod {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$
Evaluate:
$$\lim_{x\to0}\frac{x\sin{(\sin{x})}-\sin^2{x}}{x^6}$$
I have been trying to solve this for $15$ minutes but sin(sin(x)) part has me stuck.
My attempt:
I tried multiplying with $x$ inside the $\sin$ as $\sin{(\frac{x\sin{x}}{x})}$. No leads.
| $$\sin(u)=u-\frac{u^3}{6}+\frac{u^5}{120}+o(u^6)$$
Then, $$\sin^2(x)=(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6))^2 =x^2-\frac{x^4}{3} +o(x^6)$$
and
$$\sin(\sin(x))=\sin\left(x-\frac{x^3}{6}+\frac{x^5}{125}+o(x^6)\right) \\=x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6) -\frac{1}{6}\left(x-\frac{x^3}{6}+\frac{x^5}{120}+o(x^6)\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given is markov chain - Determine the probability $f_1(n)$
Given is markov chain $\left\{X_n\right\}_{n \in \mathbb{N}}$ with
transition probabilities
$$M= \begin{pmatrix}
1/2 & 1/2 & 0 & 0 & 0 & 0 \\
1/4 & 3/4 & 0 & 0 & 0 & 0 \\
1/4 & 1/4 & 1/4 & 1/4 & 0 ... | You need this:
$$
f_1(n) = \begin{pmatrix}
1, &
0, &
0, &
0, &
0, &
0
\end{pmatrix} \begin{pmatrix}
1/2 & 1/2 & 0 & 0 & 0 & 0 \\
1/4 & 3/4 & 0 & 0 & 0 & 0 \\
1/4 & 1/4 & 1/4 & 1/4 & 0 & 0 \\
1/4 & 0 & 1/4 & 1/4 & 0 & 1/4 \\
0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve cubic equation $x^3 - 15x - 4 = 0$ with Cardano's method Suppose that we wanted to solve
$$x^3 - 15x - 4 = 0$$
by using Cardano's method. In doing so, I end up with
$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$
However, by checking my answer, it seems there are two other answers I haven't accounted for: $\sqrt{3} - 2$ ... | The Cardano’s formula is normally used to calculate the real root of the a cubic equation with one real root and a pair of complex roots. In the case of three real roots, e.g. $x^3 - 15x - 4 = 0$, the formula results in the expression
$$\sqrt[3]{2 + 11i} + \sqrt[3]{2-11i}$$
which is of multiple complex values. However,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using induction to prove sum of certain number-theoretic function. For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$.
Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and
$N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number,... | I checked it and didn't find any mistakes. Nice identity!
I don't think you used any facts about primes anywhere, so if it's true it would be true for any sub-sequence of natural numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to approach this question on choosing balls from a box?
A box contains $2n$ balls of $n$ different colors, with 2 of each color. Balls are picked at random from the box with replacement until two balls of the same color have appeared. Let $X$ be the number of draws made.
a) Find a formula for $P(X > k)$, $k = 2, 3... | You interpreted the problem as if you pre-selected one of the colors before drawing any balls and require to draw two balls of that color, for example if there are red balls you might choose red and then you will stop only when you have drawn red twice. I would interpret it differently--if any color is drawn that has b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to simplify the equation $ a\sin x\cos y + b\cos x\sin y = c\sin(2x)$? I am trying to derive the relation between x and y which are two physical parameters.I am trying to express x in terms of y, a and b. Is there a way to do it?
| You can rewrite your equation as
$$b \cos x \sin y = \sin x \;( 2 \cos x - a \cos y ) $$
Squaring, we have
$$b^2 \cos^2 x \sin^2 y = \sin^2 x (2 \cos x - a \cos y )^2$$
So, writing $p:= \cos x$ and $q := \cos y$,
$$b^2 p^2 (1-q^2) = (1-p^2) (2p - a q )^2$$
This gives a quartic polynomial in $p$:
$$4 c^2 p^4 - 4 a c p^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\si... | This calls for Kepler's angle! Let
$$\sin\theta=\frac{\sqrt{1-a^2}\sin\psi}{1+a\cos\psi}$$
Then
$$\begin{align}\cos\theta&=\frac{\cos\psi+a}{1+a\cos\psi}\\
d\theta&=\frac{\sqrt{1-a^2}}{1+a\cos\psi}d\psi\\
1-a\cos\theta&=\frac{1-a^2}{1+a\cos\psi}\end{align}$$
Also when $\theta$ makes a full cycle, so does $\psi$, so
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
A simple problem with a simple and a nonsimple solution I have this seemingly simple problem to solve. Its statement is straightforward:
"If $x+y+z=3,x^2+y^2+z^2=5, x^3+y^3+z^3=7$ show that $x^4+y^4+z^4=9,x^5+y^5+z^5\neq 11$".
There is a highschool approach, that is I expanded $(x+y+z)^2, (x+y+z)^3$ and managed to get... | $$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=$$
$$=25-2((xy+xz+yz)^2-2xyz(x+y+z))=$$
$$=25-2\left(4-2\left(-\frac{2}{3}\right)3\right)=9.$$
$$x^5+y^5+z^5=(x^2+y^2+z^2)(x^3+y^3+z^3)-\sum_{sym}x^3y^2=$$
$$=35-(x^2y^2+x^2z^2+y^2z^2)(x+y+z)+xyz(xy+xz+yz)=$$
$$=35-8\cdot3-\frac{2}{3}\cdot2=\frac{29}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Tangent substitution in trigonometric substitution Find:
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}}\;dx$$
Let $u=2x$ and with tangent substitution we have $x=\frac{3}{2}\,tan\,\theta$ and now we have $dx=\frac{3}{2}\,sec^2\,d\theta$.
Also, $\sqrt{4x^2+9}=3\,sec\,\theta$.
When $x=0$, $tan\,\theta=0$, so $\theta=0$... | $$\frac{1-u^2}{u^2}=\frac{1}{u^2}-\frac{u^2}{u^2}=u^{-2}-1=-(1-u^{-2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve in $\mathbb{Z}$ the equation $x^4 + 1 = 2y^2$. Find all pairs of intergers $(x,y)$ such that $x^4 + 1 = 2y^2$.
I'm thinking of Gaussian integers, since the LHS can be factored in $\mathbb{C}$. But I don't know how to continue here.
| We consider $x, y$ positive (if $x$ is solution then $-x$ is solution, same with $y$). From $x^4+1=2y^2$ we deduce that $2y^2=t^2+1$, where $t=x^2$ must be odd. Let $t=2k+1$, so $y^2=k^2+(k+1)^2$. We can now use the fact that $y=(m+n)^2$ and $k=(m-n)^2$ and $(k+1)=2mn$ or $k=2mn$ and $(k+1)=(m-n)^2$, where $m, n\in \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
... | Here is a sneaky way to deal with one of those derivatives. Notice that $h(x)=\cosh^{-1}(x)$, therefore
$$
h’(x)=\frac{1}{\sinh\left(\cosh^{-1}(x)\right)}
= \frac{1}{\sqrt{\cosh^2\left(\cosh^{-1}(x)\right)-1}}=\frac{1}{\sqrt{x^2-1}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Dividing balls into boxes with capacity limits Problem:
In how many ways can you divide $13$ identical balls into $3$ different boxes
$k_1$, $k_2$, $k_3$, such that $k_1$ contains no more than $5$ balls, $k_2$ contains no more than $6$ balls and $k_3$ contains no more than $4$ balls?
My idea:
So my idea is to use the ... | The answer can simply be given using generating functions. We simply need to calculate the coefficient of $x^{13}$in the expansion of $$(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+x^5)(1+x+x^2+x^3+x^6)$$
$$=\frac {(1-x^5)}{(1-x)}\cdot \frac {(1-x^6)}{(1-x)}\cdot \frac {(1-x^7)}{(1-x)}$$
$$=\frac {1-x^5-x^6-x^7+x^{11}+x^{13}+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks l... | Your idea was good:
$$\sqrt[3]{(n^3 + n)^2} = n^2\sqrt[3]{(1 + 1/n^2)^2},$$
$$\sqrt[3]{n^6 - n^2} = n^2\sqrt[3]{1 - 1/n^4},$$
$$\sqrt[3]{(n^3 - n)^2} = n^2\sqrt[3]{(1 - 1/n^2)^2}.$$
So
$$
\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^6 - n^2} + \sqrt[3]{(n^3 - n)^2}\sim 3n^2
$$
and
$$
\frac{2n}{\sqrt[3]{(n^3 + n)^2} + \sqrt[3]{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Calculate the limitation:$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x $ $$I=\lim_{n\rightarrow \infty}\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x
$$
the answer is 0. This is what I've done so far.
$$ \begin{split}
\int_0^{2\pi}{\frac{\cos nx}{x+1}}\text{d}x&=\frac{1}{n}\int_0^{2\pi}{\frac{1}... | Following your intuition, an inequality is useful at the point you got to, namely $|\sin(x)| \leq 1$ Using that, we have
$$\left|\frac{1}{n}\int_0^{2\pi} \frac{\sin(nx)}{(1+x)^2}dx\right| \leq\frac{1}{n}\int_0^{2\pi}\frac{1}{(x+1)^2}dx=\frac{1}{n}\frac{-1}{x+1}\bigg|_0^{2\pi}=\frac{1}{n}\left(1-\frac{1}{2\pi+1}\right)<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad) I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:
If
$$x \cos\theta+y\sin\theta=a$$
$$x\sin\theta-y\cos\theta=b$$
then $$\tan\theta=\frac{bx+ay}{a... | $x \cos\theta + y\sin\theta = a, \,\, x \sin\theta - y\cos\theta = b$. Therefore:
$$\dfrac{x \cos\theta + y\sin\theta}{x \sin\theta - y\cos\theta} = \dfrac{a}{b}$$
$$\implies \dfrac{x + y\tan\theta}{x \tan\theta - y} = \dfrac{a}{b}$$
$$\implies bx + by\tan\theta = ax \tan\theta - ay$$
$$\implies (ax - by) \tan\theta = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
On near-Pythagorean triples $(n^5-2n^3+2n)^2 + (2n^4-2n^2+1)^2 = n^{10} + 1$ We have,
$$\begin{aligned}
(n^3-2n)^2 + (2n^2-1)^2 &= n^6 + 1\\
(n^5-2n^3+2n)^2 + (2n^4-2n^2+1)^2 &= n^{10} + 1\\
(n^7-2n^5+2n^3-2n)^2 + (2n^6-2n^4+2n^2-1)^2 &= n^{14} + 1\end{aligned}$$
and so on.
Q: What is a clever way to show that this pat... | here is one way
$n^{4k+2}+1 = (n^{2k+1}-i^{2k+1})(n^{2k+1}-(-i)^{2k+1}) \\
= (n-i)(n^{2k}+ \cdots + i^{2k})(n+i)(n^{2k}+\cdots + (-i)^{2k+1}))\\
= (n+i)(n^{2k}+ \cdots + i^{2k})(n-i)(n^{2k}+\cdots + (-i)^{2k+1}))\\
= (n^{2k+1}+2in^{2k}+\cdots +2i^{2k}n+i^{2k+1})(n^{2k+1}+2(-i)n^{2k}+\cdots +2(-i)^{2k}n+(-i)^{2k+1}) \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Generator of $SL(2,\mathbb{R})$ Find all 2x2 matrices $ A \in SL(2,\mathbb {R}) $such that $det(A) =1$ and $A\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix} $. Do these matrices form a group?
My thoughts:
I have considered the matrices \begin{pmatrix} 1... | A nonempty subset $S$ of a group $G$ forms a subgroup if for all $A,B\in S$, $AB^{-1}\in S$. In this case, $S$ is nonempty because the identity is in $S$. You then assume that $A$ and $B$ are two matrices with the given property and consider (the $\frac{1}{\sqrt{2}}$ doesn't matter in this problem since it's a scalin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Secant and Tangent identity i've been stuck on this question too long
$x = \sec A + \tan A$
show $x + \frac{1}{x} = 2\cdot \sec A$
I've been using $\tan^2 \theta + 1 = \sec^2 \theta$
and $\tan\theta = \frac{\sin \theta}{\cos\theta}$
help would be much appreciated
$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos... | $$
x=\sec A + \tan A= \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}
$$
so we have:
$$
x+\frac{1}{x}= \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=\frac{1+\sin^2 A+2\sin A+\cos^2 A}{(\cos A)(1+ \sin A)}=
$$
$$
=\frac{2(1+\sin A)}{(\cos A)(1+ \sin A)}=\frac{2}{\cos A}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Divisors and Decimals
For positive integers $n$, let $f(n)$ return the smallest positive integer $k$ such that $\frac1k$ has exactly $n$ digits after the decimal point. How many positive integer divisors does $f(2010)$ have?
I understand what the question is asking for, but I don't understand how to solve it. Here's ... | Suppose $n$ is such that the decimal expansion of ${\large{\frac{1}{n}}}$ has exactly $2010$ digits after the decimal point.
Since the decimal expansion terminates, it follows that $n=2^a5^b$ where $a,b$ are nonnegative integers.
By hypothesis,
$$\frac{1}{n}=\frac{x}{10^{2010}}$$
where $x$ is a positive integer.
Not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,12... | Hint: $y^2 = a \implies y = \color{red}{\pm}\sqrt{a}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the determinant of the following $5\times 5$ real matrix:
Let $A\in\mathbb{R^{5\times5}}$ be the matrix: $\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a\\a&a&b&a&a\\a&b&a&a&a\\b&a&a&a&a\end{array}\right)$
Find the determinant of $A$.
Hey everyone. What I've done so far: $det\left(\begin{array}{l}a&a&a&a&b\\a&a&a&b&a... | ...now take out common factor from rows $1$ to $4$ :
$$(b-a)^4\begin{vmatrix}1&0&0&0&-1\\0&1&0&0&-1\\0&0&1&0&-1\\0&0&0&1&-1\\a&a&a&a&b\end{vmatrix}\;\;(*)$$
and now take $\;aR_1\;$ from row $\;R_5\;$ , then $\;aR_2\;$ for $\;R_5\;$, etc.
$$(*)=(b-a)^4\begin{vmatrix}1&0&0&0&-1\\
0&1&0&0&-1\\
0&0&1&0&-1\\
0&0&0&1&-1\\
0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Integral $\int \sqrt{1+x^2}dx$ I was trying to do this integral
$$\int \sqrt{1+x^2}dx$$
I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form of $\int x^m(a+bx^n)^p\,dx$ and I spent a lot of time on it so I would like to see if it can ... | What don't you do the substitution $x=\sinh(x)$ and use $$\cosh^2(x)=\frac{1+\cosh(2x)}{2}\ \ ?$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$\sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)}$ How to calculate this sum
$$S_n = \sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)} ?$$
Note :
1)
$$S_n = n+\sum_{k=1}^n{\rm cotan}^2\left(\frac{(2k+1)\pi}{2n}\right)$$
2) $S_n$ seems to be equal to $\frac{n^2}{4}$.
3) I kwow how to calcula... | Take a polynomial $P(t)=t^n$ and consider the function $$F_t(z)=\frac{P(tz)-P(t)}{z-1}=t^n\frac{z^n-1}{z-1}.$$
Lagrange interpolation of $F_t$ with basis points on the roots $z_0,z_1,...,z_{n-1}$ of $z^n+1$ (so that $z_k=e^{\frac{\pi(1+2k)}{n}}$) gives:
$$
F_t(z)=\sum_{k=0}^{n-1} F_t(z_k)\frac{z^n+1}{nz_k^{n-1}(z-z_k)}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove that if $3\mid(a^2+b^2)$,then $3\mid a$ and $ 3\mid b$ I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?
Any help is greatly... | We have the fact that
If $a \equiv k \mod m$ and $b \equiv l \mod m$ then $ab \equiv kl \mod m$.
So, if $a \equiv 1 \mod 3$ then $a^2 \equiv 1^2 \mod 3$. And if $a \equiv 2 \mod3$ then $a^2 \equiv 2^2\mod3$. And same goes for $b$ as well. Using these, we don't have many cases to consider. Assuming $3|(a^2+b^2)$, we ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Leading asymptotic behaviour about irregular singular point I am trying to find the leading asymptotic behaviours, using the method of dominant balance, about the irregular singular point $x=0$ for the equation $$x^4\frac{d^2y}{dx^2}+\frac{1}{4}y=0.$$
First let $y=e^{S(x)}, y'= S'e^{S(x)}, y''=[(S')^2+S'']e^{S(x)}$. Ou... | First, I think $C \sim \log(x)$ not $-\log(x)$, because in the equation above, we have
$$ \frac{i}{x^3} + C'' -\frac{1}{4x^4} \color{red}{\text{ minus }} \frac{i}{x^2}C'+(C')^2
+\frac{1}{4x^4} = 0 $$
instead of plus.
Next, let us get the equation for $D$ and as you said. The equation is
$$\big(\color{blue}{S''} + C''... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square.
The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square. List all possible values of $c$. (The letters are the digits of the base 6 number.)
I've rewritten $abcd_6$ into $216a + 36b + 6c + d$, and I know that $d$ ca... | To look at the last two digits in base $6$, we want to look at $4k^2+4k+1$ mod $36$. Since $k^2+k$ mod $9$ repeats with period $9$, $4k^2+4k+1$ mod $36$ also repeats with period $9$. Therefore, the the last two digits of odd squares in base $6$ also repeat with period $9$.
Perhaps more concisely,
$$
\begin{align}
(2(k+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 3,
"answer_id": 2
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$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - ... | HINT.-Your equation is the same as
$$\sqrt{(x-4)(x+4)}-\sqrt{(x-4)(x-4)}=\sqrt{(x-4)(x-1)}$$ which gives besides of $x-4=0$ the equation
$$\sqrt{x+4}-\sqrt{x-4}=\sqrt{x-1}\iff4(x^2-16)=(x+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Fractions in Questions and Answers
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