Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Minimum value of $p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$ where $x,y,z\in \mathbb{R}^+$.
Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdo... | Hint: Rewrite the expression as
\begin{eqnarray*}
\frac{1}{\sqrt{5} } \left( \sqrt{3} \sqrt[4]{5} x - \frac{1}{\sqrt{3} \sqrt[4]{5} x } \right)^2 + \frac{2}{\sqrt{5}}
\\ +
5\sqrt{5} \left( \frac{y}{ \sqrt[4]{5}} - \frac{ \sqrt[4]{5}}{y } \right)^2 +10\sqrt{5} \\+
\frac{1}{6 } \left( \sqrt{6} z - \frac{1}{\sqrt{6} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n n^2 i^n}{(2n)!}$ $$\sum_{n=0}^{\infty} \frac{(-1)^n n^2 i^n}{(2n)!}$$
$i^n$ can be written as $(\frac{1+i}{\sqrt{2}})^{2n}$ but the main problem is the series :
$$\sum_{n=0}^{\infty} (-1)^n \frac{n^2 }{(2n)!}$$
I don't understand why this series can be written as a sum of sin... | As $$4n^2=2n(2n-1)+2n$$ write $y^2=-i$
$$\dfrac{(-1)^ni^nn^2}{(2n)!}=\dfrac{y^{2n}(2n(2n-1)+2n)}{4(2n)!}=\dfrac{y^2}4\dfrac{y^{2n-2}}{(2n-2)!}+\dfrac y4\dfrac{y^{2n-1}}{(2n-1)!}$$
Now $\displaystyle e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$
Find $$\dfrac{e^y+e^{-y}}2,\dfrac{e^y-e^{-y}}2$$
Replace the value of $y^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rayleigh quotient on circular region of radius 2 I' m struggling with the following problem.
We have the eigenvalue problem:
$$u'' + \lambda u = 0$$
with associated boundary condition:
$$u' + 3u = 0$$
Now by using the Rayleigh quotient for $0 \leq r \leq 2$and the trial function:
$$\psi_0(r)= \alpha - r $$
where $\alp... | \begin{equation}
J(\psi_0) \geq \lambda_1
\end{equation}
Where in this case $\psi_0$ is the trialfunction.
In order to understand the Rayleigh-quotient, now assume the same eigenvalue problem as the example above, but now by using the Rayleigh quotient method.
$$ \Delta u + \lambda u = 0$$
with associated boundary con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3523847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simple inequality proof problem of my authorship with fixed sum $a+b+c=3$. Prove that for positive numbers $a,b$ and $c$ such that $a+b+c=3$ following inequality holds:
$\sqrt[3]{\frac{1}{3a^2(8b+1)}}+\sqrt[3]{\frac{1}{3b^2(8c+1)}}+\sqrt[3]{\frac{1}{3c^2(8a+1)}}\ge 1$.
There are at least 3 different solutions that I fo... | Using AM-GM and Cauchy-Schwarz:
$$
\begin{aligned}
\frac{1}{3}\cdot LHS &= \sum \frac{1}{\sqrt[3]{9a\cdot 9a \cdot (8b+1)}} \\
&\geq \sum \frac{3}{9a+9a+8b+1} \\
&= 3\sum \frac{1}{18a+8b+1} \\
&\geq 3\cdot \frac{9}{26(a+b+c)+3} \\
&= \frac{1}{3}
\end{aligned}
$$
Equality occurs when $a=b=c=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$ I am trying to prove that $$\sum_{k=1}^{2n} \frac{(-1)^{k-1}}{k} = \sum_{k=1}^n \frac{1}{n+k}$$ for $n \in \mathbb{N}$.
My approach is to prove this by induction and this is what I got so far:
$$\sum_{k=1}^{2(n+1)} \frac{(-1)^{k-1}}{k} = (\s... | Got it, working my way backwards using Maximilian Janisch advice:
$$\sum_{k=1}^{n+1} \frac{1}{n+1+k}=\sum_{k=2}^{n+2} \frac{1}{n+k} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = (\sum_{k=1}^{n} \frac{1}{n+k}) + \frac{1}{2n+1} - \frac{1}{2n+2} = (\sum_{k=1}^n \frac{1}{n+k}) + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ ... | $1995$ I'd divisible by $3$ but not by $9$ hence $x , y $
both simultaneously are not multiple of $3$.
$$ $$ let either of $x,y$ is multiple of $3$ and other is not multiple of $3$ then $x^2-y^2$ is not multiple of $3$ where as $ 1995$. Is multiple of $3$ hence not possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Given that $f(x) = \frac{x^2}{x-2}$ show that $8 \le \int_3^4 f(x)\,dx \le 9$. Consider the function:
$$f : \mathbb{R} \setminus \{ 2\} \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{x^2}{x - 2}$$
I have to show that the following statement is true:
$$8 \le \int_3^4 f(x) \, dx \le 9$$
The first thing I did was to fi... | A very quick way to show this is by finding the area under the curve in the interval from 3 to 4
The plot of $ f(x) $ looks like this- here is the plot.
From the plot it's very clear that area under the curve in $ [3, 4] $ is bounded between 8 and 9
Also, this function can be re-written as:
$$
f(x)=\frac{x^{2}}{x-2} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
How to go from $C^2y^2+(C^2-M^2)x^2+Gx+H=0$ to an ellipse/ hiperbola? The following arrives on the context of conic sections.
(All letters denote a constant $\in \mathbb{R}$ except for $x$ and $y$).
Michael Spivak´s Calculus, arrives at
$$C^2y^2+(C^2-M^2)x^2+Gx+H=0$$
Then procedes to say that, for $\;C^2\neq M^2$, $\;... | $$C^2y^2+\underbrace{(C^2-M^2)x^2+Gx}_{\text{make this a perfect square}}+H=0$$
$$\begin{align*}
(C^2-M^2)x^2+Gx&=(C^2-M^2)\left(x^2+\frac G{C^2-M^2}x+\frac{G^2}{4(C^2-M^2)^2}-\frac{G^2}{4(C^2-M^2)^2}\right)\\[1ex]
&=(C^2-M^2)\left(\left(x+\frac G{2(C^2-M^2)}\right)^2-\frac{G^2}{4(C^2-M^2)^2}\right)\\[1ex]
&=(C^2-M^2)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3530826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to prove that when $n$ is even, and $A = (a_{ij})_{i,j}$, $1\leq i,j\leq n$ to show that $\mathrm{det} A = 1$? How to prove that when $n$ is even, and $A = (a_{ij})_{i,j}$, $1\leq i,j\leq n$ with $$a_{ij}=\begin{cases} 1& i<j,\\ 0 & i=j,\\ -1& i>j,\end{cases}$$
to show that $\mathrm{det} A = 1$ ?
| Subtract the second from the first row, the third from the second row, ..., the $n$th row from the $n-1$st row and you will get the following matrix (omitted entries are all $0$):
$$\begin{pmatrix}
1 & 1 & 0 & 0 & 0&\dots & 0 & 0&0\\
0 & 1 & 1 & 0 & 0 &\dots & 0 & 0&0 \\
0 & 0 & 1 & 1 & 0 & \dots & 0 & 0&0\\
& & & \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3532633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Tangents are drawn to the parabola $y^2=4x$ from the point P (6,5) to the touch the parabola at Q and R.
$C_1$ is the circle which touches the parabola at Q and $C_2$ is the circle which touches the parabola at R. Both circles pass through the focus of the parabola. Find the radius of circle $C_2$
The equation of tan... | Find the equation of both the circles using $S+kL=0$.
$L$ is the equation of the tangent to the parabola at points of contact. Point of contact may be considered as point circles.
Since point of contact comes out to be $(4,4)$ and $(9,6)$ and the equation of tangents are: $2y=x+4$ and $3y=x+9$
Circle through $(9,6)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induct... | This is my solution
Let
$$ \alpha_n = V_n + \beta $$
$$a_{n+1} = V_{n+1} + \beta$$
then
$$V_{n+1} = \frac{3V_n + V_n\beta + (\beta^2 - 1)}{3-V_n-\beta} $$
At here we choose $\beta = 1$ as result from $\beta^2 - 1 = 0$
So we got $$V_{n+1} = \frac{4V_n}{2-V_n}$$
Continue we let
$$V_n = \frac{1}{u_n}$$
then
$$\frac{4\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^... | L'hopital
$\lim_{x\to 7}{(\frac{(x-8)(x-10)\ln(\frac{x}{7})}{x-7})}=$
$\lim_{x\to 7}\frac{(x-8)(x-10)\frac 7x\frac 17 + (2x-18)\ln \frac x7}1=$
$\frac{(-1)(-3)}7 + (-4)*0=\frac 37$.
So $\lim_{x\to 7}e^{ln(\frac{x}{7})^{(\frac{(x-8)(x-10)}{x-7})}}=\lim_{x\to 7}e^{(\frac{(x-8)(x-10)}{x-7})ln(\frac{x}{7})}=e^{\frac 37}$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
| $$
\begin{aligned}
I &=\int \frac{1}{\left(x^{3}+1\right)^{2}} d x \\
&=-\frac{1}{3} \int \frac{1}{x^{2}} d\left(\frac{1}{x^{3}+1}\right) \\
&=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}-\frac{2}{3} \int \frac{1}{x^{3}\left(x^{3}+1\right)} d x \\
&=-\frac{1}{3 x^{2}\left(x^{3}+1\right)}-\frac{2}{3} \int\left(\frac{1}{x^{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Problem with computing $\int\frac{dx}{2x^2+5} $ by trigonometric substitution I so close to can solve this problem but I don't find the correct response:
$$\int\frac{dx}{2x^2+5}
$$
Always get the answer:
$$ \frac{\arctan{\sqrt{\frac{2}{5}}x}}{\sqrt{5}}
$$
But the correct answer have more one square root multiplying the... | Your error stems from applying the chain rule incorrectly. You basically substitute $u=x\sqrt{2}\Rightarrow du=\sqrt{2}\cdot dx$.
So
$$\int\frac{dx}{(x\sqrt{2})^2+(\sqrt{5})^2}=\int\frac{\frac{du}{\sqrt{2}}}{u
^2+(\sqrt{5})^2}=\frac{1}{\sqrt{2}}\int\frac{du}{u
^2+(\sqrt{5})^2}=$$
$$=\frac{\arctan{\frac{u}{\sqrt{5}}}}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$ I need to solve the following problem:
$$\int{\frac{5x}{(2x^2-3) \sqrt{3x^2-2x+1}}}dx$$
I tried with trigonometric substitution, but I couldn't solve it. The square root is messing me up. I searched it up and nothing appears to work well so I hope someone can help... | This is going to get ugly real quick.
$$\int{\frac{5x}{\left(2x^2-3\right)\sqrt{3x^2-2x+1}}}\,\mathrm dx\equiv 5\sqrt3\int\frac x{\left(2x^2 - 3\right)\sqrt{(3x - 1)^2 + 2}}\,\mathrm dx$$
Let $u = 3x - 1\implies\mathrm dx = \dfrac13\mathrm du$ and $x = \dfrac{u + 1}3, x^2 = \dfrac{(u + 1)^2}9$. Therefore,
$$\int\frac x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to show that this series converges? I found in a book the following exercise:
Show that the series $$\sum_{n=1}^{+\infty} \sin\left(\frac{n^2+n+1}{n+1}\right)$$ converges.
At first sight, I tried to break the fraction and rewrite the series as $$\sum_{n=1}^{+\infty} \sin\left(n+\frac{1}{n+1}\right)$$
This act... | Suppose that
$$\lim_{n\to\infty}\sin\left(\frac{n^2+n+1}{n+1}\right)=\lim_{n\to\infty}\sin\left(n+\frac{1}{n+1}\right)=0$$
We have
$$\sin\left(n+1+\tfrac{1}{n+2}\right)=\sin\left(n+\tfrac{1}{n+1}\right)\cos\left(1+\tfrac{1}{n+2}-\tfrac{1}{n+1}\right)+\cos\left(n+\tfrac{1}{n+1}\right)\sin\left(1+\tfrac{1}{n+2}-\tfrac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a... | $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a-b)^2>0.$$
We can get it also, by C-S:
$$(1^2+1^2+1^2)(a^2+b^2+c^2)\geq(a+b+c)^2,$$ which is our inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$
| Have you tried considering the Lagrangean $$\mathcal L(a, b, c) = a^2 + b^3 + c^4 + \lambda\left(a + b^2 + c^3 - \frac{325}9\right)?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2... | A Second (Magical) Solution by Cornel Ioan Valean
To get a different solution, we start from Dan Fulea's integral in this post where Cornel provided with an extremely simple solution, and by simply rearranging it, we have that
$$\frac{\pi^4}{64}$$
$$=\frac{\pi^2}{4}\underbrace{\int_0^1\frac{\operatorname{arctanh}(t)}{t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Calculate PMF of discrete random varialbe
$p(x,y)=\frac{xy^2}{30}$ be the joint PMF of discrete random variables X and Y with the support $S=\{(x,y):x\in\{1,2,3\}, y\in\{1,2\}\}$
Compute $E(X-2Y)$
Question is can I expand to be $E(X)-2E(Y)$? (by linearity)?
| We compute the marginal PMF of $X$ by summing the joint PMF over the possible values of $Y$:
\begin{align}
\mathbb P(X=1) = \mathbb P(X=1,Y=1) + \mathbb P(X=1,Y=2) = \frac{1\cdot1^2}{30} + \frac{1\cdot 2^2}{30} = \frac16\\
\mathbb P(X=2) = \mathbb P(X=2,Y=1) + \mathbb P(X=2,Y=2) = \frac{2\cdot1^2}{30} + \frac{2\cdot 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find : $\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$ Find :
$$\lim\limits_{n\to +\infty}\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}$$
My attempt : i don't know is correct or no!
I use this rule :
$$\lim\limit... | Hint:
$$\frac{\left(1+\frac{1}{n^3}\right)^{n^4}}{\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^4}}=\frac{\left(\left(1+\frac{1}{n^3}\right)^{n^3}\right)^n}{\left(\left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^3}\right)^{(n+1)}}$$
$$=\frac{\left(\left(1+\frac{1}{n^3}\right)^{n^3}\right)^n}{\left(\left(1+\frac{1}{(n+1)^3}\right)^{(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 6
} |
Verify that $4r^3+r=\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4$ Must I verify that Left hand side(LHS)= Right hand side(RHS)or can I prove that RHS= LHS?
I don’t know how to prove from LHS=RHS. How to separate the $4r^3+r$ into two terms, i.e. $\displaystyle\Bigl(r+\frac{1}{2}\Bigr)^4-\Bigl(r-\frac{1}{2}... | Using the first identity,
$$\sum_{r=1}^n(4r^3+r)=\sum_{r=1}^n\left(r+\frac{1}{2}\right)^4-\sum_{r=1}^n\left(r-\frac{1}{2}\right)^4
\\=\sum_{r=1}^n\left(r+\frac{1}{2}\right)^4-\sum_{r=0}^{n-1}\left(r+\frac{1}{2}\right)^4$$
and by telescoping, this is
$$\left(n+\frac{1}{2}\right)^4-\left(\frac{1}{2}\right)^4.$$
To estab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Integrate $\int_0^\infty \frac{1}{(1+x^2)^n} dx$ How to find $$\displaystyle \int_0^\infty \frac{1}{(1+x^2)^n}dx\quad ?$$
For $n=1$, we have $(\arctan x) |_0^\infty = \frac{\pi}{2}.$
I tried to integrate by parts to get recurrent formula:
$$
\int_0^\infty \frac{1}{(1+x^2)^n}dx
= \left. \frac{x}{(1 + x^2)^n} \right|_0^... | Hint:
$$I_n:=\int_0^\infty\frac{dx}{(x^2+1)^n}=\frac{x}{(x^2+1)^n}+2n\int_0^\infty\frac{x^2}{(x^2+1)^{n+1}}dx
\\=\left.\frac{x}{(x^2+1)^n}\right|_0^\infty+2n\int_0^\infty\frac{x^2+1-1}{(x^2+1)^{n+1}}dx=2n(I_n-I_{n+1})$$
gives the recurrence
$$I_{n+1}=\frac{2n-1}{2n}I_n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding all angles that satisfy $8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
$\text { Hence, solve the equation } 8 \cos ^{3} \theta-6 \cos \theta+1=0 \quad \text { for } \theta \in[-\pi, \pi]$
The previous part was to prove that $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta \qua... | $\cos^{-1}(x)$’s range is $[0 , \pi]$ so if we calculate $\cos^{-1}(-\frac{1}{2})$ in our calculator, we are going to get $\frac{2\pi}{3}$
However, the solution to $\cos(x)=-\frac{1}{2}$ are $\pm\frac{2\pi}{3}+2n\pi$ with integer $n$. Since $x=3\theta$, $\theta=\pm\frac{2\pi}{9}+n\frac{6\pi}{9}$.
You missed $\theta=-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Use binomial expansion to show the following inequality I wanted to show $\left(1+\frac{x}{n}\right)^n\leq\left(1+\frac{x}{n+1}\right)^{n+1}$ using binomial expansion for $n\geq 2$ and $x\geq 0$.
So my idea was to expand both using binomial expansion and try to compare term-wise. The $k^{th}$ term of $\left(1+\frac{x}{... | Let $f(x,n):=(1+x/n)^n$, then we need to show $f(x,n+1)\geq f(x,n)$ for $x\geq0,n\geq2$. Observe that the coefficient of $x^k$ in $f(x,n)$ is
$$\frac{1}{n^k}\binom{n}{k}=\frac{1}{k!}\cdot\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\dotsm\frac{n-k+1}{n}.$$
Correspondingly the coefficient of $x^k$ in $f(x,n+1)$ is$$\frac{1}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3560295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Given that $x_1, x_2, x_3$ are the roots of the polynomial $x^3-2x^2+3x+5=0$ find $(x_2-x_1)^2(x_3-x_1)^2(x_3-x_2)^2$. Consider the polynomial:
$$x^3-2x^2+3x+5=0$$
where $x_1, x_2$ and $x_3$ are the roots of the above polynomial. Now, consider the following determinant, which is defined using the above given roots:
$$\... | $\Delta$ is not symmetric, but $\Delta^2$ is, so it can be expressed in terms of $a=x_1+x_2+x_3$, $b=x_1x_2+x_2x_3+x_3x_1$ and $c=x_1x_2x_3$. Indeed, we have:
$$\Delta^2 = a^2 b^2 + 18 abc - 4 b^3 - 4 a^3 c - 27 c^2$$
The simplest way I know to prove this identity, is like this: let $x=x_1^2x_2+x_2^2x_3+x_3^2x_1$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
A formula of Ramanujan for $\cot\sqrt {w\alpha} \coth\sqrt{w\beta} $ While trying to answer this question I stumbled on a paper by Bruce C. Berndt which contains the following formula by Ramanujan $$\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta}=\frac{1}{2w}+\sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m... | I thought it would be worthwhile to at least mention how to use the Mittag-Leffler pole expansion theorem to show that $$ \begin{align}\frac{\pi}{2}\cot\sqrt{w\alpha}\coth\sqrt{w\beta} &=\frac{1}{2w}+ \frac{b-a}{6} +\sum_{m=1}^{\infty} \left(\frac{m\alpha\coth m\alpha} {w+m^2\alpha} +\frac{m\beta\coth m\beta} {w-m^2\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}... | The telescoping sum trick, with ${1\over k^2-k}={1\over k-1}-{1\over k}$, works, but you need to split off the first term:
$$\sum_{k=n}^{3n-1}{1\over k^2}={1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2}\lt{1\over n^2}+\sum_{k=n+1}^{3n-1}{1\over k^2-k}={1\over n^2}+\left({1\over n}-{1\over3n-1}\right)={5\over4n}-\left({1\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve $\cos(z) =3/4+i/4$ I need to solve the complex trinometric equation
$$\cos(z) =\frac{3}{4}+\frac{i}{4} $$
What I've done so far is:
Using the cos formula I got $e^{iz} +e^{-iz} =\frac{3}{2}+\frac{i}{2}$
Making $t=e^{iz} $ we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$
Multiplying by $t^2$ we get
$$t^2-\frac{3+i... | Note that the solutions to $t^2-\frac{3+i}{2}t+1=0$ can be simplified as,
$$t=\frac{3+i \pm \sqrt{-8+6i} } {4} =\frac{3+i \pm (1+3i) } {4}$$
or,
$$e^{iz}=1+i=\sqrt2 e^{i(\frac\pi4+2\pi n)} = e^{\frac12\ln2 +i(\frac\pi4+2\pi n)}$$
$$e^{iz}=\frac12(1-i) =\frac1{\sqrt2} e^{-i(\frac\pi4+2\pi n)} = e^{-\frac12\ln2 -i(\frac\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate the integral $\int (\sin x + 2\cos x)^5 dx$ Do I have to expand manually $(\sin x + 2\cos x)^5$?
I don't think I can do integration by parts or substitution with this function.
Edit: Try to make one term, what I know is using $\sin^2 x + \cos^2 x=1$, but I think I still have to expand the polynomial.
| One way is to recall the angle addition identity $$\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.$$ If we suppose there exists a suitable constant $k$ and angle $\theta$ such that $$\sin x + 2 \cos x = k (\sin x \cos \theta + \cos x \sin \theta),$$ then we have $$k \cos \theta = 1, \\ k \sin \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3568841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove $I=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}=\frac{1}{3} \ln 2$ How can we show that:
$$I=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}=\frac{1}{3} \ln 2$$
The logarithm in the denominator is the main trouble for me here.
I have obtained this integral from the following infinite product:
$$P=e^I=\prod_{k=1}^\... | By using the integral representation
\begin{equation}
\frac{x-1}{\ln x}=\int_0^1x^s\,ds
\end{equation}
we can express
\begin{align}
I&=-\int_0^1 \frac{(1-x)^2 dx}{(1+x^2+x^4)\ln x}\\
&=\int_0^1 \,ds\int_0^1 \frac{x^s\left( 1-x \right)}{1+x^2+x^4}\,dx
\end{align}
or, with the notation $J(s)=\int_0^1 \frac{x^s}{1+x^2+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$
I don't really know how to approach this. I was thinking doing something ... |
Applying simple substitutions: $\sqrt[3]{a}=m, \sqrt[3]{b}=n, \sqrt[3]{c}=k$, we have
$$(mnk)^3=1, m+n+k=0 \\ mnk=1, m+n+k=0 \\ m^3+n^3+k^3=m^3+n^3-(m+n)^3 =-3mn(m+n)=\frac{-3}{k} \times (-k)=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cd... | $\begin{array}\\
f(n)
&=\dfrac{n!}{n^n}\\
&=\dfrac{\prod_{k=1}^n k}{n^n}\\
&=\prod_{k=1}^n (k/n)\\
&=\prod_{k=0}^{n-1} ((n-k)/n)\\
&=\prod_{k=0}^{n-1} (1-k/n)\\
g(n)
&=\ln(f(n))\\
&=\sum_{k=0}^{n-1} \ln(1-k/n)\\
&<\sum_{k=0}^{n-1} -k/n
\qquad \ln(1-x) < -x\\
&=-\dfrac{n(n-1)}{2n}\\
&=-\dfrac{n-1}{2}\\
&=\dfrac12=\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
} |
Distance from $P(4,3)$ to tengent point on curve $x^2 + y^2 -2x-4=0$
Tangent line to curve $$x^2 + y^2 -2x-4=0$$ at A passes $P(4,3)$.
Find distance A and P.
Tangent of the line is $\frac{2-2x}{\sqrt{2x+4-x^2}}$
A(k,l) means
$l = \sqrt{2k+4-k^2}$
And
$l = \frac{2-2k}{\sqrt{2k+4-k^2}}$
So, I get $k=\frac{3\pm\sqrt{21}... | The required answer is $\sqrt{4^2+3^2-2•4-4}$ = $\sqrt{13}$
Use the formula $\sqrt{S1}$ Or proceed by using Pythagorean theorem as the other answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this limit without l'Hopital? I tried the substitution $t=x-(\pi/3)$ but it doesn't help at all. I have also tried using $\sin(\pi/3)=\sqrt{3}/2$ but couldn't do anything useful then. I tried to factor the denominator and numerator, but it didn't help either. I want a solution without l'Hopital's rule.
$... | We can rewrite the limit in the following way:
$$\lim_{x \to \frac{\pi}{3}} \frac{\sin^2{x} - \sin^2{\frac{\pi}{3}}}{x^2 - \left(\frac{\pi}{3}\right)^2} =
\frac{2\sin{\frac{\pi}{3}}}{2\frac{\pi}{3}} lim_{x \to \frac{\pi}{3}} \frac{\sin{x} - \sin{\frac{\pi}{3}}}{x - \frac{\pi}{3}}=
\frac{\sin{\frac{\pi}{3}}}{\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Compute the stabilizers and orbits of $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}1\\1\end{bmatrix}$ The natural action of $GL_2(\mathbb{R})$ on $\mathbb{R}^2$ is given by $A\cdot v=Av$, for $A\in GL_2(\mathbb{R})$ and $v\in\mathbb{R}^2$. Compute the stabilizers and orbits of $\begin{bmatrix}1\\0\end{bmatrix... | If $G$ is acting on $X$, and $x \in X$, the stabilizer is
$$
G_x=\{ g \in G | gx=x \}
$$
Hence in our case, we are looking for all the matrices $A \in \mathrm{GL}_2 (\mathbb{R})$ such that $Av=v$, where $v$ is either $\begin{pmatrix} 1 \\ 0\end{pmatrix}$ or $\begin{pmatrix} 1 \\ 1\end{pmatrix}$.
Now, if $A=\begin{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Egyptian fraction representation of $1$ where all denominators of the fractions are odd. Question: Is there an Egyptian fraction representation for $1$ where all the fractions have odd denominators?
I tried to generate one below:
$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{23}+\... | Technique: Find an odd abundant number--more precisely, an odd semi-perfect number: ($945$ is the smallest).
Write it as a sum of its factors: $945=315+189+135+105+63+45+35+27+21+7+3$.
Then divide this equation by the original number: $945$, keeping the terms on the right separated:
$1=\frac13+\frac15+\frac17+\frac19+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find game matrix and optimal strategies for each player. A, B simultaneously show a number of fingers to each other (both hands implies 0-10), money(£) is awarded as follows:
$ \bullet $ If the number of extended fingers is equal, then no exchange
$ \bullet $ If the number of extended fingers between each player differ... | In a mixed-strategy equilibrium, each player must be indifferent among the pure strategies to which she assigns non-zero probability.
In the present case, if all three strategies are assigned non-zero probabilities, this leads to (with the probability for strategy $i$ denoted by $p_i$)
$$
0p_0+3p_1-2p_2=-3p_0+0p_1+3p_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$ Prove the following inequality
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$
I tried by multying both sides by the denominator $(a+b+c)^2$ and then applying Holder for... | There is also the following proof (not mine) by SOS.
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}=\frac{\sum\limits_{cyc}\left((a^2-ab)abc+a(a-b)^2(b-2c)^2\right)}{abc(a+b+c)^2}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Ratio between the area of a region and its transformation by a change of variables. The square $T:\{a<x<a+h, b<y<b+h\}, \text{with } a>0,b>0,h>0$ is tranformed into the region $S$ of the plane $uv$ by the following change of variables:
$$
u=\frac{y^2}{x}
$$
$$
v=x^{1/2}y^{1/2}
$$
Find $\frac{area(S)}{area(T)}$.
This is... | Your solution is the same as the solution in your book.
You have
$$\frac{area(S)}{area(T)}=\frac{1}{h^2}\bigg(-\frac{6}{5}\bigg(\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}}\bigg)\bigg((b+h)^{5/2}-b^{5/2}\bigg)\bigg)$$
Firstly, we get
$$\begin{align}\frac{1}{\sqrt{a+h}}-\frac{1}{\sqrt{a}}&=\frac{\sqrt a-\sqrt{a+h}}{\sqrt a\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Double integral over square shaped region $$\iint_R(y-2x^2)dxdy$$ where $R$ is the region inside the square $|x|+|y|=1$.
So the area is:
\begin{align}
4×\left[\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}(y-2x^2)dydx\right]
&=4×\left[\int_0^1\int_{y=0}^{y=1-x}ydydx-2\int_0^1\int_{y=0}^{y=1-x}x^2dydx\right]\\
&=4×\left[\frac{1}{2}... | HINT
The proposed integral can be expressed as follows:
\begin{align*}
I = \int_{-1}^{0}\int_{-x-1}^{x+1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x + \int_{0}^{1}\int_{x - 1}^{-x + 1}(y - 2x^{2})\mathrm{d}y\mathrm{d}x
\end{align*}
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which value of x $\sum_{n=1}^\infty \frac{n^{nx}}{n!}$ converges I want to know for which values of $x$ this series
$$\sum_{n=1}^\infty \frac{n^{nx}}{n!}$$ converges.
This series is defined $ \forall x \in R$.
$a_n=\frac{n^{nx}}{n!}= \frac{(n^x)^{n}}{n!}= \frac{(n^x)}{n} \frac{(n^x)}{n-1}
\frac{(n^x)}{n-2}...\fr... | For these I always use
$n! \sim (n/e)^n$.
So
$\sum_{n=1}^\infty \frac{n^{nx}}{n!}
\sim \sum_{n=1}^\infty \frac{n^{nx}}{(n/e)^n}
= \sum_{n=1}^\infty (en^x/n)^n
= \sum_{n=1}^\infty (en^{x-1})^n
$
so we want
$en^{x-1} < 1
$
or,
taking logs,
$1+(x-1)\ln(n) < 0$
or
$(x-1)\ln(n) < -1
$.
This is false if
$x \ge 1$,
and is tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6... | This is a program for finding out. There are at least two real roots since there is one between $-1$ and $0,$ by IVT. Then by continuity and the fact that the polynomial is positive for large values of $|x|,$ there must be another real root, clearly a negative one, since for $x>0,$ we have that the polynomial always ra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $n \text{ mod } r < \frac{n}{2}$ I am trying to prove a section of the Euclidean Algorithm for greatest common factors which states:
$\gcd(m, n) = \gcd(n, r) = \gcd(r, n \text{ mod } r)$, where $r =m \text{ mod } n$
Prove that: $n \text{ mod } r < \frac{n}{2}$ (assuming all variables are integers)
My fir... | First, we must assume that $n \not \mid m$, since $n \pmod{0}$ is undefined. Also, assume that $n,m \geq 0$. Note that $r = m \pmod{n} \in \{ 1, 2, \dots , n-1\}$. To show that $ n \pmod{r} < \frac{n}{2}$, we proceed by cases. Suppose that $r < \frac{n}{2}$, then $n \pmod{r} < \frac{n}{2}$ Suppose that $r > \frac{n}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
equation with quadratic power I was thinking how to solve:
*
*If $x^{(x-1)^2}=2x+1$, find $x-\frac{1}{x}$
*Solve $x^{x-x^2+13} = x^2-12$
I noticed that in both problems, the linear part can be constructed in the quadratic exponent, I tried a few change of variable and nothing.
| Following the method of ONG SEE HAI HCI , we have $A = 2x+1$
then:
$$x^{(x-1)^2}=x^{x^2+2-(2x+1)}=2x+1$$
then $x^{x^2+2-A} = A $ with only solution $x=\sqrt{A}$.
This is because $x^{x^2+2-A}$ is increasing.
Then $x=\sqrt{A} \rightarrow x^2 = 2x+1 \rightarrow x = 2+ \frac{1}{x} \rightarrow x-\frac{1}{x} = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Using Laplace transform to solve IVP involving complex roots I am attempting to solve the following IVP :
$$ y'' + y' + \frac54 y =t -(t-\pi /2 )u_{\pi /2}(t) \quad, y(0) =0 ,\quad y'(0) = 0. \tag{1}$$
My reasoning is as follows,
\begin{align*}
\mathcal{L}[y'' + y' + \frac54 y] &= \mathcal{L}[t -(t-\pi /2 )u_{\pi /2}(t... | $$f(s)= \dfrac 1 {s^2(s^2+s+5/4)}$$
Should be decomposed as
$$f(s)= \dfrac A s + \dfrac B{s^2} +\dfrac {Cs+D} {s^2+s+5/4}$$
And both $A,B$ aren't zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Law of Cosines and Heron's Formula in inequalities I had the following question:
Suppose $a$, $b$, and $c$ are non-zero real numbers, and $x$, $y$, and $z$ satisfy the equations
$$
bx + ay = c, cx + az = b, cy + bz = a.
$$
Prove that $-1 < x, y, z < 1$ if and only if $a^4 + b^4 + c^4 < 2 (a^2b^2 + b^2c^2 + c^2a^2)... | I hope the following can help.
Since $$0<a^2b^2c^2=ab\cdot ac\cdot bc,$$ we can assume that $$bc>0.$$
Thus, since $-1<x<1$ and by your work $x=\frac{b^2+c^2-a^2}{2bc},$ we obtain:
$$-1<\frac{b^2+c^2-a^2}{2bc}<1,$$ which gives
$$(a+b+c)(b+c-a)>0$$ and $$(a+b-c)(a+c-b)>0$$ and from here
$$(a+b+c)\prod_{cyc}(a+b-c)>0$$ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find possible coordinates of a triangle that I know only the sides length I want to find the possible coordinates of each vertex of a triangle of which I know only the sides lengths like (3,4,5).
To find the first edge, I let E1(0;0) and the second E2(3;0). but I have a problem to find the 3rd vertex.How can I find the... | Let in triangle $ABC$, we know the side lengths $AB=c$, $AC=b$ and $BC=a$. We look for coordinates of vertices $A$, $B$ and $C$.
So, without loss of generality, we can take $B=(0,0)$, and then $C=(0,a)$ because $BC=a$.
We located to vertices $B$ and $C$ and we need only locate vertex $A$. To do so, suppose the circles ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Could you check my results for entropies? Could you please check my results to the following exericse:
Let $X,Y$ be independent RVs, each with values in $\{0, 1\}$ and probabilities $(1/2,1/2)$. Compute $I(X; Y )$, $I(X + Y ; X)$ and $I(X + Y ; X − Y )$.
Remark: $I$ stand for the mutual information of $X$ and $Y$. It ... | Some sanity checks that we can do without getting into the full calculation:
*
*$H(X+Y,X)$ and $H(X+Y,X-Y)$ should both be at most $2$, because $H(X,Y)$ is $2$, and the pairs $(X+Y,X)$ and $(X+Y,X-Y)$ shouldn't have more entropy than $(X,Y)$: after all, we can compute them from $(X,Y)$!
*In fact, knowing $X+Y$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove :$\tan^{-1}\left(e^{i\theta}\right)=\frac{n\pi}{2}+\frac{\pi}{4}-\frac{i}{2}\ln\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)$
Prove:
$$\tan^{-1}\left(e^{i\theta}\right)=\frac{n\pi}{2}+\frac{\pi}{4}-\frac{i}{2}\ln\left(\tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right)$$
My Attempt :
Is $\ta... | Take the derivative to get
$$\frac{d}{d\theta}\left(\tan^{-1}\left(e^{i\theta}\right)\right) = \frac{ie^{i\theta}}{e^{i2\theta}+1} = \frac{e^{i\theta}}{2}\left(\frac{1}{e^{i\theta}-i}-\frac{1}{e^{i\theta}+i}\right)$$
Now integrate the right hand side
$$\int \frac{e^{i\theta}}{2}\left(\frac{1}{e^{i\theta}-i}-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010
My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1... | Your inequality is the same as
$$\sqrt{(1 - a)(1 - b)}\sqrt{(1 - b)(1 - c)}\sqrt{(1 - a)(1 - c)}
\leq {abc \over 8}$$
By the AM-GM inequality you have
$$\sqrt{(1 - a)(1 - b)} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$\sqrt{(1 - b)(1 - c)} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$\sqrt{(1 - a)(1 - c)} \leq {1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the minimum value of $T=a^2+b^2+c^2+d^2$ Given reals $a,b,c,d$ such that $$\left\{\begin{matrix}(a+b)(c+d)=2
& & \\(a+c)(b+d)=3
& & \\ (a+d)(b+c)=4
& &
\end{matrix}\right..$$
Find the minimum value of $ T=a^2+b^2+c^2+d^2.$
I noticed that $(a+b)(c+d)+(a+c)(b+d)+(a+d)(b+c)=2ab+2ac+2ad+2bc+2bd+2cd\leq 3(a^2... | By your work and by AM-GM we obtain:
$$a^2+b^2+c^2+d^2=(a+b+c+d)^2-2-3-4\geq4(a+d)(b+c)-9=7.$$
The equality occurs for example, for $a+d=b+c=2$,
which gives that the equality indeed occurs:
$$d=2-a,$$ $$c=2-b,$$
$$(a+b)(4-a-b)=2$$ and
$$(a-b+2)(b-a+2)=3.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find a complex root of $x^{2021}=x^{2020}+1$ while this complex root also satisfies a quadratic equation with integer coefficients? How to find a complex root of $x^{2021}=x^{2020}+1$ while this complex root also satisfies a quadratic equation with integer coefficients? I have no previous experience in solving c... | Hint: $x^5 - x^4 - 1 = (x^2 - x + 1) ( x^3 - x - 1)$.
$ $
Hint: $x^3 +1 = (x^2 - x + 1 ) ( x + 1)$.
$ $
Hence, $ x^{6n+k} \equiv x^{k} \pmod{x^2-x+1}$
$ $
Hence $ x^{2021} - x^{2020} - 1 \equiv x^5 - x^4 - 1 \equiv 0 \pmod{x^2-x+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
System of 2 equations Solve the system of equations:
$$
\left\{\begin{matrix}2\sqrt{x+y} = y^2+y-x & \quad(1) \\ x(y^2+y)=(y^4-y^2)^2-2 & \quad(2) \end{matrix}\right.
$$
My attempt:
From $(1)$ I get:
$$(1) \implies (\sqrt{x+y}+y+2)(\sqrt{x+y}-y) = 0$$
So either $\sqrt{x+y}+y+2 = 0 \implies x = y^2+3y+4$ or $\sqrt{x+... | If we substitute $x=y^2-y$ we can factor $(2)$ as
$$(y^2-2)(y^2+1)(y^4-y^2+1)=0$$
and we get $y=\pm\sqrt2,\pm i,\pm w,\pm w^*$ where $w=e^{i\pi/6}$. Only $y=\sqrt2,i,w,w^*$ lead to equality in $(1)$ under the principal square root. If $x=y^2+3y+4$ we get an irreducible octic in $y$ with two real roots and three complex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}... | I like the following way.
Let $x+y=2u$ and $xy=v^2$, where $v^2$ can be negative.
Thus, $$8u^3-6uv^2=7$$ and $$4u^2-2v^2+2u+v^2=4,$$
which gives $$v^2=4u^2+2u-4.$$
But since $$v^2\leq u^2,$$ we obtain:
$$4u^2+2u-4\leq u^2$$ or
$$3u^2+2u-4\leq0$$ or
$$\frac{-1-\sqrt{13}}{3}\leq u\leq\frac{-1+\sqrt{13}}{3}.$$
Thus, $$8u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving functional equation $f(x)=3f(x+1)-3f(x+2)$ It is given that a function f(x) satisfy:
$$f(x)=3f(x+1)-3f(x+2)\quad \text{ and } \quad f(3)=3^{1000}$$ then find value of $f(2019)$.
I further wanted to ask that is there some general method to solve such equation. The method that I know to solve such questions is to... | $$ \begin{align} f(x)&=f(x-1)-\frac{f(x-2)}{3}, f(3)=3^{1000}\\
f(x)&=f(x-1)-\frac{1}{3}f(x-2)\\
f(x+1)&=f(x)-\frac{1}{3}f(x-1)=(f(x-1)-\frac{1}{3}f(x-2))-\frac{1}{3}f(x-1)\\
f(x+1) &= \frac{2}{3}f(x-1)-\frac{1}{3}f(x-2)\\
f(x+2)&=f(x+1)-\frac{1}{3}f(x)=(\frac{2}{3}f(x-1)-\frac{1}{3}f(x-2))-\frac{1}{3}(f(x-1)-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Having problems understanding the correct answer on a classic urn probability question. There are two urns, one urn $U_1$ containing $3$ black balls $B$ and $6$ white balls $B^c$, while the other urn $U_2$, contains $100$ white balls. An urn is selected uniformly at random and then a ball is drawn uniformly at random ... | If you are looking for an intuitive explanation, you can proceed this way. Suppose that the experiment is run $90$ times. $45$ times we choose urn $2$, and the first ball will be white. $45$ times we choose urn $1$. $30$ times the first ball is white, and $15$ times the first ball is black, but that didn't happen! ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
| I use the following trick.
If $f(x)$ is a polynomial, then $f(x)f(-x)$ is a polynomial with only even degree terms, so there exists a polynomial $g(x)$ such that
$$g(x^2)=f(x)f(-x).$$ Then the zeros of $g(x)$ are exactly the squares of the zeros of $f(x)$.
With $f(x)=x^4+px^3+qx^2+rx+1$ we arrive at
$$
g(x)=1 + (2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Calculating triple integral over volume common to 2 surfaces In one of my assignments, I have a question which asks you for the triple integral of $z^2$ over the volume common to a given sphere and a cylinder.
I need to transform the equation to cylindrical polar form. Also I am not able to get the limits. In the pol... | The tricky part here is figuring out how the volume is described in cylindrical coordinates (with $\rho$, $\varphi$ and $z$).
The volume is bounded below and above by the sphere. Using the equation of the sphere, $x^2+y^2+z^2=a^2$, we get:
\begin{align*}
&x^2+y^2+z^2=a^2\\
\iff\quad&z^2=a^2-x^2-y^2\\
\iff\quad&z=\pm\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $a,b,c$ be side lengths of a triangle, $a+b+c=1$. Prove that $P=a^3+b^3+c^3+3abc<\frac{1}{4}$.
Let $a,b,c$ be side lengths of a triangle such that $a+b+c=1$. Prove that $$a^3+b^3+c^3+3abc<\frac{1}{4}\,.$$
I solved this question. However, I'd like to know if there's a neater solution that doesn't involve the subst... | A solution without substitutions.
We need to prove that $$\sum_{cyc}(4a^3+4abc)<(a+b+c)^3$$ or
$$\sum_{cyc}(3a^3-3a^2b-3a^2c+2abc)<0$$ or $$-3\prod_{cyc}(a+b-c)<0,$$ which is obvios.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why does this functional equation follow from periodicity? I'm reading a proof where they say that given
$$\phi(x) = \Gamma(x)\Gamma(1-x)\sin \pi x$$
and
$$g(x) = [\log \phi(x)]''$$
then, since $g$ is periodic with period 1, it satisfies the functional equation
$$\frac{1}{4} \left(g\left(\frac{x}{2}\right) + g\left(\fr... | I figured it out myself. The only way seems to be by brute force.
$$g(x)=\frac{1}{\phi\left(\frac{x}{2}\right)^2
\phi\left(\frac{x+1}{2}\right)^2}\left[\phi\left(\frac{x}{2}\right) \phi\left(\frac{x+1}{2}\right) \left(\frac{1}{4}
\phi\left(\frac{x+1}{2}\right) \phi''\left(\frac{x}{2}\right)+\frac{1}{4}
\phi\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
3x3 system of equations I have the following system of equations:
\begin{equation}
x+y+kz = 1
\end{equation}
\begin{equation}
x+ky+z = 1
\end{equation}
\begin{equation}
kx+y+z =1
\end{equation}
where $x,y,z \in \mathbf{R}$.
For what values of $k \in \mathbf{R} $
i) the system has a single solution
ii) the system has ... | $$\Delta=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ k & 1 & 1 \end{array}\right)\right|=-k^3+3k-2=-(k+2)(k-1)^2,$$
$$\Delta_x=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & k & 1\\ 1 & 1 & 1 \end{array}\right)\right|=-(k-1)^2,$$
$$\Delta_y=\left|\left(\begin{array}{cc}1 & 1 & k\\1 & 1 & 1\\ k & 1 & 1 \end{array}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculus: $\lim_{x \to 1} \frac{x^3 - 1}{x-1}$ Is the limit $0$ or $3$?
$x^3 -1$ can be $(x-1)(x^2 +x +1)$, with the transformation, the limit will be $3$?
Why cannot we just say $x$ is to be $1$, $x^3 -1$ is going to be $0$ so the limit is $0$?
| You have shown that you can see why the limit of the expression should be $3$. Regarding why it shouldn't be $0$, consider how $\displaystyle \frac{x^3-1}{x-1}=\left(x^2+x+1\right)\frac{x-1}{x-1}$ would be evaluated at different $x$-values. At $x=3$, we have $\require{cancel}\displaystyle \left(3^2+3+1\right)\frac{\can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem... | If you are able to use theorems from complex analysis (although it is likely you can not) then this integral is easily solved using the residue theorem. Since such an answer might be useful to others, I'll put it here for posterity.
First, note that
$$\int_{-\infty}^{\infty}\frac{z^2}{(z^2+a^2)(z^2+b^2)}dz=\lim_{R\to\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is: If $a,b,c$ are the sides of a triangle, then $\dfrac{a}{b+c-a}+\dfrac{b}{c+a-b}+\dfrac{c}{a+b-c}$ is:
$A)$ $\le3$ , $B)$ $\ge3$, $(C)$ $\ge2$, $(D)$ $\le2$
My attempt is as follows:-
$$\dfrac{1}{2}\left(\dfrac{a}{s-a}+... | By C-S $$\sum_{cyc}\frac{a}{b+c-a}=\sum_{cyc}\frac{a^2}{a(b+c-a)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}a(b+c-a)}\geq3,$$
where the last inequality it's just $$\sum_{cyc}(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Umbilical points of the ellipsoid. The following is Exercise (11) of Chapter 3 of Curves and Surfaces, 2nd edition, by Montiel and Ros:
Determine the umbilical points of the ellipsoid of equation
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1
$$
where $0 < a < b < c$.
My question is:
Why the umbilica... | I think I got this.
Suppose $x = 0$. From $N(p) \cdot v = 0$ we have that
$$
v_3 = - \frac{y}{b^2} \frac{c^2}{z}v_2.
$$
Then
$$
q(v) = \left( \frac{z}{c^2} \left( \frac{1}{a^2} - \frac{1}{b^2} \right) - \frac{y}{b^2} \frac{c^2}{z} \left( \frac{1}{c^2} - \frac{1}{a^2}\right) \right) v_1 v_2 = 0 \quad \forall v_1, v_2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Pair of tangents from point $(2\sqrt2,1)$ to hyperbola $\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$
From a point $(2\sqrt2,1)$ a pair of tangents are drawn to $$\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$$ which intersect the coordinate axes in concyclic points. If one of the tangents is inclined at an angle of $\arctan\frac{1}{\... | Given the point $(2\sqrt2,1)$ and the slopes $\frac1{\sqrt2}$, $m$, the equations of the two tangent lines are
$$ y-1 =\frac1{\sqrt2}( x-2\sqrt2), \>\>\>\>\>y-1 = m(x-2\sqrt2)$$
which intersect the axes at $A(\sqrt2,0)$, $B(0,-1)$ and $C(2\sqrt2-\frac1m, 0)$, $D(0, 1-2\sqrt2m)$, respectively. Given that $A$, $B$, $C$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Restrictions on a Function If $f(x) =$
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
then for how many values of $x$ is $f(f(x)) = 5$?
I'm not sure how to really start from this question other than bashing values. I need some help on a start, thanks!
| Just do it in cases:
Case 1: $f(x) < -4$ then $f(f(x)) = 5$ means
$f(x) + 3 = 5$ so $f(x) =2$ and that's a contradiction.
So it must be that:
Case 2: $f(x) \ge -4$ so $f(f(x)) = 5$ means
$f(x)^2 -4 = 5$
$f(x)^2 = 9$
$f(x) = \pm 3$.
Case 3: $x < -4$ so $f(x) = \pm 3$ means $x+3\pm 3$ meas $x=-6, 0$ but only $-6< -4$ s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
A Question Lebesgue Dominated Convergence Theorem I have a question about the application of Lebesgue Dominated Convergence Theorem.
$\lim _{n \rightarrow \infty} \int_{0}^{1}\left(1+n x^{2}\right)\left(1+x^{2}\right)^{-n} d x=?$
Firstly i have a reference about this question http://www.ma.man.ac.uk/~mdc/old/341/solu... | You can make this easier: since $(1+x^2)^n \ge 1 +nx^2,$ we have
$$\frac{1+nx^2}{(1+x^2)^n} \le \frac{1+nx^2}{1+nx^2} =1.$$
So the constant function $1,$ which is certainly integrable on $[0,1],$ is a dominating function for your sequence.
As you said, we also have
$$\frac{1+nx^2}{(1+x^2)^n} \le \frac{1+nx^2}{1+nx^2+[n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I find $ \int \frac{6}{5\sin^{2}x+4}dx $? One of the homework questions I am working on asks to show that $ \frac{\partial }{\partial x} \left(\tan^{-1}\frac{3\tan x}{2}\right) = \frac{6}{5\sin^{2}x+4} $ which I am able to do. However, I wonder how I should approach the reverse version of this question if it... | \begin{align*}
I&=\int \frac{6}{5\sin^2x+4}\,\mathrm dx\\
&=\int \frac{6}{9\sin^2x+4\cos^2x}\,\mathrm dx\\
&=\int \frac{6}{9\tan^2x+4}\frac{1}{\cos^2x}\,\mathrm dx\\
&=(let\ t = \tan(x) )\quad\int \frac{6}{9t^2+4}\,\mathrm dt\\
&=(let\ p = \frac {3}{2}t)\quad\int \frac{1}{p^2+1}\,\mathrm dp\\
&=\arctan(p)+C\\
&=(p = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Vieta's formulas for quadratic equation problem I'm using one hack, which I never though of why it works. But now I'm curious why it's works and how I can prove it.
Here's the deal: we have quadratic equation $ax^2 + bx + c = 0$, to find roots I just multiply $c$ by $a$ and solving $y^2 + by + ca = 0$, and then I divid... | By the quadratic formula the roots of $aX^2+bX+c$ are $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
The roots of $Y^2+bY+ca$ are $y=\frac{-b\pm \sqrt{b^2-4ca}}{2}$...
We see that $x=\frac1a y$ ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Kernel and image of a product of two rectangular matrices I cannot find any concise and effective answer to the following problem :
Problem : given two matrices $A$ and $B$ of size $3\times 2$ and $2\times 3$ such that $AB = \left(\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 2 \end{matrix}\right)$ find the kernel a... | Notice that rank($AB$) $=2$
This implies $A$ and $B$ are full rank, i.e. $B$ is surjective and $A$ is injective.
Since $A$ is injective, Ker $(AB) = $ Ker $B$
Since $B$ is surjective, Im $(AB) = $ Im $A$
Since $\left(\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} , \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding residue of complex function, the result is different when using Laurent series and residue theorem. Find the residue of
$$f(z)=\dfrac{z^3+2z+1}{(z-1)(z+3)}$$
on simple pole $z=1$.
If I using residue theorem, I have
\begin{align}
\underset{z=1} {\operatorname{Res}} f(z) =
\lim\limits_{z\to 1} (z-1)\dfrac{z^3+2z+... | The "Laurent series method" had an incorrect conclusion going from the fourth to the fifth lines. There was a still an order $1$ term on the far right, it just still had to be pulled out.
We can calculate the Laurent series another way. From partial fraction decomposition we have that
$$\frac{4}{(z-1)(z+3)} = \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how to find directional derivative I am trying to find the directional derivative of the following problem
$F(x,y,z) = 4x^2+ 3y−3xz+ 2z^2$
at the point $(2,1,2)$ in the direction $i−k$;
I worked out the derivatives of $F(x,y,z)$ as
$f_x = 8x -3z$
$f_y = 3$
$f_z = -3x+4z$
But I don't know to do next; can someone expl... | Let the gradient $\nabla F$ of $F$ be
$$\nabla F = (f_x, f_y, f_z).$$
Then the gradient evaluated at $(2,1,2)$,
$$\nabla F(2, 1, 2) = (8 \cdot 2 - 3 \cdot 2, 3, -3 \cdot 2 + 4 \cdot 2) = (10, 3, 2)$$
can be dotted with the normalized direction $\frac{1}{\sqrt{2}}(1, 0, -1) = \frac{1}{\sqrt{2}}(i - k)$ to arrive at the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3648415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +... | As we can see that the term $9(abc)^{\frac{1}{3}}$ is difficult to deal with, we consider the following inequality $$-\sqrt{(a^2-ab+b^2)} -\sqrt{(b^2+c^2-bc)} -\sqrt{(a^2+c^2-ac)} +4(a+b+c)\ge 9(abc)^{1/3}.$$
Then we consider $$\frac{a+b}{2}-\sqrt{(a^2-ab+b^2)}=\frac{(\frac{a+b}{2})^2-(a^2-ab+b^2)}{\frac{a+b}{2}+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove $\log_{a} c + \log_{b} c = \log_{a+b} c$ if and only if $1 + \log_{b} a = \log_{a+b} a$ If a, b and c are positive numbers, than equality $$\log_{a} c + \log_{b} c = \log_{a+b} c$$ is true if and only if $$1 + \log_{b} a = \log_{a+b} a$$
Prove it!
I have looked at the solution but it is not clear for me.
We will ... | Note that
$$log_b (a)= \frac{ln (a)}{ln(b)}$$
where
$ln(a)$ is the natural log of $a$.
Any arbitrary base can be assumed instead of $e$.
So, if
$$log_a (c)+ log_b (c)= log_{a+b} (c) $$
$$\implies\frac{ln(c)}{ln(a)}+\frac{ln(c)}{ln(b)}=\frac{ln(c)}{ln(a+b)}$$
$$\implies\frac{1}{ln(a)}+\frac{1}{ln(b)}=\frac{1}{ln(a+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.
| Just synthetic divide and subtract and add till you reduce to something manageable.
$6x^3 -x^2 -3x + 2010 = $
$6x^3+ 2x^2 - 3x^2 - 3x + 2010=$
$2x(3x^3 + x^2) - 3x^2 - 3x+2010=$
$2x - 3x^2 - 3x + 2010=$
$-3x^2 - x + 2010=$
$-(3x^2 + x) +x - x +2010=$
$-1 + 0 + 2010=$
$2009$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
At what values of the parameter a the antiderivative of the function has at most one local minimum. Function below. At what values of the parameter a the antiderivative of the function has at most one local minimum?
Function:
(x^4 -(3+a)x^3+(2+3a)*x^2-2ax)*exp(sin(x)/(x^2+3))
or you can look here - 1
But what to do ... | let $F(x)$ be the anti derivative of the given function. Then
$$F'(x) = (x^4 -(3+a)x^3+(2+3a)*x^2-2ax)*exp(sin(x)/(x^2+3)) $$
You can see clearly that $F'(x)=0$ at x = 0.
Now when is F''(x)>0 ?
So,F"(x)>0 only when the function is "increasing" at $x$.
https://www.desmos.com/calculator/mtjvcgkkov
If you go on to this g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
ODE: $ \biggl( e^x + \frac{A\sin x}{y} \biggl) dx + \biggl(\frac{B\cos x}{y^2} + (B-1)e^x \biggl)dy =0$ Question says equation is exact, find A and B.
$$ \biggl( e^x + \frac{A\sin x}{y} \biggl) dx + \biggl(\frac{B\cos x}{y^2} + (B-1)e^x \biggl)dy =0$$
$$\frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\parti... | At your last step, it should be
$$\frac{(B-A)\sin(x)}{y^2} =(B-1)e^x$$
which holds when
$$\begin{cases}
B-A=0\\ B-1=0
\end{cases}$$
and therefore $B=1$ and $A=B=1$. Then
$$dF=\biggl( e^x + \frac{\sin(x)}{y} \biggl) dx + \biggl(\frac{\cos(x)}{y^2} \biggl)dy$$
where $F(x,y)=e^x-\frac{\cos(x)}{y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt fo... | As $x\to 0$,
$$1 - \cos^2(x) = \sin^2(x) \sim x^2$$
Use this to simplify thigs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$? Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?
Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, bu... | For integer $a ≥ 0$ define $f(a) = (4a + 1) / (2a + 1)$.
Let $g_k(a) = f(a) \cdot f(2^1a) \cdot f(2^2a) \cdot … \cdot f(2^{k-1}a)$. $g_k(a)$ is a telescopic product, and $g_k(a) = (2^{k+1}a + 1) / (2a+1)$.
We have $g_k((b-1) / 2) = (2^k b - 2^k + 1) / b$.
We have $g_k((cy - 1) / 2) = (2^k cy - (2^k - 1)) / cy = 2^k - (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Prove $\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$ with complex numbers (roots of unity) I want to prove that
Using $z^9=1$ and the fact that $1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8=0$ where $w=cis(\frac{2π}{9})$, that $$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$$
I am able to do this b... | Start out as by noticing that
$$
\omega = cis(\frac{2\pi}{9}),\quad \omega^{8}=cis(\frac{-2\pi}{9})\quad \\
\omega^2 = cis(\frac{4\pi}{9}),\quad\omega^7=cis(\frac{-4\pi}{9})
\\ etc.
$$
Now see,
$$ \omega\ + \omega^8 = 2\ cos(\frac{2\pi}{9}) \\
\omega^2\ + \omega^7= 2\ cos(\frac{4\pi}{9})\\
\omega^3 + \omega^6 = 2\ c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does this function only have 3 rational solution? Why does the function $f(x,y) = x^3 - x - y^2$ only have the three rational roots
$(0,0)$, $(1,0)$, $(-1,0)$?
| We have $f(x,y)=x^3-x-y^2$. We define a pair of roots $(x,y)$ to satisfy $f(x,y)=0$. Then:
$$x^3-x=y^2$$
Thus, we are looking for rational solutions for the above equation. Let $x=\frac{m}{n}$ where $\gcd(m,n)=1$ and $n>0$. Assume $y^2 > 0$.
$$x^3-x=x(x^2-1)=x(x-1)(x+1)=\frac{m(m-n)(m+n)}{n}=y^2$$
We have the numbers $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int _{-\infty }^{\infty }\frac{x^3\sin(x)}{x^4-b^4}\,\mathrm dx$ with real methods How can i prove with real methods that $\int_{-\infty}^\infty\frac{x^3\sin(x)}{x^4-b^4}\,\mathrm dx=\frac\pi2(e^{-b}+\cos(b))$? I was able to prove this using complex analysis but i dont know how to attack with without it.
As s... | The characteristic equation of
$$ I''''(t)-b^4I(t)=0 \tag1$$
is
$$ r^4-b^4=0. \tag2$$
(2) has roots $\pm b,\pm bi$ and hence (1) has the general solution
$$ I(t)=C_1e^{bt}+C_2e^{-bt}+C_3\sin(bt)+C_4\cos(bt). $$
Noting
$$I'\left(a\right)=b^4\int _{-\infty }^{\infty }\frac{\cos \left(ax\right)}{x^4-b^4}\:dx$$
one has $I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int\frac{4x^4}{{x^8+1}}\;dx$ I'm attempting this integral but I'm unsure how to proceed. I've begun to suspect it's actually non-elementary. Can anyone do this? I could always use Taylor series after a small bit of u-sub and integration by parts, but I wanted to know if there were a somewhat more direct way ... | Not an answer. Maybe this result inspires a by-hand method. The integral is "elementary".
\begin{align*}
&\int \frac{4x^4}{x^8+1} \,\mathrm{d}x \\
&= 4 \left(-\frac{1}{8} \cos \left(\frac{\pi }{8}\right) \log \left(x^2-2 x \sin
\left(\frac{\pi }{8}\right)+1\right)
\right. \\ &\quad \left. {}
+\frac{1}{8} \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3665818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and ... | Following the steps given in Wikipedia, we can set the problem as
$$\text{Minimize}[a^2+b^2,\{a,b\}]$$
subject to the three inequality constraints
$$P=16-3a^2 <0$$
$$D=-3 a^4+32 a^2-16 a b \leq 0$$
$$\Delta=-27 a^4-4 a^3 b^3+36 a^3 b-2 a^2 b^2+256 a^2+36 a b^3-512 a b-27 b^4+256 b^2 \leq 0$$
This is a quite simple task... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How To Prove $\int_0^1 (\ln x)^2 \frac{1+x^2}{1+x^4} \, dx=\frac {3\sqrt{2}\pi^3}{64}$ Question:-Prove that $$\int_0^1( \ln x)^2 \frac{1+x^2}{1+x^4} \, dx=\frac {3\sqrt{2}\pi^3}{64}$$
My attempt:
$$\int_0^1 \left( \ln\frac{1}{x}\right)^2 \frac{1+x^{-2}}{x^{-2}+x^2} \, dx$$
Put $\left(x-\frac{1}{x}\right)=t$, we get
$$\... | Let $f(x)=\frac{1+x^2}{1+x^4}\ln^2 x$. By the substitution $x\mapsto\frac1x$, it can proved that $I=\int^1_0f(x)dx=\int^\infty_1f(x)dx$. Hence $2I=\int^\infty_0 f(x)dx$.
Employing Feynman's technique,
$$2I=\frac{\partial^2}{\partial a^2}\underbrace{\int^\infty_0\frac{1+x^2}{1+x^4}x^adx}_{J}\bigg\vert_{a=0}$$
We rewrite... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1+\frac{1}{2^{11}}+\frac{1}{3^{11}}+\frac{1}{4^{11}}+...}{1-\frac{1}{2^{11}}+\frac{1}{3^{11}}-\frac{1}{4^{11}}+...}$ I'm not sure how to solve this summation and would like assistance. I believe the answer is ||$\frac{1024}{1023}$|| according to Wolfram. I don't know what the proper approach is (creating two s... | I will assume that the +1 in the denominator is a -1 and go on,
we can write $\frac{1}{1}+\frac{1}{2^{11}}+\frac{1}{3^{11}}+.......=S$
now we can take the even terms to one side and express it as
$\frac{1}{1}$+$\frac{1}{3^{11}}$+$\frac{1}{5^{11}}$=S +$\frac{1}{2^{11}}(S)$
$\frac{1}{1}+\frac{1}{3^{11}}+\frac{1}{5^{11}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Diophantine equation with power Find all the integer solutions of:
$2^{n+1} + 41 = m^2$
I am stuck, and I am not sure if I am going on the right path..
adding 8 to both sides:
$2^{n+1} + 41 + 8 = m^2 + 8$
$2^{n+1} + 49 = m^2 + 8$
$2^{n+1} - 8 = m^2 -49$
$2(2^n - 4) = (m+7)(m-7)$
I am not sure on how to keep goi... | Some insight can be gained by looking at the equation $\bmod 40$. Note that for $n\ge 1$, the residues of $2^{n+1} \bmod 40$ are $\{4,8,16,32,24\}$ with $4$ occurring only once and the other values cycling. Since $4$ does not yield a solution ($4+41\ne m^2$), we can ignore its single occurrence. Of course, $41\equiv 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An equivalent of Jacobi's two square theorem Jacobi's two square theorem:
The number of representations of $n$ as a sum of two squares is $4$ times the difference
between the number of divisors of $n$ congruent to $1$ modulo $4$ and the number of divisors
of $n$ congruent to $3$ modulo $4$.
This is equivalent to
... | You remark about convergence in a comment. Convergence is not relevant. These series are recording combinatorial information in their coefficients. They are formal power series, recording data in their coefficients.
Let us start with a smaller version of the last sum.
\begin{align*}
\sum_{n=-4}^4 q^{n^2}
&= q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3682346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that for all $\alpha + \beta + \gamma = \pi$, $\sum_{cyc}\frac{\sin\beta}{\cos\beta + 1} = \frac{\sum_{cyc}\cos\beta + 3}{\sum_{cyc}\sin\beta}$.
Prove that for all triangles with angles $\alpha, \beta, \gamma$, $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma +... | Using your transformation I a get sligtly different equation:
$$a+b+c=\frac{\sum_{cyc}\frac{1}{1+a^2}}{\sum_{cyc}\frac{a}{1+a^2}}$$
$$\sum_{cyc}\frac{a(a+b+c)}{1+a^2}=\sum_{cyc}\frac{1}{1+a^2}$$
$$\sum_{cyc}\frac{a^2+ab+ac-1}{(a+b)(a+c)}=0$$
$$\sum_{cyc}(a^2-bc)(b+c)=0$$
$$\sum_{cyc}a^2b+a^2c-b^2c-bc^2=0$$
Which is tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Extreme values of $|z|$ when $|z^2+1|=|z-1|$ Problem statement: Find the extreme values of $|z|$ when $$|z^2+1|=|z-1|,\ z\in \mathbb{C}-\{0\}$$
My try:
$$|z^2+1|=|z-1|\implies|z^2+1|^2=|z-1|^2\implies(z^2+1)(\overline{z}^2+1)=(z-1)(\overline{z}-1)$$
$$\implies|z|^4+(z+\overline{z})^2=3|z|^2-(z+\overline{z})\cdots\text{... | (this solution mostly is similar to what you have, which is fine, but you can avoid the quadratic equation and bounding)
Observe that
$$|z^2+1|=|z-1| \iff (z^2 + 1)(\overline{z}^2 + 1) = (z - 1)(\overline{z}-1) \iff |z|^4 - |z|^2 + (z^2 + \overline{z}^2 + z + \overline{z}) = 0$$
Note that for all pairs of reals $(a, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How $\int_1^\infty \sin \left( \frac{2}{x^{5/3}} \right)dx $ absolute converge? I have the integral $$\displaystyle{\int_1^\infty {\sin \left( \frac{2}{x^{\frac{5}{3}}} \right)}dx }$$
I know it is converge. but when I check absolute converge I did the following: according to the trigonometric identity :$\cos(2\theta)=1... | Put $y=\frac 2 {x^{5/3}}$. We get $\int_1^{\infty}| \sin (\frac 2 {x^{5/3}})|dx= 2^{-2/5}\frac 3 5 \int_0^{2} |\sin y| y^{-8/5}dy$. Now use the fact that $\frac {\sin y} y \to 1$ as $ y \to 0$ and $\int_0^{2} y^{-3/5}dy <\infty$ to see that the given inetgral is absolutely convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $g(x)$ if $f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$ $$f:[-2,2] \rightarrow \Bbb R$$
$$\text {and }f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$$
And, let $g(x)$ be equal to $|f(x)|+f(|x|)$
We need to find the value... | Your final answer appears to be correct (though I have not checked all the steps of your reasoning), however you may try to apply symmetry by dividing into three cases: $x=0$, $-2\leq x<0$, and $0<x\leq 2$. Then it would be easy to see how to get $g(x)$, since $|x|$ for $x$ in the second case implies $|x|$ belongs to t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find sum of the series $ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdots$ I have to find the sum of the given series $$ S=1 + \frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13} +\cdot... | Since $S=1+\sum_{n\ge1}\int_0^1x^{4n-2}(1-2x+x^2)dx$, by monotone convergence$$S=1+\int_0^1\frac{x^2(1-x)^2dx}{1-x^4}=1+\int_0^1\frac{x^2(1-x)dx}{(1+x)(1+x^2)}.$$You can do the rest with partial fractions:$$S=1+\left[-x+\ln(x+1)+\frac12\ln(x^2+1)\right]_0^1=\frac32\ln2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by T... | A possible way to show convergence is to rewrite
$$\cos \frac 1n = \cos \frac{2}{2n} = 1- 2\sin^2 \frac 1{2n} $$
and now use root test and the standard limits $\lim_{t\to 0}(1-t)^{\frac 1t} = \frac 1e$ and $\lim_{t\to 0}\frac{\sin t}{t}=1$:
\begin{eqnarray*}\sqrt[n]{\left(1- 2\sin^2 \frac 1{2n}\right)^{n^3}}
& = & \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find $\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$ Need solve the next integral
$$\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$$
Tried something with Laurent’s series, but i can’t conclude anything.
Thanks
| METHODOLGY $1$: CONTOUR INTEGRATION
Enforce the substitution $x\mapsto x^2$ to find that
$$\int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx=4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\tag1$$
Let $f(z)$ be given by
$$f(z)=\oint_C \frac{\log^2(z)}{(z^2+1)^2}\,dz\tag2$$
where we choose to cut the plane along the posit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
} |
there do not exist intgers $a,b,c,d,$ with $k>1$ such that $(a+bw+cw^2+dw^3)^k=1+w$
let $w=e^{\frac{2\pi\cdot i}{5}}$ be a primitive fifth root of unity,Prove that there do not exist intgers $a,b,c,d,$ with $k>1$ such that
$$(a+bw+cw^2+dw^3)^k=1+w$$
I try:let $x=a+bw+cw+dw^3(a,b,c,d\in Z)$ and note that $w+w^{-1}=\... | Too long for a comment
Remark: Proceed along the OP's approach. Not sure if it works.
From the last equation of the OP, we have $(\frac{C}{2} + \frac{D}{2}\sqrt{5})^k = \frac{3}{2} + \frac{1}{2}\sqrt{5}$
where
\begin{align}
C &= 2 a^2-a b-a c-a d+2 b^2-b c-b d+2 c^2-c d+2 d^2, \\
D &= a b-a c-a d+b c-b d+c d.
\end{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$
Prove that $0$ is the only $2\pi$-periodic solution of $\ddot{x}+3x+x^3=0$.
I don't know how to deal with this non-linear differential equation. I tried to consider $\ddot{x}(t+2\pi)+3x(t+2\pi)+x^3(t+2\pi)=0$ but with no success...
I need to p... | Letting $y=\dot{x}$ and $g(x)=3x+x^3$ we obtained the following first order planar system
$$
\begin{pmatrix} \dot{y} \\
\dot{x}
\end{pmatrix} = \begin{pmatrix} -3x - x^3\\ y
\end{pmatrix} = \begin{pmatrix} - g(x) \\ y
\end{pmatrix} \tag{1}\label{one}
$$
This is a Hamiltonian system
$$
\begin{pmatrix} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I meet a problem ,it says
Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then
\begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)}
\end{align*}
Fix $a,b,c$,then ... | Let $ab+ac+bc=1$.
Thus, we need to prove that:
$$\sum_{cyc}\sqrt{a^2+1}\geq\sqrt3+a+b+c.$$
Now, let $a=\tan x$, $b=\tan y$ and $c=\tan z$, where $\{x,y,z\}\subset\left(0^{\circ},90^{\circ}\right).$
Thus, $x+y+z=90^{\circ}$ and we need to prove that $$\sum_{cyc}f(x)\geq\sqrt3,$$ where $$f(x)=\frac{1}{\cos{x}}-\tan{x}.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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.