Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Limit, Riemann Sum, Integration, Natural logarithm For any natural number $m$, $\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots +\frac{1}{mn} \right )=\ln (m)$.
I tried to prove the statement in the following way.
Proof:
$$\lim_{n\rightarrow \infty }\left ( \frac{1}{n+1}+\frac{1}{n+2... | A simpler and more direct choice is to write
\begin{align}\lim_{n \to \infty} \left( \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{mn} \right) &= \lim_{n \to \infty} \sum_{r=n+1}^{mn} \frac{1}{r} \\&= \lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{mn} \frac{1}{r/n} \\&= \int_{x=1}^m \frac{1}{x} \, dx \\&= \log m,\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding determinant of $3\times3$ matrix $$A = \left(\begin{matrix}
\lambda - 1 & -1 & -1 \\
1 & \lambda - 3 & 1 \\
-3 & 1 & \lambda + 1 \\
\end{matrix}\right)$$
$$\det A = \begin{vmatrix}A\end{vmatrix} =
(\lambda - 1) \begin{vmatrix}
\lambda - 3 & 1 \\
1 & \lambda + 1 \\
\end{vmatrix} + 1\begin{vmatrix}
1 & 1 \\
-3 &... | $$=(\lambda - 1)\Big((\lambda - 3)(\lambda + 1)-1\Big) + ((\lambda + 1) + 3) -(1 +3\lambda - 9) $$
$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) + (\lambda + 4) -(3\lambda - 8) $$
$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) - 2\lambda +12 $$
$$=\lambda ^3 - 3\lambda^2 -2\lambda +4 - 2\lambda +12 $$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Let $aThe prove I tried is the following. I really wish someone can check if I made some logical mistake, especially the last part I found myself diffident proving $a$ can only be $1$ or $5$.
Because $b\neq c$ and $b<c$, otherwise switch the value of $b,c$. Becuase $a\mid (3b+2c)$ and $a \mid (3c+2b), \exists q_1,q_2 \... | You proof is fine, but you are over elaborating at certain points. Do you need $q_1,q_2$, for example? You do not seem to have used them.
Indeed, the simplest proof is that if $a|3b+2c$ and $a | 3c+2b$, then $a | 3(3c+2b) - 2(3b+2c) = 5c$, and $a | 3(3b+2c) - 2(3c + 2b) = 5b$.
Therefore, $a | 5c$ and $a | 5b$, which me... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding the first non-zero terms of a power series I have the function:
$f(x) = \frac{30}{(x^2 + 1)(x^2-9)}$
I need to find the first four non-zero terms of the power series centered at zero. I have not had much experience with power series so I am not sure how to start/complete this problem.
| $f(x)=\frac{-3}{x^2+1}+\frac{3}{x^2-9}$
$\frac{-3}{x^2+1}=-3(1-x^2+x^4-x^6+....)$
$\frac{3}{x^2-9}=\frac{-1/3}{1-x^2/9}=(-1/3)(1+x^2/9+x^4/81+x^6/729+.....)$
Combine to get $f(x)=-4/3+(80/27)x^2-........$
I'll let you finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{-\infty}^\infty \frac{1}{\sqrt{z^2 + 1}}\frac{1}{z - \alpha} dz$.
Evaluate $$\int_{-\infty}^\infty \frac{1}{\sqrt{z^2 + 1}}\frac{1}{z - \alpha} dz\,.$$
What is an elegant way to evaluate this integral for Im $\alpha >0$? I imagine using residue theorem will lead to an elegant solution, such as in thes... | This is probably not elegant, but you can probably find a place to use the residue theorem. Let $I$ denote the integral
$$\int_{-\infty}^{\infty}\frac{1}{\sqrt{x^2+1}\ (x-\alpha)}dx.$$
Then, $I$ equals
$$\int_0^\infty\frac{1}{\sqrt{x^2+1}}\left(\frac{1}{x-\alpha}-\frac{1}{x+\alpha}\right)dx=2\alpha\int_0^\infty\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why does trying to compute $\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}}$ result in the negative of the answer given? My textbook asks me to evaluate the limit $$\lim_{x\to-\infty} {2x-1\over \sqrt{3x^2+x+1}}$$ which evaluates to $-2\over\sqrt{3}$. The method in the book is to factor out $x^2$ from the root in the de... | Your mistake is in writing
$$\frac 1 x = \sqrt{\frac{1}{x^2}}.$$
Since $x < 0$, the correct version includes a negative sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3034808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 1,
"answer_id": 0
} |
Eigenvalues and Eigenvectors of Sum of Symmetric Matrix Question:
Let A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}
Find all eigenvalues and eigenvectors of the martrix:
$$\sum_{n=1}^{100} A^n = A^{100} +A^{99} +...+A^2+A$$
I know that the eigenvectors of A are \begin{bmatrix} 1 \\ 1 \end{bmatrix} and \begin{bmat... | Hint: If $$A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}$$then we have $$A^2 = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}=\begin{bmatrix} 2 & 2 \\ 2 & 2 \\ \end{bmatrix}\\A^3=\begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}\begin{bmatrix} 2&2 \\ 2&2 \\ \end{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3035684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
How can I find the expectation and variance of $Z=\max\{X,Y\}$ where $X$ and $Y$ are defined through joint probability distribution? Random variables $X$ and $Y$ and have the joint distribution below, and $Z=\max\{X,Y\}$
$$
\begin{array}{c|lcr}
\text{X\Y} & \text{1} & \text{2} & \text{3} \\
\hline
1 & 0.12 & 0.08 & 0.... | If $Z = \max(X, Y)$, then the following is true:
$Z = 1$ iff $X = 1$ and $Y = 1$.
$Z = 2$ iff either $X = 1$ and $Y =2$ or $X = 2$ and $Y = 1$, or $X = 2$ and $Y = 2$.
$Z = 3$ iff $X = 3$.
So $P(Z = 1) = 0.12$
$P(Z = 2) = 0.08 + 0.12 + 0.18 = 0.38$
$P(Z = 3) = 0.2 + 0.3 = 0.5$.
Thus $E(Z) = 0.12 + 2*0.38 + 3*0.5 = 0.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3038246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the $L^{p}$ norm $\|f\|_{L^{p}} := \big( \int^{b}_{a} |f(x)|^p\big)^{1/p}$ is not induced by a scalar product for $p \neq 2$.
On $X = C^0\big([a,b]\big)$, for any $p \in \mathbb{R}$, $p>1$, we define the $L^p$ norm by,
$$\|f\|_{L^{p}}:=\big(\int^{b}_{a}|f(x)|^{p}dx \big)^{1/p}.$$
Show that for $p\neq 2$,... | Your idea looks fine. However, here's a simpler approach: Let $f = \chi_A$ and $g = \chi_B$ be the indicator functions of two disjoint sets*. Then
$$\|f + g\|_p^2 + \|f - g\|_p^2 = 2 (|A| + |B|)^{2/p}$$
by a direct calculation. On the other hand,
$$2 \|f\|_p^2 + 2\|g\|_p^2 = 2 (|A|^{2/p} + |B|^{2/p}).$$
This would impl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Evaluate:$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$ Evaluate:$$\frac{1}{2^{101}}\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}$$
My Attempt:
I did try writing the series $(1+x)^{50}$and $(1+x)^{50}$ separately,then multiplied but could not determine the power of $x$ whose ... | Using the identity $$
\binom{51}{k}=\binom{50}{k}+\binom{50}{k-1},
$$ we get
$$
\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{51}{k}\binom{50}{r}=\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k}\binom{50}{r}+\sum_{k=1}^{51} \sum_{r=0}^{k-1}\binom{50}{k-1}\binom{50}{r}.$$ Observe that the former is equal to
$$
\sum_{k=0}^{50} \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Compute the limit $\lim_{n\to\infty} I_n(a)$ where $ I_n(a) :=\int_0^a \frac{x^n}{x^n+1}\,\mathrm{d}x, n\in N$. For $a>0$ we define
$$\space I_n(a)=\int_0^a\frac{x^n}{x^n+1}\,\mathrm{d}x , n\in N.$$
*
*Prove that $0\le I_n(1) \le \frac{1}{n+1}$
*Compute $\lim_{n\to\infty} I_n(a)$
My attempt:
*
*I regard $I_n(1)... | Note that we have
$$\begin{align}
\int_0^a \frac{x^n}{1+x^n}\,dx&=\int_0^1 \frac{x^n}{1+x^n}\,dx+\int_1^a \frac{x^n}{1+x^n}\,dx\\\\
&=\int_0^1 \frac{x^n}{1+x^n}\,dx+(a-1)-\int_1^a \frac{1}{1+x^n}\,dx
\end{align}$$
For $x\in [0,1]$, $0\le \frac{x^n}{1+x^n}\le x^n$ and for $x\in[1,a]$, $\frac{1}{1+x^n}\le \frac1{x^n}$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3042601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve
$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$
through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am a... | I believe the most direct way for tackling such integrals is to exploit Euler's Beta function and the reflection formula for the $\Gamma$ function: assuming $m>1$,
$$ \int_{0}^{+\infty}\frac{dx}{1+x^m}\stackrel{\frac{1}{1+x^m}\to u}{=}\frac{1}{m}\int_{0}^{1}u^{-\frac{m-1}{m}}(1-u)^{-\frac{1}{m}}\,du =\frac{\Gamma\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
If $xy$ divides $x^2 + y^2$ show that $x=\pm y$ Problem statement : Let $x,y$ be integers, show that if $xy$ divides $x^2 + y^2$ then $x=\pm y.$
What I have tried:
I can reduce this to the case where $\gcd(x,y)=1$, since if $x$ and $y$ have a common factor, $d$ say, then $d^2$ divides through both $xy$ and $x^2 + y^... | Suppose that $\gcd(x,y)=1$ and $(xy)\mid(x^2+y^2)$.
Then $y^2\equiv0\pmod x$. If $1=ax+by$ then $by\equiv1\pmod x$ and so
$1\equiv(by)^2=b^2y^2\equiv0\pmod x$. So $x=\pm1$. Likewise, $y=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Solve the differential equation $\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$ Solve the differential equation $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$\frac{dy}{dx}=\frac{1}{y^3}\frac{\left(1+2y^4\right)}{1+\frac{4x}{y^4}}$$
Multiplying both sides with $\frac{1}{y^5}$ we get:
$$\frac{1}{y... | $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)\frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=\frac{y^4}{2y^4+1}\ln(c\:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
calculating eigen values from an equation
I'm trying to use this equation(in yellow) to calculate the eigen values of B = \begin{pmatrix}
1&1&1\\
1&1&1\\
1&1&1\\
\end{pmatrix}
but I'm getting $$-λ^3+3λ^2-3λ$$
and the proper answer using the other method is
$$det( \begin{pmatrix}
1-λ&1&1\\
1&1-λ&1\\
1&1&1-λ\\
\end{pmat... | $\DeclareMathOperator\tr{tr}$The full proper formula for $n=3$ is:
$$\det(M-\lambda I_3)=(-1)^3\lambda^3 + (-1)^2\tr(M)\lambda^2 + (-1)\cdot \frac 12\big[(\tr M)^2-\tr(M^2)\big]\lambda + \det(M)$$
In this case:
$$\tr\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}^2
=\tr \begin{pmatrix}3&3&3\\3&3&3\\3&3&3\end{pmatrix}
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^... | Hint: Write your equations in the form $$\frac{3x-y}{3-x}=\frac{x+3y}{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Stuck finding the sum of two series This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.
I am completely stuck at finding the sum of the two following series:
$$\sum_{n=1}^∞(2n+1)x^n$$
$$\sum_{n=0}^∞\frac{x^n}{(n+1)2^n}$$
I... | *
*Set $f(x) = \sum_{n=1}^\infty (2n+1)x^n$. Then
$$
f(t^2)
= \sum_{n=1}^\infty (2n+1)t^{2n}
= \sum_{n=1}^\infty \frac{d}{dt}t^{2n+1}
= \frac{d}{dt} \sum_{n=1}^\infty t^{2n+1}
= \frac{d}{dt} \left( t \sum_{n=1}^\infty (t^2)^n \right)
= \frac{d}{dt} \left( \frac{t^3}{1-t^2} \right) \\
= \frac{3t^2(1-t^2)-t^3(-2t)}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Coordinates of a vector FFT I was reading a paper about DFT, at the end he got the relations $Y=\frac{1}{\sqrt N}W_n \cdot y$ and $y=\frac{1}{\sqrt N}W_n \cdot Y$, where $y$ is a vector and $Y$ is the coordinates of that vector in the basis. $W_n$ stands for the transformation matrix.
Later, the author shows how to red... | In the first step, the author is just exploiting some associativity of the matrix-vector product. For instance, the first two rows of $W_4y$ are:
\begin{align*}
\begin{bmatrix} 1&1&1&1\\ 1&\omega&\omega^2&\omega^3 \end{bmatrix} \begin{bmatrix} y_0\\y_1\\y_2\\y_3 \end{bmatrix}
&=
\begin{bmatrix} y_0+y_1+y_2+y_3\\ y_0+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsur... | I did it!
I actually have no idea whether or not this works, but this is how I did it.
$n\in\Bbb N$
Define the sequence $\{r_k^{(n)}\}_{k=1}^{k=n}$ such that
$$x^n+1=\prod_{k=1}^{n}\big(x-r^{(n)}_{k}\big)$$
We then know that $$r_k^{(n)}=\exp\bigg[\frac{i\pi}{n}(2k-1)\bigg]$$
Then we define
$$S_n=\{r_k^{(n)}:k\in[1,n]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Limit by polar coordinates: $\frac{2x(1-\cos(x-y))}{x^2+y^2}$ I need to prove that:
$$\lim_{(x,y)\to (0,0)}\frac{2x(1-\cos(x-y))}{x^2+y^2} = 0$$
I have tried using polar coordinates, but I'm not comfortable with my approach. What I did was saying:
$$|f(\rho\cos\theta,\rho\sin\theta)| = \frac{2\cos\theta(1-\cos(\rho(\co... | Solution without the use of polar coordinates, as asked by OP at the end of their post.
Rewrite $$\begin{aligned}1-\cos(x-y)&=1-\cos x \cos y - \sin x \sin y \\ &=1-\cos x + \cos x (1-\cos y)- \sin x \sin y, \end{aligned}$$
then $$\begin{aligned}\frac{2x(1-\cos(x-y))}{x^2+y^2}&= 2x\cdot \frac{1-\cos x}{x^2} \cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Tangents to parabola $y^{2}=4ax$ meet hyperbola $x^2/a^2-y^2/b^2=1$ at $A$ and $B$. Find the locus of intersections of the tangents at $A$ and $B$.
If tangents to the parabola $y^{2} = 4ax$ intersect the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ at $A$ and $B$, then find the locus of point of intersection of... | $$y^2=4ax \tag{1}$$
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \tag{2}$$
Let $P(X,Y)$ be the required locus.
*
*For hyperbola $(2)$, $(X,Y)$ is the pole of the polar $AB$ (i.e. chord $AB$ for the hyperbola).
Equation of $AB$ is
$$\frac{X x}{a^2}-\frac{Y y}{b^2}=1 \tag{3}$$
*Equation of tangent of $(1)$ at $C(x_1,y_1)$
$$y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove Nice Determinant Equations I often come across these kinds of problems in A-level exam papers:
Prove that
$$
\begin{vmatrix}
(a+b)^2 & 1 & 1 \\
a^2 & (1+b)^2 & a^2 \\
b^2 & b^2 & (1+a)^2
\end{vmatrix}
= 2ab(1+a+b)^3
$$
or
Prove that
$$
\begin{vmatrix}
1 & 1 ... | The idea is to use elementary row-column operations to have a simpler determinant. Since this is $3\times 3$, our goal is to get two successive zeros in a single row/column.
You have
$$A=\begin{bmatrix}(a+b)^2 & 1 & 1 \\a^2 & (1+b)^2 & a^2 \\b^2 & b^2 & (1+a)^2 \end{bmatrix}$$
Then
\begin{align}
\det A&=\det\begin{bm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
A Series For the Golden Ratio
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?
Some background and motivation:
Wikipedia only provides one series for the golden ratio - see also the link in the ... | We know that the approximates by partial continued fractions of the golden ratios are the quotients of successive Fibonacci numbers $1,1,2,3,5,8,13,...$. Taking
$$
\phi=\frac{1+\sqrt{5}}2\approx \frac{8}5\implies \sqrt5\approx \frac{11}{5}
$$
we get a lower approximation for $\sqrt5$. To the remainder of the square ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
How to take every third element in a series? $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$ or $\frac{\pi^2}{6}$
What if we take every 3rd term and add them up?
A = $ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$
How to take every 3rd-1 term and a... | KM101 deleted his hint... not sure why.
$\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+\dots=\frac{1}{3^2 1^2}+\frac{1}{3^2 2^2}+\frac{1}{3^2 3^3}+\dots=\frac{1}{9}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots)= \frac{\pi^2}{54}$
$\frac{1}{2^2}+\frac{1}{5^2}+\frac{1}{8^2}+\dots=\frac{1}{2^2 1^2}+\frac{1}{3^2?? 1^2}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Fredholm integral - degenerate kernel method I have started answering a fredholm integral equation of the second kind and do not know where to go from here.
The answer has to be written in the form
$$ \sum a_jx^{j-1} $$
The Fredholm integral equation is
$$ x^3+\frac16x^2+\frac15x = g(x) + \int_0^1(x^2y+xy^2) f(y) dy$$... | $$g(x) = x^3 - \frac{38}{1077} x^2 + \frac{58}{1795} x$$
By rearranging the original equation we can see that g(x) = x^3 + Ax^2 + Bx. This is because both of the integrals on the right evaluate to a constant multiple of x^2 or x respectively. By using this formula for g(x) in each of the integrals we get:
The integral ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve this simple equation $\frac{46}{y} + y = 25$? How do I solve this simple equation? $\frac{46}{y} + y = 25$
I know that the answer is $2$, but how do I arrive at that?
| Multiply both sides by $y$ and you get
$$46 + y^2 = 25y$$
$$\Leftrightarrow 0 = y^2 -25y + 46$$
Now calculating the discriminant $D = (-25)^2 -4\cdot 46 = 625-184 = 441 = 21^2$ gives
$$y = \frac{25 \pm \sqrt{21^2}}{2}$$
So $y = \frac{25+21}{2} = 23$ or $y = \frac{25-21}{2} = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Trick for Inverse Hollow Matrix Calculation (Self-Answered) Let $A$ be the hollow matrix :
$$
A=\begin{pmatrix}
0&1&1&1\\
1&0&1&1\\
1&1&0&1\\
1&1&1&0
\end{pmatrix}
$$
Find the inverse matrix $A^{-1}$ without using any elementary row/ column operations.
| Notice that
$$
A=
\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1
\end{pmatrix} - I_{4\times4}
$$
Let
$$
B=
\begin{pmatrix}
1&1&1&1\\
1&1&1&1\\
1&1&1&1\\
1&1&1&1
\end{pmatrix}
$$
then $B=A+I$. Observe that $B^2 = (A+I)^2=4B$. Expanding,
$$
A^2+2A+I = 4(A+I) \implies A^2-2A=3I
$$
Therefore $A^{-1}=\frac{1}{3}(A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that $\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$ Let $a,b,c\in \Bbb R^+$ such that $a+b+c=abc$. Prove that $$\frac{a}{c\sqrt{a^2+1}}+\frac{b}{a\sqrt{b^2+1}}+\frac{c}{b\sqrt{c^2+1}}\ge \frac{3}{2}$$
Idea 1.From $a+b+c=abc\Leftrightarrow \frac{1}{ab}+\frac{1}{bc}+\frac{... | We need to prove that
$$\sum_{cyc}\frac{a}{c\sqrt{a^2+\frac{abc}{a+b+c}}}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\sqrt{\frac{a(a+b+c)}{c^2(a+b)(a+c)}}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\sqrt{\frac{a^2b}{c(a+b)(a+c)}}\geq\frac{3}{2}$$ or
$$\sum_{cyc }\sqrt{a^3b^2(b+c)}\geq\frac{3}{2}\sqrt{abc(a+b)(a+c)(b+c)}$$ or
$$\sum_{cyc }\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the domain and range of a function The question is: find the domain and range of $f(x) = \sqrt{{(1-x)}/x}$.
For the domain, I took $x ≠ 0$ and $(1-x)/x ≥ 0$, which gave me $1 ≥ x$.
So the domain I found is $(-∞,0)∪(0,1]$. However, the proposed solution says that the domain is just $(0,1],$ without explaining, so I... | $f(x) = \sqrt{\frac {1-x}x}$ which is undefined if $x =0$ and if $\frac {1-x}x = \frac 1x - 1 < 0$. $x\ne 0$ and $\frac 1x -1 < 0 \implies \frac 1x \ge 1$.
If $x > 0$ then $\frac 1x \ge 1 \implies x \le 1$ and if $x < 0$ implies $x \ge 1$ which is a contradiction so $0 < x \le 1$ and domain is $(0,1]$.
[Alternativ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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A very strange and difficult hyperbolic integral This integral gave me serious problems, I tried to solve it by parts but it is madness! The calculations are too long and difficult, I do not think we should solve this.
$$\int _{ -\frac { 1 }{ 3 } }^{ \frac { 1 }{ 3 } }{ \sqrt { 36{ x }^{ 4 }-40{ x }^{ 2 }+4 } \cosh {... | This integral is more complicated than it looks and only requires comfort with hyperbolic trigonometric definitions. My initial instinct is to look at the expression inside the $\cosh$ to see if I can simplify it. In fact we have by the definition of the hyperbolic trigonometric functions,
$$\tanh^{-1}(x) = \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. Th... | Here $xyz=1$ gives
$$\begin{align}
x^{-1}xyz=x^{-1}1 & \iff yz=x^{-1} \\
&\iff yzx=x^{-1}x \\
& \iff yzx=1,
\end{align}$$
which gives
$$\begin{align}
y^{-1}yzx=y^{-1}1 & \iff zx=y^{-1} \\
& \iff zxy=y^{-1}y \\
&\iff zxy=1.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3069921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
$ \int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx \, \, \, (\phi>0) $ Evaluate
$$
\int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx \, \, \, (\phi>0)
$$
See my answer below for a solution using a nice substitution.
| Solution
$$ \small
\int_0^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx = \int_0^{1} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx + \int_1^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx
$$
Notice that with the substitution $y=\frac{1}{x}$,
$$\begin{aligned}
\int_1^{+\infty} \frac{1}{(1+x^2)(1+x^{\phi})}\, dx &= \int_1^0 \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Non-linear partial differential equation of order 1 Solve the PDE $$2(pq+py+qx)+x^2+y^2=0$$
where $\displaystyle p = \frac{\partial z}{\partial x}$, $\,\displaystyle q = \frac{\partial z}{\partial y}$
$f_p=2q+2x\\ f_q=2p+2x\\ f_z=0\\ f_x=2q+2x\\ f_y=2p+2y$
Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\fr... | Using Charpit's method
$$\frac{dp}{2q+2x}=\frac{dq}{2p+2y}=\frac{dz}{-4pq-2py-2qx}=\frac{dx}{-2q-2y}=\frac{dy}{-2p-2x}$$
Adding first two and last two fractions
$$\frac{dp+dq}{2p+2y+2q+2x}=\frac{dx+dy}{-2q-2y-2p-2x}$$
This implies, $a$ being an arbitrary constant $$p+x+q+y=a \tag{1}$$
Now the equation can be rewritten ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3074155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
find the probability about sum of random variables Let $X_1, X_2, X_3, Y_1, Y_2, Y_3, Z_1, Z_2, Z_3$ be random variables which have uniform distribution between 0 and 1. It means, the average of $X_1 = 0.5$
Let: $X=X_1 + X_2 + X_3,$
$Y=Y_1 + Y_2 + Y_3$, $Z=Z_1 + Z_2 + Z_3$
In this case, the probability of $\{X$ is bigg... | Thank you @gt6989b
At first, the answer of $F_X(x)$ is
$F(x)=\dfrac{1}{6}x^3 $ when $0\leq x<1$
$F(x)=-\dfrac{1}{3}x^3+\dfrac{3}{2}x^2-\dfrac{3}{2}x+\dfrac{1}{2} $ when $1 \leq x <2$
$F(x)=1-\dfrac{1}{6}(3-x)^3$ when $2\leq x<3$
and secondly,
$f(x)=\dfrac{1}{2}x^2$ when $0\leq x<1$
$f(x)=-x^2 +3x-\dfrac{3}{2}$ when $1\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For $A_i$ on $y=\sqrt{x}$ and $B_i$ on $x$-axis, with $\triangle B_{i-1}B_iA_i$ equilateral of side $\ell_i$. Find $\ell_1+\cdots+\ell_{300}$.
Let $O$ be the origin, $A_1,A_2,A_3,\ldots$ be distinct points on the curve $y=\sqrt{x}$ and $B_1,B_2,B_3,\cdots$ be points on the positive $X$-axis such that the triangles $OB... | Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
$$
\frac{y_i}{x_i-z_i}=\frac{y_i}{z_{i+1}-x_i}=\tan\frac{\pi}{3}=\sqrt{3}
$$
then
$$
z_i=y_i^2-\frac{\sqrt {3}y_i}{3}\\
z_{i+1}=y_i^2+\frac{\sqrt {3}y_i}{3}\\
l_i=\frac{2\sqrt{3}}{3}y_i
$$
Confronting them let $y_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Calculate $ a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right) $ I struggle for a while solving limit of this chain:
$
a_n\:=\:n^3\left(\sqrt{n^2+\sqrt{n^4+1}}-\sqrt{2}n\right)
$
I know from WolframAlpha result will be $ \frac{1}{4\sqrt{2}} $, but step-by-step solution is overcomplicated(28 steps). Usually I s... | From the numerator and denominator take $-n^{3}$ as common term and then rationalize the numerator again.
Once you do that you will be left with :
$$\frac{n^{3}}{(n^{2}+\sqrt{n^{4}+1})(\sqrt{n^{2}+\sqrt{n^{4}+1}}+n\sqrt{2})}$$
Divide the numerator and denominator by $n^{3}$, and once you divide the denominator by $n^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
on the quadratic form $5x^2-y^2$ Consider the following subsets of $\mathbb{Z}$:
$A=\{\frac{5p^2-y^2}{4}\, | \, p, y \,\,\text{odd positive integers}, p\,\, \text{prime}\}$ and
$B=\{\frac{5x^2-y^2}{4}\, | \, x, y \,\,\text{odd positive integers}\}$.
Are there elements in $B$ which are not in $A$?
| Note that $\frac{5\cdot9^2-3^2}{4}=99$, and if $\frac{5p^2-y^2}{4}=99$ for some prime number $p$, then
$$5p^2-y^2=396\equiv0\pmod{3},$$
from which it follows that $p\equiv y\equiv0\pmod{3}$. Because $p$ is prime it follows that $p=3$, and hence
$$y^2=5p^2-396=-351,$$
a contradiction. So $99\in B$ but $99\notin A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solution to Differential equation 2nd order, x*e^x I'm trying to solve the following differential equation:
$y'' + y = x \cdot e^x$
I already have the homogeneous solution: $y_h = c_1 \cdot cos(x) + c_2 \cdot sin(x)$ but I'm struggeling to find the particular solution.
I tried using $y_p = a \cdot x \cdot e^x$ but that... | Try $y_p = (ax+b)e^x$.
[Additional characters to bypass the 30 characters rule]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $a$ where $√a$ is area of a trapezoid which touches the circle with center $O$ (diameter is 2)?
The sides $AB, BC, CD$ of trapezoid $ABCD$ touches the circle with
center $O$ and they are equal. $AD$, goes through the point O. If
diameter is 2, then the area of the trapezoid is $√a$ . What is
... | Warning: might be a little convoluted, uses a bit of assumed knowledge, and is probably not the most elegant/efficient solution. I've triple-checked the method and arithmetic though so unless there's a fundamental flaw with my solution, I think this should be correct.
We make a duplicate trapezoid of $ABCD$, with diam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating $\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x)\,dx$
How to prove $$\int_0^{\pi/2}\operatorname{arcsinh}(2\tan x) \, dx = \frac43G + \frac13\pi\ln\left(2+\sqrt3\right),$$where $G$ is Catalan's constant?
I have a premonition that this integral is related to $\Im\operatorname{Li}_2\left(2\pm\sqrt3\right)$.
At... | Sorry in a rush, so only a PARTIAL SOLUTION:
Here I will employ Feynman's Trick:
\begin{equation}
I = \int_0^{\infty}\frac{\operatorname{arcsinh(2x)}}{1 + x^2}\:dx
\end{equation}
Let
\begin{equation}
J(t) = \int_0^{\infty}\frac{\operatorname{arcsinh(tx)}}{1 + x^2}\:dx
\end{equation}
We observe that $J(2) = I$ and $J... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 1
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Find all continuous functions in $0$ that $2f(2x) = f(x) + x $ I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( ... | This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = \frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = \frac{C(x)}{x}
$$
then
$$
a\frac{C(a x)}{a x}-\frac{C(x)}{x} = x
$$
or
$$
C(a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
convergence of a recursive sequence with parameter a How can you determine if the following recursive sequence converges:
$$x_{n+1}=\frac{1}{2}(a+x_n^2)$$
where $0\le a \le 1$ and $x_1=0$
I know that the limit x (if it exists) satisfies the following equation:
$$x=\frac{1}{2}(a+x^2)$$
since $\lim_{x\rightarrow \infty}... | By induction we can show that $0\leq x_n<1$.
Now, for $a=1$ we obtain $$x_{n+1}-x_n=\frac{1}{2}(x_n-1)^2\geq0$$ and the rest is smooth.
But for $a<1$ we obtain $0<r=\frac{2-\sqrt{1-a}}{2}<1$ and $$|x_{n+1}-1+\sqrt{1-a}|=\left|\frac{1}{2}x_n^2+\frac{1}{2}a-1+\sqrt{1-a}\right|=$$
$$=\frac{1}{2}\left|x_n^2-\left(1-\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Approximating $\pi$ with arctangent Use the fact that $\frac{\pi}{4} = \text{arctangent}(\frac{1}{2}) + \text{arctangent}(\frac{1}{3})$ to determine the number of terms summed to ensure an approximation to $\pi$ less than $10^{-3}$.
So far I've got
\begin{align*}
\pi = 4\Bigg(\sum_{i=1}^{\infty} (-1)^{i+1}\frac{1}{... | Rewrite the equality as:
\begin{array}{}
\pi = 4\left[\sum\limits_{i=0}^{n-1} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)+\sum\limits_{i=n}^{\infty} (-1)^{i}\left(\frac{1}{2^{2i+1}(2i+1)} +\frac{1}{3^{2i+1}(2i+1)}\right)\right]
\end{array}
to obtain the inequality to be solved for $n$:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Maximum value of sum with integer parts How can I find the max value of $\sum_{i}^{n} \lfloor\frac{i}{k}\rfloor (k-1)$ with $k$ integer $\in [1,100]$? I can express this sum in terms of n and k (it's quite easy) but I'm not able to find its maximum ...
$\sum_{i}^{n} \lfloor\frac{i}{k}\rfloor (k-1) = \frac{m(k-1)(2N+2-... | In order to calculate the sum we use Iverson brackets $[P]$ to cope with the floor function.
We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^n\left\lfloor\frac{i}{k}\right\rfloor}
&=\sum_{i=1}^n\sum_{j=1}^{\lfloor n/k\rfloor}j\left[j=\left\lfloor\frac{i}{k}\right\rfloor\right]\\
&=\sum_{i=1}^n\sum_{j=1}^{\lfloor n/k\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The Integral $\int \frac {dx}{(x^2-2ax+b)^n}$ Recently I came across this general integral,
$$\int \frac {dx}{(x^2-2ax+b)^n}$$
Putting $x^2-2ax+b=0$ we have,
$$x = a±\sqrt {a^2-b} = a±\sqrt {∆}$$
Hence the integrand can be written as,
$$
\frac {1}{(x^2-2ax+b)^n}
=
\frac {1}{(x-a-\sqrt ∆)^n(x-a+\sqrt ∆)^n}
$$
Resolving ... | All right, now I've got it.
The easiest way to get all the coefficients? Expand in a Laurent series around one of the roots. Substituting $z=x-a-\sqrt{\Delta}$ and later defining $D=\frac1{2\sqrt{\Delta}}$, we get
\begin{align*}\frac1{(x^2-2ax+b)^n} &= \frac1{(x-a-\sqrt{\Delta})^n(x-a+\sqrt{\Delta})^n}=\frac1{z^n(z+2\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3086878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $x^{18} \equiv 7^{99} - 7, \mod 592$ What I tried:
$x^{18} \equiv 7^{99} - 7, \mod 592 \iff \begin{cases} x^{18} \equiv 7^{99}-7 & \mod 7 \\ x^{18} \equiv 7^{99}-7 & \mod 2 \\ x^{18} \equiv 7^{99}-7 & \mod
3\end{cases} \iff x^{18} \equiv 0, \mod 7,2,3. $
I'm not sure how to proceed: is the last step equivalent ... | Hint $\bmod 37\!:\,\ x^{\large 18}\equiv \color{#c00}{7^{\large 99}}\!-7\equiv -6\,\overset{\rm square}\Longrightarrow\,x^{\large 36}\equiv -1\,$ contra little Fermat
because: $\ \ 7 \equiv 3^{\large 4}\,\Rightarrow\, \color{#c00}{7^{\large 99}}\equiv (3^{\large 4})^{\large 99}\equiv (3^{\large 36})^{\large 11}\equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3089550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $ $$\sum_ \limits{n=1}^{\infty} \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1) } $$
$$1 \cdot 3 \cdot 5 \cdots (2n+1) = \frac{1 \cdot 2 \cdot 3 \cdots (2n+2)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} = \frac{(2n+2)!}{2^{n+1} \cdot (n+1)!} $$
But the fol... | HINT:
Note that
$$\begin{align}
\frac{n}{(2n+1)!!}&=\frac12\left( \frac{2n+1-1}{(2n+1)!!}\right)\\\\&=\frac12\left(\frac{1}{(2n-1)!!}-\frac1{(2n+1)!!}\right)
\end{align}$$
Now, telescope.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Roots of polynomials and their formulae relating to coefficients Write down the cubic equation given that $\alpha + \beta + \gamma = 4$, $\alpha^2 + \beta^2 + \gamma^2 = 66$, and $\alpha^3 + \beta^3 + \gamma^3 = 280$
Ok so, the sum of roots is given and I'm able to use the sum of the roots and the sum of the roots squ... | Hint, changing to $\;a,b,c\;$ for the roots:
$$(a+b+c)^2=a^2+2a(b+c)+(b+c)^2$$
$$(a+b+c)^3=a^3+3a^2(b+c)+3a(b+c)^2+(b+c)^3$$
Play with the above and the given data...
Added Or the other way around:
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
$$a^3+b^3+c^3=(a+b+c)^3-3a^2(b+c)-3a(b+c)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3092635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluating $\lim_{\epsilon\to 0^{+}}\ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) }{\Gamma(\epsilon)}$ For $\nu \in \mathbb{C}$ and negative $y<0$ is there a way to compute the limit
$$
f(\nu,y) \equiv \lim_{\epsilon \to 0^{+}} \ \frac{ _2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \... | We may express
\begin{equation}
_2F_1\left( \tfrac{1}{2} - \nu, \tfrac{1}{2} + \nu; \epsilon; y \right) =\frac{\Gamma\left(\epsilon\right)}{\Gamma\left(\tfrac{1}{2} - \nu\right)\Gamma%
\left(\tfrac{1}{2} + \nu\right)}\sum_{s=0}^{\infty}\frac{\Gamma\left(\tfrac{1}{2} - \nu+s\right)\Gamma\left(\tfrac{1}{2} + \nu+s%
\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3093741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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What is the value of $x$ in $ABCD$ rectangle where $AE = 4, BE = 6, CE=5$ and $DE = x$?
In the diagram of rectangular $ABCD$, $AE=4, BE= 6, CE = 5$ and
$DE=x$, find the value of $x$
Source: Bangladesh Math Olympiad 2015 Junior Catagory
I can not relate these information with $DE$.
| Another way is:
$\hspace{1cm}$
$$\begin{cases}6^2=a^2+b^2\\ 5^2=b^2+c^2\\ x^2=c^2+d^2\\ 4^2=a^2+d^2\end{cases} \Rightarrow \\
x^2=(a^2+d^2)-(a^2+b^2)+(b^2+c^2)=\\
4^2-6^2+5^2=5 \Rightarrow x=\sqrt{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\sin\frac{\pi}{2}\sin\frac{\pi}{2^2}\sin\frac{\pi}{2^3}\cdots\sin\frac{\pi}{2^{11}}\cos \frac{\pi}{2^{12}}$ Evaluate :
$$\sin\frac{\pi}{2} \times \sin \frac{\pi}{2^2} \times \sin \frac{\pi}{2^3} \cdots \times \sin\frac{\pi}{2^{11}} \times \cos \frac{\pi}{2^{12}}$$
I tried to solve it by using double angle f... | Whatever you would do, the result would be a very small number very close to $0$
$$\sin\frac{\pi}{2} \times \sin \frac{\pi}{2^2} \times \sin \frac{\pi}{2^3}\times \cdots \times \sin\frac{\pi}{2^{11}} \times \cos \frac{\pi}{2^{12}}$$
We know the value of the sine for some of the angles. The first are
$$\{1 , \frac {\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$ $$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\frac{\pi^2}{8}$$
My working:
$$\int_{1}^{\infty}\frac{\ln x}{x^2-1}dx=\int_{0}^{1}\frac{\ln x}{x^2-1}dx=-\sum_{r\ge 1}\int_0^{1}x^{2r}\ln x\,dx =\sum_{r\ge 1}\frac {1}{(2r-1)^2}=
\frac{\pi^2}{8}$$
Is there any other appr... | I will provide two different methods. The first relies on properties of the polygamma function, the second converts the integral to a double integral first before evaluating it.
Let
$$I = \int_1^\infty \frac{\ln x}{x^2 - 1} \, dx.$$
Method 1 - A polygamma approach
By enforcing a substitution of $x \mapsto 1/x$ we see ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$ and $\lim_{n\to\infty} \frac{1}{\ln(n+1)}=0$
Use the definition of the limit of a sequence to prove that $\lim_{n \to \infty} (\frac{n^2-1}{2n^2+3})=\frac{1}{2}$.
We have
$$\begin{align}
\left|\frac{n^2-1}{2n^2+3}-\frac{1}{2}\right| & =\left|\frac{2... | For a) We have $\lim_{n\rightarrow \infty} \frac{n^2-1}{2n^2+3}=\lim_{n\rightarrow \infty} \frac{n^2(1-\frac{1}{n^2})}{n^2(2+\frac{3}{n^2})}= \lim_{n\rightarrow \infty} \frac{1-\frac{1}{n^2}}{2+\frac{3}{n^2}}=\frac{1}{2}$
For b) basically the same,meaning $ln$ is monotone so for $n\rightarrow \infty$ it follows that $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ I am stuck on factoring out everything properly. I feel like I am combining these fractions wrong or something because I always have an extra 1.
edit: edit: I am still stuck. Math isn't working out, I am making a mes... | Hint:
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\ldots+\frac{1}{n(n+1)}=\\\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Power series representation of $f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \left( \frac{1+x}{1-x} \right)$
We consider the power series:
$$ f(x) = \sum_{n=1}^{+ \infty} \frac{1}{n(n+1)(2n+1)}.x^{2n+2} $$
Prove that:
$$ f(x) = 3x^2 - (x^2 + 1)\ln(1 - x^2) - 2x \ln \bigg( \frac{1+x}{1-x} \biggr) $$
Starting from the... | Hint: $\frac{1}{n(n+1)(2n+1}=\frac{a}{n}+\frac{b}{n+1}+\frac{c}{2n+1}$. Find $a$, $b$ and $c$. Rewrite your series as sum of such series, justify convergence of those series and compute to get the needed function.
Recall: $\sum_{n=1}^{\infty}{\frac{t^n}{n}}=\log(1-t), \forall t, |t|<1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Binomial Theorem expansion and proving an interesting identity? In the identity
$$\frac {n!}{x(x+1)(x+2)...(x+n)} = \sum ^n_{k=0}\frac {A_k}{x+k} $$
Prove that $$A_k =(-1)^{k}\:\cdot\: ^{n}C_k$$
Also from this deduce that,
$$ \;^{n}C_0\frac 1{1.2} - \:^{n}C_1\frac1{2.3} +\; ^{n}C_2\frac1{3.4} \; ... \;{(-1)^n}\; ^{n... | There are many ways to demonstrate such an interesting identity.
a) Induction
I do not know at what level you are, so let's start with what should be the simpler: Induction
Given the thesis
$$
F(x,n) = {{n!} \over {x\left( {x + 1} \right) \cdots \left( {x + n} \right)}} = \sum\limits_{\left( {0\, \le } \right)\,k\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$ This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how you can find the answer from the original expressi... | Note that
\begin{align}
4(2-\sqrt{3}) = 8 - 4\sqrt{3} = (\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2
=
(\sqrt{6}-\sqrt{2})^2.
\end{align}
Therefore,
$$
\sqrt{6} -\sqrt{2} = \sqrt{4(2-\sqrt{3})}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solving $\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$
Solve for $x$ such that
$$\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$$
First, I use the co-function formula:
$$\sin 80^\circ = \cos 10^\circ \tag{1}$$
Also,
$$\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ \tag{2}$$
From th... | $$\dfrac{\sin(100^\circ-x)}{\sin(80^\circ-x)}=\dfrac{\sin80^\circ}{\sin20^\circ}$$
Use Componendo et Dividendo,
$$\dfrac{\sin(100^\circ-x)-\sin(80^\circ-x)}{\sin(100^\circ-x)+\sin(80^\circ-x)}=\dfrac{\sin80^\circ-\sin20^\circ}{\sin80^\circ+\sin20^\circ}$$
Use Prosthaphaeresis Formulas
$$\dfrac{\tan10^\circ}{\cot x}=\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Flux through region : paraboloid and sphere I have this region :
$D=\{(x,y,z)\mid y^2+z^2\le 3|x|,(x-2)^2+y^2+z^2 \le 4\}$
with vector field $\mathbf F=(-2x,-2y,xy)$
I can use the divergence theorem : $\mathrm{div}(\mathbf F)=-4$
Attempt :
Let's see where they intersect :
$(x-2)^2+3x=4 \implies x^2-x=0 \implies x=0,1... | I’ll skip the divergence theorem and concentrate my answer on a computation of $\int_ D (-4) dV$.
It is not clear for me why your solutions of the equalities assure that the proposed cylindrical coordinates of the integration correspond to $D$ (and below we’ll see that the proposed bounds are wrong). But I can provide ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to obtain a column of a matrix representing a homogeneous linear system by the value of the adjacent column? A homogeneous linear system of equations has a coefficient matrix A which
is row equivalent to the following matrix R in reduced echelon form:
\begin{bmatrix}
1 & 2 & 0 & 3 & 0 &5\\
0 & 0 & 1 &4&0&2 \\
... | In part (a), we find that every solution to $A\vec{x}=\vec{O}$ is of the form
\begin{align*}
\vec{x}
&= \left[\begin{array}{r}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5} \\
x_{6}
\end{array}\right]
= \left[\begin{array}{r}
-2 \, x_{2} - 3 \, x_{4} - 5 \, x_{6} \\
x_{2} \\
-4 \, x_{4} - 2 \, x_{6} \\
x_{4} \\
3 \, x_{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Fermat's last theorem short proof attempt Fermat's last theorem states:
(1) $x^n + y^n = z^n$
has no solutions for x, y, z and n positive coprime integers and n > 2. An open question is whether there exists a simple proof hinted at by Fermat. If you can spot an error in my thoughts below, please point it out.
This ... | The problem is that if two polynomials of degree 3 share a common root, their coefficients do not necessarily have to be equal (as you state in (8)-(10) comparing (5) and (7)). Look at the polynomial $(x-1)(x-2)(x-3) = x^3-6x^2+11x-6$. Its coefficients are neither the same as those of $(x-1)^3$, $(x-2)^3$ nor $(x-3)^3$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a relation between $I_n$ and $I_{n+1}$, where $I_{n}=\int_a^b(x-a)^n \sqrt{b-x}\;dx$ $n \in \Bbb{N}^*$ and $0<a<b$ and
$$I_{n}=\int_a^b(x-a)^n \sqrt{b-x}\;dx$$
I'm tasked with finding a relation between $I_n$ and $I_{n+1}$ to prove later that
$$I_n=\frac{2^{2n+2}(n+1)!n!}{(2n+3)!}(b-a)^{n+1}\sqrt{b-a}$$
Any idea... | Putting
$$
\left\{ \matrix{
y = {{x - a} \over {b - a}}\quad x = a + \left( {b - a} \right)y \hfill \cr
dy = {1 \over {b - a}}dx \hfill \cr} \right.
$$
we get the expression of the integral in terms of the Beta function, so
in terms of the Gamma, or Rising Factorial, etc.
$$
\eqalign{
& \int_{x = a}^{\;b} {\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the solutions of $x=\cot x$? Need to find intervals in which the function $y=\frac{x}{2}\cdot \cos x$ is increasing and decreasing. I tried to solve it on the way below but don't know how to continue.
$ \\
y=\frac{x}{2}\cdot \cos x,\ x\in (0,2\pi)\\
y=\frac{1}{2}\cdot x\cdot \cos x \\
{y}'=({\frac{1}{2}\cdot ... | Hint:
The roots are close to the vertical asymptotes of the cotangent, occurring at $x=k\pi$.
To get a first approximation, we can linearize the cotangent close to a root and
$$\cot x-x\approx\frac1{x-k\pi}-x=0.$$
The positive solution is
$$x=\frac{k\pi+\sqrt{k^2\pi^2+4}}{2}.$$
From this, you start Newton's iteration... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $f:I\to \mathbb{R}$ is convex and interval $I$ is bounded, prove that $f$ is bounded below.
Let $I$ be a bounded interval and $f:I\to \mathbb{R}$ be a convex function. Prove that $f$ is bounded below in $I.$
Attempt. Let $a,~b\in I$, by convexity of $f$ on $[a,b]:$
$$f(x)\leq g(x):=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$... | It is easy to show from your first inequality that the upper bound for $f$ on $[a,b]$ is $M = \max(f(a),f(b))$.
To find a lower bound, write $x = \frac{a+b}{2} + \theta$. Since $\frac{a+b}{2} = \frac{1}{2} \left(\frac{a+b}{2} + \theta \right) + \frac{1}{2} \left(\frac{a+b}{2} - \theta \right) $, we have by convexit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Finding derivative of $\sqrt[3]{\sin(2x)}$ using only definition of derivative First post here, so hello everyone.
Here's the problem:
Find the first derivative of: $$\sqrt[3]{\sin(2x)}$$
But, you can only use the difference quotient...
(i.e. the limit of $\frac{f(x+h)-f(x)}{h}$ as $h \rightarrow 0$)
This is a part... | Are you allowed to use the trigonometric limit identities of
$$\lim \limits_{u \to 0} \frac{\sin u}{u} = 1 \hspace{.5 in} \text{and} \hspace{.5 in} \lim \limits_{u \to 0} \frac{1-\cos u}{u} = 0?$$
If not, the link above gives good hints. You'll need to prove these first. If you are allowed to use them, then let's pro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3112718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
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Ramanujan's radical and how we define an infinite nested radical I know it is true that we have
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cdots}}}}=3$$
The argument is to break the nested radical into something like
$$3 = \sqrt{9}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+3\sqrt{25}}}=...=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+\cd... | As others have said, the rigorous definition of an infinite expression comes from the limit of a sequence of finite terms. The terms need to be well-defined, but in practice, we just try to make sure the pattern is clear from context.
Now let's see what goes wrong with your other example. You wrote:
$$4 = \sqrt{16}=\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple...
Suppose $a, b, c \in I$ such that greatest common divisor of $x^2 + ax + b$ and $x^2 + bx + c$ is $(x + 1)$ and the least common multiple of $x^2 + ax + b$ and $x^2 + bx + c$ is ... | We know that $x^2+ax+b$ and $x^2+bx+c$ have the same factor $x+1$.
But $$x^3-4x^2+x+6=x^3+x^2-5x^2-5x+6x+6=$$
$$=(x+1)(x^2-5x+6)=(x+1)(x-2)(x-3),$$ which gives that our polynomials they are:
$$(x+1)(x-2)=x^2-x-2$$ and $$(x+1)(x-3)=x^2-2x-3.$$
Id est, $b=-2$, $a=-1$, $c=-3$ and
$$|a+b+c|=6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum value of expression $a+b+c$
If $a,b,c$ are non negative integers such that $$2(a^3+b^3+c^3)=3(a+b+c)^2.$$
Then maximum value of $a+b+c$ is ?
My Try: Using Jensen Inequality
Let $f(x)=x^3$. Then $f''(x)>0$ for $x>0$ is convex function
So $$\frac{f(a)+f(b)+f(c)}{3}\geq f\bigg(\frac{a+b+c}{3}\bigg)$$
$$\frac{a^... | So you have the maximum possible sum and need to restrict to integers.
What is the maximum possible sum for positive integers (hint has to be less than or equal to the sum for arbitrary reals)? Call this the target sum
Check that the target sum is even (the given condition implies that).
Then is there a solution to the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Computing $\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$. The question is exactly as in the title:
$$\int_{-a}^a \int_{-b}^b \frac 1{(x^2 + y^2 +c^2)^{3/2}}\, dxdy$$
It's been so long since the last time I tried to calculate something like this. I first thought about polar coordinates but that does... | The inner integral is simple even without any substitution because ot the exponent $\frac 32$ in the denominator. Just write
$$\int \frac {dx}{(x^2 + y^2 +c^2)^{3/2}}=\frac {P(x)}{(x^2 + y^2 +c^2)^{1/2}}$$ Differentiate both sides, simplify and identify to get
$$\left(c^2+x^2+y^2\right) P'(x)-x P(x)=1$$ which is separ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factor $x^5-x+15$ It's possible to factor $x^5-x+15$. WolframAlpha gives the answer of: $$(x^2+x+3)(x^3-x^2-2x+5)$$
According to the wikipedia article on quintic functions, the general form $x^5-x+a$ is factorable only when $a=±15$, $±22440$, or $±2759640$.
Question: How would one factor such an expression? For me, it ... | I see there is an answer to 15. I tried, it appears there is no integer root to $x^5 - x \pm 22440.$ As 7 gives 16800 but 8 gives 32760. We arrive at
$$ (x^3 + A x^2 + B x + C)(x^2 + D x + E) = x^5 - x \pm 22440 $$
The point is not to solve the whole system at once, rather do one coefficient at a time and rewrite the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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$x+x^2+x^3=x^4+x^5+x^6$ implies $x^4=x$ in a ring Let $(A,+,.)$ be a ring s.t. $x+x^2+x^3=x^4+x^5+x^6$ for all $x \in A$. Prove that $x^4=x$ for all $x$ in $A$. Can somebody give me some tips, please?
| Hint 1: Plugging in $x=-1$ shows that $2=0$ in $A$.
Hint 2: The given equation can be rearranged to get
$$(x+x^2+x^3)(1-x^3)=0,$$
or equivalently $(x^4-x)(1+x+x^2)=0$.
Multiplying through by $x-x^2$ shows that
$$(x^4-x)^2=0.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What's maximum value of $x (1-x^2)$ for $0 < x <1$? Since we are being taught about AM-GM inequality, I decided to use the method however I am getting two different answers by slightly different methods.
Method 1
\begin{equation}
v=x (1-x^2)$
\implies v^2=x^2 (1-x^2)^2
\end{equation}
Using the AM-GM-inequality we obtai... | Let $x=\frac{a}{\sqrt3}.$
Thus, by AM-GM $$x(1-x^2)=\frac{1}{3\sqrt3}(3a-a^3)=\frac{1}{3\sqrt3}(2-(a^3+2-3a))\leq$$
$$\leq\frac{1}{3\sqrt3}(2-(3\sqrt[3]{a^3\cdot1^2}-3a))=\frac{2}{3\sqrt3}.$$
The equality occurs for $a=1$ or $x=\frac{1}{\sqrt3},$ which says that we got a maximal value.
We can use AM-GM also by the fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove that $\sum\limits_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac{\sqrt{3}-1}{\sqrt{2}}$ How to prove that $$\sum\limits_{k=0}^{\infty}\frac{(4k-1)!!}{(2k+1)!\cdot2^{4k+1}}=\frac{\sqrt{3}-1}{\sqrt{2}}$$
I need any hint to start to prove it.
Thanks for any help.
| Solution 1. Using the generating function for the central binomial coefficients
$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^{\infty}\binom{2n}{n}x^n,$$
we have
$$ \sum_{n=0}^{\infty} \frac{(4n-1)!!}{(2n+1)!2^{4n+1}}
= 4 \int_{0}^{\frac{1}{8}}\sum_{n=0}^{\infty} \binom{4n}{2n} x^{2n} \, \mathrm{d}x
= 2\int_{0}^{\frac{1}{8}} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality proof (strange) Given $a^2 +b^2 +c^2 +d^2 =1$ where $a,b,c,d$ are positive real numbers, prove that $a+b+c+d-1 \geq 16abcd$
How can I prove the inequality ?
My attempts:
By Cauchy-Schwarz : $ (a+b+c+d)^{2} \leq (a^2 +b^2 +c^2 +d^2 )\cdot (1^2 +1^2 +1^2 + 1^2 ) $
Or $ 0\leq a+b+c+d -1 \leq 1 $ .
By AM-GM :... | Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $u^2=tv^2$.
Thus, since $u^2\geq v^2$ it's
$$\sum_{sym}(a-b)^2\geq0,$$ we get $t\geq1$ and since by AM-GM
$$ab+ac+bc+ad+bd+cd\geq6\sqrt{abcd},$$ it's enough to prove that
$$(a+b+c+d)\sqrt{(a^2+b^2+c^2+d^2)^3}-(a^2+b^2+c^2+d^2)^2\geq16v^4$$ or
$$2u\sqrt{(4u^2-3v^2)^3}-(4u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the surface integral of scalar function $\int_{S} (x + y + z )dS$ where $S$ is the boundary of the unit ball $B$ Evaluate the surface integral of scalar function $\int_{S} (x + y + z )dS$ where $S$ is the boundary of the unit ball $B$
my attempt
$S = \int_{S} (x + y + z) dS$
Unit ball $s: x^2 + y^2 + z^2 = 1 \... | By linearity,
$$ \int_{S} (x + y + z) dS=\int_{S} xdS+\int_{S} ydS+\int_{S} zdS$$
Let $S_{x^+}$ the boundary of the unit ball in the half-plane $x\geq 0$ and define in a similar way $S_{x^-}$. Then
$$\int_{S} xdS=\int_{S_{x^+}} xdS+\int_{S_{x^-}} xdS=\int_{S_{x^+}} xdS+\int_{S_{X^+}} (-X)dS=0.$$
where in the second int... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating indefinite integrals of the form $\int \frac{x^2 \,dx}{a x^5 + b}$
Evaluate the indefinite integral
$$\int \frac{x^2 \,dx}{a x^5 + b},$$
for real parameters $a, b \neq 0$.
No apparent substitutions simplify the expression (if the exponent of $x$ in the denominator were an integral multiple of $3$, the form... | An explicit antiderivative is messy, but here's an outline for evaluating this integral by hand.
First, make a linear substitution $x = \alpha u$ for an appropriate constant $\alpha$, which transforms the integral $$\int \frac{x^2 \,dx}{a x^5 + b}$$ into some constant multiple of
$$\int \frac{u^2 \,du}{1 - u^5} .$$
Thi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given three a-triangle-sidelengths $a,b,c$. Prove that $3\left((a^{2}b(a-b)+b^{2}c(b-c)+c^{2}a(c-a)\right)\geqq b(a+b-c)(a-c)(c-b)$ . If you are interested in IMO 1983 please see: Given three a-triangle-sidelengths $a,b,c$. Prove that:
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )\geqq b(a+ b- c)(a- c)(c... | I found a nice identity to prove this!
$$3\left ( a^{2}b(a- b)+ b^{2}c(b- c)+ c^{2}a(c- a) \right )- b(a+ b- c)(a- c)(c- b)$$
$$=(a + b - c)(a + c)(a - c)^2 + (a + b - c)( c + b-a)(a - b)^2 + ( c + b-a)(2\,a - b + c)( b-c)^2 \geqq 0$$
By the way$,$ with $k=constant, k \in [0,1]$ and $a,b,c$ is three side of the triangl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
Prove that the number $7^n+1$ is divisible by $8$ if $n$ is odd. In the case where $n$ is even, give the remainder of the division of $7^n+1$ by $8$.
*
*If $n$ is od... | Yes, you are right. Another way to present it would be like so:
For $n$ odd,
$$7^n+1 \equiv (-1)^n+1 \equiv -1+1 \equiv 0 \pmod 8$$
and for $n$ even,
$$7^n+1 \equiv (-1)^n+1 \equiv 1+1 \equiv 2 \pmod 8$$
| {
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"url": "https://math.stackexchange.com/questions/3139279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integrate $\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ via trig substitution $x = a\sec\theta, dx = \sec\theta \tan\theta$
$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{(\tan^2\theta)^{\frac{3}{2}}}$ =
$ \int \frac{dx}{\tan^{\frac{7}{2}}\theta}$ =
$\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}$... | Note that $$(\tan^2 x)^{3/2}=\tan^3 x$$
Thus, yous should instead obtain
$$\int\frac1{(x^2-1)^{3/2}}dx
=\int\frac{\sec t\tan t}{\tan^3 t}dt
$$
which simplifies to
$$\int\frac{\cos t}{\sin^2 t}=\int\frac1{\sin ^2 t}d(\sin t)=-\frac1{\sin t}+C=-\csc t+C$$
Now reverse the substitution by the identity $\csc^2 t=\frac1{1-\c... | {
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"timestamp": "2023-03-29T00:00:00",
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S is the part of the cylinder $x^2+y^2=2x$ parametrize $S$ S is the part of the cylinder $x^2+y^2=2x$
with $0 \leq z \leq \sqrt{x^2 + y^2}$
how would I parameterize this ?
$x^2+y^2=2x$ can be made into
$(x-1)^2 + y^2 = 1$
| In cylindrical coordinates the surface and its limits are $r=2\cos\theta$ for $x^2+y^2=2x$ and $0\leq z\leq r$ for $0 \leq z \leq \sqrt{x^2 + y^2}$ This suggest to take $\theta$ and $z$ as parameters. But we have the restriction $0 \leq z \leq 2\cos\theta$ limiting the variation for $z$. We can use instead other parame... | {
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"url": "https://math.stackexchange.com/questions/3139627",
"timestamp": "2023-03-29T00:00:00",
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Trig Subsitution When There's No Square Root I would say I'm rather good at doing trig substitution when there is a square root, but when there isn't one, I'm lost.
I'm currently trying to solve the following question:
$$Ar \int_a^\infty \frac{dx}{(r^2+x^2)^{(3/2)}}$$
Anyway, so far, I have that:
$$x = r\tan \theta$$
$... | Firstly you made an error in the first line of working
$$(r\sec{(\theta)})^3=r^3\sec^3{(\theta)}$$
Secondly, you need to change the range of integration after performing a substitution. If $\theta=\arctan{(\frac{x}{r})}$ then the limits should change as $x=a \implies \theta=\arctan{(\frac{a}{r})}$ also $x=\infty \impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Picture proof that the area of a right triangle is $xy$ I stumbled on the following result by accident:
Let $A, B, C$ be the vertices of a right triangle, with opposite side lengths $a, b, c$ respectively, where $\angle C = 90^\circ$ and $a^2 + b^2 = c^2$.
Draw the incircle, and let $x, y, z$ be the length of the tang... | My solution uses a little bit of algebra.
If you superpose the triangles as shown in the figure, the area of the blue square is counted twice, but the green area is not counted.
The green area is (using Pythagoras) $(x-z)(y-z) = 2z^2$. One $z^2$ is the area of the triangles that we haven't count yet, and the other $z^... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove that $A^{-1} + B^{-1}$ nonsingular by showing that $(A^{-1} + B^{-1} )^{-1} = A( A + B )^{-1} B$ Let $A$, $B$, and $A + B$ be nonsingular matrices. Prove that $A^{-1} + B^{-1}$ is nonsingular by showing that
$( A^{-1} + B^{-1} )^{-1} = A( A + B )^{-1} B$
I have done progress to only knowing that $( A^{-1} + ... | To show that $A(A + B)^{-1} B$ is the inverse to $A^{-1} + B^{-1}$, you need to verify the definition of inverses. Specifically, we call the matrix $C$ the inverse of $A^{-1} + B^{-1}$ if
$$C(A^{-1} + B^{-1}) = (A^{-1} + B^{-1})C = I.$$
If such a matrix $C$ exists, then it must be unique, hence why we can call it the i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $85_b = 58_c$, what is the smallest possible value of $b$? I have searched and watched videos online and can't find a method to solve this problem:
If $85_b=58_c$ for some positive integer bases $b$ and $c$, what is the smallest possible value for $b$?
| well think it out:
$8b + 5 = 5c + 8$.
So $8b - 5c = 3$.
$b = 1; c= 1$ is a solution but it is too small as we must have $b,c > 8$. But there are an infinite number of solutions:
If $8b - 5c = 3$ then $8(b+ 5k)-(c+8k)= (8b-5c) + 40k - 40k$ will also be a solution.
$8$ and $5$ are relatively prime so all solutions are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim\limits_{n\to\infty} \prod\limits_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $ I want to calculate $$\lim\limits_{n\to\infty} \prod_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $$
Taking logarithms, it's enough to find
$$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right).$... | Probably not nicer, but still a different way is to use the facts that
$$\lim\limits_{x\rightarrow0}\frac{\tan{x}}{x}=1$$
and, as shown here
$$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$
$$\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do we compute $\sqrt[3]{x_1} +\sqrt[3]{x_2} $ using the fact that $x_1 + x_2 = 4 , x_1x_2 = -1$? Given quadratic equation
$$x^2 -4x-1 = 0$$
How do we compute $\sqrt[3]{x_1} +\sqrt[3]{x_2} $ using the fact that $x_1 + x_2 = 4 , x_1x_2 = -1$?
Regards
| Using $a^3+b^3= (a+b)(a^2-(ab)+b^2)$:
$(x_1+x_2)=$
$(x_1^{1/3}+x_2^{1/3})(x_1^{2/3} -(x_1x_2)^{1/3}+x_2^{2/3}).$
Set $X = x_1^{1/3}+x_2^{1/3}$
$4=X(x_1^{2/3}+x_2^{2/3}+1).$
Express $x_1^{2/3}+x_2^{2/3}$ in terms of $X$:
$X^2= (x_1^{1/3}+x_2^{1/3})^2=$
$x_1^{2/3}+x_2^{2/3} +2x_1^{1/3}x_2^{1/3}$;
$X^2+2= x_1^{2/3}+x_2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove/disprove that $\frac{x^2 - \sqrt{yz}}{yz - x} + \frac{y^2 - \sqrt{zx}}{zx - y} + \frac{z^2 - \sqrt{xy}}{xy - z} \ge 0$
Prove/disprove that
$$ \frac{x^2 - \sqrt{yz}}{yz - x} + \frac{y^2 - \sqrt{zx}}{zx - y} + \frac{z^2 - \sqrt{xy}}{xy - z} \ge 0$$ with $x$, $y$ and $z$ are positives.
I tried to use the Schur's i... |
Prove/disprove that $$\dfrac{x^2 - \sqrt{yz}}{yz - x} + \dfrac{y^2 - \sqrt{zx}}{zx - y} + \dfrac{z^2 - \sqrt{xy}}{xy - z} \ge 0$$with $x$, $y$ and $z$ are positives.
Choosing for example $(x,y,z)=(1,1,4)$ we get $$\frac{1^2-\sqrt{4}}{4-1}+\frac{1^2-\sqrt{4}}{4-1}+\frac{4^2-\sqrt{1}}{1-4}=-\frac{1}{3}-\frac{1}{3}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $f^{(n)}(0) \text{ for } n = 1,2,3,...$ Find $f^{(n)}(0) \text{ for } n = 1,2,3,...$ where
$$ f(x) = \begin{cases}
\frac{e^x - 1}{x}, & \text{when } x \neq 0 \\
1, & \text{when } x = 0 \\
\end{cases} $$
My approach
I decided to calculate some first $f$ derivatives. In that case I defined:
$$ g(x) = \frac{e^x - 1}... | Since$$f(x)=1+\frac x{2!}+\frac{x^2}{3!}+\cdots+\frac{x^n}{(n+1)!}+\cdots,$$then$$(\forall n\in\mathbb{Z}^+):\frac{f^{(n)}(0)}{n!}=\frac1{(n+1)!}$$and therefore$$(\forall n\in\mathbb{Z}^+):f^{(n)}(0)=\frac1{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove for all $n\in \mathbb{N}$ that $\sum_{i=0}^{n} i\cdot F_{2i} = (n+1)F_{2n + 1} - F_{2n + 2}$. $F_n$ denotes the Fibonacci sequence where $n$ is the term of the Fibonacci number in the sequence. ($F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, $F_4=3$, ... $F_n=F_{n-1} + F_{n-2}$)
I want to prove this using strong induction a... | \begin{align}
\sum_{i=0}^{k+1}iF_{2i}&=(k+1)F_{2k+2}+(k+1)F_{2k+1}-F_{2k+2}\\
&=kF_{2k+2}+(k+1)(F_{2k+3}-F_{2k+2})\\
&=(k+1)F_{2k+3}-F_{2k+2}\\
&=(k+1)F_{2k+3}-(F_{2k+4}-F_{2k+3})\\
&=(k+2)F_{2k+3}-F_{2k+4}.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3152182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find minimum of $4(a^3 + b^3 + c^3) + 15abc$ subject to $a + b + c = 2$
$a$, $b$ and $c$ are three sides of a triangle such that $a + b + c = 2$. Calculate the minimum value of $$\large P = 4(a^3 + b^3 + c^3) + 15abc$$
Every task asking for finding the minimum value of an expression containing the product of all of t... | This is particularly easy to prove by calculation.
With the substitution $a=u$, $b=u+v$, $c=u+v+w$, we get
$$4(a^3+b^3 + c^3)+15 a b c- (a+b+c)^3 = 3(u v^2 + u v w + u w^2 + 2 v w^2 + w^3)$$
$\bf{Added:}$ We have equality if and only if $w=0$ and $u v=0$, that is, if and only if $a=b=c$, or one of the $a$, $b$, $c$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Reaching upon $9=1$ while solving $x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
$x$ for $3\tan{(x-15^{\circ})}=\tan{(x+15^{\circ})}$
Substituting $y=x+45^{\circ}$, we get
$$3\tan{(y-60^{\circ})}=\tan{(y-30^{\circ})}$$
$$3\frac{\tan y - \sqrt3}{1+\sqrt3\tan y}=\frac{\tan y - 1/\sqrt3}{1+1/\sqrt3\cdot\tan y}$$ ... | Hint:
Set $\tan y=\dfrac1a$
$$\dfrac{3(1-\sqrt3a)}{a+\sqrt3}=\dfrac{\sqrt3-a}{\sqrt3a+1}$$
$$\iff3-a^2=3(1-3a^2)\iff a=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Soving PDE $u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0$ I have some problems with solving PDEs.
*
*\begin{cases}
\ u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \\[2ex]
u(x,x)=1,\quad u(1,y)=y
\end{cases}
What I've done: $$u(x,y)=\frac{1}{x^2}v(x,y)$$
$$u_{xx}-u_{yy} + \frac{4}{x}u_x+\frac{2}{x^2}u=0 \Ri... | The equation $C_1(2x)+C_2(0)=1$ shows that $C_1(2x)$ doesn't vary with $x$, i.e. $C_1$ is a constant function, say $C_1(x) = c.$
Inserting this in $C_1(1+y)+C_2(1-y)=y$ we get $C_2(1-y) = y-c$ i.e. $C_2(y) = (1-y)-c = (1-c)-y.$
Thus,
$$
\begin{cases}
C_1(x) = c \\
C_2(x) = (1-c)-y
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation $\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$.
Solve the equation: $$\sqrt{x + 2} - \sqrt{3 - x} = x^2 - 6x + 9$$
Here's what I've done.
Let $\sqrt{x + 2} = a$ and $\sqrt{3 - x} = b$
$\implies
\left\{
\begin{align}
a^2 + b^2 &= 5\\
a^2 - b^2 &= 2x - 1
\end{align}
\right.$.
We have that $a - b = (x -... | Let $3-x=t^2$, with $t>0$. Then
$$\sqrt{5-t^2}-t=t^4$$ and
$$p(t):=(t^4+t)^2-\left(\sqrt{5-t^2}\right)^2=t^8+2t^5+2t^2-5=0.$$
Now the derivative
$$8t^7+10t^4+4t$$ cancels for $t=0$ (minimum), and $8(t^3)^2+10t^3+4$ has no real root. As $p(0)<0$, the polynomial has exactly one positive and one negative root, which we di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the volume of a solid rotated around the x-axis $y=x^3, y=x, x \geq 0$
$$ \begin{align}
V &= \int_{0}^{1} A(x)dx = \int_{0}^{1} \pi (x-x^3)^2dx = \pi \int_{0}^{1}(x^2-2x^4 + x^6)dx \\
&= \pi \bigg[ \frac{1}{3}x^3 - \frac{2}{5}x^5 + \frac{1}{7}x^7\bigg]_{0}^{1}
\end{align}$$
The book answer says the answer is $\fra... | The correct setup is the difference of the squares $$\int_{0}^{1} \pi \left( x^2 - (x^3)^2 \right) dx$$
and not the square of the difference $(x-x^3)^2$.
The radius of the larger disk is $x$ and its area is $\pi x^2$. The radius of the small disk is $x^3$ thus area $\pi (x^3)^2$. The ring (annulus) is the large disk mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral with two distinct roots in them, such as: $\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx$ I'm getting familiar with basic indefinite integrals and these are the hardest ones I've met so far:
*
*$\int \frac{\sqrt{x}}{x^2(\sqrt{x+1}+\sqrt{x})}dx$
*$\int \frac{\sqrt[3]{x+2}-\sqrt[3]{x}}{x^2(\sqrt[3]{x+2}+\... | Hint: For the integral $$\int \frac{\sqrt{x^2+x}}{x^2}dx$$ substitute $$\sqrt{x^2+x}=x+t$$ it is the Eulerian substitution.
Then we get by squaring $$x=\frac{t^2}{1-2t}$$ and $$dx=-2\,{\frac {t \left( -1+t \right) }{ \left( -1+2\,t \right) ^{2}}}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y,z \in \Bbb{R} $ , $f(xf(x)+f(y))=x^2+y$
Find all functions $f:\Bbb{R} \to \Bbb{R}$ such that for all $x,y, \in \Bbb{R} $ , $f(xf(x)+f(y))=x^2+y$
We can easily get a strong condition $f(f(y))=y $ by setting $x=0$ . By this equation we know $f$ is inj... | $$ \begin{align}
&\text{As already noted, we have }\qquad\qquad\qquad\qquad\qquad f(f(y)) = y \implies f(x) = f^{-1}(x) \\
&\text {Also, using the substitution } x\to y \text{ we get}\,\,\quad f((y+1)f(y)) = y^2+y \to f(0) = 0\\
&\text{using } f= f^{-1}:\quad f(x*f(x) + f(y)) = x^2+y \implies x*f(x) + f(y) = f(x^2+y)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3170450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the remainder when the polynomial $1+x^2+x^4+x^6+....+x^{22}$ is divided by $1+x+x^2+x^3+...+x^{11}$ Find the remainder when the polynomial $$1+x^2+x^4+x^6+....+x^{22}$$ is divided by $$1+x+x^2+x^3+...+x^{11}$$
$1+x^2+x^4+x^6+....+x^{22}=\frac{x^{24}-1}{x^2-1}$
$1+x+x^2+x^3+...+x^{11}=\frac{x^{12}-1}{x-1}$
Now$$\... | Proceeding from where I left off
$$\frac{1+x^2+x^4+x^6+....+x^{22}}{1+x+x^2+x^3+...+x^{11}}=\frac{x^{12}+1}{x+1}$$$$=\frac{x^{12}-1+2}{x+1}=\frac{x^{12}-1}{x+1}+\frac{2}{x+1}=P(x)+\frac{2}{x+1}$$where $P(x)$ is a polynomial since $x^{12}-1$ is divisible by $x+1$.
So $$1+x^2+x^4+x^6+....+x^{22}=P(x)\left(1+x+x^2+x^3+...... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3171446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A nice Nesbitt inequality from a strange inequality Given $a,\,b,\,c> 0$$,$ prove that$:$
$$\frac{a}{b+ c}+ \frac{b}{c+ a}+ \frac{c}{a+ b}+ \frac{63}{5}\left [ \frac{2\,c^{\,2}}{(\,a+ b\,)^{\,2}}- \frac{c}{a+ b} \right ]\geqq 0$$
See$:$ $\lceil$ https://artofproblemsolving.com/community/c6h354642p1923888 $\rfloor$
The ... | Let $\frac{c}{a+b}=\frac{x}{2}$.
Thus, by C-S and AM-GM $$\sum_{cyc}\frac{c}{a+b}+\frac{63}{5}\left(\frac{2c^2}{(a+b)^2}-\frac{c}{a+b}\right)=\frac{a}{b+c}+\frac{b}{a+c}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\frac{x}{2}\right)=$$
$$=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{x}{2}+\frac{63}{5}\left(\frac{x^2}{2}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A probability question: Poor Alex
Alex remembers all but the last digit of his friend's telephone number. He decides to choose the last digit at random in an attempt to reach him. Given that, Alex has only enough money to make two phone calls, the probability that he dials the right number before running out of money ... | So you are choosing twice from a set of ten Without replacement. That means the first digit tried, called it A is taken from 0 through 9. The second digit, digit B, should be different, because A has been eliminated.
Let's say the correct digit is X.
What are the chances that neither A or B is X?
That question can b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k} \frac{(-1)^k}{k+1}$
Calculate $$\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k}
\frac{(-1)^k}{k+1}$$
My approach
$$\sum_{0 \le k } \binom{n+k}{2k} \binom{2k}{k}
\frac{(-1)^k}{k+1} = \\
\sum_{0 \le k } \binom{n+k}{k} \binom{n}{k}
\frac{(-1)^k}{k+1} = \\
\frac{1... | Starting from
$$\sum_{k=0}^n {n+k\choose 2k} {2k\choose k} \frac{(-1)^k}{k+1}$$
for a self-contained answer we observe that
$${n+k\choose 2k} {2k\choose k} =
\frac{(n+k)!}{(n-k)! \times k! \times k!}
= {n+k\choose k} {n\choose k}$$
so we find
$$\sum_{k=0}^n {n+k\choose k} {n\choose k} \frac{(-1)^k}{k+1}$$
which is
$$\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"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.