Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to find $\int \frac{e^{-x^2}}{x^2 + 1} dx$? I have a question about improper integrals:
How can we find $\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx$?
$\textbf{Some effort:}$
$\lim_{n \rightarrow +\infty}\int_{-n}^{n} \frac{1}{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx = \lim_{n \rightarrow +\infty} \frac{2}{n}\int_{0}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}dx $ $~~~~~~~~~\textbf{(1)}$
By setting $nx^2 = u$, we have $dx = \frac{1}{2\sqrt{n}} \times \frac{1}{\sqrt{u}}$ and $x = \frac{\sqrt{u}}{\sqrt{n}}$. So by substituting these in $\textbf{(1)}$ we have (I will not put bounds and at the end will come back to the initial bounds and also I will drop the constant in integrals)
$\textbf{(1)} = \lim_{n \rightarrow +\infty} \frac{2}{n} \int \frac{1 - e^{-u}}{\frac{u}{n}(1+u)} \times \frac{1}{2 \sqrt{n}}\times \frac{1}{ \sqrt{u}} du = \lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n}} \int \frac{1 - e^{-u}}{ u \sqrt{u}(1+u)} du $
$= \lim_{n \rightarrow +\infty} \frac{1}{\sqrt{n}} \int ( \frac{1 }{ u \sqrt{u}(1+u)} - \frac{e^{-u} }{ u \sqrt{u}(1+u)}) du$ $~~~~~~~~~\textbf{(2)}$
By setting $\sqrt{u} = v$, we have $\frac{1}{2\sqrt{u}}du = dv$. So by substituting these in $\textbf{(2)}$ we have
$\textbf{(2)} = \lim_{n \rightarrow +\infty} \frac{2}{\sqrt{n}} \int (\frac{1}{v^2(1+v^2)} - \frac{e^{-v^2}}{v^2(1+v^2)} dv) $ $~~~~~~~~~\textbf{(3)}$
$\textbf{(3)}= \lim_{n \rightarrow +\infty} \frac{2}{\sqrt{n}} \int ( \frac{1}{v^2} - \frac{1}{1+v^2} - \frac{e^{-v^2}}{v^2} + \frac{e^{-v^2}}{v^2 + 1}) dv$
Now we will calculate each term separately.
First part:
For to find $\int \frac{1}{v^2} dv $, by setting $k_1=-\frac{1}{v}$, we have $dk_1 =\frac{1}{v^2} dv$ and so we have $\int \frac{1}{v^2} dv = \int dk_1= k_1= -\frac{1}{v}= -\frac{1}{\sqrt{u}}= -\frac{1}{\sqrt{nx^2}} = -\frac{1}{\sqrt{n}|x|} $
Second part:
For to find $-\int \frac{1}{1+v^2} dv$, by setting $k_2 = \arctan(v)$, we have $dk_2 = \frac{1}{1 + v^2}dv$ and so we have $-\int \frac{1}{1+v^2} dv = -int dk_2= -k_2= -\arctan(v) = -\arctan(\sqrt{u}) = -\arctan(\sqrt{nx^2}) = -\arctan(\sqrt{n}|x|) $
Third part:
For to find $-\int \frac{e^{-v^2}}{v^2} dv $, by setting $\begin{cases}
k_3=e^{-v^2}\\
-\frac{1}{v^2}=dk_4
\end{cases}$ we will have $\begin{cases}
dk_3=-2ve^{-v^2}\\
k_4= \frac{1}{v}
\end{cases}$ and our integral will transform to $-\int \frac{e^{-v^2}}{v^2} dv = \frac{e^{-v^2}}{v} - 2 \int e^{-v^2} dv = \frac{e^{-v^2}}{v} - 2(\frac{\sqrt{\pi}}{2}) = \frac{e^{-v^2}}{v} - \sqrt{\pi} = \frac{e^{-(\sqrt{u})^2}}{\sqrt{u}} - \sqrt{\pi} = \frac{e^{-u}}{\sqrt{u}} - \sqrt{\pi} =\frac{e^{-nx^2}}{\sqrt{nx^2}} - \sqrt{\pi}=\frac{e^{-nx^2}}{\sqrt{n}|x|} - \sqrt{\pi}$
Forth part:
For to find $\int \frac{e^{-v^2}}{v^2 + 1} dv$, I cannot find it!
Can someone please help me to find $\int \frac{e^{-v^2}}{v^2 + 1} dv$?
Thanks!
| $$I_n=\int_{-n}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}\,dx=2\int_{0}^{n} \frac{1 - e^{-nx^2}}{x^2(1+nx^2)}\,dx$$ Let $$nx^2=t\implies x=\frac{\sqrt{t}}{\sqrt{n}}\implies dx=\frac{dt}{2 \sqrt{n} \sqrt{t}}$$ making
$$I_n=\sqrt{n}\int_0^1\frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt$$ You do not need to compute anything else to show that $I_n$ is just proportional to $\sqrt{n}$ and just conclude.
Now, as said in answers, the last integral cannot be computed and series expansions are required. Using Taylor, we would have
$$\frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}=\frac{1}{\sqrt{t}}-\frac{3 \sqrt{t}}{2}+\frac{5 t^{3/2}}{3}-\frac{41
t^{5/2}}{24}+\frac{103 t^{7/2}}{60}-\frac{1237 t^{9/2}}{720}+O\left(t^{11/2}\right)$$
$$\int \frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt=2 \sqrt{t}-t^{3/2}+\frac{2 t^{5/2}}{3}-\frac{41 t^{7/2}}{84}+\frac{103
t^{9/2}}{270}-\frac{1237 t^{11/2}}{3960}+O\left(t^{13/2}\right)$$ Using the bounds, we should get
$$\int \frac{ \left(1-e^{-t}\right)}{ t^{3/2} (t+1)}\,dt=\frac{1634621}{1081080}\approx 1.51203$$ while numerical integration would lead to $\approx 1.38990$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
If $x,y,z\in {\mathbb R}$, Solve this system equation: If $x,y,z\in {\mathbb R}$, Solve this system equation:
$$
\left\lbrace\begin{array}{ccccccl}
x^4 & + & y^2 & + & 4 & = & 5yz
\\[1mm]
y^{4} & + & z^{2} & + & 4 & = &5zx
\\[1mm]
z^{4} & + & x^{2} & + & 4 & = & 5xy
\end{array}\right.
$$
This is an olympiad question in Turkey (not international), which that, I could not solve it.
My idea:
$xy=a \\ yz=b \\ xz=c$
$$
\left\lbrace\begin{array}{ccccccl}
a^2c^3 & + & b^3a & + & 4b^2c & = & 5b^3c
\\[1mm]
a^{3}b^2 & + & bc^3 & + & 4c^2a & = &5c^3a
\\[1mm]
b^{3}c^2 & + & a^3c & + & 4a^2b & = & 5a^3b
\end{array}\right.
$$
Yes, I know, this is a stupid idea, because it did not work at all ( last system equation is more difficult).
| Your system can be written as follows
$$
\left\lbrace\begin{array}{clc}\tag{1}
\left( {x}^{2}-2 \right) ^{2}+ \left( y-z \right) ^{2}+ \left( x-z
\right) \left( x+z \right) +3\,({x}^{2}-\,zy)&=&0
\\ \\[1mm]
\left( {y}^{2}-2 \right) ^{2}+ \left( z-x \right) ^{2}+ \left( y-x
\right) \left( y+x \right) +3(\,{y}^{2}-\,zx)&=&0
\\ \\[1mm]
\left( {z}^{2}-2 \right) ^{2}+ \left( x-y \right) ^{2}+ \left( z-y
\right) \left( z+y \right) +3(\,{z}^{2}-\,xy) &=&0
\end{array}\right.
$$
Now one solution of the $(1)$ is $x=y=z=\sqrt{2}$ which is mentioned in the first comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_0^\infty \left(\frac{x}{\sinh x}\right)^3dx$ I need to evaluate $$\int_0^\infty \left(\frac{x}{\sinh x}\right)^3dx$$
I know that I need to use the residue theorem to solve it, but I don't understand how to choose contour properly.
Thank you for any help!
| Integrate by parts
\begin{align}
I& = \int_0^\infty \left(\frac{x}{\sinh x}\right)^3dx\\
&= -\int_0^\infty x^3 \text{csch} \>x \>d(\coth x)\\
&= -3\int_0^\infty x^2 d(\text{csch} \>x)
-\int_0^\infty x^3 \text{csch} \>x \coth^2 xdx\\
&= 6\int_0^\infty x \>\text{csch} \>x \>dx
-\int_0^\infty x^3 \text{csch} \>x \>dx - I\\
&= 3\int_0^\infty x \>\text{csch} \>x \>dx
-\frac12 \int_0^\infty x^3 \text{csch} \>x \>dx\\
&= \frac92 \int_0^\infty \frac{x}{e^x-1} dx
- \frac{15}{16}\int_0^\infty \frac{x^3}{e^x-1} dx\\
&= \frac92 \zeta(2) - \frac{45}8\zeta(4)
= \frac92 \cdot \frac{\pi^2}6- \frac{45}8\cdot \frac{\pi^4}{90}
=\frac{3\pi^2}4- \frac{\pi^4}{16}\\
\end{align}
Note $\int_0^\infty \frac{x^{n-1} }{e^x-1} dx=(n-1)! \zeta(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ Consider the solution of the ordinary differential equation $y'(t)=-y^3+y^2+2y$ subject to $y(0)=y_0 \in (0,2). $ then $\lim \limits_{t \to \infty}y(t)$ belongs to
*
*{-1,0}
*{-1,2}
*{0,2}
*{0, $+\infty $}
My Attempt: $y'(t)=-y^3+y^2+2y \\ \Rightarrow y'(t)=-y(y^2-y-2) \\ \Rightarrow y'(t)=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{dt}=-y(y-2)(y+1) \\ \Rightarrow \frac{dy}{y(y-2)(y+1)}=-dt \\ \Rightarrow \int\frac{dy}{y(y-2)(y+1)}=\int-dt \\by \;\;partial \;fraction \; we \; get \\ \int [\frac{-1}{2y}+\frac{1}{6(y-2)}+\frac{1}{3(y+1)}]dy =-\int dt \\ \Rightarrow -\frac{1}{2} log|y| +\frac{1}{6} log|y-2| + \frac{1}{3} log|y+1|=-t + logc_1 \\ \Rightarrow \frac{-3log|y|+log|y-2|+2log|y+1|}{6}=-t + logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}=-6t+6logc_1 \\ \Rightarrow log\frac{(y-2)(y+1)^2}{y^3}-logc_1^6=-6t \\ \Rightarrow log\frac{(y-2)(y+1)^2}{cy^3}=-6t \;\; where \;c_1^6=c \\ \Rightarrow \frac{(y-2)(y+1)^2}{cy^3}=e^{-6t}..........(1) \\ when \;\;t=0 \;\; \frac{(y_0-2)(y_0+1)^2}{y_0^3}=c \\ (1) \Rightarrow \frac{(y-2)(y+1)^2}{y^3}=\frac{(y_0-2)(y_0+1)^2}{y_0^3} e^{-6t}$
Now I am not getting how to apply the limit for this and how to find $\lim \limits_{t \to \infty} y(t)$
| you got $\frac{(y-2)(y+1)^2}{y^3}=\frac{(y_0-2)(y_0+1)^2}{y_0^3} e^{-6t}$ take the limit and you get $$\lim\limits_{t\to\infty}\frac{(y-2)(y+1)^2}{y^3}=0\implies \lim\limits_{t\to\infty}(y-2)(y+1)^2=0 \implies \lim\limits_{t\to\infty}y(t)=\begin{cases}2\\-1\end{cases}$$ hence option (2) is true
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving: $(x-a)(x-b)=x-c$, $(x-c)(x-b)=x-a$ and $(x-c)(x-a)=x-b$ Given three distinct real numbers $a$, $b$, and $c$, show that at least two of the three following equations
$$(x-a)(x-b)=x-c$$
$$(x-c)(x-b)=x-a$$
$$(x-c)(x-a)=x-b$$
have real solutions.
My attempt: I tried to multiply all the equations side by side to obtain
$$(x-a)^2(x-b)^2(x-c)^2=(x-a)(x-b)(x-c)$$
then what next?
| Let the three quadratics be
$$\begin{align}
p_1(x)&=(x-a)(x-b)-(x-c)\\
p_2(x)&=(x-b)(x-c)-(x-a)\\
p_3(x)&=(x-c)(x-a)-(x-b)\\
\end{align}$$
By symmetry, we may assume $a\le b\le c$. But that implies
$$p_2(b)=-(b-a)\le0\quad\text{and}\quad p_3(c)=-(c-b)\le0$$
Thus $p_2$ and $p_3$ have at least one real root (since their graphs are upward-pointing parabolas), and therefore have two real roots (since they are quadratics).
The key here is to note that any permutation of $a$, $b$, and $c$ simply permutes the three quadratics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$
Suppose that $a,b,c,x,y,z$ are all positive real numbers. Show that
$$\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}\geq \frac{3(a+b+c)}{ax+by+cz}$$
Below are what I've done, which may be misleading.
*
*I've tried to analyze when the equality holds：
1.1 Under the condition that $a=b=c$, it reduces to
$$\frac{2}{y+z}+\frac{2}{z+x}+\frac{2}{x+y}\geq \frac{9}{x+y+z}$$
1.2 Under the condition that $x=y=z$, it reduces to
$$\frac{b+c}{2a}+\frac{c+a}{2b}+\frac{a+b}{2c}\geq 3$$
Both are easy to verify. However, $ax+by+cz$ is not easy to deal with. Is there any famous inequality that I can use here?
1.3 Under the condition that $(x,y,z)$ and $(a,b,c)$ are in proportion, i.e. $x=at, y=bt, z=ct$ for some $t>0$, the inequality reduces to
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{3(a+b+c)}{a^2+b^2+c^2}$$
which can be proved by using Newton's inequality:
$$(ab+bc+ca)^2 \geq 3abc(a+b+c)$$
*I’ve also tried to construct a function
$$f(x,y,z)=\frac{b+c}{a(y+z)}+\frac{c+a}{b(z+x)}+\frac{a+b}{c(x+y)}- \frac{3(a+b+c)}{ax+by+cz}$$
and analyze its global minimum. But the first-order condition is complicated
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{c+a}{ab(z+x)^2}+\frac{a+b}{ca(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ab(y+z)^2}+\frac{a+b}{bc(x+y)^2}$$
$$\frac{3(a+b+c)}{(ax+by+cz)^2}=\frac{b+c}{ca(y+z)^2}+\frac{c+a}{bc(z+x)^2}$$
Maybe some convexity can be used here?
*Substitution has been considered. Let
$$u=\frac{a}{a+b+c},v=\frac{b}{a+b+c}, w=\frac{c}{a+b+c}$$
Then $u,v,w>0$ and $u+v+w=1$, which are just weights we assign to $x,y,z$. The inequality becomes
$$\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}\geq \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}$$
Define another function in variables $u,v,w$
$$g(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
and consider the constrained optimization problem:
$$\min g(u,v,w)\\ \text{s.t. } u>0\\ v>0 \\w>0\\ u+v+w=1$$
The corresponding Lagrangian function can be
$$L_{\lambda}(u,v,w)=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}+\lambda(u+v+w-1)$$
And the first-order conditions give
$$\frac{1}{u^2(y+z)}-\frac{3x}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{u(y+z)}-\frac{3ux}{(ux+vy+wz)^2}=\lambda u\\ \frac{1}{v^2(z+x)}-\frac{3y}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{v(z+x)}-\frac{3vy}{(ux+vy+wz)^2}=\lambda v\\ \frac{1}{w^2(x+y)}-\frac{3z}{(ux+vy+wz)^2}=\lambda\Rightarrow \frac{1}{w(x+y)}-\frac{3wz}{(ux+vy+wz)^2}=\lambda w\\ u+v+w=1$$
Summing up yields
$$\lambda=\frac{1}{u(y+z)}+\frac{1}{v(z+x)}+\frac{1}{w(x+y)}-\frac{3}{ux+vy+wz}$$
None of the methods I tried seems to work, and now I even doubt the truth of the inequality.
| This is not an answer, but putting code in a comment doesn't work well either.
Regarding your doubts about the truth of the inequality, I am still inclined to think it holds true. You can try to look for counterexamples with this code. In case python thinks the inequality doesn't hold, it outputs $a,b,c,x,y,z$ and the difference between the two sides. So far I have only found differences of $10^{-15}$ and smaller with it, which is just python lacking sufficient precision.
sta = 3.14
sto = 4.14
ste = .1
def frange(start, stop, step):
i = start
while i < stop:
yield i
i += step
for a in frange(sta,sto,ste):
for b in frange(sta,sto,ste):
for c in frange(sta,sto,ste):
for x in frange(sta,sto,ste):
for y in frange(sta,sto,ste):
for z in frange(sta,sto,ste):
if (1.*(b+c)/(a*(y+z))+1.*(c+a)/(b*(z+x))+1.*(a+b)/(c*(x+y))<3.*(a+b+c)/(a*x+b*y+c*z)):
print a,b,c,x,y,z,1.*(b+c)/(a*(y+z))+1.*(c+a)/(b*(z+x))+1.*(a+b)/(c*(x+y))-3.*(a+b+c)/(a*x+b*y+c*z)
Also, since the inequality can be reshaped to
$$\frac{a}{b(x+z)}+\frac{b}{a(y+z)}+\frac{a}{c(x+y)}+\frac{c}{a(y+z)}+\frac{b}{c(x+y)}+\frac{c}{b(x+z)}\ge\frac{3a}{ax+by+cz}+\frac{3b}{ax+by+cz}+\frac{3c}{ax+by+cz},$$
Maybe we can show that $\frac{1}{b(x+z)}+\frac{1}{c(x+y)}\ge\frac{3}{ax+by+cz}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 2
} |
Quadratic equation with different indices My maths teacher gave me a worksheet to work through as I was getting slightly bored in lessons. However, there was one question which I cannot do. The worksheet gives the answer, but you are supposed to show how you did it. Here is the question:
$729 + 3^{2x+1} = 4\times3^{x+2}$
The sheet said that the answer was $x=2$ or $x=3$
Here is my working - I 'solved' by completing the square, as you will see:
\begin{align}
3^{2x+1}-4\times3^{x+2}+729& = 0 \\
3(3^{2x})-4\times(3^2)(3^x)+729& = 0\\
3(3^x)^2-36(3^x)+729& = 0\\
y &= 3^x\\
3y^2-36y+729 & = 0\\
3(y^2-12y+243) &= 0\\
(y-6)^2-36+243 &= 0\\
(y-6)^2+207 &= 0\\
(y-6)^2&=-207\\
y-6 &= \sqrt{207}i\\
y &= 6+\sqrt{207}i\\
3^x &= 6+\sqrt{207}i\\
x &= \log_3(6+\sqrt{207}i)\\
\end{align}
Clearly, this is not the answer as stated on the sheet. Have I answered this question wrong, or is the answer on the sheet wrong? Thank you in advance.
| HINT
Set $3^x=y$
$$729 + 3y^2= 4\cdot 9 y$$
$$3y^2-36y+729=0$$
$$y^2-12y+243=0$$
$\Delta <0 \implies$ no real solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions of $2x+3y=n$ What is the smallest $m$ such that for all $n\geq m$, the equation $2x+3y=n$ has solutions with $x,y \in \mathbb{Z}$ and $x,y\geq2$?
My approach. We can write the solutions in terms of the parameter $t$ as:
$$x(t)=-n-3t$$ and
$$y(t) = n+2t$$
Setting both of those greater than or equal to $2$, we have $-n-3t\geq2$ and $n+2t\geq2$. Solving for $t$, we find $\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}$. Solving the resulting inequality, we get
$$-2n-4\geq6-3n$$
So $n\geq10$. This seems to be right. For $n=10$, take $(2,2)$. But, suppose $n=11$. Then, $2x+3y=11$ clearly has no solution where both $x,y\geq2$. Is my solution incorrect? Can someone point me in the right direction?
EDIT: Thanks for the great answers!
| You are almost there.
Solving for $t$, we find $\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}$. Solving the resulting inequality, we get
$$-2n-4\geq6-3n$$
So $n\geq10$.
You want to find $n$ such that there is at least one integer $t$ satisfying
$$\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}\tag1$$
For this, solving $\frac{-n-2}{3}\ge\frac{2-n}{2},$ i.e. $-2n-4\ge 6-3n$ is not enough.
If $\frac{-n-2}{3}-\frac{2-n}{2}\ge 1$, i.e. $n\ge 16$, then $(1)$ has at least one integer $t$.
For $n=11$, there is no $(x,y)$.
$2\cdot 3+3\cdot 2=12$.
$2\cdot 2+3\cdot 3=13$.
$2\cdot 4+3\cdot 2=14$.
$2\cdot 3+3\cdot 3=15$.
So, the answer is $12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to prove $\{n!e\} = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$ I have the definition $a_n = \frac{1}{n +1} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+1)(n+2)(n+3)} + \dots$
I need to show that:
$a)$ $0 < a_n < \frac{1}{n}$
$b)$ $a_n = n!e - \lfloor{n!e}\rfloor$
So I know that
$\frac{1}{n} = \frac{1}{n+1}+\frac{1}{(n+1)^2}+\frac{1}{(n+1)^3}+\dots$
$e = \sum_{n=0}^{+\infty}\frac{1}{n!}$
It's clear that $\{n!e\}=a_n<\frac{1}{n}$, but how to show it?
| $$n!e=n!\sum_{k=0}^\infty\frac1{k!}=\sum_{k=0}^\infty\frac{n!}{k!}$$
Since $k!$ divides $n!$ for $k\le n$, the sum
$$\sum_{k=0}^n\frac{n!}{k!}$$
is an integer, say $N$. Then
$$n!e=N+\sum_{k=n+1}^\infty\frac{n!}{k!}=\sum_{k=1}^\infty\frac1{\prod_{j=1}^k(n+k)}<\sum_{k=1}^\infty\frac1{(n+1)^k}=\frac1n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Is this a valid method of solving $3^x + 2^x = 35$? I want to know if this is a valid method of solving this equation:
$3^x + 2^x = 35$
$3^x+2^x = (7)(5)$
$3^x+2^x = (3+2^2)(2+3)$
$3^x+2^x = 3^2 + 2^3 + 18$
$3^x+2^x = 9 + 18 + 2^3 $
$3^x+2^x = 27 + 2^3 $
$3^x+2^x = 3^3 + 2^3 $
And now comes my problem. Is it correct to say that this last equation implies $x=3$? I don't think I can just compare terms just like it was a polynomial...
If this is not correct, is there a way of solving this equation algebraically?
| The claim is equivalent to
$$
f(x)=f(3)\implies x=3
$$
where $f(x)=3^x+2^x$. This is true because $f$ is injective (since it is strictly increasing for example).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How and where can I calculate constant $k\approx1,895$? We know, that
$$e=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{1}\left(1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$
If $x_{m}=m(x_{m-1}-1)$ and $x_{0}=e$, then
$$x_{1}=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{2}\left(1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$
$$x_{2}=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{3}\left(1+\frac{1}{4}\left(1+\frac{1}{5}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$
$$x_{m}=\lim\limits_{n\to\infty}^{}\left[1+\frac{1}{m+1}\left(1+\frac{1}{m+2}\left(1+\frac{1}{m+3}\left(1+\cdots\frac{1}{n}\right)\right)\cdots\right)\right]$$
$$\sum\limits_{m=0}^{n}\left(x_{m}-1\right)\approx\ln(n)+k$$
How and where can I calculate this constant? Is there relationship with this constant and $\gamma$?
If I made some mistakes, sorry for my English.
| For general $m \in \mathbb{N}$, it is easy to see
$$x_m = m!\left(e - \sum_{k=0}^{m-1}\frac{1}{k!}\right)
\quad\implies\quad x_m - 1
= m!\left(e - \sum_{k=0}^{m} \frac{1}{k!}\right)
= \sum_{k=m+1}^\infty \frac{m!}{k!}$$
This leads to
$$\begin{align}x_m - 1
= \sum_{k=0}^\infty \frac{m!}{(m+k+1)!}
&= \frac{1}{m+1} + \sum_{k=1}^\infty \frac{1}{k!}\frac{\Gamma(m+1)\Gamma(k+1)}{\Gamma(m+k+2)}\\
&= \frac{1}{m+1} + \sum_{k=1}^\infty\int_0^1 (1-t)^m \frac{t^k}{k!} dt\\
&= \frac{1}{m+1} + \int_0^1 (1-t)^m (e^t-1)dt
\end{align}
$$
Summing $m$ from $0$ to $n$, we get
$$\begin{align}
\sum_{m=0}^n (x_m - 1)
&= H_{n+1} + \sum_{m=0}^n\int_0^1 (1-t)^m (e^t-1)dt\\
&= H_{n+1} + \int_0^1 \frac{1-(1-t)^{n+1}}{1-(1-t)} (e^t-1) dt\\
&= H_{n+1} + \int_0^1 (1 - (1-t)^{n+1}) \frac{e^t-1}{t}dt
\end{align}
$$
where $H_n = \sum_{m=1}^n \frac{1}{m}$ is the $n^{th}$ harmonic number.
Notice for $t \in (0,1)$, MVT tell us for some $\xi \in (0,t)$,
$\frac{e^t - 1}{t} = e^{\xi t} \in (1, e^t) \subset (1,e)$. This allows us
to bound the $n$ dependent piece in above integral as
$$\frac{1}{n+2} \le \int_0^1 (1-t)^{n+1} \frac{e^t - 1}{t} dt \le \frac{e}{n+2}$$
We also know for large $n$,
$$H_n = \log n + \gamma + O\left(\frac1n\right)$$
Combine these, we can conclude
$$\sum_{n=0}^n (x_m - 1) = \log n + k + O\left(\frac1n\right)
\quad\text{ where }\quad
k = \gamma + \int_0^1 \frac{e^t-1}{t} dt$$
Using an CAS, we find the integral above has a closed form in terms of
exponential integral
${\rm Ei}(x)$. The end result is
$$
k = \gamma + \left( {\rm Ei}(1) - \gamma\right) = {\rm Ei}(1)
\approx 1.89511781635593675546652...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
using a base 3 decimal to express as a base 10 fraction using geometric series
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.
So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:
A quick way to get the answer by using the geometric series is: $(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots = \frac{7}{8}.$
However, I'm having a hard time understanding how to actually use the geometric series (the above answer is not clear to me).
| \begin{align}
(0.212121 \ldots)_3
&= \dfrac{21_3}{100_3}
+ \dfrac{21_3}{(100_3)^2}
+ \dfrac{21_3}{(100_3)^3} + \cdots\\
&= \frac 79 + \frac{7}{9^2} + \frac{7}{9^3} + \cdots \\
&= \frac 79\left( \dfrac{1}{1-\frac 19} \right) \\
&= \frac 79 \cdot \frac 98 \\
&= \frac 78.
\end{align}
You could also use the old shift and subtract trick...
\begin{array}{rcl}
100_3x &= &21.212121 \ldots_3 \\
x &= &\phantom{0}0.212121 \ldots_3 \\
\hline
22_3x &= &21_3 \\
8x &= &7 \\
x &= \dfrac 78
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2576843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Does $\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$ converge absolutely? Let $\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$, $x\in\mathbb{R}$. Does this series converge absolutely?
$$\sum\limits_{k=2}^\infty\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}$$ converges absolutely if $$\sum\limits_{k=2}^\infty\left|\frac{(-1)^{2^k}x^{2k}}{(2k+3)!}\right|=\sum\limits_{k=2}^\infty\frac{|x|^{2k}}{(2k+3)!}$$ converges.
I guess that it depends on $x$ if this converges or not. For $x\in [-1,1]$ by the ratio test.
However, I'm not sure about the other $x$. How to proceed further?
| Yes since $$\sinh x = \sum\limits_{k=
0}^\infty\frac{x^{2k+1}}{(2k+1)!}$$
then $$ \sum\limits_{k=
2}^\infty\frac{x^{2k}}{(2k+3)!}\le \sum\limits_{k=
2}^\infty\frac{x^{2k}}{(2k+1)!}= -1-\frac{x^2}{6}+ \sum\limits_{k=
0}^\infty\frac{x^{2k}}{(2k+1)!}\\=-1- \frac{x^2}{6}+ \frac{1}{|x|}\sum\limits_{k=
0}^\infty\frac{|x|^{2k+1}}{(2k+1)!}= =-1- \frac{x^2}{6}+ \frac{|\sinh x|}{|x|} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int \frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}dx$ Evaluate
$$\int \frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}\text dx$$
I would have given my attempt to this question but honestly, I think my attempts to solve this did nothing but only complicated it further.
Any hints or suggestions are welcome.Please verify that your method gives answer:$$\frac 1{a^2+b^2}(x+\tan^{-1} ({\frac {a^2\tan x}{b^2}}))+C$$
| Write $$\begin{align}\frac{a^2\sin^2 x+b^2\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}&=\frac1{a^2+b^2}\cdot \frac{(a^4+a^2b^2)\sin^2 x+(b^4+a^2b^2)\cos^2 x}{a^4\sin^2 x+b^4\cos^2 x}\\&=\frac1{a^2+b^2}\cdot \frac{a^4\sin^2 x+b^4\cos^2 x+a^2b^2}{a^4\sin^2 x+b^4\cos^2 x}\\&=\frac1{a^2+b^2}\cdot \left(1+\frac{a^2b^2}{a^4\sin^2 x+b^4\cos^2 x}\right)\\&=\frac1{a^2+b^2}\cdot\left(1+\frac{\frac{a^2}{\cos^2x}}{\frac{a^4}{b^2}\tan^2x+b^2}\right)\\&=\frac1{a^2+b^2}\cdot\left(1+\frac{\frac{a^2}{\cos^2x}}{\frac{a^4}{b^2}\tan^2x+b^2}\right)\end{align}$$ Now $$\frac{a^2\sec^2x}{b^2\left(1+\frac{a^4}{b^4}\tan^2x\right)}={\tan^{-1}}'\left(\frac{a^2\tan x}{b^2}\right)$$ and integrating $1$ give $x$. Combining the two with a constant yields the required answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Show that $f(x,y,z)=6x^2+4y^2+2z^2+4xz-4yz \geq 0$ for all $x, y, z$ except for $x=y=z=0$ When I try to factor the quadratic form, I end up with
$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$
which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should these kinds of problems be tackled?
| Your quadratic form is positive definite. I do not know what the eigenvalues of the matrix are. This method is the same as "completing the square." The diagonal entries of the diagonal matrix $D$ are all positive.
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
- \frac{ 1 }{ 3 } & \frac{ 1 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
6 & 0 & 2 \\
0 & 4 & - 2 \\
2 & - 2 & 2 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & - \frac{ 1 }{ 3 } \\
0 & 1 & \frac{ 1 }{ 2 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & \frac{ 1 }{ 3 } \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
\frac{ 1 }{ 3 } & - \frac{ 1 }{ 2 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
6 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & \frac{ 1 }{ 3 } \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & \frac{ 1 }{ 3 } \\
0 & 1 & - \frac{ 1 }{ 2 } \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
6 & 0 & 2 \\
0 & 4 & - 2 \\
2 & - 2 & 2 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
$a+b \sqrt{2}=(1+ \sqrt{2})^n$ when $a^2-2b^2=+1 or -1$ I want to prove this statement :
$a+b \sqrt{2}=(1+ \sqrt{2})^n$ when $a^2-2b^2=+1 or -1$
Where a,b,and n are positive integers .
Usually people consider it true saying $1+ \sqrt{2}$ is fundamental unit of $\mathbb{Z} [ \sqrt{2}]$ , but how can I proof this more rigorously ??
| Define $a_n$ and $b_n$ by
\begin{eqnarray*}
a_n+b_n \sqrt{2} =(1+\sqrt{2})^n.
\end{eqnarray*}
Their recurrence relation is easily derived by
\begin{eqnarray*}
a_{n+1}+b_{n+1} \sqrt{2} =(a_n+b_n \sqrt{2}) (1+\sqrt{2}) \\
a_{n+1}=a_n +2 b_n \\
b_{n+1} =a_n+b_n
\end{eqnarray*}
$(a_n,b_n)$ can be shown to satisfy Pell's equation inductively, as follows
\begin{eqnarray*}
a_{n+1}^2-2b_{n+1}^2 =a_n^2+4a_n b_n + 4b_n ^2 -2(a_n^2+2 a_n b_n +b_n^2) =-(a_n^2-2b_n^2) =(-1)^{n+1}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2579737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Multiplying out $(x^3 + 3x^2 + 3x + 1) \times (1 + x + x^2)$ I know this is a basic math question but still I always get stuck on this.
$(x^3 + 3x^2 + 3x + 1) \times (1 + x + x^2)$
I know this is the solution:
$1 + 4x + 7x^2 + 7x^3 + 4x^4 + x^5$
But I want to know how they got there.
This has to be solved with distributing the second term over the first one (I guess).
So I get these terms:
$x^3 + 3x^2 + 3x + 1$
$x^4 + 3x^3 + 3x^2 + 1x$
$x^5 + 3x^4 + 3x^3 + 1x^2$
And than I'd have to sum these up.
I already highly doubt this is correct, but even if it was correct I wouldn't know how to sum these up, because it's not possible to sum numbers with a different exponent.
Could someone explain step-by-step on how to solve this $(x^3 + 3x^2 + 3x + 1) \times (1 + x + x^2)$
| Yes, you are right, so far. But, why can’t you just sum up terms with same power of $x$, that is, $$(x^3+3x^2+3x+1)(1+x+x^2) $$ $$= (x^\color{red}{5})+(3x^\color{green}{4} + x^\color{green}{4}) + (x^\color{blue}{3}+3x^\color{blue}{3}+3x^\color{blue}{3})+(3x^\color{orange}{2}+3x^\color{orange}{2}+x^\color{orange}{2})+(3x^\color{cyan}{1}+x^\color{cyan}{1})+1 $$ $$= x^\color{red}{5}+4x^\color{green}{4}+7x^\color{blue}{3}+7x^\color{orange}{2}+4x^\color{cyan}{1}+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why does multiplying by $\textbf{A}^T$ make a previously unsolvable linear system solvable Consider for instance the linear system:
$$\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$
This is over determined and thus has no solution. Yet, by simply multiplying both sides by $\textbf{A}^T$:
$$\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 2 \\
3 & 4 \\
5 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
x \\
y \\
\end{array}
\right)=\left(
\begin{array}{ccc}
1 & 3 & 5 \\
2 & 4 & 6 \\
\end{array}
\right).\left(
\begin{array}{c}
1 \\
2 \\
4 \\
\end{array}
\right)$$
We find that the system now has a unique solution, which is the (x,y) that minimizes the squared error.
Now I understand the derivation of why multiplying by the transpose helps to find the pseudoinverse which then helps to perform OLS regression, but my question is perhaps a bit more fundamental.
How can multiplying both sides of an equation by a matrix change a system which previously had no solutions into one that has a unique solution? This seems to against what I assumed that the solutions to $\textbf{A}x = \textbf{B}$ were the same as the solutions to $\textbf{P}\textbf{A}x = \textbf{P}\textbf{B}$.
|
I assumed that the solutions to $\textbf{A}x = \textbf{B}$ were the same as the solutions to $\textbf{P}\textbf{A}x = \textbf{P}\textbf{B}$.
The direct implication $\textbf{A}x = \textbf{B} \implies \textbf{P}\textbf{A}x = \textbf{P}\textbf{B}$ holds for any $\textbf{P}$. However, the reverse implication $\textbf{P}\textbf{A}x = \textbf{P}\textbf{B} \implies \textbf{A}x = \textbf{B}$ only holds if $\textbf{P}$ has a left inverse $\textbf{P}^{-1}$ (because you can then multiply with $\textbf{P}^{-1}$ on the left to derive $\textbf{A}x = \textbf{B}$). But in this case, $\textbf{P}$ is a rectangular matrix with more columns than rows which has no left inverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Find: $ \lim_{x\to \infty}\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
Find: $\displaystyle \lim_{x\to \infty} f(x)=\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}$
The answer provided in the book is 0 (also checked in Wolfram Alpha), but I can't find a good argument (without L'Hopital), to prove that. I end up in a $\infty \times 0$ situation which makes me uncomfortable.
Attempt: $$f(x)=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x}(\sqrt{1+1/x}-1)}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{1+1/x}-1}{\sqrt{x+1}}$$
$$f(x)=x(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{x}\sqrt{1+1/x}}=\sqrt{x}(1-1/x)\frac{\sqrt{1+1/x}-1}{\sqrt{1+1/x}}.$$
Therefore the original limit is equivalent to
$$\lim_{x\to \infty} \sqrt{x}(1-1/x)\left(1-\frac{1}{\sqrt{1+1/x}}\right)$$
that appears to me a situation like $\infty\times 0$. How can I proceed to conclude that this limit is indeed $0$.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| $$\frac{x-1}{\sqrt{x}}-\frac{x-1}{\sqrt{x+1}}=\frac{(x-1)\sqrt{x+1}-(x-1)\sqrt{x}}{\sqrt{x}\sqrt{x+1}}=(x-1)\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x}\sqrt{x+1}}\frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}=\frac{x-1}{\sqrt{x}\sqrt{x+1}(\sqrt{x+1}+\sqrt{x})}\to0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 3
} |
Generalized Formula for the Probability of the Union of $n$ events occurring? Consider
$$
P(A \cup B) = P(A) + P(B) - P(A \cap B)
$$
What is the generalization of this formula for $n$ events occuring? That is
$$
P(\cup A_i) = \sum P(A_i) + \ldots ?
$$
| We have
\begin{align}
P(A_1\cup A_2\cup A_3)= & \;\;\;\;P(A_1)+P(A_2) +P(A_3)
\\
&
-\underbrace{P(A_1\cap A_{2})}_{1<2}
-\underbrace{P(A_1\cap A_{3})}_{1<3}
-\underbrace{P(A_2\cap A_{3})}_{2<3}
\\
&
+\underbrace{P(A_1\cap A_{2}\cap A_{3})}_{1<2<3}
\end{align}
and
\begin{align}
P(A_1\cup A_2\cup A_3\cup A_4)= & \;\;\;\;P(A_1)+P(A_2) +P(A_3)+P(A_4)
\\
&
-\underbrace{P(A_{1}\cap A_{2})}_{1<2}
-\underbrace{P(A_{1}\cap A_{3})}_{1<3}
-\underbrace{P(A_{1}\cap A_{4})}_{1<4}
\\
&
-\underbrace{P(A_{2}\cap A_{3})}_{2<3}
-\underbrace{P(A_{2}\cap A_{4})}_{2<4}
\\
&
-\underbrace{P(A_{3}\cap A_{4})}_{3<4}
\\
&
+\underbrace{P(A_{1}\cap A_{2}\cap A_{3})}_{1<2<3}
+\underbrace{P(A_{1}\cap A_{2}\cap A_{4})}_{1<2<4}
+\underbrace{P(A_{1}\cap A_{3}\cap A_{4})}_{1<3<4}
+\underbrace{P(A_{1}\cap A_{2}\cap A_{4})}_{2<3<4}
\\
&
-\underbrace{P(A_1\cap A_{2}\cap A_{3}\cap A_4)}_{1<2<3<4}.
\end{align}
In geral,
\begin{align}
P\Big( \bigcup_{i=1}^{n}A_i\Big)
=&-(-1)^1\sum_{i{_1}=1}^{n}P(A_{i_1})
\\
&-(-1)^2\sum_{1\leq i{_1}<i_{_2}\leq n}^{n}P(A_{i_1}\cap A_{i_2})
\\
&-(-1)^3\sum_{1\leq i{_1}<i_{_2}<i_{_3}\leq n}^{n}P(A_{i_1}\cap A_{i_2}\cap A_{i_3})
\\
&-(-1)^4\sum_{1\leq i{_1}<i_{_2}<i_{_3}<i_{_4}\leq n}^{n}P(A_{i_1}\cap A_{i_2}\cap A_{i_3}\cap A_{i_4})
\\
&\quad\quad\quad\vdots
\\
&\ldots-(-1)^n \sum_{1\leq i_{1}<i_{2}<i_{3}<\ldots <i_{n}\leq n}^{n}P(A_{i_1}\cap A_{i_2}\cap A_{i_3}\cap \ldots \cap A_{i_n})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$ I'm trying to solve this problem.
Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$
I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.
I factored out the $2\cos x$ and rewrote the other factor as a linear combination of cosine.
It reduces down to $$g(x)=10\cos x\cos\left(x-\tan^{-1}\frac{3}{4}\right)$$
But then I'm stuck here. Please help me. Perhaps there's a different way to approach this?
| Hint:
Let $y=A\sin^2x+B\cos^2x+C\sin x\cos x+D$
Method$\#1:$ Divide both sides by $\cos^2x$ to form a Quadratic Equation in $\tan x$
As $\tan x$ is real, the discriminant must be $\ge0$
Method$\#2:$ Divide both sides by $\sin^2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sum^m_{k=0}\binom{m}{k}(-1)^k(\frac{1}{n+k+1}) = \frac{m!n!}{(m+n+1)!}$ I am looking for a more direct proof of the following identity: $$\sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1} = \frac{m!n!}{(m+n+1)!}$$
My original proof comes from evaluating $\int^1_0{x^n(1-x)^m}{dx}$, $n, m \in \mathbb{N}_0$ in two ways:
1) Let $u = x^n$, $u' = nx^{n-1}$, $v' = (1-x)^m$, $v = -\frac{(1-x)^{m+1}}{m+1}$:
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{x^n(1-x)^{m+1}}{m+1} + \frac{n}{m+1}\int^1_0{x^{n-1}(1-x)^{m+1}}{dx}$$
$x=1\Rightarrow \frac{x^n(1-x)^{m+1}}{m+1} = 0$ and $x=0\Rightarrow \frac{x^n(1-x)^{m+1}}{m+1} = 0$, therefore
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{n}{m+1}\int^1_0{x^{n-1}(1-x)^{m+1}}{dx}$$
Repeating the process $n$ times gives:
$$\int^1_0{x^n(1-x)^m}{dx} = \frac{m!n!}{(m+n)!}\int^1_0{x^0(1-x)^{m+n}}{dx} = $$$$=\frac{m!n!}{(m+n)!}\cdot\frac{(1-1)^{m+n+1}}{m+n+1} - \frac{m!n!}{(m+n)!}\cdot\left(-\frac{(1-0)^{m+n+1}}{m+n+1}\right) = \frac{m!n!}{(m+n+1)!}$$
2)$$\int{x^n(1-x)^m}{dx} = \int{x^n\cdot\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^k\right)}{dx} = \int{\left(\sum^m_{k=0}\binom{m}{k}(-1)^kx^kx^n\right)}{dx} = $$
$$ = \sum^m_{k=0}\left(\binom{m}{k}(-1)^k\cdot\int{x^{n+k}}{dx}\right) = \sum^m_{k=0}\binom{m}{k}(-1)^k\frac{x^{n+k+1}}{n+k+1} =: F(x).$$
$$F(1) - F(0) = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k1^{n+k}}{n+k+1} - \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k0^{n+k}}{n+k+1} = \sum^m_{k=0}\frac{\binom{m}{k}(-1)^k}{n+k+1}$$
| Here is a more direct approach along the lines you were perhaps thinking.
Notice that
$$\int_0^1 x^{n + k} \, dx = \frac{1}{n + k + 1}.$$
With this result your sum can be rewritten as
$$S = \sum_{k = 0}^m \binom{m}{k} (-1)^k \int_0^1 x^{n + k} \, dx = \int_0^1 x^n \left [\sum_{k = 0}^m \binom{m}{k} (-1)^k x^k \right ] \, dx.$$
The sum appearing within the square brackets is nothing more than the binomial sum
$$(1 - x)^m = \sum_{k = 0}^m \binom{m}{k} (-1)^k x^k.$$
Thus
\begin{align*}
S &= \int_0^1 x^n (1 - x)^m \, dx\\
&= \int_0^1 x^{(n + 1) - 1} (1 - x)^{(m + 1) - 1} \, dx\\
&= \text{B}(n + 1, m + 1)\\
&= \frac{\Gamma (n + 1) \Gamma (m + 1)}{\Gamma (n + m + 2)}\\
&= \frac{n! m!}{(n + m + 1)!},
\end{align*}
as required. Here, as noted by Jack, $\text{B}(x,y)$ corresponds to Euler's Beta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Continuity And Derivative Of A Function
Let $
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$
a. Find all the points where $f(x)$ is continuous
b. Find $f'(x)$ for all the points $f(x)$ is continuous
c. Is $f'(x)$ continuous where it is defined? if not where and which discontinuity type is it?
So for a we look at $\sin^2x \sin(\frac{1}{x})$ at $x\neq 0$ this is a product of two continuous functions and therefore continuous for $x=0$ we look at $\lim_{x\to 0}\sin^2x \sin(\frac{1}{x})\leq \lim_{x\to 0}|\sin^2x \sin(\frac{1}{x})|=\lim_{x\to 0}\sin^2x |\sin(\frac{1}{x})|\leq 0\cdot 1=0$ so $$
f(x)=
\begin{cases}
\sin^2x \sin(\frac{1}{x}), x\neq 0\\
0, x=0\\
\end{cases}
$$ is continuous for all $x$
b. For $x\neq 0$ we can take $$[\sin^2x \sin(\frac{1}{x})]'=2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$$ (not sure why can we do it)
At $x=0$ we look at $$lim_{\Delta h\to 0}\frac{\sin^2(x+\Delta h) \sin(\frac{1}{x+\Delta h})-\sin^2x \sin(\frac{1}{x})}{\Delta h}\mid_{x=0}=lim_{\Delta h\to 0}\frac{\sin^2(\Delta h) \sin(\frac{1}{\Delta h})-0}{\Delta h}=lim_{\Delta h\to 0}\frac{\sin(\Delta h)}{\Delta h} \cdot sin(\Delta h)\cdot\sin(\frac{1}{\Delta h})\leq lim_{\Delta h\to 0}\mid\frac{\sin(\Delta h)}{\Delta h} \cdot sin(\Delta h)\cdot\sin(\frac{1}{\Delta h})\mid=lim_{\Delta h\to 0}\mid\frac{\sin(\Delta h)}{\Delta h}\mid \cdot \mid \sin(\Delta h)\mid \cdot \mid\sin(\frac{1}{\Delta h})\mid=1\cdot 0 \cdot 1=0$$
(When can I take the limit of a product of functions? just when I am sure the the limit of each function is finite?)
c. Now we have to check that $\lim_{x\to 0}2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$
(how do I find if this limit exist?)
In general are a and b are valid?
| First, the product of continuously differentiable maps is continuously diufferentiable. As $x \mapsto \sin^2 x$ and $x \mapsto \sin \left( \frac{1}{x} \right)$ are both continuously differentiable for $x \neq 0$, $f$ is continuously differentiable for all $x \neq 0$.
Also, for all $x \in \mathbb R$, you have $0 \le \vert x \vert \le \vert \sin x \vert$ and therefore
$$0 \le \vert f(x) \vert = \left\vert \sin^2x \sin \left(\frac{1}{x}\right) \right\vert \le x^2 \left\vert \sin \left(\frac{1}{x}\right) \right\vert \le x^2$$
for $x \neq 0$. Consequently
$$\left\vert\frac{f(x)}{x} \right\vert \le \left\vert x \right\vert$$
which enables to prove that $f$ is differentiable at $0$ with $f^\prime(0)=0$.
For $x \neq 0$, you computed
$$f^\prime(x) = 2\sin x \cos x \sin \frac{1}{x}-\frac{\cos(\frac{1}{x})\sin^2x}{x^2}$$
This answers questions a. and b. Remains questions c. and specifically continuity of $f^\prime$ at $0$.
Based on similar proofs than above you get:
$$\lim\limits_{x\to 0} 2\sin x \cos x \sin \frac{1}{x} =0$$ and you know that
$$\lim\limits_{x\to 0} \left(\frac{\sin x}{x}\right)^2 =1$$
You will conclude that $f^\prime$ is discontinuous at $0$ with an essential discontinuity at $0$ because $x \mapsto \cos\left(\frac{1}{x}\right)$ has itself an essential discontinuity at $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$ Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$
My Attempt:
Let $x=3\sec (\theta )$
$$dx=3\sec (\theta).\tan (\theta).d\theta$$
Then,
$$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$
$$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d\theta $$
$$=\int \dfrac {81 \sec^5 (\theta). 3\tan (\theta).d\theta}{3\tan (\theta)}$$
$$=81\int \sec^5 (\theta) d\theta$$
| To integrate $\int \sec^5 (\theta) d\theta$, use integration by parts.
u = $\sec^3 (\theta)$
dv = $\sec^2 (\theta) d\theta$
And so du = $\sec^3\theta\tan\theta d \theta$
and $v = \tan\theta$
Thus $$\int \sec^5 (\theta) d\theta = \sec^3 (\theta) \tan \theta - \int 3\sec^3\theta(sec^2\theta-1)d\theta = \sec^3 (\theta) \tan \theta - 3\int \sec^5\theta d\theta + 3\int \sec^3\theta d\theta$$
Move the $3\int \sec^5\theta d\theta$ to the left
$$4\int sec^5\theta d\theta = \sec^3\theta \tan \theta + 3\int sec^3 \theta d\theta$$
Now we have to find $\int \sec^3 \theta d\theta $, which can be done by using integration by parts as well. I skipped a step where you convert $\tan^2 \theta$ to $\sec^2 \theta -1$ so that one of the terms have an integral with secant to the power of the original integral you are looking for.
$$u = \sec \theta$$
$$dv = \sec^2 \theta d \theta$$
$$\int sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta d \theta$$
$$\int \sec^3 \theta d \theta = \sec \theta \tan \theta - \int \sec^3 \theta d \theta + \int \sec \theta d \theta $$
$$ 2\int \sec^3 \theta d\theta = \sec \theta \tan \theta + \int \sec \theta d \theta$$
$$ \int \sec^3 \theta d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln\left|sec \theta + \tan \theta \right| + C$$
Sub the value of $\int \sec^3 \theta d \theta $ back to the original equation, we get:
$$ \int sec^5\theta d\theta = \frac{1}{4} \sec^3 \theta \tan \theta + \frac{3}{8} \sec \theta \tan \theta + \frac{3}{8} ln\lvert \sec \theta + \tan \theta \rvert + C $$
Since $\theta = arcsec \frac{x}{3}$, we can simply plug it in. Note that according to the Pythagorean theorem, $\cos \theta = \frac{3}{x}$ implies that $\tan \theta = \frac{\sqrt{x^2-9}}{3}$.
$$ \int sec^5\theta d\theta = \frac{x^3\sqrt{x^2-9}}{4*81} + \frac{x\sqrt{x^2-9}}{8*3} + \frac{3}{8}\ln \left| \frac{x+\sqrt{x^2-9}}{3}\right| + C$$
Multiply both sides by 81
$$ 81 \int sec^5\theta d\theta = \frac{x^3\sqrt{x^2-9}}{4} + \frac{27x\sqrt{x^2-9}}{8} + \frac{243}{8}\ln \left|\frac{x+\sqrt{x^2-9}}{3} \right|t + C$$
So the answer is:
$$ \int \dfrac {x^4}{\sqrt {x^2-9}} dx = \frac{x^3\sqrt{x^2-9}}{4} + \frac{27x\sqrt{x^2-9}}{8} + \frac{243}{8}\ln \left| \frac{x+\sqrt{x^2-9}}{3}\right| + C$$
Notice that you can split $\ln \left| \frac{x+\sqrt{x^2-9}}{3} \right|$ into the difference of two natural logs, but it is probably not necessary.
I found out that some people already posted the generalized form of this method, but this is more elaborate if you want the detailed steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{\sqrt{ab+a+2}}+ \frac{1}{\sqrt{bc+b+2}}+ \frac{1}{\sqrt{ac+c+2}} \leq \frac{3}{2}$ Let $abc=1$ and $a,b,c>0$.
Prove that
$$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} \leq \frac {3}{2}.$$
I guess that $$\frac {1}{\sqrt {ab+a+2}}+ \frac {1}{\sqrt {bc+b+2}}+ \frac {1}{\sqrt {ac+c+2}} = \frac {3}{2}$$ if and only if $a=b=c=1$. I have spent two nights on this but it's just a mess and I still don't know how to apply Cauchy Schwarz or something to write a rigorous proof for this exercise.
| Let $a=\frac{y}{x}$ and $b=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{ab+a+2}}=\sum_{cyc}\sqrt{\frac{1}{\frac{z}{x}+\frac{y}{x}+2}}=\sum_{cyc}\sqrt{\frac{x}{2x+y+z}}$$
$$\leq\sqrt{\sum_{cyc}1^2\sum_{cyc}\frac{x}{2x+y+z}}=\sqrt{3\sum_{cyc}x\cdot\frac{1}{x+z+x+y}}$$
$$\leq\sqrt{3\sum_{cyc}\frac{x}{(1+1)^2}\left(\frac{1^2}{x+z}+\frac{1^2}{x+y}\right)}=\sqrt{\frac{3}{4}\sum_{cyc}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)}$$
$$=\sqrt{\frac{3}{4}\sum_{cyc}\left(\frac{y}{y+x}+\frac{x}{x+y}\right)}=\frac{3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is natural number)? How can I prove that $7+7^2+7^3+...+7^{4n} = 100*a$ (while a is entire number) ?
I thought to calculate $S_{4n}$ according to:
$$ S_{4n} = \frac{7(7^{4n}-1)}{7-1} = \frac{7(7^{4n}-1)}{6} $$
But know, I don't know how to continue for get what that rquired.
I will be happy for help or hint.
After beautiful ideas for solving this question, someone know how to do it with induction too?
| Induction proof of
$$\sum_{i=1}^{4n}7^i=100a.$$
For $n=1$, it is true:
$$7^1+7^2+7^3+7^4=2800=100\cdot 28.$$
Assuming it is true for $n$:, prove for $n+1$:
$$\begin{align}\sum_{i=1}^{4(n+1)}7^i=&\sum_{i=1}^{4n}7^i+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4} =\\ &100a+7^{4n+1}+7^{4n+2}+7^{4n+3}+7^{4n+4}
=\\ &100a+7^{4n}(7+7^2+7^3+7^4)=\\ &100a+7^{4n}\cdot 2800=\\ & (a+7^{4n}\cdot 28)\cdot 100= \\ &100\cdot b\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show the inequality involving the product of distinct primes that divide $n$. Let $\displaystyle \omega (n)=\sum_{p\mid n} 1$ be the number of distinct primes dividing $n$. Show that
$\displaystyle \phi (n) \geq n \prod_{k=2}^{\omega (n) +1} (1 - \frac{1}{k}) = \frac{n}{\omega (n) +1}$ and that
$\displaystyle \phi (n) > \frac{cn}{\text{log } n}$ (hint: show that $2^{\omega (n)} \leq \tau (n) \leq n$.)
This is a problem on a practice sheet for a section in Leveque's Number Theory. I had another question from this sheet that said to use induction, but it turned out that the solution did not need induction. Is this problem manageable with just knowing the following? If so, I am not sure on where to start and proceed.
$\displaystyle \tau (n) = \sum_{d|n} 1$.
$\tau (p^e) = e+1$.
$\tau$ is multiplicative.
$\displaystyle \sigma (n) = \sum_{d|n} d$.
$\displaystyle \sigma (p^e) = 1+p+p^2+...+p^e = \frac{p^{e+1}-1}{p-1}$.
$\sigma$ is multiplicative.
| First, suppose that $n = \prod\limits_{k = 1}^{ω(n)} p_k^{a_k}$, then\begin{align*}
\frac{φ(n)}{n} &= \prod_{k = 1}^{ω(n)} \left( 1 - \frac{1}{p_k} \right) \geqslant \prod_{k = 1}^{ω(n)} \left( 1 - \frac{1}{k + 1} \right) = \prod_{k = 2}^{ω(n) + 1} \left( 1 - \frac{1}{k} \right)\\
&= \prod_{k = 2}^{ω(n) + 1} \frac{k - 1}{k} = \frac{1}{ω(n) + 1},
\end{align*}
i.e.$$
φ(n) \geqslant n \prod_{k = 2}^{ω(n) + 1} \left( 1 - \frac{1}{k} \right) = \frac{n}{ω(n) + 1}.
$$
Now, because the positive divisors of $n$ are positive integers no greater than $n$, then $τ(n) \leqslant n$. Also,$$
τ(n) = \prod_{k = 1}^{ω(n)} (1 + a_k) \geqslant \prod_{k = 1}^{ω(n)} 2 = 2^{ω(n)}.
$$
Therefore,$$
φ(n) \geqslant \frac{n}{ω(n) + 1} \geqslant \frac{n}{\log_2 τ(n) + 1} \geqslant \frac{n}{\log_2 n + 1} > \frac{cn}{\ln n}.
$$
| {
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"url": "https://math.stackexchange.com/questions/2611655",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluating $\lim\limits_{x\to \infty}\frac{x^2}{2^x-1}$
I would like to evaluate
$$\lim_{x\to \infty}\frac{x^2}{2^x-1}$$
Without using L'HOSPITAL's rule nor series.
I tried more than one
technique such the sub
$$x=\frac{1}{y}$$
But i could not get the solution ?
| Using the hint provided by Jyrki Lahtonen in the comments, we may sneakily apply the binomial theorem in order to obtain
$$ 2^n - 1 = (1+1)^n - 1 = -1 + \sum_{k=0}^{n} \binom{n}{k} \ge \binom{n}{k} $$
for any choice of natural number $k$ less than $n$. This implies that
$$ \frac{n^2}{2^n - 1} \le \frac{n^2}{\binom{n}{k}}. \tag{1}$$
By choosing $n$ large enough, it is possible to find a value of $k$ such that $\binom{n}{k} \gg n^2$. For example, if $k = 3 < n$, then we have
$$ \binom{n}{3} = \frac{n!}{(n-3)!3!} = \frac{n(n-1)(n-2)}{3\cdot 2}. $$
Plugging this into (1), we get
$$\frac{n^2}{2^n - 1}
\le \frac{n^2}{\binom{n}{3}}
= 6 \frac{n^2}{n(n-1)(n-2)}
= \frac{6n^2}{n^3 - 3n^2 + 2n}.
$$
The expression that we want to limit is positive for any $n > 0$, thus applying the squeeze theorem we can take limits to get
$$ 0 \le
\lim_{n\to\infty} \frac{n^2}{2^n - 1}
\le \lim_{n\to\infty} \frac{6n^2}{n^3 - 3n^2 + 2n}
= \lim_{n\to \infty} \frac{\frac{6}{n}}{1 - {\frac{3}{n}} + \frac{2}{n^2}}
= 0.
$$
(The last expression, obtained via multiplication by $\frac{1/n^3}{1/n^3}$ is perhaps not strictly necessary, but I think that it aids in understanding—more generally, we might simply recall that the limit (at infinity) of a rational expression is 0 if the degree of the numerator is smaller than the degree of the denominator; $\infty$ if the numerator has greater degree, and some nonzero constant if the numerator and denominator have the same degree.)
EDIT: I left off a step of the computation that seemed obvious to me, but for the sake of completeness, all that has been shown above is that if
$$ \lim_{x\to \infty} \frac{x^2}{2^x - 1} $$
exists, then it must be equal to 0. To finish the argument, we might note that
$$ \frac{x^2}{2^x - 1}
\le \frac{\lceil x \rceil^2}{2^{\lfloor x \rfloor} - 1}
= \frac{(\lfloor x \rfloor + 1)^2}{2^{\lfloor x \rfloor} - 1}
= \frac{\lfloor x \rfloor^2}{2^{\lfloor x \rfloor} - 1} + \frac{2\lfloor x \rfloor + 1}{2^{\lfloor x \rfloor} - 1}. $$
The first term is exactly the one analyzed with $n = \lfloor x \rfloor$. The second term can by analyzed similarly.
Alternatively, with
$$ f(x) = \frac{x^2}{2^x - 1}, $$
look at $f'(x)$, and note that $f'(x) < 0$ for $x$ sufficiently large (using the same kinds of estimates as above; this requires a little work, but is not unreasonably difficult). Since the derivative is negative, the function is decreasing. Moreover, the function is nonnegative for all $x > 1$, and so bounded below by zero. As $f(x)$ is monotonically decreasing and bounded, it must have a limit. By the above argument, the limit is zero.
| {
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"url": "https://math.stackexchange.com/questions/2612735",
"timestamp": "2023-03-29T00:00:00",
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Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$
Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$?
For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down.
I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$
I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere.
Is there an algebraic proof/disproof of my above claim?
Thanks.
| Since
$$
\overbrace{n^2-n+\sqrt{n^2-n}}^{\text{$n^2-n$ is too small}}\lt n^2-\frac12\iff\overbrace{\sqrt{n^2-n}\lt n-\frac12}^{n^2-n\,\lt\,n^2-n+\frac14}
$$
and
$$
\overbrace{n^2-n+1+\sqrt{n^2-n+1}}^{\text{$n^2-n+1$ is too big}}\gt n^2+\frac12\iff\overbrace{\sqrt{n^2-n+1}\gt n-\frac12}^{n^2-n+1\,\gt\,n^2-n+\frac14}
$$
there can be no integer $x$ so that $x+\sqrt{x}$ rounds to a square.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?
| Here is another trick
$$
\begin{align}
I&=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx\\
&=\int \dfrac {1+ \sin(0)+\sin (x)}{\cos(0)+\cos (x)} dx\\
&=\int \dfrac {1+ 2\sin(x/2)\cos (x/2)}{2\cos^2(x/2)}dx\\
&=\int \dfrac {1}{2\cos^2(x/2)}dx+\int \dfrac {\sin(x/2)}{\cos(x/2)}dx\\
&=\int \dfrac {1}{2\cos^2(x/2)}dx+\int {\tan(x/2)}dx\\
I&=\tan(x/2)-2\ln|\cos(x/2)|+K
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
} |
Find $\sum_{n=1}^{\infty} n\left(2n-1\right)x^{2n}$
Question: Find $$\sum_{n=1}^{\infty} n\left(2n-1\right)x^{2n}.$$
My Approach:$$
\lim_{n\rightarrow\infty} |n\left(2n-1\right)|^{\frac{1}{n}} =1 \Longrightarrow \text{Radius
of convergence} = 1.
$$
I don't know how to find the sum. Any Hint or help will be highly appreciated.
I Solved It,Thanks To Przemysław Scherwentke
$\sum_{n=1}^{k}$$x^{2n}$=$\frac{x^{2}\left[1-\left(x^{2}\right)^{k}\right]}{1-x}$
$\Rightarrow$$\sum_{n=1}^{\infty}$$x^{2n}$=$\frac{x^{2}}{1-x}$
Differentiating twice we get the required result .
| Take $f(x)=\sum_{n=1}^{\infty}n(2n-1)x^{2n}$ therefore:$$g(x)=\int \frac{f(x)}{x^2}dx=C_1+\sum_{n=1}^{\infty}nx^{2n-1}\\g_1(x)=\sum_{n=1}^{\infty}nx^{2n-1}\\h(x)=\int \frac{g_1(x)}{x}dx^2=C_2+\sum_{n=1}^{\infty}x^{2n}=\frac{1}{1-x^2}-1+C_2$$Now by successive differentiation we obtain:$$g_1(x)=x\frac{dh(x)}{d(x^2)}=\frac{x}{(1-x^2)^2}\to g(x)=\frac{x}{(1-x^2)^2}+C_1$$and $$f(x)=x^2\frac{dg}{dx}=\frac{-3x^6+2x^4+x^2}{x^4-2x^2+1}$$ and finally:$$\sum_{n=1}^{\infty}n(2n-1)x^{2n}=x^2\frac{1+3x^2}{1-x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Is this true: $\lim_{x\to 0} \frac{1-(\cos x)^n}{x^2}=\frac n2$ Does this statement hold $\forall n\in \Bbb R$?
$$\lim_{x\to 0} \frac{1-(\cos x)^n}{x^2}=\frac n2$$
I know that it holds for $\forall n \in \Bbb N$, because $1-(\cos x)^n=(1-\cos x)(1+ \cos x + ... + (\cos x)^{n-1})$, and $\lim_{x\to 0} (1+ \cos x + ... + (\cos x)^{n-1})=(1+1+...+1)=n$, but does that stand for, for example, $n=\sqrt2$?
If it does, how do I prove it?
| Note that for $n\in \mathbb{N}$
$$1-(\cos x)^n=(1-\cos x)\stackrel{\color{red}{\text{n terms}}}{(\cos^{n-1}x+\cos^{n-2}x+...+1)}$$
thus since $\cos x\to 1$
$$\frac{1-(\cos x)^n}{x^2}=\frac{(1-\cos x)}{x^2}\stackrel{\color{red}{\text{sum up to n}}}{(\cos^{n-1}x+\cos^{n-2}x+...+1)}\to\frac12\cdot n=\frac n2$$
For $n=a\in \mathbb{R}$ note that
$$\cos x=1-\frac{x^2}{2}+o(x^2) \quad (1+x)^a=1+ax+o(x) \\\implies \left(1-\frac{x^2}{2}+o(x^2)\right)^a=1-a\frac{x^2}{2}+o(x^2)$$
then
$$\frac{1-(\cos x)^n}{x^2}=\frac{1-1+a\frac{x^2}{2}+o(x^2)}{x^2}=\frac{a}2+o(1)\to\frac{a}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Solving a biquadratic.
If $x$ is a real number and satisfies $$x+ \sqrt[4] {5-x^4}=2$$ then find the value of $$x\sqrt[4] {5-x^4}$$
My try :
The question is significantly asking for the value of $-x(x-2)$ if we get the root of the equation $$2x^4-8x^3+24x^2-32x+11=0$$
Using this I have reached till
$$-2x(x-2)(x^2-2x+8)=11$$
$$-x(x-2)=\frac {11}{2(x^2-2x-8)}$$
but I couldn't manipulate it further.
Also upon some second thought I want to ask whether could it be possible to form a quadratic polynomial with two roots $\alpha$ and $\beta$ such that $\alpha=x$ and $\beta = \sqrt[4] {5-x^4}$
but I still couldn't proceed further. Somebody please share some hints.
| Note that
$$2x^4-8x^3+24x^2-32x+11=0 \implies 2 (x^2 - 2 x)^2 + 16 (x^2 - 2 x) + 11=0$$
now take $x^2 - 2 x=y$ and solve
$$2 y^2 + 16y+ 11=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Finding a matrix with conditions on $A^{-1}$ and $\det(A)=0.5$
Find matrix $A$ (without guessing) if $A^{-1}$ $2^{\mkern1mu\text{nd}}$ column is\begin{pmatrix} 2 \\ 1 \\-1 \end{pmatrix}
And $\det(A) = 0.5$
Attempt -
I could make a progress knowing that $\det(A^{-1}) = 2$ obviously.
I could also notify that if $A = $
\begin{pmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \end{pmatrix}
Then - \begin{pmatrix} 2a_1 + a_2 - a_3 = 0 \\ 2b_1 +b_2 -b_3 = 1 \\2c_1 +c_2 -c_3 = 0\end{pmatrix}
I don't know how to procced, Any help ?
| I would proceed like this $A^{-1}=\begin{pmatrix} a & 2 & d\\ b & 1 & e \\ c & -1 & f\end{pmatrix}$
The only additional constraint we have is $\det(A^{-1})=af+ae-bd-2bf+2ce-cd=2$
Since we are required to find an instance of $A$ that works and not all possible matrices, then just select some values of $(a,b,c,d,e,f)$ that fit nicely.
For instance
*
*let start with $a=0$, it remains $-bd-2bf+2ce-cd=2$
*let choose $b=0$, it remains $2ce-cd=2$
*a trivial choice is then $d=0,f=0,c=e=1$
And we get $A^{-1}=\begin{pmatrix} 0 & 2 & 0\\ 0 & 1 & 1 \\ 1 & -1 & 0\end{pmatrix}$
But I could have started
*
*let choose $f=0$, it remains $ae-bd+2ce-cd=2$
*continue with $e=0$, it remains $-bd-cd=2$
*a trivial choice could then be $d=1,b=0,c=-2,a=0$
And we get $A^{-1}=\begin{pmatrix} 0 & 2 & 1\\ 0 & 1 & 0 \\ -2 & -1 & 0\end{pmatrix}$
But you can as well set five of the six variables arbitrarily
*
*set $a=1,b=2,c=3,d=4,e=5$, it remains $-3f+15=2$
*$f=3/13$ is forced.
And we get $A^{-1}=\begin{pmatrix} 1 & 2 & 4\\ 2 & 1 & 5 \\ 3 & -1 & \frac 3{13}\end{pmatrix}$
It is just up to you to select whichever numbers you desire, until at some point, the equation relative to $\det(A^{-1})=2$ constrains the remaining last free variable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Beckenbach Introduction to Inequalities Chapter 2: Show $(a+b)/2 \le ( (a^2 + b^2 )/2)^{1/2}$ I'm having trouble understanding the following problem
Problem
Beckenbach, Chapter 2 Pg 24 Ex 1
$$
\text{Show the following for all a, b}\quad
\frac{(a+b)}{2} \le \left(\frac{a^2 + b^2}{2}\right)^\frac{1}{2}
$$
The book provides answers in the back, the answer is shown as
$$
\text{equivalent to}\quad(a - b)^2 \ge 0
$$
My attempt
Thanks to this answer: Proving the inequality $\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}$ I was able to show the following
$$
\begin{align}
\frac{a+b}{2} &\le \sqrt{\frac{a^2 + b^2}{2}} \\
\frac{a+b}{2} &\le \sqrt{\frac{(a+b)^2 + (a-b)^2}{4}}\\
\frac{a+b}{2} &\le \sqrt{\left(\frac{a+b}{2}\right)^2\left(1 + \left(\frac{a-b}{a+b}\right)^2\right)}\\
\frac{a+b}{2} &\le \frac{a+b}{2}\sqrt{1 + \left(\frac{a-b}{a+b}\right)^2}\\
0 &\le \sqrt{1 + \left(\frac{a-b}{a+b}\right)^2} - 1 \\
\text{Since}\quad\left(\frac{a-b}{a+b}\right)^2 \ge 0\quad\text{we're done}
\end{align}
$$
This feels ugly and I don't think is the way Beckenbach intended us to solve, considering this is in Chapter 2.
Question
*
*Is my attempt valid?
*Is there a more elegant way to show this, using $(a-b)^2 \ge 0$?
Thanks
| Squaring both sides...
$\cfrac{(a+b)^2}{4}\le \cfrac{a^2+b^2}{2}$
$\cfrac{a^2+2ab+b^2}{4}\le \cfrac{a^2+b^2}{2}$
$2a^2+4ab+2b^2\le4(a^2+b^2)$
$2a^2+4ab+2b^2\le4[(a+b)^2-2ab]$
$(a+b)^2\le2(a+b)^2-4ab$
Which proves your inequality, since any numbers $0<x<1$ will produce a positive result, as $-4ab$ will equal to a value smaller than $(a+b)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Checking the differentiability of the following function
Check the differentiability of the following function $$f(x)=(x+1)|x^2-1|$$ at points $x=1$ and $x=-1$.
My approach
I have written the function in the following form:
$$f(x)=\begin{cases}
x^3-x+x^2-1 & \text{ if } x\leq-1,x\geq1 \\
x-x^3+1-x^2 & \text{ if } -1<x<1
\end{cases}$$
Now, taking derivative:
$$f'(x)=\begin{cases}
3x^2-1+2x & \text{ if } x\leq-1,x\geq1 \\
1-3x^2-2x & \text{ if } -1<x<1
\end{cases}$$
Clearly, the above derivative is continuous at $x=-1$ and discontinuous at $x=1$, hence function will be differentiable at $x=-1$ and $x=1$.
Did I do everything correctly? I am not sure about this and answer has not been provided in the answer manual.
| Not differentiable at $x= 1:$
$f(x)=(x+1)|(x+1)(x-1)| = (x+1)^2 |x-1|$
for $x>0.$
Consider : $\dfrac{f(x)-f(1)}{x-1} =$
$\dfrac{(x+1)^2|x-1| }{x-1}.$
$\lim_{x \rightarrow 1^-} \dfrac{(x+1)^2(1-x)}{x-1}=-4.$
$\lim_{x \rightarrow 1^+} \dfrac{(x+1)^2(x-1)}{x-1}= 4.$
Not differentiable at $x=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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How to prove $\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}$ Prove
$$
\tan\Big[\frac{1}{2}\sin^{-1}\frac{3}{4}\Big]=\frac{4-\sqrt{7}}{3}
$$
and justify why $\frac{4+\sqrt{7}}{3}$ is ignored.
My Attempt:
$$
\tan x=\frac{2\tan\frac{x}{2}}{1-\tan^2\frac{x}{2}}\implies\tan x-\tan x\tan^2\frac{x}{2}=2\tan\frac{x}{2}\\
\implies \tan^2\frac{x}{2}(\tan x)+2\tan\frac{x}{2}-\tan x=0\\
\implies \tan\frac{x}{2}=\frac{-2\pm2\sec x}{2\tan x}=\frac{-1\pm\sec x}{\tan x}
$$
Using this,
$$
\tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-1\pm\frac{4}{\sqrt{7}}}{\frac{3}{\sqrt{7}}}=\frac{-\sqrt{7}\pm4}{3}
$$
As $0\leq\frac{\sin^{-1}3/4}{2}\leq\frac{\pi}{4}\implies0\leq\tan(\frac{\sin^{-1}3/4}{2})\leq1$
$$
\implies \tan\bigg[\frac{\sin^{-1}\frac{3}{4}}{2}\bigg]=\frac{-\sqrt{7}+4}{3}
$$
Is my attempt correct and where is the value $\frac{4+\sqrt{7}}{3}$ to be excluded ?
| let $\sin y= \frac{3}{4}$ we need to prove
$\tan \left[\frac{1}{2}\sin^{-1} \left(\frac{3}{4}\right)\:\right]=\frac{4-\sqrt{7}}{3}$
otherwise seen as
$\tan \left[\frac{1}{2}y\:\right]=\frac{4-\sqrt{7}}{3}$
We also know the identity
$\tan \left[\frac{1}{2}y\:\right]=\frac{1-\cos\left[y\:\right]}{\sin \left[y\:\right]}$
Thus
$\tan \left[y\:\right]=\frac{1-\frac{\sqrt 7}{4}}{\frac{3}{4}}=\frac{4-\sqrt{7}}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Solve $f'_x-xf'_y=y.$
Find the solution to $$\frac{\partial}{\partial x}f(x,y)
-x\frac{\partial}{\partial y}f(x,y) = y$$
that satisfies $f(x,0)=x^2+\frac{x^3}{3},$ by using the transformation
$$\left\{ \begin{array}{rcr}
u & = & ax^2+y \\
v & = & x \\ \end{array} \right.$$
for an appropriate value of $a$.
So I have that $f'_x-xf'_y=f'_x+vf'_y=y=u-ax^2=u-av^2.$ Furthermore:
\begin{array}{lcl}
u'_x & = & 2ax \\
u'_y & = & 1 \\
v'_x & = & 1 \\
v'_y & = & 0 \\
f'_x & = & f'_uu'_x+f'_vv'_x=2axf'_u+f'_v=2avf'_u+f'_v \\
f'_y & = & f'_uu'_y+f'_vv'_y=f'_u\cdot 1+f'_v\cdot 0=f'_u
\end{array}
Plugging this into my PDE $f'_x-xf'_y=u-av^2$ I get
$$2avf'_u+f'_v-vf'_u = v(2af'_u-f'_u)+f'_v=u-av^2.$$
It's practical to choose $a=\frac{1}{2},$ then I get $$f'_v=u-\frac{v^2}{2} \ \implies f(u,v)=uv-\frac{v^3}{6} + g(u),$$
so
$$f(x,y)=\left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+y\right).$$
Using the condition $f(x,0)=x^2+\frac{x^3}{3}$ I get
$$f(x,0)=\frac{x^3}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right)= \Leftrightarrow g\left(\frac{x^2}{2}\right)=x^2.$$
Setting $t=\frac{x^2}{2}\Rightarrow x=\sqrt{2t}$ I get
$$g(t)=2t \implies g\left(\frac{x^2}{2}+y\right)= x^2+2y.$$
Finally:
$$f(x,y)=\left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+x^2+2y.$$
But apparently this is wrong answer. Can anyone see where I made the error?
| $$\begin{align}
f(x,y)= & \left(\frac{x^2}{2}+y\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+y\right)\\
f(x,0)= &\left(\frac{x^2}{2}+0\right)x-\frac{x^3}{6}+g\left(\frac{x^2}{2}+0\right) \\
=& \frac{x^3}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right) = \frac{x^3}{3}+g\left(\frac{x^2}{2}\right) \neq \frac{x^2}{2}-\frac{x^3}{6}+g\left(\frac{x^2}{2}\right) \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by
$$
u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.
$$
Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive.
Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$:
$$
u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\
u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\
u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\
u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.
$$
But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!
| Hint: $\;\require{cancel}
u_{n+1} - u_n = \left(\cancel{a} - \dfrac{1}{u_n}\right) - \left(\cancel{a} - \dfrac{1}{u_{n-1}}\right)=\dfrac{u_n-u_{n-1}}{u_nu_{n-1}}
\,$, so as long as the terms are positive the differences between consecutive terms have the same sign i.e. the sequence is monotonic, and is in fact decreasing since $\,u_1-u_0=\left(\cancel{a} - \dfrac{1}{a}\right) - \cancel{a} \lt 0\,$. For it to be positive it is necessary and sufficient that $\,0\,$ is a lower bound, so the sequence is positive iff it is convergent. In this case, passing the recurrence relation to the limit, its limit $\,A\,$ must satisfy $\,A = a - \dfrac{1}{A}$ $\iff A^2 - aA + 1 = 0\,$. The latter equation has real roots iff $\,\Delta = a^2 - 4 \ge 0 \iff a \ge 2\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability of Maximum Value
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(\max(X, Y) = k)$ for $1 \le k \le n$.
$P(\max(X, Y) = k) $
$= P(X = k \cap X > Y) + P(Y = k \cap Y > X)$
$= P(X=k)P(X>Y|X=k) + P(Y=k)P(Y>X|Y=k)$
$= \frac{1}{n} \cdot \frac{k-1}{n} + \frac{1}{n} \cdot \frac{k-1}{n}$
$= \frac{2(k-1)}{n^2}$
Textbook Answer: $\,\,\, \frac{2k-1}{n^2}$
| $P(\max(X, Y) = k)$
$= P(X = k \cap X > Y) + P(Y = k \cap Y > X) + P(X = Y = k)$
$= P(X=k)P(X>Y|X=k) + P(Y=k)P(Y>X|Y=k) + P(X = Y = k)$
$= \frac{1}{n} \cdot \frac{k-1}{n} + \frac{1}{n} \cdot \frac{k-1}{n} + \frac1{n^2}$
$= \frac{2k-1}{n^2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639134",
"timestamp": "2023-03-29T00:00:00",
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Solve $yy' + x =\sqrt{x^2 + y^2}$ - Substitution for Diff Eqs. I have this problem
$$yy' + x = \sqrt{x^{2} + y^{2}}$$
I wanted to know if my work and answer are correct.
Let $v = x^{2} + y^{2}. v' = 2x + 2yy'$
$$ \frac{v'}{2} = x + yy'$$
Substituting everything in gives me:
$$\frac{v'}{2} = \sqrt{v}$$
From here it becomes a separable equation:
$$\frac{dv}{\sqrt{v}} = 2dx$$
$$2\sqrt{v} = 2x +C$$
Subbing in $x^2 + y^2$ back gives me: $$2\sqrt{x^2 + y^2} = 2x + C$$
And now if at this point my work is correct, we just solve for $y$ right?
$$\sqrt{x^2 + y^2} = x + \frac{C}{2}$$
Square both sides:
$$x^2 + y^2 = \bigg(x+\frac{C}{2}\bigg)^{2}$$
$$y^2 = \bigg(x+\frac{C}{2}\bigg)^{2} - x^2$$
And finally,
$$y = \pm\sqrt{\bigg(x+\frac{C}{2}\bigg)^{2} - x^2}$$
| Your solution is correct.
At $$y^2 = \bigg(x+\frac{C}{2}\bigg)^{2} - x^2$$
You may simplify $$ y^2 = \bigg(x+\frac{C}{2}\bigg)^{2} - x^2=Cx+\frac {C^2}{4}$$
Which is a family of parabolas.$$y^2 =Cx+\frac {C^2}{4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How show that $40 \mid n.$ Let $n\in\mathbb N$ s.t. $2n+1$, $3n+1$ are squares. Show that $40 \mid n.$
I have shown that $8 \mid n.$ Please help me to show $5 \mid n.$
| For $5 | n$ we have:
$3n+1=a^2 ⇒ 3n=(a-1)(a+1)$
$a-1$ and $a+1$ are two consecutive numbers, therefore n must also be consecutive to 3 that is $n=5$ or $5 | n$
Also:
$2n+1 =b^2 ⇒ 2n=(b-1)(b+1)$
$b-1$ and $b+1$ are consecutive numbers therefore n must also be consecutive to 2 that is $n=4$ or $4|n$ . So n can be the least common multiplier(LCM) of 4 and 5, that is $n=20$ or $20|n$. However when $20|n$ then the last digit of numbers$3n+1$ and $2n+1$ is 1, this is possible only when they are square of numbers with 1 or 9 as their last digit, i.e numbers such as 9, 11, 19,21,29,31... that indicate minimum value of n is 40 and generally 40 must divide n.
| {
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Why does $\sqrt[4]{-2}=\frac{1+i}{\sqrt[4]{2}}$? When I type $\sqrt[4]{-2}$ into WolframAlpha, it lists $\frac{1+i}{\sqrt[4]{2}}$ as an alternate form. How did it get that? I was trying to solve for $x^{4}+2=0$, and so I guess I could just say $\sqrt[4]{-2}$ is a solution, but I want it to look like $a+bi$ for $a,b\in\mathbb{R}$.
I remember that the 4th roots of unity are $e^{\frac{2\pi k}{4}i}$ for $k=1,2,3,4$, but we want to get $-2$, not $1$.
| $$\left(\frac{1+i}{\sqrt[4]{2}}\right)^4=\frac{(1+i)^4}{2}=\frac{1}{2}\left(\sqrt{2}\left(\frac{1+i}{\sqrt{2}}\right)\right)^4=\frac{4}{2}\left(\frac{1+i}{\sqrt{2}}\right)^4=2\left(\exp\left(\frac{i\pi}{4}\right)\right)^4=$$
$$=2\exp(i\pi)=2*(-1)=-2$$
I've used these:
*
*$$\exp\left(\frac{i\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)=\frac{1+i}{\sqrt{2}}$$
*$$\exp(i\pi)=\cos(\pi)+i\sin(\pi)=-1$$
As @DonAntonio pointed it out in the comments, it's much easier this way:
$$\left(\frac{1+i}{\sqrt[4]{2}}\right)^4=\frac{1}{2}(1+i)^4=\frac{1}{2}((1+i)^2)^2=\frac{1}{2}(1^2+2i+i^2)^2=\frac{(2i)^2}{2}=-2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of the series $\csc^{-1} \sqrt{10}+ \csc^{-1} \sqrt{50}+\csc^{-1}\sqrt{170}...$
Find the sum of the series:
$\csc^{-1}\sqrt{10}+ \csc^{-1}\sqrt{50}+\csc^{-1}\sqrt{170}...\csc^{-1}\sqrt{(n^2+1)(n^2+2n+2)}$
I converted the series to $\sum^{n} _{i=0}\arcsin \dfrac{1}{\sqrt{(i^2+1)(i^2+2i+2)}}$ and then tried to use $\arcsin(x)+\arcsin(y)= \arcsin({x\sqrt{1+y^2}+y\sqrt{1+x^2}}) \forall x,y \ge 0 $ and $x^2+y^2\le1$.
There was no clear pattern in the terms obtained. How to solve this question? And what is the general trick to solve inverse trigonometric summations?
| The given sum can be represented as $$\tan ^{-1} \frac {1}{3}+\tan ^{-1} \frac {1}{7}+\tan ^{-1} \frac {1}{13}\cdots \tan ^{-1} \frac {1}{n^2+n+1}$$
This can be represented as $$\sum_{k=1}^n \tan ^{-1} \left(\frac {1}{k^2+k+1}\right)$$
$$\sum_{k=1}^n \tan ^{-1} \left(\frac {(k+1)-(k) }{1+(k+1)(k)}\right) $$
$$\sum_{k=1}^n \tan ^{-1} (k+1) -\tan ^{-1} (k)$$
Which telescopes to $$\tan^{-1} (n+1)- \frac {\pi}{4}$$
| {
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If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that $2A^2=4A+A^3$ If $A\in M_3\left(\mathbb C\right)$ is an invertible matrix such that
$$2A^2=4A+A^3$$
Then which of the following is(are)correct:
(A) $\det (A)=8;$
(B) $det\left(adj\left(\frac A2\right)\right)=1;$
(C) $tr\left(A-2I_3\right)^3=24;$
(D) $adj(A)=A^2$
My working:
From given information we get $A(A^2-2A+4I_3)=O_3$
$\implies A^2-2A+4I_3=O_3$
$\implies (A+2\omega I_3)(A+2\omega^2 I_3)=O_3$
$\implies$ Possible Eigen values of $A$ can be $-2\omega, -2\omega^2$
Now what to do
| From the given information, $A(A^2-2A+4I_3)=O_3$
$\implies A^2-2A+4I_3=O_3$
$\implies (A+2\omega I_3)(A+2\omega^2 I_3)=O_3$
$\implies$ eigen values of $A$ can be $-2\omega,-2\omega,-2\omega^2$ or $-2\omega,-2\omega^2,-2\omega^2.$ Where $\omega$ is a non real cube root of unity.
$\implies \det A=$ Product of eigen values of $A$ = $(-2\omega)\times(-2\omega)\times(-2\omega^2)$ or $(-2\omega)\times(-2\omega^2)\times(-2\omega^2)=-8\omega$ or $-8\omega^2.$
$\implies$ Option (A) is incorrect.
As in previous solution we have $A^3=-8I_3$
$\implies A^2=-8A^{-1}=-8\frac{1}{\det A}adj(A)=\omega^2 adj(A)$ or $\omega adj(A)\implies adj(A)=\omega A^2$ or $\omega^2 A^2$
$\implies$ option (D) is incorrect.
Now $adj\left(\frac A2\right)=\left(\frac 12\right)^2 adj(A)=\frac 14 \omega A^2$ or $\frac 14 \omega^2 A^2$
$\implies\det \left(adj\left(\frac A2\right)\right)=\left(\frac 14 \omega\right)^3 (\det A)^2$ or $\left(\frac 14 \omega^2\right)^3 (\det A)^2=\omega^2$ or $\omega$
$\implies$ Option (B) is incorrect.
So only correct option is (C) as demonstrated in previous solution.
| {
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How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$? How to prove that $x^2yz+xy^2z+xyz^2 \leq \frac{1}{3}$ where $x^2+y^2+z^2 = 1$ and $x,y,z >0$?
| You can write the inequality as
$$
xyz(x+y+z)\leq\frac13.
$$
On the left-hand-side, using Cauchy-Schwarz,
$$
xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz.
$$
Now we get the seemingly easier problem of maximizing $xyz$ under $x^2+y^2+z^2=1$. Here symmetry, common sense, or Lagrange multipliers show that the maximum is achieved when $x=y=z$. Then
$$
xyz(x+y+z)\leq xyz(x^2+y^2+z^2)^{1/2}(1+1+1)^{1/2}=\sqrt3\,xyz\leq\frac{\sqrt3}{(\sqrt3)^3}=\frac13.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of expression containing $2$ variables $x,y$
Finding minimum of $\displaystyle f(a,b)= (a-b)^2+\bigg(\frac{a^2}{20}-\sqrt{(17-b)(b-13)}\bigg)^2,$
where $a\in\mathbb{R^{+}},b\in(13,17)$
(Without using Geometry)
Try: Using partial derivative with r to $a$ (Treating $b$ as a constant)
$\displaystyle f'(a,b)= 2(a-b)+2\bigg(\frac{a^2}{20}-\sqrt{(17-b)(b-13)}\bigg)\cdot \frac{a}{10}$
Using partial derivative with r to $b$(Treating $a$ as a constant)
$\displaystyle f'(a,b)= -2(a-b)+2\bigg(\frac{a^2}{20}-\sqrt{(17-b)(b-13)}\bigg)\cdot (2b-30)$(Treating $b$ as a constant)
Could some help me to solve it, thanks
| As analysis teaches us, take the derivatives with respect to $a$ and $b$ separately, and converge them into a system of equations imposing them to be zero:
$$\begin{cases}
\large \frac{a^3}{100}+a \left(2-\frac{1}{5} \sqrt{-b^2+30 b-221}\right)-2 b = 0\\\\
\large \frac{a^2 (b-15)}{10 \sqrt{-b^2+30 b-221}}-2 a+30 = 0
\end{cases}$$
Now you have to solve them both for $a$ and $b$
One of the solutions (the only real one) is simple and it's
$$a = 10$$
$$b = 15 - \sqrt{2}$$
Substituting those values into the initial equation gives you the minimum value which is
$$f(10, 15 - \sqrt{2}) = \left(\sqrt{2}-5\right)^2+\left(5-\sqrt{\left(2-\sqrt{2}\right) \left(\sqrt{2}+2\right)}\right)^2 = \color{red}{54 - 20\sqrt{2}}$$
As Dr. Sonnhard Gaubner already told you.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$ $\arccos (x) +\arccos (2x)=\frac{3\pi}{4}$
$\implies \arccos (x)=\frac{3\pi}{4}-\arccos (2x)$
$\implies \cos(\arccos (x))=\cos(\frac{3\pi}{4}-\arccos (2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arccos(2x))$
$\implies x=\cos(\frac{3\pi}{4})\cos(\arccos (2x))+\sin(\frac{3\pi}{4})\sin(\arcsin(\sqrt{1-4x^2}))$
$\implies x=-\frac{2x}{\sqrt{2}}+\frac{\sqrt{1-4x^2}}{\sqrt{2}}$
$\implies \sqrt{2}x=-2x+\sqrt{1-4x^2}$
$\implies x(\sqrt{2}+2)=\sqrt{1-4x^2}$
$\implies x^2(2+4\sqrt{2}+4)=1-4x^2$
$\implies x^2(4\sqrt{2}+10)=1$
$\implies x^2=\frac{1}{4\sqrt{2}+10}$
$x=\pm\sqrt{\frac{1}{4\sqrt{2}+10}}$
But $x=-\sqrt{\frac{1}{4\sqrt{2}+10}}$ does not satisfy the given equation.
$\therefore x= \sqrt{\frac{1}{4\sqrt{2}+10}}$
I am looking for other methods to solve this equation. I applied the formula to combine two $\arccos(x)$ functions into one $\arccos(x)$ function and on simplifying I obtained a complicated equation.
| $$\arccos(y)>\dfrac\pi2$$ for $y<0$
Here what if $x<0\iff2x<0$
| {
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Finding the limit of the recursive sequence defined as $5a_{n+2}=3a_{n+1}+2a_n$
Find the limit of the recursive sequence defined as follows: $$5a_{n+2}=3a_{n+1}+2a_n\;\;\text{for $n\geq 0$, $a_1=1/2$ and $a_2=2/3$}.$$
Could someone give me a hint as to how to start this problem? I think it is increasing.
| $$5a_{n+2}-5a_{n+1}=2a_{n}-2a_{n+1}$$ or
$$a_{n+2}-a_{n+1}=-\frac{2}{5}(a_{n+1}-a_n),$$
which says that $b_n=a_{n+1}-a_n$ is a geometric progression.
Thus, $$a_{n+1}-a_n=(a_2-a_1)\left(-\frac{2}{5}\right)^{n-1}=\frac{1}{6}\left(-\frac{2}{5}\right)^{n-1}.$$
Thus, $$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)=\frac{1}{2}+\frac{1}{6}\sum_{k=1}^{n-1}\left(-\frac{2}{5}\right)^{k-1}=$$
$$=\frac{1}{2}+\frac{1}{6}\cdot\frac{\left(-\frac{2}{5}\right)^{n-1}-1}{-\frac{2}{5}-1}\rightarrow\frac{1}{2}+\frac{1}{6}\cdot\frac{-1}{-\frac{2}{5}-1}=\frac{13}{21}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum value of $5\sin x - 12 \cos x + 1$ Given that $5\sin x - 12\cos x = 13\sin (x-67.4)$
Find the maximum value of $5\sin x - 12 \cos x + 1 $ and the corresponding value of x from 0 to 360.
Maximum value = $13+1=14$
Corresponding value of $x$
$13\sin (x-67.4) + 1 = 14$
$\sin(x-67.4) = 1 $
$x = 157.4 , 337.4 $
I found the value of $x$ and there’s 2 values. However , the answer is only$157.4$ why is that the case ?
| Let $\cos\phi=\dfrac{5}{13}$, $\sin\phi=\dfrac{12}{13}$
\begin{eqnarray}
y&=&13\left(\frac{5}{13}\sin x-\frac{12}{13}\cos x\right)+1\\
&=&13(\sin(x)\cos\phi-\cos(x)\sin\phi)+1\\
&=&13\sin(x-\phi)+1
\end{eqnarray}
which has a maximum value of $14$ when $x-\phi=\frac{\pi}{2}$.
So
\begin{eqnarray}
x&=&\frac{\pi}{2}+\phi\\
&=&\frac{\pi}{2}+\arctan\left(\frac{12}{5}\right)\\
&\approx&90^\circ+67.38^\circ\\
&=&157.38^\circ
\end{eqnarray}
| {
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Block Matrix Inversion in Wikipedia Wikipedia provides two formulas for block-matrix inversion:
$$ {\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}}^{-1}={\begin{bmatrix}\mathbf {A} ^{-1}+\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}&-\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\\-(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}&(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\end{bmatrix}},$$
and
$${\begin{bmatrix}\mathbf {A} &\mathbf {B} \\\mathbf {C} &\mathbf {D} \end{bmatrix}}^{-1}={\begin{bmatrix}(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}&-(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\\-\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}&\quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\end{bmatrix}}.$$
Is it true then that all of the following equalities are true?
\begin{align}\mathbf {A} ^{-1}+\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}=&\;(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\\
-\mathbf {A} ^{-1}\mathbf {B} (\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}=&\; -(\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1} \\
-(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}\mathbf {CA} ^{-1}=&\;-\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1} \\
(\mathbf {D} -\mathbf {CA} ^{-1}\mathbf {B} )^{-1}=&\; \quad \mathbf {D} ^{-1}+\mathbf {D} ^{-1}\mathbf {C} (\mathbf {A} -\mathbf {BD} ^{-1}\mathbf {C} )^{-1}\mathbf {BD} ^{-1}\end{align}
| Provided the inverses involved exist, both terms are equal. But, generally, only one of them is used because the inverse of at least one term involved does not exist. For example, for $\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}$, the first formula cannot be used, but the second formula can be used to evaluate the inverse.
Note that both of these formulae cannot be used to evaluate $\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}^{-1}$.
| {
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Integer solutions for $x^3=y^2+5$ I have been able to solve the equation $x^3=y^2+a$ for integers where $1 \leq a \leq 4$ by splitting in $\Bbb Z [i\sqrt a]$. However as a natural continuation I would like to know whether the equation $x^3=y^2+5$ can be solved using elementary methods. Seems this case is much tougher though .I would like some hints on how to proceed.
| Working mod 4 we see that $y$ must be even and $x=1\bmod4$. Now we have $$y^2+4=x^3-1=(x-1)(x^2+x+1)$$ But $x^2+x+1=3\bmod4$, so $x^2+x+1$ is odd. It is also positive, so it must be at least 3. Hence it must have a prime factor $p=3\bmod4$. So $y^2+4=0\bmod p$, in other words -4 is a quadratic residue of $p$. But that implies that $p=1\bmod 4$. Contradiction.
| {
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How many ways are there to get a sum of $25$ when $10$ distinct dice are rolled? I have come across the following problem and have given it a good attempt below. I am wondering if I have proceeded correctly, and if not if someone could show me the correct answer or maybe a more efficient solution, thanks!
How many ways are there to get a sum of $25$ when $10$ distinct dice are rolled?
Each die can be a number from $1$ to $6$. With $10$ dice, our generating function becomes:
$$g(x) = (x+x^2+x^3+x^4+x^5+x^6)^{10}$$
We want to find the coefficient of $x^{25}$. Observe that the expression inside the brackets is a finite geometric series with $a=x, r=x, n=6$. Thus we have
$$(x+x^2+x^3+x^4+x^5+x^6)^{10}$$
$$=\left(\frac{x(1-x^6)}{(1-x)}\right)^{10}$$
$$=x^{10}(1-x^6)^{10}(1-x)^{-10}$$
Then,
$$(1-x^6)^{10}(1-x)^{-10}$$
$$=\sum \binom{10}{i}(-x^6)^i \cdot\sum\binom{-10}{j}(-x)^j$$
In order to get terms that involve $x^{15}$ there are $3$ combinations of $i$ and $j$ to consider so that $6i+j = 15$
Thus we have:
$$\left[ \binom{10}{0}(-1)^0\binom{-10}{15}(-1)^{15}\right] + \left[ \binom{10}{1}(-1)^1 \binom{-10}{9}(-1)^9\right] + \left[\binom{10}{2}(-1)^2\binom{-10}{3}(-1)^3\right]$$
$$ = \left[\binom{10}{0}\binom{-10}{15}(-1)\right]+\left[ \binom{10}{1} \binom{-10}{9} \right] + \left[ \binom{10}{2} \binom{-10}{3}(-1)\right] $$
$$=\left[ \binom{10}{0}\binom{10+15-1}{15}(-1)^{15}(-1)\right] + \left[ \binom{10}{1} \binom{10+9-1}{9}(-1)^9\right] + \left[ \binom{10}{2}\binom{10+3-1}{3}(-1)^3(-1)\right]$$
$$= \left[ \binom{10}{0}\binom{24}{15}\right] - \left[ \binom{10}{1}\binom{18}{9}\right] + \left[ \binom{10}{2}\binom{12}{3}\right]$$
$$= 831204$$
| You will enjoy reading about Polynomial Coefficients here. See specifically Theorem 2.1 in the referenced paper on the sum of discrete uniform random variables.
What you are looking for is what's referred to as multinomials. Your approach tries to distill a multinomial into its basic binomial counter parts. There is a recursive relationship between multinomials, as discussed in the referenced paper.
| {
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Real Analysis. Differentiation (Taylor's Formula). Check proof.
Show that the polynomial
$$x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \pm \frac{x^n}{n!},$$
with $n$ odd, differs from $\sin x$ by, at most, $\frac{\pi^{n+1}}{(n+1)!}$ in $[-\pi, \pi]$.
$\textbf{Solution:}$ Applying the Taylor's Formula in $f(x) = \sin x$ for $x_{0} = 0$:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \pm \frac{x^n}{n!} + R_{n+1}$$
and
$$R_{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$
for $c \in (0,x)$. Since $|f^{(n+1)}(c)| \leq 1$ and $|x|^{n+1} \leq \pi^{n+1}$
$$R_{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \leq \frac{\pi^{n+1}}{(n+1)!}.$$
The result follows.
Is this correct?
| Yes, your proof is correct.
I would change $$ R_{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \leq \frac{\pi^{n+1}}{(n+1)!}.$$
to $$ |R_{n+1}| = \frac{|f^{(n+1)}(c)|}{(n+1)!}x^{n+1} \leq \frac{\pi^{n+1}}{(n+1)!}.$$
| {
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Alternate solution to a limit without using L'Hopital's rule $$\lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)$$
I've gotten to this
$$a \cdot \lim \limits_{x \to a} \frac{x^5-2x^4+x^3-a^5+2a^4-a^3}{(x-a)(a-1)^2(x-1)^2}$$
since as far as I'm concerned it's just a matter of doing a long polynomial division between the numerator and $(x-a)$, but I'm getting a remainder that equals infinity despite knowing the answer is finite.
Upon looking up the division itself it seems I had made a mistake, but I'm still curious as to how else this could be solved other than essentially brute forcing it.
| One way is to solve with Differentiation From First Principles perspective:
$\displaystyle \lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)=\lim \limits_{x \to a} \frac{x}{x-a} \left(\dfrac{x^3(x-1)^2-a^3 (a-1)^2}{(a-1)^2 (x-1)^2}\right)=\dfrac{a}{(a-1)^4} \left( \lim_{x \to a} \dfrac{x^3(x-1)^2-a^3 (a-1)^2}{x-a} \right)=\dfrac{a}{(a-1)^4}(x^3(x-1)^2)'|_{x=a}$
Can you take it from here?
| {
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Compute $\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}$
$$\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}=?$$
Here's what I have done so far:
$y=\frac{\pi}{2}-x$ The limit becomes:$$\lim_{y\to 0}\frac{(1-\cos y)(1-\cos^2y)\dots(1-\cos^ny)}{\sin^{2n}y}=\lim_{y\to 0}\frac{1-\cos y}{\sin^2y}\frac{1-\cos^2y}{\sin^2y}\dots\frac{1-\cos^ny}{\sin^2y}$$
Also $1-\cos y=2\sin^2 \frac{x}{2}$
How should i write $1-\cos^2y ,1-\cos^3y,\dots1-\cos^ny$ in order to get the final answer.
| \begin{align}\lim_{x\to\frac\pi2}\frac{(1-\sin x)(1-\sin^2x)\dots\lim(1-\sin^nx)}{\cos^{2n}x}&=\lim_{x\to\frac\pi2}\frac{1-\sin x}{\cos^2x}\cdot\frac{1-\sin^2x}{\cos^2x}\cdot\frac{1-\sin^nx}{\cos^2x}\\&=\frac12\cdot\frac22\cdots\frac n2=\frac{n!}{2^n}\end{align}because, near $\frac\pi2$, $1-\sin^kx$ beahaves like $\frac k2\left(x-\frac\pi2\right)^2$, whereas $\cos^2x$ behaves like $\left(x-\frac\pi2\right)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Find the roots of $\mathrm{ \overline{Z} + 1 = iZ^2 + {|Z|}^2}$ I attempted the problem in two ways
Method 1
Resolved $\mathrm{ Z = X + iY}$ this led to a very big cubic equation and handling that became too difficult for me .
Method 2
I believe factorizing the equation is a far better method, I don't know how to go about it .
EDIT
The answer given is $\mathrm{Z = \frac{-i}{2} , i}$
As pointed out by @gimsui $\mathrm{\frac{-i}{2} }$ is not a solution most likely a printing error.
| The complex number $Z$ can be written as $$Z = A + iB.$$
Then, your equation becomes:
$$\overline{Z} + 1 = iZ^2 + |Z|^2 \Rightarrow \\
A-iB + 1 = i(A^2-B^2+2iAB)+A^2+B^2 \Rightarrow \\
A-iB + 1 = iA^2 - iB^2 -2AB + A^2 + B^2 \Rightarrow \\
(A+1 +2AB-A^2-B^2) + i(-B-A^2+B^2) = 0.$$
This means that both the real part $(A+1 +2AB-A^2-B^2)$ and the imaginary part $(-B-A^2+B^2)$ must be $0$. Then, you must solve the following system:
$$\begin{cases}
A+1 +2AB-A^2-B^2 & = 0\\
-B-A^2+B^2 & = 0
\end{cases}. $$
This system has the following solutions:
$$A_1 = -\frac{\sqrt{3}}{2}, B_1 = -\frac{1}{2} \Rightarrow Z_1 = -\frac{\sqrt{3}}{2} -i\frac{1}{2},\\
A_2 = \frac{\sqrt{3}}{2}, B_2 = -\frac{1}{2} \Rightarrow Z_2 = \frac{\sqrt{3}}{2}-i\frac{1}{2}, \\
A_3 = 0, B_3 = 1 \Rightarrow Z_3 = i.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Asymptotics of an integral $\mathscr{P} \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^{2} - x^{2}}$ involving principal value Consider the following integral for $x > 0$:
$$
F(x) \ = \ \mathscr{P} \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^{2} - x^{2}}
$$
The $\mathscr{P}$ is there because there is a singularity at $\omega = x$, so I suppose we can write the above function more carefully as:
$$
F(x) \ = \ \lim_{\epsilon \to 0^{+}} \left\{ \int_{0}^{x - \epsilon} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^{2} - x^{2}} \ + \ \int_{x + \epsilon}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^{2} - x^{2}} \right\}
$$
It looks like $F(0) = \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{1}{\omega}$ which is divergent (the integrand looks like $\frac{1}{\omega^2}$ near the $\omega =0$ in this case).
I am interested in obtaining asymptotics for $F(x)$ as we approach $x \to 0$. I am unsure how to do this due to the presence of the principal value $\mathscr{P}$. Is there a method for dealing of asymptotics of such integrals?
| I've figured it out. I've even got an analytic result for this integral!
$$
F(x) := \mathscr{P} \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^2 - x^2} = \frac{1}{2} \ln\left( \frac{x}{2\pi} \right) - \frac{1}{2} \mathrm{Re}\left[ \psi^{(0)}\left( \frac{ix}{2\pi} \right) \right]
$$
And as we take $x \to 0^{+}$ the above looks like:
$$
F(x) = \frac{1}{2} \ln\left( \frac{x}{2\pi} \right) \ + \ \frac{\gamma}{2} \ + \ \mathscr{O}(x^2)
$$
I've done some numerical tests in Mathematica comparing the integral using the principal value method, the full formula, and then the asymptotic form and everything works.
Proof: Note the following formula (see this post) which is derived by differentiating Binet's Second Formula:
$$
G(y) : = \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^2+y^2} \ = \ \frac{1}{2} \ln\left( \frac{y}{2\pi} \right) - \frac{\pi}{2 y} - \frac{1}{2} \psi^{(0)}\left( \frac{y}{2\pi} \right)
$$
We take an analytical continuation by considering $\lim\limits_{\epsilon \to 0^{+}}G(\epsilon+i x)$ for $x>0$. We find:
$$
\lim\limits_{\epsilon \to 0^{+}}G(\epsilon+i x) = \lim_{\epsilon \to 0^{+}} \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^2+(\epsilon + i x)^2}\to\lim_{\epsilon \to 0^{+}} \int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^2 - x^2 + i \epsilon}
$$
Then we can use the identity $\lim\limits_{\epsilon \to 0^{+}}\frac{1}{z+i \epsilon} = \mathscr{P}\frac{1}{z} - i \pi \delta(z)$. Taking $\epsilon \to 0^{+}$ on both sides gives:
$$
G(i x) = \mathscr{P}\int_{0}^{\infty} \frac{d\omega}{e^{\omega} - 1} \frac{\omega}{\omega^2 - x^2} \ - \ i \pi \int_{0}^{\infty} d\omega \frac{\omega}{e^{\omega} - 1} \delta\big(\omega^2 - x^2 \big)
$$
Where we can evaluate the imaginary part easily, and the real part is the function we're after:
$$
G(i x) = F(x) \ - \ i \frac{\pi}{2} \frac{1}{e^{x} - 1}
$$
This tells us that:
$$
F(x) = \mathrm{Re}\left[ G(i x) \right] = \mathrm{Re}\left[ \frac{1}{2} \ln\left( \frac{ix}{2\pi} \right) - \frac{\pi}{2 ix} - \frac{1}{2} \psi^{(0)}\left( \frac{ix}{2\pi} \right) \right]
$$
Which for $x>0$ simplifies to the stated result.
$\ $
BONUS: Using the above, we find that for $\alpha >0$:
$$
\mathrm{Im}\left[ \psi^{(0)}\left( i \alpha \right) \right] \ = \ \frac{1}{2 \alpha} + \frac{\pi}{2} \coth\left( \pi \alpha \right)
$$
I don't know if this is a known formula, but I certainly couldn't find it anywhere (UPDATE: It is known, I found it here on DLMF!). It works numerically, and seems to also work for $\alpha <0$. This allows us to write $F(x)$ in the form:
$$
F(x) = \frac{1}{2} \ln\left( \frac{x}{2\pi} \right) + \frac{i \pi}{2 x } + \frac{i \pi}{4} \coth\left( \frac{x}{2} \right) - \frac{1}{2} \psi^{(0)}\left( \frac{ix}{2\pi} \right)
$$
which is purely real, despite the appearance of the $i$'s.
| {
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"url": "https://math.stackexchange.com/questions/2672281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine whether $ (A_1,A_2,A_3,A_4)$ span $M_{22}$ Determine whether $\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right], \left[\begin{array}{cc} 1&1\\1&1\end{array}\right], \left[\begin{array}{cc} 1&0\\0&1\end{array}\right],\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$ spans $M_{22}$. If it does, construct the set of basis from the spanning set.
Pick an arbitrary matrix in $M_{22}=\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$.If we find the following system to be consistent, then this set will span $M_{22}$
$a_1\left[\begin{array}{cc}-1&1\\1&-1\end{array}\right]+ a_2\left[\begin{array}{cc} 1&1\\1&1\end{array}\right]+a_3 \left[\begin{array}{cc} 1&0\\0&1\end{array}\right]+a_4\left[\begin{array}{cc}0&1\\1&0\end{array}\right]=\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$ which simplifies to
$$-a_1+a_2+a_3=x$$
$$a_1+a_2+a_4=y$$
$$a_1+a_2+a_4=z$$
$$-a_1+a_2+a_3=w$$
By visually inspecting, we can see that eq #1 & 4 are the same and eq#2 & 3 are the same. So, if we perform row operations, we will end up with an inconsistent system. So, the set do no span $M_{22}$ and we cannot construct the basis.
Am I on the right track? Appreciate your feedback.
| Even if the set doesn't span $M_{22}$, it span some subspace $\subseteq M_{22}$; we need to find the basis for such subspace.
To find it you can proceed as for ordinary row/colum vectors by RREF (in this way you could solve also part 1 by this) and selecting matrices corresponding to the pivot rows.
Notably, collecting the matrices by rows with the entries ordered as for $x,y,z,w$, we have to reduce the following
$$\begin{bmatrix}-1&1&1&-1\\1&1&1&1\\1&0&0&1\\0&1&1&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\1&1&1&1\\0&-1&-1&0\\0&0&0&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\0&2&2&0\\0&-1&-1&0\\0&0&0&0 \end{bmatrix}\to \begin{bmatrix}-1&1&1&-1\\0&2&2&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}$$
from here we can conlude that the subspace spanned has dimension 2 and that we can take the first two matrices as a basis for it.
| {
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"url": "https://math.stackexchange.com/questions/2673729",
"timestamp": "2023-03-29T00:00:00",
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Easier way to identify a 3 variable determinant as symmetric polynomial How do I identify a 3 variable determinant whether it is a symmetric polynomial or not without actually expanding it ?
Examples:
$$
\begin{vmatrix}
1+a^2&ab&ac\\
ab&1+b^2&bc\\
ca&cb&1+c^2\\
\end{vmatrix}=1+a^2+b^2+c^2
$$
$$\begin{vmatrix}
x+a&y&z\\
x&y+a&z\\
x&y&z+a\\
\end{vmatrix}=a^2(x+y+z+a),\\\begin{vmatrix}
x-y-z&2x&2x\\
2y&y-z-x&2y\\
2z&2z&z-x-y\\
\end{vmatrix}=(x+y+z)^3
$$
$$
\begin{vmatrix}
3x&-x+y&-x+z\\
-y+x&3y&-y+z\\
-z+x&-z+y&3z
\end{vmatrix}=3(x+y+z)(xy+yz+zx),
\begin{vmatrix}
(y+z)^2 & x^2 & x^2 \\
y^2 & (z+x)^2 & y^2 \\
z^2 & z^2 & (x+y)^2 \\
\end{vmatrix}=2xyz(x+y+z)^3
$$
$$
\begin{vmatrix}
(y+z)^2&xy&zx\\
xy&(x+z)^2&yz\\
xz&yz&(x+y)^2
\end{vmatrix}=2xyz(x+y+z)^3
,
\begin{vmatrix}
y+z&z&y\\
z&z+x&x\\
y&x&x+y
\end{vmatrix}=4xyz\\
\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(y-z)^2+(z-x)^2\Big]^2
$$
$$
\begin{vmatrix}
1+a^2-b^2&2ab&-2b\\
2ab&1-a^2+b^2&2a\\
2b&-2a&1-a^2-b^2\\
\end{vmatrix}=(1+a^2+b^2)^3
$$
My Attempt:
Lets take the example $\Delta=\begin{vmatrix}
y+z&z&y\\
z&z+x&x\\
y&x&x+y
\end{vmatrix}$.
$$
\Delta_{x\leftrightarrow y}=\begin{vmatrix}
x+z&z&x\\
z&z+y&y\\
x&y&x+y
\end{vmatrix}=-\begin{vmatrix}
z&z+y&y\\
x+z&z&x\\
x&y&x+y
\end{vmatrix}=\begin{vmatrix}
y+z&z&y\\
z&z+x&x\\
y&x&x+y
\end{vmatrix}=\Delta
$$
$$
\Delta_{x\leftrightarrow z}=\begin{vmatrix}
y+x&x&y\\
x&z+x&z\\
y&z&z+y
\end{vmatrix}
=-\begin{vmatrix}
y&z&z+y\\
x&z+x&z\\
y+x&x&y\\
\end{vmatrix}=
\begin{vmatrix}
z+y&z&y\\
z&z+x&x\\
y&x&y+x\\
\end{vmatrix}=\Delta
$$
$$
\Delta_{y\leftrightarrow z}=\begin{vmatrix}
y+z&y&z\\
y&y+x&x\\
z&x&x+z
\end{vmatrix}
=-\begin{vmatrix}
y+z&y&z\\
z&x&x+z\\
y&y+x&x\\
\end{vmatrix}=
\begin{vmatrix}
z+y&z&y\\
z&z+x&x\\
y&x&y+x\\
\end{vmatrix}=\Delta
$$
Thus $\Delta(x,y,z)$ is a symmetric polynomial.
Is there a better and easier way to identify such determinants ?
My Observation:
It may be a stupid observation while looking for a shortcut to the mentioned problem. For each of the cases the product of diagonal terms and product of the rest are separately symmetric polynomials. If not possible then we will do row or column operation and check the same. Can I use it somehow to identify the determinant ?
Example:
$$
\begin{vmatrix}
a^2&bc&ac+c^2\\
a^2+ab&b^2&ac\\
ab&b^2+bc&c^2\\
\end{vmatrix}=4a^2b^2c^2
$$
is a symmetric polynomial of degree 6.
Here,
Product of the diagonal terms, $a^2b^2c^2$ is a symmetric polynomial.
Product of the rest,
$$
(a^2+ab)(ac+c^2)(b^2+bc).ab.bc.ca=a(a+b).b(b+c).c(a+c).a^2b^2c^2\\
=a^3b^3c^3.(a+b)(b+c)(c+a)
$$
which is also a symmetric polynomial.
Note:
The reason why I am asking this is that If I can quickly identify by just looking at it using some shortcuts rather than actually checking for each cases, atleast for a $3x3$ matrix with 3 variables, it'd help ease to factorize the determinant using factor theorem. For example, check
Prove $\Delta=\left|\begin{smallmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\-z+x&-z+y&3z\end{smallmatrix}\right|=3(x+y+z)(xy+yz+zx)$
| I don't think there's a systematic answer. But a few observations.
(A) I think it is obvious that the examples you give satisfy the $x\to y\to z\to x$ symmetry, so you only need check that swapping $x$ and $y$ is a symmetry. Or if you don't want to do that, ask what would happen if swapping $x$ and $y$ were not a symmetry: in that case we'd need to have a factor $(x-y)(y-z)(z-x)$, wouldn't we?[proof omitted]. Given the small degree of the determinant that's not going to be possible.
(B) Most of these specific examples are determinants of matrices which can be written as $\alpha I + \beta J +\gamma J^2 + D$, where $J$ is the circulant matrix and $D$ is diagonal. In this form it is easier (at least for me) to check the symmetry. (For any $M$, $\det M =\det M \det J$.)
(C) The first case you give is actually easier to tackle by regarding $a$ as the variable: it's trivial to compute the coefficients of $a^3, a^2, a, 1$. This is often a good way to tackle a determinant of the form $A+t I$ where $A$ is free of $t$.
(D) The example with the $yz-x^2$ entries is easier (as I remarked on your previous question) if we spot that it is a matrix of cofactors and so a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of a function tending to a finite number If
$$\lim_{x\to 0} \frac{ae^x - b\cos x +ce^{-x}}{x\sin x} = 2$$
then find the value of $a+b+c$.
My book has given the following solution to the above problem :-
We observe that as $x$ tends to zero , numerator tends to $a-b+c$ whereas the denominator tends to zero. Therefore for the limit to exist , we must have ,$a-b+c=0$
Now I am really confused at this point. Why would we want the numerator to attain the value of $0$ . Wouldn’t that give us an indeterminate answer? But actually it’s suposed to be two . Can you please explain ? Thank you for your help.
| Hint. By using Taylor expansions at $0$,
$$\frac{ae^x - b\cos x +ce^{-x}}{x\sin x} =\frac{a(1+x+\frac{x^2}{2}) - b(1-\frac{x^2}{2}) +c(1-x+\frac{x^2}{2})+o(x^2)}{x(x+o(x))}\\
=\frac{(a-b+c)+(a-c)x+\frac{(a+b+c)}{2}x^2+o(x^2)}{x^2+o(x^2)}.$$
In order to have the final limit $2$ we need that
$$(a-b+c)=0,\quad(a-c)=0,\quad \frac{(a+b+c)}{2}=2.$$
Can you take it from here?
P.S. If you prefer to use L'Hopital begin with your initial observation
$$\lim_{x\to 0}ae^x - b\cos x +ce^{-x}=a-b+c.$$
If $a-b+c=0$ then, by L'Hopital,
$$\lim_{x\to 0}\frac{ae^x - b\cos x +ce^{-x}}{x}=\lim_{x\to 0}\frac{ae^x + b\sin x -ce^{-x}}{1}=a-c.$$
If $a-b+c=0$ AND $a-c=0$ then, by L'Hopital (twice),
$$\begin{align}
\lim_{x\to 0}\frac{ae^x - b\cos x +ce^{-x}}{x^2}
&=\lim_{x\to 0}\frac{ae^x + b\sin x -ce^{-x}}{2x}
\\&=\lim_{x\to 0}\frac{ae^x + b\cos x +ce^{-x}}{2}=\frac{a+b+c}{2}.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Checking my proof of some inequality I would like to receive some help about the next problem:
The Problem:
Prove that it is
$$\frac{1}{n^{1 + \alpha}} < \frac{1}{\alpha} \left(\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right)$$,
if we have that the next is valid:
$$\frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = \frac{-\alpha}{(n - \theta)^{\alpha + 1}} \;, 0 < \theta < 1$$,
where $\alpha \in \Bbb{R}$, $\alpha > 0$ and $n \in \Bbb{N} \setminus \{1\} $.
My solution:
$$ n - \theta < n \Rightarrow (n - \theta)^{\alpha + 1} < n^{\alpha + 1} \Rightarrow \frac{1}{(n - \theta)^{\alpha + 1}} > \frac{1}{n^{\alpha + 1}} \Rightarrow \frac{\alpha}{(n - \theta)^{\alpha + 1}} > \frac{\alpha}{n^{\alpha + 1}} \Rightarrow \frac{-\alpha}{(n - \theta)^{\alpha + 1}} < \frac{-\alpha}{n^{\alpha + 1}}.$$
$$ \frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} = \frac{-\alpha}{(n - \theta)^{\alpha + 1}} \Rightarrow
\frac{1}{n^\alpha} - \frac{1}{(n - 1)^\alpha} < \frac{-\alpha}{n^{\alpha + 1}} \Rightarrow -\left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) < \frac{-\alpha}{n^{\alpha + 1}} \Rightarrow
\frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} > \frac{\alpha}{n^{\alpha + 1}} \Rightarrow \frac{1}{\alpha} \left( \frac{1}{(n - 1)^\alpha} - \frac{1}{n^\alpha} \right) > \frac{1}{n^{\alpha + 1}}.$$
Please, could you tell me is my solution correct and if not why?
Thank you, for your time and help!
| We can't afford to have $n=0$, as $n$ appears in the denominator. Also, by removing $n=0$, we are sure that $n-\theta >0$ and hence you can raise positive power and preserve the inequality sign in the first line of the solution.
Also, we can't have $n=1$ as $n-1$ appears in the denominator.
Other than that, that is if $n \ge 2$, the proof seems ok.
| {
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"timestamp": "2023-03-29T00:00:00",
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Symmetric Olympiad inequality Prove that for all positive reals $a, b, c$ this inequality holds:
$$\sum_{cyc}\frac{a^3}{b^2 - bc + c^2} \ge \sum_{cyc}a$$
I have proved this in a very ugly way:
I multiplied with $(a^2 - ab + b^2)(b^2 - bc + c^2)(c^2 - ca + a^2)$ and with some work i managed to prove this with Schur. I find this way boring and time-consuming. My question is - Is there a nicer solution to this?
Source: Mildorf Inequalities
| Using only algebraic manipulation and the fact that a product of similarly-signed numbers is positive:
\begin{gather*}
\sum_\text{cyc} \frac{a^3}{b^2 - bc + c^2} - \sum_\text{cyc} a
= \sum_\text{cyc} \frac{a^3(b + c) - a(b^3 + c^3)}{b^3 + c^3} \\
= \sum_\text{cyc} \frac{ab(a^2 - b^2) - ca(c^2 - a^2)}{b^3 + c^3}
= \sum_\text{cyc} bc(b^2 - c^2)\left(
\frac{1}{c^3 + a^3} - \frac{1}{a^3 + b^3}\right) \geqslant 0,
\end{gather*}
because all three terms in the final cyclic sum are non-negative.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Finding $\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ I am having some diffuculty finding the limit for this expression and would appreciate if anyone could give a hint, as to how to continue. I know the limit must be $e^{14}$ (trough an engine) and I can show it for
$(1+\frac{2}{x}+\frac{1}{x^2})^{7x}$ like
$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{1}{x^2})^{7x} = ((1+\frac{1}{x})^2)^{7x} = (1+\frac{1}{x})^{x \cdot 7 \cdot 2} = e^{7 \cdot 2} = e^{14}$
However the $\frac{3}{x^2}$ in $(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ is causing problems for me I have:
$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x} = (1+\frac{2}{x}+\frac{1}{x^2} +\frac{2}{x^2})^{7x} = ((1+\frac{1}{x})^2+\frac{2}{x^2})^{7x} =...$
but I'm not sure how to continue (how to get rid of $\frac{2}{x^2}$) I was thinking that using the binomial theorem might somehow reduce the$\frac{2}{x^2}$ for $\lim_{x\to \infty}$, but I'm not sure how to do that for an x in the exponent, or if it is useful, or if it is even allowed in this case.
If anyone could point me in the right direction, as to how to continue I would be very grateful.
| Note that
$$\left(1+\frac{2}{x}+\frac{3}{x^2}\right)^{7x}=\left[\left(1+ \frac{2}{x}+\frac{3}{x^2}\right)^{\frac1{\left(\frac{2}{x}+\frac{3}{x^2}\right)}}\right]^{{7x}{\left(\frac{2}{x}+\frac{3}{x^2}\right)}}\to e^{14}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose $A \in M_{3,3}$ with eigenvalues $1,2,3$, eigenvectors $b_1, b_2, b_3$. Let $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$. I think I did this right, but someone skim and double check?
Q: Suppose $A$ is a $3 \times 3$ matrix, with eigenvalues $1,2,3$ and corresponding eigenvectors $b_1, b_2, b_3$. Suppose that $v = b_1 - 4b_2 + 3b_3$. Compute $A^5 v$.
Let $\mathcal{B}$ be the eigenvector basis and $D$ be the diagonal eigenvalue matrix:
\begin{align*}
D &= \begin{pmatrix}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 3 \\
\end{pmatrix} \\
P_{\mathcal{BE}} &= \begin{pmatrix} \begin{pmatrix} b_1 \end{pmatrix} & \begin{pmatrix} b_2 \end{pmatrix} & \begin{pmatrix} b_3 \end{pmatrix} \end{pmatrix} \\
[v]_{\mathcal{B}} &= (1,-4,3)^T \\
A &= P_{\mathcal{BE}}^{-1} D P_{\mathcal{BE}} \\
A^5 &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} \\
A^5 v &= P_{\mathcal{BE}}^{-1} D^5 P_{\mathcal{BE}} [v]_{\mathcal{E}} \\
[A^5 v]_{\mathcal{B}} &= D^5 [v]_{\mathcal{B}} \\
[A^5 v]_{\mathcal{B}} &= (1^5 \cdot 1, 2^5 \cdot -4, 3^5 \cdot 3)^T \\
[A^5 v]_{\mathcal{B}} &= (1, -128, 729)^T \\
A^5 v &= b_1 - 128 b_2 + 729 b_3 \\
\end{align*}
| That's too compicated. You might have done just this: Since
*
*$A^5.b_1=b_1$;
*$A^5.(-4b_2)=-4A^5.b_2=-4\times2^5b_2=-128b_2$;
*$A^5.(3b_3)=3A^5.b_3=3\times3^5b_3=729b_3$,
then $A.(b_1-4b_2+3b_3)=b_1-128b_2+729b_3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Elementary Number Theory: Solving Quadratic Conguences If possible solve the following congruence,
$7x^2-4x+1 \equiv 0 \pmod {11}$.
I am using the quadratic equation but I am stuck. The following is where I got stuck:
$x \equiv 4(4\pm \sqrt{-12}\pmod {11}$
I have tried adding $11$ repeatively but I cannot get a perfect square root. Please help!
| So quadratic formula:
$7x^2 -4x + 1\equiv 0 \mod 11\implies$ (using the standard notation abuse)
$x \equiv \frac {4 \pm \sqrt {16 - 28}}{14}\mod 11$
$\frac 12 \equiv 6$ and $\frac 17 \equiv 8$ so $\frac 1{14}\equiv 48 \equiv 4$.
So $x \equiv {16 \pm \sqrt{-12} } \equiv 5\pm \sqrt{-1}$
Now $x^2 \equiv -1 \mod 11$ has no solutions as $11 \not \equiv 1 \mod 4$.
So no solutions.
Or we could try doing this directly:
$7x^2 -4x + 1\equiv 0$
$-4x^2 - 4x \equiv -1$
$4x^2 - 4x \equiv 1$
$4x^2 + 4x \equiv 1$
$4x(x+1)\equiv 1$
$x(x+1)\equiv -x(-x - 1) \equiv 3 \mod 11$.
We could simply test $6$ cases: $x=0;\pm 1$ are trivial. $\pm 2;3;4;5$ are easy.
Or we can simply list the quadratic residues mod $11$ are $0, 1,4,9,5,3$.
And completing the square:
$7x^2 - 4x + 1\equiv 0$
$\frac 17(7x^2 - 4x + 1) \equiv 0$
$8(7x^2 - 4x +1) \equiv 0$
$x^2 - 32x + 8 \equiv 0$
$x^2 - 32x \equiv -8$
$x^2 -10x + 25 \equiv 17\equiv 6$
$(x-5)^2 \equiv 6 \mod 11$ and $6$ is not a modulo class.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Find the largest cylinder inscribed inside a sphere. Is this calculation correct so far?
A right circular cylinder is inscribed in a sphere of radius $r$. Find the largest possible volume of such a cylinder.
I have that the radius of the cylinder is $r$, the radius of the sphere is $R$, and the height of the inscribed cylinder is $h$. So
$$(2r)^2+h^2=(2R)^2$$
$$4r^2 + h^2 = 4R^2$$
$$h=2\sqrt{R^2-r^2}$$
So volume of the cylinder is:
$$V=2\pi r^2\sqrt{R^2-r^2}$$
$$V'=4\pi r\sqrt{R^2-r^2}+2\pi r^2\frac d{dr}\sqrt{R^2-r^2}$$
And I think:
$$\frac d{dr}\sqrt{R^2-r^2}=\frac r{\sqrt{R^2-r^2}}$$
so
$$V'=4\pi r\sqrt{R^2-r^2}+\frac{2\pi r^3}{\sqrt{R^2-r^2}}$$
$$=4\pi r(R^2-r^2)+2\pi r^3$$
$$4\pi rR^2-4\pi r^3+2\pi r^3=4\pi rR^2-2\pi r^3$$
So finding critical values:
$$4\pi r R^2-2\pi r^3=0$$
$$4\pi r R^2=2\pi r^3$$
$$4\pi R^2=2\pi r^2$$
$$\frac{4\pi R^2}{2\pi}=r^2$$
$$r=\sqrt2R$$
Someone else who is better at math is getting $r=\sqrt{\frac23}R$. Where did I go wrong?
| You made a sign error in computing the derivative of $\sqrt{R^2-r^2}$.
$$\frac d{dr}\sqrt{R^2-r^2}=\frac{-r}{\sqrt{R^2-r^2}}$$
Thus
$$V'=4\pi r\sqrt{R^2-r^2}-\frac{2\pi r^3}{\sqrt{R^2-r^2}}=0$$
$$4\pi r(R^2-r^2)-2\pi r^3=0$$
$$2(R^2-r^2)=r^2$$
$$2R^2=3r^2$$
$$r=\sqrt{\frac23}R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685819",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 2,
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Probability that A wins a best of 7 (4 matches) $A$ and $B$ play a series of best of $4$. The probability that $A$ wins a given game is $p$. Assume that the games are independent of each other. What is the probability that $A$ wins?
Let $A_{i}$ be the event that $A$ wins the series in $i$ matches.
Then, the book states, $$P(A) = P(A_{4}) + P(A_{5}) + P(A_{6}) + P(A_{7}) = p^{4} + \binom{4}{3}p^{4}q + \binom{5}{3}p^{4}q^{2} + \binom{6}{3}p^{4}q^{3}.$$
I am having a bit of trouble understanding the calculation. I am trying to be a bit more formal here. As an example, lets concentrate on $A_{5}$. It really is $A \cap G_{5}$ where $G_{i}$ is the event that $i$ matches were played in the series. Then, $$P(A_{5}) = P(A \cap G_{5}) = P(A|G_{5})P(G_{5}) = \binom{5}{4}p^{4}qP(G_{5}).$$ Now, we equate this term to $P(A_{5})$ to get $\binom{5}{4}p^{4}qP(G_{5}) = \binom{4}{3}p^{4}q$, from which we get, $$P(G_{5}) = \frac{\binom{4}{3}}{\binom{5}{4}}.$$
Is this complete nonsense? If not, can someone give me an intuition for $P(G_{5})$? It just doesn't click in my head right now. To rephrase, $P(A_{5}) = \binom{4}{3}p^{4}q$ was given in the answer, but in my attempt, I have this $P(G_{5})$ term which I am not sure how to find without the equation $\binom{5}{4}p^{4}qP(G_{5}) = \binom{4}{3}p^{4}q$.
| I don't see any way to go with your approach without utilizing the book's approach but I will do so anyways.
For $G_5$ to occur, either team $A$ needs to win $3$ or the first $4$ games and then win game $5$ with probability $$\binom{4}{3}p^{4}q$$ or team $B$ needs to win $3$ or the first $4$ games and then win game $5$ with probability $$\binom{4}{3}q^{4}p$$
so $$P(G_5)=\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p$$
and $$P(A\mid G_5)=\frac{\binom{4}{3}p^{4}q}{\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p}$$
Then
$$P(A|G_{5})P(G_{5})=\frac{\binom{4}{3}p^{4}q}{\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p}\cdot\left(\binom{4}{3}p^{4}q+\binom{4}{3}q^{4}p\right)=\binom{4}{3}p^{4}q$$
Of course it's easier to just note that team $A$ must win $3$ of the first $4$ games and then win the $5^{th}$ game. This is a negative binomial with $n$ trials given $k$ successes where $n=5$ and $k=4$.
We have
$$P(X=n)={n-1 \choose k-1}p^kq^{n-k}={4 \choose 3}p^4q$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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prove that for any nonsingular matrix $A$ there exist $X$ such that $X^2=A$ Prove that given any matrix A, where $$\det(A)\neq0$$ $$A\in M_{n,n}(\mathbb C)$$
the following equation
$$X^2=A$$
always has a solution.
Should I do something with Jordan Normal form?
Any help will be appreciated
| Swear i have done this one fairly recently...
$$
\left(
\begin{array}{rr}
t & \frac{1}{2t} \\
0 & t \\
\end{array}
\right)^2 =
\left(
\begin{array}{rr}
t^2 & 1 \\
0 & t^2
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{rrr}
t & \frac{1}{2t} & \frac{-1}{8 t^3} \\
0 & t & \frac{1}{2t} \\
0 & 0 & t
\end{array}
\right)^2 =
\left(
\begin{array}{rrr}
t^2 & 1 & 0\\
0 & t^2 & 1 \\
0 & 0 & t^2
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{rrrr}
t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5} \\
0 & t & \frac{1}{2t} & \frac{-1}{8 t^3}\\
0 & 0 & t & \frac{1}{2t}\\
0 & 0 & 0 & t
\end{array}
\right)^2 =
\left(
\begin{array}{rrrr}
t^2 & 1 & 0 & 0\\
0 & t^2 & 1 & 0 \\
0 & 0 & t^2 & 1 \\
0 & 0 & 0 & t^2
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{rrrrr}
t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5}& \frac{-5}{128 t^7} \\
0 & t & \frac{1}{2t} & \frac{-1}{8 t^3} & \frac{1}{16 t^5}\\
0 & 0 & t & \frac{1}{2t} & \frac{-1}{8 t^3}\\
0 & 0 & 0 & t & \frac{1}{2t} \\
0 & 0 & 0 & 0 & t \\
\end{array}
\right)^2 =
\left(
\begin{array}{rrrrr}
t^2 & 1 & 0 & 0 & 0\\
0 & t^2 & 1 & 0 & 0 \\
0 & 0 & t^2 & 1 & 0\\
0 & 0 & 0 & t^2 & 1 \\
0 & 0 & 0 & 0 & t^2
\end{array}
\right)
$$
$$ $$
And
$$ \sqrt{t^2 + 1} \; \; = \; \; t \; \; \sqrt{1 + \frac{1}{t^2}} \; \; = \; \; t + \frac{1}{2t} - \frac{1}{8 t^3} + \frac{1}{16 t^5} -\frac{5}{128 t^7} + \frac{7}{256 t^9} -\frac{21}{1024 t^{11}} \cdots $$
The resemblance is not cosmetic or accidental. We have, in a Jordan block of size $n,$ an identity matrix $I$ and a nilpotent matrix $N$ with $N^n=0.$ We are asking for $\sqrt{t^2 I + N}.$ As with other real analytic functions, we can use the facts that $IN=NI$ commute to give a power series for the resulting matrix, and the series is finite because $N$ is nilpotent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
I need help with integrating $\int\frac{1}{(x^2-1)^2}dx$ The problem is:
$$\int\frac{1}{(x^2-1)^2}dx$$
I tried using substitution $u=x^2-1$ but that does not bring me got results. I get an integral:
$$\frac{1}{2}\int\frac{1}{u^2\sqrt{u+1}}$$
And that does not really make things any more simple. From here I tried using partial decomposition but didn't really get anywhere.
Any help with this would be much appreciated.
| $$\begin{align*}
\int \frac{1}{(x^2 - 1)^2}\,\mathrm{d}x &\equiv \int \frac{1}{\left((x + 1)\cdot(x - 1)\right)^2}\,\mathrm{d}x \\
&= \int \frac{1}{(x + 1)^2\cdot(x - 1)^2}\,\mathrm{d}x\tag{1}
\end{align*}$$
Decompose $(1)$ into partial fractions,
$$\int \frac{1}{(x - 1)^2\cdot(x + 1)^2}\,\mathrm{d}x = \frac 14\int \frac{1}{x + 1} + \frac{1}{(x + 1)^2} - \frac{1}{x - 1} + \frac{1}{(x - 1)^2}\,\mathrm{d}x\tag{2}$$
Apply linearity to split the integral and then use the substitution $u = x + 1$ for the first two terms and $u = x - 1$ for the last two terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
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Find out how many invertible and diagonal solutions $X^2-2X=0 $ has when $ X \in\Bbb{R}^{3\times 3}$ I want to find out how many invertible solutions and how many diagonal solutions matrix equation
has when
$ X \in\Bbb{R}^{3\times 3}.$
I have researched everywhere and can't seem to find the solution to this problem. I would be very thankful for a lead in the right way.
| If $X$ is invertible, then multiplying each side of
$$
X^2 - 2X = 0
$$
by the inverse of $X$ gives us that
$$
X - 2I = 0
$$
and so
$$
X = 2I.
$$
For the case where $X$ is diagonal, let
$$
X = \begin{pmatrix} a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c \end{pmatrix}.
$$
Then we have that
$$
0 = X^2 - 2x = \begin{pmatrix} a^2 & 0 & 0 \\
0 & b^2 & 0 \\
0 & 0 & c^2 \end{pmatrix}
- 2 \begin{pmatrix} a & 0 & 0 \\
0 & b & 0 \\
0 & 0 & c \end{pmatrix}
= \begin{pmatrix} a^2 - 2a & 0 & 0 \\
0 & b^2 - 2b & 0 \\
0 & 0 & c^2 - 2c \end{pmatrix}
$$
and so we have a solution if and only if $a^2 - 2a = b^2 - 2b = c^2 - 2c = 0$. This is equivalent to each of $a$, $b$, and $c$ being either $0$ or $2$, and so we see that there are $8$ solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$ is a rational number Question
Find all the rational values of $x$ at which $y=\sqrt{x^2+x+3}$
My attempt
Since we only have to find the rational values of $x$ and $y$, we can assume that
$$ x \in Q$$
$$ y \in Q$$
$$ y-x \in Q $$
Let$$ d = y-x$$
$$d=\sqrt{x^2+x+3}-x$$
$$d+x=\sqrt{x^2+x+3}$$
$$(d+x)^2=(\sqrt{x^2+x+3})^2$$
$$d^2 + x^2 + 2dx =x^2+x+3$$
$$d^2 +2dx = x +3$$
$$x = \frac{3-d^2}{2d-1}$$
$$d \neq \frac{1}{2}$$
So $x$ will be rational as long as $d \neq \frac{1}{2}$.
Now
$$ y = \sqrt{x^2+x+3}$$
$$ y = \sqrt{(\frac{3-d^2}{2d-1})^2 + \frac{3-d^2}{2d-1} + 3}$$
$$ y = \sqrt{\frac{(3-d^2)^2}{(2d-1)^2} + \frac{(3-d^2)(2d-1)}{(2d-1)^2} + 3\frac{(2d-1)^2}{(2d-1)^2}}$$
$$ y = \sqrt{\frac{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}{(2d-1)^2}} $$
$$ y = \frac{\sqrt{(3-d^2)^2 + (3-d^2)(2d-1) + 3(2d-1)^2}}{(2d-1)}$$
$$ y = \frac{\sqrt{d^4-2d^3+7d^2-6d+9}}{(2d-1)}$$
I know that again $d \neq \frac{1}{2}$ but I don't know what to do with the numerator. Help
| Hint: the numerator will be $(d^2-d+3)^2$
| {
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"question_score": "3",
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"answer_id": 1
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Evaluating Riemann Summation Limit: $\lim\limits_{n\to \infty}\sum\limits_{k=1}^n\frac{6(k-1)^2}{n^3}\sqrt {1+2\frac{(k-1)^3}{n^3}}$ $$\lim_{n\to \infty}\sum_{k=1}^n\frac{6(k-1)^2}{n^3}\sqrt {1+2\frac{(k-1)^3}{n^3}}$$
I have no idea how to approach this problem apart from trying to convert into a definite integral using the left Riemann sum formula, but I have failed.
| Hint: By Riemann sum we have
$$\lim_{n\to \infty}\sum_{k=1}^n\frac{6(k-1)^2}{n^3}\sqrt {1+2\frac{(k-1)^3}{n^3}}
\\=6\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\left(1-\frac{k}{n}\right)^2\sqrt {1+2\left(1-\frac{k}{n}\right)^3}\\=6\int_0^1 (1-x)^2\sqrt{1+2(1-x)^3}dx
\\=6\int_0^1x^2\sqrt{1+2x^3}dx$$
| {
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"url": "https://math.stackexchange.com/questions/2701164",
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"source": "stackexchange",
"question_score": "1",
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Throw $4$ red and $4$ black balls into $4$ pots (probability question)
Suppose we have $4$ red and $4$ black balls.
We throw these balls into $4$ pots.
Find the probability:
(a) each pot contains $1$ red and $1$ black ball.
(b) each pot contains $2$ balls.
Attempt. (a) Let $R_i,~B_i$ be the events that pot $i$ contains exactly $1$ red, $1$ black ball, respectively. The desired event is $E=R_1B_1R_2B_2R_3B_3R_4B_4$ and so:
$$P(E)=P(R_1B_1)\,P(R_2B_2|R_1B_1)\,P(R_3B_3|R_1B_1R_2B_2)\,\,P(R_4B_4|R_1B_1R_2B_2R_3B_3)$$
where:
$$P(R_1B_1)=P(R_1)P(B_1)=\frac{1}{4}\frac{1}{4},~
P(R_2B_2|R_1B_1)=P(R_2|R_1)P(B_2|B_1)=\frac{1}{3}\frac{1}{3}$$
etc so the desired probability is $1/(4!)^2.$
(b) I am not sure how to approach this.
Am I on the right path regarding (a)? How should I approach (b)?
Thanks in advance for helping.
|
Suppose we have $4$ red and $4$ black balls. We throws these balls into $4$ pots. Find the probability that each pot receives one red and one black ball.
Method 1: Assuming each ball lands in one of the pots, there are four places each of the eight balls could land. Therefore, there are $4^8$ ways to distribute the balls to the pots.
For the favorable cases, each red ball and lands in a different pot, as does each black ball. There are $4!$ ways to distribute each red ball to a different pot and $4!$ ways to distribute each black ball to a different pot. Hence, the number of favorable cases is $4!4!$.
Therefore, the probability that each pot receives one red and one black ball is
$$\Pr(\text{each pot receives one red and one black ball}) = \frac{4!4!}{4^8} = \frac{9}{1024}$$
Method 2: Assume the first red ball lands in some pot. The probability that the second red ball lands in a different pot is $3/4$. The probability that the third red ball lands in a pot different from the first two is $2/4$. The probability that the fourth red ball lands in a pot different from the first three is $1/4$. Hence, the probability that each red ball lands in a different pot is
$$\Pr(\text{each pot receives one red ball}) = 1 \cdot \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{3}{32}$$
Since the probability that each pot receives one black ball is the same, the probability that one red and one black ball lands in each pot is
$$\Pr(\text{each pot receives one red and one black ball}) = \frac{3}{32} \cdot \frac{3}{32} = \frac{9}{1024}$$
Suppose we have $4$ red and $4$ black balls. We throws these balls into $4$ pots. Find the probability that each pot receives one red and one black ball.
Line the balls up in some order to throw them in the pots. The number of favorable cases is the number of ways we can select two of the eight balls to be placed in the first pot, two of the remaining six balls to be placed in the second pot, two of the remaining four balls to be placed in the third pot, and both of the remaining two balls to be placed in the fourth pot. Hence, the number of favorable outcomes is
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
Therefore, the probability that two balls land in each pot is
$$\Pr(\text{each pot receives two balls}) = \frac{\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{4^8} = \frac{315}{8192}$$
| {
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"source": "stackexchange",
"question_score": "2",
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Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent? Here is Prob. 10, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:
Establish the convergence or the divergence of the sequence $\left( y_n \right)$, where
$$ y_n \colon= \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \ \mbox{ for } n \in \mathbb{N}. $$
My Attempt:
For each $n \in \mathbb{N}$, we have
$$
\begin{align}
y_{n+1} - y_n &= \left( \frac{1}{ (n + 1) +1} + \frac{1}{ (n+1) +2 } + \cdots + \frac{1}{2(n+1)} \right) - \left( \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right) \\
&= \left( \frac{1}{ n + 2} + \cdots + \frac{1}{2n+ 2 } \right) - \left( \frac{1}{ n+1} + \cdots + \frac{1}{2n} \right) \\
&= \frac{ 1 }{2n+1} + \frac{ 1}{2n+2} - \frac{1}{n+1} \\
&= \frac{1}{2n+1} - \frac{1}{2n+2} \\
&= \frac{1}{ (2n+1) (2n+2) } > 0,
\end{align}
$$
which shows that our sequence is monotonically increasing.
Also, for each $n \in \mathbb{N}$, we have
$$ y_n = \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} < \underbrace{\frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n} }_{ \mbox{ $n$ times } } = 1. $$
Thus our sequence, being monotonically increasing and bounded, is convergent.
Is what I've done so far all correct?
How to determine the limit of this sequence?
| What you did is fine.
Now, as far as the imit is concerned, note that$$y_n=\frac1n\left(\frac1{1+1/n}+\frac1{1+2/n}+\cdots+\frac1{1+n/n}\right).$$This is a Riemann sum.
| {
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"url": "https://math.stackexchange.com/questions/2704730",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Find derivative of $y=\sin^{-1}\frac{2x}{1+x^2}$
Find $\frac{dy}{dx}$ if $y=\sin^{-1}\frac{2x}{1+x^2}$
The solution is given as $\frac{2}{1+x^2}$. But is it a complete solution ?
My Attempt
$$
2\tan^{-1}x=\begin{cases}\sin^{-1}\frac{2x}{1+x^2},\quad |x|\leq 1\\
\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\; x>0\\
-\pi-\sin^{-1}\frac{2x}{1+x^2},\quad |x|>1 \;\&\;x>0\\
\end{cases}\\
\sin^{-1}\frac{2x}{1+x^2}=\begin{cases}2\tan^{-1}x,\quad |x|\leq 1\\
\pi-2\tan^{-1}x,\quad |x|>1 \;\&\; x>0\\
-\pi-2\tan^{-1}x,\quad |x|>1 \;\&\;x>0\\
\end{cases}\\
$$
Thus,
$$
\frac{dy}{dx}=\frac{d}{dx}\bigg[\sin^{-1}\frac{2x}{1+x^2}\bigg]=\begin{cases}
\frac{d}{dx}[2\tan^{-1}x]=\frac{2}{1+x^2},\quad |x|\leq 1\\
\frac{d}{dx}[\pm\pi-2\tan^{-1}x]=\frac{-2}{1+x^2},|x|>1
\end{cases}
$$
Is it correct ?
| Note that
*
*$(\sin^{-1}x)'=\frac{1}{\sqrt{1-x^2}}$
then apply chain rule $f(g(x))'=f'(g(x))g'(x)=\frac{\frac{2(1+x^2)-4x^2}{(1+x^2)^2}}{\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}}=\frac{2(1-x^2)}{(1+x^2)|1-x^2|}$
| {
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For which positive integers $k$ dividing $4p^3$, does the equation $x^3+px=k^2$ has an integer solution in $x$? Let $p$ be an odd prime. For which positive integers $k$ dividing $4p^3$, does the equation $x^3+px=k^2$ has an integer solution in $x$ ?
Since for integer $x$, $px+x^3$ is even, so $k$ is even, so $4|x^3+px$. Now if $x$ is even, then this implies $4|x$.
I am unable to deduce anything else. Please help.
| The question as stated is really just 12 different Diophantine equations:
\begin{align*}
&x^3+px=1 & &x^3+px=4 & &x^3+px=16 \\
&x^3+px=p^2 & &x^3+px=4p^2 & &x^3+px=16p^2 \\
&x^3+px=p^4 & &x^3+px=4p^4 & &x^3+px=16p^4 \\
&x^3+px=p^6 & &x^3+px=4p^6 & &x^3+px=16p^6
\end{align*}
As already noted, the LHS of each equation is even, so the 4 equations on the left have no solutions. Since $p$ is an odd prime, there can be no solutions to any of the remaining 8 Diophantine equations involving non-positive values for $x$, since the LHS would be non-positive while the RHS is positive.
Thus, the equation $x^3+px=4$ has only one solution $(x,p)=(1,3)$, and the equation $x^3+px=16$ has no solutions.
For the remaining 6 Diophantine equations, it is clear that any solution must have $x$ divisible by $p$. It is obvious that no solution to $x^3+px=4p^2$ or $x^3+px=16p^2$ can have $x\geq 2p$, and that there are only finitely many primes $p$ to check; we then see that $x^3+px=4p^2$ has only one solution $(x,p)=(3,3)$ and that $x^3+px=16p^2$ has no solutions.
Now consider the equation $x^3+px=4p^4$. Any solution must not only have that $x$ is divisible by $p$, but in fact that $x$ is divisible by $p^3$. Thus, if there is a solution, defining $y=p^3x$ and substituting into our equation yields that $p^9y^3+p^4y=4p^4$ must also have a solution. This is clearly absurd.
The exact same argument works for the equation $x^3+px=16p^4$, and the final 2 equations are handled with similar arguments (instead of divisibility by $p^3$ you get divisibility by $p^5$).
Therefore, the only solutions to any of the 12 Diophantine equations in consideration are $(x,p)=(1,3)$ for $x^3+px=4$ and $(x,p)=(3,3)$ for $x^3+px=4p^2$.
| {
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Prove that the sequence $b_n = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges. I need to prove that the sequence $b_1 = 1$, $b_{n+1} = \frac{1}{1+b_n}$ converges.
If $\lim_{n\to\infty} b_n = L$ then $\lim_{n\to\infty} b_{n+1} = L$, so $\lim_{n\to\infty} b_{n+1} = \frac{1}{1+\lim_{n\to\infty} b_n} \implies L = \frac{1}{1+L} \implies L^2 + L - 1 = 0 \implies L = \frac{-1+\sqrt{5}}{2}$.
So I know that if the sequence converges to $L$ then $L = \frac{-1+\sqrt{5}}{2}$. How can I prove that the sequence converges?
EDIT:
With the help of the answers I got my resolution:
Let's prove that $b_n$ converges by proving that the subsequences $b_{2n}$ and $b_{2n-1}$ are monotonic and bounded.
I want to prove that $\forall n \in \mathbb N$, $b_{2n} < b_{2(n+1)} < L$ and $b_{2n-1} > b_{2(n+1)-1} > L$.
I'm going to prove using induction.
First Step:
$\frac{1}{2} < \frac{3}{5} < \frac{-1 + \sqrt{5}}{2} \implies b_2 < b_4 < L $
and
$1> \frac{2}{3} > \frac{-1 + \sqrt{5}}{2} \implies b_1 > b_3 > L $
Second Step:
Let's prove that $b_{2n-1} > b_{2n+1} > L \implies b_{2n+1} > b_{2n+3} > L$
and $b_{2n} < b_{2n+2} < L \implies b_{2n+2} < b_{2n+4} < L$
$b_{2n-1}>b_{2n+1}>L \implies 1+b_{2n-1} > 1+b_{2n+1}>1+L \implies$
$\frac{1}{1+b_{2n-1}} = b_{2n} < \frac{1}{1+b_{2n+1}} = b_{2n+2} < \frac{1}{1+L} = L \implies $
$1+b_{2n} < 1+b_{2n+2}<1+L \implies \frac{1}{1+b_{2n}} = b_{2n +1} > \frac{1}{1+b_{2n+2}} = b_{2n+3} > \frac{1}{1+L} = L \implies$
$1+b_{2n+1} > 1+b_{2n+3} > 1+L \implies \frac{1}{1+b_{2n+1}} = b_{2n +2} < \frac{1}{1+b_{2n+3}} = b_{2n+4} < \frac{1}{1+L} = L$.
Thus $b_n$ converges.
| Here's another interesting way of proving its convergence (And might provide some nice insights into the limit): Define
$$b_n=\frac{h_n}{g_n}$$
where these are the numerator and denominator of $b_n$ (with no reductions to a simpler form after appyling the recursion formula). Then from your definition
$$b_{n+1}=\frac{h_{n+1}}{g_{n+1}}=\frac{1}{1+\frac{h_n}{g_n}}=\frac{g_n}{h_n+g_n}.$$
Hence,
$$h_{n+1}=g_n$$
and
$$g_{n+1}=h_n+g_n.$$
Rearranging slightly, we get
$$h_{n+1}=h_n+h_{n-1}$$
and
$$g_{n+1}=g_n+g_{n-1}.$$
Of course, we can find just by computing the first few terms that $h_1=h_2=1$, $g_1=1$, and $g_2=2$. However, this is simply the recursion formula for the fibonacci numbers! Thus, we know $h_n=F_n$ and $g_n=F_{n+1}$ (where $F_n$ is the $n$th fibonacci number). Therefore,
$$b_n=\frac{F_n}{F_{n+1}}.$$
At this point, there are many ways to show that the limit exists (And also show why you got the limit of $\frac{-1+\sqrt{5}}{2}$.).
| {
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Proving that $ \mid z+1\mid^{2}=2\mid z\mid^{2} \Leftrightarrow \mid z-1\mid^{2}=2 $ Let z be a complex number.
How does one prove, using the properties of the conjugate and the modulus, that:
$$ \mid z+1\mid^{2}=2\mid z\mid^{2}
\Leftrightarrow
\mid z-1\mid^{2}=2 $$
| An elementary solution:
First, note that $|x+iy|^2=x^2+y^2,|x+1+iy|^2=x^2+2x+1+y^2$, and $|z-1 |^2=(x-1)^2+y^2.$
We have then
$$ x^2+2x+1+y^2=2x^2+2y^2$$
or $$x^2-2x+1-2+y^2=0$$
which is equivalent with the following
$$(x-1)^2+y^2=2$$
and again, this is what we needed
$$|z-1 |^2=2.$$
| {
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Last three digits of $6^{2002}$ Find the last three digits of $6^{2002}$. I did some work and figured out that the last two digits is 36. Can anyone help me with the hundredth digit? By the way, I used modular arithmetic and the recursion method for the tens digit, but it fell short when I attempted to do the hundreds digit. Thank you in advance!
How I figured out the last two digits:
I used the formula $\frac {1}{10^k} [n-a_k (n)]$. The last digit is obviously 6. I obtained the tens digit this way: $\frac {1}{10}(6^{2002}-6)=\frac {6}{10}(6^{2001}-1)=\frac {3}{5} (6^{2001}-1)=\frac {3}{5}(6-1)(6^{2000}+6^{1999}+...+6^2+6+1)=3(6^{2000}+6^{1999}+...+6^2+6+1)\equiv3(6+6+...+6+6+1)=3(2000\cdot6+1)\equiv3 (mod 10)$
Therefore, the last two digits of $6^{2002}$ are $36$.
| $$6=1+5,6^{25n}=(1+5)^{25n}\equiv1\pmod{5^3}$$
$$\implies6^{25n-1}\equiv6^{-1}\equiv21$$
$$6^{25n+2}\equiv6^3(21)\pmod{5^36^3}$$
$$\equiv216\cdot21\pmod{2^35^3}\equiv?$$
| {
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Center manifold theorem example. Does somebody know an easy way to tackle these center manifold problems?
Consider the system
\begin{align}
\dot x&=ax^3+x^2y\\
\dot y &=-y+y^2+xy-x^3
\end{align}
(a) Determine an approximation for the Center Manifold of this system.
(b) Use Center Manifold theory to investigate the stability of the origin $(x,y)=(0,0)$ depending on the parameter $a$ (so for $a<0,a>0$ and $a=0$).
It has to be written in some standard form I guess:
\begin{align}
\dot{x}&=Cx + F(x,y)
\\
\dot{y}&=Py+G(x,y)
\end{align}
In this case this leads to:
$C=0$, $\quad$ $P=-1$, $\quad$ $F(x,y)=ax^{3}+x^{2}y$ $\quad$ and
$G(x,y)=y^2+xy-x^3$.
Next we define an $h(x)$ as:
$$h(x)=a_2x^2+a_3x^3+a_4x^4+...$$
We take $\frac{\partial}{\partial x}h(x) = 2a_2x+3a_3x^2+4a_4x^3+...$ and substitute this into:
$$\frac{\partial}{\partial x}h(x)(Cx+F(x,h(x)))=Ph(x)+G(x,h(x))$$
which gives:
$$(2a_2x+3a_3x^2+4a_4x^3+...)(ax^3+x^2h(x)) \\ =-(a_2x^2+a_3x^3+a_4x^4+...)+(h(x)^2+xh(x)-x^3)$$
Which leads to:
$$2aa_2x^4+3aa_3x^5+4aa_4x^6+2a_2h(x)x^3+3a_3h(x)x^4+4a_4h(x)x^5
\\
=-a_2h(x)^2x^2-a_2h(x)x^3+a_3h(x)^2x^3+a_3h(x)x^4+a_4h(x)^2x^4-a_2x^5+a_4h(x)x^5-a_3x^6-a_4x^7$$
And then take $O(x^2)$ and $O(x^3)$ and leave out the other terms.
I really doubt if this is the way to do it.
| For small $x$, the first equation tells us that $x$ moves slowly compared to the exponential decay of $y$. Now the (moving) equilibrium of the second equation is at the roots of that quadratic equation, giving roots at $y\approx 1-x$ and $y\approx -x^3$. The first is unstable and far away from $(0,0)$, the second tells us that the series for $h(x)$ starts with $-x^3$.
$$
h(x)=-x^3+c_4x^4+c_5x^5+...
$$
Inserted into the DE $h'(x)F(x,h(x))=-h(x)+G(x,h(x))$ results then in the lower power terms
\begin{align}
&x^5(-3+4c_4x+5c_5x^2+...)(a-x^2+c_4x^3+c_5x^4+....)
\\&=x^3-c_4x^4-c_5x^5-...+x^6(-1+c_4x+c_5x^2+...)^2+x^4(-1+c_4x+c_5x^2+...)-x^3
\end{align}
The third power terms on the right cancel. The 4th power terms give
$$
0=-c_4-1\implies c_4=-1.
$$
In the fifth power we get terms on both sides
$$
-3a=-c_5+c_4\implies c_5=3a-1
$$
Continuing to the 6th degree
$$
4c_4a=-c_6+1+c_5\implies c_6=7a
$$
and so on.
The slow flow along the center manifold follows approximately the differential equation
$$
\dot x = ax^3+x^2h(x)=ax^3-x^5-x^6+(3a-1)x^7+7ax^8+...
$$
as the fast flow in $y$ direction will quickly revert any deviations towards the curve $y=h(x)$.
For medium to large $a$ the stability is thus determined by the sign of $a$ while for small $a$ the picture can be refined to contain the next term
$$
\dot x=x^3(a-x^2)
$$
which gives a pitchfork bifurcation in direction of positive $a$ with stable equilibria at about $x=\pm\sqrt{a}$ and an unstable point at the origin.
| {
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Evaluate coordinate of turning point : $y={1-\sin x\over 1+\cos x}$ How do I find the coordinate of the turning of $(1)$
$$y={1-\sin x\over 1+\cos x}={u\over v}\tag1$$
Using the Quotient rule:
$${dy\over dx}={vu^{'}-v^{'}u\over v^2}$$
$u=1-\sin x$, $u^{'}=-\cos x$
$v=1+\cos x$, $v^{'}=-\sin x$
$${dy\over dx}={-\cos x(1+\cos x)+(1-\sin x)\sin x\over (1+\cos x)^2}$$
$$={-1-\cos x+\sin x\over (1+\cos x)^2}$$
Turning point happens ${dy\over dx}=0$.
$$1+\cos x=\sin x\tag2$$
How do I solve for $x$?
@King tut:
$$\sqrt{2}\sin(x-{\pi\over 4})=1$$
$$x={\pi\over 2}$$
$$y={1-\pi\over 1+\pi}$$
Turning point $({\pi\over 2},{1-\pi\over 1+\pi})$
| $y={1-\sin x\over 1+\cos x} = \frac{1}{1+\cos{x}}-\frac{\sin{x}}{1+\cos{x}} = \frac{1}{2} \left(1+\tan^2{\frac{x}{2}} \right) - \tan{\frac{x}{2}}$
$y' = \frac{1}{2}\tan{\frac{x}{2}}(1+\tan^2{\frac{x}{2}}) - \frac{1}{2}(1+\tan^2{\frac{x}{2}}) \stackrel{!}{=} 0 \Rightarrow \tan{\frac{x}{2}} = 1 \rightarrow x_t = \frac{\pi}{2}+2k\pi \; k \in \mathbb{Z}$
$y''(x) = \left( \frac{1}{2}(1+\tan^2{\frac{x}{2}})(\tan{\frac{x}{2}}-1) \right)' = \frac{1}{2}(\ldots)(\tan{\frac{x}{2}}-1) + \frac{1}{2}(1+\tan^2{\frac{x}{2}})\frac{1}{2}(1+\tan^2{\frac{x}{2}})$
$\Rightarrow y''(x_t) \gt 0$
| {
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Solve: $5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$ Solve to find the general value of $x$:
$$5\cos^2 (x) - 4\sin (x)\cos (x)+3\sin^2 (x)=2$$
My Attempt:
$$5(1-\sin^2 (x))-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$5-5\sin^2 (x)-4\sin (x)\cos (x)+3\sin^2 (x)=2$$
$$2\sin^2 (x)+4\sin (x)\cos (x)=3$$
| Guide:
$$2\sin^2(x)+4\sin(x)\cos(x)=3$$
$$2\sin(2x)=2+1-2\sin^2(x)$$
$$2\sin(2x) = 2+\cos(2x)$$
$$2\sin(2x)-\cos(2x)=2$$
The trick from here should help in solving the problem.
| {
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Find the horizontal asymptotes of $f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$ Taken from Thomas' Calculus 12e
Find the horizontal asymptote of the graph of:
$$f(x) = \frac{x^3-2}{\lvert x\rvert^3+1}$$
Solution: We calculate the limits as ${x \to \pm \infty}$
For $x\ge0$: $$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}=1$$
I understand all except why the answer is one. I actually cannot find out how to get the answer.
| I assume the function's domain is $\mathbb R$ except where $f$ is not defined namely $x=-1$.
Now if $\lim_{x \to \infty} f(x) = a$ for some $a \in \mathbb R$, then $y=a$ is a horizontal asymptote.
Similarly, if $\lim_{x \to -\infty} f(x) = b$, for some $b \in \mathbb R$, then $y=b$ is a horizontal asymptote.
For $x\to\infty$, we have
$$\lim\limits_{x\to\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to\infty}\frac{x^3-2}{x^3+1}$$
$$=\lim\limits_{x\to\infty}\frac{x^3/x^3-(2/x^3)}{x^3/x^3+(1/x^3)}=\lim\limits_{x\to\infty}\frac{1-(2/x^3)}{1+(1/x^3)}$$
$$=\frac{\lim\limits_{x\to\infty} [1-(2/x^3)]}{\lim\limits_{x\to\infty} [1+(1/x^3)]}=\frac{ [\lim\limits_{x\to\infty}1-\lim\limits_{x\to\infty}(2/x^3)]}{ [\lim\limits_{x\to\infty}1+\lim\limits_{x\to\infty}(1/x^3)]}$$
$$=\frac{ [1-\lim\limits_{x\to\infty}(2/x^3)]}{ [1+\lim\limits_{x\to\infty}(1/x^3)]} = \frac{ [1-0]}{ [1+0]} = \frac11 = 1$$
Thus, $y=1$ is a vertical asymptote
Similarly, for $x\to-\infty$, we have
$$\lim\limits_{x\to-\infty}\frac{x^3-2}{\lvert x\rvert ^3+1} =\lim\limits_{x\to-\infty}\frac{x^3-2}{(-x)^3+1}$$
$$= \cdots$$
$$= \cdots$$
$$= \frac{ [1-\lim\limits_{x\to-\infty}(2/x^3)]}{ [-1+\lim\limits_{x\to-\infty}(1/x^3)]} = \frac{ [1-0]}{ [-1+0]} = \frac1{-1} = -1$$
Thus, $y=-1$ is a vertical asymptote
Soooo...do you know why
$$0 = \lim\limits_{x\to-\infty}(2/x^3) = \lim\limits_{x\to\infty}(2/x^3) = \lim\limits_{x\to-\infty}(1/x^3) = \lim\limits_{x\to\infty}(1/x^3)$$
?
| {
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Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$ Show that there are no rational solutions on the ellipse $2x^2 + 3y^2 = 1$.
Want to make sure that my proof is correct.
Suppose there are rational solutions.
Note $2x^2 + 3y^2 = 1 \iff 2x^2 + 3y^2 = z^2\ |\ x,y.z \in \mathbb{Z}$
Since taking $(\frac{x}{z}, \frac{y}{z})$ as our rational solution, $2(\frac{x}{z})^2 + 3(\frac{y}{z})^2 = 1 \implies 2x^2 + 3y^2 = z^2$.
Now reducing $mod\ 3$, we have $2x^2 + 3y^2 \equiv 1 \ mod\ 3 \implies 2x^2 \equiv 1 \ mod\ 3 \implies -1x^2 \equiv 1\ mod\ 3 \implies x^2 \equiv -1\ mod\ 3 $, which is impossible.
| One minor, easily fixed, problem with your proof is that you are using $x$ and $y$ first to denote rational numbers in the equation $2x^2+3y^2=1$, and then to denote the numerators of those rational numbers. It would be better to write $x=a/c$ and $y=b/c$ and then consider the equation $2a^2+3b^2=c^2$ with $a,b,c\in\mathbb{Z}$ (and $c\not=0$).
What's missing is the tacit assumption $\gcd(a,c)=1$. As José Carlos Santos and Isaac Browne point out, from $2a^2\equiv c^2$ mod $3$, it follows that $3\mid c$ implies $3\mid a$ (and vice versa), so you can assume $c\equiv\pm1$ mod $3$ (since those are the only other residue classes). It follows that $2a^2\equiv(\pm1)^2=1$, which now falls in line with the rest of your proof.
Just to give an alternative proof that there are no integer solutions to $2a^2+3b^2=c^2$ (with $c\not=0$), the (tacit) assumption $\gcd(b,c)=1$ tells us $b$ and $c$ are not both even, so $3b^2\equiv c^2$ mod $2$ tells us both are odd. Working mod $8$ now tell us $c^2\equiv1$ mod $8$, while
$$2a^2+3b^2\equiv
\begin{cases} 3\text{ if }a\text{ is even}
\\5\text{ if }a\text{ is odd}\end{cases}$$
| {
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A closed form for infinite series $ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$ It is well known that $$ \sum _1 ^\infty \frac {1}{n^2} = \frac {\pi ^2}{6}$$ and $$ \sum _1 ^\infty \frac {1}{n^4} = \frac {\pi ^4}{90}$$
We also know that $$ \sum _1 ^\infty \frac {1}{n^3} $$
$$=1.202056903159594285399738161511449990764986292340498881.....$$
My question is:
Do we have a closed form for this series besides $$ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$$
| In addition to robojohn's answer, there are some formulas expressing $\zeta(3)$ (and other odd zeta values) in terms of powers of $\pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:
$$
\begin{aligned}
\zeta(3)&=\frac{7\pi^3}{180}-2\sum_{n=1}^\infty \frac{1}{n^3(e^{2\pi n}-1)},\\
\sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} &= -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right).
\end{aligned}
$$
Moreover, in this Math.SE post we have:
$$
\frac{3}{2}\,\zeta(3) = \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}.
$$
You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.
| {
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"url": "https://math.stackexchange.com/questions/2727593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving determinant with variables I have a problem that asks:
Prove that det$\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$
I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$
What I did:
det$\begin{pmatrix}
b && c \\
b^2 && c^2 \end{pmatrix}$ -
det$\begin{pmatrix}
a && c \\
a^2 && c^2 \end{pmatrix}$ +
det$\begin{pmatrix}
a && b \\
a^2 && b^2 \end{pmatrix}$
which gives
$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$
which simplifies to
$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$
and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.
| If a=b, b=c, c=a then determinant value is 0 so
(a-b)(b-c)(c-a) are factor of the determinant now comparing the leading coefficient so the Value of the determinant is (a-b)(b-c)(c-a)
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Explaining $\int_{-1}^1\frac{1}{1+x^2}\,dx = \frac{\pi}{2}$. How would you explain to a student that
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) = \frac{\pi}{2} $$
and not
$$ \int_{-1}^1\frac{1}{1+x^2}\,dx = \arctan(1) - \arctan(-1) \neq \frac{\pi}{4} - \frac{3\pi}{4} = -\frac{\pi}{2}$$
besides the obvious fact that $\arctan x$ cannot map to two distinct values?
| Well, you can also use the even function 'card',
i.e, since $f(x) = f(-x)$,
$$ \int_{-1}^{1}\frac{1}{1+x^2}=2\int_{0}^{1}\frac{1}{1+x^2}=2(\arctan 1 +\arctan 0) = 2 \, \left(\frac{\pi}{4} \right) =\frac{\pi}{2} $$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "10",
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Find Jordan canonical form and basis of a linear operator. Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be a linear operator such that: $T(x,y,z)=(-y-2z,x+3y+z,x+3z)$, I need to find a Jordan canonical form and a basis.
This is what i did:
In the first place, I found the associated matrix to this linear operator in the canonical basis which is this one:
$$A=\begin{pmatrix}
0 & -1 & -2 \\
1 & 3 & 1 \\
1 & 0 & 3 \\
\end{pmatrix}$$
After that i found the characteristic polynomial which is: $(\lambda-2)^3=0$ so we have this polynomial that has only one root with multiplicity 3.
After finding the eigenvalue I found the eigenvector associated to 2, which is $V_3=(-1,0,1)$.
Now the Jordan canonical form should be this one(If I have done it correctly):
$$J=\begin{pmatrix}
2 & 0 & 0 \\
1 & 2 & 0 \\
0 & 1 & 2 \\
\end{pmatrix}$$
We know the a Jordan basis is formed with these vectors: $B=v_1,v_2,v_3$
First I found $v_2$ :
$A.v_2=2.v_2+1.v_3$, which is equal to :
$$
\begin{align*}
\begin{cases}
2x+y+2z &=1\\
x+y+z&=0\\
x+z&=1
\end{cases}
\end{align*}
$$So $v_2=(-1,-1,1)$
The problem is here with the last vector: After doing the same operation I get to this point:
$A.v_1=2.v_1+1.v_2$, which is equal to:
$$
\begin{align*}
\begin{cases}
2x+y+2z &=1\\
x+y+z&=0\\
x+z&=1
\end{cases}
\end{align*}
$$ and clearly this system does not have solution... what am I doing wrong?
| Usually the Jordan normal form is as follow
$$J=\begin{pmatrix}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2 \\
\end{pmatrix}$$
By Jordan theorem we know that a matrix $P$ exists such that $$P^{-1}AP=J$$
let $$P=[v_1,v_2,v_3,v_4]$$
then P has to satisfy the following system: $$AP=PJ$$ that is in this case $$Av_1=2v_1\implies (A-2I)v_1=0$$ $$Av_2=v_1+2v_2\implies (A-2I)v_2=v_1$$ $$Av_3=v_2+2v_3\implies (A-2I)v_3=v_2$$
Once we have $v_1$ we can find $v_2$ and finally $v_3$, that is
$$(A-2I)v_1=0 \implies\begin{pmatrix}
-2 & -1 & -2 \\
1 & 1 & 1 \\
1 & 0 & 1 \\
\end{pmatrix}v_1=0 \implies v_1=(1,0,-1)$$
$$(A-2I)v_2=v_1 \implies\begin{pmatrix}
-2 & -1 & -2 \\
1 & 1 & 1 \\
1 & 0 & 1 \\
\end{pmatrix}v_2=v_1 \implies v_2=(-1,1,0)$$
$$(A-2I)v_3=v_2 \implies\begin{pmatrix}
-2 & -1 & -2 \\
1 & 1 & 1 \\
1 & 0 & 1 \\
\end{pmatrix}v_3=v_2 \implies v_3=(0,1,0)$$
| {
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"source": "stackexchange",
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Find the limit of sequence $\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}$ without using of derivatives and etc. I need to find limit of sequence
$$
\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)
$$
I tried to solve it and stopped here
$$
f(n+1) = \frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}+\frac{2n+1}{2^{n+1}}
$$
$$
2f(n+1) = 1+\frac{3}{2}+\frac{5}{2^{2}}+\cdots+\frac{2n-1}{2^{n-1}}+\frac{2n+1}{2^{n}}
$$
$$
2f(n+1) -f(n) = 1+ \left(1 + \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}\right) = 1 + g(n)
$$
I can find the limit of $g$, but what to do with the other parts?
| $$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+...+\frac{2n-1}{2^{n}}\right)=\sum _1^{\infty} \frac {2n-1}{2^n}=2\sum _1^{\infty} \frac {n}{2^n} -\sum _1^{\infty} \frac {1}{2^n}=2\sum _1^{\infty} \frac {n}{2^n} -1$$
Note that
\begin{align}
\sum _1^{\infty} \frac {n}{2^n}&=\left(\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\cdots\right)\\&=\left(\frac12+\frac14+\frac18 +\cdots\right) +\left(\frac14+\frac18+\frac1{16}+\cdots\right)+\left(\frac18+\frac1{16}+\frac1{32}+\cdots\right)+\cdots\\&=1+\frac12+\frac14+\frac18+\cdots=2
\end{align}
Thus we have $$\lim_{n \to \infty }\left(\frac{1}{2}+\frac{3}{2^{2}}+\frac{5}{2^{3}}+\cdots+\frac{2n-1}{2^{n}}\right)= 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Solve system:$yz(y+z-x)= a(x+y+z)\\ zx(z+x-y)= b(x+y+z)\\ xy(x+y-z)= c(x+y+z)$ Solve system:
$$yz(y+z-x)= a(x+y+z)\\
zx(z+x-y)= b(x+y+z)\\
xy(x+y-z)= c(x+y+z)$$
$$a, b, c>0$$
I can only solve the system at $a= 2, b= 5, c= 10$. I have tried all things but it's hard with me. Somebody help me?
| Solve equation 1 for $x$:
$$ x=-{\frac { \left( y+z \right) \left( -yz+a \right) }{yz+a}}$$
(unless $yz+a=0$, which I'll ignore for now).
Substitute this into the other two equations, and simplify. I get
$$ \eqalign{yz \left( y+z \right) \left( y{z}^{3}-ayz-a{z}^{2}-byz+{a}^{2}-ab
\right) &= 0\cr
yz \left( y+z \right) \left( z{y}^{3}-a{y}^{2}-ayz-cyz+{a}^{2}-ac
\right) &= 0\cr} $$
So either $y=0$, in which case $x = -z$, or $z=0$, in which case $x=-y$, or $y+z=0$, in which case $x=0$, or $$\eqalign{y{z}^{3}-ayz-a{z}^{2}-byz+{a}^{2}-ab &= 0\cr z{y}^{3}-a{y}^{2}-ayz-cyz+{a}^{2}-ac &= 0\cr}$$
Solve the first of these for $y$, substitute into the second, and simplify.
$$ \eqalign{y &= \frac{a (a-b-z^2)}{z(a+b-z^2)}\cr
\left( -{z}^{2}+a \right) & \left(c{z}^{6}+ \left( ab-2\,ac-2\,bc \right) {z}^{4}- \left( a+b \right)
\left( 2\,ab-ac-bc \right) {z}^{2}+ab(a-b)^2\right)
=0\cr}$$
Note the second factor in the left side of the last equation is a cubic in $z^2$.
Things may get somewhat messy if you solve this explicitly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let A be a matrix then $A^{50}$ is?
If $$A=\begin{bmatrix}1&0&0\\1&0&1\\0&1&0\\ \end{bmatrix}$$ then ${A^{50}}$ is?
How to calculate easily? What is the trick behind it? Please tell me.
| Although there are easier ways in this particular case, I will outline a trick involving the Cayley-Hamilton theorem.
We have $\det\begin{pmatrix}
1-X & 0&0\\
1& -X& 1\\
0&1&-X
\end{pmatrix}=-(X+1)(1-X)^2=Q(X)$. Hence there are two eigenvalues, namely $-1$ and $1$. We know that $A$ is diagonalizable if there are two independent eigenvectors belonging to the eigenvalue $1$, but that's not the case here.
By the Cayley-Hamilton theorem, we have that $-(A+1)(1-A)^2=0$. Now consider the polynomial $P(X)=X^{50}$. We can divide this polynomial by $Q(X)$ and obtain $$P(X)=S(X)Q(X)+R(X)$$
where $R(X)$ is a polynomial of degree at most $2$. Notice that $P(A)=R(A)$ by the Cayley-Hamilton theorem. Write $R(X)=a+bX+cX^2$. Now notice that $$50X^{49}=P'(X)=S'(X)Q(X)+S(X)Q'(X)+R'(X).$$
It follows that $50=P'(1)=R'(1)=b+2c$, here I used that $1$ is a double root and hence $Q'(1)=0$. Also, $1=P(1)=R(1)=a+b+c$ and $P(-1)=R(-1)=a-b+c$. Now you can solve for $a,b$ and $c$. You should find that $a=-24,b=0$ and $c=25$. It follows that $$A^{50}=R(A)=-24I+25A^2=-24I+25\begin{pmatrix}
1&0&0\\
1&1&0\\
1&0&1
\end{pmatrix}=\begin{pmatrix}
1&0&0\\
25&1&0\\
25&0&1
\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The square of the solution of the equation... The square of the solution of the equation
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
is equal to: ...
$x\sqrt{7} + \sqrt{8 - 3\sqrt{7}} - \sqrt{8 + 3\sqrt{7}} = 0$
$\implies x\sqrt{7} = \sqrt{8 + 3\sqrt{7}} - \sqrt{8 - 3\sqrt{7}}$
$\implies 7x^2 = \left(8 + 3\sqrt{7}\right) + \left(8 - 3\sqrt{7}\right) - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}}$
$\implies 7x^2 = 16 - 2\cdot\sqrt{64-63} = 14$
$\implies x^2 = 2$
I know how to do all the steps up to the part where the underlined section comes into play, can someone please explain where does this come from. Thanks!
| Note that $$(a-b)^2 = a^2 + b^2 - 2ab $$
Your confusion is the $$-2ab = - \underline{2\sqrt{\left(8+3\sqrt{7}\right)\cdot\left(8-3\sqrt{7}\right)}} $$
| {
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"source": "stackexchange",
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For what real values of $x$ does the series $\sum_{n=1}^{\infty} \frac{n}{5^{n-1}}(x+2)^n $ converge?
For what real values of $x$ does the series $\sum_{n=1}^{\infty} \frac{n}{5^{n-1}}(x+2)^n$ converge?
My attempt: $-1< (x+2)^n < 1$. Now, $-3 < x < -1$, so by the Leibniz test, the given series will converge in $-3 <x<-1$.
Is this correct?
| HINT
By ratio test
$$\left| \frac{(n+1)(x+2)^{n+1}}{5^{n}}\frac{5^{n-1}}{n(x+2)^n}\right|=\frac{n+1}{5n}\left|x+2\right|\to\frac{1}{5}\left|x+2\right|$$
then the series converges for $|x+2|<5 \implies -7<x<3$.
Then check separetely the cases $x=-7$ and $x=3$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Finding a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ I am trying to find a tight lower bound for $\left(\frac{1+x}{(1+x/2)^2}\right)^n$ as a function of $x$ and $n$ and for large $n$,
where $x$ changes with $n$ such that $\lim_{n\to\infty}x=0$.
I am not sure wether my approach to solve this is right or not, but this is what I did:
\begin{align*}\left(\frac{1+x}{(1+x/2)^2}\right)^n&=e^{n(\ln({1+x})-2\ln{(1+x/2)})}\\
&=e^{n(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots-2(\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{24}-\cdots))}\\
&=e^{n(-\frac{x^2}{4}+\frac{x^3}{4}-\frac{15x^4}{64}+\cdots)}\\
&\geq e^{n(-\frac{x^2}{4})}\\
&=1-(n\frac{x^2}{4})+(n\frac{x^2}{4})^2-\cdots \\
&\geq 1-(n\frac{x^2}{4})
\end{align*}
We know $\lim_{n\to\infty}x=0$, but we don't know whether $\lim_{n\to\infty}nx^2=0$ . Hence, the last inequality is not necessarily correct,
because the sum of the terms after $1-(n\frac{x^2}{4}) $ may not be greater than zero.
| Write $y = x/2$.
Then
$\left(\frac{1+x}{(1+x/2)^2}\right)^n
=\left(\frac{1+2y}{(1+y)^2}\right)^n
=\frac{(1+2y)^n}{(1+y)^{2n}}
$.
Since
$(1+2y)^n
=\sum_{j=0}^n \binom{n}{j}2^jy^j
$
and
$\frac1{(1+y)^{2n}}
=\sum_{k=0}^{\infty} \binom{2n+k-1}{k}(-1)^ky^k
$,
$\begin{array}\\
\frac{(1+2y)^n}{(1+y)^{2n}}
&=\sum_{j=0}^n \binom{n}{j}2^jy^j\sum_{k=0}^{\infty} \binom{2n+k-1}{k}(-1)^ky^k\\
&=\sum_{j=0}^n\sum_{k=0}^{\infty}y^{j+k} \binom{n}{j}2^j \binom{2n+k-1}{k}(-1)^k\\
&=\sum_{m=0}^{\infty}y^m\sum_{j=0}^n\binom{n}{j}2^j \binom{2n+m-j-1}{m-j}(-1)^{m-j}\qquad j+k = m, k = m-j\\
&=\sum_{m=0}^{\infty}y^m(-1)^m\sum_{j=0}^n\dfrac{n!(2n+m-j-1)!}{j!(n-j)!(m-j)!(2n-1)!}2^j (-1)^{j}\\
&=\dfrac{n!}{(2n-1)!}\sum_{m=0}^{\infty}y^m(-1)^m\sum_{j=0}^n\dfrac{(2n+m-j-1)!}{j!(n-j)!(m-j)!}2^j (-1)^{j}\\
\text{so}\\
\frac{(1+x)^n}{(1+x/2)^{2n}}
&=\dfrac{n!}{(2n-1)!}\sum_{m=0}^{\infty}(-1)^m2^{-m}x^m\sum_{j=0}^{\min(m, n)}\dfrac{(2n+m-j-1)!}{j!(n-j)!(m-j)!}2^j (-1)^{j}\\
\end{array}
$
With this,
you can get the power series.
Note:
Wolfy says this starts like
$1-\dfrac{nx^2}{4}+\dfrac{nx^3}{4}+\dfrac{n(n-7)x^4}{32}
-\dfrac{n(n - 3) x^5}{16} - \dfrac{n (n^2 - 33 n + 62) x^6}{384}+O(x^7)
$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Elementary proofs for Taylor expansion for natural logarithm For $|x|<1$, we have that
$$ \ln(1+x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \cdots $$
Is there any elementary(try not to use integration or differantiation) proof for the equality above?
Edit: I have changed my definition of elemantary.
| Here is a sketch, which if extended will find any particular coefficient without calculus, on the assumption a polynomial expression exists
Let's start with the definitions:
*
*$\exp(x)= 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$ and
*for positive $x$ we have $\exp(\ln(x))=x$
Suppose $\ln(1+x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots$
Clearly $a_0=0$ since $\exp(0)=1$ and so $\ln(1+0)=0$
Then we need $1+x = 1 + \frac{1}{1!}(a_1x+a_2x^2+a_3x^3+\cdots)+\frac{1}{2!}(a_1x+a_2x^2+\cdots)^2+\frac{1}{3!}(a_1x+\cdots)^3+\cdots$
which by
*
*matching coefficients of $x$ will give $a_1=1$, and
*matching coefficients of $x^2$ will give $a_2+\frac{a_1^2}{2!}=0$ so $a_2=-\frac12$, and
*matching coefficients of $x^3$ will give $a_3+\frac{2a_1a_2}{2!}+\frac{a_1^3}{3!}=0$ so $a_3=+\frac13$, and
*matching coefficients of $x^4$ will give $a_4+\frac{2a_1a_3+a_2^2}{2!}+\frac{3a_1^2a_2}{3!}+\frac{a_1^4}{4!}=0$ so $a_4=-\frac14$, and
*we can do something similar for later coefficients
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex integral $I=\int_{|z|=2} \frac{z^3 e^{\frac{1}{z}}}{z+1}dz$ I have this integral from someone who told me it has a nice answer.$$I=\int_{|z|=2} \frac{z^3 e^{\frac{1}{z}}}{z+1}dz$$ I tried to evaluate this using residues theorem, and expanded into series to find the residue at $\infty$$$f(z)=z^3[\sum_{n=0}^{\infty}(-1)^nz^n(\sum_{n=0}^{\infty} \frac{1}{(n!)z^n})]$$ And with Cauchy product:$$f(z)=z^3(\sum_{n=0}^{\infty}\sum_{k=0}^n \frac{(-1)^k z^{2k-n}}{(n-k)!})$$ Now I am lost, how can I find the coefficient of $z^{-1}$ in this double series? It seems that is not just a simple number... Or shall I use another method?
| Substitute $u = \frac 1z$, so the integral is over the circle $|u| = 1/2$, and is now clockwise. $z = \frac 1u$, so $dz = -\frac{1}{u^2} du$, and so the minus signs cancel
$$\begin{align}I &= \int_{|u| = \frac{1}{2}} \frac{\left(\frac{1}{u}\right)^3 e^u}{\frac{1}{u} + 1} \frac{1}{u^2}\, du\\
&=\int_{|u| = \frac{1}{2}} \frac{e^u}{u^4(u+1)}\,du.\end{align}$$
Then we have to find the residue of $f(z) = \frac{e^u}{u^4 (u+1)}$ at $0$, which can be done using a Taylor series as standard.
$$\begin{align} \frac{e^u}{u^4 (u+1)} &= \frac{1}{u^4}\left(1+u+\frac{1}{2}u^2 + \frac{1}{6}u^3+O(u^4)\right)\left(1-u+u^2-u^3 + O(u^4)\right) \\
&=\frac{1}{u^4}\left(1+u(1-1)+u^2\left(1-1+\frac{1}{2}\right)+u^3\left(\frac{1}{6}-\frac{1}{2}+1-1\right)+O(u^4)\right) \\
&=\frac{1}{u^4}\left(1+\frac{1}{2}u^2 - \frac{1}{3}u^3 + O(u^4) \right) \\
&=\frac{1}{u^4}+\frac{1}{2u^2}-\frac{1}{3u}+O(1).\end{align}$$
So, the residue of $f(z)$ at $0$ is $-\frac{1}{3}$, and the integral is $-\frac{2\pi i}{3}$.
| {
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"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$
Find $$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}$$
without L'Hôpital's rule.
My work:
1) I know that $$\lim_{x\rightarrow0}\frac{\sin x}{x}=1$$
2) Let $x=t-\frac{\pi}{6}$. Then
$$\lim_{x\rightarrow\frac{\pi}{6}}\frac{1-2\sin x}{\cos3x}=\lim_{t\rightarrow 0}\frac{3t}{\sin3t}\cdot\frac{1-2\sin \left(t-\frac{\pi}6\right)}{3t}$$
| Hint:
$1-2\sin(t-\frac{\pi}{6})=1-2(\sin t\cos\frac{\pi}{6}+\cos t\sin\frac{\pi}{6})=2\sin^2\frac{t}{2}-\sqrt{3}\sin t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
attempting to solve the diff equation $y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$ $$y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$$
$$y'\sqrt{(x^2+1)}-\frac{2xy}{\sqrt{(x^2+1)}}=4\sqrt{y}$$
$$\frac{y'\sqrt{(x^2+1)}-\frac{2xy}{\sqrt{(x^2+1)}}}{x^2+1}= \frac{4\sqrt{y}}{x^2+1}$$
I tried to solve this but I can't get rid of the 2 which would have enabled me to wrtie $$(\frac{y}{x^2+1})'$$
how do I solve this?
| It's Bernouilli's equation
$$y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$$
do this rather
$$y'y^{-1/2}(x^2+1)-2xy^{1/2}=4\sqrt{(x^2+1)}$$
$$2(y^{1/2})'(x^2+1)-2x(y^{1/2})=4\sqrt{(x^2+1)}$$
Substitute $z=y^{1/2}$
$$z'(x^2+1)-xz=2\sqrt{(x^2+1)}$$
Now you can solve it easily...
Edit
$$(\frac{y}{x^2+1})'= \frac{4\sqrt{y}}{x^2+1}$$
You could do this starting from where you were stuck
$$(\frac{y}{x^2+1})'= 4\sqrt{\frac{{y}}{x^2+1}}\frac{1}{\sqrt{x^2+1}}$$
$$\frac {d(\frac{y}{x^2+1})}{\sqrt{\frac{{y}}{x^2+1}}}= \frac{4dx}{\sqrt{x^2+1}}$$
And simply integrate...
$$\sqrt{\frac{{y}}{x^2+1}}= 2 \int \frac{dx}{\sqrt{x^2+1}}$$
But Bernouilli's method is easier
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the Equation of the Ellipse By Completing the Square I have the following equation of an ellipse:
$$5x^2+9y^2+40x=100$$
I need to put it in the form:
$$\frac{(x-h)^2}{m^2} + \frac{(y-k)^2}{n^2} =1 $$
I was trying to complete the square with the coefficients that have an $x$ variable.
$$5x^2 + 40x + 9y^2 = 100$$
$$5(x^2 + 8x) + 9y^2 = 100$$
$$5(x^2 + 8x + 16) + 9y^2 = 100 + 16$$
$$5(x+4)^2 + 9y^2 = 116$$
I then divided both sides of the equation by $116$.
$$\frac{5(x + 4)^2}{116} + \frac{9y^2}{116} = 1$$
However, when I graph the equation I do not get the same ellipse that was represented with the original equation of $5x^2 + 9y^2 + 40x = 100$. From the graph, it seems like they are two similar ellipses. Where am I making the mistake? Any help will be appreciated.
| We have
$$5x^2+9y^2+40x=100\iff 5(x+4)^2+9y^2=180\iff (x+4)^2+\frac{y^2}{\frac59}=36\\\iff\frac{(x+4)^2}{36}+\frac{y^2}{20}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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