Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving $ \frac{dy}{dx} = y^2 - 9$ $ \frac{dy}{dx} = y^2 - 9$
This is separable so I rewrite it as $ \frac{1}{(y^2 - 9)}dy = dx$ then I get
$$\int \frac{1}{(y^2 - 9)}dy = \int dx = \int 1dx = x + c,$$ for some $c \in \mathbb{R} $
The left hand side:
$$\int \frac{1}{(y^2 - 9)}dy = \int \frac{1}{(y+3)(y-3)}dy = \int \lef... | Simplifying from where you stopped:
$\displaystyle \ln|\frac{y-3}{y+3}| = 6x + \underbrace{6c}_{=\,c}\Longrightarrow|\frac{y-3}{y+3}|=e^{6x+c}=e^{6x}\cdot \underbrace{e^c}_{=\,c}=Ce^{6x}$
$$|\frac{y-3}{y+3}|=Ce^{6x} \Longrightarrow \frac{y-3}{y+3}=-Ce^{6x} \Longrightarrow y-3=-Ce^{6x}(y+3) \Longrightarrow y(1+Ce^{6x})=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$ I have tried parts which ended up into a more complex form, any suggestions?
$$\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}dx$$
| Edit: I decided that the previous solution admits a minor but nice generalization, so here goes.
Let $$t = \dfrac{x}{\sqrt{x^4+1}}, \quad dt = \frac{1-x^4}{(x^4+1)^{3/2}} \, dx.$$ Then $$\frac{1}{1 \pm 2t^2} = \frac{x^4+1}{x^4 + 1 \pm 2x^2} = \frac{x^4+1}{(1 \pm x^2)^2},$$ so that
$$ \begin{align*} \int \frac{x^2 \mp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/922882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
The minimum of $x^2+y^2$ under the constraints $x+y=a$ and $xy=a+3$ I solved the following problem: If $x,y,a \in \mathbb{R}$ such that $x+y=a$ and $xy=a+3$, find the minimum of $x^2+y^2$
Here is my solution. $x^2+y^2=(x+y)^2 -2xy= a^2-2a-6$. The minimum value is obtained when $a=1$ and it is $-7$. Where did I go wrong... | Another way:
$$(x+y)^2 \ge 2xy \implies a^2 \ge 2(a+3) \implies a \in (\infty, -2] \cup [3,\infty)$$
$$(x^2+y^2)(1+1)\ge(x+y)^2 \implies x^2+y^2 \ge \tfrac12 a^2 \ge 2$$
Equality is achieved in both inequalities when $x=y=-1$, so this is the minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/925301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrating Trig Functions How you I solve the following integral?
$\int_{-\pi}^\pi \cos(5x)\cos(nx)dx$
I know I need to use the ${1\over 2}(\cos(u-v)+\cos(u+v))$ but I keep getting zero.
| Firstly, $\cos(nx)\cos(5x)=\frac{1}{2}(\cos((n-5)x)+\cos((n+5)x))$.
If $n\neq\pm5$, we split the integral up as follows:
\begin{align*}
\int_{-\pi}^{\pi} \cos(nx)\cos(5x)dx&=\frac{1}{2}\left(\int_{-\pi}^{\pi}\cos((n-5)x)dx+\int_{-\pi}^{\pi}\cos((n+5)x)\right) \\
&= \frac{1}{2}\left[\frac{1}{n-5}\sin((n-5)x)+\frac{1}{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/925854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_{1}^{\infty} \frac{\log^3 x}{x(x-1)} dx$ How do I arrive at the closed form expression of the integral $$\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx$$
Most probably the closed form is $\dfrac{\pi^4}{15}$
| Using a CAS, $$I_1=\displaystyle\int\dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \text{Li}_4(x)+3 \text{Li}_2(x) \log ^2(x)-6 \text{Li}_3(x) \log (x)-\frac{1}{4}
\log ^4(x)+\log (1-x) \log ^3(x)$$ $$I_2=\displaystyle\int_{1}^{a} \dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \text{Li}_4(a)+3 \text{Li}_2(a) \log ^2(a)-6 \text{Li}_3(a) \log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/927315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solving $y^3=x^3+8x^2-6x+8$ Solve for the equation $y^3=x^3+8x^2-6x+8$ for positive integers x and y.
My attempt- $$y^3=x^3+8x^2-6x+8$$
$$\implies y^3-x^3=8x^2-6x+8$$
$$\implies (y-x)(y^2+x^2+xy)=8x^2-6x+8$$
Now if we are able to factorise $8x^2-6x+8$ then we can compare LHS with RHS.Am I on the... | Using Gerry Myerson's clever inequality observation, we have two possibilities.
Case 1: $y=x+1$. Then by substitution and simplification, we must solve
$$5x^2-9x+7=0,$$
which has no integer solutions, as seen by using the quadratic equation.
Case 2: $y=x+2$. This yields the [factored] quadratic
$$2x(x-9)=0.$$
Evidently... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Can anybody help me solve this combinatorial identity? While trying to derive some physical equation, I noticed that the following identity was needed:
$\sum^{4a \leq 2k}_{a=0}{2k \choose 4a} + \sum^{4a+1 \leq 2k}_{a=0} {2k \choose 4a+1} = \left\{ \begin{array}{ll} \frac{2^k(2^k +1)}{2} & (k=4l+1, 4l+4)\\
... | A tool: By the Binomial Theorem, we have
$$(1+x)^n=1+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\binom{n}{4}x^4+\binom{n}{5}x^5+\cdots. \tag{1}.$$ Putting $x=1$ we get a familiar identity, and putting $x=-1$ we get something almost as familiar.
Using addition and subtraction, we get the sum of the binomial coeffic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to find $n$-th value in a series Let $(x_n, y_n, z_n) = (3, 1, 0)$ for $n=0$
For $n \ge 1$,
$$\begin{align}
x_n &= x_{n-1} +3 z_{n-1}\\
y_n &= x_{n-1} +2 z_{n-1}\\
z_n &= 5 y_{n-1}
\end{align}$$
Please let me know the formula to find $x_n,y_n,z_n$ values of any integer $n$.
The following would be the series:
$$\beg... | As pointed out by GerryMyerson, you can rewrite your problem as
\begin{align}
X_n=A\cdot X_{n-1} =\ldots=A^nX_0
\end{align}
With $X_n =(x_n,y_n,z_n)^T$ and $X_0=(3,1,0)$, we only need the appropriate matrix $A$.
Your recurrence relation is
\begin{align}
x_n &=x_{n-1}+3z_{n-1}\\
y_n &=x_{n-1}+2z_{n-1}\\
z_n &= 5 y_{n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to Solve $ \int \frac{dx}{x^3-1} $ I am having quite a difficult time integrating
$$
\int \frac{\mathrm{d}x}{x^3-1}
$$
My first approach was to apply a partial fraction decomposition
$$
\int \frac{\mathrm{d}x}{x^3-1} = \int \frac{\mathrm{d}x}{(x-1)(x^2+x+1)} = \frac{1}{3} \int \frac{\mathrm{d}x}{x-1} - \frac{1}{3}... | Hint: Complete the square in the denominator: $x^2+x+1 = (x + 1/2)^2 + 3/4$. Now write the fraction as a linear combination of two fractions whose integral you can do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Special values $\psi \left(\frac12\right)$ and $\psi \left(\frac13\right)$ I wonder if it is easy to prove that
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac13\right) & = -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3,
\end{align}
$$
where $\psi$ is the digamma function-the logarithmi... | You might start from the sum form:
$$ \psi(z) = -\gamma + \sum_{k=1}^\infty \left( \dfrac{1}{k} - \dfrac{1}{k+z-1}\right)$$
For $z=1/2$ the partial sum up to $k=n$ (let's say for convenience that $n$ is even) is
$$ \eqalign{\sum_{k=1}^n \left( \dfrac{1}{k} - \dfrac{2}{2k-1} \right) &=
\dfrac{1}{1} + \dfrac{1}{2} + \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$ Evaluation of $\displaystyle \int\frac{1}{\sin^2 x\cdot \left(5+4\cos x\right)}dx$
$\bf{My\; Solution::}$ Given $\displaystyle \int\frac{1}{\sin^2 x\cdot (5+4\cos x)}dx = \int \frac{1}{(1-\cos x)\cdot (1+\cos x)\cdot (5+4\cos x)}dx$
Now Using Partial ... | Decompose the integrand as follows\begin{align}\int\frac{1}{\sin^2 x\left(5+4\cos x\right)}dx
=& \ \frac19 \int 5\csc^2x-\frac{4\cos x}{\sin^2 x}- \frac{16}{5+4\cos x}\ dx\\
= &\ \frac19\bigg(-5\cot x+\frac4{\sin x} -\frac{32}3\tan^{-1}\frac{\tan\frac x2}3 \bigg)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Two methods to integrate? Are both methods to solve this equation correct?
$$\int \frac{x}{\sqrt{1 + 2x^2}} dx$$
Method One:
$$u=2x^2$$
$$\frac{1}{4}\int \frac{1}{\sqrt{1^2 + \sqrt{u^2}}} du$$
$$\frac{1}{4}log(\sqrt{u}+\sqrt{{u} +1})+C$$
$$\frac{1}{4}log(\sqrt{2}x+\sqrt{2x^2+1})+C$$
Method Two
$$u=1+2x^2$$
$$\frac{... | In method 1, you have a mistake. If $$u=2x^2$$ $$x=\frac{\sqrt{u}}{\sqrt{2}}$$ $$dx=\frac{1}{2 \sqrt{2} \sqrt{u}}$$ and $$\int \frac{x}{\sqrt{1 + 2x^2}} dx=\frac{1}{4} \int \frac{du}{\sqrt{u+1}}=\frac{\sqrt{u+1}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Using Inclusion/Exclusion to solve $x_1+x_2+x_3=15$ with $x_1,x_2\leq 5$ and $x_3\leq 7$ for non negative integers $x_1,x_2,x_3$ I want to solve for:
Number of integers solutions to the equation $x_1+x_2+x_3=15$ with $x_1,x_2\leq 5$ and $x_3\leq 7$ for non negative integers $x_1,x_2,x_3$
How can this be done using th... | If $N$ denotes the total number of ways $x_1+x_2+x_3 =15$ without any restrictions (of course, non-negative integers) and $M$ is the number of ways with desired constraints, then
$$
M = N - \left(N_{x_1>5} + N_{x_2>5} + N_{x_3>7}\right) + \left(N_{x_1>5\ and\ x_2>5} + N_{x_1>5\ and\ x_3>7} + N_{x_2>5\ and\ x_3>7}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
parametric integral relating to hyperbolic function Suppose that $a$ is real number such that $0<a<1$, how can we calculate
$$ I(a)=\int_0^\infty \big(1-\frac{\tanh ax}{\tanh x}\big)dx .$$
As for some speical cases, I can work out $I(1/2)=1$. Any suggestion to the integral above? Tks.
| If you are interested in particular cases:
$$\begin{align}
I\left( \frac15 \right) & = \sqrt{2-\frac{2}{\sqrt{5}}} \, \, \pi \\
I\left( \frac14 \right) & = \frac{\pi}{2} + 1 \\
I\left( \frac13 \right) & = \frac{\sqrt{3}}{3} \, \pi \\
I\left( \frac25 \right) & = \left(\sqrt{2+\frac{2}{\sqrt{5}}} - \frac54 \right) \pi \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/938664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Prove by induction $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{n}\ge\frac{2}{3}n\sqrt{n}$ for all positive integers Assumption: $\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}\ge\frac{2}{3}k\sqrt{k}$
Prove true for $n=k+1$ $$\sqrt{1}+\sqrt{2}+\sqrt{3}+...+\sqrt{k}+\sqrt{k+1}\ge\frac{2}{3}(k+1)\sqrt{k+1}$$
I'm upto :$$\sqrt{1}+\sqr... | For the induction step, you want to show that:
$$
\frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\
2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\
$$
Working backwards:
$$
2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\
2k\sqrt{k} \geq (2k-1)\sqrt{k+1} \\
4k^2 \times k \geq (4k^2 - 4k+1)(k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/939578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Convergent or divergent $\sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}}$ \begin{align}
& \sum\limits_{n=0}^{\infty }{\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}} \\
& \text{ordering} \\
& a_{n}=\frac{1\cdot 3\cdot 5...(2n-1)}{2\cdot 4\cdot 6...(2n+2)}=\frac{1\... | You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)} &=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I find 2x2 orthonormal diagonalizing matrices using only trigonometry? I have a matrix $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where all values are known), and I eventually want to diagonalize it into:
$$
A=UDV^T
$$
for orthonormal U and V. If I represent U and V as:
$$
U=\begin{bmatrix} cos(\theta) & ... | Problem statement and restrictions
The problem you propose is this
$$
\begin{align}
\mathbf{A} &= \mathbf{U} \, \mathbf{S} \, \mathbf{V}^{*} \\
% A
\left[
\begin{array}{cc}
a & b \\
c & d \\
\end{array}
\right]
%
&=
%
% U
\left[
\begin{array}{cr}
\cos \theta & -\sin \theta \\
\sin \theta & \cos \theta \\
\end{array... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/943254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to rationalize this root form? Suppose that we have a equation like this:
$$\sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a+b-2\sqrt{ab}}$$
In order to rationalize it, we can apply the formula:
$$\sqrt{a} + \sqrt{b} = \sqrt{a+b+2\sqrt{ab}}$$ or $$\sqrt{a} - \sqrt{b} = \sqrt{a+b -2\sqrt{ab}}$$, where $a>b$
My question is:... | Notice that there are $3$ terms with $2$ so probably it will be of form $(a+b+c)^2$
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)\\a^2+b^2+c^2=10\\ab=\sqrt{15}\\ac=\sqrt{10}\\bc=\sqrt{6}\\b^2+c^2=10-a^2\\a^2b^2+a^2c^2=25\\a^2(b^2+c^2)=25\\a^2(10-a^2)=25\\10a^2-a^4-25=0\\a^4-10a^2+25=0\\(a^2-5)^2=0\\a=\sqrt{5}\\ab=\sqrt{15}\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/944682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What exactly happens in the algebraic steps here? $$ \frac{n(n+1)}{2} + (n+1) = (n+1)(\frac{n}{2} + 1) = \frac{(n+1)(n+2)}{2} $$
I don't understand what happens from the first to the second and from the second to the third one.
| Note that $$\frac{n(n+1)}{2} = \frac{n}{2}(n+1)$$ so in the original expression, we have the sum of two terms both of which contain a factor of $(n+1)$. Taking this common factor out we obtain
$$\frac{n(n+1)}{2} + (n+1) = \frac{n}{2}(n+1) + 1(n+1) = (n + 1)\left(\frac{n}{2}+1\right)$$
which is the second expression. As... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving equation involving binomial function Solve for $x$ in terms of $i$ and $j$:
$$
\binom{x}{i} = j
$$
where $x$ is Real; $i$ and $j$ are Integers: $x \geqslant i$, $i \geqslant1$, $j \geqslant 0$.
I came across this problem while trying to unrank combinations $\binom{n}{k}$, pp. 51, Algorithm 2.12. I tried the fol... | As a partial answer I generated some solutions using CAS.
For $i=1$: $x_1 = j.$
For $i=2$:
$$x_1 = 1/2+1/2\,\sqrt {1+8\,j},$$
$$x_2 = 1/2-1/2\,\sqrt {1+8\,j}.$$
For $i=3$:
$$x_1 = 1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}+{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1,$$
$$x_2 = -1/6\,\sqrt [3]{81\,j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I show that $Var(Y) = n\frac{(\theta _{1} - \theta _{2})^{2}}{(n+1)^{2} (n+2)}$? My original pdf is $f(y) = \frac{n (y_{n} - \theta_{1})^{n-1}}{(\theta_{2} - \theta_{1})^{n}}$ for $\theta_{1} < y < \theta_{2}$.
After using U-substitution, I obtain $E(Y) = \frac{n \theta_{2} + \theta_{1}}{(n+1)}$.
For variance o... | The conceptual way to do this is to recognize that $Y$ is a location-scale transformation of a beta distribution. If $f_Y(y) \propto (y-\theta_1)^{n-1}$ for $y \in [\theta_1, \theta_2]$, then this suggests the linear, order-preserving transformation $$X = g(Y) = \frac{Y-\theta_1}{\theta_2 - \theta_1}.$$ Then $Y = g^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/947692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
summation algebra for $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1}$ Why does $\sum_{n=0}^\infty x^n + \sum_{n=0}^\infty x^{n+1} = 1 + 2\sum_{n=1}^\infty x^n$? Shouldn't this be $1 + x + 2\sum_{n=1}^\infty x^n$ because of the $n+1$ in the second summation?
| This may convince you.
\begin{array}\
\displaystyle\sum_{n=0}^{\infty} x^n \hphantom{+ \displaystyle\sum_{n=0}^{\infty} x^{n+1}} &= 1 + \hphantom{2}x + \hphantom{2}x^2 + \dots\\
&\\
\hphantom{\displaystyle\sum_{n=0}^{\infty} x^n +} \displaystyle\sum_{n=0}^{\infty} x^{n+1} &= \hphantom{1 + 2}x + \hphantom{2}x^2 + \dots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/948790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$ Does the series
$$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$
converges?
My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n!
$$\frac{(2n)!}{n!n!4^n} \sim (\frac{1}{4^n} \frac{\sqrt{4\pi n}(\frac{2n}{e})^{2n}}{\sqrt{2 n \p... | Another approach: Note that $\frac{(2n)!}{n!n!}$ equals the binomial coefficient $\binom{2n}{n}$. The binomial coefficients $\binom{2n}{0}$, $\binom{2n}{1}$, ..., $\binom{2n}{2n}$ have sum $2^{2n} = 4^n$, thus their average is $\frac{4^n}{2n+1}$. Since $\binom{2n}{n}$ is the largest of these binomial coefficients, it i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
How prove this$\lfloor \sqrt{2x-\lfloor\sqrt{2x}\rfloor}\rfloor=\lfloor\frac{\sqrt{8x+1}-1}{2}\rfloor$ Question:
let $x\ge 0$, show that
$$\lfloor \sqrt{2x-\lfloor\sqrt{2x}\rfloor}\rfloor=\lfloor\dfrac{\sqrt{8x+1}-1}{2}\rfloor$$
My idea: let $\lfloor \sqrt{2x}\rfloor =m$
then
$$\sqrt{2x}-1<m\le \sqrt{2x}$$
so
$$m^2... | Note that $\frac{\sqrt{8x+1}-1}{2}=\sqrt{2x+\frac 14}-\frac 12$.
We know that the LHS equals $m$ or $m-1$. Suppose that the LHS equals $m$. Then
$$m\le \sqrt{2x-m}< m+1$$
Thus
$$m^2+m\le 2x < (m+1)^2+m$$
So the RHS satisfies
$$\sqrt{m^2+m+\frac 14}-\frac 12\le \sqrt{2x+\frac 14}-\frac 12<\sqrt{(m+1)^2+m+\frac 14}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$ \cos {A} \cos {B} \cos {C} \leq \frac{1}{8} $ In an acute triangle with angles $ A, B $ and $ C $, show that
$ \cos {A} \cdot \cos {B} \cdot \cos {C} \leq \dfrac{1}{8} $
I could start a semi-proof by using limits: as $ A \to 0 , \; \cos {A} \cos {B} \cos {C} $ becomes big (as we want), but $ A+B+C $ becomes too small... | Homage a lab
$2\cos A\cos B\cos C=(\cos(A+B)+\cos(A-B))\cos C=(\cos(A-B)-\cos C)\cos C=\frac{1}{4}\cos^2(A-B)-(\cos C-\frac{1}{2}\cos(A-B))^2\le\frac{1}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Number of possible eight digit number divisible by 9 An eight digit number divisible by 9 is o be formed by using 8 digits out of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 without repetition. Find the number of ways in which it can be done.
I know divisible rule of 9 is sum of all the digits should be multiple of nine. But don't kn... | Eight digit no divisible by $9$ i.e. sum of digits divisible by $9$
$(1.)$ Total no formed by $(1,2,3,4,5,6,7,8) = 8!$
$(2.)$ Total no formed by $(0,2,3,4,5,6,7,9) = 7\times7!$
$(3.)$ Total no formed by $(1,0,3,4,5,6,9,8) = 7×7!$
$(4.)$ Total no formed by $(1,2,0,4,5,9,7,8) = 7×7!$
$(5.)$ Total no formed by $(1,2,3,0,5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Dividing the function by the largest power of x to rationalize(infinity minus infinity)
Calculating the limits of the above when x approaches infinity, why can't we just divide the function by the largest power of x and then say inside of the square root should be equal to "1" when x is at infinity level?
This seems... | Since $\lim_{x\to \infty}|x|\sqrt{1\pm\frac 2x}\to \infty$, we have $+\infty-\infty$. We cannot find the limit in this form.
Multiplying
$$\sqrt{x^2+2x}-\sqrt{x^2-2x}$$
by
$$\frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\ (=1)$$
gives you
$$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-2x})(\sqrt{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x^2+y^2+z^2=3/2$, then $\sum\frac{x^2}{x(4x-3)+z^2+y^2} \le1+\frac{\sqrt2}2$ If $x$, $y$, and $z$ are real numbers satisfying
$$
x^2 +y^2 +z^2 = 3/2
$$
then prove that $1+\frac{\sqrt{2}}{2} \geq$ the cyclic sum of
$$
\frac{x^2}{x(4x-3)+z^2+y^2} .
$$
I've tried Cauchy-Schwarz, but I don't know how to restructure t... | I think your constant should be $1+\frac{\sqrt2}{2} \lt 1+\frac{2}{\sqrt2}$ which is stronger and still true. Since the denominator $4x^2+y^2+z^2-3x$ is bigger if $x$ is negative it is enough to prove that the inequality holds for positive real numbers $x,y,z$. Therefore, to make it a bit more beautiful, you can make t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/958394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integrating $\cos^3(x)$ My attempt at integrating $\cos^3(x)$:
$$\begin{align}\;\int \cos^3x\mathrm{d}x &= \int \cos^2x \cos x \mathrm{d}x
\\&= \int(1 - \sin^2 x) \cos x \mathrm{d}x
\\&= \int \cos x dx - \int \sin^2x \cos x \mathrm{d}x
\\&= \sin x - \frac {1}{3}\sin^3x + C\end{align}$$
My question is how does integra... | If you let $u = \sin x$, then $du = \cos x \ dx,$ and
$\int \sin^2x \cos x \ dx $ becomes $$\int u^2 du = \frac{1}{3} u^3 + C.$$
Then you substitute $u=\sin x$ back in.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding $\sum_{k=1}^{\infty} \left[\frac{1}{2k}-\log \left(1+\frac{1}{2k}\right)\right]$ How do we find $$S=\sum_{k=1}^{\infty} \left[\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right]$$
I know that $\displaystyle\sum_{k=1}^{\infty} \left[\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right]=\gamma$, where $\gamma$ is... | Observe that, by absolute convergence:
$$\begin{align}
\sum_{k=1}^{\infty} \left(\frac{1}{2k} -\log\left(1+\dfrac{1}{2k}\right)\right)& =\sum_{k=1}^{\infty} \dfrac{1+(-1)^k}{2}\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\right)\right)\\\\
&=\frac{1}{2}\sum_{k=1}^{\infty} \!\left(\frac{1}{k} -\log\left(1+\dfrac{1}{k}\!\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$ I have no idea where to even start. WolframAlpha cant compute it either.
$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$
I think it can be done with series, but I am not sure, can someone help a little so I can get a start??
Thanks!
| Using the series of $\ln^2(1-x)$,
\begin{align}
\int^1_0\ln^2{x}\ln^2(1-x) \ {\rm d}x
&=\sum^\infty_{n=1}\frac{2H_n}{n+1}\int^1_0x^{n+1}\ln^2{x}\ {\rm d}x\\
&=\sum^\infty_{n=1}\frac{4H_n}{(n+1)(n+2)^3}
\end{align}
Then integrate $f(z)=\dfrac{(\gamma+\psi_0(-z))^2}{(z+1)(z+2)^3}$ along an infinitely large square. The in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/959701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Area of the intersection of four circles of equal radius
This picture basically shows a rearrangement of four quarters of a circle of radius 1. It asks for the shaded area.
I got the answer to be $\frac{2\pi + 6}{13}$. But then it is incorrect.
The way I did it is by rearranging the four pieces into respectively a ci... | Note that we could use the first quadrant and then get the final answer by symmetry. We have three quarter-circles in this region with the equations$$\begin{array}{l}{C_1}:{\left( {x - \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\\{C_2}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Interesting use of remainder theorem I am asking this question in reference to this post Write an Efficient Method to Check if a Number is Multiple of 3
In the proof of the method the author writes that any 2 digit number(AB) can be written in the form $$AB=11A-A+B$$Any 3 digit number(ABC) can be written as $$ABC = 99A... | Considering congruences modulo $11$ is easier:
\begin{align}
10^0 & \equiv 1 \pmod{11} \\
10^1 & \equiv 10 \equiv -1\\
10^2 & \equiv -1\cdot 10\equiv -1\cdot(-1)\equiv 1\\
10^3 & \equiv 1\cdot 10\equiv 1\cdot(-1)\equiv -1\\
&\dots\\
10^{2k} & \equiv -1\cdot 10\equiv -1\cdot(-1)\equiv 1\\
10^{2k+1} & \equiv 1\cdot 10\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximise the volume of an open triangular prism An open container is to be constructed out of 200 square centimeters of cardboard. The two end pieces are equilateral triangles. The open top is a horizontal rectangle. Find the lengths of the sides of the triangle for maximum volume of the container.
So effectively we ha... | Let's let our variable x represent the side of the equilateral triangle. In that case, the area of the equilateral triangle is $$\frac{x^2\sqrt{3}}{4}$$
Therefore, the remaining area for the cardboard to be used by the three rectangles is $$200-
\Big(\frac{x^2\sqrt{3}}{4}\Big)2$$ Because we have three sides, the area t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve this exponential equation? $2^{2x}3^x=4^{3x+1}$. I haven't been able to find the correct answer to this exponential equation:
$$\eqalign{
2^{2x}3^x&=4^{3x+1}\\
2^{2x} 3^x &= 2^2 \times 2^x \times 3^x\\
4^{3x+1} &= 4^3 \times 4^x \times 4\\
6^x \times 4 &= 4^x \times 256\\
x\log_6 6 + \log_6 4 &= x\log_64 ... | $$2^{2x}3^x=4^{3x+1},$$
$$(e^{\log2})^{2x}(e^{\log3})^x=(e^{\log4})^{3x+1},$$
$$e^{2x\log2}e^{x\log3}=e^{(3x+1)\log4},$$
$$e^{2x\log2+x\log3}=e^{(3x+1)\log4}.$$
Take the log,
$$2x\log2+x\log3=(3x+1)\log4,$$and
$$x=\frac{\log4}{2\log2+\log3-3\log4}=\frac1{\frac{\log3}{\log4}-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/964962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Show that for any integer a, a^2 + 5 is not divisible by 4. My solution is:
Assume by contradiction that there is at least one number a such that $a^2$ + 5 is divisible by 4. Then a is either odd or even.
Consider the case when a is odd. Then a= 2k+1 for some integer k. Then $a^2$ = $4k^2$ + 4k + 1 so $a^2$ is odd. Th... | Your idea is fine. More compactly, any number is congruent to $0,1,2,3$ modulo $4$. Squaring gives $0,1,0,1$. But $5=-1$ modulo $4$, but it cannot be the case $a^2=-1\mod 4$ for any $a$, by the above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Continuous Functions such that $f(0) = 1$ and $f(3x) - f(x) = x$ Just had a midterm with the following problem:
Find all continuous functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(0) = 1$ and $f(3x) - f(x) = x$.
I was just curious how this would end up.
During the exam I tried setting $f(1) = c$, and end up gettin... | Notice that:
\begin{align*}
f(3x) - f(x) &= x \\
f(9x) - f(3x) &= 3x \\
f(27x) - f(9x) &= 9x \\
&~~\vdots \\
f(3^kx) - f(3^{k-1}x) &= 3^{k-1}x
\end{align*}
Summing them together, the LHS telescopes, giving us:
$$
f(3^kx) - f(x) = x\sum_{i=0}^{k-1}3^i = x \cdot \frac{3^k - 1}{2}
$$
Now consider what happens when we take... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integrals: Partial Fractions $$ \int \frac{x^2-x+12}{x^3+3x} $$
I factored the denominator to get $ x(x^2+3) $. I then seperated the x and the $x^2+3$ into the partials $\frac{A}{x}$ and $\frac{Bx+C}{x^2+3}$.
After combining the two, I came up with $$ \frac{A(x^2+3)+ x(Bx+C)}{x(x^2+3)}$$
This is where I'm stuck. I'm as... | Note that $x^2-x+12=12$ when $x=0$.
Note also that $A(x^2+3)+x(Bx+C)=3A$ when $x=0$. Thus $A=\frac{12}{3}=4$.
The coefficient of $x^2$ in $x^2-x+12$ is $1$. In $A(x^2+3)+x(Bx+C)$ it is $A+B$. Since $A=4$, we have $B=-3$.
We leave finding $C$ to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/976023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\... | There have already been enough answers, but none addresses why your proposed answer is incorrect. Mine will do just that.
The most basic mistake is that you use what you want to prove as if it was true. This is a logical fallacy called circular reasoning. There is also a number of other mistakes:
*
*From $b^2=4-a^2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.
Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$
Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and... | First Proof
$a+b=x$ , $b+c=y$ , $c+a=z$
$\therefore 2 \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-6\\=\underbrace{\dfrac{x}{y}+\dfrac{y}{x}}_{\geq 2}+\underbrace{\dfrac{y}{z}+\dfrac{z}{y}}_{\geq 2}+\underbrace{\dfrac{z}{x}+\dfrac{x}{z}}_{\geq 2}-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/980751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
... | $$
\frac{3}{\left(\frac 3 2\right)} = 3\cdot\frac 2 3 \ne 3\cdot \frac 3 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Inverse Trigonometric Integrals How to calculate the value of the integrals
$$\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx,$$
$$\int_0^1\left(\frac{\arctan x}{x}\right)^3\,dx
$$
and
$$\int_0^1\frac{\arctan^2 x\ln x}{x}\,dx?$$
| Here is a simple and a nice way to evaluate the first and the second integral.
Evaluation of $1^{\mbox{st}}$ Integral :
Making substitution $x=\tan\theta\,$ followed by integration by parts, we get
\begin{align}
\int_0^1\left(\frac{\arctan x}{x}\right)^2\,dx&=\color{red}{\int_0^{\Large\frac{\pi}{4}}\frac{\theta^2}{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
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Closed-form of $\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx$ Inspired by this question, is there a closed-form of
$$\int_0^1\left(\frac{\arctan x}{x}\right)^n\,dx\,?$$
Here $n \in \mathbb{N_+}$. In the answers to the question above we could find proofs of cases $n=2,3$.
I state here some specific cases.
$$\begin{ali... | Using substitution and integration by parts,
\begin{align}&\int_0^1\left(\frac{\tan^{-1}x}{x}\right)^n\,dx=\int_0^{\pi/4}\frac{x^n}{\tan^n x}\sec^2 x\,dx\\&=\left[\frac{x^n}{\tan^n x}\tan x\right]_0^{\pi/4}-\int_0^{\pi/4}\frac{nx^{n-1}}{\tan^{n-1}x}-\frac{nx^n}{\tan^n x}\sec^2x\,dx\end{align}
So we have $$\int_0^1\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/984725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 1,
"answer_id": 0
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Finding $\sum \frac{1}{n^2+7n+9}$ How do we prove that $$\sum_{n=0}^{\infty} \dfrac{1}{n^2+7n+9}=1+\dfrac{\pi}{\sqrt {13}}\tan\left(\dfrac{\sqrt{13}\pi}{2}\right)$$
I tried partial fraction decomposition, but it didn't work out after that. Please help me out. Hints and answers appreciated. Thank you.
| $$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+7n+9}
&=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}-\frac1{n+\frac72+\frac{\sqrt{13}}2}\right)\\
&=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}+\frac1{-n-\frac72-\frac{\sqrt{13}}2}\right)\\
&=\frac1{\sqrt{13}}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
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Square root and principal square root confusion A few months ago I asked a question about the $\pm$ symbol because I was confused about it... I still carry the same confusion (which really bugs me) but I think the real confusion has to do with the square root and principal square root. I hope I can finally grasp the co... | *
*We have to keep in mind that $x$ might be negative because we don't know what $x$ is and it might by negative.
But we don't have to keep anything in mind about $12$ because we know exactly what $12$ is.
But perhaps we need to understand why we are allowed to split $\sqrt {ab}=\sqrt{a}\sqrt{b}$ in the first place.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/990394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$ How to find the minimum distance between the curves $y^2-xy-2x^2 =0$ and $y^2=x-2$
Let $y^2-xy-2x^2 =0...(1)$ and $y^2=x-2...(2)$
In equation (1) coefficient of $x^2 =-2; y^2=1, 2xy =\frac{-1}{2}$
We know that a second degree equation where $a... | $y^2-xy-2x^2=0$, factorize it we get $(y-2x)(y+x)=0$, so the graph is two straight line $-2x+y=0$ or $x+y=0$.
Points on $y^2=x-2$ can be written as $(t^2+2,t)$, the distance from $(t^2+2,t)$ to $-2x+y=0$ is $\frac{|-2(t^2+2)+t|}{\sqrt{(-2)^2+1^2}}= \frac{|2t^2-t+4|}{\sqrt{5}}=\frac{|2(t-1/4)^2+31/8|}{\sqrt{5}}$.
The di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$
Find a matrix transformation mapping $\{(1,1,1),(0,1,0),(1,0,2)\}$ to $\{(1,1,1),(0,1,0),(1,0,1)\}$.
Is the answer
$$
\begin{bmatrix}1& 0& -1\\0& 1& 1\\0& 0& 1\end{bmatrix}?
$$
I understand the concept of Matrix Tran... | There is a very simple method to solve the problem described in "Beginner's guide to mapping simplexes affinely" and "Workbook on mapping simplexes affinely."
Consider the formula
$$
\vec{L}(\vec{p}) = (-1)
\frac{
\det
\begin{pmatrix}
0 & \vec{x} & \vec{y} & \vec{z} \\
p_1 & a_1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Tricky Decomposition Decompose the following polynomial:
$A=x^4(y^2+z^2) + y^4(z^2+x^2) + z^4(x^2+y^2) +2x^2 y^2 z^2 $
(by the way I'm not quite sure about the tags and the title feel free to edit them if you wish)
| $$
A = x^4(y^2+z^2) + x^2(y^4+2y^2 z^2 +z^4) +y^2z^2(y^2+z^2) \\
= (y^2+z^2)\left(x^4+(y^2+z^2)x^2 +y^2z^2\right) \\
=(y^2+z^2)(x^2+y^2)(z^2+x^2)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/998667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{x^2+y^3}$ and $\sqrt{x^3+y^2}$ are rational Are there infinitely many pairs of different positive rational numbers $x,y$ such that $\sqrt{x^2+y^3}$ and $\sqrt{x^3+y^2}$ are rational?
Consider such a pair. Then we have $x^2+y^3=a^2$ and $x^3+y^2=b^2$ for some rationals $a,b$. So $(a^2-y^3)^3=x^6=(b^2-y^2)^2$. Doe... | Yes, there are infinitely many solutions in integers.
In particular, we have a solution with $y = 2x$ whenever $x + 4$ and $8 x + 1$ are both squares (i.e. $x^2 + (2x)^3 = x^2 (1 + 8 x)$ and $(2x)^2 + x^3 = x^2 (4 + x)$). If $x + 4 = u^2$ and $8 x + 1 = v^2$, eliminating $x$ gives us the
equation $8 u^2 - 31 = v^2$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that $X^4+8X^3+X^2+2X+5$ is irreducible? How to prove that $X^4+8X^3+X^2+2X+5$ is irreducible on $\mathbb Q[X]$?
I can try to find $a,b,c,d$ such that
$$X^4+8X^3+X^2+2X+5=(X^2+aX+b)(X^2+cX+d)$$
and prove that $a,b,c,d\notin\mathbb Q$ and find $a,b,c,d$ such that
$$X^4+8X^3+X^2+2X+5=(X-a)(X^3+bX^2+cX+d)$$
... | Let $P(X)=A(X)B(X)$ a decomposition of your polynomial into quadratics. Observe that
$$
P(X)\equiv X(X^3+3X^2+X+2) \bmod 5
$$
and the deg 3 factor is irreducible mod 5 because it has no roots mod 5. Thus one of the quadratic factors $A, B$ has only one root mod 5, but this is not possible. Hence there's no factorizatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find basis for subspace given by $x_1/3=x_2/4=x_3/2$ Find a basis for the subspace W: $$
W = \left\{ \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3}\end{bmatrix} \in \mathbb{R}^3 \left| \dfrac{x_{1}}{3}=\dfrac{x_{2}}{4}=\dfrac{x_{3}}{2} \right.\right\}
$$
The two equation signs are really throwing me off. If it was a linear eq... | What you have here are two equation:
$$
\frac{x_{1}}{3}=\frac{x_{2}}{4}
$$
$$
\frac{x_{2}}{4}=\frac{x_{3}}{2}
$$
from the first one you get $x_{2}=\frac{4}{3}x_{1}$ and from the
second one you get $x_{3}=\frac{1}{2}x_{2}$, now using $x_{2}=\frac{4}{3}x_{1}$
we get $x_{3}=\frac{1}{2}\cdot\frac{4}{3}x_{1}=\frac{2}{3}x_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$ for $y$ I have the equation:
$\frac{1}{2}kx^{2}-cx=\frac{1}{2}ky^{2}+cy$,
where $k$ and $c$ are arbitrary constants.
How do I go about simplifying this and solving for $y$ in terms of $x$, excluding the obvious solution $y=-x$
| $$\frac{ky^2}{2}+cy=\frac{kx^2}{2}-cx \iff y^2+\frac{2cy}{k}=\frac2k \left (\frac{kx^2}{2}-ck \right ) \iff y^2+\frac{2cy}{k}+\frac{c^2}{k^2}=\frac2k \left (\frac{kx^2}{2}-ck \right )+\frac{c^2}{k^2} \iff \left (y+\frac{c}{k}\right )^2 =\frac{(c-kx)^2}{k^2}$$
I think you can continue from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ I'm trying to determine $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}}$ using L'Hopital's Rule.
I can clearly see that $\lim_{x\rightarrow\infty}(\frac{x+1}{x-1})^{\sqrt{x^2-1}} = \frac{\infty}{\infty},$ so we can use L'Hoptial's Rule. I'm having ... | Note that:
$$\left(\frac{x+1}{x-1}\right)^{\sqrt{x^2-1}} = \left(1 + \frac{2}{x-1}\right)^{\sqrt{x^2-1}} = e^{\sqrt{x^2-1} \log \left(1 + \frac{2}{x-1}\right)}$$
Now we work with the exponent:
$$\lim_{x\rightarrow \infty} \sqrt{x^2-1} \cdot \log \left(1 + \frac{2}{x-1}\right)
\overset{(1)}{=}
\lim_{x\rightarrow \infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
A limit-determinant question
Let $d_n$ be the determinant of the $n\times n$ matrix whose entries, from left to right and then from top to bottom, are $\cos1,\cos2,\ldots,\cos n^2$. (For example,
$$d_3=
\begin{vmatrix}
\cos1&\cos2&\cos3\\
\cos4&\cos5&\cos6\\
\cos7&\cos8&\cos9\\
\end{vmatrix}$$
The argument of $\cos$ i... | $$\begin{array}{l} \left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \cos 4 & \cos 5 & \cos 6 \\ \cos 7 & \cos 8 & \cos 9 \end{array}\right| \\ =\left|\begin{array}{ccc}\cos 1 & \cos 2 & \cos 3 \\ \cos 3\cos 1 - \sin 3\sin 1 & \cos 3\cos 2 - \sin 3\sin 2 & \cos 3\cos 3 - \sin 3\sin 3 \\ \cos 7 & \cos 8 & \cos 9 \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving Complex Quadratic equations After working a few exercises on the topic, the questions become progressively harder. In this particular exercise I was asked to solve the equations. However I can't quite seem to break this problem as I did previously and I'll explain why.
The equation is as follows: $$z^2 + (1-i... | $$
z^2 + (1-i)z + (-6 + 2i) = 0\\
\begin{align}
\implies z &= \frac{-(1-i) \pm \sqrt{(1-i)^2 - 4(-6 + 2i)}}{2}\\
&= \frac{1}{2}\left( i - 1 \,\pm \sqrt{1+ 2i^2 - i^2 +24 - 8i}\right)\\
&= \frac{1}{2}\left(i-1 \pm 2i\sqrt{2i - 6}\right)\\
&= \left(\frac{1}{2}\pm\sqrt{2(i - 3)} \right)i - \frac{1}{2}
\end{align}
$$
Alwa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sin^2(x), \cos^2(x),$ and $\sin^4(x)$ and linear dependence Since
$$
\sin^{4}x=\sin^{2}x\sin^{2}x = \sin^{2}x(1-\cos^{2}x) = \sin^{2}x-\sin^{2}x\cos^{2}x
$$
does this mean that
$$
\sin^{2}x,\cos^{2}x, \text{ and } \sin^{4}x
$$
are linearly dependent?
| You can once again use the Wronskian: $$W(x) = \begin{vmatrix} \sin^2x & \cos^2x & \sin^4x \\ \sin(2x) & -\sin(2x) & 4\sin^3x \cos x \\ 2\cos(2x) & -2\cos(2x) & 12\sin^2x \cos^2x + -4\sin^4x\end{vmatrix},$$ and see if it is identically zero, or not. But that's quite a pain here, so you can take a leap of faith: $$W(\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\prod_{n=1}^{\infty} (1+\frac{(-1)^{n+1}}{2n-1})=\sqrt2$ This question came up while I was revising the Gamma function.
$$
\prod_{n=1}^{\infty} (1+\frac{(-1)^{n+1}}{2n-1})=\sqrt2
$$
Please prove me prove this infinite product, and explain the steps involved. I'm really having trouble with this one. Many thanks.... | Let $P_m = \displaystyle\prod_{n=1}^{m}\left[1+\dfrac{(-1)^{n+1}}{2n-1}\right]$. Then, $P_{2m} = \dfrac{2}{1}\dfrac{2}{3}\dfrac{6}{5}\dfrac{6}{7} \cdots \dfrac{4m-2}{4m-3}\dfrac{4m-2}{4m-1}$.
This can be rewritten as $P_{2m} = \dfrac{2^{2m}(2m)!^3}{(4m)!(m!)^2} = \dfrac{2^{2m}\binom{2m}{m}}{\binom{4m}{2m}}$.
The Cent... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Indefinite integral of $\frac{\arctan x}{x^2+1}$ EDIT: I was studying from a site that uses really ambiguous notation so I misread $\arctan\ (x)^2$ as $\arctan\ (x^2)$. Now I can see why the integral is actually $\frac{1}{2} \arctan^2\ x + c $. Thanks to everyone who answered and corrected me!
Why is $$\int\frac{\arcta... | Your notation may be a bit off?
Doing the $u$-substitution gives
$$\int \frac{\arctan x}{1+x^2}dx = \int u\text{ }du = \frac{u^2}{2} + C = \frac{(\arctan x)^2}{2} + C.$$
Note that $(\arctan x)^2 \neq \arctan \left(x^2\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Help with Mathematical Induction $$1^3+2^3+\cdots+n^3=\left[\frac{n(n+1)}2\right]^2$$
so far I have..
$$1^3+2^3+\cdots+k^3+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2$$
then..
$$\left[\frac{k(k+1)}2\right]^2+(k+1)^3=\left[\frac{(k+1)(k+2)}2\right]^2$$
where do I go from here so that the rhs equals the lhs?
| Inductive step:
$$\begin{align*}\underbrace{1^3+2^3+\ldots+n^3}_{=\left(\dfrac{n(n+1)}{2}\right)^2}+(n+1)^3&=\left(\dfrac{n(n+1)}{2}\right)^2+(n+1)^3=(n+1)^2\cdot\left(\frac{n^2}{4}+(n+1)\right)=\\&=(n+1)^2\cdot\left(\frac{n^2+4n+4}{4}\right)=(n+1)^2\cdot\left(\frac{(n+2)^2}{2^2}\right)=\\&=\left(\frac{(n+1)(n+2)}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lipschitz continuity of a function Show that $g(x) = \frac{1}{x^{2} +1}$ is Lipschitz conitnuous.
From the definition, we must show that $\forall x,y \in \mathbb{R}$, $|f(x)-f(y)| \leq K|x-y|$, for some real constant $K$.
First, I tried to find:
$|f(x)-f(y)|=|\displaystyle\frac{1}{x^{2} +1} - \frac{1}{y^{2} +1}|=|\fra... | \begin{align}
\left|\frac{x^2-y^2}{(1+x^2)(1+y^2)}\right|&=\left|\frac{x+y}{(1+x^2)(1+y^2)}\right| |x-y|\\
&\le \max\left(\frac{x+y}{(1+x^2)(1+y^2)}\right)|x-y|\\
&={3\sqrt{3}\over 8}\,|x-y|
\end{align}
where the last equality follows from a little calculus.
Of course, we do not need such a sharp $K$, so we can of cour... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Parametric solutions to $(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square$ Let $a,b,c$ and $d$ be rational.Find a rational parametric solutions for $a,b,c$ and $d$ so that
$$(4/3)b^2c^2+(4/3)a^2d^2-(1/3)a^2c^2-(4/3)b^2d^2=\square.$$
| Rewrite it as
$$4b^2(c^2-d^2) = a^2(4d^2-c^2) = a^2(4d^2-4c_0^2)$$
where $c_0 = c/2$. Then
$$b^2(c^2-d^2) = a^2(d^2-c_0^2)$$
We can assign: $b^2 = d^2-c_0^2$ and $4d^2-c^2 = a^2$.
For the first equation we use the famous Pythagorean triple: $b = 3, d = 5, c_0 = 4 \rightarrow c = 8$.
Then it brings us to the last varia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I find the determinant of a 4x4 matrix when the diagonal is made up of variables? Evaluate: $\det(A)$, where $A=
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a\end{bmatrix}$
| When computing the determinant, you are allowed to add any linear combination of the other rows (or columns) to any particular row (or column). Also recall that the determinant is linear in the rows (or columns), and this allows you to break out common factors.
One way of doing it might look like
$$
\begin{vmatrix}
a &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Integral (close form?) I'm struggling to evaluate the following integral:
$\displaystyle \int_{-1}^{1}\frac{1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12}}{\sqrt{\left ( 1+x^2 \right )\left ( 1+x^4 \right )\left ( 1+x^6 \right )}}\,dx$.
I see that the integrand is even, thus the integral can be re-written as: $\displaystyle \in... | I'll play and see if
I can find anything interesting.
$I
=\int_{0}^{1}\dfrac{1+2x^2+3x^4+4x^6+5x^8+6x^{10}+7x^{12}}{\sqrt{( 1+x^2 ) ( 1+x^4 ) ( 1+x^6 )}}\,dx$
Let $y=x^2$.
Then
$dy = 2x\, dx
$
or
$dx
=\frac{dy}{2x}
=\frac{dy}{2\sqrt{y}}
$.
Then
$\begin{array}\\
I
&=\frac12\int_{0}^{1}\dfrac{1+2y+3y^2+4y^3+5y^4+6y^5+7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How find this sum $\sum_{k=1}^{\infty}\frac{1}{1+a_{k}}$
Let $\{a_{n}\}$ be the sequence of real numbers defined by $a_{1}=3$ and for all $n\ge 1$,
$$a_{n+1}=\dfrac{1}{2}(a^2_{n}+1)$$ Evaluate
$$\sum_{k=1}^{\infty}\dfrac{1}{1+a_{k}}$$
My idea 1:
since
$$2a_{n+1}=a^2_{n}+1$$
so we have
$$2(a_{n+1}-1)=(a_{n}+1)(... | First, observe that
$$
a_{n+1} -1= \frac{a_n^2-1}{2} = \frac{(a_n+1)(a_n-1)}{2},
$$
and consequently, (one can easily show that $a_n >1$ for all $n\ge 1$)
$$
\frac{1}{a_{n+1}-1}=\frac{1}{a_n-1} - \frac{1}{a_n+1}.
$$
Rewriting this as
$$
\frac{1}{a_n+1} = \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}
$$
Summing up this identity f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A sequence is given by $a_1 = 1$, $a_{n+1} = \sqrt{1+2a_n}$ for $n=1,2,3,4...$ Show that the sequence is increasing and bounded by 3.
Find out if it converges, and find its limit.
So far i think the limit is
$$
1+\sqrt{2},
$$
and that i should use induction to find that its bounded by 3.
I have solved everything excep... | Assume that, as $n\to\infty$, the limit of $a_n$ is $u$, hence $a_{n+1}=a_n=u$.
$$\begin{align}
a_{n+1}&=\sqrt{1+2a_n}\\
u&=\sqrt{1+2u}\\
u^2-2u-1&=0\\
\because u>0\therefore u&=1+\sqrt{2}
\end{align}$$
NB:
$$\begin{align}
a_1&=1\\
a_2&=\sqrt{1+2}=\sqrt{3}\\
a_3&=\sqrt{1+2\sqrt{3}}\\
a_4&=\sqrt{1+2\sqrt{1+2\sqrt{3}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1019974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\frac{n}{(n+r)^2}$ $$
\lim_{n\ \to\ \infty}\left[\,%
{n \over \left(\, n + 1\,\right)^{2}}+
{n \over \left(\, n + 2\,\right)^{2}}+\cdots +
{n \over \left(\, 2n\,\right)^{2}}\,\right]
$$
How can I deal with this limit in a reasonable way ?.
EDIT: I am STILL looking for a... |
I am looking for a solution without use of calculus
I'm interpreting that as "without integration", since limits are calculus.
Write
$$\begin{aligned}
\frac{1}{(n+r)^2} &= \frac{1}{(n+r)(n+r+1)} + \left(\frac{1}{(n+r)^2}-\frac{1}{(n+r)(n+r+1)}\right)\\
&= \left(\frac{1}{n+r} - \frac{1}{n+r+1}\right) + \frac{1}{(n+1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$? is $f(x)=x^4 +x^3 +x^2 +x +1$ irreducible in $R=(\mathbb Z/7\mathbb Z)[x]$?
I know that $f(x)$ cannot be expressed as a product of degree 1 polynomial and degree 3 polynomial since it has no roots in $R$ but what about two degree 2 polynimials?... | If $f=(ax^2+bx+c)(dx^2+ex+f)$ with $a,d \neq 0$ mod $7$ one can divide the first factor by $a$ to make it monic, then remultiply by $a$ to compensate. So resetting the coefficients at this point $f=(x^2+bx+c)(dx^2+ex+f).$ [these are "reset" $d,e,f$] Then the second factor is also monic to make the coefficient of $x^4$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1023344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$ Let $a,b,c,d,e$ be integers such that $a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0$. Prove that $a+b+c+d+e$ divides $a^5+b^5+c^5+d^5+e^5-5abcde$.
I'm reminded of the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. But for $5$th degree, how can I find a factoriza... | If you use the following notations:
$$s_1=a+b+c+d+e$$
$$s_2=ab+ac+ad+ae+bc+bd+be+cd+ce+de$$
$$s_3=abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde$$
$$s_4=abcd+abce+abde+acde+bcde$$
$$s_5=abcde$$
Then we have:
$$a^5+b^5+c^5+d^5+e^5=\left(\left(s_1\left(s_1^2-2s_2\right)-s_1s_2+3s_3\right)s_1-s_2\left(s_1^2-2s_2\right)+s_3s_1-4s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Product identity for $n^n$ I came across the rather nice identity
\begin{align}
&&\frac{(-n)^{n-1} \Gamma (n+1)}{(1-n)_{n-1}}&&\tag{1}&\\
\\
&=&\prod _{k=1}^{n-1} \frac{(k+1) n^2}{n^2-k n}&&\tag{2}&\\
\\
&=&\frac{2 n^2}{n^2- n}\cdot\frac{3 n^2}{n^2-2 n}\cdot\frac{4 n^2}{n^2-3 n}
\cdots\frac{n^3- 2n^2}{3 n}\cdot\frac{n^... | Steps (1) to (2)
\begin{align}
f(n)
&= \frac{(-n)^{n-1} \Gamma(n+1)}{(1-n)_{n-1}} \\
&= \frac{(-n)^{n-1} n!}
{\underbrace{(1-n)(1-n+1)(1-n+2)\cdots -2 \cdot -1}_{n-1 \tiny \mbox{ factors}}} \\
&= \frac{(-n)^{n-1} n!}
{(1-n)(2-n)(3-n)\cdots -2 \cdot -1} \\
&= \frac{n^{n-1} n!}
{(n-1)(n-2)(n-3)\cdots 2 \cdot 1} \\
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Determining the asymptotics of the Summatory function of an Arithmetic Function We define the arithmetic function: $\displaystyle f(n) = \max\limits_{p^{\alpha} || n} \alpha$, that is if $\displaystyle n = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ (prime factorization of $n$) then $f(n) = \max\limits_{1 \le i \le k} \alpha ... | If we group the numbers by value of $f(n)$, we have
$$S(x) = \sum_{k \leqslant \log_2 x} k\cdot\operatorname{card} \{ n \leqslant x : f(n) = k\} = \sum_{k \leqslant \log_2 x} \operatorname{card} \{ n \leqslant x : f(n) \geqslant k\}.$$
If $p(k,x)$ denotes the count of numbers $\leqslant x$ that are multiples of a $k$-t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the area of a triangle if its two sides measure $6 in.$ and $9 in.$, and the bisector of the angle between the sides is $4\sqrt{3}$ in. Find the area of a triangle if its two sides measure $6$ in. and $9$ in., and the bisector of the angle between the sides is $4\sqrt{3}$ in. I'm thinking of using the formula $A$=... | If you know the lengths of two sides $a$ and $b$, and the angle $\theta$ between them, then the area of the triangle is $\frac{1}{2}ab\sin \theta$. Let's label the sides $a = 6$, $b = 4\sqrt 3$, and $c = 9$, and the half-angle as $\theta$. The sum of each sub triangle area must equal the total triangle area:
$$A = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Show that $\sum_{i=1}^{r} i^2 = \binom{r+1}{3} + \binom{r+2}{3}$ by finding generating function Find the generating function for the sequence $c_r$ where $c_0 = 0$ and $ c_r = \sum_{i=1}^{r} i^2 $ for $r \in \mathbb N$. Hence show that $\sum_{i=1}^{r} i^2 = \binom{r+1}{3} + \binom{r+2}{3}$
EDIT:
Sorry, I forgot to men... | Let $C(x)$ be the GF for the sequence $\langle c_n \rangle_{n\ge 0}$, where $c_n$ is the $n^\text{th}$ partial sum of the sequence of squares. If $f(x)$ is the GF for the sequence of squares, then $\frac{f(x)}{1-x}$ is the GF for the sequence of the partial sums of squares.
So, we first find the GF for $\langle n^2 \r... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Prove, using the method of mathematical induction that the following holds true For natural numbers $n\ge1$ show the following inequality using induction.
$$n^{1/n}\le 1+\sqrt{\frac{2}{n}}$$
| To solve this problem, merely binomial theorem will be enough.
The base case $n = 1$ is obviously true. Then we assume that $$n^{\frac1n}\le 1+\sqrt{\frac{2}{n}} \iff n \le \left(1+\sqrt{\frac2n}\right)^n,$$ and we try to prove $$(n+1)^{\frac1{n+1}}\le 1+\sqrt{\frac{2}{n+1}} \iff n+1 \le \left(1+\sqrt{\frac2{n+1}}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $13\vert(3^{n+1} +3^{n} +3^{n-1})$
Prove that $3^{n+1} +3^{n} +3^{n-1}$ is divisible by $13$ for all positive integral values of $n$
| Besides the obvious factorization approach, you can reason by induction.
$$3^{(n+1)+1}+3^{(n+1)}+3^{(n+1-1)}=3\cdot 3^{n+1}+3\cdot 3^n+3\cdot 3^{n-1}=3 \cdot (3^{n+1}+3^n+3^{n-1}),$$
and
$$3^{1+1}+3^{1}+3^{1-1}=13.$$
Hence
$$S_{n+1}=3\cdot S_n\text{, and }S_1=13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$ I would like to understand why $(x-5)^2-4$ can be factorised as $(x-5-2)(x-5+2)$
I am particularly concerned with the term, $-4$.
| Note that
$$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b)$$
Now put $a=(x-5)$ and $b=2$
$$(x-5)^2-4=(x-5)^2-2^2=((x-5)+2)((x-5)-2)=(x-5+2)(x-5-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Proving $x^2 - y^2 + z^2 \gt (x - y + z)^2$ Prove that
$$x^2 - y^2 + z^2 > (x - y + z)^2$$
where: $x < y <z$ for all natural numbers.
Thank for help.
| I think the formula $A^2 − B^2 = (A− B)(A+ B)$, will be perfect.
$x^2 − y^2 > (x − y + z)^2 − z^2$
$(x − y)(x + y) > (x − y + z − z)(x − y + z + z)$,
$(x − y)(x + y) > (x − y)(x − y + 2z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1040772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Help me, a doubt $f(x)=\cot^{-1} \frac{1-x}{1+x}$ I have a doubt
$$f(x)=\cot^{-1} \frac{1-x}{1+x}$$
$$f´(x)=\frac{1}{(\frac{1-x}{1+x})^2}\cdot\frac{(-1)(1+x)-(1-x)}{1+\frac{(1-x)^2}{(1+x)^2}}$$
mm this could to be really easy but I do not understand in the first denominator gives one, someone who can explain,
| $$
f'(x)= \frac{-1}{1+{(\frac{1-x}{1+x}})^2}.\frac{d}{dx}(\frac{1-x}{1+x})
$$
$$f'(x)=\frac{-(1+x)^2}{2(1+x^2)}.\frac{-1(1+x)-(1-x)}{(1+x)^2}$$
$$f'(x)=\frac{-1(-1-x-1+x)}{2(1+x^2)}$$
$$f'(x)=\frac{1}{1+x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1041715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Use the definition of infinite limit to prove $\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$
Prove
$$\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$$
And I was given the solution like this: but I could not understand how it removes the complicated terms.
Let $\delta=\min(0.5,\frac{1}{5M})$.
$$\frac{x}{x^2-1}=\frac{x}{(x+1)(x-1... | We can first use partial fraction decomposition to $\frac{x}{x^2-1},$ which leads to
\begin{gather*}
\frac{x}{x^2-1}=\frac{1}{2(x+1)}+\frac{1}{2(x-1)},
\end{gather*}
which implies that
\begin{gather*}\tag{1}
\frac{x}{x^2-1}>\frac{1}{2(x-1)}, \qquad \text{provided } x>-1.
\end{gather*}
Let $M>0.$ Set
$$\delta=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1043682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Surface integrals: Find the area of the portion of the cone $x^2+y^2=z^2$ above the $xy$ plane and inside the cylinder $x^2+y^2=ax$ I need to find the area of the portion of the cone $x^2+y^2=z^2$ above the $xy$ plane and inside the cylinder $x^2+y^2=ax$ .
For this, I used cylindrical coordinates to parametrize the reg... | The final double-integral you have doesn't evaluate to an elementary function, but it can be recognized as a combination of Elliptic Integrals.
Take the inner integral first. By putting $u=1+r^2$, It is easy to show that
$$
\int_{0}^{a\cos\theta} r\sqrt{1+r^2}dr = \frac{1}{3} \sqrt{a^2 \cos ^2\theta +1} \left(a^2 \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Evaluting $\tan15°$ using difference formula
Evalute $\tan15°$ using difference formula
Steps I took:
$$\begin{align}
\tan(45-30)&=\frac { \tan(45)-\tan(30) }{ 1+\tan(45)\tan(30) }\\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \sqrt { 3 } }{ 3 } } \\
&=\frac { 1-\frac { \sqrt { 3 } }{ 3 } }{ 1+\frac { \... |
Hint: Go on
$$\frac { 1-\frac { \sqrt { 3 } }{ 3 } -\frac { \sqrt { 3 } }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } }=\frac { 1-\frac { \sqrt { 3 } }{ 3 } -\frac { \sqrt { 3 } }{ 3 } +\frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } }\cdot\frac33=
\frac {3-2\sqrt3+1 }{3-1}=
\frac {4-2\sqrt3}{2}=2-\sqrt3$$
$$\large\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1047924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$.. Question :
Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$
What I have done :
nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2... | The $r$th term is
$$\frac12 \left (\frac1{r}-\frac1{r+1} \right )-\frac12 \left (\frac1{r+1}-\frac1{r+2} \right )$$
so the sum telescopes. The result is
$$\sum_{r=1}^n \frac{2 r-1}{r (r+1) (r+2)} = \frac12\left (1-\frac1{2} \right )-\frac12 \left (\frac1{n+1}-\frac1{n+2} \right ) = \frac14 - \frac1{2 (n+1) (n+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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For fixed $m$, find all positive integer solutions to $a^m+b^m = p^n$ where $p$ is prime The problem:
For fixed $m$,
find all solutions to
$a^m+b^m = p^n$
where $p$ is prime,
all variables are positive integers,
$a \le b$,
and
$m \ge 2$.
This is a generalization
of this question:
Solve $a^{2013}+b^{2013}=p^n$ for ... | The only positive integer solutions of $p^n=a^m+b^m$ with $p$ prime, $m\ge 3$ not a power of $2$ are $(p,n,a,b,m)=(3,2,2,1,3),(2,km+1,2^k,2^k,m),\forall k\ge 1$.
If $a=b$, then $(p,n,a,b,m)=(2,km+1,2^k,2^k,m), k\ge 1$. If $(a,b,m)=(2,1,3)$, then $(p,n,a,b,m)=(3,2,2,1,3)$. Otherwise: let $m=2^kt$ with $t\ge 2$ odd. Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1048646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Number of solutions to equation, range restrictions per variable Find the number of solutions of the equation $x_1+x_2+x_3+x_4=15$ where variables are constrained as follows:
(a) Each $x_i \geq 2.$
(b) $1 \leq x_1 \leq 3$ , $0 \leq x_2$ , $3 \leq x_3 \leq 5$, $2 \leq x_4 \leq 6$
I believe I understand part A. I can fix... | Assume $a$ as a real number, such that $0<a <1$
The number of integral solutions should be the same as:
The coefficient of $a ^ {15}$ in
$ \left(a + a^{2} + a^{3} \right)\left(1+a + a^{2} + a^{3} + ... \right)\left(a^{3} + a^{4} + a^{5} \right)\left(a^{2} + a^{3} + a^{4} +a^{5} + a^{6}\right) $
This is same as
the co... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How prove this inequality $\prod_{1\le i
let $z_{1},z_{2},z_{3},z_{4},z_{5}$ are complex numbers,and such
$$|z_{1}|^2+|z_{2}|^2+|z_{3}|^2+|z_{4}|^2+|z_{5}|^2=5$$
$$A=\begin{bmatrix}1&1&1&1&1\\z_1&z_2&z_3&z_4&z_5\\z_1^2&z_2^2&z_3^2&z_4^2&z_5^2\\z_1^3&z_2^3&z_3^3&z_4^3&z_5^3\\z_1^4&z_2^4&z_3^4&z_4^4&z_5^4\\\end{bmatrix... | The required inequality is a straightforward corollary of Hadamard’s inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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Prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent How do I prove that $ \sum_{n=1}^{\infty} f_n(x)= \sum_{n=1}^{\infty} \frac {1} {n^2 x^2 +1} $ is convergent for every x in real numbers except for $x=0$?
I tried using the ratio test, but it doesn't seem to be conclusi... | Assuming you want to know whether $\sum_{n = 1}^\infty\frac{1}{n^2x^2 + 1}$ converges.
Two cases:
*
*$|x| \geq 1$. In this case, we have $$\frac{1}{n^2x^2 + 1} \leq \frac{1}{n^2 + 1} \leq \frac{1}{n^2}$$
and $\sum_{n = 1}^\infty \frac{1}{n^2}$ converges.
*$0<|x| < 1$. In this case, we have
$$
\frac{1}{n^2 x^2 + 1} ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$? As you can see from the title, I need help proving that $\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) = 0$. I first looked for $N$ by using $|\sqrt{n^2 + 1} - n - 0| = \sqrt{n^2 + 1} - n < \varepsilon$ and solving for $n$. I got $n > \frac{1 - \varepsilon ... | $$\lim _{n\to \infty} (\sqrt{n^2 + 1} - n) =\lim _{n\to \infty} \frac{(\sqrt{n^2 + 1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2 + 1} + n}\\ = \lim _{n\to \infty} \frac{n^2 + 1-n^2}{\sqrt{n^2 + 1} + n}=\lim _{n\to \infty} \frac{1}{\sqrt{n^2 + 1} + n}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1050741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Calculate a lim $\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2} $ $$
\lim_{x\to \infty } \left ( \frac{x^2+2}{x^2-4} \right)^{9x^2}
$$
Can you help with it?
| Hint: $\left(\dfrac{x^2+2}{x^2-4}\right)^{9x^2} = \left(\left(1+\dfrac{6}{x^2-4}\right)^{x^2-4}\right)^9\cdot \left(1+\dfrac{6}{x^2-4}\right)^{36}$
| {
"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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Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
| Multiply out the right hand side and confirm that the two expressions are equal.
$$\frac{x^6-1}{x-1}=(1+x+x^2+x^3+x^4+x^5)\iff x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 12,
"answer_id": 8
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Double substitution when integrating. I need to integrate
$$f(x) = \cos(\sin x)$$
my first thought was substituting so that $u = \cos(x)$, but that doesn't seem to do the trick. Is there any way to do a double substitution on this?
Any ways on how to proceed would be appreciated.
| You can express in terms of Incomplete Bessel Functions:
Consider $$\begin{align}J_0(1,w)&=\dfrac{2}{\pi}\int_0^w\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\cos\left(x-\dfrac{\pi}{2}\right)~d\left(x-\dfrac{\pi}{2}\right)
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\sin x~dx
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1055847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Weak principle of induction for $5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$
Show that
$$5+10+15+\ldots+5n= \frac{5n(n+1)}{2}$$
Proving the base case $n(1)$:
$5(1)= \frac{5(1)(1+1)}{2}$
$5 = \frac{5(2)}{2}$
$5 = 5$
Induction hypothesis:
$n = k$
$5+10+15+\ldots+5k = \frac{5k(k+1)}{2}$
Induction step (adding $k+1$):
$5+10+15... |
Since you have your answer now! Let's see it with a different approach
$$S=\color{red}{5}+\color{blue}{10}+\cdots+\color{green}{5(n-1)}+\color{orange}{5n}$$
$$S=\color{red}{5n}+\color{blue}{5(n-1)}+\cdots+\color{green}{10}+\cdots+\color{orange}{5}$$
$$2S=\underbrace{\color{red}{5(n+1)}+\color{blue}{5(n+1)}+\cdots+\co... | {
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"url": "https://math.stackexchange.com/questions/1055959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Sum of squares and $5\cdot2^n$ Does anyone know of a proof of the result that $5\cdot2^n$ where $n$ is a nonnegative integer is always the sum of two squares?
That is, nonzero integers $x,y$ must always exist where:
$x^2+y^2=5\cdot2^n$
when $n$ is $0$ or a positive integer?
For example:
$$1^2+2^2=5\cdot 2^0$$
$$1^2+3... | Note that $5$ is a sum of two squares, for $5=2^2+1^2$.
If $n$ is even, say $n=2k$, then $5\cdot 2^n=(2^k\cdot 2)^2+(2^k\cdot 1)^2$.
Also, $10=3^2+1^2$. So if $n$ is odd, say $n=2k+1$, then $5\cdot 2^{n}=10\cdot 2^{2k}=(2^k\cdot 3)^2+(2^k\cdot 1)^2$.
| {
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"url": "https://math.stackexchange.com/questions/1058568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with $\int \frac{1}{(\sin x + \cos x)}$ Kindly solve this question
$$\int \frac{1}{(\sin x + \cos x)} dx$$
I reached up to
$$\frac{(1+\tan^2x)}{1-\tan^2x + 2\tan x}$$
| HINT:
$$\sin x+\cos x=\sqrt2\sin\left(x+\frac\pi4\right)$$ or $$\sin x+\cos x=\sqrt2\cos\left(x-\frac\pi4\right)$$
More generally, set $a=r\cos\phi,b=r\sin\phi$ to find
$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin\left(x+\arctan2(b,a)\right)$
Similarly, set $a=r\sin\psi,b=r\cos\psi$ to find
$a\sin x+b\cos x=\sqrt{a^2+b^2}\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1059185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Just a proof of algebra If $a+b+c=0$,
Show that
$\left[\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right]\left[\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\right]=9.$
I am struck with this problem but can't find a solution. Please help me.
| If you expand the expression and factor you get $$\frac{(a^3+b^3+c^3-ab^2-a^2b-ac^2-a^2c-bc^2-b^2c+3abc)(b-c)(c-a)(a-b)}{abc (b-c)(c-a)(a-b)}$$ and if you then use $c=-a-b$ you get $$\frac{-9a^2b-9ab^2}{-a^2b-ab^2}=9$$ providing you have not divided by zero (so the $a,b,c$ must be distinct and non-zero as Suzu Hirose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x$ $\forall x\neq -1$ Given function $y=f(x)$
such that
$$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{x+3}{x+1}\right)=x \quad\forall x\neq -1$$
find $f(x)$ and $f(2007)$.
| With
$$f(x) = \frac{x+3}{x-1}-\frac{3}{2}$$
we have
$$a := f\left(\frac{x-3}{x+1}\right) = -x-\frac{3}{2}$$
and
$$b := f\left(\frac{x+3}{x+1}\right)=2x+\frac{3}{2}$$
Therefore the sum is $a+b=x$ and $f(2007)=-\frac{999}{2006}.$
Edit regarding the comments: My function $f(x)\,$ is a solution of the functional equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1060737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Prove by induction that an expression is divisible by 11
Prove, by induction that $2^{3n-1}+5\cdot3^n$ is divisible by $11$ for any even number $n\in\Bbb N$.
I am rather confused by this question. This is my attempt so far:
For $n = 2$
$2^5 + 5\cdot 9 = 77$
$77/11 = 7$
We assume that there is a value $n = k$ such th... | This is the same as proving that $2^{6n-1}+5\cdot3^{2n}$ is divisible by $11$ for all $n$. The case $n=1$ is obvious.
By induction hypothesis, you can assume $2^{6n-1}+5\cdot3^{2n}=11k$, which can be written
$$
2^{6n-1}=11k-5\cdot3^{2n}
$$
Now
\begin{align}
2^{6(n+1)-1}+5\cdot3^{2(n+1)}
&=2^6\cdot2^{6n-1}+45\cdot3^{2n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 4
} |
ordered pairs $(A,B)$ of subsets of $X$ >such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
Let $X$ be a set of $5$ elements. Then the number of ordered pairs $(A,B)$ of subsets of $X$
such that $A\neq \phi\;,B\neq \phi$ and $A\cap B = \phi\;,$ is
$\bf{My\; Try::}$ Let $X = \left\{1,2,3,4,5\right\}\;,$
th... | First let us try to get the number of ordered pairs $(A,B)$ such that $A\cup B = \phi$, without any restrictions on $A$ and $B$. Clearly, each element can EITHER be in only $A$ OR only $B$ OR neither. So, that is $3^5$. Now, remove the cases when $A = \phi$. This is $2^5-1$. Similarly for $B$. So, the total is $3^5 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1061919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Let $x$ and $y$ be two positive real numbers with $x < y$. Using only the axioms for real numbers, show that $0 < \frac{1}{y} < \frac{1}{x}$.
Let $x$ and $y$ be two positive real numbers with $x < y$. Using only the axioms for real numbers, show that $0 < \frac{1}{y} < \frac{1}{x}$.
How can I prove this?
This is what... | $y>x$ means $y=x+M$ for some $M>0$.
$1/y>0$ is obvious: it's a ratio between positive numbers.
Then
$$
\frac1{y}=\frac1{x+M}=\frac1{x(1+\frac{M}{x})}=\frac1x\cdot\frac{1}{1+\frac{M}{x}}
$$
Next observe that $M,x>0$ hence $M/x>0$, thus $1+\frac{M}{x}>1$.
Now, by axioms of real numbers, the sign of an inequality is prese... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1064013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Probabilistic Proof of $\prod\limits_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$ Using probability methods (characteristic function?) prove
$$\prod_{i=1}^\infty\cos\left(\frac t{2^i}\right)=\frac{\sin t}t$$
I know what is characteristic function but I have no idea how use it in this task. I will grateful... | I don't know if it answers your question since I don't use characteristic function. By using
$$ \sin(2\theta)=2\sin\theta\cos\theta $$
one has
\begin{eqnarray}
\prod_{i=1}^n\cos\left(\frac t{2^i}\right)&=&\cos\left(\frac t{2}\right)\cos\left(\frac t{2^2}\right)\cdots\cos\left(\frac t{2^{n-1}}\right)\cos\left(\frac t{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Simplifying Permutations Could someone explain the process of simplifying the following permutation in $S_6$
(1,3,5)(2,4,5)(2,3,6)
An explanation on how you arrived at the simplified form would also be greatly appreciated.
Thanks
| In the book I read this from they started at the right. Having this in mind we will now simplify $(1,3,5)(2,4,5)(2,3,6)$. To do this we start with $1$. Where does $1$ end up after this permutation? to see this look for the first appearence of $1$ (starting from the right), it only appears once and it goes to $3$, so wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int_{0}^{\infty}\frac{(x^2-1)\ln{x}}{1+x^4}dx$? How to evaluate the following integral
$$I=\int_{0}^{\infty}\dfrac{(x^2-1)\ln{x}}{1+x^4}dx=\dfrac{\pi^2}{4\sqrt{2}}$$
without using residue or complex analysis methods?
| Noting that
$$ \int_0^1x^n\ln x\,dx=-\frac{1}{(n+1)^2} $$
we have
\begin{eqnarray}
\int_0^\infty\frac{(x^2-1)\ln x}{1+x^4}dx&=&2\int_0^1\frac{(x^2-1)\ln x}{1+x^4}\,dx\\
&=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^2-1)x^{4n}\ln x\,dx\\
&=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n+2}-x^{4n})\ln x\,dx\\
&=&2\sum_{n=0}^\infty(-1)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.