Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$y(x)$ is a solution of the differential equation Given that $y(x)$ is a solution of the differential equation
$$
x^2 y^{\prime \prime}+x y^{\prime}-4 y=x^2
$$
on the interval $(0, \infty)$ such that $\lim _{x \rightarrow 0^{+}} y(x)$ exists and $y(1)=1$. I need to find the value of $y^{\prime}(1)$.
$$
\begin{aligned}
... | You know that $\displaystyle\lim_{x\to 0^+}y(x)$ exists.
It is equivalent to say that, $\displaystyle\lim_{x\to 0^+}\left(c_1x^2+c_2\frac{1}{x^2}+\frac{\ln(x)}{4}x^2\right)$ exists.
You know that $\displaystyle\lim_{x\to 0^+}\frac{\ln(x)}{4}x^2=0$.
Now, you know that $\displaystyle\lim_{x\to 0^+}\left(c_1x^2+c_2\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expected value to get all possible outcomes for coin flipping trials Consider the experiment where we keep flipping two fair coins. If the random variable X is defined to be the number of trials to get all possible outcomes, find E(X) and Var(X). Since the coins are distinct (H,T) and (T,H) would count as distinct outc... | For one coin
Let's call the number of flips as $n$.
$$
P(X=1) = 0 \\
P(X=n) = \dfrac{1}{2^{n-1}}; n = 2,3,\cdots
$$
$$
\begin{align}
\text{E}[X] &= \sum_{n=2} n\times P(n) = \sum_{n=2} \dfrac{n}{2^{n-1}} = 3 \\
\text{var}[X] &= \text{E}[X^2] - \text{E}[X]^2 \\
&= \sum_{n=2} n^2\times P(n) - 3^2 \\
&= \sum_{n=2} \dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$ I encountered the following binomial equality:
$$\sum_{i=0}^K(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}=\frac{1}{2}(1+(-1)^K)$$
which I know it's true, but I don't know how to prove it directly. I entered the left-hand-side to Mathemat... | Here is a starter. We transform the sum and separate one part which can be simplified. We start with the left hand side of OPs identity and obtain
\begin{align*}
\sum_{i=0}^K&(-1)^i\binom{2n+1-i}{i}\binom{2n-2i}{K-i}\\
&=\sum_{i=0}^K(-1)^i\frac{(2n+1-i)!}{i!(2n+1-2i)!}\,\frac{(2n-2i)!}{(K-i)!(2n-K-i)!}\\
&=\sum_{i=0}^K... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
To prove $1^1\cdot2^2\cdot 3^3...\cdot n^n<(\frac{2n+1}{3})^{\frac{n(n+1)}{2}} $ So we have to prove the following for $n\in N $ $$1^1\cdot 2^2\cdot 3^3...\cdot n^n<\left(\frac{2n+1}{3}\right)^{\frac{n(n+1)}{2}} $$
So I used concept of weighted means (arithmetic and geometric) used AM GM inequality.
$$AM=\frac{a_1w_1+a... | Use the AM-GM inequality [where there are $\frac{n(n+1)}{2}$ terms; indeed for each positive integer $i \le n$, there are $i$ terms of $i$]:
$$\frac{\sum_{1=1}^n i^2}{n(n+1)/2} \ \ge \ \sqrt[\frac{n(n+1)}{2}]{1^12^2 \cdots n^n},$$
where in the above inequality, the LHS represents the arithmetic mean of the above terms ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4598722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
If $\frac{\sin2x}{\cos x}-1=0$, find $x$ given $\ \frac{\pi}{2}\le x\le\pi$ (cancelling out vs multiplying out the $\cos x$) To solve the equation
$$\frac{\sin2x}{\cos x}-1=0 \tag1$$
I manipulated it into the following:
$$\frac{2\sin x\cos x}{\cos x}-1=0 \tag2$$
Then, by multiplying $\cos x$ out, I obtained:
$$2\sin x\... | The confusion is due to the violation of equivalence conditions between mathematical steps.
Problem statement:
$$\frac{\sin2x}{\cos x}-1=0$$
where, $\frac {\pi}{2}\leqslant x\leqslant\pi$.
$\rm Step-1.$
$$\frac{2\sin x\cos x}{\cos x}-1=0$$
There is no issue here. Because: $\sin 2x=2\sin x\cos x$.
$\rm Step-2.$
$$2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
$f(x)=\sqrt[3]{x^2}$ derivative by defintion
Find $f(x)=\sqrt[3]{x^2}$ derivative by definition
my first try was
$\lim_\limits{h \to 0}\frac{\sqrt[3]{(x+h)^2}-\sqrt[3]{x^2}}{h}$
I tried multiplying by $\frac{\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2}+\sqrt[3]{x^2}}$ and got$\lim_\limits{h \to 0}\frac{(x+h)(\... | First of all, I would suggest you to apply the following identity:
\begin{align*}
a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})
\end{align*}
According to it, one concludes the desired result:
\begin{align*}
f'(x_{0}) & = \lim_{x\to x_{0}}\frac{f(x) - f(x_{0})}{x - x_{0}}\\
& = \lim_{x\to x_{0}}\frac{x^{2/3} - x^{2/3}_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A pen-and-paper proof for a matrix implication. Suppose $A = \begin{bmatrix} x & 1\\ y & 0\end{bmatrix}, B = \begin{bmatrix} z & 1\\ w & 0\end{bmatrix}$, for $x,y,z,w \in \Bbb{R}$.
I have observed by considering many examples of $x,y,z,w$ that:
If all the eigen values of $A^2B$ and $AB^2$ are less than one in absolute... | The claim in the question can be rephrased by taking the inverse formulation:
$\det(AB+A+I) < 0 \qquad $ AND $\qquad \det(BA+B+I) < 0$
$\implies$ At least one eigenvalue of $(A^2B$ ; $AB^2)$ is greater or equal than one in absolute value.
As a boundary case, there are indeed situations with
$\det(AB+A+I) = 0$ ; $\det... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Proving the a matrix is bounded for every power I want to show that the following matrix satisfies $\|A^n(z)\| \leq C$ for every $z \in \mathbb{C}$ with $Re(z) \leq 0$, for $n= 0,1,2,...$, and some constant $C$. How do I approach this kind of problems?
$$A(z) = \begin{bmatrix} \frac{z^2+6z+8}{2(2-z)^2} & -\frac{z^2+10z... | Given matrix $M=\{M_{ij}\}$ and $N=\{N_{ij}\}$ where $M_{ij}\in\Bbb C$ and $N_{ij}\in\Bbb R_{>0}$, write $M\prec N$ if $|M_{ij}|<N_{ij}$ for all $i,j$. It is straightforward to verify that if $M\prec N$ and $M'\prec N'$, then $MM'\prec NN'$.
It is enough to prove the following claim.
Claim: Let $z \in \mathbb{C}$ with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Assuming x is small, expand $\frac{\sqrt{1-x}}{\sqrt{1+2x}}$ up to and including the term in $x^{2}$ I have tried this many times but can't quite land on the correct answer.
The correct answer:
$1-\frac{3x}{2}+\frac{15x^{2}}{8}$
These are the steps I took:
*
*Re wrote it as: $\left ( 1-x \right )^{\frac{1}{2}}\left (... | $$\frac{\sqrt{1-x}}{\sqrt{1+2x}}=(1-x)^{1/2}(1+2x)^{-1/2}$$
Consider that $(1+t)^{\alpha}\sim 1+\alpha t+\frac{1}{2}(\alpha -1)\alpha t^2$, hence you obtain:
$$ (1-x)^{1/2}\sim 1-\frac{x}{2}-\frac{x^2}{8} \quad \text{and}\quad (1+2x)^{-1/2}\sim 1-x+\frac{3x^2}{2}$$
It follows that:
$$\left(1-\frac{x}{2}-\frac{x^2}{8}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluating a double integral I was trying to evaluate the following integral
$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx.$$
I have a guess that the value of this integral is $\frac{\pi^4}{8}$. But I am unable to prove it.
Could someone please help me in evaluating this integra... | $$I=\int\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy $$ is not bad since
$$A=\frac{y }{(x^2+ y^2)( 1+y^2)}=\frac{y}{(y-i) (y+i) (y-i x) (y+i x)}$$ Using partial fraction decomposition
$$A=\frac 1{2(x^2-1)}\left(\frac 1{y-i} +\frac 1{y+i}-\frac 1{y-ix}-\frac 1{y+ix}\right)$$ and
$$\int \frac {\log(y)}{y+a}\,dy=\text{Li}_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How to find the mass of a paraboloid region underneath a plane given a non-constant density Given the paraboloid $z=4x^2 + 4y^2$, the plane $z=4$, and the density function $\sigma(x, y, z) = z + 2y$, find the mass of the region bounded by the paraboloid under the plane.
Intuitively, this makes the most sense using cyl... | You almost got it right. It should be$$\int_0^{2\pi}\int_0^1\int_{4r^2}^4\bigl(\color{red}{zr}+2r^2\sin(\theta)\bigr)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\frac{16\pi}3,$$since $\sigma(r\cos(\theta),r\sin(\theta),z)=z+2r\sin(\theta)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\int_0^{\frac{\pi}{2}}x\ln\tan xdx = \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^3}=\frac78\zeta(3)$. Prove that $\int_0^{\frac{\pi}{2}}x\ln\tan xdx = \sum_{n=0}^{+\infty}\frac{1}{(2n+1)^3}=\frac78\zeta(3)$.
I think it's natural to use series, but if I take $u=\tan x$ the integral becomes: $\int_{0}^{+\infty}\frac... | Let's consider a more general case and denote
$\displaystyle I(\alpha)=\int_0^\frac{\pi}{2}\tan^\alpha (x)\,x\,dx=\int_0^\infty t^\alpha\arctan t\,\frac{dt}{1+t^2};\,\alpha\in(-2;1)\tag*{}$
Then
$$I_k=\int_0^{\frac{\pi}{2}}x\ln^k(\tan x)dx=\int_0^\infty \ln^kt\,\arctan t\,\frac{dt}{1+t^2}=\frac{d^k}{d\alpha^k}I(\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4609446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do we prove $x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0$? Question
How do we prove the following for all $x \in \mathbb{R}$ :
$$x^6+x^5+4x^4-12x^3+4x^2+x+1\geq 0 $$
My Progress
We can factorise the left hand side of the desired inequality as follows:
$$x^6+x^5+4x^4-12x^3+4x^2+x+1=(x-1)^2(x^4+3x^3+9x^2+3x+1)$$
However, after... | $$x^4+3x^3+9x^2+3x+1=\left(x^2+\frac32x+1\right)^2+\frac{19}4x^2$$ is always positive.
There is a general theorem that if a real polynomial is always non-negative, it can be written as a sum of squares.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
Proving $\left\lfloor(\frac{1+\sqrt{5}}{2})^{4n+2}\right\rfloor-1$ is a perfect square for $n=0,1,2,\ldots$ Let $$S_n = \left \lfloor\left(\frac{1+\sqrt{5}}{2}\right)^{4n+2}\right\rfloor-1$$ ($n=0, 1, 2, \ldots$).
Prove that $S_n$ is a perfect square.
In Art of Problem Solving website, there is a hint
$$
\begin{align}
... | Let $$X_n=(\frac{1+\sqrt{5}}{2})^{4n-2}=(\frac{3+\sqrt{5}}{2})^{2n-1}=I+f, I \in I^+, 0 < f <1,0 <f'=(\frac{3-\sqrt{5}}{2})^{2n-1}<1$$
$$X_n+f'=I+f+f'
=(\frac{3+\sqrt{5}}{2})^{2n-1}+(\frac{3-\sqrt{5}}{2})^{2n-1}$$
$$= (7+3\sqrt{5})^{n}\frac{2}{3+\sqrt{5}}+(7-3\sqrt{5})^{n}\frac{2}{3-\sqrt{5}}=3A_n-5B_n=K$$
Where $(7+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $y''(x) + \epsilon y'(x) + 1 = 0$ using power series We are given
\begin{align}
\begin{cases}
y''(x) + \epsilon y'(x) + 1 =0, \ 0 < \epsilon <<1\\
y(0)=0, \ y'(0)=1
\end{cases}
\end{align}
and asked to solve this using the solution form $y(x) = \sum_{n=0}^{\infty}\epsilon^{n}y_n(x)$.
Doing the known method for ... | One can solve directly
$$
y''+(ϵy'+1)=0\implies ϵy'(x)+1=(ϵy'(0)+1)e^{-ϵx}=(1+ϵ)e^{-ϵx}
\\
\implies
ϵy(x)+x=(1+ϵ)\frac{1-e^{-ϵx}}{ϵ}=(1+ϵ)\left(x-ϵ\frac{x^2}{2}+ϵ^2\frac{x^3}{3!}-ϵ^3\frac{x^4}{4!}+\dots\right)
\\
y(x)=x-\frac{x^2}{2}-ϵ\left(\frac{x^2}{2}-\frac{x^3}{3!}\right)+ϵ^2\left(\frac{x^3}{3!}-\frac{x^4}{4!}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the real part of $\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}$?
What is the real part of this complex number?
$$\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}$$
I am trying to times conjugate of denominator which will be $$\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}\cdot\frac{e^{-i\theta}-1}{e^{-i\theta}-1}$$
but it make... | The real part of $\large\frac{a+bi}{c+di}$ is $\large\frac{ac+bd}{c^2+d^2}$ so the real part of $\large\frac{e^{(n+1)i\theta}-1}{e^{i\theta}-1}=\frac{\cos(n+1)\theta-1+i\sin(n+1)\theta}{\cos\theta-1+i\sin\theta}$ is
$$\frac{(\cos(n+1)\theta-1)(\cos\theta-1)+\sin(n+1)\theta\sin\theta}{(\cos\theta-1)^2+(\sin\theta)^2}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Determining the sum of $\frac{a_{n+1}}{a_n}$ where $a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$
Let $a_0=1, a_1=\frac{1}{2}, a_{n+1}=\frac{na_n^2}{1+(n+1)a_n}$, then find $\lim_{n\to\infty} \sum_{k=0}^{n}\frac{a_{k+1}}{a_k}=\dots$
We have $(a_n)$ is strictly decreasing as $a_{n+1}-a_n=\frac{-a_n(a_n+1)}{1+(n+1)a_n}<0, a_n>0$ ... | You can rearrange your equation in the following manner
$a_{n+1} (1 + (n+1) a_n) = n a_n^2$
$ na_n^2-(n+1)a_{n+1}a_n = a_{n+1}$
$\frac{a_{n+1}}{a_n} = na_n-(n+1)a_{n+1}$
The terms on the right hand side cancel out once you apply summation on both sides leaving
$\Sigma_{n=1}^{\infty} \frac{a_{n+1}}{a_n} = a_1$.
Note: L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Probability of a sum being divisible by $4$ Roll $n$ six-sided dice.
What is the probability that the sum of the results is divisible by $4$?
This is an 11th grade problem, and I really don't know -after a lot of searching-what tools do I need to solve this problem using only high school knowledge.
| Let $S_n$ denotes the random variable corresponding to the sum modulo $4$ after $n$ throws. You can model $S_n$ as a Markov Chain with $4$ states $\{0,1,2,3\}$ and initial value $S_0 = 0$
The matrix associated with this chain is :
$$M = \frac{1}{6}\begin{pmatrix}
1 & 2 & 2 & 1 \\
1 & 1 & 2 & 2 \\
2 & 1& 1 & 2 \\
2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Without Calculator find $\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$
Without Calculator find $$\left\lfloor 2 \cos \left(50^{\circ}\right)+\sqrt{3}\right\rfloor$$
Where $\left \lfloor x \right \rfloor $ represents floor function.
My Try:
Let $x=2\cos(50^{\circ})+\sqrt{3}$. We have
$$\begin{al... | A calculus-based solution:
As you noticed, we only need to prove that $ \cos 50^{\circ}>\frac{3-\sqrt 3}{2}$.
Because of the identity $\cos 3x=4\cos^3x -3\cos x $, we conclude that $\cos 50^{\circ}$ is a root of the equation:
$$\cos 150 ^{\circ}=\frac{-\sqrt 3}{2}=4t^3-3t.$$
However, it is very easy to see that $4t^3-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
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Showing that the power series converges at the endpoints of its interval of convergence The problem is the following:
Find the power series for $\frac{1}{2}\arctan^2(x)$ and show it converges in the endpoints of the interval of convergence.
My difficulty is in the convergence part.
The power series that I got:
$\displa... | Let $a_n=\frac1{2n+2}\sum_{k=0}^n\frac1{2k+1}.$
You will easily check that $(a_n)$ is decreasing, and the hint serves to prove it tends to $0.$
Then apply the alternating series test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Inequality $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5} $ I have trouble with solving this inequality:
Prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5}$ for a,b,c>0.
Using Cauchy-Schwartz I got this: $\frac{a^3}{3b^3+c^3} +\frac{b... | By Am-Gm inequality we have $$3ab^2\leq a^3+b^3+b^3$$ so $$3ab^2 +2c^3\leq a^3+2b^3+2c^3$$ so $${a^3 \over 3ab^2+2c^3} \geq {a^3\over a^3+2b^3+2c^3}$$ Let $x=a^3$, $y=b^3$ and $z=c^3$. We can assume that $x+y+z=1$, and so we have to prove $$f(x) +f(y) +f(z)\geq 3/5$$ where $$f(x) ={x\over 2-x}$$ The tangent at $x=1/3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4636361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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When is it valid to convert "$\le$" or "$\ge$" to "$=$"? I don't understand why the conversion of ≤ to = in this proof in Spivak's Calculus is legitimate. (The conversion in the previous line from = to ≤ makes sense to me.) Could someone please explain or elaborate on how $(2)$ follows from $(1)$ below?
this proof is... | In case the existing answers are still not clear...
*
*The presentation $$
\begin{align*}
{}& (|a+b|)^2\\
={}& (a+b)^2 \\
={}& a^2+2 a b+b^2 \\
\leq {}& a^2+2|a| \cdot|b|+b^2 \tag{1} \\
={}& |a|^2+2|a| \cdot|b|+|b|^2 \tag{2} \\
={}& (|a|+|b|)^2
\end{align*}
$$ is asserting the conjunction of these 5 statements:
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4638246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solve over natural numbers: $m^3=2n^3+6n^2$. Functional equation gives rise to a diophantine equation! My question is basically to find all natural numbers $(m,n)$ such that
$m^3=2n^3+6n^2$
First for some background (this is not really that relevant but anyways):
I was trying to solve an olympiad functional equation
$f... | This doesn't answer the diophantine equation $ m^3 = 2n^3 + 6n^2$ (which is hard), but answers the functional equation $f(x+y) = f(x) + f(y) + 3(x+y) \sqrt[3]{f(x) f(y) } $ when defined over all integers $ f: \mathbb{Z} \rightarrow \mathbb{Z}$.
With $ x = y = 0$, $f(0) = 0$.
Now, suppose there is some $ a \neq 0 $ suc... | {
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Evaluate $\int \sqrt{\frac{1+x^2}{x^2-x^4}}dx$
Evaluate $$\large{\int} \small{\sqrt{\frac{1+x^2}{x^2-x^4}} \space {\large{dx}}}$$
Note that this is a Q&A post and if you have another way of solving this problem, please do present your solution.
| Recall the half-angle identity
$$\tan \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}}. \tag{1}$$ This suggests the substitution $$x^2 = \cos 2\theta, \quad x \, dx = -\sin 2\theta \, d\theta$$ which yields for $0 < x < 1$
$$\begin{align}
\int \sqrt{\frac{1+x^2}{x^2-x^4}} \, dx
&= \int \frac{1}{x^2} \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find integral of $\int\frac{x^2+3}{\sqrt{x^2+6}}dx$ from Cambridge IGCSE Additional Mathematics $$\int\frac{x^2+3}{\sqrt{x^2+6}}dx$$
I tried to $$\int\frac{x^2+6-3}{\sqrt{x^2+6}}dx=\int\frac{x^2+6}{\sqrt{x^2+6}}dx-\int\frac{3}{\sqrt{x^2+6}}dx= \int\sqrt{x^2+6}dx - 3\int\frac{1}{\sqrt{x^2+6}}dx$$
Then got stuck
In exam-... | You can enforce an Euler substitution,
$$t=\frac{\sqrt{x^2+6}-\sqrt6}x \implies x=\frac{2\sqrt6\,t}{1-t^2} \implies dx = 2\sqrt6\frac{1+t^2}{(1-t^2)^2} \, dt$$
to transform the starting integral to
$$\int \frac{x^2+3}{\sqrt{x^2+6}}\,dx = 2\sqrt6 \int \frac{\frac{24t^2}{(1-t^2)^2}+3}{\frac{2\sqrt6\,t^2}{1-t^2}+\sqrt6} \... | {
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"source": "stackexchange",
"question_score": "2",
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Valid proof for integral of $1/(x^2+a^2)$ I'm trying to prove some integral table formulae and had a concern over my proof of the following formula:
$$\int\frac{1}{x^2+a^2}\;dx=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C$$
Claim:
$$\frac{1}{x^2+a^2}=\frac{1}{a^2}\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{x}{a}\right)^{2k}... | This is what I call an almost immediate antiderivative, based on the fact that if $\;\int f(x)\,dx = F(x)\;$ , then we can conclude that $\;\int f(g(x)) g'(x)\,dx = F(g(x))\;$, which of course it's just the basis of substitution, and thus:
$$\int\frac{dx}{a^2+x^2}=\frac1a\int\frac{\frac1adx}{1+\left(\frac xa\right)^2}=... | {
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Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,... | There are actually two "more direct" proofs of the fact that this limit is $\ln (2)$.
First Proof Using the well knows (typical induction problem) equality:
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \,.$$
The right side is $\frac{1}{n} ... | {
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"question_score": "60",
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What is $\sqrt{i}$? If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?
Is it used or considered often in mathematics? How is it notated?
| $$\sqrt{i}=\left|\sqrt{i}\right|e^{\arg\left(\sqrt{i}\right)i}$$
First we look to $\left|\sqrt{i}\right|$:
$$\left|\sqrt{i}\right|=\left|\sqrt{\frac{1}{2}+i-\frac{1}{2}}\right|=\left|\sqrt{\frac{1+(0+2i)-1}{2}}\right|=\left|\sqrt{\frac{1+2(0+1i)+(0+1i)^2}{2}}\right|=$$
$$\left|\sqrt{\frac{(1+(0+1i))^2}{2}}\right|=\left... | {
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Summing the series $(-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ How does one sum the series $$ S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} - \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots $$
This was asked to me by a high school student, and I am embarrassed that I couldn't solve it. Can anyon... | You can use the formula
$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$
This is called Wallis's product.
So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$
Interchanging the sum and t... | {
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"source": "stackexchange",
"question_score": "8",
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sangaku - a geometrical puzzle
Find the radius of the circles if the size of the larger square is 1x1.
Enjoy!
(read about the origin of sangaku)
| Let $r$ be the length the radius of the circles, and let $\theta$ be the measure of the (smaller) angle made at the corner of the big square.
The width of the square is equal to two radii and the projection of a double diameter (a quadruple-radius), so that
$(1)\hspace{1.0in}4r\cos\theta=1-2r$
Looking at the four righ... | {
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"source": "stackexchange",
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basic combinatorics question Each person from a group of 3 people can choose his dish from a menu of 5 options. Knowing that each person eats only 1 dish what are the number of different orders the waiter can ask the chef?
| (5+2 choose 3)=(7 choose 3)=35 not (5 choose 3). The 35 different orders orders are {{1, 1, 1}, {1, 1, 2}, {1, 1, 3}, {1, 1, 4}, {1, 1, 5}, {1, 2, 2}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 3}, {1, 3, 4}, {1, 3, 5}, {1, 4, 4}, {1, 4, 5}, {1, 5, 5}, {2, 2, 2}, {2, 2, 3}, {2, 2, 4}, {2, 2, 5}, {2, 3, 3}, {2, 3, 4}, {2, ... | {
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Evaluate $\int \frac{1}{\sin x\cos x} dx $ Question: How to evaluate $\displaystyle \int \frac{1}{\sin x\cos x} dx $
I know that the correct answer can be obtained by doing:
$\displaystyle\frac{1}{\sin x\cos x} = \frac{\sin^2(x)}{\sin x\cos x}+\frac{\cos^2(x)}{\sin x\cos x} = \tan(x) + \cot(x)$ and integrating.
However... | Taking log of $\rm\ sin^2(x)\ =\ 1 - cos^2(x)\ = (1-cos(x))\ (1+cos(x))\ $ shows both answers identical
| {
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Bug in mathematica: computing the sum of the ratios $(n-k+1)/(n-k+2)$ While experimenting with certain sums, I came to the following sum:
$$S_n = \sum_{k=0}^n \frac{n-k+1}{(n+1) (n+2) (n-k+2)}.$$
After rewriting the summand as
$$\frac{n-k+1}{(n+1) (n+2) (n-k+2)} = \frac{n^{\underline{k}}}{(n+2)^{\underline{k}}(n-k+2)^... | Here is the same answer with a variation on the calculations
$\begin{eqnarray*}
S_{n} &=&\sum_{k=0}^{n}\dfrac{n-k+1}{(n+1)(n+2)(n-k+2)} \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\dfrac{n-k+1}{n-k+2}\qquad (n+1)(n+2)%
\text{ is independent of }k \\
&=&\dfrac{1}{(n+1)(n+2)}\sum_{m=2}^{n+2}\frac{m-1}{m}\qquad \text{su... | {
"language": "en",
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Finding the least positive root How to find the least positive root of the equation $\cos 3x + \sin 5x = 0$?
My approach so far is to represent $\sin 5x$ as $\cos \biggl(\frac{\pi}{2} - 5x\biggr)$ then the whole equation reduces to $$2\cos \biggl(\frac{\pi}{4} - x\biggr)\cdot \cos \biggl(\frac{\pi}{4} - 4x\biggr) = 0$$... | EDIT: So apparently you want a full solution...
A product of two real numbers is $0$ precisely when a factor is $0$, so you must have that
$$ \begin{align}
& \cos\left(\tfrac{\pi}{4} - x\right) = 0 \qquad\;\; (1) \text{, or} \\\\
& \cos\left(\tfrac{\pi}{4} - 4x\right) = 0 \qquad (2).
\end{align} $$
Now we use tha... | {
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Generating function for an Arithmetic mean I want to find out, how many ways I can produce an average 4.6 of 13 numbers, when I have only numbers {1,2,3,4,5}.
I thought, that it could be done by generating functions, but since my knowledge on them isn't very deep yet, I'm a little lost. Also I have a hard time dealing ... | If you are adding 13 numbers and the average is $4.6$, this is equivalent to their sum being $4.6\times 13$. Unfortunately, that gives $59.8$, which is impossible to achieve if your original numbers were all integers. Which suggests (to me, at least) that you are in fact approximating the average, and that the actual a... | {
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How to know that $a^3+b^3 = (a+b)(a^2-ab+b^2)$ Is there a way of go from $a^3+b^3$ to $(a+b)(a^2-ab+b^2)$ other than know the property by heart?
| There is a 'trick' for odd powered exponents as follows;
recall the identity
$ (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}) = x^n - y^n $ (*)
Now, there is a real number, a, such that y = (-a) [really this real number is -y but it is not important for the purposes of searching for a possible factorization].
For n even:... | {
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Calculation of inverse of chi-square's expectation I don't know where to begin to calculate the expectation value of the random variable $1/V$, where $V$ is a random variable with chi-square distribution $\chi^2(\nu)$.
Could somebody help me?
| I try to help this question.
A random variable $X$ with inverse chi-square distribution has p.d.f
$$\frac{1}{2^{\frac{v}{2}} \Gamma(\frac{v}{2})} x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big), x>0$$
Since it is a proper distribution, we have
$$\int_0^{\infty} x^{-\frac{v}{2}-1} exp\Big(-\frac{1}{2x}\Big)dx=1 \righta... | {
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How to calculate the total number of possible triangles where the perimeter is given, and all sides are integers? How would I go about calculating the total number of possible triangles where the perimeter is given, and all sides are integers?
Exempli gratia:
*
*$a + b + c = 15$
*$d + e + f = 16$
*$x + y + z = 100... | Let ${\cal S}_p$ be the set of ordered triples $(a,b,c) \in \mathbb{N}^3$ such that $c \ge b \ge a \ge 1$, $a+b+c=p$, and $a+b > c$ (the triangle inequality). For instance, ${\cal S}_3 = \{ (1,1,1) \}$ and ${\cal S}_4 = \{ \}$. Each triple in ${\cal S}_{p-2}$ gives rise to a triple in ${\cal S}_{p}$ under the transfo... | {
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Second-Order Linear Differential Equation I have the following differential equation:
$$y''+y=\cos(t)\cos(2t)$$
Maybe something can be done to $\cos(t)\cos(2t)$ to make it easier to solve. Any ideas?
Thanks in advance.
| I don't see an elegant solution (and I fail to see how Arturo's comment helps immediately), but it can be done by brute force using the method of variation of parameters.
The homogeneous differential equation $y''(t) + y(t) = 0$ has the two fundamental solutions $y_{1}(t) = \sin{(t)}$ and $y_{2}(t) = \cos(t)$.
We must ... | {
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Proving $(2n+1) + (2n+3) + \cdots + (4n-1) = 3n^{2}$ for all positive integers $n$
Prove $(2n+1) + (2n+3) + (2n+5) +
\cdots + (4n-1) = 3n^{2}$ for all
positive integers $n$.
So the provided solution avoids induction and makes use of the fact that $1 + 3 + 5 + \cdots + (2n-1) = n^{2}$ however I cannot understand th... | The first item, $(2n+1) + (2n+3) + (2n+5) + \cdots + (4n-1)$
$ = (1 + 3 + 5 + \cdots + (4n-1)) -(1 + 3 + 5 + \cdots + (2n-1))$
comes because you are just adding and subtracting the same set of terms on the RHS.
Your base case for the induction has just one term on the left side, which I would write $2*1+1=3*1^2$ For t... | {
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How to solve $x^3 + 2x + 2 \equiv 0 \pmod{25}$? My attempt was:
$x^3 + 2x + 2 \equiv 0 \pmod{25}$
By inspection, we see that $x \equiv 1 \pmod{5}$. is a solution of $x^3 + 2x + 2 \equiv 0 \pmod{5}$. Let $x = 1 + 5k$, then we have:
$$(1 + 5k)^3 + 2(1 + 5k) + 2 \equiv 0 \pmod{25}$$
$$\Leftrightarrow 125k^3 + 75k^2 + 25k ... | Hint by others, we use Hensel's lemma. By Hensel's lemma we solve $x^{3}+2x+2=0$ (mod 5). This has solution $1$ and $3$. But $1$ satisfies $f'(1)=0$, by Chan's previous work it should be cast aside. Hensel's lemma suggests solutions of the form $3+5k$ with $3(3+5k)^{2}+2\not=0$(mod 5). Hensel's lemma give $k$ to be:
$k... | {
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Factorization of sum of two square The difference is $a^2 - b^2 = (a - b).(a + b)$
But what about when I have $a^{25} + 1$ ? According to wolfram alpha, the alternate form is:
*
*$(a+1) (a^4 -a^3 + a^2 -a + 1)( a^{20} - a^{15} + a^{10} -a^5 +1)$
However, the square root of 25 is a rational number 5.
But If I had 5... | Such factorizations are special cases of general formulas for factorizations of cylotomic polynomials, e.g. see wikipedia. These formulas follow by Möbius inversion.
Such polynomial factorizations come in handy for integer factorization. For example, Aurifeuille, Le Lasseur and Lucas discovered so-called Aurifeuillian ... | {
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Prove $\sin(\pi/2)=1$ using Taylor series
Prove $\sin(\pi/2)=1$ using the Taylor series definition of $\sin x$,
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
It seems rather messy to substitute in $\pi/2$ for $x$. So we have
$$\sin(\pi/2)=\sum_{n=0}^{\infty} \frac{(-1)^n(\pi/2)^{2n+1}}{(2n+1)!}.$$
I'm not too ... | [ A sketch: ]
Let $f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$. You can easily show that
$$\begin{eqnarray}
f(x + c) & = & f(c) + f'(c)x + f''(c)\frac{x^2}{2!} + \cdots \quad(1) \\
f''(x) & = & -f(x) \quad(2)\\
f'(x) & = & 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \quad(3)
\end{eqnarray}$$
Then it is st... | {
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Show that $x^2 - 3y^2 = n$ either has no solutions or infinitely many solutions I have a question that I have problem with in number theory about Diophantine,and Pell's equations. Any help is appreciated!
We suppose $n$ is a fixed non-zero integer, and suppose that $x^2_0 - 3 y^2_0 = n$, where $x_0$ and $y_0$ are bigge... | There are an infinite number of solutions to $x^2-3y^2=1$, which you can generate from the base solution $2^2-3\cdot 1^2 = 1$ as follows:
Given $(x,y) \in \mathbb{Z}^2$ such that $x^2-3y^2 = 1$:
\begin{align*}
x^2-3y^2 & =(x-\sqrt{3}y)(x+\sqrt{3}y)\\
&= 1 \\
(x-\sqrt{3}y)^n(x+\sqrt{3}y)^n &= (X - \sqrt{3}Y)(X+\sqrt{3}Y... | {
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Where's the trick in this partial fractions question? Here is a partial fractions question from page 287, q22 of Schaum's Calculus 5e:
$$ \int{ \frac{x^6 + 7\, x^5 + 15\, x^4 + 23\, x^2 + 25\, x - 3}{{\left(x^2 + 1\right)}^2\, {\left(x^2 + x + 2\right)}^2} dx} $$
Their answer: (verified)
$$\ln\!\left(\frac{x^2 + 1}{x^2... | The term $\rm\:32\ x^3\:$ is mistakenly omitted in the numerator of the integrand. Once you add that in then the partial fraction is
$$\rm - \frac{2\:x+1}{(x^2+x+2)^2} - \frac{2\:x+1}{x^2+x+2} + \frac{6\:x}{(x^2+1)^2} + \frac{2\:x}{x^2+1} $$
from which the claimed result follows easily.
| {
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Mathematical Induction (product of $n$ consecutive numbers) Assumption:
$$(n+1)(n+2) \cdots (2n) = (2^n)\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)$$
Prove for $n+1$:
$$(n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
Using the assumption, I divide both sides by $(n+1)$ and substitute RHS in... | Assumption
$$C(n) := (n+1)(n+2) \cdots (2n) = (2^n)\cdot 1 \cdot 3 \cdot 5 \cdots (2n-1)$$
Proof
Basis:
$$2=2^1$$
Inductive step:
$$C(n+1):= (n+2)(n+3) \cdots (2(n+1)) = (2^{n+1}) \cdot 1 \cdot 3 \cdot 5 \cdots (2(n+1)-1)$$
$$We\ have\ to\ prove: C(n) \Rightarrow C(n+1)$$
$$\vdots$$
$$Little\ substitution:$$
$$(n+2)(n+... | {
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Explaining an algebra step in $ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4)$ I have encountered this step in my textbook and I do not understand it, could someone please list the intermediate steps?
$$ \frac{n^2(n+1)^2}{4} + (n+1)^3 = \frac{(n+1)^2}{4}(n^2+4n+4). $$
Thanks,
| Take out the common factor $(n+1)^{2}$ and simplify. So you get $$ (n+1)^{2} \cdot \biggl[ \frac{n^{2}}{4} + (n+1)\Bigr] = (n+1)^{2} \cdot \biggl[ \frac{n^{2}+4n + 4}{4}\biggr]$$
| {
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Are these fractions all equal? Are the following expressions all equal to one another?
(2ab+c)/y
(2ab)/y + c/y
(2a)/y * (2b/y) + c/y
$$\dfrac{2ab+c}{y}=\dfrac{2ab}{y}+\dfrac{c}{y}=\dfrac{2a}{y}×\dfrac{2b}{y}+\dfrac{c}{y}$$
| By starting with the first, and applying the distributive law, we get the second: $$\frac{2ab+c}{y} = (2ab+c) \frac{1}{y} = 2ab \frac{1}{y} + c \frac{1}{y} = \frac{2ab}{y} + \frac{c}{y}$$
The last one is not equal to the other two, because
$$\frac{2a}{y} \cdot \frac{2b}{y} + \frac{c}{y}= \frac{2a\cdot 2b}{y\cdot y} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/43385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why can ALL quadratic equations be solved by the quadratic formula? In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these t... | The proof of the quadratic formula takes advantage of completing the square.
$$ax^2 + bx + c = 0, \ a \neq 0, \ a, \ b, \ c \ \in \mathbb R$$
$$ax^2 + bx = -c$$
$$x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$$
$$x^2 + \dfrac{b}{a}x + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$$
$$\left(x+\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "294",
"answer_count": 22,
"answer_id": 4
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Integration of powers of the $\sin x$ I have to evalute
$$\int_0^{\frac{\pi}{2}}(\sin x)^z\ dx.$$
I put this integral in Wolfram Alpha, and the result is
$$\frac{\sqrt{\pi}\Gamma\left(\frac{z+1}{2}\right)}{2\Gamma\left(\frac{z}{2}+1\right)},$$
but I don't know why. If $z$ is a positive integer, then one can do integrat... | Just, following Theo's hint
$$
\begin{align*}
\int_{0}^{\frac{\pi}{2}}{(\sin\psi)^x}d\psi&= \int_{0}^{\frac{\pi}{2}}{(\sin\psi)^{2\cdot \frac{1}{2}(x+1)-1}(\cos\psi)^{2\cdot \frac{1}{2}-1}}d\psi\\
&=\frac{1}{2}B\left( \frac{x+1}{2},\frac{1}{2} \right)\\
&= \frac{1}{2}\cdot \frac{\Gamma\left(\frac{x+1}{2}\right)\Gamma\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/50447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
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Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$ Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$.
| Square and add the two to get $$2 + 2 \cos(x-y) = a^2 + b^2$$
$$2 \cos(x-y) = a^2 + b^2 - 2 $$
Square and subtract the two to get $$\cos(2x) + 2 \cos(x+y) + \cos(2y) = b^2 - a^2$$
Now, $$\cos(2x) + \cos(2y) = 2 \cos(x+y) \cos(x-y) = \cos(x+y) (a^2+b^2-2)$$
Hence, we get
$$\cos(x+y) (a^2 + b^2) = b^2 - a^2$$
Hence,
$$\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/51070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Rules for Factorisation? Basically presented with this, simplify
\begin{aligned}
{\Bigl(\sqrt{x^2 + 2x + 1}\Big) + \Bigl(\sqrt{x^2 - 2x + 1}\Big)}
\end{aligned}
Possible factorisations into both
\begin{aligned}
{\Bigl({x + 1}\Big)^2}, {\Bigl({x - 1}\Big)^2}
\end{aligned}
\begin{aligned}
{\Bigl({1 + x}\Big)^2} , {\B... | We know $\sqrt{x^2} = |x|$, so
$$\sqrt{x^2 + 2x + 1} + \sqrt{x^2 - 2x + 1} = |x + 1| + |x-1|.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$
Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$.
One possible approach could be by first writing $$ \left(\frac{x^n}{n} + \frac{1} {x}\right) = \left( \frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx} + \text{upto ... | For $x \lt 0$ there is no minimum. As $x$ gets very close to $0$, the $\frac{1}{x}$ term gets very large and negative, faster than the $\frac{x^n}{n}$ can get positive (assuming $n$ is even-if $n$ is odd this term is negative, too).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Local minimum and maximum of the function Can anyone help me to solve the following question?
maximize and minimize the function $(10-x)(10-\sqrt{9^2-x^2})$ over $x\in[0,10]$
This is a high school question, so is there any simple trick help solve it? Thanks!
| We give a short argument for both maximum and minimum.
To bring out the symmetry, let $y=\sqrt{81-x^2}$.
We study the behaviour of $(10-x)(10-y)$, that is, of
$$xy-10(x+y)+100.\qquad\qquad\text{(Expression $1$)}$$
We will find the maximum and minimum values of Expression $1$, given that
$x^2+y^2=81$ and $x \ge 0$, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/55152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:
Find the largest number $n$ such that $(2004!)!$ is divisible by
$((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?
For the second one the suggested solution is like this : $ 2^8 + 2^{11... | The second one is true. Because $$(2^{4} + 2^{6})^{2} = 2^{8} + 2 \cdot 2^{4} \cdot 2^{6} + 2^{12}$$ so your $n=12$.
As far as I know, the main idea is to write $2^{8}+2^{11}+2^{n}$ as $(2^4)^{2} + 2 \cdot 2^{4} \cdot x + x^{2}$. Then you will have to manipulate what $x$ is and intuition say $x=2^{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/57177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Hard simultaneous equation problem $5x^2y-4xy^2+3y^3-2(x+y)=0$, $xy(x^2+y^2)+2=(x+y)^2$ How to solve this system of equations:
$$\begin{align*}
5x^2y-4xy^2+3y^3-2(x+y) &=0 \\
xy(x^2+y^2)+2 &=(x+y)^2
\end{align*}
$$
| Notice that for every solution $(x, y)$, $(-x, -y)$ is also a solution.
The second equation admits factorization:
$$
(-1 + x y) (-2 + x^2 + y^2) = 0
$$
Now solve for $y$ and substitute into the other equation, and solve for $x$. Positive solutions resulting from this are $x=y=1$ and $x = 2 y = 2 \sqrt{\frac{2}{5}}... | {
"language": "en",
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"source": "stackexchange",
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How to find inverse of a composite function? I am stuck with this question,
Let $A=B=C=\mathbb{R}$ and consider the functions $f\colon A\to B$ and $g\colon B\to C$ defined by $f(a)=2a+1$, $g(b)=b/3$. Verify Theorem 3(b): $(g\circ f)^{-1}=f^{-1}\circ g^{-1}.$
I have calculated $f^{-1}$, $g^{-1}$, and their compositio... | You find the inverse of $g\circ f$ by using the fact that $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$.
In other words, what gets done last gets undone first.
$f$ multiplies by 2 and then adds 1.
$g$ divides by 3.
Dividing by 3 is done last, so it's undone first.
The inverse first multiplies by 3, then undoes $f$.
Later not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/59929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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On a property of the binomial coefficient Let $n$ be a positive integer and $p$ a prime, how can I prove that the highest power of $p$ that divides $\binom{2n}{n}$ is exactly the number of $k\geq1$ such that $\lfloor 2n/p^k\rfloor$ is odd?
| Let us write out this binomial coefficient a bit more
$$\binom{2n}{n} = \frac{2n \cdot (2n-1) \cdot \ldots \cdot (n+1)}{n \cdot (n-1) \cdot \ldots \cdot 1}$$
Above we have all numbers between $n + 1$ and $2n$, while below we have all numbers from $1$ to $n$. Now suppose $\lfloor 2n/p\rfloor = 2m + 1$. Then there are ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/60403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Ramanujan's identity involving cube roots I have found this pretty identity involving cube roots, first stated by Ramanujan:
$$
\sqrt{m(4m-8n)^{\frac13} + n(4m+n)^{\frac13}}
= (4m+n)^{\frac23} + (4(m-2n)(4m+n))^{\frac13}- (2(m-2n))^{\frac23}.
$$
So I ask if there is any method to create similar identities?
| The identity seems incorrect; instead it should read $(\ast)$
$$
\color{Red}{3} \cdot \sqrt{m(4m-8n)^{\frac13} + n(4m+n)^{\frac13}}
= (4m+n)^{\frac23} + (4(m-2n)(4m+n))^{\frac13}- \color{Red}{(2(m-2n)^2)^{\frac13}}.
$$
I might be asking for trouble by being the heretic =), but I find this particular identity quite u... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$ I came up with this equation during my homework :
$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
My algebra is weak and I can't seem to find a way to solve for x nicely
Could someone please show me a decent way of doing this?
Thanks alot, Jason
| Generally the best way is to just plough through the algebra (and algebra gets quite a bit more advanced than this!) :
*
*$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
*$4=x(1-\sqrt{5})+(1-x)(1+\sqrt{5})$ (dividing through by $2$ simplifies a lot of the subsequent terms)
*$4=x-x\sqrt{5}+1-x+\sqrt{5}-x\sqrt{5}$ (multi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| [All: This answer came here when I merged duplicates. Please take that into account when voting, JL]
You should prove before
$$
\sum_{i=0}^n i = \frac{n(n+1)}{2}
$$
Cases $n=0,1$ are trivial. Suppose it's true for $n-1$ then
\begin{align}
\sum_{i=0}^n i^3 &= n^3 + \sum_{i=0}^{n-1}i^3 \\
&=n^3 + \left(\sum_{i=0}^{n-1} i... | {
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"timestamp": "2023-03-29T00:00:00",
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For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? For vectors in three-dimensional space, if $a \cdot b$ and $a \cdot c$ are equal, and $a \times b$ and $a \times c$ are equal, are b and c equal? I tried looking for counter-... | We can conclude that $\mathbf b= \mathbf c$ provided $\mathbf a \neq \mathbf 0$.
Write $\mathbf d = \mathbf b - \mathbf c$. The given equations can be written as
$$
\mathbf a \cdot \mathbf d = 0, \ \ \
\mathbf a \times \mathbf d = \mathbf 0.
$$
Now, suppose $\theta$ is the angle between $\mathbf a$ and $\mathbf d$... | {
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Twin primes of form $2^n+3$ and $2^n+5$ How to prove that $2^n+3$ and $2^n+5$ are both prime for only finitely many integers $n$?
And how to prove that there are infinitely many primes of the form $2^n+3$ and $2^m+5$
| Settling the twin prime part of the question seems to go as follows:
1) If $2\mid n$, then $3\mid 2^n+5$, so it suffices to study odd values of $n$.
2) If $n\equiv 1\pmod 4$, then $5\mid 2^n+3$, so it suffices to study the case $n\equiv 3\pmod 4$.
3) If $n\equiv 1\pmod 3$, then $7\mid 2^n+5$. If $n\equiv 2\pmod 3$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/70682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Easy system in five equations An interesting little problem:
I have found solutions $[a=3,b=2,c=1,d=5,e=4]$ but not able to find proof that these are all that exist.
Find, with proof, all integers $a$, $b$, $c$, $d$ and $e$ such that:
$a^2 = a + b - 2c + 2d + e - 8$
$b^2 = -a - 2b - c + 2d + 2e - 6$
$c^2 = 3a + 2b + c ... | Calculate the sum $(a-3)^2+(b-2)^2+(c-1)^2+(d-5)^2+(e-4)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/71041",
"timestamp": "2023-03-29T00:00:00",
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divisibility of an expression by 30 let $g$ be a natural number how to show that $30$ divides $(-8 g^5+20 g^4-50 g^3+115 g^2-167 g+90)$?
my guess:
$30$ divides $90$ so it is enough to show that $30|-8 g^5+20 g^4-50 g^3+115 g^2-167 g = g(-8 g^4+20 g^3-50 g^2+115 g-167) $ now if $30|g$ we are done, otherwise we have to ... | Note that for any $n\in\mathbb{Z}$,
$$30\mid n\iff 2\mid n\text{ and }3\mid n\text{ and }5\mid n.$$
Then note that
$$-8g^5+20g^4-50g^3+115g^2-167g+90\equiv0+0+0+g^2+g+0\equiv g+g\equiv 0\bmod 2$$
because, by Fermat's Little Theorem, $g^2\equiv g\bmod 2$. Similarly,
$$-8g^5+20g^4-50g^3+115g^2-167g+90\equiv g^5+2g^4+g^3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The limit of truncated sums of harmonic series, $\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}$ What is the sum of the 'second half' of the harmonic series?
$$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$
More precisely, what is the limit of the above sequence of partial sums?
| For every $x\in [n,n+1]$, with $n\ge1$, we have $\displaystyle \frac{1}{n+1} \le \frac{1}{x} \le \frac{1}{n}$. If we integrate over $[n,n+1]$, we get
\begin{align}
\int_{n}^{n+1}\frac{1}{n+1}dx &\le& \int_{n}^{n+1}\frac{1}{x}dx &\le& \int_{n}^{n+1}\frac{1}{n}dx\\
\frac{x}{n+1}\Big|_{n}^{n+1} &\le& \ln(x)\Big|_{n}^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/73550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "76",
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"answer_id": 0
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Ring such that $x^4=x$ for all $x$ is commutative Let $R$ be a ring such that $x^4=x$ for every $x\in R$. Is this ring commutative?
| Here's an old post of mine from Yahoo! Answers:
First, note $-x = (-x)^4 = x^4 = x$, so $x+x = 0$ for any $x$ in $R$. Then
$(x^2+x)^2 = x^2 + x + x^3 + x^3 = x^2+x$. Thus $x^2+x$ is idempotent, and it is easy to see idempotent elements are central in this ring. [I give a proof of this at the end.]
Now let $x=a+b$, wher... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$ I want to find all odd primes $p$ for which $15$ is a quadratic residue modulo $p$.
My thoughts so far: I want to find $p$ such that $ \left( \frac{15}{p} \right) = 1$. By multiplicativity of the Legendre symbol, this is equivalent to $ \left( \fr... | I think the law of quadratic reciprocity hasn't been fully applied. From Quadratic Reciprocity,
$$
\left(\frac{15}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{5}{p}\right)=(-1)^{(p-1)/2}\left(\frac{p}{3}\right)\left(\frac{p}{5}\right).
$$
There are now two cases. If $p\equiv 1\pmod{4}$, you have $(15|p)=(p|3)(p|5)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/77351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
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Asymptotics of $1^n + 2^{n-1} + 3^{n-2} +\cdots + (n-1)^2 + n^1$ Suppose $n\in\mathbb{Z}$ and $n > 0$.
Let $$H_n = 1^n + 2^{n-1} + 3^{n-2} +\cdots + (n-1)^2 + n^1.$$
I would like to find a Big O bound for $H_n$. A Big $\Theta$ result would be even better.
| My approach is very much along the same line as that of leonbloy.
Let $f(k) = k^{n+1-k}$. Use Euler-Maclaurin formula:
$$
\begin{eqnarray}
\sum_{k=1}^n f(k) &=& \int_1^n f(x) \mathrm{d} x + \frac{1}{2} \left( f(1) + f(n) \right) + \sum_{m=1}^q \frac{B_{2m}}{(2m)!} \left( f^{(2m-1)}(n) - f^{(2m-1)}(1) \right) \\
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 0
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Prove by induction: $2^n + 3^n -5^n$ is divisible by $3$ Let $P(n) =
2^n + 3^n - 5^n
$.
I want to prove that $P(n)$ is divisible by $3$ for all integers $n\geq 1$.
The basis step for this proof is easy enough: $P(1)$ is divisible by $3$.
For the inductive step, I let $k$ be an arbitrary integer, then assume $P(k)$ is d... | As an alternative to Rankeya's answer, you have:
$$P(k+1) = 2^{k+1} + 3^{k+1} - 5^{k+1}.$$
Then, proceeding as you do, we have:
$$P(k+1) = 2\cdot 2^k + 3\cdot 3^k - 5\cdot 5^k.$$
At this point, you want to use your induction hypothesis. Notice that you have enough $2^k$s, $3^k$s and $5^k$s for two $2^k+3^k-5^k$, with s... | {
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"question_score": "6",
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The equation $(1+x)^2\frac{dy}{dx}-xy=x^2y^2$ I am very grateful for all your comments and answers to my previous question concerning ODEs ( The equation $(x-2xy-y^2)\frac{dy}{dx}+y^2=0$ ). Now I am struggling with this one
$$(1+x)^2\frac{dy}{dx}-xy=x^2y^2.$$
It seems not to be hard but nevertheless all the tricks I kn... | Step 1.
Let $y(x) = \frac{1}{w(x)}$. Then $y^\prime(x) = -\frac{1}{w(x)^2}\cdot w^\prime(x)$, so the differential equation becomes $ \frac{(1+x)^2 w^\prime(x) + x w(x) + x^2}{w(x)^2} = 0$, that is $ w^\prime(x) + \frac{x}{(1+x)^2} w(x) + \frac{x^2}{(1+x)^2}=0$.
Step 2. Integrating factor, let $w(x) = f(x) g(x)$, and w... | {
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"timestamp": "2023-03-29T00:00:00",
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probability of number of draws A box contains 8 tickets. Two are marked 1, two marked 2, two marked 3, and two marked 4.
Tickets are drawn at random from the box without replacement until a number appears that has appeared before. Let X be the number of draws that are made.
so the probability when X = 1 is: $4*\frac{2... | You are right, the probabilities are not the same, although I am a little puzzled by your case "$X=1$." The smallest possible value of the number of draws until we get our first match is $2$. But perhaps you are not counting the last draw.
So let $Y$ be the number of draws until the first match, including the draw tha... | {
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square squares with diagonals also squares The numbers reading across and down in these squares are square:
$\begin{array}{ccc}
1 & 4 & 4\\
4 &8&4\\
4&4&1
\end{array}$
$\begin{array}{ccc}
5&2&9\\
2&5&6\\
9&6&1
\end{array}$
$\begin{array}{cccc}
2&1&1&6\\
1&2&2&5\\
1&2&9&6\\
6&5&6&1
\end{array}$
Are there an... | Those matrices are certainly hard to find, I tried it a lot and found that there are no $4\times4$ or $3\times3$ matrices that do what you want in base $10$ or below. However watch this one in base $11$:
$$\begin{array}{ccc}
1 & 9 & 5\\
9 & 6 & 1\\
5 & 1 & 9
\end{array}$$
You have
$169_{11}=196=14^2$
$195_{11}=225=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/80483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How to get closed form from generating function? I have this generating function:
$$\frac{1}{2}\, \left( {\frac {1}{\sqrt {1-4\,z}}}-1 \right) \left( \,{
\frac {1-\sqrt {1-4\,z}}{2z}}-1 \right)$$
and I know that $\frac {1}{\sqrt {1-4\,z}}$ is the generating function for the sequence $\binom {2n} {n}$, and $\frac {1-\... | Expanding the product, and after some algebra:
$$
g(z) = \frac{1}{2} \left(\frac{\sqrt{1-4 z}-1}{z}+\frac{1}{\sqrt{1-4 z}}+1\right)
$$
Using generalized binomial theorem:
$$
[z]^n g(z) = \frac{1}{2} \left( \delta_{n,0} + \binom{1/2}{n+1} (-4)^{n+1} + \binom{-1/2}{n} (-4)^n \right)
$$
Using
$$ \begin{eqnarra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/87805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to prove :If $p$ is prime greater than $3$ and $\gcd(a,24\cdot p)=1$ then $a^{p-1} \equiv 1 \pmod {24\cdot p}$? I want to prove following statement :
If $p$ is a prime number greater than $3$ and $\gcd(a,24\cdot p)=1$ then :
$a^{p-1} \equiv 1 \pmod {24\cdot p}$
Here is my attempt :
The Euler's totient function ca... | Hint: The claim follows from proving the following facts separately: $a^{p-1}\equiv 1\pmod 8$, $a^{p-1}\equiv 1\pmod 3$, and $a^{p-1}\equiv 1\pmod p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}$ I want to prove that$$2 \arctan\sqrt{x} = \arcsin \frac{x-1}{x+1} + \frac{\pi}{2}, x\geq 0$$
I have started from the arcsin part and I tried to end to the arctan one but I failed.
Can anyone help me solve it?
| One way would be to notice that both functions have the same derivative and then find out what the "constant" is by plugging in $x=0$.
Here's another way. Look at
$$
\frac\pi2 - 2\arctan\sqrt{x} = 2\left( \frac\pi4 - \arctan\sqrt{x} \right) = 2\left( \arctan1-\arctan\sqrt{x} \right).
$$
Now remember the identity for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/89759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Find a closed term for $f(n) = n + 2 f(n-1)$, $f(1)=1$ I cannot help myself, but I don't get the closed term for: $f(n) = n + 2 f(n-1)$, where f(1) = 1. I tried to find the pattern when looking at some iterations, and I think I see the pattern very clearly:
$...(6 + 2 ( 5 + 2 ( 4 + 2 ( 3 +2 ( 2 \cdot 1 ) ) ) ) ... )$
... | Just for fun, here is a solution using generating functions.
Let $F(x)=\sum_{n=1}^\infty f(n)x^n$. Then,
$$\begin{align*}
F(x)&=\left(\sum_{n=2}^\infty (n+2f(n-1))x^n\right)+x\\
&=\left(\frac{x}{(1-x)^2}-x+2x\sum_{n=1}^\infty f(n)x^n\right)+x\\
&=\frac{x}{(1-x)^2}+2xF(x)
\end{align*}$$
So,
$$F(x)=\frac{x}{(1-x)^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/90023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to remember the trigonometric identities I have a test tomorrow and I am having trouble remembering those pesky trigonometrical identities (such as $1-\cos x=2\sin^2(\frac{x}{2})$ )
Do you guys have any tips on how I can remember these?
Thanks :)
| My favourite trick: I don't remember any of them. :-) The only thing I have in mind is that this matrix
$$
\begin{pmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{pmatrix}
$$
rotates vectors in the plane by an angle $\theta$ and matrix multiplication is the same as composition. Hence, you have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 2
} |
On computing: $ \gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right)$ I would like to calculate
$$ d=\gcd \left({2n \choose 1}, {2n \choose 3},\cdots, {2n \choose 2n-1}\right) $$
We have:
$$ \sum_{k=0}^{n-1}{2n \choose 2k+1}=2^{2n-1} $$
$$ d=2^k, 0\leq k\leq2n-1 $$
...
Any idea?
| Let $q = 2^i$ where $2^i | 2n$ and $2^{i+1} \not|2n$. Claim: $d=q$.
First we'll show that each term ${2n \choose 2k+1}$, for $0 \leq k \leq n-1$ has $q$ as a factor. Consider,
$$\begin{align*}
{2n \choose 2k+1} &= \frac{(2n)(2n-1)(2n-2) \cdots (2n- [2k+1] + 1)}{(2k+1)(2k)(2k-1)\cdots(1)}\\
&= \frac{2n}{2k+1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 2,
"answer_id": 1
} |
How to find the derivative of $f(x)$? This is not a homework problem.
How to find $f'(x)$ if $f(x)=2x^{1/2}\log(2)$?
Thanks in advance for your help! I just can't figure it out...
What rule should I use?
| Note that $\frac{d}{dx} (c \cdot f(x)) = c \cdot f'(x)$. Hence you just want to find the derivative of $f(x) = \sqrt{x}$. Using the definition of the derivative we have
\begin{align*}
f'(x) & = \lim_{h \to 0} \: \frac{f(x+h)- f(x)}{h} \\ &= \lim_{h \to 0} \: \frac{\sqrt{x+h} - \sqrt{x}}{h} \\ &= \lim_{h \to 0} \: \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/93958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Combinatorial interpretation of sum of squares, cubes Consider the sum of the first $n$ integers:
$$\sum_{i=1}^n\,i=\frac{n(n+1)}{2}=\binom{n+1}{2}$$
This makes the following bit of combinatorial sense. Imagine the set $\{*,1,2,\ldots,n\}$. We can choose two from this set, order them in decreasing order and thereby o... | We give a combinatorial interpretation of the formula
$$
2^2+4^2+6^2+\cdots +(2n)^2=\binom{2n+2}{3} \qquad\qquad (1)
$$
for the sum of the first $n$ even squares.
There are $\binom{2n+2}{3}$ ways to choose $3$ numbers from the numbers $1, 2, \dots, 2n+2$. We organize and count the choices in another way.
Maybe the lar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/95047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric expressions possible pattern While solving trigonometric problems I've noticed possible pattern that reminded me of Fibonacci numbers and Pascal triangle. So I tried to find next "element" of this pattern (8 degree exponent ) and failed to do so. So can someone please tell me is there a pattern or just "c... | There is a pattern but I do not think it is as simple as what the first three terms might suggest.
Let me use the abbreviation $s := \sin x$ and $c := \cos x$. In this notation, we have
$$
(s^6 + c^6) = (s^6 + c^6)(s^2 + c^2) = (s^8 + c^8) + s^2c^2(s^4+c^4).
$$
Now plugging in the expressions you obtained for $s^6+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/95417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Converting $\tanh^{-1}{x}$ to an expression involving the natural logarithm I know how to convert $\sinh^{-1}{x}$ and $\cosh^{-1}{x}$ to $\ln{|x+\sqrt{x^2 \pm 1}|}$, but for some reason I am struggling to do the same for the following statement:
$$\tanh^{-1}{\frac{x}{2}}$$
Can someone please show me how to convert it ... | $$\tanh x=\frac{e^{2x}-1}{e^{2x}+1} \Rightarrow \tanh \frac{x}{2}=\frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanh\frac{x}{2}$ we have to solve following equation :
$$\frac{x}{2}= \frac {e^{2artanh \frac{x}{2}}-1}{e^{2artanh \frac{x}{2}}+1} \Rightarrow x \cdot e^{2artanh \frac{x}{2}}+x=2e^{2artanh \frac{x}{2}}-2 \R... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Limit $\lim_{n\to+\infty} (1-\frac1{2^2})(1-\frac1{3^2})\cdot \cdots \cdot(1-\frac{1}{n^2})$ and series $\sum_{n=2}^{\infty} \ln(1-\frac1{n^2})$
Possible Duplicate:
Finding Value of the Infinite Product $\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)$
Compute:
\begin{align*}
\lim_{n\to+\infty} (1-\frac{1}{2^2})(1-\frac{1}{3^2}... | Actually there is a neat form for the partial products.
Let
$$f(n)=(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})\cdot \cdots \cdot(1-\frac{1}{n^2})=\prod _{k=2}^n \left(1-\frac{1}{k^2}\right)$$
We show by induction that $f(n)=\frac{n+1}{2 n}$. $$f(n)=\left(1-\frac{1}{n^2}\right)f(n-1)=\frac{n^2-1}{n^2}\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/99537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration - Partial Fraction Decomposition I was wondering if someone would be kind enough to walk me through the logic used in solving the following integral. I have been to class, and have read the section (9.4 of Swokowski's Classic), and have studied the answer in the solution manual, but I can't quite seem to ma... | Procedure of decomposition
So:
$$\frac{x^2+3x+1}{(x^2+4)(x^2+1)} = \frac{ax+b}{x^2+4}+\frac{cx+d}{x^2+1} \Rightarrow$$
$$ \Rightarrow x^2+3x+1 =(x^2+1)(ax+b)+(x^2+4)(cx+d) \Rightarrow$$
$$\Rightarrow x^2+3x+1 =(a+c)x^3+(b+d)x^2+(a+4c)x+(b+4d)$$
Therefore , you have to solve following system of equations :
$\begin{cases... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/103625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how to work out a closed form of a sequence
Consider the following linear recurrence sequence.
$x_1 = 11$, $x_{n+1} = -0.8x_n + 9,\quad n = 1,2,3, \ldots.$
Find a closed form for this sequence.
| To get an idea of what the closed form might look like, let's iterate the relation a few times:
$$
\begin{align}
x_n
&=9-.8x_{n-1}\\
&=9-.8(9-.8x_{n-2})\\
&=9-.8\cdot9+.8^2x_{n-2}\\
&=9-.8\cdot9+.8^2(9-.8x_{n-3})\\
&=9-.8\cdot9+.8^2\cdot9-.8^3x_{n-3}\tag{1}
\end{align}
$$
Looking at $(1)$, it appears that
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/104691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Probability of All Distinct Faces When Six Dice Are Rolled If six fair dice are rolled what is probability that each of the six numbers will appear exactly once?
| Imagine you throw one after the other. You consider a throw as a success if the number is different from all previous numbers. You start with one. This is always a succes so $P(\text{first}) = 1 = \frac{6}{6}$. Your second throw is a success if one of the remaining $5$ numbers shows, so $P(\text{second}) = \frac{5}{6}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/105300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Does $x^2\equiv x\pmod p$ imply $x^2\equiv x\pmod {p^n}$ for all $n$? Suppose $x^2\equiv x\pmod p$ where $p$ is a prime, then is it generally true that $x^2\equiv x\pmod {p^n}$ for any natural number $n$? And are they the only solutions?
| Way 1: Let us compute a bit. Let $p=2$. Note that $x^2\equiv x \pmod 2$ for any $x$. Is it true that always $x^2\equiv x\pmod 4$? No, let $x=2$.
Way 2: We do more work, but will get a lot more information. Rewrite the congruence $x^2\equiv x \pmod p$ as $x^2-x\equiv 0 \pmod p$, and then as $x(x-1)\equiv 0\pmod p$.
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/109961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
What is the probability that 5 digit number divisible by 6? The main constraint is that each digit can only take digits from $\{1, 2, 3, 4, 5\}$. So the sample space will be $5^{5}$.
What is the probability that a random number taken from this sample space will be divisible by $6$?
Thanks.
| $$\color{red}{416/3125}=0.13312.
$$
The last digit must be $2$ or $4$, this happens with probability $2/5$.
The sum of the four other digits must be $\pm1\pmod{3}$, according to the last digit being $2$ or $4$. Since both events have the same probability, the answer is $2/5$ times the probability that the sum $s$ of f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/110324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int\csc^3{x} \ dx$ I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.
How did we go from:
$$\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$$
to
$$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$
Thank you for... | The question seems to be why the following are equal:
$$(\csc^2 x - \csc x \cot x)\,dx = d(-\cot x + \csc x)$$
The answer is that
$$
\frac{d}{dx} \cot x = -\csc^2 x\quad \text{ and }\quad\frac{d}{dx} \csc x = -\csc x \cot x.
$$
The quotient rule for derivatives can establish both of these identities if you know how t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/112918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Integrate using Partial Fraction decomposition, completing the square The given problem is $\int{x\over x^3-1}dx$.
I know this equals
$${1\over3}\int {1\over x-1}-{x-1\over x^2+x+1}dx,$$
which can be separated into
$${1\over3}\int {1\over x-1}dx - {1\over3}\int{x+(1/2)-(3/2)\over x^2+x+1}dx.$$ This can further be s... | Use the substitution
$$\frac{\sqrt{3}}{2}u=x+\frac{1}{2}.$$
Or else, if you want to do it in two steps, make the substitution $u=x+\frac{1}{2}$.
You will end up with an expression that includes $u^2+\frac{3}{4}$. Then let $u=\frac{\sqrt{3}}{2}v$. The substitution that I proposed is a little faster.
Remark: Suppose th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/114918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplifying an expression $\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$ if we know $x+y+z=0$ The following expression is given:
$$\frac{x^7+y^7+z^7}{xyz(x^4+y^4+z^4)}$$
Simplify it, knowing that $x+y+z=0$.
| Exploit the innate symmetry! Using Newton's identities to rewrite the power sums as elementary symmetric functions is very simple because $\rm\:e_1 = x+y+z = 0\:$ kills many terms.
Write $\rm\ \ c = e_2 = xy + yz + zx,\ \ \ d = e_3 = xyz,\ \ \ p_k =\: x^k + y^k + z^k.$
$\rm\qquad\qquad p_1\ =\ e_1 = 0$
$\rm\qquad\qqua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Find the minimum value.
Find the minimum value of $4^x + 4^{1-x}$ , $x\in\mathbb{R}$.
In this I used the property that $a + \frac{1}{a}\geq 2$.
So I begin with
$$ 4^x + \left(\frac{1}{4}\right)^x + 3\left(\frac{1}{4}\right)^x \geq 2 + 3\left(\frac{1}{4}\right)^x$$
So I think the minimum value be between 2 to 3.
Bu... | $$4^{x} + 4^{1-x} = (\sqrt{4^x} - \frac{2}{\sqrt{4^x}})^2 + 4 \ge 4$$
with equality occuring when $\sqrt{4^x} = \frac{2}{\sqrt{4^x}}$ and so $x = \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If we know the GCD and LCM of two integers, can we determine the possible values of these two integers? I know that $\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab$, so if we know $\gcd(a,b)$ and $\operatorname{lcm}(a,b)$ and we want to find out $a$ and $b$, besides factoring $ab$ and find possible values, can we find th... | Given gcd(a,b) and lcm(a,b), the possible values for a (which are the same as the possible values for b, by symmetry) are the unitary divisors of lcm(a,b)/gcd(a,b) multiplied by gcd(a,b). So suppose gcd(a,b) = 3 and lcm(a,b) = 6300. Factoring 6300/3 gives $2^2\cdot3\cdot5^2\cdot7$ and so the possible values are the mem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Help with the line integral $\int x^2dS$ where C is a upper-half of a circle $x^2+y^2=r^2$ $$\int_C x^2dS$$ where C: $$x^2+y^2=r^2$$
So, $$ x=r\cos(\theta) $$ $$ y=r\sin(\theta) $$ $$ 0 \leq \theta \leq \pi $$
How would arc length of this curve go? $$ dS = \sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}d\theta $$ What sho... | $r$ is constant and $> 0$ in a circle. Therefore:
$
\frac{dr}{d\theta}=0
$
And
$
ds = \sqrt{(0)^2+(r)^2} d\theta = r d\theta
$
From there, your integral should be straightforward.
Another way to look at it:
\begin{align}
ds &= \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \ d\theta \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/116337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
interesting square of log sin integral I ran across this challenging log sin integral and am wondering what may be a good approach.
$$
\int_{0}^{\frac{\pi}{2}}x^{2}\ln^{2}(2\cos(x))dx=\frac{11{{\pi}^{5}}}{1440}
$$
This looks like it may be able to be connected to the digamma or incomplete beta function somehow.
I tr... | As I remarked in a comment, I have spent a while trying to derive this identity by means of contour integration. Though I have not succeeded at that, (I take great pleasure in striking that out; see my other answer below!) I have found some related identities in my search which surprised me quite a bit, and I think the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
"answer_id": 1
} |
For complex $z$, find the roots $z^2 - 3z + (3 - i) = 0$
Find the roots of:
$z^2 - 3z + (3 - i) = 0$
$(x + iy)^2 - 3(x + iy) + (3 - i) = 0$
$(x^2 - y^2 - 3x + 3) + i(2xy -3y - 1) = 0$
So, both the real and imaginary parts should = 0. This is where I got stuck since there are two unknowns for each equation. How do I p... | $$9y^2+6y+1-4y^4-18y^2-6y+12y^2=0$$
$$-4y^2+3y^2+1=0`×(-1)$$
$$4y^2-3y^2-1=0$$
$$(4y^2+1)(y^2-1)=0$$
$$y^2=1$$
$$y=+1, x=(3+1)/2= 2$$
$$y=-1, x=(-3+1)/-2=1$$
$$z=x+yi$$
$$z1=2+i$$
$$z2=1-i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what... | You don't need to use induction here. Just reduce everything modulo 9:
$10 \equiv 1 \mod 9$, so $10^n \equiv 1 \mod 9$, so $10^n+5 \equiv 6 \mod 9$.
To find the possible values of $4^n \mod 9$, note that $4^3 \equiv 1 \mod 9$, so $4^n$ must be $1, 4,$ or $7 \mod 9$. Thus $3 \cdot 4^n$ is equal to $3 \mod 9$.
Henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 5
} |
How to compute $\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}$? I got stuck at the computation of the sum
$$
\sum\limits_{k=0}^n (-1)^k{2n-k\choose k}.
$$
I think there is no purely combinatorial proof here since the sum can achieve negative values. Could you give me solution, it seems to involve generating functions.
| Let
$$
a_n=\sum_{k=0}^n(-1)^k\binom{n-k}{k}
$$
Noting that
$$
\sum_{n=k}^\infty\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}\tag{1}
$$
we can compute the generating function of $a_n$:
$$
\newcommand{\cis}{\operatorname{cis}}
\begin{align}
\sum_{n=0}^\infty a_nx^n
&=\sum_{n=0}^\infty x^n\sum_{k=0}^n(-1)^k\binom{n-k}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 2
} |
Using antiderivative to calculate complex integral $$\int_{1}^{3}(z-2)^3 dz $$
I get the following -
$$\frac{1}{4}[(3-2)^3 - (1-2)^3] = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$
However the answer sheet I have just show it reduced to $\frac{1}{4} - \frac{1}{4} = 0$
Cant see how they are getting a - instead of a +...wha... | You should increase the power by 1 first to get:
$\frac{1}{4}[(3-2)^4 - (1-2)^4] = \frac{1}{4} - \frac{1}{4} = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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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.