Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solution to the $\displaystyle \phi'' + 4\phi' + 3\phi = x \cos {3x}. $ I came across the following differential equation while practicing $ \displaystyle \phi'' + 4\phi' + 3\phi = x \cos {3x} .$
Now, I used VARIATION OF PARAMETERS to solve it. I found general solution the equation $ \displaystyle \phi'' + 4\phi' + 3... | You can also apply the method of undetermined coefficients to find the particular solution. The solution to the complementary differential equation $\phi'' + 4\phi' + 3\phi=0$ is
$$\phi_c(x) =c_1e^{-3x}+c_2e^{-x}.$$
Thus to find the particular solution to $\phi'' + 4\phi' + 3\phi = x \cos {3x}$ we try a particular so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Why do the rounded solutions of $n^x=x!$ follow this $2, 3, 3$ pattern as $n$ increases? Let $y$ be the solution to the following equation:
$n^x = x!$ where $n$ is an integer. I have observed that if you round $y$ to the nearest integer it follows a certain pattern. Let $c$ be the nearest integer to $y$. $c$ increases ... | Your $n^x = x!$ is equivalent to $n = (x!)^{1/x}$
Stirling's approximation gives $x! \approx \sqrt{2\pi x}\left(\frac x e\right)^x$
so $(x!)^{1/x} \approx (2\pi x)^{1/(2x)}\dfrac x e$
For large $x$ the first term heads to $1$ so for large $n$ you have $n \approx \dfrac x e$ and $x \approx n e$
Your $2,3,3$ pattern has... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Finding number of solutions of $f(x)=f^{-1}(x)$ where $f(x)=x^3+x-1$
Let $f:\mathbb R\to\mathbb R, f(x)=x^3+x-1$. Find the number of solutions of the equation $f(x)=f^{-1}(x)$.
$f'(x)=3x^2+1\gt0\implies f(x)$ is an increasing function. From this can we say anything about the nature of $f^{-1}(x)$?
I know that a funct... | If $f(x)=f^{-1}(x)$ then $f(f(x))=x$, where
\begin{eqnarray*}
f(f(x))&=&f(x)^3+f(x)-1\\
&=&(x^3+x-1)^3+(x^3+x-1)-1\\
&=&x^9+3x^7-3x^6+3x^5-6x^4+5x^3-3x^2+4x-3.
\end{eqnarray*}
So you want to find the real roots of
$$x^9+3x^7-3x^6+3x^5-6x^4+5x^3-3x^2+3x-3=0.$$
The above is equivalent to
$$(x^3+x-1)^3+(x^3+x-1)-1=x,$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ Let $A,B,C$ be the three angles of a triangle. Prove that:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin... | Put $x=$sin $(\frac {A}{2}) $ and similarly define $y $ and $z $. Then required inequality becomes
$2 (xy+yz+zx)\leq (x+y+z) $ which translates to $s^2\leq s+x^2+y^2+z^2$ where $s=x+y+z $. Since $\sum_{cyc}x^2\geq \frac {s^2}{3} $, it is enough to prove that $3s^2\leq 3s+s^2$. Sunce $s>0$ it is enough to prove that $s\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In the figure, the polygons shown are regular. Find $"x"$. For reference:
my progress:
I marked the angles I could find but couldn't finish
$a_{i9} = \frac{180(n-2)}{n}=\frac{180(9-2}{9}=\angle140^\circ\\a_{i6}= \frac{180(6-2)}{6}=\angle120^\circ\\\angle ADB = \frac{360^\circ-2(140^\circ)}{2} = \angle 40^\circ\\\angle... | Hint: $\triangle BAN$ is isosceles (because $ABCN$ is a rhombus), and therefore so is $\triangle NAL$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see: Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y... | We can use the $uvw$'s technique here.
Indeed, let $\dfrac{9(63+5\sqrt{105})}{32}=k$, $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Thus, we need to prove that:
$$ku^4\geq\sum_{cyc}x^3y$$ or
$$2ku^4-\sum_{cyc}(x^3y+x^3z)\geq\sum_{cyc}(x^3y-x^3z)$$ or
$$2ku^4-(27u^2v^2-18v^4-3uw^3)\geq3u(x-y)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0}\frac{1-\cos(x^2)}{x^3(4^x-1)}$
I need to calculate
$$\lim_{x\rightarrow 0}\frac{1-\cos(x^2)}{x^3(4^x-1)}$$
and the options are:
(a) $\frac 12 \ln 2\quad$ (b) $\ln 2\quad$ (c) $\ln 4\quad$ (d) $1 - \frac 12 \ln \left( \frac{e^2}{4}\right)$.
The answers would are given to be $b$ and $d$
I trie... | As @Joe said, it would be simpler to write it as $\frac 1{2 \ln (4)}$, but there is nothing wrong with your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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$(A^{-1}+B^{-1})^{-1}=A-A(A+B)^{-1}A$
Let $A, B$ and $A+B$ be invertible matrices. Show that
$(A^{-1}+B^{-1})^{-1}=A-A(A+B)^{-1}A$
Here's my assumption:
\begin{align}
A^{-1} + B^{-1} &= B^{-1} + A^{-1} \\
&= B^{-1}(A + B)A^{-1} \\
\end{align}
Then
\begin{align}
(A^{-1} + B^{-1})^{-1} &= [B^{-1}(A + B)A^{-1}]^{-1} \\
... | From $(A^{-1}+B^{-1})^{-1}=A(A+B)^{-1}B$, you get
$A(A+B)^{-1}B=A(A+B)^{-1}(B+A-A)=A-A(A+B)^{-1}A$
Note: Besides invertibility of $A$ and $B$, you may need to add the assumption that $(A+B)$ is invertible too. From that, following your reasoning, it follows that $(A^{-1}+B^{-1})$ is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Getting strict inequality in $|f(x)|\le \frac 1x$, $x\gt 0$, where $f(x)=\int_x^{x+1} \sin t^2 \, dt$ Given that $f(x)=\int_x^{x+1} \sin t^2 \, dt$
Substituting $u=t^2$ gives:
$f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}... | You correctly derived that
$$
f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du \, .
$$
It follows that
$$
|f(x)| \le\frac{|\cos x^2|}{2x}+\frac{|\cos (x+1)^2|}{2(x+1)}+\int_{x^2}^{(x+1)^2}\frac{|\cos u|}{4u^{3/2}}\, du ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the general solution and the specific solution of the following differential equation: $(3y - 1)^2) (y'^2) = 4y$ Calculate the general solution and the specific solution of the following differential equation: $(3y - 1)^2 (y')^2 = 4y.$
The general solution is $(x + C)^2 = y(y - 1)^2$ and the specific is $y =... | You have got $\displaystyle y'=\frac{2\sqrt y}{3y-1}$
Therefore $\displaystyle \frac{dy}{dx}=\frac{2\sqrt y}{3y-1}$
$\displaystyle\implies\int \frac{3y-1}{2\sqrt y}dy=\int dx$
$\implies\displaystyle \frac{3}{2}\int\sqrt y-\frac{1}{2}\int\frac{dy}{\sqrt y}=x+C$
$\implies \displaystyle y^{\frac{3}{2}}-y^{\frac{1}{2}}=x+C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$(1+x+x^2)^n=P_0+P_{1}x + P_{2}x^2+ \cdots +P_{2n}x^{2n}$ Prove that,$ P_0+P_{3}+P_{6}+ \cdots =3^{(n-1)}$ Let's say $$ S_n = (1+x+x^2)^n $$
n=1 $$S_1=1+x+x^2$$
n=2 $$S_2=1+2x+3x^2+2x^3+x^4$$
n=3 $$S_3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6$$
n=4 $$S_4=1+4x+10x^2+16x^3+19x^4+16x^5+10x^6+4x^7+x^8$$
By taking coefficients of the $... | Let define
$$(1+x+x^2)^{n}=P_0(n)+P_{1}(n)x + P_{2}(n)x^2 \pmod {x^3}$$
with
*
*$P_0(n)= P_0+P_3+\cdots$
*$P_1(n)= P_1+P_4+\cdots$
*$P_2(n)= P_2+P_5+\cdots$
therefore it is easy to check that
$$(1+x+x^2)^{n+1}=P_0(n+1)+P_{1}(n+1)x + P_{2}(n+1)x^2=$$
$$=\left[P_0(n)+P_{1}(n) + P_{2}(n)\right]+\left[P_0(n)+P_{1}(n) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Expressing the $n$–th derivative of $y=\frac{1}{(1+x^2)^2}$ In an attempt to find a way to express the n-th derivative of $y=\frac{1}{(1+x^2)^2}$ using the binomial theorem I got stuck in the last computation.
I'll explain what I did:
$$f(x)=\frac{1}{(1+x^2)^2}=\frac{1}{(x+i)^2(x-i)^2}=\\
\frac{i}{4}\left( \frac{1}{x+i... | I am sorry, i haven't comprehend everything in detail, but you have to be very careful, when you would like to use $(a\cdot b)^n=a^n\cdot b^n$ for complex numbers $a,b$ instead of positive real numbers. There is a popular example here Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$?
However, you can find the power series. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find the derivative matrix $g\circ f$ at $\dbinom uv$ Vector function $f$ and $g$ are defined by:
$$f \dbinom uv=\begin{pmatrix}u\cos(v)\\u\sin(v)\end{pmatrix},\begin{pmatrix}0<u<\infty\\-\frac{\pi}{2}<v<\frac{\pi}{2}\end{pmatrix}$$
$$g \dbinom xy=\begin{pmatrix}\sqrt{x^2+y^2}\\\arctan(\frac{y}{x})\end{pmatrix}, 0<x<\i... | As noticed in the comments by chain rule we have
$$(g \circ f)' \dbinom uv = g' \bigg( f \dbinom uv \bigg) f' \dbinom uv = g' \dbinom{u \cos v}{u \sin v} f' \dbinom uv$$
that is in this case
$$\begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}}\\-\frac{y}{x^2+y^2}&\frac{x}{x^2+y^2}\end{pmatrix}_{(u\cos v,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit at (0,0) of $\frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$ $$\text{Let}\ f(x,y) = \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4}$$
WolframAlpha tells me that $\lim_{(x,y) \to (0,0)} f(x,y)$ does not exist. To prove the non-existence of the limit, I tried three different paths ($y=kx, y=kx^2, y=kx^3)$ and they all equal zero. From th... | As an alternative, by $y^2=v$ and polar coordinates, with $x=\rho\cos \theta$ and $v=\rho|\sin \theta|$, we obtain
$$ \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4} =\frac{xv\sqrt{x^2+v}}{x^2 +v^2}=\sqrt \rho \cos \theta |\sin \theta| \sqrt{\rho\cos^2 \theta|\sin \theta|}\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Finding and proving $\bigcap\limits_{n=1}^{\infty} \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$ Letting $A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$, I am trying to find $\bigcap\limits_{n=1}^{\infty} A_n$. I believe the answer is $\left[0, \frac{1}{3}\right)$. I don't have any intuition for why this is the case o... | Let $A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$. We want to find $\bigcap_{n=1}^\infty A_n = \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$.
*
*Suppose $x\in \bigcap_{n=1}^\infty A_n$. Then, $x\in A_n$ for all $n\ge 1$. That is, $x > -\frac1n$ for all $n\ge 1$, and taking limits as $n\to\infty$ gives you $x\ge 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, ... | We avoid using induction in this approach.
You have $$f(n)=\frac{n}{2n+1}$$
Hence, $$f(n)=\frac{1-\frac{1}{2n+1}}{2}$$
Now, as $n$ increases, $\frac{1}{2n+1}$ decreases and hence,
$f(n)$ increases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Prove two inequalities. How can we prove these two inequalities, I proved the first question which I guess will be used to prove the following inequalities, but I don't know how to start. Let $a,b,c$ be there real numbers. Prove that if :
$\sin a+ \sin b + \sin c\ge 2
\implies \cos a+ \cos b + \cos c\le \sqrt 5$
and
$... | Part 1:
Let $0\le k\le 3$ be any given real number then $\begin{align} & \sin a+ \sin b + \sin c\ge k\\
\implies & (\sin a+ \sin b + \sin c)^2 \ge k^2\\
\implies & 3(\sin^2 a + \sin^2 b +\sin^2 c) \ge (\sin a+ \sin b + \sin c)^2 \ge k^2 \tag{1}\\
\implies & 3(1-\cos^2 a + 1-\cos^2 b +1- \cos^2 c) \ge k^2\\
\implies & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the diophantine equation $ax^2+by^2=cz^2+d$ always have solutions? Let $a,b,c,d,x,y,$ and $z \in \mathbb{N}$ where $a,b,c,$ and $d$ are constants but d is allowed to be zero .
$ax^2+by^2=cz^2+d$
First example :
when $a=b=c=1 $ and $d=0$ we have the equation : $x^2+y^2=z^2$ which I know its general solution .
Sec... | This is the partial solution.
$ax^2+by^2-cz^2=d\tag{1}$
Substitute $ x=1, y=pt+1, z=qt+1$ to above equation.
$p,q$ are arbitrary.
If $a+b=c+d$ then we get $t = \large\frac{-2(bp-cq)}{-cq^2+bp^2}.$
If $bp^2-cq^2=\pm1$, we can get the integral solution $(x,y,z).$
Thus, equation $bp^2-cq^2=\pm1$ is reduced to Pell's equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$ and $\frac{A_{19}}{A_5}=\frac{1}{2^7},$ then $c$ is? $(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$. The problem is that we can't express $(3x^2+2x+c)^{12}$ as a perfect square
There was a hint in my book, put $x=\frac{c}{3x}$ but that isn't leading anywhere either
Let $... | Let $x\rightarrow\frac{c}{3x}$
$$(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$$
$$\left(\frac{c^2}{3x^2}+\frac{2c}{3x}+c\right)^{12}=\sum_{r=0}^{24}A_rc^r3^{-r}x^{-r}$$
$$\frac{c^{12}}{3^{12}x^{24}}\left(3x^2+2x+c\right)^{12}=\sum_{r=0}^{24}A_rc^{r}3^{-r}x^{-r}$$
$$\left(3x^2+2x+c\right)^{12}=\sum_{r=0}^{24}A_rc^{r-12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$
For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$
I managed to show this with a 3 instead of a 5 :
The inequality is symmetric in $a$, $b$, $c$ so... | Let $a\geq b\geq c$, $b=c+u$ and $a=c+u+v$, where $u$ and $v$ are non-negatives.
Thus, we need to prove that:
$$5\min\{u^2,v^2,(u+v)^2\}\leq c^2+(c+u)^2+(c+u+v)^2$$ or
$$5\min\{u^2,v^2\}\leq3c^2+2c(2u+v)+2u^2+2uv+v^2,$$
for which it's enough to prove that
$$2u^2+2uv+v^2\geq5\min\{u^2,v^2\},$$ which is obvious.
| {
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"url": "https://math.stackexchange.com/questions/4247909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $\frac{x^{11}+x}{x^7+x^5}=\frac{205}{16}$
Blockquote
Find $x$ satisfy the equation$$\frac{x^{11}+x}{x^7+x^5}=\frac{205}{16}$$
Here is my work
$$\frac{x^{10}+1}{x^6+x^4}=\frac{205}{16},\quad\quad t:=x^2$$
$$\frac{t^5+1}{t^3+t^2}=\frac{205}{16}$$
$$16t^5-205t^3-205t^2+16=0$$
$t=-1$ is a root of the polynomial b... | As an alternative from here we have
$$\frac{t^5+1}{t^3+t^2}=\frac{205}{16} \iff \frac{(t+1)(t^4-t^3+t^2-t+1)}{t^2(t+1)}=\frac{205}{16}$$
and since $t=x^2\ge 0$
$$\iff \frac{t^4-t^3+t^2-t+1}{t^2}=\frac{205}{16} \iff t^2-t+1-\frac1 t+\frac1{t^2}=\frac{205}{16}$$
$$\iff t^2+2+\frac1{t^2}-t-\frac1 t-\frac{221}{16}=0 \iff \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How many $6$ digit numbers are there that doesn't have number $2$ in them and have two $1$? Is this the right way to calculate it?
*
*If the first number is $1$, then there are $5$ ways to choose another $1$ and $4$ digits left that can be equal to $8$ different numbers
So in total $5\cdot8^4=20480$
*If the first n... | $\mathbf{\text{Alternative Approach:}}$
By using exponential generating functions:
İf there are two $1's$ then the exponential generating function of it : $$\bigg(\frac{x^2}{2!} \bigg)$$
If there is not any two and do not have any restriction over the others then the exponential generating function of others : $$\bigg(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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The summation $\sum_{k=0}^n \frac{{2n \choose k}}{k+1}$ Mathematica gives this summation
$$\sum_{k=0}^n \frac{{2n \choose k}}{k+1}$$
in terms of a regularized Gauss hypergeometric function $~_2F_1$. Can the hypergeometric function be eliminated to have a simple answer?
| If you're looking for a (somewhat) closed form solution, observe that
\begin{aligned}
\frac{\binom{2n}{k}}{k+1} &= \frac{(2n)!}{(k+1)! \cdot (2n - k)!} \\
&= \frac{1}{2n+1} \frac{(2n+1)!}{(k+1)! \cdot ((2n+1) - (k+1))!} \\
&= \frac{1}{2n+1}\binom{2n+1}{k+1}.
\end{aligned}
So, by the Binomial theorem, we have
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the general expression for the n-th derivative I want to find the $n$-th derivatives of the function $$f(x)=\frac{1}{(1-x^2)^b}$$ with respect to x. Here $b$ is a positive constant. By using the chain rule, I can get
the first derivative is $$f'(x)=(-b)(1-x^2)^{-b-1}(-2x).$$ By using the chain rule and product ru... | I computed the $n^{th}$ derivative with Mathematica. Not sure how you would derive it but since you just wanted the formula here it is.
\begin{align*}
f(x)&=\frac{1}{(1-x^2)^b}\\ f^{(n)}(x)&=n!(1-x^2)^{-b}(x+1)^{-n}\binom{-b}{n} \, _ 2F_1 \left(b,-n;-b-n+1;\textstyle{\frac{x+1}{x-1}} \right)
\end{align*}
Over here $\,_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Inequality with strange sum of cubic roots For positive numbers $a$, $b$, $c \geq 0$ and $a+b+c=1$ show that:
$\sqrt[3]{4+17a^2b}+\sqrt[3]{4+17b^2c}+\sqrt[3]{4+17c^2a}+10 \Big(\frac{1}{27}-abc \Big) \geq 5$
I tried to use $AM-GM$ with $\frac{17}{27}+17a^2b$ and tried after that use Holder, but after that stucked.
Cubin... | Just AM-GM and Holder help here!
Let $abc=\frac{x}{27}$.
Thus, by AM-GM $$abc\leq\left(\frac{a+b+c}{3}\right)^3=\frac{1}{27}$$ and by AM-GM and Holder we obtain: $$\sum_{cyc}\sqrt[3]{4+17a^2b}\geq3\sqrt[9]{\prod_{cyc}(4+17a^2b)}\geq3\sqrt[3]{4+17abc}=3\sqrt[3]{4+\frac{17x}{27}}=\sqrt[3]{108+17x}$$ and it's enough to pr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Existence of $\int_0^{\infty} \frac{x\sin(x)}{x^2+1}\ \mathrm{d}x$ I am currently reading some lecture notes where the calculation of
\begin{align}
\int_{-\infty}^{\infty} \frac{x\sin(x)}{x^2+1}\ \mathrm{d}x
\end{align}
using residues is explained. The existence of $\lim_{R\to \infty}\int_0^{\infty} \frac{x\sin(x)}{x^2... | Note that the derivative of $x \mapsto \frac{x}{x^2+1}$ is pointwise equal to $\frac{1-x^2}{(1+x^2)^2}$, hence by integration by parts $$ \int_0^\infty \frac{x}{x^2+1}\sin(x)dx = -\cos(x)\frac{x}{x^2+1}\Big|_{x=0}^{x=+\infty} + \int_0^\infty \frac{1-x^2}{(1+x^2)^2}\cos(x) dx $$
The "boundary" term is $0$ (i.e this term... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$
If inverse of the function $y=\sqrt{x+\sqrt{2x-1}}$ be equal to
$y=ax^2+bx+c$, then what is the value of $a^2+b^2+c^2$?
$A) 3$
$B) 4$
$C) 5$
$D) 6$
I tried to isolate $x$ from the equation $y=\sqrt{x+\sqrt{2x-1}}$:
$$y^2=x+\sqrt{2x-1}$$
$$(y^2-x)^2=2x-1$$
$$y^4-2xy... | You have\begin{align}\require{cancel}y=\sqrt{x+\sqrt{2x-1}}&\implies y^2=x+\sqrt{2x-1}\\&\implies y^2-x=\sqrt{2x-1}\\&\implies(y^2-x+1)^2=(y^2-x)^2+1+2(y^2-x)\end{align}But then\begin{align}(y^2-x+1)^2=2x-\cancel1+\cancel1+2\sqrt{2x-1}=2y^2,\end{align}and therefore $y^2-x+1=\pm\sqrt2y$. So, $x=y^2\pm\sqrt2y+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Solve a non-homogeneous recurrence relation I need to solve this equation. I know how to solve the homogeneous side, but when I get to the non-homogeneous it becomes a problem.
$$ a_n = 7a_{n-1}- 10a_{n-2}+5^n-6$$
So far I have this
$$ a_n = 7a_{n-1}- 10a_{n-2}$$
$$ \frac{r^n}{r^{n-2}} = \frac{7r^{n-1}}{r^{n-2}}- \frac... | The original recurrence relation is
$$a_n = 7a_{n-1} - 10a_{n-2} + 5^n - 6 \tag1$$
Assume particular solution
$$a_n^p = An5^n + B \tag2$$
Substitute $(2)$ into $(1)$:
\begin{align}
An5^n + B &= 7(A(n-1)5^{n-1} + B) - 10(A(n-2)5^{n-2} + B) + 5^n - 6 \\
&= 7An5^{n-1} - 7A5^{n-1} + 7B - 10An5^{n-2} + 20A5^{n-2} -10B + 5^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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the sum of the reciprocal of the squares of roots $$x^3-6x^2+5x-7=0$$
Find $\frac{1}{r^2}$+$\frac{1}{s^2}$+$\frac{1}{t^2}$ where $r,s,t$ are roots of the equation.
*
*First I got the reciprocal of the equation above and I got $-7x^3+5x^2-6x+1=0$
*Using newtons identity/sum the sum of all the roots of this equation i... | $$x^3-6x^2+5x-7=0$$
Let $y=\frac1x.$ Then $$\frac1{y^3}-\frac6{y^2}+\frac5{y}-7=0\\7y^3-5y^2+6y-1=0.$$
So, by Vieta's Formulae, $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}\\=y_1^2+y_2^2+y_3^2\\=(y_1+y_2+y_3)^2-2(y_1y_2+y_2y_3+y_3y_1)\\=\left(\frac57\right)^2-2\left(\frac67\right)\\=-\frac{59}{49}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve: $x^2\equiv 1 \pmod{20},x^2\equiv 6 \pmod{15},x^2\equiv 9 \pmod{18}.$ I want to solve: $x^2\equiv 1 \pmod{20}, x^2\equiv 6 \pmod {15}, x^2\equiv 9\pmod{18}.$ This is a system of congruence equations, but these are not linear and moduli are not coprime. So,we cannot apply chinese remainder theorem here. However, I... | $$x^2 \equiv 1 \pmod{5} \iff x \equiv \pm 1 \pmod{5}$$
$$x^2 \equiv 1 \pmod{4} \iff x \equiv 1,3 \pmod{4}$$
$$x^2 \equiv 0 \pmod{9} \iff x \equiv 0 \pmod{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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If $n$ is a natural number, then $4$ divides $5^n-1$. The question is Statement 3.38.
If $n$ is a natural number, then $4$ divides $5^n-1$.
So far all I have is:
*
*$5^k\times5-1$
*$5^k(4+1)-1$
*$4\times5^k+5^k-1$
*$4\times5^k+4m$
I do not get how this equals what we are trying to get to using proof of induction... | Another solution is using the algebraic identity
$$x^n - y^n = (x - y) \left( x^{n - 1} + x^{n - 2} y + x^{n - 3} y^2 + \cdots x y^{n - 2} + y^{n - 1} \right)$$
In our case, $x = 5$ and $y = 1$ so
$$5^n - 1 = (5 - 1) \left( 5^{n - 1} + 5^{n - 2} + \cdots 5 + 1 \right) = 4 \sum_{k = 0}^{n - 1} 5^k$$
which is a multiple ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Matrix congruences If A, B are two integer square matrices of the same size such that $A\equiv B\pmod n$, is $A^p\equiv B^p \pmod{pn} $ for a prime p dividing n?
| let $n=p=2$
let $A=\begin{pmatrix}0&0\\1&0\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\1&0\end{pmatrix}$
$A^2$ is the zero matrix and $B^2=\begin{pmatrix}2&0\\0&2\end{pmatrix}$
$A\equiv B\pmod{2}$ but $A^2\not\equiv B^2\pmod{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $\lim_{x\to 2} f(x) = 3$.
Let $f(x) = \begin{cases}x^2 + 2x - 5; &x\ne 2\\ 0 &x=2.\end{cases}$. Prove that $\lim\limits_{x\to 2}f(x) = 3$.
By theorem,
$$\lim\limits_{x\to a}f(x) = L \iff \lim_{x\to a+}f(x) = L = \lim_{x\to a-}f(x).$$
Demostración: Let $\epsilon > 0$. Note that $|x^2 +2x - 5 - 3| = |x+4||x-2|$.... | Let $0 < |x - 2| < \delta_{\varepsilon}$. Then we have:
\begin{align*}
|f(x) - 3| & = |x^{2} + 2x - 5 - 3|\\\\
& = |x^{2} + 2x - 8|\\\\
& = |(x^{2} - 4x + 4) + (6x - 12)|\\\\
& = |(x-2)^{2} + 6(x-2)|\\\\
& \leq |x - 2|^{2} + 6|x - 2|\\\\
& < \delta^{2}_{\varepsilon} + 6\delta_{\varepsilon} := \varepsilon
\end{align*}
C... | {
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"url": "https://math.stackexchange.com/questions/4290545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\tan\frac\pi{12}-\cot\frac\pi{12}+\tan\frac{5\pi}6-\cot\frac{5\pi}6$ knowing the roots $3x^4+4\sqrt{3}x^3-18x^2-4\sqrt{3}x+3=0$ I know that the roots of this equation
$$3x^4+4\sqrt{3}x^3-18x^2-4\sqrt{3}x+3=0$$ are $x=\tan{\frac{\pi}{12}}, \tan{\frac{\pi}{3}}, \tan{\frac{7\pi}{12}}, \tan{\frac{5\pi}{6}}$.
I ... | Another way :$\cot x-\tan x=\cdots=2\cot2x$
$$p=\implies\tan\dfrac\pi{12}-\cot\dfrac\pi{12}+\tan\dfrac{5\pi}6-\cot\dfrac{5\pi}6=-2\cot\dfrac\pi6-2\cot\dfrac{5\pi}3$$
$$\cot\dfrac{5\pi}3=\cot\left(2\pi-\dfrac\pi3\right)=\cot\dfrac\pi3=-\tan\dfrac\pi6$$
$$p=-2\cot\dfrac\pi6+2\tan\dfrac\pi6=-2\left(2\cot\dfrac\pi3\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294230",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Calculating $B^{10}$
Calculate $B^{10}$ when $$B = \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$$
The way I did it was
$$ B = I + A $$
where
$$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
and $A^2=-I$. Since $A$ and $I$ are commutative,
$$\begin{aligned} B^2 &= (I+A)^2 = 2A \\ B^3 &= (I+A)2A = 2A-2I\\ B^4 &= ... | $$
B^2 = 2A, A^2=-I \implies
B^{100}= (B^2)^{50} = (2A)^{50} = 2^{50} A^{50} = 2^{50} (A^{2})^{25} = 2^{50} (-I)^{25} = -2^{50} I
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ Given the function $f: \mathbb{R}\to \mathbb{R}$ satisfies:
\begin{cases}
f(1)=1 \\
f(x+y)=f(x)+f(y)+2xy,\quad&\forall x,y \in \mathbb{R}\\
f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4},&\forall x\neq0
\tag{1}\label... | The easiest way to go is Taylor expansion.
Then the limit becomes
$$L=\lim_{x\to 0} \frac{e^{-2x^2}-\sqrt[3]{1+x²}}{\ln(1+x^2) } $$
$$=\lim_{x\to 0} \frac{(1-2x^2+2x^4-\cdots +\cdots )-(1+\frac{x^2}{3}-\frac{x^4}{9}+\cdots -\cdots)}{x^2-\frac{x^4}{2}+\frac{x^6}{3}+\cdots} $$
Simplifying this gives
$$L=\lim_{x\to 0} \f... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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What's the measure of the radius of the circle inscribed in the triangle ATB? For reference: Calculate the radius of the circle inscribed
in the triangle ATB
My progress: I put the trigonometric solution...would a geometric solution be possible?
|
$\triangle AO_1T: \triangle BO_2T\rightarrow \text{isosceles}$
$\triangle ATB(\text{right})\\\angle O_1AT = x \implies \angle ABT = x\\AO_1T = 180-2x$
By law of cosines,
$AT^2=2+2\cos(2x)=2(1+\cos(2x))=4\cos^2(x)$
$BT^2=4+4-8\cos(2x)=8(1-\cos(2x))=16\sin^2(x)$
From triangle $\triangle ABT$,
$\displaystyle \tan^2(x)=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why do I get incorrect solutions $ x = 0 $ and $ x = 2 $ for $ x + 1 + \sqrt{4x + 1} = 0 $? Here is my incorrect attempt at solving
$$ x + 1 + \sqrt{4x + 1} = 0. $$
Subtracting $ \sqrt{4x + 1} $ from both sides,
$$ x + 1 = -\sqrt{4x + 1} $$
Squaring both sides,
$$ x^2 + 2x + 1 = 4x + 1 $$
Subtracting $ 4x + 1 $ from bo... | You can easily track where the extra solutions were introduced.
$$ x + 1 + \sqrt{4x + 1} = 0\to\color{red}{0+1+\sqrt1=0}$$
$$ x + 1 = -\sqrt{4x + 1}\to\color{red}{0+1=-\sqrt{0+1}}$$
$$ x^2 + 2x + 1 = 4x + 1\to\color{green}{0+0+1=0+1}$$
$$ x^2 - 2x = 0\to\color{green}{0-0=0}$$
$$ x(x - 2) = 0\to\color{green}{0(0-2)=0}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove that $\cos^{-1}(\frac{2}{\sqrt 7})+\tan^{-1}(\frac{\sqrt 3}{5}) = \frac{\pi}{3}$ I started off by making two right angled triangles. One having an angle $\alpha$ such that $\cos \alpha = \frac{2}{\sqrt 7}$ and the other having an angle $\beta$ such that $\tan \beta = \frac{\sqrt 3}{5}$.
Now from here I must show ... | Observe that$$\cos^{-1}\left(\frac{2}{\sqrt 7} \right) = \tan^{-1}\left(\frac{\sqrt3}{2} \right)$$
So we have
$$\tan^{-1}\left(\frac{\sqrt3}{2} \right) + \tan^{-1}\left(\frac{\sqrt3}{5} \right) = \tan^{-1}\left(\frac{\frac{\sqrt3}{2} + \frac{\sqrt3}{5}}{1 - \frac{\sqrt3}{2} \cdot \frac{\sqrt3}{5}} \right) = \tan^{-1}(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$ Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$.
I wanted to approach this using the reduction formula. Here's what I have tried
$$\int x^2e^{-x^3}dx = -\frac{1}{3}e^{-x^3}$$
Then extending this to $n$
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx $$
Using su... | a very simple way is to use Gamma function.
$$\int_0^{\infty}x^{11} e^{-x^3}dx$$
Let's set $y=x^3$ obtaining
$$\frac{1}{3}\int_0^{\infty}y^3 e^{-y}dy=\frac{1}{3}\Gamma(4)=\frac{3!}{3}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Help with finding the equation of line tangent to the semi circle with the equation: $y=\sqrt{1-x^2}$
Ok so I need to show that the equation of the tangent above to the semi-circle with the equation: $y=\sqrt{1-x^2}$, is $y=-\frac{1}{\sqrt{3}}x$ + $\frac{2}{\sqrt{3}}$
What we from the question:
The tangent intersects... | Your original equation can be restated as $y^2 = 1 - x^2$. Using implicit differentiation,
$$ 2y y' = -2x x' \text{.} $$
Specializing the independent variable to $x$, we obtain
$$ 2y \frac{\mathrm{d}y}{\mathrm{d}x} = -2x \text{,} $$
so
$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-x}{y} \text{.} $$
So let's run... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $\int_0^\infty \frac{\sin^2 x-x\sin x}{x^3} \, dx= \frac{1}{2} - \ln 2$
Prove that
$$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$
Integration by parts gives
\begin{align*}
&\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\
&= \lim_{R\to \infty} \biggl(... | Here is a correct solution that you might want to compare with:
\begin{align*}
&\int_{0}^{\infty}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x \\
&= \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \int_{\varepsilon}^{R}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x \\
&= \lim_{\substack{R\to\infty \\ \varepsilon \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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What's the measure of the $\angle BAC$ in the triangle below? For reference: In the right triangle $ABC$, right at $B$, the corner $AF$ is drawn such that $AB = FC$ and $\angle ACB = 2 \angle BAF$. Calculate $\angle BAC$.
My progress:
$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: L... | If $y=BF$ and $t=\tan\frac{C}{2}=\frac{y}{x}$, then $\tan C=\frac{x}{y+x}=\frac{1}{1+t}$
Also $\tan{C}=\frac{2t}{1-t^2}$ from the double angle formula. It follows that:
$$\frac{2t}{1-t^2}=\frac{1}{1+t}$$
And you cand find $t=\frac{1}{3}$ by solving this equation. It follows that $\tan C=\frac{3}{4}$, $\tan\angle BAC=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Find the maximum value of $(x+2) \cdot (12-3x)$ , x is a Real number Question:- Find the maximum value of $(x+2) \cdot (12-3x)$. $x$ is a Real number.
The solution given in the textbook applies AM-GM to solve this
$$3 \cdot (x+2) \cdot (4-x)$$
Now, $x+2 = a$ and $4-x = b$
$a+b=6$
Objective to maximize $3 \cdot a \cdot ... | $(c+x)·(d-x) = \left({c+d\over2}+\left(x+{c-d\over2}\right)\right)·\left({c+d\over2}-\left(x+{c-d\over2}\right)\right) ≤ \left({c+d\over2}\right)^2 $
Above identity is always true, thus no need to check where x is.
$3·(2+x)·(4-x) ≤ 3·\left({2+4\over2}\right)^2 = 3^3 = 27$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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$a_1=\sqrt{6}$ , $a_{n+1} = \sqrt{6+a_n}$ A sequence $(a_n)$ is defined by $a_n > 0$ and $a_{n+1} = \sqrt{6+a_n}$ for $n\ge1$. Show that
$1)$ the sequence $(a_n)$ is monotone increasing if $0 < a_1 < 3$
$2)$ the sequence $(a_n)$ is monotone decreasing if $a_1 > 3$
I know how to verify it by substituting values for $n$ ... | If $a_n > 3$ then
$$a_n = \sqrt{a_n \cdot a_n} > \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} > \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} > \sqrt{6 + 3} = \sqrt{9} =3 $$
and if $0 < a_n < 3$ then
$$a_n = \sqrt{a_n \cdot a_n} < \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} < \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} < \sqrt{6 + 3} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculating limit using sub sequences $\lim\limits_{n\to\infty }\frac{(-5)^{n}+2\cdot(-2)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$
Find the following limit if it exists , if it does not exist (even not infinity) explain if it does exist then also find all the Sub sequential limits. $\lim\limits_{n\to\infty }\frac{(-5)^{n}+2\... | There's a theorem that you may or may not have proved yet that goes something like the following:
If $(a_n)$ converges to $\ell$, then any subsequence $(a_{n_k})$ converges to $\ell$ as well.
So, if there are two subsequences that converge to two different values, that must mean that the original sequence does not conv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the recursive integral of $I\left(n\right)=\int \frac{1}{\left(1-x^2\right)^n}\:dx$ Using only simple calculus and rewriting the integral how could I find the recursive integral of
$I\left(n\right)=\int \frac{1}{\left(1-x^2\right)^n}\:dx$
I tried using trig substitution like so:
$x = \sin(u)$ with $dx = \cos(u) du... | For $n \ge 2$ we have
$$\begin{align}\int\frac{1}{(1-x^2)^n}dx&=\int\frac{1-x^2+x^2}{(1-x^2)^n}dx\\
&=\int\frac{1}{(1-x^2)^{n-1}}dx+\int x\cdot\frac{x}{(1-x^2)^n}dx\end{align}.$$
For the second integral, use integration by parts with $u=x, dv=\frac{x}{(1-x^2)^n}dx$:
$$\int x\cdot\frac{x}{(1-x^2)^n}dx=x\cdot\frac{1}{2(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{ali... | Let $t=\frac{x}{\sqrt{1-x^2}} $, then $x=\frac{t}{\sqrt{1+t^2}}$ and $d x=\frac{1}{\left(1+t^2\right)^{\frac{3}{2}}} d t$, which transforms the integrand into a rational function.
$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} &=\int \frac{d t}{(1+t)\left(1+t^2\right)} \\
&=\frac{1}{2} \int\left(\frac{1}{1+t}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Finding the vectorial expression for the mutual slant of two cones with a common vertex I am looking for a closed-form solution to find the mutual slant of two cones with a common vertex shown in the image. Any insight on the problem-solving approach would be useful.
If $\hat{k}_1$ and $\hat{k}_3$, are known unit vect... | Start by parameterizing $\mathbf{k_2}$ in terms of $\mathbf{k_1}$. Let $\mathbf{u_1}, \mathbf{u_2}$ be two mutually orthogonal unit vectors that are also orthogonal to $\mathbf{k_1}$, then any vector $\mathbf{k_2}$ on the cone with axis $\mathbf{k_1}$ and semi-vertical angle $\alpha_{12} $ can be written as
$\mathbf{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Alternative way of finding a basis for a vector space The problem:
Let $U$ be a subspace of $\mathbb{R}^4$. $U$ is given by the span of the vectors $\begin{Bmatrix} \begin{pmatrix}2\\0\\0\\-2\end{pmatrix},\begin{pmatrix}1\\-3\\3\\-1\end{pmatrix},\begin{pmatrix}-2\\0\\-4\\-2\end{pmatrix},\begin{pmatrix}-1\\-3\\-1\\-3\e... | The standard way of solving this problem is to leave the five vectors listed from top to bottom, that is, as columns of $4 \times 5$ matrix. Then use Gauss-Jordan elimination in the standard way. At the end, the independent vectors (from the original set) are the ones that correspond to leading $1$'s in the (reduced)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to tackle the integral $\int_{0}^{\infty} \frac{\ln x}{x^{n}-1} d x$? In my post, I started to investigate the integral $\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^{2}-1} d x$. Fortunately,
$$\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^{2}-1} d x =2 \int_{0}^{1} \frac{\ln x}{x^{2}-1} d x.$$
So we only need ... | In order to make use of power series, I split the integration interval of $I_n$ as:
$$
I_n=\underbrace{\int_{0}^{1} \frac{\ln x}{x^{n}-1} d x}_{J}+\underbrace{\int_{1}^{\infty} \frac{\ln x}{x^{n}-1} d x}_{K} $$
$$
\begin{aligned}
J &=-\sum_{k=0}^{\infty} \int_{0}^{1} x^{n k} \ln x d x \\
&=-\sum_{k=0}^{\infty} \int_{0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4319023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the minimum of $\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$ Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$
A) $\frac 23$
B) $\frac 43$
C) ... | Let $t:=x-y$ and $u:=x+y$. We have
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}=2\cos(t)(\csc(2y)+\csc(2x))=\frac{4\sin(u)\cos^2(t)}{\cos^2(t)-\cos^2(u)}.$$
As $u$ is known to be constant, we have a univariate function in $t$. It has a stationary point at $t=0$, giving
$$\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$ is convergent if $x_0 > 0$ and $x_1 > 0$ I have stumbled across this problem in an olympiad-material sent by my teacher. Prove that the sequence $$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$$ is convergent when $x_0 > 0$ and $x_1 > 0$. My only idea which has absolutely failed so far is tr... |
$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$ is convergent if $x_0 > 0$ and $x_1 > 0$.
Proof: an initial guess for the limit of $x_n$, is $1$ by replacing $x_n$, $x_{n+1}$ and $x_{n+2}$ with $l$ in the recurrence relation and solving for $l$. We hence prove that $x_n\to 1$. Note that if $\epsilon<x_n<\frac{1}{\epsilon}$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4321285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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The equation $y = x^2 + \frac{1}{2}ax + a$ represents a parabola for all real values of $a$ The task is to show that any parabola with the value a passes through shared point.
I graphed the parabola and understand that the shared point is $(-2,4)$. My current issue is generalizing why this is and how to prove it.
| Just a note that $$x^2+\frac{1}{2}ax+(a-4)\\=x^2+2x-2x+\frac{1}{2}ax+a-4\\=x(x+2)-\left(2-\frac{1}{2}a\right)(x+2)\\=(x+2)\left(x-2+\frac{1}{2}a\right)$$
And hence, you get 2 points which equal 4 for the original function: $x=-2$ and $x=2-\frac{1}{2}a$.
In general, there is such common points when you have some expres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4325252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Rather strange Inverse Laplace transform I want to find the inverse Laplace transform of:
\begin{equation}
\frac{s}{s^2+4s+5}
\end{equation}
So I rewrite it to:
\begin{equation}
\frac{s}{(s+2)^2+1^2}
\end{equation}
I see that I can extract $s+2 = p$, and thus add $+2$ and $-2$ to the numerator:
\begin{equation}
\frac{(... | Note that
$$
H(s)
=
\frac{s}{s^2+4s+5}
=
\frac{(s+2)-2}{(s+2)^2+1^2}
=
G_1(s+2)- 2G_2(s+2)
$$
where
$$
G_1(s) = \frac{s}{s^2+1^2},
G_2(s) = \frac{1}{s^2+1^2}
$$
We deduce
$h(t)=
[\cos(t)-2\sin(t)]e^{-2t}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find all functions $f:\mathbb N\to \mathbb N$ such that $\frac{4f(x)f(y-3)}{f(x)f(y-2)+f(y)f(x-2)}$ is an integer for all $x>2$ and $y>3$. The following problem is from an online contest that ended today:
Find all functions $f:\mathbb N\to \mathbb N$ such that $$\frac{4f(x)f(y-3)}{f(x)f(y-2)+f(y)f(x-2)}$$ is an intege... | I assume that $\Bbb N$ does not contain zero.
As you have worked out, we have $f(x + 1) \mid 2f(x)$ for all $x$.
This means that the number of odd prime divisors of $f(x)$, counted with multiplicity, never increases. As a result, there is an odd number $m$ such that $f(x) = 2^{v_x}m$ for all sufficiently large $x$ (whi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evalute $\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}}$ Evaluate the given expression $$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}$$ The given answer is $\dfrac{1}{4}$. My attempt:
$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}... | Using the identity $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$, we find that
$$
\begin{align*}
\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}} &= \sqrt[n]{\frac1{2^n}}\cdot\sqrt[n]{\frac{20}{2^4+2^2}}\\
&= \sqrt[n]{\frac1{4^n}}\cdot\sqrt[n]{\frac{20}{16+4}}\\
&=\sqrt[n]{\left(\frac14\right)^n}\cdot\sqrt[n]{\frac{20}{20}}\\
&=\frac14.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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find $(f^{-1})^{\prime} (\frac{ \sqrt 2 }{2})$ let $f:\left[0, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$ and $f(x)=\int_{0}^{x} \max \{\sin t, \cos t\} d t$ .
find $(f^{-1})^{\prime} (\frac{ \sqrt 2 }{2})$
we have $f(x)=\int_{0}^{x} \max \{\sin t, \cos t\} d t =\int_{0}^{x} \cos t d t $ for $ 0 \le x \le \fra... | As you said, the first thing to notice is that :
*
*if $0 \leq x \leq \displaystyle{\frac{\pi}{4}}$, then $\cos(x) > \sin(x)$, so
$$f(x) = \int_0^x \cos(t) dt = \sin(x)$$
*if $ \displaystyle{\frac{\pi}{4} \leq x \leq \frac{\pi}{2}}$, then $\cos(x) \leq \sin(x)$, so
$$f(x) = \int_0^{\frac{\pi}{4}} \cos(t) dt + \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The polynomial $A(x)=x^4+2x^3-5x^2-22x-24$ is divisible by $B(x)=x^2+ax+b$. Find for which values of $a$ and $b$ the polynomial $A(x)=x^4+2x^3-5x^2-22x-24$ is divisible by $B(x)=x^2+ax+b$.
We can write $$A(x)=B(x)Q(x)+R(x)$$ where $Q(x)$ is the quotient and $R(x)$ is the remainder. We want to show that $R(x)=0$. The de... | By plugging in small values of $x$ (for example $x=-2,-1,0,1,2,\ldots$), we can note that for $x=3$ and $x=-2\;$, $A(x)=0$. Hence by Rational root theorem $(x-3)(x+2)=x^2-x-6$ is a factor of $A(x)$. The other factor can be found by for example long division or other methods. The factorization will be ,
$$x^4+2x^3-5x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solve for $x$: $\frac{x^2-10x+15}{x^2-6x+15}=\frac{3x}{x^2-8x+15}$ Solve the equation: $\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{3x}{x^2-8x+15}$.
When $x\ne3$ and $x\ne5$ we get $$(x^2-10x+15)(x^2-8x+15)=3x(x^2-6x+15)\\(x^2-9x+15-x)(x^2-9x+15+x)-3x(x^2-6x+15)=0\\(x^2-9x+15)^2-x^2-3x(x^2-6x+15)=0.$$ I am stuck here. The Rat... | Taking from where you left off, and rewrite the left hand side of your last line:
$\begin{align}(x^2+15)^2 - 18x(x^2+15) + 81x^2 - x^2 -3x(x^2+15)+18x^2&=0\\ \implies (x^2+15)^2-21x(x^2+15)+98x^2&=0\\ \implies (x^2+15)^2 - 2\cdot 10x\cdot (x^2+15)+(10x)^2 - x(x^2+15)-2x^2&=0\\ \implies (x^2+15-10x)^2-x(x^2+15-10x)-12x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
limit of $\lim_{n\to \infty} (-1)^{n-1} \sin(\pi \sqrt{n^2 + 0.5n + 1})$ what am I doing wrong? I have to evaluate $$\lim_{n\to \infty} (-1)^{n-1} \sin(\pi \sqrt{n^2 + 0.5n + 1})\,\,where \,\,n\,\epsilon\,N$$
what I would do is just factor out $n^2$ in the sqrare root, then just make the $\frac{0.5}{n}$ and $\frac{1}{n... | The problem is that as $n\to\infty$, $\sqrt{n^2 +0.5n + 1}$ diverges, and $\sqrt{n^2 +0.5n + 1} -n \neq 0$. We have
\begin{align*}
\lim_{n\to\infty} \sqrt{n^2 +0.5n + 1} -n &= \lim_{n\to\infty} \sqrt{(n + 0.25)^2 + 1 - 0.25^2} - n\\
&= \lim_{n\to\infty} \frac{(\sqrt{n^2 +0.5n + 1} - n)(\sqrt{n^2 +0.5n + 1} + n)}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4348056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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For $a, b, c$ with $a+b+c=0$ prove $\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$ and $\frac17\sum a^7=\frac15\sum a^5\cdot\frac12\sum a^2$ Consider the following problem:
Problem. Suppose that real numbers $a$, $b$ and $c$ satisfy the condition $a+b+c=0$. Prove the following identities:
$$
\frac{a^5+b^5+c^5}{5... | I'll tackle the first identity. The idea is to use Newton's sums. Note that $P_2=a^2+b^2+c^2=S_1P_1-2S_2=-2S_2$.
Additionally, $P_3=S_1P_2-S_2P_1+3S_3=3S_3$. Also,
$$P_5=S_1P_4-S_2P_3+S_3P_2-S_4P_1+5S_5=S_3P_2-S_2P_3+5S_5$$
So we want to prove
$$-S_2S_3=S_5+\frac{1}{5}(S_3P_2-S_2P_3)$$
$$-(ab+bc+ac)(abc)=\frac{1}{5}(-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4348852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Without using approximation, prove that $ \ln(4)<\sqrt{2}$. I have been asked to prove the inequality $$\ln(4)<\sqrt{2}$$
without using the fact that
$$\ln(4)\approx 1.38 \text{ and }\;\sqrt{2}\approx 1.41$$
I defined at $ [1,2] $ the function
$$f(x)=x^3\ln(x)-1$$
and tried to see if $ f(\sqrt{2})<0 $ but i need to kno... | The function $f(x) = \frac{1}{x}$ is convex. Therefore, we have $\frac{(f(\sqrt{2}) + f(1))(\sqrt{2} - 1)}{2} > \int\limits_1^{\sqrt{2}} \frac{1}{x} dx$ by trying to approximate the region of integration as a trapazoid.
Evaluating both sides gives us $\frac{1}{2}(\frac{\sqrt{2}}{2} + 1)(\sqrt{2} - 1) > \ln \sqrt{2}$. M... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 0
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Limit of $(1-\frac{1}{2})*(1-\frac{1}{4})*\cdot\cdot\cdot*(1-\frac{1}{2^{n}})$? Using Monotone convergence theorem, we know the sequence ${x_{n}} = (1-\frac{1}{2})*(1-\frac{1}{4})*\cdot\cdot\cdot*(1-\frac{1}{2^{n}})$ has limit, but I can not derive the exactly limit, so how to calculate the limit of $x_{n}$?
| Write
$$P=\prod_{k=1}^\infty \left(1-\frac{1}{2^n}\right)=\prod_{k=1}^4 \left(1-\frac{1}{2^n}\right)\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right)=\frac{315}{1024}\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right)$$ Now
$$\log\left(\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right) \right)=\sum_{k=5}^\infty \log\left(1-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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When is $2^2+...+n^2=p^k$. Find all $n>1$ such that $$\sum_{i=2}^ni^2=p^k$$
where $p$ is a prime and $k\ge 1$.
I tried to use the formula for the sum of the first $n$ perfect squares, $$\frac{n(n+1)(2n+1)-6}{6}=p^k$$
$$(n-1)(2n^2+5n+6)=6p^k$$
Furthermore $d=\gcd(n-1,2n^2+5n+6)=13$ or $1$.
Let’s see when is equal $1$, w... | If $d=1$, then you've done the computation correctly.
If $d = 13$, then note that $p=13$ is forced, so dividing by $13^2$ on both sides,
$$
\frac{(n-1)}{13}\frac{(2n^2+5n+6)}{13} = 6 \times 13^{k-2}
$$
Since the quantities on the left are co-prime, we get that one of $\frac{n-1}{13}$ or $\frac{2n^2+5n+6}{13}$ is a di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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$a,b,c>0$, prove: $\sum_{cyc}{\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}}\le2$
Let $a,b,c>0$. Prove that: $$\!\!\frac{a+\sqrt{bc}}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}+ \frac{b+\sqrt{ca}}{\sqrt{(b+c)(b+a)}+\sqrt{ca}}+ \frac{c+\sqrt{ab}}{\sqrt{(c+a)(c+b)}+\sqrt{ab}}\le2$$
AOPS link
My approach using AM-GM: $\sqrt{(a+b)(... | Here is SOS form I get. I used new algorithm for MatLab to kill it:
The OP is equivalent to:
$$\sum\frac{a^3\left(\sqrt{b}-\sqrt{c}\right)^4}{\left(\sqrt{(a+b)(a+c)}+\sqrt{bc}\right)(X_a+Y_a)}\ge0$$
Where :$X_a=a(2a+b+c)\sqrt{(a+b)(a+c)}$ and $Y_a=a\sqrt{bc}(a+b+c)+2a^2(a+b+c)-a^2\sqrt{bc}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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From homogeneous recursive relation to Matrices and Linear Algebra After the lesson in recursive relations in the university, I realised that we can transform a homogeneous recursive relation to a linear algebra problem.
Let $f_{n}-2f_{n-1}+f_{n-2}=0$, with $f_{0}=7, f_{1}=15$.
We can write this recursive relation in t... | In fact, dealing with infinite matrices is cumbersome.
A much simpler linear algebra approach amounts to describe the relationship as being between consecutive terms in this way:
$$\begin{pmatrix}f_n\\f_{n-1}\end{pmatrix}=\begin{pmatrix}2&-1\\1&0\end{pmatrix}\begin{pmatrix}f_{n-1}\\f_{n-2}\end{pmatrix} \ \text{with} \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\sqrt{x^2}=|x|$ or $\sqrt{x^2}=x$ in an indefinite integral Question:
Find the following integral: $\int{\sqrt{1+\cos(x)}}dx$
My attempt:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\ \left|\cos\frac{x}{2}\right|}dx$$
$$=\sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx\tag4$$
$$=\... | Note that $\sqrt{t^2}=\text{sgn}(t)\cdot t$. Then, piecewise integrate
\begin{align}
\int\sqrt{1+\cos x}\>dx=
&\>\sqrt2\int\sqrt{\cos^2\frac x2}\>dx
=\>\sqrt2 \>\text{sgn}(\cos \frac x2)\int\cos \frac x2 \>dx\\
= &\>2\frac{ \sqrt{2\cos^2\frac x2}}{\cos \frac x2}\sin \frac x2
=\>2\sqrt{1+\cos x}\>\tan \frac x2
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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How do you solve the equation $2^{|x+1|}-|2^x-1| = 2^x+1$? | ... | is absolute value
$2^{|x+1|}-|2^x-1| = 2^x+1$, move $|2^x-1|$ to RHS: $2^{|x+1|} = 2^x+1+|2^x-1|$
Log base 2 each side:
$|x+1| = log_2 (2^x+1+|2^x-1|)$
Here we quickly observe: $-1$ is not a solution but $0$ is a solution. If you try negative numbers $-... | For $x\geq0$ the equation becomes
$$2^{x+1}-(2^x-1)=2^x+1$$
and the left hand side simplifies to $2^x+1$ so that every $x\geq 0$ is a solution.
For $x<0$ the equation becomes
$$2^{|x+1|}+(2^x-1)=2^x+1$$
so we get $2^{|x+1|}=2$ so that $|x+1|=1$ so that $x=-2$ is another solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Prove algebraically that the line $y=x+c$ intersects the curve $y=f(x)$ if $|a| \geq 1$
The function $f(x)$ is defined by $$f(x) = \frac{x(x-2)(x-a)}{x^2-1}$$
Prove algebraically that the line $y=x+c$ intersects the curve
$y=f(x)$ if $|a| \geq 1$, but there are values of $c$ for which there
are no points of intersecti... |
I started off by letting $f(x) = x+c$ and then rerranging produces $$(a+c+2)x^2-(2a-1)x - c = 0$$
You have a typo here. It should be
$$(a+ c+ 2) x^2 - (2 a \color{red}{+} 1)x-c=0\tag1$$
I think that you have correctly showed that if $|a|\geqslant 1$, then $\Delta\geqslant 0$.
There is another way.
$$\begin{align}\Del... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Solving $\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$
I've been asked to solve the limit.
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Here's my approach:
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Using the identity, $... | $$
\cos\left(x\right)-1 \underset{(0)}{=}-\frac{x^2}{2}+o\left(x^3\right)
$$
Hence combined with what you did
$$
\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))} \underset{(0)}{\sim}\frac{\pi}{2x^2}\left(-\frac{x^2}{2}\right) = -\frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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Show that U,V and H are colinear We are given a regular icosagon as below:
I wanna prove that the red line exists.
I know that $U$ is the incenter of $\triangle TLB$ ($T,U,G$ are collinear)
I know that $V$ is the incenter of $\triangle GMC$ ($M,V,E$ are collinear)
I tried to use Pascal theorem but it probably has to b... | Here is a solution using complex numbers, i was waiting for the bounty to be given, myself having no simple synthetic solution.
I am still posting it, since there was an interesting experience of computing in the cyclotomic field $\Bbb Q(\zeta_{20})$.
Let $z=\zeta_{20}=\cos\frac{2\pi}{20}+i\cos\frac{2\pi}{20}$ be this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{... | $$
\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\theta\quad\iff\quad\sin(\theta)=\frac{2x}{1+x^2}\quad\land\quad-\frac\pi2\le\theta\le\frac\pi2
$$
Note that
$$
\cos(\theta)=\left|\frac{1-x^2}{1+x^2}\right|
$$
When $|x|\le1$,
$$
\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}=\frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Hard Integral without Partial Fraction Decomposition So I was trying to use partial fraction decomposition on this problem, and I realized that it didn't work, as it is already in partial fraction decomposition form.
$\int{\frac{3x+4}{(x^2+5)^2}dx}=$
Well, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^... | hint
Write it as the sum $ I+J$ with
$$I=\frac 32\int \frac{2xdx}{(x^2+5)^2}$$
easy
and
$$J=\frac 45\int \frac{x^2+5-x^2}{(x^2+5)^2}dx$$
$$=K+L$$
where $ K =\frac 45\int \frac{1}{x^2+5}dx$ is easy and
$$L=-\frac 25\int x \frac{2x}{(x^2+5)^2}dx$$
which we integrate by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\sec x=\frac{a}{b}$, find the exact value of $\cot x-\frac1{\sin x}$, for positive $a$ and $b$, and for $x\in(\frac{3\pi}{2},2\pi)$
If $\sec(x) = \frac{a}{b}$, find the exact value of
$$\cot(x) - \frac{1}{\sin(x)}$$ and $a$, $b$ are positive real values and $x\in(\frac{3\pi}{2},2\pi)$
*
*I know that $\sec(x) = ... | $$\cos x=\frac ba$$
$$\frac 1{\sin x}=\frac 1 {\sqrt{1-\frac{b^2}{a^2}}}=\frac a{\sqrt {a^2-b^2}}$$
$$\cot (x) =\frac {\cos x}{\sin x}=\frac{\frac ba}{\frac{\sqrt{a^2-b^2}}a}=\frac b{\sqrt {a^2-b^2}}$$
$\Rightarrow$ $$\cot (x)-\frac 1{\sin x}=\frac {a-b}{\sqrt{a^2-b^2}}$$
Multiplying numerator and denominator by $\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.
My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^... | $$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2} \bigg( (a-b)^2 + (b-c)^2 + (c-a)^2 \bigg)=1$$
$$ \implies (a-b)^2 + (b-c)^2 + (c-a)^2 =2$$
So, two between $a,b,c$ are equal and the other has the difference of $1$ from the others.
WLOG, assume $b=a$ and $c=a \pm 1$ (Ignoring $a\le b\le c$ here.).
$$a+b+c=3a\pm1 = 2017 = 3\times672 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Equalities with 2 given substitions Let $\alpha$ and $\beta$ be real numbers such that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ Which of the following equalities hold?
A. $\alpha^2 + \beta^2 = 2$
B. $\alpha^3 + \beta^3 = 3$
C. $\alpha^4 + \beta^4 = 6$
D. $\alpha^5 + \beta^5 = 12$
E. $\alpha^6 + \beta^6 = 18... | Let $x=\alpha$ and $y=\beta$. We have:
$$x^4+y^4=(x^2+y^2)^2-2x^2y^2=3^2-2=7$$
And:
$$x^6+y^6=(x^3+y^3)^2-2x^3y^3=4^2-2\cdot(-1)=16+2=18$$
So, the correct answer is E.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4378168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrate $\int \frac{1}{1+\sin x}\: dx$ using substitution $u=1+\sin x$ I'm trying to work out what went wrong with the following method and how I can precisely correct it.
MY ATTEMPT:
\begin{align*}
\text{Let } u&=1+\sin x\\
\therefore du&=\cos x\:dx\\
&= \pm\sqrt{1-\sin^2x}\:dx\\
&= \pm\sqrt{1-(u-1)^2}\:dx\\
&=\pm \... | By the way, we can evaluate the integral by rationalisation.
$$
\begin{aligned}
\int \frac{1-\sin x}{1-\sin ^2 x} d x &=\int \frac{1-\sin x}{\cos ^2 x} d x \\
&=\int \sec ^2 x d x-\int \tan x \sec x d x \\
&=\tan x-\sec x+C\\\textrm{ OR }&= -\frac{\cos x}{1+\sin x}+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\left(\frac{a-b}{a+b}\right)^{11}+\left(\frac{b-c}{b+c}\right)^{11}+\left(\frac{c-a}{c+a}\right)^{11}\leq 1$ where $a,b,c >0$ I have to prove:
$ \displaystyle \tag*{} \left(\frac{a-b}{a+b}\right)^{11}+\left(\frac{b-c}{b+c}\right)^{11}+\left(\frac{c-a}{c+a}\right)^{11}\leq 1$ where $a,b,c >0$
My approach:
$ \disp... | You can consider the expression
$$f(a,b,c)=\left(\frac{a-b}{a+b}\right)^{n}+\left(\frac{b-c}{b+c}\right)^{n}+\left(\frac{c-a}{c+a}\right)^{n}\leq 1$$
for any odd $n$. $f(a,b,c)$ is invariant under cyclic permutations. Hence, you only have to consider the cases (1) $a<b<c$ and (2) $a>b>c$. Introducing $x=a/b>0$ and $y=b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Green-Ruzsa covering lemma I am trying to understand the proof of Green-Ruzsa covering lemma from Tao-Vu book but in my humble opinion the proof is written so unclearly that I was not able to comprehend some moments even after 10 times of reading.
*
*Can anyone explain how does this algorithm work? What is the essen... | Let's consider the example where $A = \{1,2,4,8\}$ and $B = \{1,2,3\}$.
Step 1. Initially, $X = \emptyset$. Then $X + A - A = \emptyset$ because $X + A - A$ is formally defined as $$\{x + a - a' : x \in X, a \in A, a' \in A\}.$$ Since there are no possible choices of $x \in X$, there are no possible elements of the for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Squaring $\sqrt{(x-2)^{2}+(y-3)^{2}}=\frac{|5x+6y+5|}{\sqrt{5^{2}+6^{2}}}$ does not lose information? Scenario 1:
$$y=3x+5\tag{1}$$
$$y^{2}=\left(3x+5\right)^{2}\tag{2}$$
When both sides are squared in $(1)$, we lose some information and get $(2)$: the graphs of $(1)$ and $(2)$ aren't the same.
Scenario 2:
$$\sqrt{\lef... | The equation "$A^2=B^2$" is equivalent to "$A=B$ or $A=-B$".
In your first example, both $y = 3x+5$ and $y = -(3x+5)$ are perfectly possible, and $y^2 = (3x+5)^2$ combines those.
In your second example, both $\sqrt{\left(x-2\right)^{2}+\left(y-3\right)^{2}}$ and $\frac{\left|5x+6y+5\right|}{\sqrt{5^{2}+6^{2}}}$ are gu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finding the radius of incircle of a triangle The question is -
AB and AC are the circular arcs of center O' and O of radius 5 units respectively. OO' = 6 unit. In $\Delta$ABC, AB = AC and a circle is inscribed in the triangle. The radius of the circle can be written as $\frac{a}{\sqrt b +1}$ where a, b are integers, ... | $AO = AO' = 5$ so perp from $A$ to $OO'$ will meet at the midpoint of $OO'$, say $H$. Then, $OH = O'H = 3$
So, $AH = \sqrt{AO^2 - OH^2} = \sqrt{5^2 - 3^2} = 4$
Now, $BH = BC/2 = 2$ and $AB = AC = \sqrt{AH^2 + BH^2} = \sqrt{4^2 + 2^2} = 2 \sqrt5$
Sub-perimeter of $\triangle ABC$,
$s = \frac 12 (4 + 2 \cdot 2 \sqrt5) = 2... | {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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If $y=y(x)$ and $\frac{2+\sin x}{y+1}(\frac{dy}{dx})=-\cos x,y(0)=1,$ then $y(\frac\pi2)=$
If $y=y(x)$ and $\dfrac{2+\sin x}{y+1}\left(\dfrac{dy}{dx}\right)=-\cos x,y(0)=1,$ then $y(\frac\pi2)=$
My solution:$$\frac{dy}{y+1}=-\frac{\cos x}{2+\sin x}dx\\\implies\ln|y+1|=-\ln(2+\sin x)+\ln c\\\implies|y+1|=\frac{c}{2+\s... | Since the denominator $2+\sin x$ is always positive, the absolute value can be absorbed into the $c$
$$y + 1 = \frac{c}{2+\sin x}$$
$y(0) = 1$ fixes the branch of the function to be the positive branch, so there is only one answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What's the area of the shaded regions in the triangle below? For reference: In triangle ABC, $S_1$ and $S_2$ are areas of the shaded regions. If $S_1 \cdot{S}_2=16 cm^4$, calculate $MN$.
My progress:
$\frac{AM.DM}{2}.\frac{CN.FN}{2}=16 \implies AM.DM.CN.FN=64\\
\frac{S1}{S2} = \frac{AM.MD}{CN.FN}\\
\frac{S1}{\frac{MI.... | Making a few observations does reduce the work considerably. Say $DM = x, FN = y$
First, observe that $~\displaystyle \frac{x}{AM} = \frac{CN}{y} \implies AM \cdot CN = xy$
$s_1 \cdot s_2 = \frac 12 AM \cdot x \cdot \frac 12 CN \cdot y \implies xy = 2 \sqrt{s_1 s_2} = 8$
Next observation: Given $BDIF$ is a rectangle, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4390855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate indefinite integral using susbtitution I have the following integral to evaluate. I think it should be done by substitution but I get stuck midway when I use $u=x^5$ and $du=5x^4dx$
$$\int x^{14}\sqrt{x^5+2}\,dx$$
| Let $u=\sqrt{x^{5}+2}$, then $2 u d x=5 x^{4} d x$
$$
\begin{aligned}
I &=\int\left(u^{2}-2\right)^{2} u \cdot \frac{2}{5} u d u \\
&=\frac{2}{5} \int\left(u^{6}-4 u^{4}+4 u^{3}\right) d u \\
&=\frac{2}{5}\left(\frac{u^{7}}{7}-\frac{4 u^{5}}{5}+\frac{4 u^{3}}{3}\right)+C \\
&=\frac{2 u^{3}}{525}\left(15 u^{4}-84 u^{2}+... | {
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"url": "https://math.stackexchange.com/questions/4395433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1... | The form you chose
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
is not general enough - your calculation proves this. It is clear apriori, since why should $1$ be expressible in the form $A(1-x)^2 + B(1+x)$? There are $3$ constraints (the coefficient of $x^2$ should be $0$, as well as that of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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An upper bound for the ratio of one random variable to the square root of the sum of squares of random variables Let $X_1, X_2, X_3$ be positive, iid, non-degenerate random variables with finite variances.
I wish to show that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{2}{\sqrt{3}... | Consider the distribution of $(X_1, X_2, X_3)$. Because $X_1, X_2, X_3$ are iid, it follows that they are exchangeable, hence the distribution of $(X_1, X_2, X_3)$ is the same as $(X_2, X_1, X_3)$.
By linearity of expectation, we have that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = \mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $(1+\frac{1}{n^3})^{n^2}$ I have been trying to solve the limit of $y_n = (1+\frac{1}{n^3})^{n^2}$. Through graphical analysis, I have found that $$\lim_{n \to \infty} y_n = 1$$ Which can also be intuitively be understood as $n^3 \geq n^2$. Using Bernoulli's inequality, you can easily find that $$y_n \geq (1+\... | We have
$$y^n =(1+\frac{1}{n^3})^{n^3} \to e$$
as $ n \to \infty.$ Hence there is $N$ such that
$$2 \le y^n \le 3$$
for $n>N.$
Can you proceed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4402975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Minimize $\frac{x^2}{x-9} $ given that $x > 9$ Its given that $ x > 9 $ and I have to find minima of
$$ y = \frac{x^2}{(x-9) } $$
I did this using three methods.
Method 1). Let
$$ f(x) = \frac{x^2}{(x-9) } $$
Using, calculus, we get
$$ f'(x) = \frac{2x}{(x-9)} - \frac{x^2}{(x-9)^2} $$
Using this, we can find that... | Yet another method:
$$ \frac{x^2}{x-9}=\frac{(x-18)^2+36x-18^2}{x-9}
=36+\frac{(x-18)^2}{x-9}\ge 36$$
with equality iff $x=18$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Finding the value of an expression using the roots of a given polynomial
Let $a,$ $b,$ $c,$ $d,$ and $e$ be the distinct roots of the equation $x^5 + 7x^4 - 2 = 0.$ Find
\begin{align*} &\frac{a^4}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^4}{(b - a)(b - c)(b - d)(b - e)} \\ &\quad + \frac{c^4}{(c - a)(c - b)(c - d)(c - ... | a Mobius transformation takes polynomials into rational functions. The roots of the polynomial are mapped to the roots of the resulting numerator. with polynomial $f(x) = x^5 + 7 x^4 - 2,$ I then wrote
$$x = \frac{28t}{-5t+1} $$
The resulting rational function $f(x)$ in $t$ was
$$ g(t) = \frac{-4296342t^5 + 429... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k... | The proof using twice integration by parts requires differentiability of $f.$ Instead we can use summation by parts as follows. The integral in question is the limit of Riemann sums
$$S_n:={1\over n}\sum_{j=1}^n f\left ({2\pi j\over n}\right )\cos {2\pi k j\over n}$$ It suffices to show that $\lim S_n\ge 0.$
Observe t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
Finding the symbolic expression for a particular matrix raised to the $n^{\text{th}}$ power Consider the following matrix, with $a$ being a real number between $0$ and $1$.
\begin{equation}
\text{M}=\left[
\begin{array}{cc}
1-\frac{a^2}{3} & \frac{1}{3} \left(2 a-a^2\right) \\
\frac{2 a^2}{3} & 1-\frac{2}{3} \left(2... | $
\def\a{\alpha}\def\b{\beta}\def\l{\lambda}
\def\o{{\tt1}}\def\p{\partial}
\def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)}
\def\qiq{\quad\implies\quad}
\def\grad#1#2{\frac{\p #1}{\p #2}}
\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
$Given a rank-one matrix $A=uv^T$, multiply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $4\leq |z+3|+|z^2-z+4|\leq 11$ where $|z|=2$ Let $z\in \mathbb{C}$ and$|z|=2$.Show that $$4\leq |z+3|+|z^2-z+4|\leq 11$$
Is there a non-calculus way to do this
My Attempt:
I used method which involved calculus.
Let $z=2(\cos t+i\sin t)$
$\Rightarrow|z+3|=\sqrt{13+12\cos t}$
$\Rightarrow|z^2-z+4|=2|4\cos t-1|$... | From the OP's proof by calculus:
Let $f(x )= \sqrt{13 + 12x} + 2|4x - 1|$ where $x \in [-1, 1].$
The expression $u = \sqrt{13 + 12x}$ is an increasing function of
$x$ in $[-1, 1],$ and $12x = u^2 - 13,$ therefore:
$$
f(x) = \begin{cases}
u + 2 - \tfrac23(u^2 - 13) = 11 - \tfrac13(u - 1)(2u - 1) &
(-1 \leqslant x \leq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Prove that $(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$
Let, $x,y,z>0$ such that $xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(x+y+z)^2$$
My progress:
Using the Cauchy-Schwars inequality I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥2(xy+yz+xz)(x+y+z)≥2(x+y+z)^2 \implies xy+yz+xz≥x+y+z$$
But, this is not always tr... | This is not an answer! But this is what I have so far. One can consider symmetric polynomials $\sigma_1=x+y+z,\ \sigma_2=xy+yz+zz,\ \sigma_3=xyz=1\ $.
We see that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yx+xz)=\sigma_1^2-2\sigma_2.$ Using that we can rewrite your inequality in the following way.
$$\sigma_2(\sigma_1^2-2\sigma_2+\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Seeking nice proof of $ \frac {(ab-1)(bc-1)(ca-1)} {(a^2-1)(b^2-1)(c^2-1)} < \frac 9 8$
The inequality $$ 1 < \frac {(ab-1)(bc-1)(ca-1)} {(a^2-1)(b^2-1)(c^2-1)} < \frac 9 8 $$ is true if $3 \le a<b<c$.
The lower bound is quite easy.
For the upper bound let $a=x+3,b=y+3,c=z+3$ and expand 9 $\times$ denominator $-$ 8 $... | Per your request, for the right inequality:
Using $3 \le a < b < c$, we have
$$ab - \frac{(bc - 1)(ca - 1)}{c^2 - 1} = \frac{bc + ac - ab - 1}{c^2 - 1} > 0.$$
Using $ab > \frac{(bc - 1)(ca - 1)}{c^2 - 1} > 0$, it suffices to prove that
$$\frac{ab - 1}{(a^2 - 1)(b^2 - 1)} \cdot ab \le \frac{9}{8}$$
or
$$a^2b^2 - 9a^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4426247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum $n$ such that $x^3-\frac{2\sqrt2+1}{2}x^2+\frac{2\sqrt2+1}{2}x-1\mid x^n-1$ Set $f(x)= x^3-\frac{2\sqrt2+1}{2}x^2+\frac{2\sqrt2+1}{2}x-1$. Next I found the roots of this polynomial that $x=1$ and $\frac{1}{2}(\frac{2\sqrt2 -1}{2} \pm i \frac{\sqrt7+1}{2})$. Here the $n$ is a natural number.
But I can't... | The question is wrong, in that the $f$ given in the question does not divide $x^n -1$ for any positive $n$. Let $f$ be a monic cubic of the form
$$f(x) = (x-1)(x^2-2cx+1)=x^3 - (1+2c)x^2 + (1+2c)x - 1$$ such that $c$ is quadratic over $\mathbb{Q}$ and such that $f\mid(x^n-1)$.
Since $c$ is quadratic over $\mathbb{Q}$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4428173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Point $D$ in $\triangle{ABC}$ with $\angle{ABD}=5^{\circ}, \angle{DBC}=20^{\circ}, \angle{DCB}=65^{\circ}, \angle{DAC}=40^{\circ}$, find $\angle{ACD}$ Point $D$ in $\triangle{ABC}$ with $\angle{ABD}=5^{\circ}, \angle{DBC}=20^{\circ}, \angle{DCB}=65^{\circ}, \angle{DAC}=40^{\circ}$, find $\angle{ACD}$.
Trigonometric Ce... | Introduction: I will use the parameter $x=\widehat{ABD}$ instead of the variable $a$ that appears in the comment of Ivan Kaznacheyeu.
It turns out that it is convenient to use also
$$
y := 30^\circ-2x\ .
$$
We try thus to show in a geometrical way a geometrical substitute for the trigonometric identity (Ceva):
$$
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Conjecture: Every $n \geq 20 \in \mathbb{N}$ can be written as a sum of three integers $(\geq 2)$ that are pairwise coprime This question on the sum of pairwise prime numbers piqued my interest, and I started looking at what numbers can be written as the sum of three pairwise coprime numbers (excluding $1$):
$$
\begin{... | If $n \ne 2 \!\!\! \mod 3$ and $n \ne 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=3+5+(n-8)$ provided $n \ge 10$.
If $n \ne 2 \!\!\! \mod 3$ and $n = 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=9+5+(n-14)$ provided $n \ge 16$.
If $n =2 \!\!\! \mod 3$ and $n \ne 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=3+25+(n-28)$ provided $n \ge 30$. $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I prove that, if $x,y,z>0$ and $xyz=1$, then $2(x^2+y^2+z^2)+9\geq 5(x+y+z)$
How can I prove that, if $x,y,z>0$ and $xyz=1$, then
$$2(x^2+y^2+z^2)+9\geq 5(x+y+z)$$
I used the famous inequality
$$x^2+y^2+z^2+3\geq 2(x+y+z)$$
I got $$2(x^2+y^2+z^2)+9\geq 4(x+y+z)+3\geq 5(x+y+z)$$
But, the last inequality gives ... | pqr method:
Let $p = x + y + z,\, q = xy + yz + zx,\, r = xyz = 1$.
It suffices to prove that $2(p^2 - 2q) + 9 \ge 5p$ or
$$2p^2 - 4q + 9 - 5p \ge 0.$$
Using $p^3 - 4pq + 9r \ge 0$ (three degree Schur), we have
$$q \le \frac{p^3 + 9}{4p}.$$
It suffices to prove that
$$2p^2 - 4\cdot \frac{p^3 + 9}{4p} + 9 - 5p \ge 0$$
o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the probability that at least one of the committee members is in upper 6? One of the questions in my workbook read:
The school bowling club has $20$ members from upper 6 and $10$ from lower 6.The club wishes to form a committee consisting of $3$ members. Find the probability that at least one of the committee ... | Your answer is not correct because your calculation does not reflect the fact that the committee selection is not ordered. So for instance, $$\frac{20}{30} \cdot \frac{10}{29} \cdot \frac{9}{28}$$ represents the probability that the first member selected is from upper 6, then the following two members are selected fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Fractions in Questions and Answers
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