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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\phi = 0 $ to be $\displaystyle C_1 e^{-3x} + C_2 e^{-x}.$
Now we need to find two functions $\lambda \ \text{and} \ \nu $ such that $\displaystyle \lambda'e^{-3x} + \nu'e^{-x} = 0 $ and $\displaystyle -3\lambda'e^{-3x} - \nu'e^{-x} = x \cos {3x} .$
Then the general solution would be : $\displaystyle \lambda e^{-3x} + \nu e^{-x} .$
On solving I got : $$\displaystyle \lambda = -\frac {1}{2} \int x \cdot \cos{3x} \cdot e^{3x} \mathrm {d}x \\ \nu = \frac {1}{2} \int x \cdot \cos{3x} \cdot e^{x} \mathrm {d}x $$
These integrals took me a long time to solve. I used the traditional DI method to solve for it.
Finally getting the answer as : $$ \displaystyle C_1 e^{-3x} + C_2 e^{-x} + \frac {x}{30} ( 2 \sin {3x} - \cos {3x} ) + \frac {47}{1800}\cos {3x} +\frac {19}{200}\sin {3x} . $$
I want my answer to get verified and am open to some good way to the integral quickly.
Thank you in advance.
| 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 solution of the form
$$\phi_p(x)=(Ax+B)\cos(3x)+(Cx+D)\sin(3x).$$
Hence
$$\phi_p'(x)=(A + 3Cx + 3D)\cos(3 x)+(-3 A x - 3 B + C) \sin(3 x), $$
$$\phi_p''(x)= (-9 A x - 9 B + 6 C)\cos(3 x) + (-6 A -9Cx-9D)\sin(3 x).$$
Plugging these into the differential equation gives
\begin{equation}
\begin{split}
(-6A+12C)x\cos(3x) +(-6C-12A)x\sin(3x)+(4A-6B+6C+12D)\cos(3x)\\
+(-6A-12B+4C-6D)\sin(3x)=x\cos(3x).
\end{split}
\end{equation}
Therefore by relating coefficients
$$-6C-12A =0 \implies C=-2A,$$
$$-6A+12C=1 \implies \boxed{A=-\frac{1}{30}}, \quad \boxed{C=\frac{1}{15}},$$
$$4A-6B+6C+12D=0 \implies -6B+12D = -\frac{4}{15}\implies D = \frac{B}{2} -\frac{1}{45},$$
$$-6A-12B+4C-6D=0 \implies -15B+\frac{2}{15}=-\frac{7}{15}\implies \boxed{B=\frac{1}{25}}, \quad \boxed{D= -\frac{1}{450}}.$$
So the particular solution becomes
$$\phi_p(x)=\left(-\frac{1}{30}x+\frac{1}{25}\right)\cos(3x)+\left(\frac{1}{15}x-\frac{1}{450}\right)\sin(3x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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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 by $2, 3, 3$ as $n$ increases.
For example:
$$\begin{align}
1^x = x!, &\quad c = \phantom{1}1 \\
2^x = x!, &\quad c = \phantom{1}3 \quad(=\phantom{1}1+\color{red}{2}) \\
3^x = x!, &\quad c = \phantom{1}6 \quad(=\phantom{1}3+\color{red}{3}) \\
4^x = x!, &\quad c = \phantom{1}9 \quad(=\phantom{1}6+\color{red}{3}) \\
5^x = x!, &\quad c = 11 \quad(=\phantom{1}9+\color{red}{2})\\
6^x = x!, &\quad c = 14 \quad(=11+\color{red}{3})
\end{align}$$
Is this pattern true as $n$ goes to infinity? And, if so, why is this the case?
| 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 an average of about $2.667$. As $n\to \infty$ the average is in fact $e \approx 2.718$ so slightly more than two-thirds of the steps will be $3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 function and its inverse are mirror images in $y=x$. Not sure how to use that here.
Generally, we find inverse function by writing $x$ in terms of $y$, where $y=f(x)$. Here, $$y=x^3+x-1$$
Not able to obtain $f^{-1}(x)$ from this.
| 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
$$(x^3+x-1)^3+(x^3+x-1)-1=x,$$
which shows that the polynomial is divisible by $x^3-1$. Then a quick check shows that
$$x^9+3x^7-3x^6+3x^5-6x^4+5x^3-3x^2+3x-3=(x^3-1)(x^6+3x^4-2x^3+3x^2-3x+3).$$
Now to show that $x=1$ is the unique real solution, it suffices to show that the sextic factor has no real roots. It is not hard to express it as a sum of squares:
\begin{eqnarray*}
x^6+3x^4-2x^3+3x^2-3x+3&=&(x^3-1)^2+3x^4+3x^2-3x+2\\
&=&(x^3-1)^2+3x^4+3(x-\tfrac12)^2+\tfrac54.
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin\dfrac{C}{2}+\sin\dfrac{A}{2}\sin\dfrac{C}{2})$
Here all I did:
$p=\dfrac{a+b+c}{2}$
$\sin \dfrac{A}{2} = \sqrt { \dfrac {(p-b)(p-c)}{bc}} $
$\Rightarrow \sin \dfrac{A}{2} \sin \dfrac{B}{2} = \dfrac{(p-c)}{c} \sqrt { \dfrac {(p-a)(p-b)}{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {2p-a-b}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {c}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2}\le \dfrac{(p-c)}{\sqrt{ba}}$
So we need to prove that : $\sum \dfrac{(p-c)}{\sqrt{ba}} \le \sum\sqrt { \dfrac {(p-b)(p-c)}{bc}} $. But I still have no idea (I'm not sure that's true either). I hope to get help from everyone. Thanks a lot
| 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\leq \frac {3}{2} $. To prove this consider a $\triangle XYZ $ such that $2X=180^{\circ}-A $ and similarly. Then $s=$cos$X+$cos $Y+$cos $Z\leq 1+\frac {r}{R} $ wrt $\triangle XYZ $, so Euler's formula finishes the job.
Credit to https://artofproblemsolving.com/community/c6h1405767p7873009
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"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
$a_{i9} = \frac{180(n-2)}{n}=\frac{180(9-2}{9}=\angle140^\circ\\a_{i6}= \frac{180(6-2)}{6}=\angle120^\circ\\\angle ADB = \frac{360^\circ-2(140^\circ)}{2} = \angle 40^\circ\\\angle BJI = \frac{\angle 180^\circ-\angle 120^\circ}{2}=30^\circ\\a_{i5}= \frac{180(5-2)}{5}=\angle108^\circ\\\\
S_{ai5} = 180(n-2) = 180(5-2) =\angle 540^\circ \\
\angle HDE = \frac{540 - 3(140)}{2}=\angle 60^\circ$
| Hint: $\triangle BAN$ is isosceles (because $ABCN$ is a rhombus), and therefore so is $\triangle NAL$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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 $x, y, z \ge 0$.
But for real $x,y,z$ I can't solve it
| 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
$$2ku^4-(27u^2v^2-18v^4-3uw^3)\geq3u(x-y)(x-z)(y-z).$$
We'll prove that $2ku^4-(27u^2v^2-18v^4-3uw^3)\geq0.$
Indeed, this inequality is a linear inequality of $w^3$, which by $uvw$ says that it's enough to prove this inequality for equality case of two variables.
Let $y=z=1$.
We need to prove that $$\frac{2k}{81}(x+2)^4-2x^3-2x-2\geq0$$ and since $2k>27$, it's enough to prove that
$$(x+2)^4\geq6(x^3+x+1)$$ or $$x^4+2x^3+24x^2+26x+10\geq0$$ or
$$x^2(x+1)^2+23x^2+26x+10\geq0,$$ which is obvious.
Id est, it's enough to prove that:
$$(2ku^4-(27u^2v^2-18v^4-3uw^3))^2\geq9u^2\prod_{cyc}(a-b)^2$$ or
$$(2ku^4-(27u^2v^2-18v^4-3uw^3))^2\geq243u^2(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6),$$
which is a quadratic inequality of $w^3$.
Now, prove that $\Delta\leq0.$
Can you end it now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 tried to solve it in the following manner:
\begin{align}\lim_{x\rightarrow 0}\frac{1-\cos(x^2)}{x^3(4^x-1)} &=\lim_{x\rightarrow 0}(\frac{2\sin^2(\frac{x^2}{2})}{x^4}\cdot\frac{x}{4^x-1})\\
&=\lim_{x\rightarrow 0}(\frac{2\sin^2(\frac{x^2}{2})}{(\frac{x^2}{2})^2\cdot 4}\cdot\frac{x}{4^x-1})\\
&=\lim_{x\rightarrow 0}(\frac{1}{2}\frac{\sin^2(\frac{x^2}{2})}{(\frac{x^2}{2})^2})\lim_{x\rightarrow 0}(\frac{x}{4^x-1})\\
&=\frac{1}{2}\frac{1}{\ln(4)}\\
&=\frac{1}{4}\log_2(e).
\end{align}
Is my solution correct? Or am I missing something?
| As @Joe said, it would be simpler to write it as $\frac 1{2 \ln (4)}$, but there is nothing wrong with your solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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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} \\
&=A(A+B)^{-1}B
\end{align}
\begin{align}
(A^{-1} + B^{-1})^{-1}B^{-1} &= A(A+B)^{-1}
\end{align}
\begin{align}
-(A^{-1} + B^{-1})^{-1}B^{-1}A &=- A(A+B)^{-1}A
\end{align}
\begin{align}
A-(A^{-1} + B^{-1})^{-1}B^{-1} A&=A- A(A+B)^{-1}A
\end{align}
But I'm stuck. I would be very appreciate if someone could help.
| 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.
| {
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"timestamp": "2023-03-29T00:00:00",
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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:
$f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du$
It follows that $|f(x)|\le\frac 1{2x}+\frac 1{2(x+1)}+\frac 1{2x}-\frac 1{2(x+1)}=\frac 1x$
But how to refute the possibility that $|f(x)|=\frac 1x$ for some $x\gt 0$?
I tried to prove the strict inequality using contradiction as follows:
Suppose on the contrary that there exists a $y\gt 0$ such that $|f(y)|=\frac 1y$, then $|f(y)|=\frac 1y=|\int_y^{y+1}\sin t^2\,dt|\le1\implies y\ge 1$
$y\ne 1$ as $y=1\implies |f(1)|=1=\int_1^2\sin t^2\, dt$, which is not true as RHS is approximately $0.50$. So we must have $y\gt 1$.
Let $g(x)=\frac 1x-f(x)$ and then I try to see if $g$ is monotonic on $(1,\infty)$ or not but $g'(x)=-\frac 1{x^2}+\sin x^2-\sin(x+1)^2\ge-1+\sin x^2-\sin (x+1)^2$.
I am stuck here. Please help. Thanks.
| 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
$$
|f(x)| \le\frac{|\cos x^2|}{2x}+\frac{|\cos (x+1)^2|}{2(x+1)}+\int_{x^2}^{(x+1)^2}\frac{|\cos u|}{4u^{3/2}}\, du \\
\overset{(*)}{<} \frac{1}{2x}+\frac{1}{2(x+1)}+\int_{x^2}^{(x+1)^2}\frac{1}{4u^{3/2}}\, du = \frac 1x \, .
$$
We have a strict inequality at $(*)$ because $|\cos u| = 1$ does not hold for all $u \in [x, x+1]$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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 $y = 0$.
I have tried rewriting it like $y' = \dfrac{2y}{3y + 1}$ but I don't know what to do next.
| You have got $\displaystyle y'=\frac{2\sqrt y}{3y-1}$
Therefore $\displaystyle \frac{dy}{dx}=\frac{2\sqrt y}{3y-1}$
$\displaystyle\implies\int \frac{3y-1}{2\sqrt y}dy=\int dx$
$\implies\displaystyle \frac{3}{2}\int\sqrt y-\frac{1}{2}\int\frac{dy}{\sqrt y}=x+C$
$\implies \displaystyle y^{\frac{3}{2}}-y^{\frac{1}{2}}=x+C$
Squaring both sides we get
$y^3+y-2y^2=(x+C)^2\implies y(y-1)^2=(x+C)^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $S_n$ we can form this type of triangle similar to Pascal's Trinagle
$$\begin{matrix}
&&&&&&&&&1\\
&&&&&&&1&&1&&1\\
&&&&&1&&2&&3&&2&&1\\
&&&&1&&3&&6&&7&&6&&3&&1\\
&&1&&4&&10&&16&&19&&16&10&&4&&1
\end{matrix}$$
| 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
$$(1+x+x^2)^{n+1}=P_0(n+1)+P_{1}(n+1)x + P_{2}(n+1)x^2=$$
$$=\left[P_0(n)+P_{1}(n) + P_{2}(n)\right]+\left[P_0(n)+P_{1}(n) + P_{2}(n)\right]x+\left[P_0(n)+P_{1}(n) + P_{2}(n)\right]x^2$$
and since $P_0(1)=P_1(1)=P_2(1)=1$, we easily obtain that
*
*$P_0(1)=1$
*$P_0(2)=3P_0(1)=3$
*$P_0(3)=3P_0(2)=3^2P_0(1)=3^2$
*$\cdots$
*$P_0(n)=3^{n-1}$
| {
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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:
$$f(x)=\frac{1}{(1+x^2)^2}=\frac{1}{(x+i)^2(x-i)^2}=\\
\frac{i}{4}\left( \frac{1}{x+i} \right)-\frac{1}{4}\left( \frac{1}{(x+i)^2} \right)-\frac{i}{4}\left( \frac{1}{x-i} \right)-\frac{1}{4}\left( \frac{1}{(x-i)^2} \right)$$
Then I used $y=\frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^nn!a^n}{(ax+b)^{n+1}}$ and $y=\frac{1}{(ax+b)^2} \Rightarrow y_{n}=\frac{(-1)^n(n+1)!a^n}{(ax+b)^{n+2}}$ to get:
$$\frac{(-1)^n}{4}\left( in!\left( \frac{(x-i)^{n+1}-(x+i)^{n+1}}{(x^2+1)^{n+1}} \right)-(n+1)!\left( \frac{(x-i)^{n+2}+(x+i)^{n+2}}{(x^2+1)^{n+2}} \right) \right)$$
Then I stopped for a second and I analized $(x-i)^{n+1}-(x+i)^{n+1}$ and $(x-i)^{n+2}+(x+i)^{n+2}$ using the binomial theorem:
-
$(x-i)^{n+1}-(x+i)^{n+1}$ = $$\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}\cdot i^k((-1)^k-1)$$
$i^k((-1)^k-1) =0$ for $k=2n$ ,whereas for $k=2n+1$ $\longrightarrow$$-2i(-1)^{\frac{k-1}{2}}$
-
$(x-i)^{n+2}+(x+i)^{n+2}$ = $$\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}\cdot i^k((-1)^k+1)$$
$i^k((-1)^k+1) =0$ for $k=2n+1$ , whereas for $k=2n$ $\longrightarrow$$2(-1)^{\frac{k}{2}}$.
And now I got stuck sorting out the indeces $n,k$ of summations; what I obtain is:
$$\frac{d^n}{dx^n}(f(x))=\\
\frac{(-1)^nn!}{2(x^2+1)^{n+1}}\cdot \sum_{k=1}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} - \frac{(-1)^n(n+1)!}{2(x^2+1)^{n+2}}\cdot \sum_{k=2}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}$$
But testing it I can see it’s wrong.
Can anyone help me please sorting the $n,k$ indices and perhaps have a look at my computations also?
Thank you
| 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. We know $\frac{1}{1-t}=\sum_{k=0}^\infty t^k$ for $|t|<1$. If we calculate the derivative, we get
$$
\frac{1}{(1-t)^2}=\sum_{k=0}^\infty kt^{k-1}=\sum_{k=0}^\infty (k+1)t^{k}.
$$
Now, let $t=-x^2$, then
\begin{align}
\frac{1}{(1+x^2)^2}
&=\sum_{k=0}^\infty (k+1)(-x^2)^{k}\\
&=\sum_{k=0}^\infty (-1)^k(k+1)x^{2k}\\
&=\sum_{2\mid n}^\infty (-1)^{n/2}\left(\frac{n}{2}+1\right)x^{n}\\
&=\sum_{2\mid n}^\infty \frac{n!(-1)^{n/2}(\frac{n}{2}+1)}{n!}x^{n}.
\end{align}
Hence by taylor's formula the $n$-th derivative of $f$ is $$f^{(n)}(0)=\begin{cases}0&2\nmid n\\ n!\cdot (-1)^{n/2}\cdot \left(\frac{n}{2}+1\right)&2\mid n \end{cases}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the derivative matrix $g\circ f$ at $\dbinom uv$ Vector function $f$ and $g$ are defined by:
$$f \dbinom uv=\begin{pmatrix}u\cos(v)\\u\sin(v)\end{pmatrix},\begin{pmatrix}0<u<\infty\\-\frac{\pi}{2}<v<\frac{\pi}{2}\end{pmatrix}$$
$$g \dbinom xy=\begin{pmatrix}\sqrt{x^2+y^2}\\\arctan(\frac{y}{x})\end{pmatrix}, 0<x<\infty$$
Find the derivative matrix of $g\circ f$ at $\dbinom uv$.
My workings:
$$f' \dbinom uv=\begin{pmatrix}\cos(v) & -u\sin(v)\\\sin(v)&u\cos(v)\end{pmatrix}$$
$$g' \dbinom xy=\begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}}\\-\frac{y}{x^2+y^2}&\frac{x}{x^2+y^2}\end{pmatrix}$$
I had thought I would have to multiply $f'$ with $g'$ and replace $x,y$ with $u,v$, however I do not get the final expected result which is:
$$\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
| 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
$$\begin{pmatrix}\frac{x}{\sqrt{x^2+y^2}}& \frac{y}{\sqrt{x^2+y^2}}\\-\frac{y}{x^2+y^2}&\frac{x}{x^2+y^2}\end{pmatrix}_{(u\cos v, u \sin v)}\cdot \begin{pmatrix}\cos v & -u\sin v\\\sin v&u\cos v\end{pmatrix}$$
that is
$$\begin{pmatrix}\cos v& \sin v\\-\frac{\sin v}u&\frac{cos v}u\end{pmatrix}\cdot \begin{pmatrix}\cos v & -u\sin v\\\sin v&u\cos v\end{pmatrix}=\begin{pmatrix}1 & 0\\ 0&1\end{pmatrix}$$
Refer also to the related
*
*Derivation of the multivariate chain rule
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 the graph of the function it looks like the limit is indeed zero. Also I thought I was able to prove the limit does indeed exist at $(0,0)$ with the epsilon-delta definition:
$$y^2 \leq x^2+y^2 \leq x^2 + y^4 \iff \frac{y^2}{x^2+y^4} \leq 1 \overset{|x|<1}{\iff} \frac{|x|y^2}{x^2+y^4} \leq 1 \iff \frac{|x|y^2\sqrt{x^2+y^2}}{x^2+y^4} \leq \sqrt{x^2+y^2} \iff \left|\frac{xy^2\sqrt{x^2+y^2}}{x^2+y^4} - 0\right|\leq \left|\sqrt{(x-0)^2+(y-0)^2}\right| \lt δ := ε $$
So for $δ = ε$ we have that $\|x -(0,0)\| < δ \implies |f(x,y) - 0| < ε$
I'm confused
| As an alternative, by $y^2=v$ and polar coordinates, with $x=\rho\cos \theta$ and $v=\rho|\sin \theta|$, we obtain
$$ \frac{xy^2\sqrt{x^2+y^2}}{x^2 +y^4} =\frac{xv\sqrt{x^2+v}}{x^2 +v^2}=\sqrt \rho \cos \theta |\sin \theta| \sqrt{\rho\cos^2 \theta|\sin \theta|}\to 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 other than attempting to prove that the answer is $\left[0, \frac{1}{2}\right)$. Approacing the left and right endpoints as sequences and taking $n \to \infty$ didn't help.
I'm going to attempt to prove this.
Let $x \in \left[0, \frac{1}{3}\right)$. Then $0 \leq x < \frac{1}{3}$. We must show that $x \in A_n$ for all $n$. So fix $n \geq 1$. Certainly $- \frac{1}{n} < 0 \leq x$. Now we show that $x < \frac{n}{2n+1}$. Since $n \geq 1$, $2n \geq 2$ and $2n + 1 \geq 3$, so $\frac{1}{2n+1} \leq \frac{1}{3}$. Certainly, because $n \geq 1$, $\frac{n}{2n+1} \geq \frac{1}{2n+1}$, but that doesn't help, so at this point I'm stuck.
For the reverse direction, let $x \in \bigcap\limits_{n=1}^{\infty} A_n$, so $x \in \left(- \frac{1}{n}, \frac{n}{2n+1}\right)$ for all $n$. So $- \frac{1}{n} < x < \frac{n}{2n+1}$. Outside of taking $n \to \infty$, I don't know how to proceed.
Any help or hints would help.
| 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 $x\ge 0$. Also, $x < \frac{n}{2n+1}$ for all $n\ge 1$. This is where you have to be careful - simply taking limits as $n\to\infty$ will not help. Here's why:
$$x < \frac{n}{2n+1} = \frac12\cdot \left(1 - \frac{1}{2n+1} \right) \quad (\forall n\ge 1)$$
and $\frac12\cdot \left(1 - \frac{1}{2n+1} \right)$ is an increasing function of $n$. As a result, if $x < \frac12\cdot \left(1 - \frac{1}{2k+1} \right)$ for some $k\in \mathbb N$, then we certainly have $x < \frac12\cdot \left(1 - \frac{1}{2m+1} \right)$ for all $m \ge k$. Thus, put $n=1$ and see that $x < \frac13$. As discussed, $x < \frac12\cdot \left(1 - \frac{1}{2n+1} \right)$ for all $n\ge 1$ is then automatically satisfied. Hence, $x \in [0, \frac13)$.
*On the other hand, suppose $x\in [0,\frac13)$. You have already shown $x > -\frac1n$ for all $n\ge 1$. $\frac12\cdot \left(1 - \frac{1}{2n+1} \right)$ is an increasing function of $n \in \mathbb N$, and its minimum (over natural numbers) is attained at $n=1$. In particular, this minimum is $\frac13$. Thus $x < \frac13$ ensures that $x < \frac12\cdot \left(1 - \frac{1}{2n+1} \right)$ for all $n\ge 1$. Consequently, $x\in \bigcap_{n=1}^\infty A_n$, as desired.
Let me know if you have any questions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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, because $n$ was arbitrary, this holds for all $n$? The only thing left out would be the base case, but. I'm not fully sure why I need it here.
Regardless, here is my attempt at the induction:
Let $f: \mathbb{N} \to \mathbb{R}$ be defined by $f(n) = \frac{n}{2n+1}$. We prove by induction on $n$ that $f$ is increasing in $n$. If $n = 1$, we notice that
\begin{align*}
f(1) = \frac{1}{3} \leq \frac{2}{5} = f(2).
\end{align*}
Suppose inductively that we have $f(n) \leq f(n+1)$ for some $n \geq 1$. So we have
$\frac{n}{2n+1} \leq \frac{n+1}{2n+3}$. First, we have
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3}.
\end{align*}
Furthermore, $2n + 5 \geq 2n + 3$, so $\frac{1}{2n + 5} \leq \frac{1}{2n+3}$, so $\frac{n+3}{2n + 3} \leq \frac{n+3}{2n+5}$. Therefore, it follows that
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3} \leq \frac{n+3}{2n + 5} = \frac{(n+2) + 1}{2(n+2) + 1},
\end{align*}
so $f(n+1) \leq f(n+2)$, which closes the induction
| 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",
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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
$\sin a+ \sin b + \sin c\ge \frac 32\implies \sin (a-\pi/6)+ \sin(b-\pi/6) + \sin (c-\pi/6)\ge 0.$
The first question was to prove that if
$x,y,z$ are real numbers, then $(x+y+z)^2\le 3(x^2+y^2+z^2)$. I proved this part, it is a direct result using $x^2+y^2\ge 2xy$ and by expanding $(x+y+z)^2$. Thanks for your help.
| Part 1:
Let $0\le k\le 3$ be any given real number then $\begin{align} & \sin a+ \sin b + \sin c\ge k\\
\implies & (\sin a+ \sin b + \sin c)^2 \ge k^2\\
\implies & 3(\sin^2 a + \sin^2 b +\sin^2 c) \ge (\sin a+ \sin b + \sin c)^2 \ge k^2 \tag{1}\\
\implies & 3(1-\cos^2 a + 1-\cos^2 b +1- \cos^2 c) \ge k^2\\
\implies & 9-3(\cos^2 a +\cos^2 b + \cos^2 c) \ge k^2\\
\implies & 3(\cos^2 a +\cos^2 b + \cos^2 c) \le 9-k^2\\
\implies & (\cos a + \cos b +\cos c)^2\le 3(\cos^2 a +\cos^2 b + \cos^2 c) \le 9-k^2 \tag{2}\\
\implies &-\sqrt{9-k^2} \le \cos a + \cos b +\cos c \le \sqrt{9-k^2} \tag{3}\end{align}$
where in $(1)$ and $(2)$ we used $(x+y+z)^2\le 3(x^2+y^2+z^2)$
For your first question put $k=2$ and you'll get $\sin a+ \sin b + \sin c\ge 2 \implies \cos a + \cos b +\cos c \le \sqrt{9-2^2}=\sqrt{5}$
Part 2:
Given that $\sin a+ \sin b + \sin c\ge \frac{3}{2} \tag{4}$
We get, $\cos a + \cos b +\cos c \le \sqrt{9-\left(\frac{3}{2}\right)^2}={3\sqrt{3}\over {2}}$ using $(3)$
This implies $-\frac{1}{2} (\cos a + \cos b +\cos c) \ge -{3\sqrt{3}\over {4}} \tag{5}$
Now use $(4)$ and $(5)$ to prove $\sin (a-\pi/6)+ \sin(b-\pi/6) + \sin (c-\pi/6)\ge 0$
| {
"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 .
Second example:
$a=1 ,b=4,c=1$ and $d=0$ we have the equation : $x^2+4y^2=z^2$ . it has solutions.
one solution of it is: $x=3,y=2,$ and $z=5$
Third example:
$a=2 ,b=3,c=1$ and $d=0$ we have the equation : $2x^2+3y^2=z^2$ .
this example I tried to find solutions among small numbers but I didn't find any solution.
So does $2x^2+3y^2=z^2$ have solutions but I didn't find any ?
or is there proof that $2x^2+3y^2=z^2$ doesn't have any solution?
| 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 equation $X^2-bcY^2=\pm b\tag{2}.$
If equation $(2)$ has a solution , then equation $(1)$ has infinitely many positive integer solutions.
In particular, let $b=1$ then if $c$ is not a perfect square, Pell's equation $X^2-cY^2=1$ always has infinitely many distinct integer solutions.
Example for $(a,b,c,d)=(4,1,2,3): p^2-2q^2 = 1.$
$(x,y,z)=(1, 7, 5),(1, 239, 169),(1, 8119, 5741),(1, 275807, 195025)$...
Example for $(a,b,c,d)=(14, 1, 13, 2): p^2-13q^2 = 1.$
$(x,y,z)=(1, 2194919, 608761),(1, 3698003921039, 1025641750321)$...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $\displaystyle x=\frac{c}{3x}$ Then $\displaystyle\left(\frac{c^2}{3x^2}+\frac{2c}{3x}+c\right)^{12}*\frac{x^{24}}{x^{24}}=\left(\frac{x^2+\frac{2cx}{3}+\frac{c^2}{3}}{x^2}\right)^{12}$
We can also do this through the general term of multinomial coefficient, but that would be a very long method and I only have $5$ minutes in the exam. Can someone help, thanks $:)$
| Let $x\rightarrow\frac{c}{3x}$
$$(3x^2+2x+c)^{12}=\sum\limits_{r=0}^{24}A_rx^r$$
$$\left(\frac{c^2}{3x^2}+\frac{2c}{3x}+c\right)^{12}=\sum_{r=0}^{24}A_rc^r3^{-r}x^{-r}$$
$$\frac{c^{12}}{3^{12}x^{24}}\left(3x^2+2x+c\right)^{12}=\sum_{r=0}^{24}A_rc^{r}3^{-r}x^{-r}$$
$$\left(3x^2+2x+c\right)^{12}=\sum_{r=0}^{24}A_rc^{r-12}3^{12-r}x^{24-r}$$
Then the coefficient of $x^5$ on both sides should be the same which gives
$$A_5=A_{19}c^{7}3^{-7}\implies c^{7}=3^{7}\left(\frac{A_{19}}{A_5}\right)^{-1}\implies c=6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246819",
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"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 we can assume $0<c\leq b \leq a$, so $M=\min\{(b-c)^2,(c-a)^2,(a-b)^2\} = \min\{(b-c)^2,(a-b)^2\}$ and $a-c=a-b+b-c \geq 2\min\{(a-b),(b-c)\}$ so $(a-c)^2\geq 4\min\{(a-b),(b-c)\}^2=4M$. Thus we have $6M \leq (b-c)^2+(c-a)^2+(a-b)^2=2(a^2+b^2+c^2-(ab+ac+bc)) \leq 2(a^2+b^2+c^2)$ so $M\leq \frac{a^2+b^2+c^2}{3}$.
| 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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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 but not sure how to find the other roots.
Another approach I tried is dividing numerator and denominator of the original fraction by $x^6$,
$$\frac{x^5+\frac1{x^5}}{x+\frac1x}=\frac{205}{16}$$
$$x^4-x^2+1-\frac1{x^2}+\frac1{x^4}=\frac{205}{16}$$
| 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 $t=x^2\ge 0$
$$\iff \frac{t^4-t^3+t^2-t+1}{t^2}=\frac{205}{16} \iff t^2-t+1-\frac1 t+\frac1{t^2}=\frac{205}{16}$$
$$\iff t^2+2+\frac1{t^2}-t-\frac1 t-\frac{221}{16}=0 \iff \left(t+\frac1t\right)^2-\left(t+\frac1t\right)-\frac{221}{16}=0$$
and by $u= t+\frac1t$ we obtain
$$u^2-u-\frac{221}{16}=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "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 number isn't $1$, then it can have $7$ different values. We can choose two $1$ in $5\cdot4$ ways and there are left $3$ digits that can be equal to $8$ different values.
In total $7\cdot5\cdot4\cdot8^3=71680$.
Adding all of this I get $92160$ different numbers.
| $\mathbf{\text{Alternative Approach:}}$
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 : $$\bigg(1 + x +\frac{x^2}{2!}+\frac{x^3}{3!} +\frac{x^4}{4!} \bigg)^8$$ , because of there are two $1's$ , we have maximum $4$ times same number.Moreover , because of any two and ones will not be used again , the exponential is $8$
Now find the coefficient of $x^6$ in the expansion of $$\bigg(1 + x +\frac{x^2}{2!}+\frac{x^3}{3!} +\frac{x^4}{4!} \bigg)^8 \times \bigg(\frac{x^2}{2!} \bigg)$$ and multiply it by $6!$. (you can also find the coefficient of $\frac{x^6}{6!}$)
However , realize that this calculation counts the strings that start with zero , so we must subtract them from the result. Now, lets assume that it start with zero and have two $1's$.To do that , find the expansion of $$\bigg(1 + x +\frac{x^2}{2!}+\frac{x^3}{3!} +\frac{x^4}{4!} \bigg)^8 \times \bigg(\frac{x^2}{2!} \bigg)$$
, after that find the coefficient of $x^5$ and multiply it by $5!$.(you can also find the coefficient of $\frac{x^5}{5!}$)
Now , it is the time for subtracting them..
Calculation by wolfram
$$6! \times \frac{256}{3} - 5! \times \frac{128}{3}=61440-5120=56320$$
Unfortunately, your answer is wrong..
| {
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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{aligned}
\sum_{k=0}^n \frac{\binom{2n}{k}}{k+1} &= \frac{1}{2n+1} \sum_{k=0}^n \binom{2n+1}{k+1} \\
&= \frac{1}{2n+1} \left(\sum_{k=0}^{n+1} \binom{2n+1}{k} - 1 \right) \\
&= \frac{1}{2n+1} \left(\frac{2^{2n+1}}{2} + \binom{2n+1}{n+1} - 1 \right) \\
&= \frac{1}{2n+1} \left(2^{2n} + \binom{2n+1}{n}- 1 \right).
\end{aligned}
| {
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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 rule, I can get
the second derivative which is $$f''(x)=(-b)(-b-1)(1-x^2)^{-b-2}(-2x)^2+(-b)(1-x^2)^{-b-1}(-2).$$ Again, the third derivative is $$f'''(x)=(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}(-2x)^3+(-b)(-b-1)(1-x^2)^{-b-2}2(-2x)(-2)+(-b)(-b-1)(1-x^2)^{-b-2}(-2x)(-2).$$ The fouth derivative is $$f^{(4)}(x)=(-b)(-b-1)(-b-2)(-b-3)(1-x^2)^{-b-4}(-2x)^4+(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}3(-2x)^2(-2)+(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}2(-2x)^2(-2)+(-b)(-b-1)(1-x^2)^{-b-2}2(-2x)(-2)+(-b)(-b-1)(-b-2)(1-x^2)^{-b-3}(-2x)^2(-2)+(-b)(-b-1)(1-x^2)^{-b-2}(-2)(-2).$$
My question is: Is there a general formula or a pattern for the $n$th derivative of $f(x)$? Any suggestions or comments would be very welcome. Thanks in advance.
| 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 $\,_2F_1(a,b;c;d)$ denotes the Ordinary Hypergeometric function.
This closed form matches with the examples you provided.
| {
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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.
Cubing both sides doesn't work too.
| 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 prove that
$$\sqrt[3]{108+17x}+10\left(\frac{1}{27}-\frac{x}{27}\right)\geq5$$ or
$$27\sqrt[3]{108+17x}\geq125+10x,$$ where $0\leq x\leq1,$ which is true because
$$27\sqrt[3]{108+17x}=9\sqrt[3]{27\cdot108+27\cdot17x}\geq9\sqrt[3]{(14+x)^3}=9(14+x)\geq125+10x.$$
| {
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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 $\lim_{R\to \infty}\int_0^{\infty} \frac{x\sin(x)}{x^2+1}\ \mathrm{d}x$ is only briefly discussed by stating that it can be shown using integration by parts. I do not see that.
How can I use integration by parts to proof the existence of this limit?
| 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 without integral, which can be easily seen). So the convergence of our integral is equivalent with convergence of $$\int_0^\infty \frac{1-x^2}{(1+x^2)^2}\cos(x)dx $$
And the latter is even absolutelly convergent, because (by $|\cos(x)| \le 1$ and triangle inequality) $\frac{|1-x^2|
}{(1+x^2)^2}|\cos(x)| \le \frac{1}{(1+x^2)^2} + \frac{x^2}{(1+x^2)^2}$
| {
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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 $y=\sqrt{x+\sqrt{2x-1}}$:
$$y^2=x+\sqrt{2x-1}$$
$$(y^2-x)^2=2x-1$$
$$y^4-2xy^2+x^2=2x-1$$
$$y^2(y^2-2x)=-x^2+2x-1$$
But I still have $-2x$ in the parenthesis of the LHS.
| 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, $x=y^2\pm\sqrt2y+1$.
| {
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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
$$ a_n = 7a_{n-1}- 10a_{n-2}$$
$$ \frac{r^n}{r^{n-2}} = \frac{7r^{n-1}}{r^{n-2}}- \frac{10r^{n-2}}{r^{n-2}}$$
Then the characteristic equation for the homogeneous part is
$$ r^2 = 7r - 10 $$ where $ r_1=2, r_2=5$.
So the homogeneous part is
$$ a_n^h=\alpha_1r_1^n + \alpha_2r_2^n = \alpha_1(2)^n+\alpha_2(5)^n$$
and I think the equation of the non-homogeneous part should be
$$ a_n^p = A(5^n)n+B$$
$$An(5^n)+B = 7(A(5^{n-1})(n-1)+B) -10(A(5^{n-2})(n-2)+B)$$
$$An(5^n)+B = 7(A(5^{n-1})(n-1))+7B-10(A(5^{n-2})(n-2))-10B$$
And I know there's a way to find $A$ and $B$ easily with like terms, but I don't know if I can use it here or how to use it.
| 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 - 6 \\
&= \left(\frac{7}{5}-\frac{10}{25}\right)An5^n + \left(-\frac{7}{5}+\frac{20}{25}\right)A5^n -3B + 5^n - 6 \\
&= An5^n + \left(1-\frac{3A}{5}\right)5^n - 3B - 6
\end{align}
Comparing like terms yields:
\begin{align}
0 &= 1-\frac{3A}{5} \\
B &= -3B - 6
\end{align}
So $A=5/3$ and $B=-3/2$, which implies particular solution
$$a_n^p = \frac{5}{3}n5^n - \frac{3}{2} = \frac{n5^{n+1}}{3} - \frac{3}{2}$$
and general solution
$$a_n = a_n^h + a_n^p = \alpha_1 2^n+\alpha_2 5^n + \frac{n5^{n+1}}{3} - \frac{3}{2}$$
Use any two provided initial conditions (which you did not specify) to solve for $\alpha_1$ and $\alpha_2$.
| {
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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 is $5/7$. The sum of the squares of the equations can be found by solving $-7s+5(5/7)-12=0$, for $s$ (s is the sum of the squares of the roots). I get $59/-49$
Question: Am I correct or are my steps wrong. I know that if you simplify $\frac{1}{r^2}$+$\frac{1}{s^2}$+$\frac{1}{t^2}$ you get $\frac{s^2t^2+r^2t^2+r^2s^2}{r^2s^2t^2}$ but I don't know how to calculate the numerator of the fraction.
Thanks in advance.
| $$x^3-6x^2+5x-7=0$$
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, $$\frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}\\=y_1^2+y_2^2+y_3^2\\=(y_1+y_2+y_3)^2-2(y_1y_2+y_2y_3+y_3y_1)\\=\left(\frac57\right)^2-2\left(\frac67\right)\\=-\frac{59}{49}.$$
| {
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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 think I can solve for $y\equiv 1 \pmod{20}, y\equiv 6 \pmod{15}, y\equiv 9 \pmod{18}$ by using extended chinese remainder theorem,then we have to search for the square numbers out of them.How to do that?
| $$x^2 \equiv 1 \pmod{5} \iff x \equiv \pm 1 \pmod{5}$$
$$x^2 \equiv 1 \pmod{4} \iff x \equiv 1,3 \pmod{4}$$
$$x^2 \equiv 0 \pmod{9} \iff x \equiv 0 \pmod{3}$$
| {
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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, which is $k+1$. If we factor that out we get $4(5^k+m)$. What do I do with that remaining $m$? And is $4(5^k+m)$ equal to the $k+1$ was this all that I need for my final answer.
| 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 of 4.
| {
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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 $A^2\not\equiv B^2\pmod{4}$
| {
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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 $|x^2 +2x - 5 - 3| = |x+4||x-2|$.
Showing right hand limit. Let $0 < x-2 \le 1$. Then $|x+4| = |x-2+6| \leq 1 + 6$. Set $\delta_1 = \min\{1,\frac \epsilon 7\}$. Then if $0 < x-2<\delta_1$, this implies that $|x+4||x-2| \le 7|x-2|<\epsilon$.
Showing left hand limit. Let $-1 < x-2 < 0$. Then $|x+4| \le |x-2| + 4 < 4$. Then set $\delta_2 = \min\{1,\frac \epsilon 4\}$. Then if $-\delta_2 < x-2<0$, this implies $|x+4||x-2|\le 4|x-2|<\epsilon$.
Because the $\lim_{x\to 2+}f(x) = 3 = \lim_{x\to 2-}f(x)$, we conclude that $\lim_{x\to 2}f(x) = 3$.
I am a bit iffy on proving left hand limits. Please help.
| 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*}
Can you take it from here?
| {
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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 need to use this information to find the value of
$$\tan{\frac{\pi}{12}}-\cot{\frac{\pi}{12}}+\tan{\frac{5\pi}{6}}-\cot{\frac{5\pi}{6}}$$
I know that I can rearrange $\tan{\frac{\pi}{12}}-\cot{\frac{\pi}{12}}$ into $-2\left(\frac{2\tan{\frac{\pi}{12}}}{1-\tan{\frac{\pi}{12}}^2}\right)^{-1}$ and then use double angle to find the value but this doesn't use the part beforehand.
Also from playing on the calculator I know $-\cot{\frac{\pi}{12}}=\tan{\frac{7\pi}{12}}$ and that $-\cot{\frac{5\pi}{6}}=\tan{\frac{\pi}{3}}$ so if I can prove these two results I can just use sum of roots but I don't know how.
| Another way :$\cot x-\tan x=\cdots=2\cot2x$
$$p=\implies\tan\dfrac\pi{12}-\cot\dfrac\pi{12}+\tan\dfrac{5\pi}6-\cot\dfrac{5\pi}6=-2\cot\dfrac\pi6-2\cot\dfrac{5\pi}3$$
$$\cot\dfrac{5\pi}3=\cot\left(2\pi-\dfrac\pi3\right)=\cot\dfrac\pi3=-\tan\dfrac\pi6$$
$$p=-2\cot\dfrac\pi6+2\tan\dfrac\pi6=-2\left(2\cot\dfrac\pi3\right)$$
$$p^2=\dfrac{16}3$$
| {
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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,
$$\begin{aligned} B^2 &= (I+A)^2 = 2A \\ B^3 &= (I+A)2A = 2A-2I\\ B^4 &= (I+A)(2A-2I) = -4I\\ \vdots \\ B^{10} &= 32A \end{aligned}$$
Is there a simpler method or a smarter approach if you want to do this for, e.g., $B^{100}$?
| $$
B^2 = 2A, A^2=-I \implies
B^{100}= (B^2)^{50} = (2A)^{50} = 2^{50} A^{50} = 2^{50} (A^{2})^{25} = 2^{50} (-I)^{25} = -2^{50} I
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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{eqn1}
\end{cases}
Then find the limit of:
\begin{align}
L=\lim_{x \to 0} \frac{\frac{1}{e^{2f(x)}}-\sqrt[3]{1+f(x)}}{\ln\bigl(1+f(x)\bigr)}
\end{align}
My attempts
I can easily guess that $f(x)=x^2$, which totally satisfies \eqref{eqn1}. What I did was that I set $g(x)=f(x)-x^2$, then I got:
\begin{cases}
g(x)+g(y)=g(x+y)\\
g\left(\frac{1}{x}\right)x^4=g(x)
\tag{2}\label{eqn2}
\end{cases}
From \eqref{eqn1}, substitute $x=y=0$ we have $f(0)=0$
Now $g(1)=g(0)=0$, then I substitute $y=1-x$ in \eqref{eqn2} and got that
\begin{align}
g\left(\frac{1}{x}\right)x^4+g\left(\frac{1}{1-x}\right)(1-x)^4=0
\end{align}
Because the values of either $\frac1x$ or $\frac1{1-x}$ must lie between $0$ and $1$, I suspect that we can prove $g \equiv 0$.
Even if I have $f$ then, I'm still clueless about how to calculate the limit.
Any idea or solution or suggestion on the tools?
Any help is appreciated!
| 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
$$L=\lim_{x\to 0} \frac{-\frac{7x^2}{3}+\frac{19x^4}{9}-\cdots +\cdots }{x^2-\frac{x^4}{2}+\frac{x^6}{3}+\cdots} $$
Dividing both numerator and denominator by $x^2$ gives the result
$$L=-\frac{7}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $\triangle ABT$,
$\displaystyle \tan^2(x)=\frac{AB^2}{BT^2}=\frac{\cos^2(x)}{4\sin^2(x)} \implies \tan^4(x)=\frac{1}{4} \implies \tan(x)=\frac{1}{\sqrt{2}}$
Therefore $\cos(x)=\frac{\sqrt{6}}{3}$ and $\sin(x)=\frac{\sqrt{3}}{3}$.
$AT^2=4\cdot\dfrac{6}{9}=\dfrac{8}{3} \rightarrow AT=\dfrac{2\sqrt{6}}{3}\\BT^2=16\sin^2(x)=16\cdot\dfrac{1}{3} \rightarrow BT=\dfrac{4\sqrt{3}}{3}\\AB^2=\dfrac{8}{3}+\dfrac{16}{3}=8 \rightarrow AB=2\sqrt{2}$
But, $\displaystyle A=pr \rightarrow r=\frac{A}{p}\\\displaystyle \therefore r=\frac{\frac{2\sqrt{6}}{3}.\frac{4\sqrt{3}}{3}}{\frac{2\sqrt{6}}{3}+\frac{4\sqrt{3}}{3}+2\sqrt{2}}= 0.5569$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 both sides,
$$ x^2 - 2x = 0 $$
We have obtained
$$ x(x - 2) = 0 $$
It has two solutions: $ x = 0 $ and $ x = 2 $.
But if we substitute $ x = 0 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 1 + \sqrt{1} = 2 $$
If we substitute $ x = 2 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 3 + \sqrt{9} = 6 $$
Apparently I have made a mistake in some step that has led to this contradiction. Which step is incorrect in my solution above? It must be the squaring step that changes the minus sign to positive sign.
But squaring both sides is an often used step in many equations. What rules I need to keep in mind while solving such equations so that I do not get incorrect solution after squaring both sides?
| You can easily track where the extra solutions were introduced.
$$ x + 1 + \sqrt{4x + 1} = 0\to\color{red}{0+1+\sqrt1=0}$$
$$ x + 1 = -\sqrt{4x + 1}\to\color{red}{0+1=-\sqrt{0+1}}$$
$$ x^2 + 2x + 1 = 4x + 1\to\color{green}{0+0+1=0+1}$$
$$ x^2 - 2x = 0\to\color{green}{0-0=0}$$
$$ x(x - 2) = 0\to\color{green}{0(0-2)=0}$$
Repeat with $x=2$.
If you turn $a=b$ to $a^2=b^2$, you also allow $a=-b$.
| {
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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 that $\alpha + \beta = \frac{\pi}{3}$.
However I am unable to find a relation between these two triangles.
It would be appreciated if someone could help me with the proof or give some kind of hint.
Thank you in advance.
| Observe that$$\cos^{-1}\left(\frac{2}{\sqrt 7} \right) = \tan^{-1}\left(\frac{\sqrt3}{2} \right)$$
So we have
$$\tan^{-1}\left(\frac{\sqrt3}{2} \right) + \tan^{-1}\left(\frac{\sqrt3}{5} \right) = \tan^{-1}\left(\frac{\frac{\sqrt3}{2} + \frac{\sqrt3}{5}}{1 - \frac{\sqrt3}{2} \cdot \frac{\sqrt3}{5}} \right) = \tan^{-1}(\sqrt 3) = \frac{\pi}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 substitution I can get
$$u = x^{n-2}; du = (n-2)x^{n-3}; dv = x^2e^{-x^3}; v = -\frac{1}{3}e^{-x^3}$$
Plugging this in
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx = \left[-\frac{x^{n-2}}{3}e^{-x^3} \right]_0^{\infty}+\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx $$
$$\implies I_n = \int_0^{\infty}x^ne^{-x^3}dx =\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx$$
$$=\frac{n-3}{3}I_{n-3}$$
Then starting from $n=11$ I get
$$\frac{8}{3} \cdot \frac{5}{3} \cdot\frac{2}{3} \int_0^{\infty} x^2e^{-x^3}dx$$
However this does not produce the answer which is $2$. How should I go about the reduction formula for this?
| 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
$$\frac{1}{3}\int_0^{\infty}y^3 e^{-y}dy=\frac{1}{3}\Gamma(4)=\frac{3!}{3}=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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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 the $x$-axis at $(2,0)$, (assume that you do not know any other points of intersection)
What I tried to do so far:
Since its a tangent, I decided to differentiate the semi-circle function using the chain rule, getting:
$\frac{dy}{dx}$ = $\frac{-x}{\sqrt{1-x^2}}$
Then using the equation of a line formula:
$y=mx+c$
--> $y=\frac{-x}{\sqrt{1-x^2}}x +c$
Substituting the point in:
-->$0=\frac{-2}{\sqrt{1-2^2}}(2) +c$
Now clearly something is wrong because I will get the square root of a negative, which does not make any sense. Need help from here.
*Note: I know that there are other ways to solve this, but I would prefer if calculus was used to solve this question.
| 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 our finger, the variable $a$, along the semicircle as $x$ ranges from $0$ to $1$ and see where the tangent line meets the $x$-axis. We have the point $(a,\sqrt{1-a^2})$ and the slope $\frac{-a}{\sqrt{1-a^2}}$, so the line is
$$ y - \sqrt{1-a^2} = \frac{-a}{\sqrt{1-a^2}}(x-a) \text{.} $$
To pass through the point $(2,0)$, we must satisfy
$$ 0 - \sqrt{1-a^2} = \frac{-a}{\sqrt{1-a^2}}(2-a) \text{,} $$
which we can solve for $a$. \begin{align*}
- \sqrt{1-a^2} &= \frac{-a}{\sqrt{1-a^2}}(2-a) \\
- (1-a^2) &= -a(2-a) \\
a^2 - 1 &= a^2 - 2a \\
1 &= 2a \\
a &= 1/2 \text{.}
\end{align*}
Therefore, the equation of the line is
$$ y - \sqrt{1-\frac{1}{4}} = \frac{-1/2}{\sqrt{1-\frac{1}{4}}}\left( x-\frac{1}{2} \right) \text{.} $$
(This simplifies a little, which I leave to the reader.)
| {
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"timestamp": "2023-03-29T00:00:00",
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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( \int_{0}^{R} \frac{\sin^2x}{x^3} \, dx - \int_{0}^{R} \frac{\sin x}{x^2} \, dx \biggr)\\
&= \lim_{R\to\infty} \biggl( \frac{\sin^2 x}{-2x^2}\Biggr\rvert_{0}^{R} - \int_{0}^{R} \frac{\sin (2x)}{-2x^2} \, dx - \biggl(-\frac{\sin x}{x} \Biggr\rvert_{0}^{R} + \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr) \biggr) \\
&= \lim_{R\to \infty} \biggl(\frac{1}{2} + \int_{0}^{2R} \frac{\sin u}{u^2/2} \, \Bigl(\frac{1}2 \, du\Bigr) - \biggl( 1 + \int_{0}^{R} \frac{\cos x}{x} \, dx \biggr)\biggr) \\
&\hspace{22em}\text{(using the substitution $u\mapsto 2x$)}\\
&= -\frac{1}{2} + \lim_{R\to \infty} \biggl(\int_{0}^{2R} \frac{\sin u}{u^2} \, du - \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr)\\
&= \frac{1}{2} + \lim_{R\to\infty} \biggl(\int_{R}^{2R} \frac{\cos x}{x} \, dx \biggr)
\end{align*}
Thus it suffices to show that $\lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, dx = \ln 2$. The Taylor series expansion of $\cos x$ is given by $\cos x = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{(2i)!}$.
(If the step below (the one involving the interchanging of an infinite sum and integral) is valid, why exactly is it valid? For instance, does it use uniform convergence?)
The limit equals
$$
\lim_{R\to\infty} \int_{R}^{2R} \biggl( \frac{1}{x} + \sum_{i=1}^{\infty} \frac{(-1)^i x^{2i-1}}{(2i)!} \biggr) \, dx
= \ln 2 + \lim_{R\to\infty} \sum_{i=1}^{\infty} \biggl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\biggr]_{R}^{2R} . $$
But I don't know how to show $\lim_{R\to\infty} \lim_{R\to\infty} \sum_{i=1}^{\infty} \Bigl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\Bigr]_{R}^{2R} = -2 \ln 2$.
| 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 \to 0^+}} \biggl( \left[ -\frac{\sin^2 x}{2x^2} + \frac{\sin x}{x} \right]_{\varepsilon}^{R} + \int_{\varepsilon}^{R} \left( \frac{\sin(2x)}{2x^2} - \frac{\cos x}{x} \right) \, \mathrm{d}x \biggr) \\
&= -\frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \int_{2\varepsilon}^{2R} \frac{\sin x}{x^2} \, \mathrm{d}x - \int_{\varepsilon}^{R} \frac{\cos x}{x} \, \mathrm{d}x \biggr) \\
&= -\frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \left[ -\frac{\sin x}{x} \right]_{2\varepsilon}^{2R} + \int_{2\varepsilon}^{2R} \frac{\cos x}{x} \, \mathrm{d}x - \int_{\varepsilon}^{R} \frac{\cos x}{x} \, \mathrm{d}x \biggr) \\
&= \frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x - \int_{\varepsilon}^{2\varepsilon} \frac{\cos x}{x} \, \mathrm{d}x \biggr).
\end{align*}
Now by noting that
\begin{align*}
\left|\int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x\right|
&= \left| \frac{\sin R}{R} - \frac{\sin (2R)}{2R} + \int_{R}^{2R} \frac{\sin x}{x^2} \, \mathrm{d}x \right| \\
&\leq \frac{1}{R} + \frac{1}{2R} + \int_{R}^{2R} \frac{1}{x^2} \, \mathrm{d}x \\
&= \frac{2}{R},
\end{align*}
it follows that
$$ \lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x = 0. $$
On the other hand, by substituting $x = \varepsilon u$,
$$ \int_{\varepsilon}^{2\varepsilon} \frac{\cos x}{x} \, \mathrm{d}x
= \int_{1}^{2} \frac{\cos(\varepsilon u)}{u} \, \mathrm{d}u
\xrightarrow{\varepsilon \to 0^+} \int_{1}^{2} \frac{1}{u} \, \mathrm{d}u = \log 2 $$
by the dominated convergence theorem. Therefore
$$ \int_{0}^{\infty}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x = \frac{1}{2} - \log 2. $$
| {
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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:
$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: Law ~of~ cosines:\\
AF^2 = x^2+AC^2-2.x.AC.cosC\\
\triangle ABC:\\
cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\
x^2+(x+BF)^2 = AC^2\\
Th.Stewart \triangle ABC:\\
AC^2.BF+x^3=AF^2BC+BC.x.BF$
...??
| 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 $\tan C=\frac{3}{4}$, $\tan\angle BAC=\frac{4}{3}$, and finally $\angle BAC=\arctan\frac{4}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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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 $3 \cdot a \cdot b$
by AM-GM inequality,
$$\frac{a+b}{2} \geq \sqrt{ab}$$
$$a \cdot b \leq 9$$
Therefore $3 \cdot a \cdot b \leq 27$
But is the solution correct? as we know that while applying AM-GM, the terms should be positive real but $a$ and $b$ are not positive for every $x$ as Real number , if I am correct the range of $(-\infty,-2)$ to $(4, \infty)$ should be given for $a$ and $b$ to be both positive and then apply AM-GM, please suggest if some other way is possible too to solve this.
| $(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.
$3·(2+x)·(4-x) ≤ 3·\left({2+4\over2}\right)^2 = 3^3 = 27$
| {
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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 $n$ but to prove generally, I don't know.
I know that the limit of the sequence is $3$ which is obtained by the positive root of $x^2 - x -6 =0$.
We usually use $a_{n+1} - a_n$ to check if a sequence is monotone increasing our monotone decreasing. But here, since intervals are involved, I don't know how to proceed.
Thank you in advance.
| 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
$$a_n = \sqrt{a_n \cdot a_n} < \sqrt{3 a_n} = \sqrt{2 \cdot a_n + a_n} < \underbrace{\sqrt{6 + a_n}}_{= a_{n+1}} < \sqrt{6 + 3} = \sqrt{9} =3 $$
by strict monotony of the square root (in each non-trivial step an $a_n$ is replaced by a $3$). The desired statement follows by complete induction on $n$.
| {
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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. $\lim\limits_{n\to\infty }\frac{(-5)^{n}+2\cdot(-2)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$
Before I show what I tried , Sorry if some words and terms are not translated right as I could not find the terms after searching , would appreciate if someone can help edit if there are mistakes.
My try:
First let $a_n=\frac{(-5)^{n}+2\cdot(-2)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$ then I wrote it in a more "comfortable" way so $\frac{5^n\cdot(-1)^{-n}+(2)^{n+1} \cdot(-1)^{n}+3}{5^{n+1}+2\cdot(-3)^{n}+3}$ and then $\frac{(-1)^{n}\cdot(5^n+(2)^{n+1})+3}{5^{n+1}+2\cdot(-3)^{n}+3}$
if we look at the subsequences $(a_{2n})$ and $(a_{2n+1})$ then they cover the sequence $(a_n)$
and if two sub sequences converge to the same limit and they cover the sequence $(a_n)$ then $(a_n)$ also converges.
*
*$(a_{2n})$ = $\frac{(5^n+(2)^{2n+1})+3}{5^{2n+1}+2\cdot(3)^{2n}+3}$
so $\lim\limits_{n\to\infty }\frac{(5^n+(2)^{2n+1})+3}{5^{2n+1}+2\cdot(3)^{2n}+3}$ and using basic arithmetic limit rules $\lim\limits_{n\to\infty }\frac{5^{2n}}{5^{2n+1}}\cdot\frac{(1+\frac{2^{2n+1}}{5^{2n}}+\frac{3}{5^{2n}})}{(1+\frac{2\cdot(3)^{2n}}{5^{2n+1}}+\frac{3}{5^{2n+1}})}$ = $\frac{1}{5}$
*$(a_{2n+1})$ = $\frac{(-1)^{2n+1}\cdot(5^{2n+1}+(2)^{2n+2})+3}{5^{2n+2}+2\cdot(-3)^{2n+1}+3}$ so we get $(a_{2n+1})$ = $\frac{-5^{2n+1}-(2)^{2n+2}+3}{5^{2n+2}-2\cdot(3)^{2n+1}+3}$
then to calculate the limit I also used basic arithmetic rules
$\lim\limits_{n\to\infty}\frac{-5^{2n+1}-(2)^{2n+2}+3}{5^{2n+2}-2\cdot(3)^{2n+1}+3}$
$\lim\limits_{n\to\infty}\frac{5^{2n+1}}{5^{2n+2}}\cdot\frac{(-1-\frac{2^{2n+2}}{5^{2n+1}}+\frac{3}{5^{2n+1}})}{(1+\frac{2\cdot(3)^{2n+1}}{5^{2n+2}}+\frac{3}{5^{2n+2}})} = -\frac {1}{5}$
so I got that the limits of the subsequences are different , can I conclude from that that the limit of $(a_n)$ does not exist? is there a different way to approach this?
| 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 converge.
Double check your class notes and the textbook to see if you've already proved this (or another similar theorem), but from there you've already done the work to prove that it does not converge.
| {
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} |
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 $dx = \cos(u) du$
writing $1 = \sin^2(u) + \cos^2(u)$
and also $(1-\sin(u)^2)=\cos^2(u)$
Eventually I reach that $I(n) = \int \frac{\sin^2\left(u\right)}{\cos^{n-1}du}du + \int \cos^{1-n}\left(u\right)du$
I stopped here as I do not know how to continue or if the route I chose is even ok.
| 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 $u=x, dv=\frac{x}{(1-x^2)^n}dx$:
$$\int x\cdot\frac{x}{(1-x^2)^n}dx=x\cdot\frac{1}{2(n-1)(1-x^2)^{n-1}}-\frac{1}{2(n-1)}\int\frac{1}{(1-x^2)^{n-1}}dx $$
so we get
$$\begin{align}I(n)&=I(n-1)+\frac{x}{2(n-1)(1-x^2)^{n-1}}-\frac{1}{2(n-1)}I(n-1)\\
&=\frac{2n-3}{2(n-1)}I(n-1)+\frac{x}{2(n-1)(1-x^2)^{n-1}}\end{align} $$
| {
"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:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{aligned}
$$
I find the solution is
$$\frac{\ln{\left(x + \sqrt{1 - x^{2}} \right)}}{2} + \frac{\sin^{-1}{\left(x \right)}}{2}+C$$
How can I get this without trigonometric substitution?
| 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.
$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} &=\int \frac{d t}{(1+t)\left(1+t^2\right)} \\
&=\frac{1}{2} \int\left(\frac{1}{1+t}+\frac{1-t}{1+t^2}\right) d t \\
&=\frac{1}{2}\left(\ln |1+t|+\int \frac{d t}{1+t^2}-\int \frac{t d t}{1+t^2}\right) \\
& =\left.\frac{1}{2}[\ln |1+t|+\tan ^{-1} t-\frac{1}{2} \ln \left(1+t^2\right)\right]+C \\
&=\frac{1}{2}\left[\ln \left|1+\frac{x}{\sqrt{1-x^2}}\right|+ \sin ^{-1} x-\frac{1}{2} \ln \left|1+\frac{x^2}{1-x^2}\right|+C\right.\\
&= \frac{1}{2}\left[\ln \left|x+\sqrt{1-x^2}\right|+\sin ^{-1} x\right]+C
\end{aligned}
$$
| {
"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 vectors and the angle between an unknown unit vector $\hat{k}_2$ and , $\hat{k}_1$ and $\hat{k}_3$, are given as $\alpha_{12}$ and $\alpha_{23}$ such that $$\hat{k}_1.\hat{k}_2=\cos\alpha_{12}$$
and $$\hat{k}_2.\hat{k}_3=\cos\alpha_{23}$$, the angle between $\hat{k}_1$ and $\hat{k}_3$ is $\beta$. I am looking for the vectors where the cones intersect.
| 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_2} = \cos \alpha_{12} \mathbf{k_1} + \sin \alpha_{12} (cos \phi \mathbf{u_1} + \sin \phi \mathbf{u_2}) $
Now, you want $\mathbf{k_2}$ to lie on the cone that has axis $\mathbf{k_3}$, and semi-vertical angle $\alpha_{23}$, so the angle between $\mathbf{k_2}$ and $\mathbf{k_3}$ will be $\alpha_{23} $
So you need to solve the equation
$ \mathbf{k_2} \cdot \mathbf{k_3} = \cos \alpha_{23} $
for $\phi$. Upon plugging in the expression of $\mathbf{k_2}$ the above equation becomes
$ \cos \alpha_{12} (\mathbf{k_1} \cdot \mathbf{k_3}) + \sin \alpha_{12} \left(cos \phi (\mathbf{u_1} \cdot \mathbf{k_3} ) + \sin \phi (\mathbf{u_2} \cdot \mathbf{k_3})\right) = \cos \alpha_{23} $
which is of the form
$ a \cos \phi + b \sin \phi = c $
and can be solved easily (see below). It will result in two solutions if $\alpha_{12} + \alpha_{23} \gt \beta $, one solution if $\alpha_{12} + \alpha_{23} = \beta $, and no solutions if $\alpha_{12} + \alpha_{23} \lt \beta $.
Now for the step-by-step solution of $ a \cos \phi + b \sin \phi = c $, divide through by $\sqrt{a^2 + b^2} $, resulting in
$\dfrac{a}{\sqrt{a^2 + b^2}} \cos \phi + \dfrac{b}{\sqrt{a^2 + b^2}} \sin \phi = \dfrac{c}{\sqrt{a^2 + b^2}} $
Define angle $\psi$ such that $\cos \psi = \dfrac{a}{\sqrt{a^2 + b^2}} $ and $\sin \psi = \dfrac{b}{\sqrt{ a^2 + b^2}} $, then
$ \cos \phi \cos \psi + \sin \phi \sin \psi = \dfrac{c}{\sqrt{a^2 + b^2}} $
The left hand side is simply $\cos(\phi - \psi) $, hence
$ \phi = \psi \pm \cos^{-1} \dfrac{c}{\sqrt{a^2 + b^2}} $
There will two solutions if $| c | \lt \sqrt{a^2 + b^2} $ , one solution if $ |c| = \sqrt{a^2 + b^2} $ and no solutions if $|c| \gt \sqrt{a^2 + b^2} $
| {
"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 $\begin{Bmatrix} \begin{pmatrix}2\\0\\0\\-2\end{pmatrix},\begin{pmatrix}1\\-3\\3\\-1\end{pmatrix},\begin{pmatrix}-2\\0\\-4\\-2\end{pmatrix},\begin{pmatrix}-1\\-3\\-1\\-3\end{pmatrix},\begin{pmatrix}1\\-3\\1\\-3\end{pmatrix}\end{Bmatrix}$.
I need to find a basis of $U$.
My attempt:
I wrote the vectors into a matrix horizontally like this:
$\begin{pmatrix}2&0&0&-2\\1&-3&3&-1\\-2&0&-4&-2\\-1&-3&-1&-3\\1&-3&1&-3\end{pmatrix}$
Here, I tried to find linear combinations and was able to reduce the vectors down to the form:
$\begin{pmatrix}1&1&0&0\\0&1&0&1\\0&0&1&1\\0&0&0&0\\0&0&0&0\end{pmatrix}$
This leaves me with a basis for $U$ as follows: $B = \begin{Bmatrix} \begin{pmatrix}1\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\1\end{pmatrix},\begin{pmatrix}0\\0\\1\\1\end{pmatrix}\end{Bmatrix}$.
By definition, a basis of a vector space is a set of linearly independent vectors, which span the vector space (in this example $U$). Since this is the case, $B$ is a basis for $U$.
Is my reasoning correct? If so, please let me know if there is a more standard way of writing down this procedure.
| 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) row echelon from.
Applying this, the reduce row echelon form is
$\begin{bmatrix} 1 && 0 && 0 && 0 && 0.5 \\ 0 && 1 && 0 && 1 && 1 \\ 0 && 0 && 1 && 1 && 0.5 \\ 0 && 0 && 0 && 0 && 0 \end{bmatrix} $
Therefore, a basis for the space spanned by the $5$ vectors is the set of the first $3$ vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 to evaluate the integral $J$ using series and integration by part.
$\displaystyle \begin{aligned} J\displaystyle & = \int_{0}^{1} \frac{\ln x}{1-x^{2}} d x =\sum_{k=0}^{\infty} \int_{0}^{1} x^{2 k} \ln x d x=\sum_{k=0}^{\infty}\left(\left[\frac{x^{2 k+1} \ln x}{2 k+1}\right]_{0}^{1}-\frac{1}{2 k+1} \int_{0}^{1} x^{2 k+1} \cdot \frac{1}{x} d x\right) \\\displaystyle &=-\sum_{k=0}^{\infty}\frac{1}{(2 k+1)^{2}}=-\frac{\pi^{2}}{8} \end{aligned} \tag*{} $
$$\therefore \displaystyle \int_{0}^{\infty} \frac{\ln x}{x^{2}-1} d x =-2J=\frac{\pi^{2}}{4} $$
However, when I began to increase the power $n$, I found, in Wolframalpha, that there is a pattern for the integral$$
I_{n}=\int_{0}^{\infty} \frac{\ln x}{x^{n}-1} d x
$$
$$
\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline I_{n} & \text { Diverges } & \frac{\pi^{2}}{4} & \frac{4 \pi^{2}}{27} & \frac{\pi ^2}{8} & \frac{8 \pi^{2}}{25(5-\sqrt{5})} & \frac{\pi^{2}}{9} & \frac{\pi^{2}}{49} \csc ^{2}\left(\frac{\pi}{7}\right) & \frac{\pi^{2}}{64} \csc ^{2}\left(\frac{\pi}{8}\right) \\
\hline
\end{array}
$$
By the pattern, let’s guess the formula for $I_n$ as $$
I_{n}=\left(\frac{\pi}{n}\right)^{2}\csc ^{2}\left(\frac{\pi}{n}\right).
$$
How to prove it? Is it difficult or interesting? Looking forward to your suggestions and proofs.
| In order to make use of power series, I split the integration interval of $I_n$ as:
$$
I_n=\underbrace{\int_{0}^{1} \frac{\ln x}{x^{n}-1} d x}_{J}+\underbrace{\int_{1}^{\infty} \frac{\ln x}{x^{n}-1} d x}_{K} $$
$$
\begin{aligned}
J &=-\sum_{k=0}^{\infty} \int_{0}^{1} x^{n k} \ln x d x \\
&=-\sum_{k=0}^{\infty} \int_{0}^{1} \ln x d\left(\frac{x^{n k+1}}{n k+1}\right) \\
& \stackrel{IB P}{=}-\sum_{k=0}^{\infty}\left(\left[\frac{x^{n k+1}\ln x}{n k+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{n k}}{n k+1} d x\right) \\
&=\sum_{k=0}^{\infty} \frac{1}{(n k+1)^{2}} \\
&=\frac{1}{n^{2}} \sum_{k=0}^{\infty} \frac{1}{\left(\frac{1}{n}+k\right)^{2}}
\end{aligned}
$$
Similarly, $$
\begin{aligned}
K &=\int_{1}^{\infty} \frac{\ln x}{x^{n}\left(1-\frac{1}{x^{n}}\right)} d x \\
&=\sum_{k=0}^{\infty} \int_{1}^{\infty} \ln x \cdot x^{-n-n k} d x\\ &\stackrel{IBP}{=}\sum_{k=0}^{\infty}\left(\left[\frac{x^{-n-n k+1}\ln x}{-n-n k+1}\right]_{1}^{\infty}-\int_{1}^{\infty} \frac{x^{-n-n k}}{-n-n k+1} d x\right)\\& =\sum_{k=0}^{\infty} \frac{1}{[-n (k+1)+1]^{2}}\\
\end{aligned}
$$
Reindexing by replacing $k$ by $-k-1$ gives $$
\begin{aligned}
K&=\sum_{k=-\infty}^{-1} \frac{1}{(n k+1)^{2}}=\frac{1}{n^{2}} \sum_{k=-\infty}^{-1} \frac{1}{\left(\frac{1}{n}+k\right)^{2}}
\end{aligned}
$$
Finally, adding $J$ and $K$ gives $$
\begin{aligned}
I_n &=\frac{1}{n^{2}} \sum_{k=-\infty}^{-1} \frac{1}{\left(\frac{1}{n}+k\right)^{2}}+\frac{1}{n^{2}} \sum_{k=0}^{\infty} \frac{1}{\left(\frac{1}{n}+k\right)^{2}} =\frac{1}{n^{2}} \sum_{k=-\infty}^{\infty} \frac{1}{\left(\frac{1}{n}+k\right)^{2}}
\end{aligned}
$$
By the theorem, $$
\sum_{k=-\infty}^{\infty} \frac{1}{z+k}=\pi \cot (\pi z)
$$
Differentiating w.r.t. $z$ yields
$$
\sum_{k=-\infty}^{\infty} \frac{1}{(z+k)^{2}}=\pi^{2} \csc^{2}(\pi z)
$$
Putting $ \displaystyle z=\frac{1}{n}$ yields the conclusion that
$$\boxed{I_n=\left(\frac{\pi}{n}\right)^{2} \csc ^{2}\left(\frac{\pi}{n}\right)}.$$
Wish you enjoy the proof. Your comments and alternate proofs are warmly welcome.
| {
"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
} |
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) $\frac 89$
D) $\frac {16}{9}$
E) $\frac{32}{27}$
My attempts:
I think that the all possible answers are wrong. Because, by Am-Gm inequality we have
$$\frac{\sin x}{\cos y}+\frac{\cos y}{\sin x}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}≥2+2=4.$$
But, Wolfram Alpha gives us a different result : The Global Minimum doesn't exist. However, the local minimum must be $6.$
But, still the problem is not solved. Because Wolfram's graph shows that the minimum can be less than $6$.
I also tried
Let $$\sin x=a,\cos y=b,\cos x=c,\sin y=d$$ with
$$ab+cd=\frac 23≥2\sqrt{abcd}\implies abcd≤\frac 19\\ a^2+c^2=b^2+d^2=1 $$
then I need
$$\min \left(\frac ab+\frac ba+\frac cd+\frac dc\right)$$
But, I can't do anything from here. Finally, I attach the graph drawn by WA.
| 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
$$\sin(x+y)=\sin(2x)=\frac23$$ and
$$2\tan(2x)+2\cot(2x)=6.$$
| {
"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
} |
$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$ 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 trying to prove that the sequence is monotone and bounded, but after some examples, I found that it isn't monotone. Could you help and maybe suggest other strategies for proving the convergence of a sequence other than the one I tried?
|
$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$ 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 $\epsilon<x_{n+1}<\frac{1}{\epsilon}$ for some $0<\epsilon<1$, then
$$
\epsilon<
\frac{2+\epsilon}{2+\frac{1}{\epsilon}}<
x_{n+2}=\frac{2+x_{n+1}}{2+x_n}
<\frac{2+\frac{1}{\epsilon}}{2+\epsilon}<\frac{1}{\epsilon},
$$
where the inequalities $\frac{2+\frac{1}{\epsilon}}{2+\epsilon}<\frac{1}{\epsilon}$ and $\epsilon<
\frac{2+\epsilon}{2+\frac{1}{\epsilon}}$ follow from $0<\epsilon<1$ immediately. Hence, by defining $\epsilon=\min\{x_0,x_1,\frac{1}{2}\}$, we have proven that there exists $0<\epsilon<1$ such that $x_n>\epsilon$ for all $n$.
The next steps follow as below:
$${
x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}\implies
\\
x_{n+2}-1 = \frac{x_{n+1}-x_n}{2+x_{n}}\implies
\\
|x_{n+2}-1| = \frac{|x_{n+1}-x_n|}{2+x_{n}}=\frac{|x_{n+1}-1-(x_n-1)|}{2+x_{n}}\implies
\\
|x_{n+2}-1| \le \frac{|x_{n+1}-1|+|x_n-1|}{2+x_{n}}\implies
\\
|x_{n+2}-1| \le \frac{|x_{n+1}-1|+|x_n-1|}{2+\epsilon}\implies
\\
|x_{n+2}-1|+\frac{1}{2}|x_{n+1}-1| \le \frac{(2+\frac{\epsilon}{2})|x_{n+1}-1|+|x_n-1|}{2+\epsilon}\implies
\\
|x_{n+2}-1|+\frac{1}{2}|x_{n+1}-1| \le \frac{|x_{n+1}-1|+\frac{1}{2+\frac{\epsilon}{2}}|x_n-1|}{\frac{2+\epsilon}{2+\frac{\epsilon}{2}}}\implies
\\
|x_{n+2}-1|+\frac{1}{2}|x_{n+1}-1| \le \frac{|x_{n+1}-1|+\frac{1}{2}|x_n-1|}{\frac{2+\epsilon}{2+\frac{\epsilon}{2}}}\implies
\\
}$$
By defining $y_n=|x_{n+1}-1|+\frac{1}{2}|x_n-1|$ and $r=\frac{2+\frac{\epsilon}{2}}{2+\epsilon}$, we have proven that $|y_{n+1}|\le r|y_n|$ and $0<r<1$. Hence $y_n\to 0$ which means that $x_n$ is convergent to $1$ and the proof is complete $\blacksquare$
| {
"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
} |
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 expression $x^2+f(a) x+g(a)$.
| {
"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{(s+2)}{(s+2)^2+1^2}-\frac{2}{(s+2)^2+1^2}
\end{equation}
The first form, with $p=s+2$ gives $\mathscr{L}^{-1}\frac{p}{p^2+1^2}=\cos t$. The second, with $p=s+2$, $\mathscr{L}^{-1}\frac{2}{p^2+1^2}=\sin(2t)$, so I would be tempted to write:
\begin{equation}
\mathscr{L}^{-1}\frac{s}{s^2+4s+5}=\cos t+\sin(2t)
\end{equation}
But this is wrong! So how can I get this right?
Thanks!
| 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",
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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 integer for all $x>2$ and $y>3$.
I have made the following progress:
Taking $x=y\geq4$, the constraint becomes $$\frac{2f(x-3)}{f(x-2)}$$ which we want to be an integer for all $x\geq4$.
Now one case is that $\frac{2f(x-3)}{f(x-2)}=c$ for all $x\geq4$, where $c$ is a positive integer constant. We consider this case.
Let $\frac{2f(n)}{f(n+1)}=c$ for all $n\geq1$. We have $$f(n+1)=\frac{2f(n)}{c}\implies f(n)=\left(\frac 2 c\right)^{n-1}f(1)$$
If $c>2$, then $f(n)$ can not always be integer. So we have $c=1$ or $c=2$.
If $c=2$, then we have $f(n)=f(1)=k$ for all $n$. Plugging it in the given expression, we have $$\frac{4k^2}{2k^2}=2$$
which is an integer. So $f(n)=k$ works.
If $c=1$, then we have $f(n)=2^{n-1}f(1)=2^{n-1}k$. We can confirm that this function works by plugging in the expression.
But there is other case when $\frac{2f(n)}{f(n+1)}$ is not a constant. I can't find a way to solve this case.
| 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 $x$ (which I will abbreviate to "for s.l. $x$"). Moreover, we have $v_{x + 1} \leq v_x + 1$ for s.l. $x$.
Now in the original condition, we have $2^{v_x + v_{y - 2}} + 2^{v_y + v_{x - 2}} \mid 2^{v_x + v_{y - 3} + 2}$ for s.l. $x, y$. This is only possible when $v_x + v_{y - 2} = v_y + v_{x - 2}$, i.e. $v_x - v_{x - 2} = v_y - v_{y - 2}$ for s.l. $x, y$. We denote by $k$ this common value.
It follows that $v_x = v_{x + 2} - k = v_{x + 4} - 2k = \cdots = v_{x + 2s} - sk$ for any $s > 0$. But $v_{x + 2s}$ is nonnegative, hence we get $k \geq -\frac{v_x}s$. Taking $s \rightarrow \infty$ shows that $k \geq 0$.
On the other hand, we have $v_x \leq v_{x - 1} + 1 \leq v_{x - 2} + 2$, namely $k \leq 2$. This leaves only three possibilities: $k = 0, 1, 2$.
Case $k = 2$: the only possibility is that $v_x = v_{x - 1} + 1$ for s.l. $x$, namely $f(x) = 2f(x - 1)$. This implies that there exists $r \in \Bbb Q$ such that $f(x) = 2^xr$ for s.l. $x$.
We will prove that $f(x) = 2^xr$ holds for all $x$. Suppose it is not the case. Let $N$ be the largest integer such that $f(N) \neq 2^Nr$. From $f(N + 1)\mid 2f(N)$, we know that $f(N) = c \cdot 2^Nr$ for some $c\in \Bbb N$. Taking $x = N + 2$ and s.l. $y$ in the original condition, we get $1 + c \mid 2$ which forces $c = 1$. This contradicts the assumption that $f(N) \neq 2^Nr$.
Thus in this case we have $f(x) = 2^xr$ for some $r$.
This technique will be repeatedly used below and the details are sometimes omitted.
Case $k = 1$: there are two subcases: either $v_x = v_{x - 1}$ for s.l. even $x$, or $v_x = v_{x - 1}$ for s.l. odd $x$. I will treat the latter case, the previous one being similar.
Thus we have $v_x = v_{x - 1}$ for s.l. odd $x$ and $v_x = v_{x - 1} + 1$ for s.l. even $x$. This means that there exists $r \in \Bbb Q$ such that $f(x) = 2^{\lfloor \frac x 2\rfloor}r$ for s.l. $x$.
We can again show that $f(x) = 2^{\lfloor \frac x 2\rfloor}r$ for all $x$, by taking $N$ to be the largest integer that doesn't satisfy this identity.
For the subcase that $v_x = v_{x - 1}$ for s.l. even $x$, we get similarly $f(x) = 2^{\lceil \frac x 2\rceil}r$.
Case $k = 0$: there are three subcases:
i. $v_x = v_{x - 1}$ for s.l. $x$;
ii. $v_x = v_{x - 1} + 1$ for s.l. even $x$ and $v_x = v_{x - 1} - 1$ for s.l. odd $x$;
iii. $v_x = v_{x - 1} - 1$ for s.l. even $x$ and $v_x = v_{x - 1} + 1$ for s.l. odd $x$.
In (ii) and (iii), the same argument as above (considering largest $N$ for which the formula is not valid) shows that the function $f$ is an alternating sequence $r, 2r, r, 2r, \dots$ or $2r, r, 2r, r, \dots$.
It remains to consider (i). In this case, there exists $r$ such that $f(x) = r$ for s.l. $x$. Let $N$ be the largest integer such that $f(N) \neq r$.
We will show by induction that, there exists a sequence of nonnegative integers $w_n$ such that
*
*$f(x) = 3^{w_x}r$ for all $x$;
*$w_x = 0$ for all $x > N$;
*$w_x - w_{x + 1} \in \{0, 1\}$ and $w_x - w_{x + 2} \in \{0, 1\}$.
In other words, the sequence $(w_n)_n$ decreases by $0$ or $1$ at each step, and does not have two consecutive decreases by $1$.
The claim is clearly true for all $x > N$. Suppose it is true for all $x > M$ and let us prove it for $x = M$.
Taking s.l. $y$ in the original condition, we have $f(x) + f(x - 2) \mid 4f(x)$ for all $x$.
Taking $x = M + 2$, we get $t + f(M) \mid 4t$ where $t = f(M + 2)$.
By induction hypothesis, we have either $f(M + 1) = t$ or $f(M + 1) = 3t$.
If $f(M + 1) = t$, then we have $f(M) = \frac t2 c$ for some $c \in \Bbb N$. This implies $c = 2$ or $c = 6$, i.e. $f(M) = t$ or $f(M) = 3t$.
If $f(M + 1) = 3t$, then we have $f(M) = \frac{3t}2c$ for some $t \in \Bbb N$. This implies $c = 2$, i.e. $f(M) = 3t$.
We have thus proved our claim.
Conversely, let $(w_x)$ be a sequence of integers such that
*
*$w_x = 0$ for s.l. $x$;
*$w_x - w_{x + 1} \in \{0, 1\}$ and $w_x - w_{x + 2} \in \{0, 1\}$;
and we define $f(x) = 3^{w_x}r$. Let us verify that this function $f$ satisfies the original condition.
Since we have $f(y - 2) \mid f(y - 3)$, it suffices to show that $$\frac{4f(x)f(y - 2)}{f(x)f(y - 2) + f(y)f(x - 2)} \in \Bbb Z.$$ Since both $f(x)/f(x - 2)$ and $f(y)/f(y - 2)$ belong to $\{1, 3\}$, there are only four cases and a direct check shows that the quotient is an integer in all four cases.
In conclusion, $f$ is one of the following functions:
*
*$f(x) = 2^x r$;
*$f(x) = 2^{\lfloor\frac x2\rfloor} r$;
*$f(x) = 2^{\lceil\frac x 2\rceil} r$;
*$f(x) = 2^{x \mod 2} r$;
*$f(x) = 2^{(x + 1) \mod 2}r$;
*$f(x) = 3^{w_x}r$ for some sequence of integers $w$ such that $w_x - w_{x + 1}, w_x - w_{x + 2} \in \{0, 1\}$ for all $x$, and $w_x = 0$ for all sufficiently large $x$.
| {
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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:
$$\sqrt[n]{\dfrac{20}{2^{2n+4}+2^{2n+2}}}=\sqrt[n]{\dfrac{20}{2^{2n}\cdot2^4+2^{2n}\cdot2}}\\=\sqrt[n]{\dfrac{20}{2^{2n}\cdot18}}=\sqrt[n]{\dfrac{10}{9\cdot2^{2n}}}$$ This is as far I as I am able to reach. Thank you!
PS I don't see how one can get $\dfrac14$. For that we have to get something like $\sqrt[n]{A^n}$.
| Using the identity $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$, we find that
$$
\begin{align*}
\sqrt[n]{\frac{20}{2^{2n+4}+2^{2n+2}}} &= \sqrt[n]{\frac1{2^n}}\cdot\sqrt[n]{\frac{20}{2^4+2^2}}\\
&= \sqrt[n]{\frac1{4^n}}\cdot\sqrt[n]{\frac{20}{16+4}}\\
&=\sqrt[n]{\left(\frac14\right)^n}\cdot\sqrt[n]{\frac{20}{20}}\\
&=\frac14.
\end{align*}
$$
| {
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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 \frac{\pi}{4}$ and $f(x)=\int_{\frac{\pi}{4}}^{x} \sin t d t$ for $ \frac{\pi}{4} \le x \le \frac{\pi}{2}$
| 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
$$f(x) = \int_0^{\frac{\pi}{4}} \cos(t) dt + \int_{\frac{\pi}{4}}^{x} \cos(t) dt = \sqrt{2} - \cos(x)$$
It is easy to see that the function $f$ is strictly increasing and differentiable from $\displaystyle{\left[ 0, \frac{\pi}{2}\right]}$ to $\displaystyle{\left[ 0, \sqrt{2}\right]}$, hence $f$ is a bijection and $f^{-1}$ is well-defined. Moreover, one has
$$f \left(\frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}, \quad \quad \text{so } f^{-1} \left( \frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}.$$
Differentiating $f^{-1} \circ f = \text{Id}$, you get that $f'(x) \left(f^{-1}\right)'(f(x)) = 1$ for every $x$. In particular, for $x = \displaystyle{\frac{\pi}{4}}$, this gives
$$ \left(f^{-1}\right)'\left( \displaystyle{\frac{\sqrt{2}}{2}}\right) = \frac{1}{f' \left( \displaystyle{\frac{\pi}{4}}\right)}$$
and because $f' \left( \displaystyle{\frac{\pi}{4}}\right) = \cos \left( \displaystyle{\frac{\pi}{4}}\right) = \displaystyle{\frac{\sqrt{2}}{2}}$, you get finally that $$ \boxed{ \left(f^{-1}\right)'\left( \displaystyle{\frac{\sqrt{2}}{2}}\right) = \sqrt{2}}$$
| {
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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 degree of $Q(x)$ is equal to $4-2=2$, so $Q(x)=mx^2+nx+p$.
The degree of $R(x)$ is at most $1$, so we can consider that $R(x)=qx+r$. What to do next?
| 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 ,
$$x^4+2x^3-5x^2-22x-24=(x^2-x-6)(x^2+3x+4)$$
Hence there are two set of values for $a$ and $b$.
| {
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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 Rational Root Theorem won't be useful as the equation does not have such roots.
I got $-9x$ by averaging $-10x$ and $-8x$.
I don't know if it makes sense.
| Taking from where you left off, and rewrite the left hand side of your last line:
$\begin{align}(x^2+15)^2 - 18x(x^2+15) + 81x^2 - x^2 -3x(x^2+15)+18x^2&=0\\ \implies (x^2+15)^2-21x(x^2+15)+98x^2&=0\\ \implies (x^2+15)^2 - 2\cdot 10x\cdot (x^2+15)+(10x)^2 - x(x^2+15)-2x^2&=0\\ \implies (x^2+15-10x)^2-x(x^2+15-10x)-12x^2&=0\\ \implies (x^2-10x+15 -4x)(x^2-10x+15 + 3x) &= 0 \\ \implies (x^2-14x+15)(x^2-7x+15)&=0 \end{align}$
From this you can use quadratic formula to calculate the roots.
| {
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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 $\frac{1}{n^2}$ term zero.
$$\sin{\left(\pi\sqrt{n^2(1+\frac{0.5}{n}+\frac{1}{n^2})}\,\right)}$$
n goes to infinity so this becomes :-
$$\lim_{n\to\infty} (-1)^{n-1} \sin{\left(n\pi\right)}$$
now $\sin n\pi$ is just $0$ so the answer should be $0$
But the answer is $\sin{\left(-\frac{\pi}{4}\right)}$
Could someone please explain to me what I am doing wrong
| 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{n^2 +0.5n + 1} + n}\\
&= \lim_{n\to\infty} \frac{0.5n + 1}{\sqrt{n^2 +0.5n + 1} + n} = \frac{1}{4}
\end{align*}
| {
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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:
$$
\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2},
\\
\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}.
$$
Perhaps, the shortest solution I can think of is as follows: plug $c=-a-b$ into the equation but instead of expanding everything use the following identities
$$
(a+b)^3-a^3-b^3=3ab(a+b),
\\
(a+b)^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2),
\\
(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2.
$$
However, these identies are coming out of nowhere and moreover, in order to prove them one still needs to do some computations.
Question. Is it possible to solve this problem in a "smart" way (i.e. avoiding computations and preferrably elementary since it is almost a high school problem)? Any other solutions are also welcome.
Comment. It should be also noted that it is unclear (at least for me) how those identies were invented. It seems there is no nice similar identities for $\frac{a^p+b^p+c^p}{p}$ for $p$ other than 2, 3, 5, 7.
| 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
$$-S_2S_3=S_5+\frac{1}{5}(S_3P_2-S_2P_3)$$
$$-(ab+bc+ac)(abc)=\frac{1}{5}(-2abc(ab+bc+ac)-3abc(ab+bc+ac))$$
Since $S_5=0$. This is clearly true, as desired. The second can be solved similarly, but might require a bit more work.
| {
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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 know where the sign of $ f(x) $ changes.
Thanks in advance.
| 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 $\frac{1}{2}(\frac{\sqrt{2}}{2} + 1)(\sqrt{2} - 1) > \ln \sqrt{2}$. Multiply both sides by 4 to get $(\sqrt{2} + 2)(\sqrt{2} - 1) > \ln 4$.
The left-hand side simplifies to $\sqrt{2}$. So we have $\sqrt{2} > \ln 4$.
| {
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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
$$\log\left(\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right) \right)=\sum_{k=5}^\infty \log\left(1-\frac{1}{2^n}\right)\sim -\sum_{k=5}^\infty \frac{1}{2^n}=-\frac 1 {16}$$
$$\log\left(\prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right) \right)\sim -\frac 1 {16}\implies \prod_{k=5}^\infty \left(1-\frac{1}{2^n}\right)\sim e^{-1/16}$$ So
$$P \sim \frac{315}{1024}e^{-1/16} =0.288980\cdots$$ while the "exact" value is $0.288788\cdots$
Edit
Using q-Pochhammer symbols, the formal expression of the infinite product
$$P=\prod_{n=1}^\infty \left(1-\frac{1}{2^n}\right)=\left(\frac{1}{2};\frac{1}{2}\right)_{\infty }$$ If you look at equation $(9)$ in this page, you will find an approximation which, for your case, write
$$P \sim 2^{13/24} \sqrt{\frac{\pi }{\log (2)}} \exp\Bigg[-\frac{\pi ^2}{6 \log (2)} \Bigg]$$ which is a a relative error of $2.22\times 10^{-14}$.
| {
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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 $1$, we get these subcases $$\cases{n-1=1 \\ 2n^2+5n+6=6p^k}, \cases{n-1=2\\ 2n^2+5n+6=3p^k}, \cases{n-1=3\\ 2n^2+5n+6=2p^k} $$
$${\cases{n-1=6\\ 2n^2+5n+6=p^k}}$$
There are some restrictions on $p$ in the last $3$ subcases, but they’re obvious. from these cases we get these solutions $$(p,q)\in\{(2,2),(13,1),(29,1),(139,1)\} $$
But the problem lies in the case when $d=13$, I can’t get a handle on the subcases.
| 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 divisor of $6$, and the other is a multiple of $13^{k-2}$. However, for $n \geq 5$, $$2n^2+5n+6 \geq 2\times 5^2+5 \times 5 +6 \geq 81 > 78 = 13 \times 6$$
Therefore, $\frac{2n^2+5n+6}{13}$ can be a divisor of $6$ only if $n<5$. In this case, however, $\frac{n-1}{13}$ isn't even an integer (we want $n \geq 2$ so $n=1$ is ruled out).
It follows that $\frac{n-1}{13}$ must be a divisor of $6$ i.e. that $n-1 = 13,26,39,78$ i.e. that $n= 14,27,40,79$.
We will diverge from the approach in the AoPS post here, to show a different method. Indeed, one can substitute the values above and check that no more solutions are found , but we can do something that minimizes numerical computation.
We've already seen that $\frac{2n^2+5n+6}{13}>6$ for $n>5$. So if this expression is to be a multiple of a power of $13$ times a divisor of $6$ ,that power of $13$ must be at least $13$. Therefore, $2n^2+5n+6$ must be a multiple of $169$.
The idea now is to use the fact that we only need to test numbers belonging to the arithmetic progression $13k+1$. So we compute the following polynomial modulo $169$:
$$
2(1+13k)^2 + 5(1+13k)+6 \pmod{169} = 2(26k+1)+65k+5+6 = 13(9k+1)
$$
Therefore, we need $9k+1$ to be a multiple of $13$ : this doesn't occur for $k=1,2,3,6$. So no more solutions exist.
| {
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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: $\sqrt{(a+b)(a+c)}\ge2\sqrt[4]{a^2bc}$ ; so we need to prove that: $$\sum_{cyc}{\frac{a+\sqrt{bc}}{2\sqrt[4]{a^2bc}+\sqrt{bc}}}\le2$$ Due to homogenious, I denote $abc=1$ which implies the new one variable inequality: $$\sum_{cyc}{\frac{a+\dfrac{1}{\sqrt{a}}}{2\sqrt[4]{a}+\dfrac{1}{\sqrt{a}}}}\le2$$
I am trying to find suitable term to finish my idea.
Is there any good way to full of my approach or other better idea? Thanks for help.
| 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
} |
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 terms of matrices as:
\begin{align*} \left[\begin{array}{ccc}
1 &0 &0 &0 &... \\
0 &1 &0 &0 &... \\
1 &-2 &1 &0 &... \\
0 &1 &-2 &1 &... \\
... &...&...&...&...\\
\end{array}\right]
\begin{bmatrix}
f_{0} \\
f_{1} \\
\vdots \\
\vdots \\
\end{bmatrix} \\
{} = \begin{bmatrix}
7 \\
15 \\
\vdots\\
\vdots\\
\end{bmatrix}
\end{align*}
Therefore, our initial problem has now the form of $$A\vec{v}=\vec{w}$$
My question here is what can we obtain if we calculate the inverse of the matrix $A$ and what is the general formula for the inverse of $A$ for any homogeneous recursive relation.
In this specific problem I think that it might be the matrix
\begin{bmatrix}
1 & 0 & 0 & 0 & ...\\
0 & 1 & 0 & 0 & ...\\
-1 & 3 & 1 & 0 & ...\\
-3 & 8 & 3 & 1 & ...\\
... & ... & ... & ... & ...
\end{bmatrix}
| 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:
$$\begin{pmatrix}f_n\\f_{n-1}\end{pmatrix}=\begin{pmatrix}2&-1\\1&0\end{pmatrix}\begin{pmatrix}f_{n-1}\\f_{n-2}\end{pmatrix} \ \text{with} \ \begin{pmatrix}f_1\\f_{0}\end{pmatrix}=\begin{pmatrix}15\\7\end{pmatrix}$$
finally giving :
$$\begin{pmatrix}f_n\\f_{n-1}\end{pmatrix}=\begin{pmatrix}2&-1\\1&0\end{pmatrix}^{n-1}\begin{pmatrix}15\\7\end{pmatrix}$$
| {
"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:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\ \left|\cos\frac{x}{2}\right|}dx$$
$$=\sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx\tag4$$
$$=\sqrt{2}\int{\pm\cos\frac{x}{2}}dx$$
$$=\pm\sqrt{2}\int{\cos\frac{x}{2}}dx$$
$$=\pm2\sqrt{2}\sin\frac{x}{2}+C$$
My book's attempt:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\cos\frac{x}{2}}dx$$
$$=\sqrt{2}\int{\cos\frac{x}{2}}dx$$
$$=2\sqrt{2}\sin\frac{x}{2}+C$$
Basically, my book didn't put $\pm$ sign, while I did. My book did this essentially: $\sqrt{x^2}=x$, while I did this: $\sqrt{x^2}=|x|$. Is my process more correct?
| 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{align}
A global ante-derivative can be concatenated as
$$ \int\sqrt{1+\cos x}\>dx= 2\sqrt{1+\cos x}\>\tan \frac x2 +4\sqrt2\lfloor \frac {x+\pi}{2\pi}\rfloor
$$
| {
"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
} |
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 $-2$ is also a solution.
How do I find the value of $x$ (not intuitively like I did previously) for which the equation is true?
Thanks in advance for any help.
edit: Solved, thanks to all who replied. Split $2^{|x+1|} = 2^x+1+|2^x-1|$ into three cases: $x >= 0$ and $x < -1$ and $-1 <= x < 0$. Solution: $[0, \infty]$ ∪ ${-2}$
| 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,
"answer_id": 0
} |
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 intersection if $|a| < 1$.
Workings: I started off by letting $f(x) = x+c$ and then rerranging produces $$(a+c+2)x^2-(2a-1)x - c = 0$$
In order for the line $x+c$ and $f(x)$ intersect, then the discriminant $\Delta \geq 0 \iff (2a-1)^2-4(a+c+2)(-c) \geq 0$
Rerranging produces $$4a^2+4a+1+4ac+4c^2+8c \geq 0$$
This is the point where I start having difficulties. We have a discriminant in two variables. I notice that we can rewrite as a quadratic in $c$, so $$4c^2 + (8+4a)c + 4a^2+4a+1 \geq 0$$. If we want this quadratic in $c$ to be $\geq 0$ for all $c$ then this second discriminant must $\leq 0$. $$\Delta_2 = (8+4a)^2 - 16(4a^2+4a+1) \leq 0$$ Hence, $$\Delta_2 \leq 0 \iff 48-48a^2 \leq 0$$
So we see that when $|a| \geq 1$, the line intersects the curve.
However, I do not know how to approach the second part.
I.e.
but there are values of $c$ for which there are no points of
intersection if $|a| < 1$
Some additional questions, that I would like to ask.
Q1: Does the line $x+c$ intersect $f(x)$ for whatever $c$ value as long as $|a|\geq 1$. What if we viewed the first discriminant as a quadratic in $a$ instead of $c$, and got a range of values for $c$ for which the curves would not intersect? Would this mean that the curves would intersect only when $|a| \geq 1$ and not within some range of values of $c$
|
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.
$$\begin{align}\Delta&=(-(2a+1))^2-4(a+c+2)(-c)
\\\\&=4c^2+(4a+8)c+(2a+1)^2
\\\\&=4\bigg(c^2+(a+2)c\bigg)+(2a+1)^2
\\\\&=4\bigg(\bigg(c+\frac{a+2}{2}\bigg)^2-\bigg(\frac{a+2}{2}\bigg)^2\bigg)+(2a+1)^2
\\\\&=4\bigg(c+\frac{a+2}{2}\bigg)^2-4\bigg(\frac{a+2}{2}\bigg)^2+(2a+1)^2
\\\\&=(2c+a+2)^2+3(a^2-1)\end{align}$$
*
*If $|a|\geqslant 1$, i.e. $a^2-1\geqslant 0$, then $\Delta\geqslant 0$.
*If $|a|\lt 1$, i.e. $a^2-1\lt 0$, then there are values of $c$ for which $\Delta\lt 0$ since $$\begin{align}\Delta\lt 0&\iff (2c+a+2)^2\lt 3(1-a^2)
\\\\&\iff -\sqrt{3(1-a^2)}\lt 2c+a+2\lt \sqrt{3(1-a^2)}
\\\\&\iff \frac{-a-2-\sqrt{3(1-a^2)}}{2}\lt c\lt \frac{-a-2+\sqrt{3(1-a^2)}}{2}\end{align}$$
Does the line $x+c$ intersect $f(x)$ for whatever $c$ value as long as $|a|\geq 1$
No, consider the following examples :
*
*If $a=1$, then $y=x-3$ does not intersect $y=f(x)$.
*If $a=-1$, then $y=x-1$ does not intersect $y=f(x)$.
So, in the first place, it is not true that the line $y=x+c$ intersects the curve $y=f(x)$ if $|a|\geqslant 1$.
(To notice the above examples, note that we have $x\not=\pm 1$ in the first place since the denominator of $f(x)$ is $x^2-1$. $(1)$ has a solution $x=1$ iff $a=1$. $(1)$ has a solution $x=-1$ iff $a=-1$. For $a=1$, $(1)$ can be written as $(x-1)((c+3)x+c)=0$. For $a=-1$, $(1)$ can be written as $(x+1)((c+1)x-c)=0$.)
So, we can say the followings :
*
*The line $y=x+c$ intersects $y=f(x)$ for whatever $c$ value except $c=-3,-1$ as long as $|a|\geqslant 1$.
*The line $y=x+c$ intersects $y=f(x)$ for whatever $c$ value as long as $|a|\color{red}{\gt} 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
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, $\cos(x) =\sin(90^{\circ} - x)$
\begin{aligned}\implies \lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
& = \lim_{x\to0}\frac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
\\& = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)\cdot\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}{\sin(\sin(x^2))}
\\ & = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}\cdot \underbrace{\lim_{x\to0}\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}_{1}
\\ & = \dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(\sin(x^2))}{\sin(x^2)}}_1\cdot\sin(x^2)}
\\ & =\dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(x^2)}{x^2}}_1\cdot x^2}
\\ & = \color{blue}{\boxed{\lim\limits_{x\to0}\dfrac{\pi}{2x^2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}}
\end{aligned}
Now, I'm unable to think of anything to do with this boxed part. Can anyone check my above method and tell me what to do further with this question? Any other shorter method is also most welcomed!
| $$
\cos\left(x\right)-1 \underset{(0)}{=}-\frac{x^2}{2}+o\left(x^3\right)
$$
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
} |
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 be used more than once and I have no clue which one would be useful... Maybe letting line $UV$ hit the circle in two random points may help but I could not develop it. There are 6 pairs of points to choose from in order to get the Pascal hexagons.
| 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 primitive root of order $20$, a root of the cyclotomic polynomial $\Phi_{20}=Z^8-Z^6+Z^4-Z^2+1$.
We identify the vertices $A,B,\dots,T$ of the given $20$-gon with their affixes $a, b, \dots,t$ which are of the shape $z^k\in \Bbb C$, $k\in\Bbb Z/20$, so that $a=1$, $b=z$, $c=z^2$, and so on.
Then $V$ is the orthocenter of $\Delta JRE$, and the centroid is between zero and $V$ at one third from zero, so
$v=3\cdot\frac 13(j+r+e)=z^9+z^{17}+z^4 = z^4+z^9-z^7$. (Or use the parallelogram $VEHJ$.)
Then $U$ is the incenter of the right, isosceles $\Delta TLB$, it can be obtained as a $+90^\circ$-rotation of $B$ around $A$, so $u = 1 + z^5(z-1)$.
Then $U,V,H$ are collinear, iff the fraction $(u-h)/(v-h)$ is real, so iff the product $(u-h)(\bar v - \bar h)$ is invariated by complex conjugation. Now we compute:
$$
\begin{aligned}
(u-h)
&=(z^6-z^5+1-z^7)=(z^6+1)-z^5(z^2+1)\\
&=(z^2+1)(z^4-z^2+1-z^5)=-(z^2+1)(z-1)(z^4+z+1)
\\[2mm]
(\bar v-\bar h)
&=\overline{z^9 - 2z^7 + z^4}
=z^5(z-1)(z^5+1)(z^3-z-1)\ ,
\\[2mm]
(u-h)(\bar v-\bar h)
&=-z^5\cdot (z-1)^2\cdot (z^2+1)\cdot (z^5+1)
\cdot \underbrace{(z^3-z-1)(z^4+z+1)}_{=-\frac 1z(z^3+z^2+z+1)}\\
&= z^4\cdot (z-1)\cdot (z^2+1)\cdot (z^5+1)\cdot (z^4-1)
\ ,
\end{aligned}
$$
and the last expression is real, i.e. invariated by complex conjugation, because for $z^k\ne -1$
$$
\frac{1+z^k}{\overline{1+z^k}}=z^k\text{ , so }
\frac{\ z -1\ }{\overline{z-1}}=-z\ ,\
\frac{\ z^2+1\ }{\overline{z^2+1}}=z^2\ ,\
\frac{\ z^5+1\ }{\overline{z^5+1}}=z^5\ ,\
\frac{\ z^4-1\ }{\overline{z^4-1}}=-z^4\ .
$$
$\square$
| {
"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 $$\sin^{-1}\left(\frac{2x}{x^2+1}\right)=2\tan^{-1}(x)$$
(a)
We wish to show that these are equal for $-1\leq x \leq 1$.
For this domain $\displaystyle -1 \leq\frac{2x}{x^2+1} \leq 1 \implies -\frac{\pi}{2} \leq \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
$$\frac{2x}{x^2+1} = \sin(2\tan^{-1}(x))$$
The task is now to show $\sin(2\tan^{-1}(x))=\frac{2x}{x^2+1}$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
For $-1 \leq x \leq 1 \implies -\frac{\pi}{4}\leq\tan^{-1}(x)\leq\frac{\pi}{4} \implies -\frac{\sqrt{2}}{2}\leq \sin(\tan^{-1}(x)) \leq \frac{\sqrt{2}}{2}$
Also, $\frac{\sqrt{2}}{2}\leq\cos(\tan^{-1}(x)) \leq 1$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x)) \iff 2(\frac{x}{\sqrt{x^2+1}})(\frac{1}{\sqrt{x^2+1}})= \frac{2x}{x^2+1}$$
Which was to be shown.
(b)
$\displaystyle x \geq 1 \implies 0<\frac{2x}{1+x^2}\leq 1 \implies 0< \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
Hence,
We must show that $$\sin^{-1}(\frac{2x}{x^2+1}) = \pi - 2\tan^{-1}(x) \iff \frac{2x}{x^2+1} = \sin(\pi - 2\tan^{-1}(x))$$ for $x\geq 1$
$$\sin(\pi - 2\tan^{-1}(x))=\sin(\pi)\cos(2\tan^{-1}(x)) - \cos(\pi)\sin(2\tan^{-1}(x))=\sin(2\tan^{-1}(x))$$
$$\sin(2\tan^{-1}(x)) = 2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
Which was to be demonstrated.
Note: This problem didn't flow like I thought it would. I had imagined that during some of the intermediate steps, I would be presented with the option of choosing an $f(x)$ or trig function value that would only be true in one of the intervals. But no such situation presented itself. Did I do something wrong? Did I overlook something?
| $$
\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\theta\quad\iff\quad\sin(\theta)=\frac{2x}{1+x^2}\quad\land\quad-\frac\pi2\le\theta\le\frac\pi2
$$
Note that
$$
\cos(\theta)=\left|\frac{1-x^2}{1+x^2}\right|
$$
When $|x|\le1$,
$$
\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}=\frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}}=x
$$
Therefore,
$$
\theta=2\tan^{-1}(x)
$$
When $|x|\gt1$,
$$
\tan(\theta/2)=\frac{\sin(\theta)}{1+\cos(\theta)}=\frac{\frac{2x}{1+x^2}}{1-\frac{1-x^2}{1+x^2}}=1/x
$$
Therefore,
$$
\theta=2\tan^{-1}(1/x)
$$
Thus,
$$
\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\left\{\begin{array}{}
2\tan^{-1}(x)&\text{if }|x|\le1\\
2\tan^{-1}(1/x)&\text{if }|x|\gt1
\end{array}\right.
$$
and
$$
\tan^{-1}(1/x)=\left\{\begin{array}{rl}
\frac\pi2-\tan^{-1}(x)&\text{if }x\gt0\\
-\frac\pi2-\tan^{-1}(x)&\text{if }x\lt0\\
\end{array}\right.
$$
| {
"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, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}$
$(x^2+5)^2[\frac{3x+4}{(x^2+5)^2}]=(x^2+5)^2[\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}]$
$3x+4=(Ax+B)(x^2+5)+(Cx+D)$
$3x+4=Ax^3+5Ax+Bx^2+5B+Cx+D$
$3x+4=(A)x^3+(B)x^2+(5A+C)x+(5B+D)$
So, $A=0$, $B=0$, $5A+C=3$, and $5B+D=4$
So, $A=0$, $B=0$, $C=3$, and $D=4$
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{(0)x+(0)}{x^2+5}+\frac{(3)x+(4)}{(x^2+5)^2}}dx$
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{3x+4}{(x^2+5)^2}}dx.$
As this clearly doesn't work, I am wondering what the integral would be. WolframAlpha tells me that the integral is as follows:
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\frac{1}{50}(\frac{5(4x-15)}{x^2+5}+4\sqrt{5}\tan^{-1}(\frac{x}{\sqrt{5}}))+c$
I am not sure how to get here. Any help would be greatly appreciated.
| 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) = \frac{1}{\cos(x)}, $ so: $\cos(x) = \frac{b}{a}$
*I used the right angled triangle to find the third side, which is $\sqrt{a^2-b^2}$
*I used reciprocal functions of $\tan$ and $\sin$ to obtain the values:
$cosec(x) = \frac{a}{(\sqrt{a^2-b^2})}$ and $\cot(x) = \frac{b}{\sqrt{a^2-b^2}}$
and finally getting $\frac{(b-a)}{\sqrt{a^2-b^2}}$ as my answer. But that is not correct.
The correct answer is: $\frac{\sqrt{a^2-b^2}}{a+b}$. Now I am wondering how.
| $$\cos x=\frac ba$$
$$\frac 1{\sin x}=\frac 1 {\sqrt{1-\frac{b^2}{a^2}}}=\frac a{\sqrt {a^2-b^2}}$$
$$\cot (x) =\frac {\cos x}{\sin x}=\frac{\frac ba}{\frac{\sqrt{a^2-b^2}}a}=\frac b{\sqrt {a^2-b^2}}$$
$\Rightarrow$ $$\cot (x)-\frac 1{\sin x}=\frac {a-b}{\sqrt{a^2-b^2}}$$
Multiplying numerator and denominator by $\sqrt{a^2-b^2=(a-b)(a+b)}$ we get:
$$\cot (x)-\frac 1{\sin x}=\frac {\sqrt{a^2-b^2}}{a+b}$$
| {
"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
} |
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, $$a+b+c=2017$$ $$a^2+b^2+c^2-ab-bc-ca=1$$
How should I proceed after this?
| $$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2} \bigg( (a-b)^2 + (b-c)^2 + (c-a)^2 \bigg)=1$$
$$ \implies (a-b)^2 + (b-c)^2 + (c-a)^2 =2$$
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.).
$$a+b+c=3a\pm1 = 2017 = 3\times672 \,+1$$
Thus $c=a+1$, $a=b=672$ and $c=673$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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. $\alpha^6 + \beta^6 = 18$
My attempts of solving this problem:
A.$$ \alpha^2 + \beta^2 = 2 $$
$$\alpha^2 + (-2\ \alpha \cdot \beta) + \beta^2 + 2\ \alpha\cdot\beta$$
$$=\left(\alpha + \beta \right)^2 + \left( -2\ \alpha\beta \right) $$
We already know that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ which implies the following equation :
$1^2 + 2 \cdot 1= 3$
So we can conclude A isn't the answer
B.
$$\alpha^3 + \beta^3 = \left(\alpha +\beta \right)\left(\alpha^2 - \alpha\beta + \beta^2 \right)$$
$$1\cdot \left(3 + 1)\right) =4 $$
So B is neither the right answer.
For the following multiple choices I can't get to the final answer. Could someone help me with this?
Kind regards in advance.
| 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",
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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 \sqrt{2u-u^2}\:dx\\
\therefore dx&= \dfrac{du}{\pm \sqrt{2u-u^2}}
\end{align*}
\begin{align*}
\therefore \int \dfrac{1}{1+\sin x}\: dx &= \pm \int \dfrac{1}{u} \times \dfrac{1}{\sqrt{2u-u^2}} \: du\\
&= \pm \int \dfrac{1}{u^2\sqrt{2/u -1}} \: du\\
&= \pm \int \dfrac{1}{u^2} \times \left( \dfrac{2}{u}-1 \right)^{-1/2}\: du\\
&=\pm \dfrac{1}{2} \int \dfrac{2}{u^2} \times \left( \dfrac{2}{u}-1 \right)^{-1/2}\: du\\
&= \mp \dfrac{1}{2} \times 2\times \sqrt{\left(\dfrac{2}{u}-1\right)} + C\\
&= \mp \sqrt{\dfrac{2}{1+\sin x}-1}+C
\end{align*}
My final answer seems to be the absolute value of one of the primitives of $\dfrac{1}{1+\sin x}$. I know that having the $\pm$ in the $u$-sub is not ideal, but I left it there because I wasn't sure how to address it correctly. From my experience, these problems usually occur because the substitution was of the form $x=f(u)$ instead of $u=g(x)$. But in my method, the latter is used...
Question: How can I properly address this issue?
P.S. Yes, I know there are better ways (Weierstrass sub and clever manipulations) to approach this integral. I just want to know how to properly use this approach.
| 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",
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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:
$ \displaystyle \tag*{}x \mapsto a-b \\\\ y\mapsto b-c \\\\ z \mapsto c-a$
We get, $x+y+z=0$ and then the inequality becomes:
$\displaystyle \tag*{} \left(\frac{x}{x+2b}\right)^{11}+\left(\frac{y}{y+2c}\right)^{11}+\left(\frac{z}{z+2a}\right)^{11}\leq 1$
I don't know how to proceed from this; Can we use AM-GM inequality to prove? Any hints would be appreciated, thanks.
| 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 $y=b/c>0$, the expression becomes
$$f(x,y)=\left(\frac{x-1}{x+1}\right)^{n}+\left(\frac{y-1}{y+1}\right)^{n}+\left(\frac{1-xy}{1+xy}\right)^{n}\leq 1 \,.$$
The function $\frac{x-1}{x+1}$ is strictly increasing from $-1$ to $1$. Therefore, in the first case $a<b<c$, $x<1$ and $y<1$, the first two terms are negative and the last one positive and $\leq 1$. Overall it is clear that $f(x,y)\leq 1$. In the second case $a>b>c$, $x>1$ and $y>1$, the first two terms are positive and the last one negative. However, since $xy>y$ it is clear that $$\left(\frac{y-1}{y+1}\right)^{n}-\left(\frac{xy-1}{xy+1}\right)^{n} < 0$$
and the result follows.
| {
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"url": "https://math.stackexchange.com/questions/4380924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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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 essence? He writes that $X+A-A$ is also the empty set but it cannot be an empty.
*What does "the size of $|X+A|$ increases by at least $|A|/2$, by construction, and at the first stage it increases by $|A|." mean? What is the mathematical language of that sentence?
Actually I have much more questions but firstly I would like to understand these questions.
Please help to understand!
| 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 form $x + a - a'$.
So inially, we may choose any element $y \in B$ to add to $X$. Let's add $1$.
Step 2. Now, $X = \{1\}$. With this $X$, $X + A - A$ is no longer empty; it is the set $$\{-6,-5,-3,-2,-1,0,1,2,3,4,5,7,8\}.$$ If we are keeping track of multiplicities, then $1$ appears $4$ times (because it is $1+1-1 = 1+2-2 = 1+4-4 = 1+8-8$) while all other elements appear only once (for example, $5$ only appears as $1+8-4$).
We are looking for $y \in B$ such that $|(y + A) \cap (X+A)| \le |A|/2$, or in other words $$\big| \{y+1, y+2, y+4, y+8\} \cap \{2,3,5,9\} \big| \le 2.$$ Taking $y=1$ does not work, because then $\{y+1,y+2,y+4,y+8\} = \{2,3,5,9\}$, intersecting $\{2,3,5,9\}$ in four elements. However, either $y=2$ or $y=3$ would work:
*
*If $y=2$, then $\{y+1,y+2,y+4,y+8\} = \{3,4,6,10\}$, intersecting $\{2,3,5,9\}$ in only one element.
*If $y=3$, then $\{y+1,y+2,y+4,y+8\} = \{4,5,7,11\}$, also intersecting $\{2,3,5,9\}$ in only one element.
Let's pick $y=3$.
Step 3. Now, $X = \{1,3\}$. With this $X$, we get
$$
X+A-A = \{-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10\}
$$
where many elements appear with multiplicity, but in particular $1$ appears with multiplicity $5$ ($1 = 1+1-1 = 1+2-2 = 1+4-4 = 1+8-8 = 3+2-4$), $2$ appears with multiplicity $2$ ($2 = 1+2-1 = 3+1-2$), and $3$ appears with multiplicity $5$ ($3 = 1+4-2 = 3+1-1 = 3+2-2 = 3+4-4 = 3+8-8$). We have found a set $X$ satisfying the conclusion of the theorem
When the proof says
the size of $|X+A|$ increases by at least $|A|/2$, by construction, and at the first stage it increases by $|A|$
it is referring to the size of $X+A$ at the beginning of each step. In this example:
*
*At the beginning of step 1, $X = \emptyset$ so $X+A = \emptyset$ as well, and $|X+A| = 0$.
*At the beginning of step 2, $X = \{1\}$, so $X+A = \{2,3,5,9\}$, and $|X+A| = 4$. From step 1 to step 2, this size has increased by $|A|=4$, as promised. This happens because at this point, $X$ is a singleton set, so $X+A$ is a translate of $A$, and $|X+A| = |A|$.
*At the beginning of step 3, $X = \{1,3\}$, so $X+A = \{2,3,4,5,7,9,11\}$, and $|X+A| = 7$. From step 2 to step 3, this size has increased by $3$. The proof promised us that this increase would be at least $|A|/2 = 2$, and it is.
Why does the size increase by at least $|A|/2$ after every step? Because the new element $y$ we introduce satisfies $|(y + A) \cap (X+A)| \le |A|/2$. In the next step, the new value of $X+A$ will be $(X \cup \{y\}) + A$, which can be rewritten as $(X + A) \cup (y + A)$. Since $y+A$ has $|A|$ elements, and at most $|A|/2$ of them are already in $X+A$, at least $|A|/2$ are new.
The essence of the proof is that:
*
*When there is an element we can add to $X$ that increases $|X+A|$ by a lot, we do it.
*Once there is no longer any such element, we're done and have the set $X$ the lemma wanted.
*Because we increase $|X+A|$ by a lot at every step, but it can never exceed $|B+A|$, we must stop quickly. This means that when we're done, $|X|$ is still small.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$\sqrt{\left(x-2\right)^{2}+\left(y-3\right)^{2}}=\frac{\left|5x+6y+5\right|}{\sqrt{5^{2}+6^{2}}}\tag{3}$$
$$\left(x-2\right)^{2}+\left(y-3\right)^{2}=\frac{\left(5x+6y+5\right)^{2}}{5^{2}+6^{2}}\tag{4}$$
The graphs of $(3)$ and $(4)$ are exactly the same! When both sides are squared in $(3)$, we get $(4)$, and no information is lost in the process: the two graphs are identical!
Question
*
*Why is information preserved in scenario 2, i.e. the graphs of $(3)$ & $(4)$ are identical, when information is lost in scenario 1, i.e. the graphs of $(1)$ & $(2)$ aren't identical?
| 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 guaranteed to be positive, so one cannot be the negative of the other; only the $A=B$ case is a possibility to begin with.
In other words: by squaring both sides, we are adding in all points which satisfy $$\sqrt{\left(x-2\right)^{2}+\left(y-3\right)^{2}} = -\frac{\left|5x+6y+5\right|}{\sqrt{5^{2}+6^{2}}}$$ but there are no such points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4384242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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, a+b=?
My try:
At first,
*
*AO = 5
*AO' = 5
*OC = 5
*O'B = 5
*OO' = 6
*OB = OO' - O'B = 6-5 = 1
*O'C = 1
*BC = OO' - (OB+O'C) = 6-2 = 4
So,BC = 4,
*
*In $\Delta$AOO' ,(using Law of cosines) AO$^2$ = AO'$^2$ + OO'$^2$ - 2AO'$*$OO'$*$$\cos$$\angle$AO'O
So $\angle$AO'O = 53.13°
*$\Delta$AO'B is Isosceles , i know AO' = 5, BO' = 5 and $\angle$AO'B = 53.13°
So base AB = 2×AO'×$\sin$(0.5×$\angle$AO'B) = 4.4721.
So AC = 4.4721
Now i know the length of AB, AC and BC, and i can find the radius of incircle of $\Delta$ABC
so the radius= $\frac{\sqrt{s(s-AB)(s-AC)(s-BC)}}{s}$
$s = (AB+AC+BC)/2 = (4.4721+4.4721+4)/2 = 6.4721.$
Also
${\sqrt{s(s-AB)(s-AC)(s-BC)}} = 8$ but the value of s is 6.4721, how can I express $\frac{8}{6.4721} $ as $\frac{a}{\sqrt b +1}$ ??
| $AO = AO' = 5$ 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 $\triangle ABC$,
$s = \frac 12 (4 + 2 \cdot 2 \sqrt5) = 2 (1 + \sqrt5)$
Area of $~\triangle ABC, [ABC] = \frac 12 \cdot BC \cdot AH = 8$
If $r$ is the inradius of $\triangle ABC$, we know $[ABC] = r\cdot s$ $$\implies r = \dfrac{4}{1 + \sqrt5}$$
$ \therefore ~a + b = 9$
| {
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"timestamp": "2023-03-29T00:00:00",
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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:$$\frac{dy}{y+1}=-\frac{\cos x}{2+\sin x}dx\\\implies\ln|y+1|=-\ln(2+\sin x)+\ln c\\\implies|y+1|=\frac{c}{2+\sin x}$$$y(0)=1\implies c=4.$ And putting $x=\frac\pi2,$ we get, $|y+1|=\frac43\implies y=\frac13,-\frac73$
Are both the answers acceptable? The answer given in the book is only $y(\frac\pi2)=\frac13$.
| 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.
| {
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"url": "https://math.stackexchange.com/questions/4388150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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:
$\frac{AM.DM}{2}.\frac{CN.FN}{2}=16 \implies AM.DM.CN.FN=64\\
\frac{S1}{S2} = \frac{AM.MD}{CN.FN}\\
\frac{S1}{\frac{MI.DM}{2}}=\frac{AM}{MI}\implies S1 = \frac{AM.DM}{2}\\
\frac{S2}{\frac{NI.FN}{2}}=\frac{CN}{NI}\implies S2 = \frac{CN.FN}{2}$
.....????
| 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, diagonals $BI$ and $FD$ are equal and $O$ is the midpoint of both diagonals. We also have, $DM \parallel OI \parallel FN$
That leads to $~OI = \dfrac{x+y}{2} \implies FD = BI = x + y$
Finally using Pythagoras, $MN^2 = (x+y)^2 - (y-x)^2 = 4xy = 32~$ and $~MN = 4 \sqrt2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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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 $2 u d x=5 x^{4} d x$
$$
\begin{aligned}
I &=\int\left(u^{2}-2\right)^{2} u \cdot \frac{2}{5} u d u \\
&=\frac{2}{5} \int\left(u^{6}-4 u^{4}+4 u^{3}\right) d u \\
&=\frac{2}{5}\left(\frac{u^{7}}{7}-\frac{4 u^{5}}{5}+\frac{4 u^{3}}{3}\right)+C \\
&=\frac{2 u^{3}}{525}\left(15 u^{4}-84 u^{2}+140\right)+C \\
&=\frac{2\left(x^{5}+2\right)^{\frac{3}{2}}}{525}\left[15\left(x^{5}+2\right)^{2}-84\left(x^{5}+2\right)+140\right]+C \\
&=\frac{2\left(x^{5}+2\right)^{\frac{3}{2}}}{525}\left(15 x^{10}-24 x^{5}+32\right)+C .
\end{aligned}
$$
Wish it helps!
| {
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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 $x = -1$, $$A = \frac{1}{4}$$
However, $$\frac{1}{4(1 + x)} + \frac{1}{2 (1 - x)^2}$$ doesn't seem to equal $$\frac{1}{(1 + x)(1 - x)^2}$$
How do we decompose this into partial fractions?
Thanks
| 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 $x$, and the constant term should be $1$), but you've allowed only $2$ degrees of freedom $A,B$.
The general form which allows decomposing any fraction into partial fractions can be found e.g. in Wikipedia (see specifically the examples where it is the easiest to understand it).
Specifically for your problem, the form you should try to decompose into is
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{1 - x} + \frac{C}{(1-x)^2}.$$
| {
"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
} |
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
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{2}{\sqrt{3}}\text{.}$$
My efforts. We can start off by observing that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = 2 \cdot \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]$$
so it is sufficient to prove that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{1}{\sqrt{3}}\text{.}$$
I am not sure how to go about this step.
One could try to handwave replacing $X_2$ and $X_3$ with $X_1$, but I can't think of a way of theoretically justifying that (it's probably a bad idea).
| 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
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] + \mathbb{E}\left[\dfrac{X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]\text{.}$$
The two terms of the right-hand side of the above equation are equal by exchangeability, hence we know that
$$\mathbb{E}\left[\dfrac{X_1 + X_2}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] = 2 \cdot \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right]$$
so it suffices to prove that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] < \dfrac{1}{\sqrt{3}}\text{.}$$
By Cauchy-Schwarz, observing that $\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} > 0$ with probability $1$, and therefore $\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] > 0$, we have that
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}}\right] = \mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \cdot 1 \right] \leq \sqrt{\mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right]}\text{.}$$
Observe that
$$\mathbb{E}\left[\dfrac{X_1^2 + X_2^2 + X_3^2}{X_1^2 + X_2^2 + X_3^2} \right] = 1 = \sum_{i=1}^{3}\mathbb{E}\left[\dfrac{X_i^2}{X_1^2 + X_2^2 + X_3^2} \right]\text{.}$$
Exploiting exchangeability again, we have that
$$1 = \sum_{i=1}^{3}\mathbb{E}\left[\dfrac{X_i^2}{X_1^2 + X_2^2 + X_3^2} \right] = 3 \cdot \mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right]$$
hence
$$\mathbb{E}\left[\dfrac{X_1^2}{X_1^2 + X_2^2 + X_3^2} \right] = \dfrac{1}{3}$$
hence
$$\mathbb{E}\left[\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} \right] \leq \dfrac{1}{\sqrt{3}}\text{.}$$
However, equality holds if and only if $\dfrac{X_1}{\sqrt{X_1^2 + X_2^2 + X_3^2}}$ is constant (i.e., proportional to $1$) almost surely, so we can drop the equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $$y_n \geq (1+\frac{n^2}{n^3}) \geq 1$$ I have also found that $$y_n - y_{n+1} \geq \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} - \left(1+\frac{1}{(n+1)^3}\right)^{(n+1)^2} = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left(1 - \left( 1+\frac{1}{(n+1)^3} \right) ^{2n+1} \right) = \left( 1+\frac{1}{(n+1)^3}\right)^{n^2} \left( \frac{1}{(n+1)^3} \right) \left( 1 + \left( 1 + \frac{1}{(n+1)^3} \right) + \cdots + \left( 1+\frac{1}{(n+1)^3}\right)^{2n} \right) \geq 1*0*2n\geq 0 $$$$\implies yn \geq y_{n+1}$$ Thus, by using the monotone convergence theorem, we know $y_n$ converges and has a lower bound of $1$. I am however stuck at showing that $\inf{\{y_n | n \geq 1\}} = 1$, which would show that $\lim_{n \to \infty} y_n = 1$. Could I get a hint or a nudge in the right direction ?
PS: I cannot use exponential and logarithmic properties, nor l'hopital's rule, as we have not defined all these things in class
| 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
} |
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 the critical point is $ x = 18$ and $ f'(x) < 0 $ for $ 9 < x < 18 $ and $ f'(x) > 0 $ for $ x > 18 $. So, using first derivative test, $ f(x) $ has absolute minimum at $x = 18$. So, we can find that $f(18) = 36 $ is the absolute minimum.
Method 2). Since we have
$$ y = \frac{x^2}{(x-9) } $$
$$ \Longrightarrow x^2 - xy +9y = 0 $$
This is a quadratic equation. Now $x > 9$, so it follows that $ y > 0 $. Now the discriminant of this quadratic equation is
$$ D = y^2 - 36 y $$
Since roots of this quadratic equation are values of $x$ and since $x > 9$, we know that the roots are real and so we have discriminant $ D \geqslant 0 $, which means that
$$ y ( y-36) \geqslant 0 $$
Since $ y > 0 $, it follows that $ y \geqslant 36 $. So, 36 is the minimum value of $y$.
Method 3.
Now, we have
$$ \frac{(x-9) }{x^2} = \frac{1}{x} - \frac{9}{x^2} $$
$$ \frac{(x-9) }{x^2} = \Big( \frac{1}{9} \Big) \Big( \frac{9}{x} \Big) \Big( 1- \frac{9}{x} \Big) $$
We should note that since $ x > 9$, all quantities above are positive. So, we can now use AM-GM inequality
$$ \frac{9}{x} + \Big( 1- \frac{9}{x} \Big) \geqslant 2 \bigg[ \frac{9}{x}\Big( 1- \frac{9}{x} \Big) \bigg]^{1/2} $$
$$ 1 \geqslant 6 \bigg [ \frac{1}{x} \Big( 1- \frac{9}{x} \Big) \bigg]^{1/2} $$
$$ \therefore \; \bigg[ \frac{1}{x} - \frac{9}{x^2} \bigg ] \leqslant \frac{1}{36} $$
$$ \therefore \; \frac{(x-9) }{x^2} \leqslant \frac{1}{36} $$
$$ \therefore \; \frac{x^2}{x-9} \geqslant 36 $$
So, the minimum value is $36$
Are these good approaches ?
| 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
} |
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 - e)} + \frac{d^4}{(d - a)(d - b)(d - c)(d - e)} \\ &\quad + \frac{e^4}{(e - a)(e - b)(e - c)(e - d)}. \end{align*}
I started by using Vieta's formulas to obtain the following system of equations:
\begin{cases}
&\sum_{\text{cyc}} a = -7, \\
&\sum_{\text{cyc}} ab = 0, \\
&\sum_{\text{cyc}} abc = 0, \\
&\sum_{\text{cyc}} abcd = 0, \\
&abcde = 2.
\end{cases}
Now let $f_a(x) = (x - b)(x - c)(x - d)(x - e)$, then $f_a(x) = x^4 - x^3\sum_{\text{cyc}} b + x^2 \sum_{\text{cyc}} bc -x \sum_{\text{cyc}}bcd + bcde$. By using the above system of equations, we can eventually simplify it down and see that $f_a(a) = a^4 + (7a^3 + a^4)4$. Now there are a ton of options for what we can do, but now the sum we desire to find is given by $$\sum_{\text{cyc}} \frac{a^4}{a^4 + (7a^3 + a^4)4}.$$
Using that $a^5 + 7a^4 - 2 = 0$, we can simplify further to find that $f_a(a)a = 10 - \frac{14}{a+7}$ and $\frac{2a}{a+7} = a^5$ so $$\sum_{\text{cyc}} \frac{a^4}{a^4 + (7a^3 + a^4)4} \cong \sum_{\text{cyc}} \frac{a}{5a+28}.$$
Now this should be simpler, but after thinking about it I don't really see a good way to proceed. What can I do? Did I drive into the weeds, or am I still on the misty road and I just don't see the end?
| 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
$$ g(t) = \frac{-4296342t^5 + 4296342t^4 + 2500t^3 - 500t^2 + 50t - 2}{-3125t^5 + 3125t^4 - 1250t^3 + 250t^2 - 25t + 1} $$
Now, $g(t)$ is zero when
$$ h(t) = -4296342t^5 + 4296342t^4 + 2500t^3 - 500t^2 + 50t - 2 $$
is zero.
If $t$ is a root of $h(t) $ and
$$t = \frac{x}{5x+28} , $$
then $x$ is a root of your $x^5 + 7 x^4 - 2.$ The sum of the five roots of $h$ gives the sum of your desired $ \frac{x}{5x+28} $ with $x$ the five roots...
To get monic, let $h(t) = -4296342 h_1(t),$ so that
$$ h_1(t) \; = \; \; t^5 \; - \; t^4 \; - \;\frac{1250t^3}{2148171} + \; \frac{250t^2}{2148171} - \; \frac{25t}{2148171} + \; \frac{1}{2148171} $$
and the sum of the roots of $h_1$ is ...
| {
"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
} |
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 $k=1$, and for that value of $k$ try to write the integral from $0$ to $2\pi$ as a sum of four integrals from $0$ to $\dfrac{\pi}{2}$ and use the the theorem for first derivative monotony. No luck so far.
Any help would be much appreciated.
Edit 1: I saw the link here about a similarly asked topic. However, this process gets the general case as I perceive it and I am really supposed to use the method described above. I will try this and come back with a definitive answer.
Edit 2: I cannot use the $f''(x) \geq 0$ argument due to the simple fact that I have not been formally taught it as part of the class.
Edit 3: Final proof, with thanks to the contributors below.
Let's start by setting $A = \frac{\pi}{2}$ and $B = \frac{3\pi}{2}$. It follows from basic trigonometry that:
\begin{align}
&\cos x \geq 0, \ x \in [A,B] \ \text{and}\\
&\cos x \leq 0, \ x \in [0, A] \cap [B, 2\pi].
\end{align}
And we also set $L$ to be the line segment such that $L(A) = f(A), \ L(B) = f(B)$. We will prove a basic property of said line in regards to the convex function $f$.
*
*I can take for granted that (we proved this in class)
\begin{align}
L(x) = \dfrac{x-A}{B-A}f(B) + \dfrac{B-x}{B-A}f(A)
\end{align}
so for $x \in [A,B]$ there exists $\lambda \in [0,1]$ such that: $x = \lambda A + (1-\lambda) B$. Taking the aforementioned expression and replacing it on $L(x)$ we get (I omit trivial algebra)
\begin{align}
L(x) = (1-\lambda) f(B) + \lambda f(A).
\end{align}
Since $f$ is convex, we can write
\begin{align}
&f(\lambda A + (1-\lambda) B) \leq \lambda f(A) + (1-\lambda) f(B)\\
\implies &f(x) \leq L(x), \ \forall \ x \in [A,B] \ \text{and} \ \lambda \in [0,1].
\end{align} $\blacksquare$
*We assume that there exists $x \in [0,A]: \ f(x) < L(x)$. Then there exists $A \in [x, B] \ \text{and} \ \lambda \in [0,1]: \ A = \lambda x + (1-\lambda) B$. Then
\begin{align}
&L(A) = \dfrac{A-x}{B-x}f(B) + \dfrac{B-A}{B-x}f(x), \ A \in [x,B]\\
\implies &L(A) = (1-\lambda)L(B)+\lambda L(x)\\
\implies &L(A) = (1-\lambda)f(B) + \lambda L(x).
\end{align}
Then assuming that $L(x) > f(x)$ we get
\begin{align}
L(A) > (1-\lambda) f(B) + \lambda f(x) \geq f(A), \ \text{assuming convexity}.
\end{align}
Because $L(A) = f(A)$ the above inequality becomes $L(A) > f(A)$ which is a contradiction.
$\blacksquare$
In the same spirit, for $x \in [B, 2\pi]$ doing the exact same replacements and applying the exact same principles we also get
\begin{align}
&f(\lambda B + (1-\lambda) 2 \pi) \leq \lambda f(B) + (1-\lambda) f(2\pi)\\
\implies &f(x) \leq L(x).
\end{align}
$\blacksquare$
We then set $g(x) = f(x) - L(x)$ and from (1) and (2) above it is safe to assume that $\cos x$ and $g(x)$ will have the same sign in the whole domain, that is:
\begin{align}
&g(x) \geq 0, \ \text{where} \ \cos x \geq 0\\
&g(x) \leq 0, \ \text{where} \ \cos x \geq 0.
\end{align}
*We have
\begin{align}
\int_0^{2\pi} g(x) \cos x dx = \int_0^{\frac{\pi}{2}} g(x) \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} g(x) \cos x dx + \int_{\frac{3\pi}{2}}^{2\pi}g(x) \cos x dx \geq 0,
\end{align}
because
\begin{align}
&\text{in} \left[0, \frac{\pi}{2} \right], \ \cos x \geq 0 \implies g(x) \geq 0\\
&\text{in} \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right], \ \cos x \leq 0 \implies g(x) \leq 0\\
&\text{in} \left[\frac{3 \pi}{2}, 2 \pi \right], \ \cos x \geq 0 \implies g(x) \geq 0.\\
\end{align}
*We have that
\begin{align}
&\int_0^{2 \pi}\cos x dx = 0 \ \text{trivial}\\
&\int_0^{2 \pi}x \cos x dx = \int_0^{2 \pi}x (\sin x)' dx = \left[ x \sin x \right]_0^{2\pi} - \int_0^{2 \pi} \sin x dx = 0.
\end{align}
It then follows that
\begin{align}
\int_0^{2 \pi} f(x) \cos x dx = \int_0^{2 \pi} g(x) \cos x dx \geq 0
\end{align}
which was previously proven. For the case $k=1$, the proof is over.
For $k>1$, we have:
\begin{align}
\int_0^{2\pi}f(x) \cos (kx) dx = \sum_{i=0}^{k-1} \int_{\frac{2\pi i}{k}}^{\frac{2\pi (i+1)}{k}} f(x) \cos (kx) dx.
\end{align}
We perform the change of variable
\begin{align}
x = \dfrac{y+2\pi i}{k}\\
\implies \begin{cases}dx = \dfrac{1}{k}dy\\ x = \dfrac{2\pi i}{k} \to y=0\\ x = \dfrac{2\pi (i+1)}{k} \to y = 2\pi \end{cases}.
\end{align}
So the above sum becomes
\begin{align}
\sum_{i=0}^{k-1} \dfrac{1}{k} \int_{0}^{2\pi}f \left(\dfrac{y+2 \pi i}{k} \right) \cos (y+2\pi i)dy.
\end{align}
Because $f$ is convex in $[0, 2\pi]$ there exists $\lambda \in [0,1]$ such that:
\begin{align}
\theta f\left(\frac{y_1 + 2\pi i}{k}\right)
+ (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right)
&\ge f\left(\theta\frac{y_1 + 2\pi i}{k}
+ (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\
&= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right).
\end{align}
Having performed the change of variables:
\begin{align}
x_1 = \dfrac{y_1 + 2\pi i}{k} \ \text{and} \ x_2 = \dfrac{y_2 + 2\pi i}{k}.
\end{align}
So $f \left( \dfrac{y + 2\pi i}{k}\right)$ convex on $[0, 2\pi]$. Using the result from $k=1$ we have
\begin{align}
\int_0^{2\pi} f \left( \dfrac{y + 2\pi i}{k} \right)\cos y dy \geq 0.
\end{align}
$\blacksquare$
| 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 that $$2\sin{\pi k \over n}\cos{2\pi kj\over n}=\sin {2\pi k(j+{1\over 2}) \over n}-\sin {2\pi k(j-{1\over 2}) \over n}$$
Therefore denoting $a_j=f\left ({2\pi j\over n}\right )$ and
$b_j=\sin {2\pi k(j+{1\over 2}) \over n}$
leads to
$$2n\sin\ {\pi k \over n}\ S_n=\sum_{j=1}^n a_j(b_j-b_{j-1})
=\sum_{j=0}^{n-1} (a_j-a_{j+1})b_j +a_{n}b_n-a_0b_0\\
=\sum_{j=0}^{n-1} (a_j-a_{j+1})b_j+[f(1)-f(0)]\,\sin{\pi k\over n}
$$
We have
$$ 2\sin {\pi k \over n}\cdot b_j=\left [1-\cos{2\pi k (j+1)\over n}\right ]-\left [1-\cos{2\pi k j\over n}\right ]$$
Hence denoting $\Delta a_j=a_j-a_{j+1},$ $c_j=1-\cos{2\pi k j\over n}$ and observing that $c_0=c_n=0$ gives
$$ 4n\sin^2{\pi k\over n} \ S_n= \sum_{j=0}^{n-1}\Delta a_j\,[c_{j+1}-c_{j}]+ 2[f(1)-f(0)]\,\sin^2{\pi k\over n}\\
= \sum_{j=1}^{n-1} [\Delta a_{j-1}-\Delta a_j]c_j+\Delta a_{n-1}\,c_n-\Delta a_0\,c_0 + 2[f(1)-f(0)]\,\sin^2{\pi k\over n}\\
= \sum_{j=1}^{n-1} [\Delta a_{j-1}-\Delta a_j]c_j+ 2[f(1)-f(0)]\,\sin^2{\pi k\over n}$$
Summarizing
$$S_n={1\over 4n\sin^2{\pi k\over n}} \sum_{j=1}^{n-1} [\Delta a_{j-1}-\Delta a_j]c_j+{1\over 2n}[f(1)-f(0)]$$
The sum is nonnegative as $c_j\ge 0$ and $\Delta a_{j-1}-\Delta a_j=a_{j-1}+a_{j+1}-2a_j\ge 0$, by the convexity of $f$. Hence $\lim S_n$ is nonnegative.
| {
"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 a-a^2\right) \\
\end{array}
\right].
\end{equation}
For a general integer $n$, I am trying to compute the closed form expression for the entries of:
\begin{equation}
\text{FinalVector}= \text{M}^{n} \left(
\begin{array}{c}
1\\
1 \\
\end{array}
\right).
\end{equation}
It is very easy to compute what $\text{FinalVector}$ is like for a few specific values of $n$, but is there a way to find a formula for a general $n$?
Here are the values for $\text{FinalVector}$ for the first few $n$ (from Mathematica.)
$n = 1$
$$\left(
\begin{array}{c}
-\frac{2 a^2}{3}+\frac{2 a}{3}+1 \\
\frac{4 a^2}{3}-\frac{4 a}{3}+1 \\
\end{array}
\right)$$
$n = 2$
$$\left(
\begin{array}{c}
-\frac{2 a^4}{9}+\frac{10 a^3}{9}-\frac{20 a^2}{9}+\frac{4 a}{3}+1 \\
\frac{4 a^4}{9}-\frac{20 a^3}{9}+\frac{40 a^2}{9}-\frac{8 a}{3}+1 \\
\end{array}
\right)$$
$n = 3$
$$\left(
\begin{array}{c}
-\frac{2 a^6}{27}+\frac{2 a^5}{3}-\frac{22 a^4}{9}+\frac{122 a^3}{27}-\frac{14 a^2}{3}+2 a+1 \\
\frac{4 a^6}{27}-\frac{4 a^5}{3}+\frac{44 a^4}{9}-\frac{244 a^3}{27}+\frac{28 a^2}{3}-4 a+1 \\
\end{array}
\right)$$
$n = 4$
$$\left(
\begin{array}{c}
-\frac{2 a^8}{81}+\frac{26 a^7}{81}-\frac{16 a^6}{9}+\frac{440 a^5}{81}-\frac{812 a^4}{81}+\frac{308 a^3}{27}-8 a^2+\frac{8 a}{3}+1 \\
\frac{4 a^8}{81}-\frac{52 a^7}{81}+\frac{32 a^6}{9}-\frac{880 a^5}{81}+\frac{1624 a^4}{81}-\frac{616 a^3}{27}+16 a^2-\frac{16 a}{3}+1 \\
\end{array}
\right)$$
| $
\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 by $u$ and note that
$$\eqalign{
Au &= u\LR{v^Tu} = \l u
\qiq \l=u^Tv\quad\big({\rm eigenvalue}\big)
}$$
Similarly, mulitply by any vector $q$ orthogonal to $v$
to find that it is also an eigenvector corresponding to an eigenvalue of $0$
$$\eqalign{
Aq &= u\LR{v^Tq} \;=\; 0
}$$
Multiplying the matrix by itself reveals an unusual property
$$\eqalign{
A^2 &= \l A,\quad A^3 &= \l^2 A,\quad\ldots\quad A^k &= \l^{k-1}A \\
}$$
which allows any analytic function of the matrix to be reduced to a
linear polynomial, i.e.
$$ f(A) = \a A + \b I $$
Combining this with the fact that the eigenvalues of $f(A)$ are given by $f(\l_k)$
generates a system of scalar equations for our two eigenvalues which can be trivially solved
for the unknown coefficients
$$\eqalign{
\a 0 + \b &= f(0) &\qiq \b = f(0) \\
\a\l + \b &= f(\l) &\qiq \a = \frac{f(\l)-f(0)}{\l} \\
}$$
Now we can write an expression for the function of a rank-one matrix,
including the limiting case when $\l\to 0\,$ (via l'Hopital's Rule)
$$\eqalign{
f(A) &= f(0)\,I &+\; \fracLR{f(\l)-f(0)}{\l}A \\
\lim_{\l\to 0}\;f(A) &= f(0)\,I &+\; f'(0)\;A \\
}$$
and multiply it by the all-ones vector
$$\eqalign{
f(A)\,\o &= f(0)\,\o &+\; \fracLR{f(\l)-f(0)}{\l}\LR{v^T\o}u \\
\lim_{\l\to 0}\;f(A)\,\o &= f(0)\,\o &+\; f'(0)\LR{v^T\o}u \\
\\
}$$
Substituting the parameters for this particular problem
$$\eqalign{
u &= \m{+\o \\ -2},
\quad &v = \frac 13\m{-a^2 \\ 2a-a^2 },
\quad A = uv^T \\
\l &= \frac{a^2-4a}3,\quad &\LR{\o+\l} = \frac{\LR{a-1}\LR{a-3}}3 \\
f(\l) &= \LR{\o+\l}^n,\quad &f'(\l) = n\LR{\o+\l}^{n-1} \\
f(0) &= \o, &f'(0) = n \\
\\
}$$
$$\eqalign{
M^n &= f(A) \\
&= f(0)\,I \;+\; \fracLR{f(\l)-f(0)}{\l}A \\
&= I \;+\; \fracLR{\LR{\o+\l}^n-\o}{\l}A
\qquad
\qquad
\qquad
\qquad
\quad
\\
\\
\lim_{\l\to 0}\,M^n
&= f(0)\,I \;+\; f'(0)\;A \\
&= I \;+\; nA \\
\\
}$$
Another interesting limit $($assuming $a\ne 0)\,$
is $\,n\to\infty$
$$\eqalign{
&0\lt a\le\o \qiq -\o\le\l\lt 0 \qiq 0\le\LR{\o+\l}\lt\o \\\\
&\lim_{n\to\infty} f(\l) = \lim_{n\to\infty}\LR{\o+\l}^n = 0 \\
&\lim_{n\to\infty} M^n = I - \fracLR{\o}{\l}A \\
}$$
| {
"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 $z=2(\cos t+i\sin t)$
$\Rightarrow|z+3|=\sqrt{13+12\cos t}$
$\Rightarrow|z^2-z+4|=2|4\cos t-1|$
Let $f(x)=\sqrt{13+12x}+2|4x-1|$ where $x\in [-1,1]$
Now here $f(-1)=f(1)=11$
$f'(x)=\frac{6}{\sqrt{13+12x}}-8<0$ for $x\in [-1,\frac{1}{4})$
and $f'(x)=\frac{6}{\sqrt{13+12x}}+8>0$ for $x\in (\frac{1}{4},1]$
So, clearly $x=\frac{1}{4}$ is a critical point as $f(x)$ is not differentiable at $x=\frac{1}{4}$.
$f(\frac{1}{4})=4$
So, $4\leq f(x)\leq 11$
| 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:
$$
f(x) = \begin{cases}
u + 2 - \tfrac23(u^2 - 13) = 11 - \tfrac13(u - 1)(2u - 1) &
(-1 \leqslant x \leqslant \tfrac14, \ 1 \leqslant u \leqslant 4), \\
u - 2 + \tfrac23(u^2 - 13) = \tfrac13(u + 1)(2u + 1) - 11 &
(\tfrac14 \leqslant x \leqslant 1, \ 4 \leqslant u \leqslant 5).
\end{cases}
$$
No calculus is needed to deduce that $f(x)$
decreases from $11$ to $4$ for $x \in [-1, \tfrac14]$ and
increases from $4$ to $11$ for $x \in [\tfrac14, 1].$
| {
"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 true.
I also tried
$$x^2+y^2+z^2≥\frac{(x+y+z)^2}{3}$$
I got
$$(xy+yz+xz)(x^2+y^2+z^2+x+y+z)≥(xy+yz+xz)\left(\frac{(x+y+z)^2}{3}+x+y+z\right)≥2(x+y+z)^2 \implies (xy+yz+xz)(x+y+z+3)≥6(x+y+z)$$
But, again I failed.
| 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.
$$\sigma_2(\sigma_1^2-2\sigma_2+\sigma_1)\geq2\sigma_1^2 \Leftrightarrow$$
$$\sigma_1^2\sigma_2-2\sigma_2^2+\sigma_1\sigma_2-2\sigma_1^2\geq0.$$
However, you can show that the following inequalities are true: i) $\sigma_1^2\geq 3\sigma_2;$ 2) $\sigma_2^2\geq 3\sigma_2$; 3) $\sigma_1\sigma_2\geq 9$.
I believe that using these inequalties one can show that your inequality is also true.
| {
"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 $\times$ numerator to get a positive constant and 20 terms each a positive coefficient times some power of $x$, $y$ and $z$.
Question: Is there a "nicer" proof?
| 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
$$a^2b^2 - 9a^2 + 8ab - 9b^2 + 9 \ge 0$$
or
$$(a^2 - 9)(b^2 - 9) + 8(ab - 9) \ge 0$$
which is true.
We are done.
| {
"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 proceed the next step. What should I do to find the minimal value of the $n$ without any just calculation (long division)? The method not finding roots also welcomed.
| 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 $\mathbb{Q}$, the corresponding roots of unity are algebraic of degree $4$ over $\mathbb{Q}$. But by the totient function formula for the degree of cyclotomic polynomials, such is the case only if the roots are $n$th roots of unity for $n=5,8,10,12$, so that the only polynomials of the prescribed form that divide some $x^n-1$ are
$$\begin{align}
f_{5\pm}(x)&=(x-1)(x^2\mp(\tfrac{\sqrt{5}-1}{2})x+1) \\
f_{8\pm}(x) &= (x-1)(x^2\mp\sqrt{2}x+1) \\
f_{10\pm}(x) &= (x-1)(x^2\mp(\tfrac{\sqrt{5} + 1}{2})x+1) \\
f_{12\pm}(x) &= (x-1)(x^2 \mp\sqrt{3}x+1)\text{.}
\end{align}$$
But the $f$ of the problem statement
$$f(x)=(x-1)(x^2-(\sqrt{2}-\tfrac{1}{2})x + 1)$$
is not on this list.
| {
"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
} |
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 Ceva theorem approach is quite straight forward:
$$
\begin{multline}
\shoveleft \dfrac{\sin(50^{\circ}-x)}{\sin40^{\circ}}\dfrac{\sin x}{\sin65^{\circ}}\dfrac{\sin20^{\circ}}{\sin5^{\circ}}=1 \\
\shoveleft \implies \sin(50^{\circ}-x)\sin x=\dfrac{\sin5^{\circ} \cdot \sin40^{\circ}\cdot \sin65^{\circ}}{\sin20^{\circ}}=2\sin5^{\circ}\cdot \cos20^{\circ} \cdot \sin65^{\circ}\\
\shoveleft \implies \cos(50^{\circ}-2x)-\cos50^{\circ}=4\sin5^{\circ} \cdot \cos20^{\circ} \cdot \cos 25^{\circ} \\
\shoveleft \implies \cos(50^{\circ}-2x)=4\sin5^{\circ} \cdot \cos20^{\circ} \cdot \cos 25^{\circ}+\sin40^{\circ}\\
\shoveleft \qquad =4\sin5^{\circ} \cdot \cos20^{\circ} \cdot \cos 25^{\circ}+2\sin20^{\circ}\cdot \cos20^{\circ}\\
\shoveleft \qquad =2\cos20^{\circ}(2\sin5^{\circ}\sin65^{\circ}+\cos70^{\circ})\\
\shoveleft \qquad =2\cos20^{\circ}(\cos60^{\circ}-\cos70^{\circ}+\cos70^{\circ})\\
\shoveleft \qquad =\cos20^{\circ}\\
\shoveleft \implies 50^{\circ}-2x=\pm 20^{\circ}\\
\shoveleft \implies x=\boxed{15^{\circ}} \text{ or } x= \boxed{35^{\circ}}
\end{multline}
$$
When $x=35^{\circ}$ the $\angle{ACB}>90^{\circ}$:
As usual, any pure geometric approach to this problem? Thanks.
| 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):
$$
\begin{aligned}
1
&=
\frac{\sin 2y}{\sin (30^\circ+x)}\cdot
\frac{\sin x}{\sin y}\cdot
\frac{\sin (60^\circ+x)}{\sin 3x}\ ,
\\[3mm]
&
\qquad\text{ with the given particular form for $x=5^\circ$, $x=20^\circ$ :}
\\[3mm]
1
&=
\frac{\sin 40^\circ}{\sin \color{red}{35^\circ}}\cdot
\frac{\sin 5^\circ}{\sin 20^\circ}\cdot
\frac{\sin 65^\circ}{\sin \color{red}{15^\circ}}\ .
\end{aligned}
$$
The three fractions correspond to proportions of the sine values, as they appear in the first picture in the posted question. The problem is to find (geometrically) two angles (maybe like $\color{red}{35^\circ}$ and $\color{red}{15^\circ}$) adding to
$50^\circ$ and making the corresponding product equal to one.
Note that exchanging the red values, we obtain again a solution!
So it is enough to show geometrically that (either) one of the two configurations of points is valid.
If we are already using this argument, then why not also permute the factors in the numerator, so that the new obtained configuration of angles comes with a better situation / symmetry. I will show instead geometrically:
$$
\begin{aligned}
1
&=
\frac{\sin 2y}{\sin (30^\circ+x)}\cdot
\frac{\sin x}{\sin 3x}\cdot
\frac{\sin (60^\circ+x)}{\sin y}\ ,
\\[3mm]
&
\qquad\text{ with the given particular form for $x=5^\circ$, $y=20^\circ$ :}
\\[3mm]
1
&=
\frac{\sin 40^\circ}{\sin \color{red}{35^\circ}}\cdot
\frac{\sin 5^\circ}{\sin \color{red}{15^\circ}}\cdot
\frac{\sin 65^\circ}{\sin 20^\circ}\ .
\end{aligned}
$$
For such changes there is also a simple geometric argument.
Exchanging numerator and denominator corresponds to passing from one intersection point of cevians to its isogonal conjugate. Else, permutations are generated by transpositions. (Below we use in fact only a transposition.) For a transposition of angles, like passing from a triangle $\Delta ABC$ and cevians for the tuple $(\color{blue}{x},x';\color{red}{y},y';z,z')$, where $\hat A=x+x'$, $\hat B=y+y'$ , $\hat C=z+z'$ to a cousin triangle with cevians for the tuple $(\color{red}{y},x';\color{blue}{x},y';z,z')$ we have the following simple observation:
Observation:
Let $\Delta ABC$ be a triangle, let $S$ be a(n interior) point such that we have the following angle separation determined by the cevians $AS$, $BS$, $CS$:
$$
\hat A=x+x'\ , \ \hat B=y+y'\ ,\ \hat C=z+z'\ .
$$
Denote such a configuration of cevians by $(\color{blue}{x},x';\color{red}{y},y';z,z')$. Consider an other triangle, determined up to similarity, with three cevians realizing the separation of angles for the tuple $(\color{red}{y},x';\color{blue}{x},y';z,z')$. Then these cevians are also concurrent.
Picture proof: Let $A'\in AC$ be such through $BA'$ is building the angle $x$ with $BD$. Then $ABDA'$ is cyclic. So $\widehat {BA'D}=\widehat {BAD}=x'$.
Use the one "other triangle" $\Delta A'BC$ realizing the tuple $(y,x';x,y';z,z')$ with cevians $A'D$, $BD$, $CD$ concurrent in the same point $D$ to conclude.
$\square$
After this introduction we are in position to start the answer.
Claims and Proofs:
Our task explicity is to show the following:
Proposition:
Let $x$ be an angle between $0^\circ$ and $15^\circ$. Denote by $y$ the angle $y=30^\circ-2x$. Consider the triangle $\Delta ABC$ with angles $90^\circ-3x$ in $A$, and $4x$ in $B$, and $90^\circ-x$ in $C$. Draw three cevians as in the picture, that separate the angles as follows:
*
*in $A$: $90^\circ-3x=2y+(30^\circ+x)$,
*in $B$: $4x=x+3x$,
*in $C$: $90^\circ-x=(60^\circ+x)+y$.
Then the three cevians are concurrent in a point $D$.
Notes: In the picture, $Q:=AD\cap BC$, $I$ is the incenter in $\Delta ABQ$,
the marked angle in $I$, exterior to $\Delta IAB$ has measure $2x+y=30^\circ$,
and the marked angles in $Q$ follow because $IQ$ is the third angle bisector through $I$, and we know the angle in $Q$ exterior to $\Delta QAB$ with measure $4x+2y=2(2x+y)=60^\circ$.
It is because of realizing the angles $3y$ in $A$ and $4x$ in $B$ why i was choosing to state and proof this variation, among the many other obtained by the one or other permutation of angles.
For the proof of the proposition, we show first the following Lemma, which isolate a part of the figure.
Lemma: Let $x,y$ be angles as above, $2x+y=30^\circ$. Consider a triangle $\Delta ABQ$ with angles $2y$ in $A$, $4x$ in $B$, and the rest of $180^\circ - 2(2x+y)=2\cdot 60^\circ$ in $Q$. Let $I$ be its incenter. Reflect $I$ w.r.t. angle bisector of $\widehat{IBQ}$ in a point $I'\in BQ$.
Construct the equilateral triangle $II'D$. (Here $B$ and $D$ are in different half-planes w.r.t. the bisector $AI$ of $\hat A$.) Then $D\in AQ$.
Proof of the Lemma:
The quadrilateral $II'QD$ is cyclic because of $\hat I+\hat Q=60^\circ + 2\cdot 60^\circ=180^\circ$. Then:
$$
\widehat{IQD} =
\widehat{II'D} =
60^\circ=
\widehat{IQA}\ .
$$
So $QD$, $QA$ build the same angle w.r.t. the reference line $IQ$.
$\square$
Corollary: Let $ABQ$ be the triangle from the above Lemma. Then the angle bisector of $\widehat{IBQ}$ intersects $AQ$ in a point $D$ with
$\widehat{IDB} = 30^\circ$.
$\square$
We are now in the position to conclude:
Proof of the Proposition:
Define $D$ rather as the intersection of the cevians from $A$ and $B$.
(We show $D$ is on the cevian from $C$ from the Proposition.)
We compare $\Delta IAQ$ and $\Delta SAQ$, so
$S$ is the reflection of $I$ w.r.t. $AQ$.
From the Corrolary, we know the angle $\widehat{IDB}=30^\circ$ in the picture:
From this, we can compute all angles around $S$. We need:
$$
\begin{aligned}
\widehat{ASC} &= 180^\circ-\hat A-\hat C=180^\circ - 3x - (90^\circ-x)=2x\text{ in }\Delta ASC\ ,\\
\widehat{ASD} &=
\widehat{AID} = 30^\circ + (30^\circ+x)=60^\circ +x\ ,\\
\widehat{DSB} &=
180^\circ - \widehat{ASC}-\widehat{DSA}
\\
&=30^\circ+x=y+3x
\\
&=\widehat{DAC}\ .
\\[3mm]
&\qquad\text{ So $ADSC$ is cyclic, giving }
\\[3mm]
\widehat{DCS} &= \widehat{DAS} =y\ .
\end{aligned}
$$
So the cevian $CD$ separates $C$ in the angles $y$ and $(60^\circ+x)$, what we wanted to show.
$\square$
| {
"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
} |
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 $1$):
$$
\begin{align*}
10&=5+3+2\\
12&=7+3+2\\
14&=7+5+2\\
15&=7+5+3\\
16&=9+5+2\\
18&=11+4+3\\
19&=11+5+3\\
20&=11+7+2\\
21&=11+7+3\\
22&=11+7+4\\
&\ldots
\end{align*}
$$
It's fairly clear why $10$ is the smallest number possible, but I couldn't find any pattern until $20$ where it seems like all the rest follow suit.
It seems after $20$, there are so many possible combinations that every following number can be written as a sum of three integers $(\geq 2)$ which are pairwise coprime. Is this a known result, and, can it be (dis)proved?
EDIT:
Thanks for the responses all, I believe I have devised a nice case proof (inspired by Yuval Peres' case answer!)
Proof
Case 1: $n$ is even
$$
\begin{align*}
n&=6k=2+3+(6k−5)\\
n&=6k+2=4+3+(6k−7)\\
n&=6k+4=2+3+(6k−5)
\end{align*}
$$
Case 2: $n$ is odd
$$
\begin{align*}
n&=12k+1=3+(6k−7)+(6k+5) \text{ with } k≥2\\
n&=12k+3=9+(6k−5)+(6k−1) \text{ with } k≥2\\
n&=12k+5=3+(6k−5)+(6k+7) \text{ with } k≥2\\
n&=12k+7=3+(6k−1)+(6k+5)\\
n&=12k+9=3+(6k−1)+(6k+7)\\
n&=12k+11=3+(6k+1)+(6k+7) \hspace{35pt}\blacksquare
\end{align*}
$$
| 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 $n \ge 30$. $\quad$ Observe that 26=2+5+19.
If $n =2 \!\!\! \mod 3$ and $n = 3 \!\!\! \mod 5$: $\;\;$ Write $\;n=9+25+(n-34)$ provided $n \ge 36$. $\quad$ Observe that 23=3+7+13.
| {
"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 $x+y+z\leq3$ which is not correct.
| 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
$$2p^2 - 4\cdot \frac{p^3 + 9}{4p} + 9 - 5p \ge 0$$
or
$$(p - 3)(p^2 - 2p + 3)/p \ge 0$$
which is true since $p\ge 3\sqrt[3]{r} = 3$ (AM-GM).
We are done.
Remark: Three degree Schur inequality is
$$a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b) \ge 0.$$
In pqr language (or substitution), it is $p^3 - 4pq + 9r \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4435064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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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 members is from upper 6.
My answer is $311/609$ while the book's answer is $197/203$. Please explain to me why I did not get the same answer as the book and how to get the correct answer.
My working:
At least one of the committee members is from upper 6 means either one of the committee members is from upper 6 or two of the committee members are from upper 6 or three of the committee members are from upper 6.
Probability that one of the committee members is from upper 6
$$= \frac{20}{30} \cdot \frac{10}{29} \cdot \frac{9}{28} = \frac{15}{203}$$
Probability that two of the committee members are from upper 6 $$= \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{10}{28}= \frac{95}{609}$$
Probability that three of the committee members is from upper 6
$$= \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{18}{28} = \frac{57}{203}$$
Probability that at least one of the committee members is from upper 6
$$=\frac{15}{203}+\frac{95}{609}+\frac{57}{203}=\frac{311}{609}$$
| 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 from lower 6. Since selected members do not have an ordering, the actual probability is $3$ times this value, since for any such selection, there are $3$ positions for the member that is from upper 6.
Similarly, the second probability you write needs to be multiplied by $3$. But the last probability, in which all members are from upper 6, does not need to be multiplied by $3$ since all of them are from the same group.
Thus the corrected calculation would look like this:
$$3 \cdot \frac{20}{30} \cdot \frac{10}{29} \cdot \frac{9}{28} + 3 \cdot \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{10}{28} + \frac{20}{30} \cdot \frac{19}{29} \cdot \frac{18}{28} = \frac{197}{203}.$$
However, none of this is necessary. By far the easiest solution is to compute the complementary probability; i.e., the probability that none of the selected members are from upper $6$. This occurs with probability
$$\frac{10}{30} \cdot \frac{9}{29} \cdot \frac{8}{28} = \frac{6}{203},$$ thus the desired probability that at least one member is from upper 6, is $$1 - \frac{6}{203} = \frac{197}{203}.$$
| {
"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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