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If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ If $\sin^2x+\sin^2y<1 \forall x,y \in R$, then prove that $\sin^{-1}(\tan x\cdot\tan y)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
My attempt is as follows:-
$$f(x)=\tan x\tan y$$
$$f(x)=\dfrac{\sin x\sin y}{\cos x\cos y}\tag{1}$$
Let's find out upper bound of $f(x)$, for that we need to find upper bound of $\sin x\sin y$ and lower bound of $\cos x\cos y$
$$\sin^2x+\sin^2y<1$$
$$(\sin x-\sin y)^2>=0$$
$$\sin^2x+\sin^2y-2\sin x\sin y>=0$$
$$\dfrac{\sin^2x+\sin^2y}{2}>=\sin x\sin y$$
$$\sin x\sin y<\dfrac{1}{2}\quad\forall x,y\tag{2}$$
$$(\cos x-\cos y)^2>=0$$
$$\cos^2x+\cos^2y-2\cos x\cos y>=0$$
$$1-\sin^2x+1-\sin^2y-2\cos x\cos y>=0$$
$$\dfrac{2-(\sin^2x+\sin^2y)}{2}>=\cos x\cos y$$
$$1-\dfrac{\sin^2x+\sin^2y}{2}>=\cos x\cos y$$
Now here I am stuck as I am not finding the way to calculate lower bound of $\cos x\cos y$
|
Using this, $$-\dfrac\pi2\le\sin^{-1}z\le\dfrac\pi2$$ for $-1\le z\le1\iff z^2\le1$
$$\sin^2y<1-\sin^2x=\cos^2x\implies\dfrac{\sin^2y}{\cos^2x}<1$$
$$\text{Similarly, }\dfrac{\sin^2x}{\cos^2y}<1$$
$$\implies\dfrac{\sin^2x\sin^2y}{\cos^2x\cos^2y}=\dfrac{\sin^2x}{\cos^2y}\cdot\dfrac{\sin^2y}{\cos^2x}<1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Suppose that $x^5$ and $20x+\frac {19}x$ are rational numbers. Then $x$ is also rational Let $x\neq0$ be a real number such that $x^5$ and $20x+\frac {19}x$ are rational. How can we prove that $x$ is also rational? (This was a question from the RMO 2019 in India.)
My attempt: Let $a,b,c,d$ be integers such that $20x+\frac {19}{x}=\frac ab$ and $x^5 = \frac{c}{d}$.
Then we have $$x=\frac{a\pm\sqrt{a^2-1520 b^2}}{40b}$$ so $x$ is rational iff $\sqrt{a^2-1520 b^2}$ is rational.
However, I don't know how to prove that $\sqrt{a^2-1520 b^2}$ is rational using that $$x=\frac{\sqrt[5]{c}}{\sqrt[5]{d}}$$
|
It seems that one answer has been posted online. The argument is elementary and straightforward. I will repeat it here.
Let $r=20x+\frac{19}{x}\in\mathbb{Q}$. It follows that
$$
20x^2-rx+19=0
$$
and the quadratic formula implies that
$$
x=r_1\pm\sqrt{r_2},\quad
r_1=\frac{r}{40},\quad
r_2={\frac{r^2-4\cdot 20\cdot 19}{40^2}}\geq 0.
$$
The binomial theorem implies that
$$
x^5=(r_1\pm\sqrt{r_2})^5=r_1^5+r_2^2\sqrt{r_2}
+5r_1^4\sqrt{r_2}+5r_1r_2^2
+10r_1^3r_2+10r_1^2r_2\sqrt{r_2}\\
=q\pm(r_2^2+5r_1^4+10r_1^2r_2)\sqrt{r_2},\quad q\in\mathbb{Q}.
$$
But $x^5\in\mathbb{Q}$, and $r_2^2+5r_1^4+10r_1^2r_2\neq 0$ (since $r_2\geq 0$ and $x\neq 0$). It follows that $\sqrt{r_2}\in\Bbb{Q}$. We are done.
|
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|
How to determine a limit without L'Hospital's rule
How to solve this type of limit without L'Hospital rule.
$$\lim_{x\to a}\frac{a^x-x^a}{x-a}$$
|
To make life easier, let $x=a+y$
$$\lim_{x\to a}\frac{a^x-x^a}{x-a}=\lim_{y\to 0}\frac{a^{a+y}-(a+y)^a}{y}$$ Factor out $a^a$ to make
$$a^{a+y}-(a+y)^a=a^a\left(a^y-\left(1+\frac{y}{a}\right)^a \right)$$
Now
$$a^y=e^{y \log(a)}=1+y \log (a)+\frac{1}{2} y^2 \log ^2(a)+O\left(y^3\right)$$
$$b=\left(1+\frac{y}{a}\right)^a\implies \log(b)=a \log\left(1+\frac{y}{a}\right)=y-\frac{y^2}{2 a}+O\left(y^3\right)$$
$$b=e^{\log(b)}=1+y+\frac{1}{2} \left(1-\frac{1}{a}\right) y^2+O\left(y^3\right)$$
$$a^y-\left(1+\frac{y}{a}\right)^a=y (\log (a)-1)+\frac{1}{2} y^2 \left(\frac{1}{a}+\log ^2(a)-1\right)+O\left(y^3\right)$$
$$\frac{a^{a+y}-(a+y)^a}{y}=a^a \left(\log (a)-1+\frac{1}{2} y \left(\frac{1}{a}+\log ^2(a)-1\right)+O\left(y^2\right) \right)$$ which not only shows the limit but also how it is approached.
|
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|
Solve Exponential Equation with square Solve equality with square
$2^{2x}=7\cdot 2^{x+\sqrt{x-1}}+8\cdot 2^{2\sqrt{x-1}}$
$x-1\ge0 \\\sqrt{x-1}=t\ge0\Rightarrow x-1=t^2\Rightarrow x=t^2+1\\2^{2(t^2+1)}=7\cdot2^{t^2+1+t}+2^{3+2t}$
It looks very complicated and I don't know how to move it.
Is there a better way to approach this task?
|
Hint
If $$2^x=a,2^{\sqrt{x-1}}=b$$
Now for real $y,2^y>0$
we have $$0=a^2-7ab-8b^2=(a-8b)(a+b)$$
So, $$2^x=2^{3+\sqrt{x-1}}$$
Set $\sqrt{x-1}=p\ge0\implies x=p^2+1$
Like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
as $2\ne\pm1$
$$\implies p^2+1=3+p\iff(p-2)(p+1)=0$$
|
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|
How many solutions does the equation $x_1+x_2+x_3+x_4+x_5 = 8$ have? $x$$1$+$x$$2$+$x$$3$+$x$$4$+$x$$5$ = 8
where $x$$i$'s can take values $\{0,1,2,3\}$.
|
In this case, it's easy to check by hand that the only solutions are
\begin{align*}
0+0+2+3+3&=8\\
0+1+1+3+3&=8\\
0+1+2+2+3&=8\\
0+2+2+2+2&=8\\
1+1+1+2+3&=8\\
1+1+2+2+2&=8,
\end{align*}
up to permutations of the terms in the different sums. Thus the number of solutions is
$$\frac{5!}{2!\,2!}+\frac{5!}{2!\,2!}+\frac{5!}{2!}+\frac{5!}{4!}+\frac{5!}{3!}+\frac{5!}{2!\,3!} = \boxed{155}.$$
It's easier to order the set of quintuples (say lexicographically like I did) when coming up with these.
Alternatively, you can construct a generating function for the problem, in this case $(1+x+x^2+x^3)^5$ is suitable, and study the coefficient of $x^8$, which is $\boxed{155}$.
If you are not familiar with generating functions, read this short example of how they can be used to solve counting problems. They are a very useful tool when it comes to solving counting problems like this one.
|
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|
If $n\geq 3$, $9^n \equiv a (\mod 100)$ and $9^{n+1} \equiv b (\mod 100)$, then $a+b=90$. I noticed a pattern in the powers of 9 modulo 100.
$9^1 \equiv 9 \pmod{100}$
$9^2 \equiv 81 \pmod{100}$
$9^3 \equiv 29 \pmod{100}$
$9^4 \equiv 61 \pmod{100}$
.
.
.
and conjectured the following:
If $n\geq 3$ is an odd integer where $9^n \equiv a \pmod{100}$ and $9^{n+1} \equiv b \pmod{100}$, then $a+b=90$.
How do I prove this?
|
$$a+b\equiv9^n(1+9)\pmod{100}$$
Now $9^n\equiv(-1)^n\pmod{10}\implies10\cdot9^n\equiv10\cdot(-1)^n\pmod{10\cdot10}$
|
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|
Differentiation under integration $\int_0^{\frac{\pi}{2}}\frac{\ln(1+\cos\alpha\cos x)}{\cos x}dx$
Question: Discuss the method of differentiation under the sign of integration. Hence evaluate following integrals: $$(i)\int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx\quad(ii)\int_0^{\frac{\pi}{2}}\frac{\ln(1+\cos\alpha\cos x)}{\cos x}dx\quad\cdots$$
First of all I am confused which parameter I take. Although I tried to take both parameter individually to get some intuition.
\begin{align*}
I(b) &= \int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx \\
I'(b)&=\frac{d}{db}\int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx\\
I'(b)&=\int_0^{\infty}\frac{\partial}{\partial b}\left(\frac{\ln(1+a^2x^2)}{1+b^2x^2}\right)dx\\
I'(b)&=\int_0^{\infty}-\frac{2x^2b\ln \left(1+a^2x^2\right)}{\left(1+x^2b^2\right)^2}dx
\end{align*}
Now I feel I choose wrong parameter. Then try again,
\begin{align*}
I(a) &= \int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx \\
I'(a)&=\frac{d}{da}\int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx\\
I'(a)&=\int_0^{\infty}\frac{\partial}{\partial a}\left(\frac{\ln(1+a^2x^2)}{1+b^2x^2}\right)dx\\
I'(a)&=\int_0^{\infty}\frac{2ax^2}{\left(1+x^2b^2\right)\left(1+a^2x^2\right)}dx
\end{align*}
Now I am frustrated and think I am missing something or didn't understand the concept.
questions:
$(1)$ Actually how to deal with them$?$
$(2)$ How to choose parameter wisely or assure that I am going right direction because after that partial derivatives those look more ugly to integrate/handle
and backtracking isn't possible in Exam Hall$?$
Thanks in advance and thanks for your time .
|
Using Nyssa's comment I tried like that:
\begin{align*}
I(a) &= \int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx \\
I'(a)&=\frac{d}{da}\int_0^{\infty}\frac{\ln(1+a^2x^2)}{1+b^2x^2}dx\\
&=\int_0^{\infty}\frac{\partial}{\partial a}\left(\frac{\ln(1+a^2x^2)}{1+b^2x^2}\right)dx\\
&=\int_0^{\infty}\frac{2ax^2}{\left(1+x^2b^2\right)\left(1+x^2a^2\right)}dx\\
&=\int_0^{\infty}\left(\frac{2a}{(a^2-b^2)(1+x^2b^2)}-\frac{2a}{(a^2-b^2)(1+x^2a^2)}\right)dx\\
&=\frac{2a}{a^2-b^2}\frac{1}{b}\arctan(bx)\bigg|_0^\infty -\frac{2a}{a^2-b^2}\frac{1}{a}\arctan(ax)\bigg|_0^\infty\\
&=\frac{\pi}{a^2-b^2}\left(\frac{a}{b}-1\right)=\frac{\pi}{(a-b)(a+b)}\frac{a-b}{b}=\frac{\pi}{b(a+b)}
\end{align*}
Now we integrate back:
\begin{align*}
I(a)&=\frac{\pi}b\int\frac{1}{a+b}da\\
&=\frac{\pi}b\ln(a+b)+C\\
I(0)=0\implies C=-\frac{\pi}{b}\ln b \implies I(a)=\frac{\pi}{b}\ln\left(\frac{a+b}{b}\right)
\end{align*}
For $(2)$ I personally now think to choose parameter such a way that it remove $\ln,\sin^{-1}\cdots$ stuff. Please correct me If I am wrong. I put my answer as community so that anyone can edited it.
|
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|
$p_1^2 + p_2^2 = q_1^2 + q_2^2 = r_1^2 + r_2^2 = 2$, $p_1^2 + q_1^2 + r_1^2 = 3$, positive distinct rational solution I would like to know if equations
$p_1^2 + p_2^2 = 2$
$q_1^2 + q_2^2 = 2$
$r_1^2 + r_2^2 = 2$
$p_1^2 + q_1^2 + r_1^2 = 3$
have a solution where $p_1,p_2,q_1,q_2,r_1,r_2$ are pairwise distinct and positive rational numbers.
I will appreciate any help.
|
Yes, the possible solutions in the rationals are $x,y = \pm1, \pm1$, so
$$p = [1,1]$$
$$q = [-1,1]$$
$$r = [-1,-1]$$
is a solution
Edit:
Based on comments.
We seek a solution $x,y$ such that $x^2 +y^2=2$ where $x,y \in \mathbb{Q}$. To fulfill this (and without loss of generality within the specified domain) we require that $0 \le x \le 1$ and $1\le y\le 2$. Let's then define two new variables $a,b$ and require $0 \le a,b \le 1$ and reexpress our equation as
$$(1+a)^2 + (1-b)^2 = 2$$
Expanding this equation
$$ a^2 + 2 a + 1 + b^2 - 2b + 1 = 2$$
Simplifying,
$$ b^2 - 2b + a^2 + 2 a= 0$$
Therefore,
$$ a^2 + b^2 = 2 ( b - a )$$
so $b \ge a$
If $b = a + \epsilon$, for some non-negative $\epsilon$
$$ a^2 + (a+\epsilon)^2 = 2 ( a+\epsilon - a )$$
$$ 2 a^2 + 2a\epsilon + \epsilon^2 = 2 \epsilon $$
Solving for $a$
$$a = \frac{-2 \epsilon \pm \sqrt{(2 \epsilon)^2 - 4 (2) (\epsilon^2 - 2 \epsilon ) }}{2 \cdot 2}$$
$$a = \frac{-\epsilon \pm \sqrt{4 \epsilon - \epsilon^2}}{2}$$
so for $a$ to be real, $0 \le \epsilon \le 4$. And for $a$ to be rational, there must be some coprime integers $m,n$ such that
$$4 \epsilon - \epsilon^2 = \left( \frac{m}{n} \right)^2$$
Or equivalently
$$ n^2( 4 \epsilon - \epsilon^2 ) = m^2$$
The only way for this to be the case is if $4 \epsilon - \epsilon^2$ is a square number, which within the domain limits on $\epsilon$ means $\epsilon=0,4$
This corresponds to $a=0$ and $b=0$ or $b=2$, thus the solutions are $x,y= 1,\pm1$ and cannot be distinct and positive.
Note:
Because of the definition of $a$, we lose the $[-1,-1]$ solution
|
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|
Solve differential equations $y' = \frac{x-y^2}{2y(x+y^2)}$ Help me!
I'm can try solve this:
$$y' = \frac{x-y^2}{2y(x+y^2)}$$
I'm try to change variable
$$ t = y^2$$
but what is next?
am I need to solve this?
$$t'= \frac{x-t}{x+t}$$
|
$$ 2y(x+y^2)dy=(x-y^2)dx $$ $\Rightarrow $
$$ xdx-y^2dx-2yxdy-2y^3dy=0$$ We can make a replacement (1)
$$ 1=2\alpha=\alpha+1+\alpha-1=3\alpha+\alpha-1 $$
$ \alpha=\frac{1}{2}$ $\Rightarrow $ $ y=\sqrt{z}$
$$\frac{dy}{dx}=\frac{dz}{2\sqrt{z}dz} $$ $\Rightarrow $
$$(x-z)dx-(x+z)dz=0 $$
$$\frac{x-z}{x+z}=\frac{dz}{dx} $$
Replacement (2)
$ \frac{z}{x}=m $ $\Rightarrow$ $$ \frac{dz}{dx}=\frac{dm}{dx}x+m$$
$$ \frac{1-m}{1+m}-m=\frac{dm}{dx}x$$
$$C+\int\frac{dx}{x}=\int\frac{dm}{\frac{1-m}{1+m}-m} $$
But we know that $$ \int\frac{(1+m)dm}{-m^2-2m+1}=\frac{-1}{2}\int\frac{d(m^2+2m-1)}{m^2+2m-1}=\frac{-1}{2}log(m^2+2m-1)$$ $$\Rightarrow $$
$$ \frac{C}{x^2}=m^2+2m-1$$ $m=\frac{y^2}{x} $
Final answer:
$$ C=y^4+2y^2x-x^2$$
|
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|
Find all $x$ such that determinant is zero . $$\begin{vmatrix}x-a & a-b & b-c \\ c-a & x-b & a-c \\ b-a & c-b & x-c \end{vmatrix} =0 $$
I began expanding but quickly gave up, i couldn't factor it.Adding or subtracting rows/columns got me nowhere because i made no zeroes.
Any idea would be helpful. Hopefuly there is a more elegant solution than expansion and brute force,if that is the only way how to factor(apart from using cubical formula) .
|
Without constraints or additional information, your determinant is:
$$-a^3+a^2 (b+c+x)-a \left(b^2+2 b (c-x)-c^2+2 c x+x^2\right)+b^3-b^2 (c+x)+b \left(c^2+2 c x-x^2\right)-c^3+c^2 x-c x^2+x^3$$
and brute force is the only way forward. The three solutions (using the cubic equation) are:
$$\{x\to a-b+c\} \\
\left\{x\to b-\sqrt{-a^2+2 a c-c^2}\right\} \\
\left\{x\to
\sqrt{-a^2+2 a c-c^2}+b\right\}$$
|
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|
Solve a diophantine equation The questions is: Solve in the positive integers $$3^n=m^4+m+1.$$
I can prove that $n$ is odd, and $m\equiv 4(\text{mod }9)$, but i dont know why $n$ and $m$ are $1$.
|
Probably not of much use but some elementary observations about the numbers $m^4+m+1$ are as follows.
There is precisely one solution modulo $3^n$ of the equation
$$m^4+m+1\equiv 0\pmod {3^n}$$
The solution for $3^{n+1}$ can be obtained from the solution for $3^n$ as follows:-
If $m^4+m+1\equiv 3^nd\pmod {3^{n+1}}$, where $d=1$ or $2$, then
$$(m+3^nd)^4+(m+3^nd)+1\equiv 0\pmod {3^{n+1}}$$
The first few solutions are then
$m=1:$ $m^4+m+1\equiv 3\pmod {3^2}$
$m=1+3=4:$ $m^4+m+1\equiv 2\times 3^2\pmod {3^3}$
$m=4+2\times 3^2=22:$ $m^4+m+1\equiv 3^3\pmod {3^4}$
$m=22+3^3=49:$ $m^4+m+1\equiv 2\times 3^4\pmod {3^5}$
$m=49+2\times3^4=211:$ $m^4+m+1\equiv 3^7\pmod {3^8}$
$m=211+3^7=2398:$ $m^4+m+1\equiv 2\times 3^8\pmod {3^9}$
|
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|
Prove that in any triangle
Prove that in any triangle with side lengths $a,b,c$ inradius $r$, and
circumradius $R$, we have
$\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} >
\frac{5}{3}$
I think it is $\frac{a}{b+c}+ \frac{b}{c+a}+ \frac{c}{a+b}+ \frac{r}{R} \leq 2 $.
with $a+b+c=2s,\ ab+b+ca=(4R+r)r+s^2,\ abc=4Rrs$ rewrites as$$\frac{2\left(s^2-Rr-r^2\right)}{s^2+2Rr+r^2}+\frac{r}{R}-\frac{5}{3}\ge 0$$which boils down to showing$$(R+3r)s^2-\left(16R^2+5Rr-3r^2\right)r\ge 0$$which by Gerretsen $s^2\ge 16Rr-5r^2$ is just$$2(19R-6r)r^2\ge 0$$true by Euler $R\ge 2r$.
|
In the standard notation we need to prove that
$$\sum_{cyc}\frac{a}{b+c}+\frac{\frac{2S}{a+b+c}}{\frac{abc}{4S}}\leq2$$ or
$$\sum_{cyc}\frac{a}{b+c}+\frac{16S^2}{2abc(a+b+c)}\leq2$$ or
$$\sum_{cyc}\frac{a}{b+c}+\frac{\sum\limits_{cyc}\left(-a^3+a^2b+a^2c-\frac{2}{3}abc\right)}{2abc}\leq2$$ or
$$\sum_{cyc}(a^5b+a^5c-2a^3b^3-2a^4bc+a^3b^2c+a^3c^2b)\geq0$$ or
$$\sum_{cyc}ab(a^2-b^2)^2-abc\sum_{cyc}(a+b)(a-b)^2\geq0$$ or
$$\sum_{cyc}ab(a+b)(a-b)^2(a+b-c)\geq0$$ and we are done!
|
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|
How to calculate the determinant of a fully/negative symmetric matrix? I have two similar questions.
*
*How to calculate the determinant of the following $n \times n$ matrix?
$$\begin{vmatrix}
a & b & \dots & b & b\\
-b & a & \dots & b & b\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
-b & -b & ... & a & b\\
-b & -b & ... & -b & a
\end{vmatrix}$$
*How to calculate the determinant of the following $n \times n$ matrix?
$$\begin{vmatrix}
a+b & a & \dots & a\\
a & a + b & \dots & a\\
\vdots & \vdots & \ddots & \vdots \\
a & a & \dots & a + b
\end{vmatrix}$$
For the second one, I have seen this, but I don't understand the part where $$\det M_n = \det M_{n - 1} - r(n-1) N_{n - 1}$$ Why minus shouldn't it be something like $(-1)^{i+j}$?
|
Matrix 1
Consider the determinant polynomials
$$
\begin{align}
p_n(x)
&=
\det\begin{bmatrix}
1&x&x&\cdots&x\\
-x&1&x&\cdots&x\\
-x&-x&1&\cdots&x\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-x&-x&-x&\cdots&1
\end{bmatrix}\tag{1a}\\[6pt]
&=
\det\begin{bmatrix}
1-x&0&0&\cdots&1+x\\
-x&1&x&\cdots&x\\
-x&-x&1&\cdots&x\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-x&-x&-x&\cdots&1
\end{bmatrix}\tag{1b}\\[6pt]
&=
\det\begin{bmatrix}
\color{#C00}{2}&0&0&\cdots&\color{#090}{1+x}\\
0&1&x&\cdots&x\\
0&-x&1&\cdots&x\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
\color{#090}{1-x}&-x&-x&\cdots&1
\end{bmatrix}\tag{1c}\\[30pt]
&=\color{#C00}{2p_{n-1}(x)}+\color{#090}{\left(x^2-1\right)p_{n-2}(x)}\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: definition
$\text{(1b)}$: add column $n$ to column $1$
$\text{(1c)}$: add row $n$ to row $1$
$\text{(1d)}$: expand the determinant on column $1$:
$\phantom{\text{(1d):}}\quad$ the $(1,1)$ term is $2$ and the $(1,1)$ minor is $p_{n-1}(x)$
$\phantom{\text{(1d):}}$ the $(n,1)$ term is $1-x$; expand the $(n,1)$ minor on row $1$:
$\phantom{\text{(1d):}}\quad$ the $(1,n-1)$ term is $1+x$ and the $(1,n-1)$ minor is $p_{n-2}(x)$
Start with $p_0(x)=p_1(x)=1$ and compute the first several polynomials:
$$
\begin{align}
p_0(x)&=1\\
p_1(x)&=1\\
p_2(x)&=1+x^2\\
p_3(x)&=1+3x^2\\
p_4(x)&=1+6x^2+x^4\\
p_5(x)&=1+10x^2+5x^4\\
p_6(x)&=1+15x^2+15x^4+x^6
\end{align}\tag2
$$
From the binomial coefficients in $(2)$, we can start to see a pattern:
$$
p_n(x)=\frac{(1+x)^n+(1-x)^n}2\tag3
$$
Equation $(3)$ can be verified using $\text{(1d)}$ and induction. Setting $x=\frac ab$ and multiplying the whole matrix by $b$ gives
$$
\det\begin{bmatrix}
b&a&a&\cdots&a\\
-a&b&a&\cdots&a\\
-a&-a&b&\cdots&a\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
-a&-a&-a&\cdots&b
\end{bmatrix}
=\frac{(b+a)^n+(b-a)^n}2\tag4
$$
Matrix 2
Note that any vector orthogonal to $\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}$ is multiplied by $b$ (on any subspace orthogonal to a given vector, that vector can be subtracted from each column of the matrix without changing the action on that subspace). $\begin{bmatrix}1&1&1&\cdots&1\end{bmatrix}$ is multiplied by $an+b$. Thus,
$$
\det\begin{bmatrix}
a+b&a&a&\cdots&a\\
a&a+b&a&\cdots&a\\
a&a&a+b&\cdots&a\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
a&a&a&\cdots&a+b
\end{bmatrix}
=b^{n-1}(an+b)\tag5
$$
|
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|
How to find singular solutions and determine their types for system of equations.
Given equations $x^\prime=x(1-x-2y)$ and $y^\prime=y(1-y-4x)$ find all the singular solutions in the upper quadrant, $x\geq 0, y\geq 0$ and determine the type and stability.
So for singular solutions I believe I want points such that:
$x(1-x-2y)=y(1-y-4x)=0$
Which gives me $x=0,y=0$ or $(1-x-2y)=(1-y-4x)\implies y=3x$
So I have solutions $(0,0),(0,1),(1,0),(1/7,3/7)$
So I have the Jacobin matrix, $J=\begin{pmatrix} 1-2x-2y & -2x\\-4y&1-2y-4x\\ \end{pmatrix}$
And I need to evaluate the eigenvalues at different critical points.
$J(0,0)=\begin{pmatrix} 1&0\\0&1\\ \end{pmatrix}$
Which is already diagonal and has $1$ repeated real positive eigenvalue.
$J(0,1)$ has $1$ repeated negative eigenvalue
$J(1,0)$ has eigenvalues $-3,0$
$J(1/7,3/7)$ has one positive and one negative eigenvalue.
|
We want to simultaneously solve
$$x(1-x-2y) = 0 \\y (1-y-4x)=0$$
Clearly, $(x, y) = (0, 0)$.
When $x = 0$, we have $y(1 - y) = 0 \implies y = 0, y = 1$.
When $y = 0$, we have $x(1 - x) = 0 \implies x = 0, x = 1$.
Then we have
$$(1-x-2y) = 0 \\ (1-y-4x)=0$$
This gives us $x = \dfrac{1}{7}, y = \dfrac{3}{7}$.
So, our critical points are
$$(x, y) = (0, 0), (0, 1), (1, 0), \left(\dfrac{1}{7}, \dfrac{3}{7}\right)$$
Here is a contour plot that verifies this
Can you continue?
Hints:
Here is a phase portrait as a guide
Find the Jacobin matrix and evaluate the eigenvalues at each critical point while keeping track of any degenerate cases.
Update: For the Jacobian, I get
$$J(x, y) = \begin{pmatrix} -2 x-2 y+1 & -2 x \\ -4 y & -4 x-2 y+1 \\ \end{pmatrix}$$
My answer might also be helpful.
|
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|
Inequality with Fibonacci numbers $ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1} > \text{arccot} \: 2$
Prove that $$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1} > \text{arccot} \: 2$$ holds for all $n \in \mathbb{N}$.
(The Fibonacci sequence, defined by the recurrence $F_1 = F_2 = 1$ and $\forall n \in \mathbb{N},$ $F_{n+2} = F_{n+1} + F_n$)
My work. I proved that the sequence $a_n=\sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1}$ is decreasing. Therefore inequality cannot be proved by induction.
|
Show that
$$\text{arccot} F_{2n+1} = \text{arccot} L_{2n} + \text{arccot} L_{2n+2}, \quad (1)$$ where $L_n$ is the Lucas series with $L_0 = 2, L_1 = 1 , L_3 = 3, L_{n+2} = L_{n+1} + L_n$.
Hence,
$$ \sum \limits_{k=1}^{2n+1} (-1)^{k+1} \: \text{arccot} F_{2k-1} = \text{arccot} L_0 + \text{arccot} L_{4n+2} > \text{arccot} 2 $$
Proof of (1):
A necessary condition for integer solutions to $\tan^{-1} \frac{1}{x} = \tan^{-1} \frac{1}{y} + \tan^{-1} \frac{1}{z}$ is $ 1/x = \frac{ 1/y + 1/z} { 1 + 1/(yz)}$, or that $1+yz = xy+xz \Rightarrow (y-x)(z-x) = x^2 - 1$.
So, with $L_n = F_{n-1} + F_{n+1}$, we can verify that
$(L_{2n} - F_{2n+1} ) ( L_{2n+2} - F_{2n+1} ) = F_{2n-1} F_{2n+3} = F_{2n+1} ^2 - (-1)^{2n+1} = F_{2n+1} ^2 + 1$
Finally, show that this is a sufficient condition because the angles are small enough.
To de-construct what I did, I stumbled upon the observation that the cotangent of the difference was an integer, and started to hunt down what it was. Having the Lucas number $L_{4n+2}$ appear was surprising. Given that it's an olympiad problem, it suggests that there is a nice solution, which led to the guess of $(1)$.
|
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|
Solve in Z. $y^2+y=x^4+x^3+x^2+x$ Solve in Z. $$y^2+y=x^4+x^3+x^2+x$$
In my attempt to solve this, I used the fact that the left side is even.I write the eqyation in the following form: $$(x+1)(x^2+1)=0 \pmod 2$$. This clearly means x must be an odd number, as a result $$4|y(y+1)$$. This is as far as I could go.
Question from Jalil Hajimir
|
Hint: Bound between 2 squares
For all but finitely many values of $x$, we can bound $(2y+1)^2$ between 2 consecutive perfect squares, hence it is never a perfect square.
$(2x^2 + x)^ 2 < (2y+1) ^2 = 4y^2 + 4y + 1 = 4x^4 + 4x^3 + 4x^2 + 4x + 1 < 4x^4 + 4x^3 + 5x^2 + 2x + 1 = (2x^2 + x + 1)^2 $
Hence, we only need to check those values when the inequalities fail:
RHS inequality is false when: $x \in [0, 2 ] $.
$x = 0$ yield $ y = -1, 0$.
$x = 1$ yields no integer solution for $y$.
$x = 2$ yield $ y = -6, 5$.
LHS inequality is false when $x \in [-1, -\frac{1}{3} ]$.
$x = -1$ yields $y = -1, 0$.
|
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|
Split into partial fractions $\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2}$ This is from "Calculus Made Easy", Exercises 10, Question 15 (page 147). I've worked this one over and over and still haven't made progress.
This is my initial setup:
$$\frac{3x^2+2x+1}{(x+2)(x^2+x+1)^2} = \frac{A}{(x+2)} + \frac{Bx+C}{(x^2+x+1)} + \frac{Dx+E}{(x^2+x+1)^2}$$
then:
$$3x^2+2x+1 = A(x^2+x+1)^2 + (Bx+C)(x+2)(x^2+x+1) + (Dx+E)(x+2)$$
and I can solve for A by setting $x=-2$, which yields $A=1$.
I know the final answer is this:
$$\frac{1}{x+2} - \frac{x-1}{x^2+x+1} - \frac{1}{(x^2+x+1)^2}$$
But I've worked this problem many ways and cannot make progress on the numerators for the other fractions.
|
I'll expand all the things from the comment.
Method 1:
By expanding we get
$$
\begin{cases}
A+B=0&&\hbox{ for }x^4\\\
2A+3B+C=0&&\hbox{ for }x^3\\\
3 A + 3 B + 3 C + D - 3=0&&\hbox{ for }x^2\\\
2 A + 2 B + 3 C + 2 D + E - 2=0&&\hbox{ for }x\\\
A + 2 C + 2 E - 1=0&&\hbox{ for }1
\end{cases}
$$
which boils down to $A = 1, B = -1, C = 1, D = 0, E = -1$.
Method 2 "substitution":
with $x=-\frac12\pm\frac{\sqrt{3}}{2}i$ we get rid of $(x^2+x+1)$, can have equations for $D,E$ and after finding $D,E$ and bringing them to the LHS we can cancel by $(x+2)$ which boils down to Mohammad Riazi-Kermani's answer
|
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How do I solve recurrence relation without characteristic equation? Question:
Solve the recurrence relation
$\ a_n = 3a_{n-1} - 2a_{n-2} + 1 $, for all $\ n \ge 2$
$\ a_0 = 2 $
$\ a_1 = 3 $
Write $\ a_n $ in terms of n
I tried to solve this by finding the characteristic equation, $\ r^2 - 3r + 2 - 1 = 0 $ which is $\ r^2 - 3r + 1 $. However, I can't simplify that further because of the "+ 1" unless I use the quadratic general formula... but the roots will be in fractions and they are definitely not correct compared to the answers..
So I tried to find $\ a_2, a_3, a_4 $ and so on... like this:
$\ a_2 = 3a_1 - 2a_0 + 1 = 3(3) - 2(2) + 1 = 6 $
$\ a_3 = 3a_2 - 2a_1 + 1 = 3(6) - 2(3) + 1 = 13 $
$\ a_4 = 3a_3 - 2a_2 + 1 = 3(13) - 2(6) + 1 = 28 $
and so on...
But it leads me to nowhere as I couldn't find any common pattern between $\ a_2, a_3, a_4 $ and so on, to derive $\ a_n $...
How do I solve recurrence relations like this?
|
A general way to solve this is given by generating functions. Define:
$\begin{equation*}
A(z)
= \sum_{n \ge 0} a_n z^n
\end{equation*}$
Take the recursion, shift so there are no subtractions in indices, multiply by $z^n$ and sum over $n \ge 0$. Recognize the resulting sums, use initial values:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= 3 \sum_{n \ge 0} a_{n + 1} z^n
- 2 \sum_{n \ge 0} a_n z^n
+ \sum_{n \ge 1} z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
&= 3 \frac{A(z) - a_0}{z} - 2 A(z) + \frac{1}{1 - z} \\
\frac{A(z) - 2 - 3 z}{z^2}
&= 3 \frac{A(z) - 2}{z} - 2 A(z) + \frac{1}{1 - z}
\end{align*}$
Now solve for $A(z)$, write as partial fractions:
$\begin{align*}
A(z)
&= \frac{2 - 5 z + 4 z^2}{1 -4 z + 5 z^2 - 2 z^3} \\
&= \frac{2 - 5 z + 4 z^2}{(1 - z^2) (1 - 2 z)} \\
&= \frac{2}{1 - 2 z} + \frac{1}{1 - z} - \frac{1}{(1 - z)^2}
\end{align*}$
We want the coefficient of $z^n$ in the above:
$\begin{align*}
[z^n] A(z)
&= [z^n] \frac{2}{1 - 2 z}
+ [z^n] \frac{1}{1 - z}
- [z^n] \frac{1}{(1 - z)^2} \\
&= 2 \cdot 2^n
+ 1^n
- (-1)^n \binom{-2}{n} \cdot 1^n \\
&= 2^{n + 1} + 1
- \binom{n + 2 - 1}{2 - 1} \\
&= 2^{n + 1} + 1
- (n + 1) \\
&= 2^{n + 1} - n
\end{align*}$
|
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|
Given $2018 \times 4$ grids and tint them with red and blue. So that each row and each column...
Given $2018 \times 4$ grids and tint them with red and blue. So that each row and each column has the same number of red and blue grids, respectively. Suppose there're $M$ ways to tint the grids with the mentioned requirement. Determine $M \pmod {2018}$.
Solution?: Each column can be colored such as $$(R,R,B,B),(R,B,R,B),(R,B,B,R),$$$$(B,B,R,R),(B,R,B,R),(B,R,R,B)$$ Say the column colorings appears $a,b,c,a',b',c'$ respectively.
Clearly we must have $a+b+c= a'+b'+c'$ since $R$ appears the same times as $B$ in first row. Simillary we have $a+b'+c' = a'+b+c$ which implies $a=a'$. The same is true for $b=b'$ and $c=c'$. So we have that $a+b+c=1009$ and thus $$M =\sum_{a+b+c=1009}\frac{2018!}{a!^2b!^2c!^2} \equiv ?\pmod{2018}$$
Edit: after Ross Millikan answer.
My question here is: is this is correct and seeking for an alternative solution via generating functions.
|
This is only a response to if your way of thinking about the problem is correct.
Let's consider this a slightly different way:
$$\hat{v_1} = \begin{pmatrix}R \\ R \\ B \\ B\end{pmatrix}, \hat{v_2} = \begin{pmatrix}R \\ B \\ R \\ B\end{pmatrix}, \hat{v_3} = \begin{pmatrix}R \\ B \\ B \\ R\end{pmatrix}, \hat{v_4} = \begin{pmatrix}B \\ R \\ R \\ B\end{pmatrix}, \hat{v_5} = \begin{pmatrix}B \\ R \\ B \\ R\end{pmatrix}, \hat{v_6} = \begin{pmatrix}B \\ B \\ R \\ R\end{pmatrix}$$
Then let $v_i$ be the number of columns of type $\hat{v_i}$.
We have:
$$\sum_{i=1}^6 v_i = 2018$$
$$v_1+v_2+v_3=v_4+v_5+v_6$$
$$v_1+v_4+v_5=v_2+v_3+v_6$$
$$v_2+v_4+v_6=v_1+v_3+v_5$$
$$v_3+v_5+v_6=v_1+v_2+v_4$$
This is five equations in six variables.
In augmented matrix form:
$$\left[\begin{array}{cccccc|c}1 & 1 & 1 & 1 & 1 & 1 & 2018 \\ 1 & 1 & 1 & -1 & -1 & -1 & 0 \\ 1 & -1 & -1 & 1 & 1 & -1 & 0 \\ -1 & 1 & -1 & 1 & -1 & 1 & 0 \\ -1 & -1 & 1 & -1 & 1 & 1 & 0\end{array} \right]$$
And according to Wolframalpha, in Row Echelon form:
$$\left[ \begin{array}{cccccc|c}1 & 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 1009 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]$$
So, this looks conclusive that your assumption was correct.
I do not have a solution involving generating functions, though.
Edit: Wolframalpha did find a closed form for this, though:
$$\sum_{a+b+c=n} \dfrac{(2n)!}{(a!)^2(b!)^2(c!)^2} = \dfrac{16^n\Gamma\left(n+\dfrac{1}{2}\right)^2 {_3}F_2\left(-n,-n,-n;1,\dfrac{1}{2}-n;\dfrac{1}{4}\right)}{\pi \Gamma (n+1)^2}$$
|
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|
How can I prove analytically that the golden ratio is less then $\frac{\pi^2}{6}$. Or stated in other terms, prove that
$$\frac{1+\sqrt{5}}{2} < \sum_{n=1}^{\infty}\frac{1}{n^2}$$
|
Without needing to know the numerical value of $\pi^2/6$, you can use the infinite series representation of the golden ration and some very crude bounds to arrive at the conclusion. Recall $$\varphi = \sum^\infty_{n=0} \frac{(-1)^n}{F_n F_{n+1}}$$ where $F_0 = 1, F_1 = 1$ and $F_n = F_{n-1} + F_{n-2}, \,\,\, n \ge 2$ is the Fibonacci sequence. Now you can prove the inequality term-by-term but using each pair of consecutive terms in the sum: $$\frac{1}{F_nF_{n+1}} - \frac{1}{F_{n+1}F_{n+2}} < \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2}, \,\,\,\, n = 0,2,4,\cdots. \,\,\,\,\,\,\,\,\,\,\, (1)$$ Indeed, $$\frac{1}{F_nF_{n+1}} - \frac{1}{F_{n+1}F_{n+2}} = \frac{F_{n+2} - F_n}{F_nF_{n+1}F_{n+2}} = \frac{1}{F_n{F_{n+2}}},$$ and we have $F_n \ge n$, so $$\frac{1}{F_n F_{n+2}} \le \frac{1}{n(n+2)} \le \frac{1}{2n^2} + \frac{1}{2(n+2)^2} < \frac {1}{(n+1)^2} + \frac{1}{(n+2)^2}.$$ Note, the last inequality only necessarily holds when $2n^2 \ge (n+1)^2$ (i.e, $n \ge 3$). You can verify $(1)$ when $n=0,2$ separately.
|
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Understanding the equation $|x+1|=x^2 -1$ I want to understand the equation
$$|x+1|=x^2 -1$$
$$\Leftrightarrow x^2 - |x+1| - 1 = 0$$
Case $1$:
$$x+1 \geq 0 \Rightarrow x^2 - x-2 = 0$$
$$x_{1,2} = \frac{1}{2} \big( 1\pm \sqrt{1-4\cdot(-2)} \big) = \frac{1}{2}(1\pm3) \Rightarrow x_1 = 2, x_2 = -1$$
Case $2$:
$$x+1 < 0 \Rightarrow x^2 + x = 0 \Rightarrow x_3= 0 \text{, (but doesn't fulfill } x+1 < 0), x_4 = -1$$
$$\Rightarrow L = \{2,-1\}$$
What I don't get is how $x+1 < 0 \Rightarrow x^2 + x = 0$. How do we get $x^2 + x = 0$?
|
For $x+1 \geq 0$, $|x+1| = x+1$, and hence the equation reads as:
$$x+1-x^2 +1 = 0 \Rightarrow x^2 -x - 2=0.$$
For $x+1 < 0$, $|x+1| = -(x+1)$, and hence the equation reads as:
$$-(x+1)-x^2 +1 = 0 \Rightarrow x^2 +x=0.$$
|
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|
How to prove that $ \arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2} $ I want to prove that: $$\arctan x + 2 \arctan(\sqrt{1 + x^2} - x) = \frac{\pi}{2}, \forall x \in \mathbb{R} $$
I know that $$ \arctan x + arctan \frac{1}{x} = \frac{\pi}{2} $$
But that doesn't seem to be helping. How should I proceed?
|
We have
\begin{eqnarray*}
\tan^{-1}(A) + \tan^{-1}(B) + \tan^{-1}(C) = \tan^{-1} \left( \frac{A+B+C-ABC}{1-AB-BC-CA} \right).
\end{eqnarray*}
If the RHS is to give $ \pi/2$ then we require the denominator to be zero, so
\begin{eqnarray*}
1-2x( \sqrt{1+x^2} -x) -(\sqrt{1+x^2} -x)^2=0
\end{eqnarray*}
Which is easily verified.
|
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|
$a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number? Let $a,b,c,d$ be natural numbers such that $a^{2} + b^{2}+ ab = c^{2} + d^{2} + cd$. Is it possible that $a+b+c+d$ be prime number?
Now assume that $a+b+c+d = p > 2$ for some choice of $a,b,c,d$. Notice that we cannot have $a=b=c=d$. Also notice that $(a+b+c+d)^{2}$ must be odd and only have three factors: $1, p, p^{2}$.
$$ p^{2} = a^{2} + b^{2} + c^{2} + d^{2} + 2(ab+ac+ad + bc+bd + cd) $$
$$ = 2(c^{2}+d^{2}+ac+ad++bc+bd) + (ab + 3cd) $$
$$ =2(c^{2}+d^{2}+ac+ad++bc+bd + cd) + (ab + cd) $$
So $(ab+cd)$ must be odd.
Now if $a+b+c+d$ is prime $>2$ then either 3 of them is odd and 1 is even, or 3 of them is even and one is odd.
$WLOG$, let $a,b,c$ be even and $d$ is odd, then $a^{2} + b^{2}+ ab$ is even and $c^{2} + d^{2}+ cd$ is odd, so we can't have 3 even and 1 odd.
But it is possible for 3 odd and 1 even.
|
Another solution, whose approach is distinct from the rest.
Suppose that for some prime $ p \geq 4$, $ a + b + c + d = p$.
Then $ a+ b \equiv - (c+d) \pmod{p}$,
$ a^2 + 2ab + b^2 \equiv c^2 + 2cd + d^2 \pmod{p}$,
$ab \equiv cd \pmod{p}$.
$a^2 - 2ab + b^2 \equiv c^2 - 2cd + d^2 \pmod{p}$
WLOG $ a - b \equiv c - d \pmod{p}$
Hence $ 2a \equiv (a-b) + (a+b) \equiv (c-d) - (c-d) \equiv - 2d \pmod{p}$.
Since $ p \neq 2$, so $ a + d \equiv 0 \pmod{p}$.
Then, $ a + b + c + d > a + d \geq p $ which is a contradiction.
|
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|
Limit of $\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$ I tried using symbolab to get the limit of
$$\lim_{x \to 0} \frac{(1+x^5)^{10} -1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1) }$$
but it couldn't solve it. With WolframAlpha I got $100$ for the limit.
Can someone show me how it's done "per hand"? I tried, but couldn't figure it out.
|
Are you sure you got $200$ and not $100$? Is there a typo here or in your input to WolframAlpha?
We have, either applying binomial theorem, or L'Hospital, or the fundamental exponential limit $\frac{e^x-1}{x}$ that
$\frac{(1+x^5)^{10}-1}{10x^5}\to1$
$\frac{\sqrt{1+x^3}-1}{\frac{1}{2}x^3}\to1$
$\frac{\sqrt[5]{1+x^2}-1}{\frac{1}{5}x^2}\to1$
Therefore, $$\frac{(1+x^5)^{10}-1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1)}=\left[\frac{(1+x^5)^{10}-1}{(\sqrt{1+x^3}-1)(\sqrt[5]{1+x^2}-1)}\frac{\frac{1}{2}x^3\frac{1}{5}x^2}{10x^5}\right]\frac{10x^5}{\frac{1}{2}x^3\frac{1}{5}x^2}\to 100$$
|
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|
What's wrong in my calculation of $\int \frac{\sin x}{1 + \sin x} dx$? I have the following function:
$$f: \bigg ( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg ) \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sin x}{1 + \sin x}$$
And I have to find $\displaystyle\int f(x) dx $. This is what I did:
$$\int \dfrac{\sin x}{1 + \sin x}dx=
\int \dfrac{1+ \sin x - 1}{1 + \sin x}dx =
\int dx - \int \dfrac{1}{1 + \sin x}dx =
$$
$$ = x - \int \dfrac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx$$
$$= x - \int \dfrac{1 - \sin x}{1 - \sin ^2 x} dx$$
$$= x - \int \dfrac{1 - \sin x}{\cos^2 x} dx$$
$$= x - \int \dfrac{1}{\cos^2x}dx + \int \dfrac{\sin x}{\cos^2 x}dx$$
$$= x - \tan x + \int \dfrac{\sin x}{\cos^2 x}dx$$
Let $u = \cos x$
$du = - \sin x dx$
$$=x - \tan x - \int \dfrac{1}{u^2}du$$
$$= x - \tan x + \dfrac{1}{u} + C$$
$$= x - \tan x + \dfrac{1}{\cos x} + C$$
The problem is that the options given in my textbook are the following:
A. $x + \tan {\dfrac{x}{2}} + C$
B. $\dfrac{1}{1 + \tan{\frac{x}{2}}} + C$
C. $x + 2\tan{\dfrac{x}{2}} + C$
D. $\dfrac{2}{1 + \tan{\frac{x}{2}}} + C$
E. $x + \dfrac{2}{1 + \tan{\frac{x}{2}}} + C$
None of them are the answer I got solving this integral. What is the mistake that I made and how can I find the right answer? By what I've been reading online, you can get different answers by solving an integral in different ways and all of them are considered correct. They differ by the constant $C$. I understand that, but I don't see how to solve this integral in such a way to get an answer among the given $5$. And, even more importantly, how can I recognize the right answer in exam conditions if the answer provided by my solution is not present among the given options? Is solving in a different manner my only hope?
|
Required answer is $E$. Observe that
$$\frac{1}{\cos x}-\tan x=\frac{1-\sin x}{\cos x}$$
$$=\frac{(\cos\frac{x}{2}-\sin\frac{x}{2})^2}{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}$$
$$=\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}$$
$$=\frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}}$$
$$=\frac{2}{1+\tan\frac{x}{2}}-1$$
Also, note that you could have directly got this answer if you have integrated $\frac{1}{1+\sin x}$ in a different manner, as follows.
$$\int\frac{1}{1+\sin x}dx=\int\frac{1}{(\cos\frac{x}{2}+\sin\frac{x}{2})^2}dx$$
$$=\int \frac{1}{\cos^2\frac{x}{2}(1+\tan\frac{x}{2})^2}dx$$
Now substitute $\tan\frac{x}{2}$ and you are done.
|
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|
Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2a}
$$
I think I am getting stuck here, is there a better substitution that I can chose so that the integral becomes more simple to evaluate ?
Solution as per my reference: $\dfrac{2(\sqrt{x}-1)}{\sqrt{1-x}}$
Note: I'd prefer to choose a substitution which does not make use of partial fractions, as there seems to be 4 terms for the substitution $\sqrt{x}=y\implies \frac{dx}{2\sqrt{x}}=dy$.
$$
I=\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}\\
=\int\frac{2dy}{y(1+y)(1-y^2)}=\int\frac{2dy}{y(1+y)^2(1-y)}\\
\frac{2}{y(1+y)^2(1-y)}=\frac{A}{y}+\frac{B}{1-y}+\frac{C}{1+y}+\frac{D}{(1+y)^2}
$$
|
You might have better luck with $y=\sqrt{x}$, provided you work with partial fractions. Something similar is achieved by continuing your current approach with $t=\tan a$, which amounts to starting with $\sqrt{x}=\frac{1-t^2}{1+t^2}$.
|
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|
Prove that $f(x) = \sqrt{x^2 + x}$ is uniform continuity at the interval $[0, \infty)$. Prove that $f(x) = \sqrt{x^2 + x}$ is uniform continuity at the interval $[0, \infty)$.
There is a hint to the exercise:
Prove that $\forall x,y \geq 1: \frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} \leq 2$
How does the hint help me?
|
$f$ is clearly uniformly continuous on $[0,1]$.
To show uniform continuity on $[1, +\infty\rangle$ notice that for $x \ge 1$ we have
$$f'(x) = \frac{2x+1}{2\sqrt{x(x+1)}} \le \frac{2\cdot 1+1}{2\sqrt{1(1+1)}}= \frac{3}{2\sqrt2}$$
so for $x,y \ge 0$ the mean value theorem gives that there exists $\theta$ between $x$ and $y$ such that
$$|f(x) - f(y)| = |f'(\theta)||x-y| \le \frac{3}{2\sqrt2}|x-y|.$$
|
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|
How to solve this quadratic equation (with $x$ represented by a fraction containing a square root)? If $x-\frac{4}{5} = \pm \frac{\sqrt{31}}{5}$, how can I find the values of a and b in the following equation?
$$5x^2+ax+b=0?$$
I've tried substituting $(\frac{4}{5} \pm \frac{\sqrt{31}}{5})$ for $x$ and got totally confused there.
|
So, I have: $$x=\frac{-a\pm\sqrt{b^2-20a}}{10}=-\frac{a}{10}\pm\frac{\sqrt{b^2-20a}}{5}$$
Now $x=\frac{4}{5}\pm\frac{\sqrt{31}}{5}$, so: $-\frac{a}{10}=\frac{4}{5}$ and $\frac{\sqrt{b^2-20a}}{5}=\frac{\sqrt{31}}{5}$
From the first $a=-8$ and substituing in the second: $b=-3$
|
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|
Theta Functions and Partitions I am reading some papers by Ramanujan on congruence properties of the partition function. At one point he says that he will be using "theta functions" and introduces the following:
It can be shewn that
$$
\begin{align}
&\dfrac{(1-x^5)(1-x^{10})(1-x^{15})\dots}{(1-x^{1/5})(1-x^{2/5})(1-x^{3/5})\dots} = \dfrac{1}{\xi^{-1}-x^{1/5}-\xi x^{2/5}}\\
&= \dfrac{\xi^{-4}-3x\xi+x^{1/5}(\xi^{-3}+2x\xi^2)+x^{2/5}(2\xi^{-2}-x\xi^3)+x^{3/5}(3\xi^{-1}+x\xi^4)+5x^{4/5}}{\xi^{-5}-11x-x^2\xi^5}
\end{align}
$$
where
$$
\xi = \dfrac{(1-x)(1-x^4)(1-x^6)(1-x^9)\dots}{(1-x^2)(1-x^3)(1-x^7)(1-x^8)\dots}
$$
the indices of the powers of x, both in the numerator and denominator of $\xi$, forming two arithmetical progressions with common difference 5. It follows that:
$$
(1-x^5)(1-x^{10})(1-x^{15})\dots\{p(4)+p(9)x+p(14)x^2\dots\} = \dfrac{5}{\xi^{-5}-11x-x^2\xi^5}
$$
Written a little cleaner he is saying that:
$$
\left(\prod_{n=1}^\infty(1-x^{5n})\right)\left(\sum_{n=0}^\infty p(5n+4)x^{n}\right) = \dfrac{5}{\xi^{-5}-11x-x^2\xi^5}
$$
I don't have any experience with this function or theta functions. I would appreciate some references to read more about these theta functions in general,
some understanding of why he uses this identity with powers of 1/5 and how it is derived, and help understanding how this is connected to the partition function for these particular values.
|
The general Ramanujan theta function
is defined by
$$ f(a,b) := 1 + (a+b) + ab(a^2+b^2) + (ab)^3(a^3+b^3) + \dots. \tag{1} $$
which factors according to the Jacobi triple product as
$$ f(a,b) = (-a;ab)_\infty(-b;ab)_\infty(ab;ab)_\infty. \tag{2} $$
An important special case is the single variable theta function
$$ f(-x) := f(-x,-x^2) = (1-x)(1-x^2)(1-x^3)\cdots. \tag{3} $$
For convenience define the variable $\, q := x^{1/5} \,$
so that $\, x = q^5.\,$ Define the functions
$$ r := \frac{f(-x,-x^4)}{f(-x^2,-x^3)} =
\frac{(1-x)(1-x^4)(1-x^6)(1-x^9)\cdots}
{(1-x^2)(1-x^3)(1-x^7)(1-x^8)\cdots}, \tag{4} $$
$$ y := f(-x^5)/f(-q),\;\;\text{ and }
\;\; z := (f(-x^5)/f(-x))^6. \tag{5} $$
For technical reasons introduce the variants
$$ R := q\,r, \quad Y := q\,y, \quad Z := x\,z. \tag{6} $$
Somehow Ramanujan has proved that
$$ R\,Y^{-1} = 1 - R - R^2 \tag{7} $$
(which is a series multisection) and also proved that
$$ R^5Z^{-1} = 1 - 11\,R^5 - R^{10}. \tag{8} $$
This implies that
$$ R^4\,Y\,Z^{-1} = (1-11\,R^5-R^{10})/(1-R-R^2). \tag{9} $$
Dividing the two polynomials gives the result
$$ R^4YZ^{-1}=R^8-R^7+2R^6-3R^5+5R^4+3R^3+2R^2+R+1.\tag{10} $$
Divide both sides by $\,R^4\,$ and pair up the
powers of $\,R\,$ and $\,q\,$ to get
$$ YZ^{-1} \!=\! 5 \!+\! (R^{-4}\!-\!3R) \!+\!
(R^{-3}\!+\!2R^2) \!+\!
(2R^{-2}\!-\!R^3) \!+\! (3R^{-1}\!+\!R^4) \tag{11} $$
which is a series multisection.
Use equations $(5),(6)$ to rewrite this as
$$ f(-x^5)q^{-4}f(-q)^{-1}z^{-1} \!=\! A_0\!+\! A_1 \!+\!A_2 \!+\! A_2 \!+\! A_4 \;\text{ where } \;A_0 \!=\! 5. \tag{12} $$
Now also use series multisection to get
$$ f(-q)^{-1} = p_0 + p_1 + p_2 + p_3 + p_4 \tag{13} $$
where $$ p_k := \sum_{n=0}^\infty p(5n+k)\,q^{5n+k} =
q^k\sum_{n=0}^\infty p(5n+k)\,x^n. \tag{14} $$
Use the multisection equation $(12)$
to select integer powers of $\,x\,$ to get
$$ f(-x^5)\,p_4\, z^{-1} = 5. \tag{15} $$
Use equations $(5),(6),(8)$ to get
$$ z^{-1} = r^{-5} - 11\,x - x^2\,r^5. \tag{16} $$
The final result is
$$ f(-x^5)\,p_4 = \frac5{r^{-5} -11\,x -x^2\,r^5} \tag{17}$$
where $\,r\,$ is denoted by $\,\xi\,$ in Ramanujan's paper.
|
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|
Find real $a$, $b$, $c$, $d$ satisfying $(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$
If real numbers $a$, $b$, $c$, $d$ satisfy
$$(1-a)^2+(a-b)^2+(b-c)^2+c^2=\frac{1}{4}$$
then find $(a,b,c,d)$.
What I try:
$$1+2a^2+2b^2+2c^2-2a-2ab-2bc=\frac{1}{4}$$
$$8a^2+8b^2+8c^2-8a-8ab-8bc+7=0$$
How do I solve it? Help me, please.
|
Calculalte discriminant twice and the discriminat tells the value is only one.
$2a^2-2(b+1)a+b^2+(b-c)^2+c^2+1-1/4=0 $
$D/4=(b+1)^2-2{2b^2-2bc+2c^2+3/4}\geq0$
We get,
$3b^2-2(1+2c)b+4c^2+1/2=0 $
$D/4=(1+2c)^2-3(4c^2+1/2)\geq0 $
$-8c^2+4c-1/2\geq0 $
$16(c-1/4)^2\leq0 $
$c=1/4 $
$3b^2-3b+3/4<=0 $
$(2b-1)^2\leq0 $
$b=1/2$
$2a^2-3a+1/4+1/16+1/16+1-1/4=0 $
$16a^2-24a+9=0 $
$(4a-3)^2=0$
$a=3/4$
|
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|
Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.
$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$
Any tips will be helpful.
Sorry, but I don't want to use the Taylor series as well.
|
Result 1: $\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} = \frac{1}{3}$
Proof. Note that $\sin x = x - \frac{x^3}{3!} + o(x^5).$
Thus, $\sin^2x = x^2 - 2x\frac{x^3}{3!} + o(x^5).$
This, gives $x^2 - \sin^2 x = \frac{x^4}{3} + o(x^5),$ and the result follows.
Result 2: $\displaystyle\lim_{x\to0} \dfrac{\sin^4x}{x^4} = 1$
Proof. Follows trivially from $\displaystyle \lim_{x\to0} \frac{\sin x}{x} = 1.$
Result 3: $\ln(1 - x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} + o(x^4).$ (Expansion is valid near $0$)
Proof. Standard result. This is the Taylor expansion of $\ln(1-x)$ near $0$.
Solution.
$\displaystyle\lim_{x\to0}\dfrac{x^2 + 2\ln(\cos x)}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 + \ln(\cos^2 x)}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 + \ln(1 - \sin^2 x)}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 + (-\sin^2x - \frac{\sin^4x}{2} + o(x^6))}{x^4}$
$=\displaystyle\lim_{x\to0}\dfrac{x^2 - \sin^2x}{x^4} - \dfrac{1}{2}\displaystyle\lim_{x\to0}\dfrac{\sin^4x}{x^4} + 0$
$=\dfrac{1}{3} - \dfrac{1}{2}$
$=\boxed{-\dfrac{1}{6}}$
|
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|
Find a closed formula for the following recursive function How can I express a closed-form formula for the following equation?
$$f(n)=f(n-1)+\frac{C}{f(n-1)} $$
Where $C>0$ and $f(0)=\sqrt{C}$.
|
This is the best I can do. It seems like
\begin{align}
\frac{f(n)-f(n-1)}{n-(n-1)} = \frac{C}{f(n-1)}.
\end{align}
Let us solve the differential equation
\begin{align}
f' = \frac{C}{f} \ \ \implies \ \ \frac{d}{dx}(f)^2 = C
\end{align}
which means $f^2 = Cx$ or $f(x) = \sqrt{C}\sqrt{x}$. Observe
\begin{align}
\sqrt{C}\sqrt{n}-\sqrt{C}\sqrt{n-1} = \frac{\sqrt{C}}{\sqrt{n}+\sqrt{n-1}}
\end{align}
then it follows
\begin{align}
\frac{\sqrt{C}}{2f(n)}=\frac{\sqrt{C}}{2\sqrt{n}}\leq f(n)-f(n-1) \leq \frac{\sqrt{C}}{2\sqrt{n-1}} = \frac{\sqrt{C}}{2f(n-1)}
\end{align}
|
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|
sum of coefficient of all even power of $x$
The sum of all Coefficient of even power of $x$ in
$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$
what i try
for $n=1,$ we have
$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$
for $n=2,$ we have
$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$
$=(1+1+1)-(1+1)+(1+1+1)-(1+1)+(1+1+1)=5$
so in this way , get sum of coefficient in original expression is $2n+1$
but How do i solve it without substituting value of $n$, Help me
|
These are two standard geometric series,
one with ratio $-x$,
the other with ratio $x$.
Their product is
$(1-(-x)^{2n+1})(1-x^{2n+1})/((1+x)(1-x))\\
=(1+x^{2n+1})(1-x^{2n+1})/(1-x^2)\\
=(1-x^{4n+2})/(1-x^2)
$
At $x=1$, applying Hoppy,
we get
$(4n+2)x^{4n+1}/(2x)
=(2n+1)x^{4n}
\to 2n+1
$.
|
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|
$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}$$
I really don't want to divide the equations by $x-y$ and $x+y$, respectively. I am taught to divide by expressions containing variables only in special cases. Is there any other way here?
|
After considering of cases $x=y$ or $x=-y$ use
$$7(x^2+xy+y^2)=19(x^2-xy+y^2).$$
|
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|
Interval Length Needed For Interpolation Let $f(x)=\sqrt{x}$ defined on $[1,2]$, What is the length needed between the sampling points such that the approximation error by interpolation polynomial of order $2$ will not exceed $5*10^{-8}$
We know that $|f(x)-P_n(x)|=|\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)...(x-x_n)|$
So to bound $f^{(n+1)}(c)$ we take the $3$rd derivative of $x^{\frac{1}{2}}$ which is $$\frac{3}{8x^{\frac{5}{2}}}$$ so the Max is obtained at $x=1$
So we are looking for:
$$|\frac{\frac{3}{8}}{(3)!}(x-x_0)(x-x_1)(x-x_2)|\leq 5*10^{-8}\iff |\frac{3}{48}(x-x_0)(x-x_1)(x-x_2)|\leq 5*10^{-8}$$
Which is:
$$|(x-x_0)(x-x_1)(x-x_2)|\leq 8*10^{-7}$$
Assuming each $\Delta x_i=h$ we have:
$$h^3\leq 8*10^{-7}\iff h\leq 9.28*10^{-3}$$
So we need to take $$\Delta x_i = 9*10^{-3} $$
Is it correct?
|
If the quadratic interpolation is used for $|x-x_i|\le\frac12h$, then the maximum of
$|(x-x_{i-1})(x-x_i)(x-x_{i-1})|$ is at $x=x_i\pm \frac12h$, so that the maximum of the error bound is in fact $\frac{3/8}{3!}⋅\frac38⋅h^3=3⋅2^{-7}⋅h^3$, and to get this smaller than the prescribed error $5\cdot10^{-8}$ gives
$$
h^3\le \frac{2^6}3⋅10^{-7}\iff h\le 0.04\sqrt[3]{0.1/3}=0.01287..
$$
Even taking the full interval $|x-x_i|\le h$, the maximum of the product is at $x=x_i\pm\frac1{\sqrt3}h$, so that the bound is $\frac{3/8}{3!}\frac{2}{3^{1.5}}⋅h^3=\frac1{8⋅3^{1.5}}$, which compared to the tolerance gives
$$
h^3\le 4⋅3^{1.5}⋅10^{-7}\iff h\le 0.01\sqrt{3}\sqrt[3]{0.4}=0.0127618..
$$
which allows to chose $h=0.01$.
|
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|
show that the following recursive series is bounded and monotonic I need to show that the following recursive series $$a_1=1, \quad a_{n+1}=\frac{a_n+3}{5}$$ is bounded and monotonic.
At first, I want to show that the series is bounded from below with $\frac{3}{4}$. By induction:
For $n=1$ we have $a_1=1>\frac{3}{4}$. We assume it is true for $n$, i.e. $a_n>\frac{3}{4}$ and we check for $a_{n+1}$:
$$a_{n+1}=\frac{a_n+3}{5}>\frac{\frac{3}{4}+3}{5}=15>\frac{3}{4}.$$
We conclude that $a_n>\frac{3}{4}$ for all n.
The series is increasing because
$a_{n+1}-a_{n}=\frac{-4a_n+3}{5}=\frac{-4(a_n-\frac{3}{4})}{5}<0$$
so we have also $a_n \leq a_1=1$.
Is my solution ok? Thank you for your help.
|
You can solve the $a_n$. Actually, for $a_{n+1}=\frac{a_n+3}5$
$$\therefore a_{n+1}+k=\frac{a_n+3+5k}5 $$
Make $k=3+5k$, so $k=-\frac 34$, now
$$a_{n+1}-\frac 34=\frac{a_n+3-\frac{15}4}5=\frac{a_n-\frac 34}5 $$
Let $b_n=a_n-\frac 34$, now $b_1=\frac 14$, and
$$b_{n+1}=\frac{b_n}5$$
So we know $a_n-\frac 34=b_n=\frac 1{4\cdot 5^{n-1}}$, so $a_n=\frac 34+\frac 1{4\cdot 5^{n-1}}$, it is bounded and monotonic.
|
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|
Prove that : $a,b,c\in\mathbb{Z}$ then : $abc\equiv 0\pmod{3} \implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$ Prove that :
$a,b,c\in\mathbb{Z}$ then :
$$abc\equiv 0\pmod{3} \implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
My try :
Let $a\not\equiv 0\pmod{3}$
Then :
$$a\equiv 1,2\pmod{3}$$
So :
$$a^{3}\equiv ±1\pmod{9}$$
Similarly :
$$b^{3}\equiv ±1\pmod{9}$$
$$c^{3}\equiv ±1\pmod{9}$$
$$\implies a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
Is my solution correct ??
And we can generalized or no ?
$$abc\equiv 0\pmod{3} \iff a^{3}+b^{3}+c^{3}\equiv 0\pmod{9}$$
|
As pointed in the comments , the equivalence is false but it's converse is always true . $$a^3+b^3+c^3 \equiv 0 \mod9\implies abc\equiv 0\mod3$$
Since a cube can be written as $x^3\equiv 0,1,8 \mod 9$ , the only way to add these residues to be divisible by $9$ is $(0,0,0)$ and $(0,1,8)$ and their permutations.
In either the case we get at least one cube divisible by $9$ implying that the number is divisible by $3$ and the conclusion follows.
The problem with your attempt is that you assumed $$a^3 \equiv b^3\equiv c^3 \equiv \pm 1\mod 9$$ which implies that $3\not\mid a,b,c$ , which means that $abc\equiv 0\mod 3$ is false.
|
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|
Can a non-right triangle be solved with only two sides if it is inside a parallelogram? In this triangle
I am given 2 side lengths for one triangle and two side lengths for the parallelogram. I am asked to find the length of m (FE) and n (DE)
I am given the lenghts:
*
*h (AC) = 9
*k (AF) = 15
*f (AB) = 16
I don't see how to use Law of Sines because I don't have any angles and I don't see how to use Law of Cosines to solve triangle ACF because I am missing a side length.
Am I missing a concept or is this problem missing given information?
|
I don't believe there's enough information. Let's let
$$A=(0,0)$$
$$B=(16,0)$$
$$C=(9\cos\theta,9\sin\theta)\text{ for some }0<\theta<90^{\circ}$$
$$D=B+C=(16+9\cos\theta,9\sin\theta)$$
$$F=\left(\sqrt{15^2-9^2\sin^2\theta},9\sin\theta\right)$$
Then we have a parallelogram $ABDC$ with a point $F$ on $CD$, such that $|AC|=9$, $|AF|=15$, and $|AB|=16$. Our next step is to find $E$. We'll be done if we can show that $|AE|$ is a non-constant function of $\theta$.
$E$ is the intersection of the lines determined by segments $AF$ and $BD$. To find the coordinates of $E$, we'll first find the equations of these lines. Using the point-slope form, we have that the equation for the line determined by segment $AF$ is:
$$y=\frac{9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}}\cdot x$$
Using the point-slope form, we have that the equation for the line determined by segment $BD$ is:
$$y=\frac{9\sin\theta}{9\cos\theta}\cdot (x-16)$$
Hence we can find the $x$-coordinate of $E$ by solving
$$\frac{9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}}\cdot x=\frac{9\sin\theta}{9\cos\theta}\cdot (x-16)$$
This gives us that
$$x=\frac{16\cdot\sqrt{15^2-9^2\sin^2\theta}}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
We can then plug this into the equation for the line determined by segment $AF$ to obtain that
$$y=\frac{16\cdot9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
Hence
$$E=\left(\frac{16\cdot\sqrt{15^2-9^2\sin^2\theta}}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta},\frac{16\cdot9\sin\theta}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}\right)$$
It follows that
$$|AE|=\frac{16\cdot15}{\sqrt{15^2-9^2\sin^2\theta}-9\cos\theta}$$
Note that if $\theta=30^{\circ}$, then $|AE|\approx36.8$, but if $\theta=60^{\circ}$, then $|AE|\approx28.9$. So $|AE|$ is a non-constant function of $\theta$.
Finally, note that $m=|AE|-15$. So $m$ is a non-constant function of $\theta$. We need more information.
|
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|
A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as follows:-
Let the equation of circle be $x^2+y^2+2gx+2fy+c=0$
As center lies on the line $x+y=2$
$$-g-f=2$$
$$g+f=-2\tag{1}$$
As circle passes through the point $(0,1)$
$$1+2f+c=0$$
$$2f+c=-1\tag{2}$$
As $4x – 3y + 4 = 0$ is tangent to the circle
$$\dfrac{\left|-4g+3f+4\right|}{5}=\sqrt{g^2+f^2-c}$$
Squaring both sides
$$16g^2+9f^2-24gf+16+8(-4g+3f)=25g^2+25f^2-25c$$
$$9g^2+16f^2+24gf+32g-24f-25c-16=0$$
Eliminating $g$ with the help of equation $(1)$
$$9(-2-f)^2+16f^2+24(-2-f)f+32(-2-f)-24f-25c-16=0$$
$$9(4+f^2+4f)+16f^2-48f-24f^2-64-32f-24f-25c-16=0$$
$$f^2-68f-44-25c=0$$
Eliminating $c$ with the help of equation $(2)$
$$f^2-68f-44-25(-1-2f)=0$$
$$f^2-18f-19=0$$
$$f^2-19f+f-19=0$$
$$f=19,-1$$
$$(g,f,c)\equiv (-1,-1,1),(-21,19,-39)$$
So equations are $x^2+y^2-2x-2y+1=0$, $x^2+y^2-42x+38y-39=0$
But this got too long, any shorter method?
|
hint...let the centre of the circle be $(p, 2-p)$ so you can use the distances to form a quadratic in $p$
|
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|
How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(4(1-(\cos x)^2 + 2\cos x -1) & = \cos 9x + \cos 8x + 1
\end{align*}
I don't see any way to get rid of $8x$ and $9x$ as arguments to have same angles from there
|
Using Euler's formula $e^{ix} = \cos x+i\sin x$ we can simplify this a bit:
\begin{align}
\sin 2x\sin x + \cos^2x &= \frac1{2i}\left(e^{2ix}-e^{-2ix}\right)\frac1{2i}\left(e^{ix}-e^{-ix} \right) + \frac14\left(e^{ix} + e^{-ix}\right)^2\\
&= \frac14\left(-e^{3ix} + e^{ix} +e^{-ix}-e^{-3ix} + e^{2ix} + 2 + e^{-2ix} \right)
\end{align}
and
\begin{align}
\sin 5x\sin 4x + \cos^24x &= \frac1{2i}\left(e^{5ix}-e^{-5ix}\right)\frac1{2i}\left(e^{4ix}-e^{-4ix}\right)+\frac14\left(e^{4ix}+e^{-4ix}\right)^2\\
&= \frac14\left(-e^{9ix} + e^{ix} +e^{-ix} -e^{-9ix} + e^{8ix} + 2 + e^{-8ix} \right).
\end{align}
Factoring and subtracting out common terms, we have
$$
-(e^{3ix} +e^{-3ix}) +e^{2ix}+e^{-2ix} = -(e^{9ix}+e^{-9ix}) +e^{8ix}+e^{-8ix},
$$
and hence
$$
\cos9x + \cos2x = \cos8x +\cos3x.
$$
From the identity
$$
\cos\theta + \cos\varphi = 2\cos\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta-\varphi}2\right)
$$
this becomes
$$
\cos\left(\frac{11}2x\right)\left(\cos\left(\frac72x\right) - \cos\left(\frac52x\right) \right) = 0.
$$
From the identity
$$
\cos\theta - \cos\varphi = -2\sin\left(\frac{\theta+\varphi}2\right)\sin\left(\frac{\theta-\varphi}2\right)
$$
this becomes
$$
\cos\left(\frac{11}2x\right)\sin(3x)\sin\left(\frac12x\right)=0.
$$
Hence the solutions are
$$
x = \frac{2\pi\left(n+\frac12\right)}{11},\ n=-5,-4,-3,-2,-1,0,1,2,3,4
$$
and
$$
x = \frac{n\pi}3,\ n=-2,-1,0,1,2,3.
$$
|
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|
Prove that $\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$ I need to prove that: $$\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$$
I am new to inequalities so I only tried C-S and AM-GM but none of those work.
Any hints on how to proceed here?.
|
$(9x^4 + y^4)(x^2 + y^2)\ \ge\ (3x^3 + y^3)^2\ \ge\ |6x^3y^3|$
|
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|
System of equations modulo $n$ Suppose we have the following system of equations module $n$
\begin{align*}
2x-y & \equiv 1\mod{n}\\
4x+3y & \equiv 2 \mod{n}
\end{align*}
Determine the integers $n$ for which the system has solutions.
I have posted my solution as an answer below.
|
Let $2x-y=1+an$ and $4x+3y=2+bn$ where $a,b$ are integers
$2+bn=4x+3(2x-1-an)\iff5(2x-1)=(3a+2b)n$
As $(3,2)=1, $ any integer can be expressed in terms of $3a+2b$
Let us find $a,b$ such that $3a+2b=5c\iff3(a-c)=2(c-b)$
$\dfrac{2(c-b)}3=a-c$ which is an integer
$\implies3|2(c-b)\iff3|(c-b),c-b=3d$ (say for some integer $d$)
$\implies b=c-3d,a=c+2d$
$\implies2x-1=cn$
As the the left hand side is odd, the above equation will have a solution iff $n$ is odd
Please check for $y$
|
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|
Search for a Rationalizing Factor Just something that I've been thinking about lately, and can't figure how to generalize efficiently:
Say we have a surd of the form $\frac{1}{\sqrt2+\sqrt3+\sqrt5+\sqrt7}$, or something along those lines. How can I find the rationalizing factor, without brute force? Does there exist one?
We could add more numbers of the form $\sqrt p$ in the denominator, where $p$ is prime, and even multiply by certain coefficients for the very general case -
$$\frac{1}{(a_1\sqrt2+a_2\sqrt3+a_3\sqrt5+a_4\sqrt7+...a_i\sqrt p +...)}$$, where $a_i \in Z$.
Any ideas? This seems like an interesting problem!
|
You have for example
$$(a+b\sqrt 2+c\sqrt 3 +d\sqrt 6 )(a-b\sqrt 2 +c\sqrt 3 -d\sqrt 6)(a+b\sqrt 2-c\sqrt 3 -d\sqrt 6 )(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)=a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4$$
Note that the second term has the signs on the terms with $\sqrt 2$ flipped, the third on the ones with $\sqrt 3$ and the fourth in both simultaneously. So you can easily rationalize this:
$$\frac 1{a+b\sqrt 2+c\sqrt 3 +d\sqrt 6}=\frac{(a-b\sqrt 2 +c\sqrt 3 -d\sqrt 6)(a+b\sqrt 2-c\sqrt 3 -d\sqrt 6 )(a-b\sqrt 2-c\sqrt 3+d\sqrt 6)}{a^4 - 4 a^2 b^2 - 6 a^2 c^2 - 12 a^2 d^2 + 48 a b c d + 4 b^4 - 12 b^2 c^2 - 24 b^2 d^2 + 9 c^4 - 36 c^2 d^2 + 36 d^4}$$
You can always do this, no matter how many square roots you have. But in general if you have the square roots of n primes you have to multiply $n^2$ terms, which is a lot of effort. So it's better to choose one of the primes to remove from the expression and repeat, for example:
$$
\begin{split}\frac 1{3\sqrt 2-\sqrt 3+2\sqrt 5+\sqrt 7}&=\frac{3\sqrt 2-\sqrt 3+2\sqrt 5-\sqrt 7}{34 - 6\sqrt 6 + 12 \sqrt{10} - 4 \sqrt{15}}\\ &=\frac{(3\sqrt 2-\sqrt 3+2\sqrt 5-\sqrt 7)(34-6\sqrt 6 -12\sqrt{10}+4\sqrt{15})}{72\sqrt 6 -308}\end{split}$$
Note that in the second line I flipped the signs on the $\sqrt{10}$ and $\sqrt{15}$ terms, which are the multiples of 5. Lots of questions like this can be solved with abstract algebra.
|
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|
Logarithm Subtraction and Division with Same Bases I'm rusty on logarithms. What is the approach to a problem like this? Any hints would be appreciated.
I'm thinking the subtraction on the numerator and denominator can become division since the bases are the same?
|
The identities $\log_a b + \log_a c \equiv \log_a(bc)$ and
$b\log_a c \equiv \log_a c^b$ will be needed here. You can then do
\begin{align*}
\frac{\log_2 24 - \frac 12 \log_2 72}
{\log_3 18 - \frac 13 \log_3 72}
&=
\frac{\log_2 24 - \log_2 \sqrt{72}}
{\log_3 18 - \log_3 \sqrt[3]{72}} \\
&=
\frac{\log_2 \frac{24}{\sqrt{72}}}
{\log_3 \frac{18}{\sqrt[3]{72}}} \\
&=
\frac{\log_2 \frac{24}{3\sqrt 8}}
{\log_3 \frac{18}{2\sqrt[3]{9}}} \\
&=
\frac{\log_2 \frac{8}{\sqrt 8}}
{\log_3 \frac{9}{\sqrt[3]{9}}} \\
&=
\frac{\log_2 8 - \log_2 \sqrt 8}
{\log_3 9 - \log_3 \sqrt[3]{9}} \\
&=
\frac{\log_2 2^3 - \log_2 2^{3/2}}
{\log_3 3^2 - \log_3 3^{2/3}} \\
&=
\frac{3 - \frac 32}
{2 - \frac 23} \quad \text{by definition} \\
&= \frac 98
\end{align*}
|
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|
Is $x^{14}+x^7+1$ irreducible over $Q[x]$? Is $x^{14}+x^7+1$ irreducible over $\mathbb{Q}[x]$?
I think it is, but I'm not able to justify it using any existing criterion (e.g. Eisenstein). Any help?
Indeed, this is a question I encounter on a linear algebra one. The original question gives a $8 \times 8$ real matrix satisfying $A^{21}=I$ and asks to prove that $\mathbb{R}^8$ can be decomposed into the direct sum of 4 2-dimensional vector subspace invariant w.r.t. $A$.
My attempt was to find the minimal polynomial then discuss several cases, which need the factor of $x^{21}-1$. Any hint on the original question is also appreciated.
|
$x^{14}+x^7+1=(x^7-\omega)(x^7-\omega^2)$
$$=(x^7-\omega^7)(x^7-\omega^{14}) ----(1)$$
Factorizing each factors further we get
$$(x^7-\omega^7)=(x-\omega)(x^6+x^5\omega +x^4\omega^2 +x^3 +x^2\omega +x \omega^2+1)$$
$$=((1+x^3+x^6)+(x^2+x^5)\omega +(x+x^4)\omega^2 ---(2)$$
Similarly $(x^7-\omega^{14})=(x-\omega^2)((1+x^3+x^6)+(x^2+x^5)\omega^2+(x+x^4)\omega) -----(3)$
$$let\,p(x)=(1+x^3+x^6)\,,\,q(X)=(x^2+x^5)\,,\,r(x)=(x+x^4)$$
Using (1) , (2) , and (3) we get
$$x^{14}+x^7+1=(x-\omega)(x-\omega^2)(p(x)+q(x)\omega +r(x)\omega^2)(p(x)+q(x)\omega^2 +r(x)\omega)$$
$$=(1+x+x^2)((p(x))^2+(q(x))^2+(r(x))^2-p(x)q(x)-q(x)r(x)-r(x)p(x))$$
|
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|
Why the wrong answer in integral calculus? Calculate the volume of the solid obtained by rotating the region about the $x$-axis, and bounded by $y=x^2+1, y=3-x^2$.
My calculations:
$V=\pi \int_a^b f^2(x) dx; \; V = V_1 + V_2 = \pi \int_{-1}^1 (3-x^2)^2 dx - \pi \int_{-1}^1 (x^2+1)^2 dx;$
$V_1 = \pi \int_{-1}^1 (3-x^2)^2 dx = \pi \int_{-1}^1 (9-3x^2-3x^2+x^4) dx = \pi (9x-2x^3+\frac{x^5}{5} \; |_{-1}^1 )=14.4 \pi;$
\begin{align*}
V_2 & = \pi \int_{-1}^1 (x^2+1)^2 dx\\
& = \pi \int_{-1}^1 (x^2+1)(x^2+1) dx\\
& = \pi \int_{-1}^1 (x^4+x^2+x^2+1) dx\\
& = \pi\left(\frac{x^5}{5}+\frac{2}{3} x^3+x \; \bigg|_{-1}^1\right)\\
& = \pi\left(\frac{1}{5}+\frac{2}{3}+1+\frac{1}{5}+\frac{2}{3}+1\right)\\
& = 2\pi\left(\frac{1}{5}+\frac{2}{3}+1\right)\\
& =2\pi\left(\frac{3}{15}+\frac{10}{15}+1\right)\\
& =\left(2+\frac{26}{15}\right)\pi;
\end{align*}
My answer: $V=14.4\pi - (2+\frac{26}{15})\pi;$, but this answer does not fit.
|
I think it's worth noting that your method, while completely correct (both in result and in detail as far as anyone has found), is far more laborious (and hence more error-prone) than it needs to be.
In this case, the "washer method" is easier.
For the washer method, we note that a given value of $x$ the disk inside the curve $y=3-x^2$ has area $\pi(3-x^2)^2$ and the disk inside the curve $y = x^2 + 1$ has area $\pi(x^2 + 1)^2$; then we remove the smaller disk from the larger disk, leaving a "washer" (also called an annulus), which is the shape we get when we take the line segment from $(x,x^2 + 1)$ to $(x,3-x^2)$ and rotate it around the $x$-axis.
The area of the washer is
$$\pi(3-x^2)^2 - \pi(x^2 + 1)^2 = \pi ((3-x^2)^2 - (x^2 + 1)^2)$$
(in general, $\pi(r_1^2 - r_2^2)$ where $r_1$ is the larger radius and $r_2$ is the smaller),
which we integrate from $-1$ to $1.$ Note that this is exactly equivalent to your method, since
$$ \int_{-1}^1 (\pi(3-x^2)^2 - \pi(x^2 + 1)^2)\,\mathrm dx
= \int_{-1}^1 \pi(3-x^2)^2\,\mathrm dx - \int_{-1}^1 \pi(x^2 + 1)^2\,\mathrm dx. $$
As with the disk method you used, the integral is usually simplified by taking the constant factor $\pi$ outside:
$$ V = \pi \int_{-1}^1 (r_1^2 - r_2^2)\,\mathrm dx
= \pi \int_{-1}^1 ((3-x^2)^2 -(x^2 + 1)^2)\,\mathrm dx. $$
Now here's the neat thing about this method for this problem. Note that
\begin{align}
(3-x^2)^2 - (x^2 + 1)^2
&= (3^2 - 2(3)(x^2) + (x^2)^2) - ((x^2)^2 + 2(1)(x^2) + 1^2) \\
&= (9 - 6x^2 + x^4) - (x^4 + 2x^2 + 1) \\
&= 9 - 6x^2 + x^4 - x^4 - 2x^2 - 1 \\
&= 8 - 8x^2.
\end{align}
So we just need to integrate
$$ \pi\int_{-1}^1 (8 - 8x^2)\,\mathrm dx, $$
which is a lot less work than two integrals with both $x^4$ and $x^2$ terms in them as well as constant terms.
|
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|
Prove $\frac{b^{4} - 24 a^{4}}{ab} \ge 3b^{2} + 6ab - 28a^{2} $ I made up a simple problem. I wonder if there are other ways to solve this problem? or maybe whether or not the inequality seems very obvious?
If $a,b$ are positive real numbers and $b$ is at least twice of $a$, prove that
$$ \frac{b^{4} - 24 a^{4}}{ab} \ge 3b^{2} + 6ab - 28a^{2} $$
Solution:
We want to prove
$$ b^{4} - 3ab^{3} - 6a^{2}b^{2} + 28a^{3}b - 24a^{4} \ge 0 $$
Now let $b = at$ with $t \ge 2$. So the inequality becomes:
$$ t^{4} - 3t^{3} - 6t^{2} + 28t - 24 \ge 0 $$
Notice equality when $t=2$, so $(t-2)$ is a factor:
$$ (t-2)(t^{3} - t^{2} -8t + 12) \ge 0 $$
$$ (t-2)(t-2)(t^{2}+t-6) = (t-2)^{3}(t+3) \ge 0 $$
|
Your inequality equivalent to
$b^4 -24a^4 \geq ab(3b^2 +6ab-28a^2)$
We have: $LHS-RHS = (b-2a)^3 (3a+b) \geq 0$ (because: $b$ is at least twice of $a$)
|
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|
I proved that $\sqrt{z}$ is a continuous function. My proof is not smart. Please show me a smarter proof. Let $U:=\{z \in \mathbb{C} - \{0\} | -\pi < \arg z < \pi\}$.
For $z \in U$, the equation $w^2=z$ has two solutions $w=\pm(u + i v)$, where $u>0$.
Let $f : U \to \mathbb{C}$ be a function such that $f(z):=u+iv$, where $u>0$.
We define $\sqrt{z}:=f(z)$.
Prove that $\sqrt{z}$ is continuous.
Proof:
Let $z=x+iy\in U$ and $w^2 =(u+iy)^2= x+iy$ and $u>0$.
$u^2-v^2=x$ and $2uv=y$.
*
*If $y=0$, then $x>0$.
And $u=0$ or $v=0$.
And $u^2-v^2=x>0$.
So, $v=0$ and $u=\sqrt{x}$.
*Suppose $y>0$.
$v=\frac{y}{2u}$.
$u^2-\frac{y^2}{4u^2}=x$.
$4u^4-4xu^2-y^2=0$.
$u^2=\frac{2x+\sqrt{4x^2+4y^2}}{4}=\frac{x+\sqrt{x^2+y^2}}{2}$.
$u=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$.
$v^2=u^2-x=\frac{-x+\sqrt{x^2+y^2}}{2}$.
$v=\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
*Suppose $y<0$.
$v=\frac{y}{2u}$.
$u^2-\frac{y^2}{4u^2}=x$.
$4u^4-4xu^2-y^2=0$.
$u^2=\frac{2x+\sqrt{4x^2+4y^2}}{4}=\frac{x+\sqrt{x^2+y^2}}{2}$.
$u=\sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$.
$v^2=u^2-x=\frac{-x+\sqrt{x^2+y^2}}{2}$.
$v=-\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
So, for $z=x+iy \in U \cap \{x+iy \in \mathbb{C} | y\geq0\}$, $\sqrt{z} = \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} + \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} i$.
So, for $z=x+iy \in U \cap \{x+iy \in \mathbb{C} | y<0\}$, $\sqrt{z} = \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}} - \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}} i$.
Obviously $(x, y) \mapsto \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto \sqrt{\frac{x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\{(x, y) | x+yi \in U\}$.
Obviously $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y>0\}\}$.
Obviously $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ is continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y<0\}\}$.
Let for $(x, y) \in U\cap\{x+iy \in \mathbb{C} | y\geq 0\}$ $g(x, y) := \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
Let for $(x, y) \in U\cap\{x+iy \in \mathbb{C} | y< 0\}$ $g(x, y) := -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$.
Obviously $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ and $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ are continuous on $\mathbb{R}^2$.
So, $(x, y) \mapsto \sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ and $(x, y) \mapsto -\sqrt{\frac{-x+\sqrt{x^2+y^2}}{2}}$ are continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y=0\}\}$ and $\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}} = -\sqrt{\frac{-a+\sqrt{a^2+b^2}}{2}} = 0$ for $(a, b) \in U\cap\{x+iy \in \mathbb{C} | y=0\}$.
So, $(x, y) \mapsto g(x,y)$ is continuous on $\{(x, y) | x+yi \in U\cap\{x+iy \in \mathbb{C} | y=0\}\}$.
|
It's easier to note that if $z = x + iy = re^{i\theta}$ then $\sqrt{z} =
r^{1 \over 2}\cos{\theta \over 2} + i r^{1 \over 2}\sin{\theta \over 2}$, where
$r^{1 \over 2} = \sqrt{x^2 + y^2}$ and $\theta$ (suitably defined) are both continuous functions of $x$ and $y$. Here you can use branches of $\theta = \arctan({y \over x}), \theta = \arccos({x \over \sqrt{x^2 + y^2}})$, or
$\theta = \arcsin({y \over \sqrt{x^2 + y^2}})$ depending on the region.
Thus the real and imaginary parts of $\sqrt{z}$ are expressible as products of compositions of continuous functions and as a result $\sqrt{z}$ is continuous
|
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|
If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ or $y$ is divisible by $3$, but not both, then:
$\begin{align}x \equiv \pm 1 \pmod{3} \land y \equiv 0 \pmod{3}
&\iff x^2 \equiv 1 \pmod{3} \land y^2 \equiv 0 \pmod{3}
\\
&\iff x^2 - y^2 \equiv 1 \pmod{3}\end{align}$
So if $x$ or $y$ is divisible by $3$ but not both then no values for $x$ and $y$ will ever make $x^2 - y^2$ divisible by $3$.
*If $x$ and $y$ are divisible by $3$:
Since both $x$ and $y$ are divisible by $3$ they can be expressed $x = 3m$ and $y = 3n$ for some integers $m$, $n$.
$(3m)^2 - (3n)^2 = 1995 \iff 9m^2 - 9n^2 = 1995 \iff 3(m^2-n^2) = 665$
And it's easy to see that there are no whole number solutions for m and n, and thus there are no whole number solutions for x and y such that $x^2 - y^2 = 1995$ where x or y (or both) is divisible by 3.
Is this a reasonable proof? Are there more elegant ways to do it?
|
Your argument is fine and well expressed. More succinctly you could paraphrase parts of your argument as follows.
If precisely one of $x,y$ is divisible by $3$ then $x^2-y^2$ is not divisible by $3$.
If both of $x,y$ are divisible by $3$ then $x^2-y^2$ is divisible by $9$.
|
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|
Coefficients that make a quartic trivariate polynomial non-negative Let
$$f(x,y,z)=a_1 x^4+a_2 x^3+a_3 x^2+a_4 x^2y+a_5 x^2z+a_6 y^2+a_7 z^2+a_8 xy+a_9 xz+a_{10} yz+a_{11}x+a_{12}y+a_{13}z$$
be a polynomial of degree $4$. Can we determine coefficients $a_1, \dots, a_{13}$ such that $f(x,y,z) \ge 0$ for all $x,y,z\in\mathbb{R}$?
Any reference (in particular, a systematic approach to solve such problems), suggestion, idea, or comment is welcome. Thank you!
|
We can use the Newton polytope. A positive polynomial should have a convex hull defined by even degree monomials. Our polynomial has a polytope representation as can be observed in the following plot.
In black the convex hull monomials which are
$$
\left[
\begin{array}{ccc}
x & y & z\\
4 & 0 & 0 \\
0 & 2 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2 \\
0 & 0 & 1 \\
1 & 0 & 0 \\
\end{array}
\right]
$$
and in red the inner monomials which are.
$$
\left[
\begin{array}{ccc}
x & y & z\\
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
2 & 0 & 0 \\
2 & 0 & 1 \\
2 & 1 & 0 \\
3 & 0 & 0 \\
\end{array}
\right]
$$
Accordingly we should have $a_{11}=a_{12}=a_{13}=0$. After that, the polytope is represented as
Now the convex hull is formed by
$$
\left[
\begin{array}{ccc}
x & y & z\\
4 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2 \\
2 & 0 & 0 \\
\end{array}
\right]
$$
and this arrangement has chance of positivity. To proceed we can obtain this convex hull
$$
\{x^4,\ x^2,\ y^2,\ z^2\}
$$
using a convenient monomial basis as
$$
\{x^2,\ x,\ y, \ z\}
$$
so our polynomial form can be represented as
$$
f(x,y,z) = Z^{\dagger}B Z
$$
with $Z = \{x^2,\ x,\ y, \ z\}^{\dagger}$ and $B$ a $4\times 4$ matrix of coefficients. At this point we can follow with the analysis performed on the comment references.
NOTE
After the $B$ determination (definite positiveness) we have the relationships
$$
\left\{
\begin{array}{rcl}
b_{4,4}&=&a_1 \\
b_{3,4}+b_{4,3}&=&a_2 \\
b_{2,4}+b_{4,2}&=&a_4 \\
b_{1,4}+b_{4,1}&=&a_5 \\
b_{3,3}&=&a_3 \\
b_{2,3}+b_{3,2}&=&a_8 \\
b_{1,3}+b_{3,1}&=&a_9 \\
b_{2,2}&=&a_6 \\
b_{1,2}+b_{2,1}&=&a_{10} \\
b_{1,1}&=&a_7 \\
\end{array}
\right.
$$
|
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|
Evaluating $\int_3^7x^3dx$ using $\sum_{j=1}^Nj^3=\left(\frac12{N(N+1)}\right)^2$ Question:
$$\int_3^7x^3dx$$
I know how to solve it using the Power Rule, but I wanted to know how to solve it using:
$$1^3+2^3+\cdots+N^3=\sum_{j=1}^Nj^3=\left({N(N+1)\over2}\right)^2$$
So I'm just confused on how we use that formula to find an answer to that integral. Thank you!
|
Recall $$\int_a^bf(x)dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i)\Delta x$$
where $\Delta x=\frac{b-a}{n}$ and $x_i=a+i\Delta x$ (I'm using right-endpoints in our Riemann-sum). Thus $$\int_3^7x^3dx=\lim_{n\to\infty}\sum_{i=1}^n\left(3+\frac{4i}{n}\right)^3\cdot\frac{4}{n}$$ $$=\lim_{n\to\infty}\frac{4}{n}\sum_{i=1}^n\left(3^3+3\cdot3^2\cdot\frac{4i}{n}+3\cdot3\cdot\frac{(4i)^2}{n^2}+\frac{(4i)^3}{n^3}\right)$$ $$=\lim_{n\to\infty}\frac{4}{n}\sum_{i=1}^n\left(27+\frac{108}{n}\cdot i+\frac{144}{n^2}\cdot i^2+\frac{64}{n^3}\cdot i^3\right)$$ $$=\lim_{n\to\infty}\left(\frac{4}{n}\sum_{i=1}^n27+\frac{432}{n^2}\sum_{i=1}^n i+\frac{576}{n^3}\sum_{i=1}^ni^2+\frac{256}{n^4}\sum_{i=1}^n i^3\right)$$
Then note $$\sum_{i=1}^nc=nc,\space\space \sum_{i=1}^n i=\frac{n(n+1)}{2},\space\space \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6},\space\space\sum_{i=1}^ni^3=\left(\frac{n(n+1)}{2}\right)^2$$ Thus $$\int_3^7x^3dx=$$ $$=\lim_{n\to\infty}\left(\frac{4}{n}\cdot27n+\frac{432}{n^2}\cdot\frac{n(n+1)}{2}+\frac{576}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+\frac{256}{n^4}\cdot\left(\frac{n(n+1)}{2}\right)^2\right)$$ $$=\lim_{n\to\infty}\left(108+216\left(1+\frac{1}{n}\right)+96\left(2+\frac{3}{n}+\frac{1}{n^2}\right)+64\left(1+\frac{2}{n}+\frac{1}{n^2}\right)\right)$$ $$=108+216+96\cdot2+64=580$$
|
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|
Calculate definite integral using the definition (Riemann Sum) I would like to know how to use the Riemann Sum (With Unequal width) to calculate $\displaystyle f(x) = \sqrt{x}$ from $[1,2]$
I know how to calculate it when the interval starts with 0, for example [0,1], I will do the following way:
*
*$\displaystyle c_i = \frac{i^2}{n^2}$ and $\displaystyle \Delta x_i = \frac{i^2}{n^2}-\frac{(i-1)^2}{n^2}=\frac{2i-1}{n^2}$
*$\displaystyle \lim_{n\to\infty}\sum_{i=1}^nf(c_i)\Delta x_i = \lim_{n\to\infty}\sum_{i=1}^n = \sqrt{\frac{i^2}{n^2}}\left(\frac{2i-1}{n^2} \right) = \frac{2}{3}$
I understand that by choosing a partition of $\displaystyle\frac{i^2}{n^2}$, it will facilitate our calculation of summation because we have the following formula
*
*$\displaystyle \sum_{i=1}^n i = \frac{n(n+1)}{2}$
However, if the partition starts with 1 or other numbers, we would have something like $\displaystyle \sum_{i=1}^n \sqrt{1+\frac{i^2}{n^2}}$ which we don't have a directly formula to use.
Is it possible for me to do such Riemann Sum ($\displaystyle\sqrt{x}$ when interval doesn't start with 0) with what I have learned from First Year Calculus in university?
|
Let $$c_i=\left(1+ \frac{(\sqrt2-1)i}{n}\right)^2$$
Hence,
\begin{align}\Delta x_i&=\left(1+ \frac{(\sqrt2-1)i}{n}\right)^2-\left(1+ \frac{(\sqrt2-1)(i-1)}{n}\right)^2\\
&=\left( 2+\frac{(\sqrt2-1)(2i-1)}{n}\right)\left(\frac{\sqrt{2}-1}{n} \right)\end{align}
and the integral is
\begin{align}
\lim_{n \to \infty} \sum_{i=1}^n f(c_i) \Delta x_i &= \lim_{n \to \infty} \sum_{i=1}^n \left(1+ \frac{(\sqrt2-1)i}{n}\right)\left( 2+\frac{(\sqrt2-1)(2i-1)}{n}\right)\left(\frac{\sqrt{2}-1}{n} \right) \\
&= \lim_{n \to \infty} \frac{\sqrt2-1}{n^3} \sum_{i=1}^n [n+(\sqrt2-1)i][2n+(\sqrt2-1)(2i-1)] \\
&= \lim_{n \to \infty} \frac{\sqrt2-1}{n^3} \sum_{i=1}^n(2n^2+n(\sqrt2-1)(2i-1)+2(\sqrt2-1)in + (\sqrt2-1)^2i(2i-1))\\
&= \lim_{n \to \infty} \frac{\sqrt2-1}{n^3}\left(2n^3 +n \sum_{i=1}^n 4(\sqrt2-1)i + 2\sum_{i=1}^n (\sqrt2-1)^2i^2 \right) \\
&=\lim_{n \to \infty } \frac{\sqrt2-1}{n^3} \left(2n^3 + 2n^3(\sqrt2-1)+\frac23(\sqrt2-1)^2n^3 \right) \\
&= (\sqrt2-1)\left(2\sqrt2+ \frac23(3-2\sqrt2) \right) \\
&= \frac13(\sqrt2-1)(2\sqrt2+6) \\
&= \frac23(\sqrt2-1)(\sqrt2+3) \\
&= \frac23 (2\sqrt2-1)
\end{align}
|
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|
Prove that two random variables are independent distributed normally Let $M$ and $N$ are independent random variables distributed Uniform$[0, 1]$.
Define $(M_n)_{n\geq 1}$ and $(N_n)_{n\geq 1}$ which are two independent sequences of iid random variables distributed uniformly over $[−1, 1]$. Let $Z = \inf \{ n ≥ 1, 0 < M^2_n + N^2_n< 1 \}$ and
$X = M_Z \sqrt{\frac{-2 \log (M_Z^2 + N_Z^2)}{M_Z^2 + N_Z^2}}~~$ and $Y = N_Z \sqrt{\frac{-2 \log (M_Z^2 + N_Z^2)}{M_Z^2 + N_Z^2}}$
What is the distribution of $Z$? show that $X$ and $Y$ are two independant random variables distributed $N(0,1)$.
Addition A previous required question related to this exercise is asking to show that $X$ and $Y$ are independent random variables distributed $N(0, 1)$, knowing that:
$ X = \sqrt{-2 \log(M)} \cos(2 \pi N)$ and $Y = \sqrt{-2 \log(M)} \sin(2 \pi N)$
I already proved this part using the change of variables transformation. Then to show that they are independent, the joint density of $X,Y$ can factor into separate densities of $X$ and $Y$.
|
$M, N$ are uniformly distributed on the unit disk, so $f_{M,N}(m,n) = \pi^{-1}$ in the unit disk and zero outside. We have new variables $X,Y$ expressed in terms of $M,N.$ Recall the Change of Variables theorem, which states that if we have some bijective function $g : (x,y) \ \mapsto \ (m, n)$ with continuous partial derivatives, then the joint density of $X,Y$ is given by
$$f_{X,Y}(x,y) = f_{M,N}( g_1(x,y), g_2(x,y) ) \cdot |J_g|$$
where $J_g$ is the determinant of the Jacobian matrix of $g,$ and $|J_g|$ is the absolute value of that.
For this specific question, rearranging the defining equations for $X,Y,$ gives $$g(x,y) = h(x,y)\begin{pmatrix}x\\y\end{pmatrix}, \ \ \text{ where }\ h(x,y) = \frac{\exp \left(-\frac{x^2+y^2}{4}\right)}{\sqrt{x^2+y^2}}$$
One could start routinely calculating the partial derivatives and simplifying the resulting terms to compute $J_g,$ but the computation is considerably easier if we continue with the special form that $g$ has.
\begin{align}
J_g &= \frac{ \partial g_1}{\partial x} \frac{ \partial g_2}{\partial y} - \frac{\partial g_1}{\partial y} \frac{\partial g_2}{\partial x}\\
&=\left(h + x \frac{\partial h}{\partial x}\right)\left(h + y \frac{\partial h}{\partial y}\right) -\left(x \frac{\partial h}{\partial y}\right)\left(y \frac{\partial h}{\partial x}\right)\\
&=h^2 + \left( x \frac{\partial h}{\partial x} + y \frac{\partial h}{\partial y}\right)h\\
&= h^2 \left( 1 + \bigg\langle \nabla \log h \ , \begin{pmatrix}x\\y\end{pmatrix} \bigg\rangle \right)
\end{align}
We have $\displaystyle\nabla \log h(x,y) = \left( \frac{-1}{x^2+y^2} - \frac{1}{2} \right) \begin{pmatrix}x\\y\end{pmatrix},$ so
$\displaystyle 1 + \bigg\langle \nabla \log h \ , \begin{pmatrix}x\\y\end{pmatrix} \bigg\rangle = - \frac{x^2+y^2}{2}.$
Thus, $$J_g = - \frac{1}{2} \exp \left( - \frac{x^2+y^2}{2}\right)$$
and $$f_{X,Y}(x,y) = \frac{1}{2\pi} \exp \left( - \frac{x^2+y^2}{2}\right)$$
Therefore, the joint distribution of $X$ and $Y$ is that of the standard bivariate normal distribution, which implies that $X$ and $Y$ are independent identically distributed standard normal variables.
|
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|
Conjecture about a continued fraction
Conjecture:
$$\large 2^{n-1}+\frac{1}{2+\cfrac{1}{2^{n}-1+\cfrac{1}{2+\cfrac{1}{2^{n}-1+\cfrac{1}{2+\ddots}}}}}=\frac{1+\sqrt{3a_n}}{2}\tag*{[1]}$$ such that $a_n=4a_{n-1}+1$ and $a_0=0$.$\quad(n\geqslant 1)$
Ex. If $n=1$, then $a_n=4a_0+1=4\times 0 + 1 = 1$. $$\therefore 1+\frac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\ddots}}}}}=\frac{1+\sqrt{3}}{2}\tag*{[2]}$$ This can be proven using the formula $x=a+\dfrac{1}{b+\dfrac{1}{x}}=a+\dfrac{1}{b+\dfrac{1}{a+\dfrac{1}{b+\dfrac{1}{a+\ddots}}}}$
Solving for $x$ results in a quadratic equation for which $x=\dfrac12\left\{a+\sqrt{a\left(a+\dfrac4b\right)}\right\}$.
Substituting $a=1$ and $b=2$ yields $[2]$ as required.
Problem is, I am unsure on how to (dis)prove this conjecture given the recursive sequence involved. How do I appropriately go about this? Any suggestions or counter-examples?
Thanks :)
|
Claim:
For all $n\ge1$, we have $[2^{n-1},2,2^n-1,2,2^n-1,\cdots]=(1+\sqrt{3a_n})/2$ with $a_n=4a_{n-1}+1$ and $a_0=0$.
Proof:
As $a_n=4a_{n-1}+1=4^2a_{n-2}+4^1+4^0=\cdots=4^ka_{n-k}+\sum\limits_{i=0}^{k-1}4^i$, choosing $k=n$ gives $$a_n=4^na_0+\sum_{i=0}^{n-1}4^i=\frac{4^n-1}3\implies\frac{1+\sqrt{3a_n}}2=\frac{1+\sqrt{4^n-1}}2.$$ We know that the continued fraction converges by the Seidel-Stern theorem, as the sum of the convergents diverges. Thus, writing the continued fraction as $y=2^{n-1}+x=2^{n-1}+[0,2,2^n-1+x]$, $$2+\frac1{2^n-1+x}=\frac1x\implies2x(2^n-1+x)+x=2^n-1+x$$ giving the quadratic $$2x^2+2(2^n-1)x-(2^n-1)=0\implies x=-\frac{2^n-1}2\pm\frac{\sqrt{4^n-1}}2.$$ Evidently, the positive root must be chosen so that $y=(1+\sqrt{4^n-1})/2$ as required.
|
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limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
Now it is $e^{(0\cdot \infty)}$ which we can not conclude about the limit
|
Hint:
\begin{align}
\lim_{x \to 7} (x-8)(x-10)\frac{\ln x - \ln 7}{x-7} = 3 \lim_{x \to 7} \frac{\ln x - \ln 7}{x-7}
\end{align}
|
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|
Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
|
$$I=\int{\frac{1}{(x^3+1)^2}}dx$$
$$I=-\int \frac {1}{3x^2}\color {red}{{\frac{-3x^2}{(x^3+1)^2}}}dx$$
The function in red is a derivative.
$$I=-\int \frac {1}{3x^2}\color {red}{ \left ({\frac{1}{x^3+1}} \right )'}dx$$
The integral is now of the form:
$$ \color {blue}{I=\int f(x) g'(x)dx}$$
Integrate by part
$$ \color {blue}{\int f(x) g'(x)dx=f(x)g(x)-\int f'(x)g(x) dx}$$
$$I=- \frac {1}{3x^2}{\frac{1}{(x^3+1)}}-\frac {2}{3}\int \frac {1}{x^3}{\frac{1}{(x^3+1)}}dx$$
$$I_2=\int \frac {1}{x^3}{\frac{1}{(x^3+1)}}dx$$
$$I_2=\int \frac {dx}{x^3}-\int {\frac{dx}{(x^3+1)}}$$
$$I_2=- \frac {1}{2x^2}-\int {\frac{1}{(x^3+1)}}dx$$
So that we have :
$$I=\frac x{3(x^3+1)}+\frac23\int\frac1{x^3+1}dx$$
Then use fraction decomposition method for that integral.
|
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|
Evaluate $\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
An approach I can think about is to expand $\cos$ using taylor series, is there another approach?
|
Taylor expand
$$\frac x{1-x^2}=x+x^3+x^5+O(x^7),\>\>\>\>\>\cos t = 1-\frac{t^2}2+\frac{t^4}{24}+O(t^6)$$
to get
$$\cos\frac{x}{1-x^2}=1-\frac{1}2x^2-\frac{23}{24}x^4+O(x^6)$$
Thus,
$$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
$$=\lim _{x\to 0}\frac{1-\frac{x^2}{2}-(1-\frac{x^2}2-\frac{23}{24}x^4+O(x^6))}{x^4} $$
$$=\lim _{x\to 0}\frac{\frac{23}{24}x^4+O(x^6)}{x^4}=\frac{23}{24}$$
|
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|
Limit of integral $\lim_{n \to \infty} n\int_0^{\frac{\pi}{4}} \tan^n x\,dx$ Compute
$$\lim_{n \to \infty}n \int_0^{\frac{\pi}{4}}\tan^n x\,dx$$
I tried to define a recurrence with $I_n=\int_0^{\frac{\pi}{4}} \tan^n x$ :
$I_0 = \frac{\pi}{4}, I_1=\ln\sqrt{2}$
and $I_{n}=\frac{1}{n-1}-I_{n-2}$, but I can't complete it.
|
Define $f : \mathbb{R} \to \mathbb{R}, f(x)=\dfrac{1}{1+x^2}$. Now, integrating by parts:
$$
\begin{aligned}
\int_0^1 (n+1)x^nf(x)\,dx &= \left[x^{n+1}f(x)\right]_0^1-\int_0^1x^{n+1}f'(x)\,dx \\
&= f(1)+2\int_0^1 \frac{x^{n+2}}{(1+x^2)^2}dx
\end{aligned}
$$
Using $1\leq 1+x^2\leq 2$ over $[0,1]$ we get
$$\frac{1}{2(n+3)}=\frac{1}{2}\int_0^1 x^{n+2}dx\leq 2\int_0^1\frac{x^{n+2}}{(1+x^2)^2}dx \leq 2\int_0^{1} x^{n+2}dx=\frac{2}{n+3}$$
and squeezing, we can easily see that:
$$2\int_0^1\frac{x^{n+2}}{(1+x^2)^2}dx \to 0$$
and therefore:
$$\lim_{n\to \infty}n\int_0^1 x^nf(x)\,dx=\lim_{n\to \infty}\frac{n}{n+1}\int_0^1 (n+1)x^nf(x)\,dx=f(1)=\frac{1}{2}$$
Now substitute $x \to \tan x$ to get that the limit equals $\dfrac{1}{2}$.
|
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|
How many integer solutions are there to $x+y+z=8$
How many integer solutions are there to $x+y+z=8$ When $x,y,z>0$? When $x,y,z\geq -3$?
So I know there is a formula for computing the number of nonnegative solutions
${8+3-1 \choose 3-1}={10\choose 2}$
So I then just subtracted cases where one or two integers are $0$.
If just $x=0$ then there are $6$ solutions where neither $y,z=0$.
So I multiplied this by $3$, then added the cases where two integers are $0$
$3\cdot 6+3=21$. So I get ${10 \choose 2}+21=66$
For the last problem where $x,y,z\geq -3$ I'm not sure how to deal with it.
|
Your answer to the question of how many solutions the equation $$x + y + z = 8 \tag{1} $$ has in the positive integers is incorrect. As a sanity check, observe that there must be fewer solutions to the equation in the positive integers than there are in the nonnegative integers since we are not allowed to substitute $0$ for any of the variables.
How many integer solutions are there to the equation $x + y + z = 8$ when $x, y, z > 0$?
If $x, y, z$ are positive integers, then $x' = x - 1$, $y' = y - 1$, and $z' = z - 1$ are nonnegative integers. Substituting $x' + 1$ for $x$, $y' + 1$ for $y$, and $z' + 1$ for $z$ in equation 1 yields
\begin{align*}
x' + 1 + y' + 1 + z' + 1 & = 8\\
x' + y' + z' & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}$$
solutions. Notice that there are fewer solutions to equation 1 in the positive integers than the nonnegative integers, as we would expect.
How many integer solutions are there to the equation $x + y + z = 8$ when $x, y, z \geq -3$?
If $x, y, z \geq -3$, then $x' = x + 3$, $y' = y + 3$, and $z' = z + 3$ are nonnegative integers. Substitute $x' + 3$ for $x$, $y' + 3$ for $y$, and $z' + 3$ for $z$ in equation 1, then proceed as above.
|
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|
Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$
Question: Show that
$$\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$$
From Wolfram alpha, it seems that the equality above is indeed correct.
But I do not know how to prove it.
Any hint is appreciated.
|
Famously, $\displaystyle \int_0^{\pi/2} \cos^{2n}{x}\,\mathrm{d}x = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$ (e.g. see here); and $\displaystyle \frac{1}{1+n} = \int_0^1 y^n \, \mathrm{d} y$, thus:
$\displaystyle \begin{aligned} \sum_{n \ge 0} \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} & = \frac{1}{\pi} \sum_{n \ge 0}\int_0^{\pi/2}\int_0^1 y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} y^n \cos^{2n}{x}\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi} \int_0^{\pi/2}\int_0^1\sum_{n \ge 0} (y\cos^2{x})^n\,\mathrm{d}y\,\mathrm{d}x\, \\& = \frac{1}{\pi}\int_0^{\pi/2}\int_0^1\frac{1}{1-y \cos^2{x}}\mathrm{d}y\,\mathrm{d}x \\& = \frac{1}{\pi} \int_0^{\pi/2}\sec^2{x} \cdot \log\left({\csc^2{x}}\right)\,\mathrm{d}x\, \\& = \frac{1}{\pi} \cdot \bigg[2x+\tan{x}\log(\sec^2{x})\bigg]_{x \to 0}^{x \to \pi/2} \\& = \frac{1}{\pi}\cdot \pi \\& = 1. \end{aligned} $
|
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|
Square root property proof Can anyone provide a link to a proof of the following square root property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$. Could not find it online anywhere.
|
By the definition of square-root, $x=\sqrt{y}\hspace{.15cm}$ means that $x^2=y$.
Assuming that $a,b\in\mathbb{R}$ with $a\geq 0$ and $b>0$, then we can simply compute:
\begin{align*}
\left(\frac{\sqrt{a}}{\sqrt{b}}\right)^2&=\frac{\sqrt{a}}{\sqrt{b}}\cdot \frac{\sqrt{a}}{\sqrt{b}}
=\frac{\sqrt{a}\cdot \sqrt{a}}{\sqrt{b}\cdot\sqrt{b}}=\frac{a}{b}
\end{align*}
Therefore, $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\hspace{.25cm}$
Note that I used the fact that $\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$
|
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|
calculating total number of allowable paths from $(0,0)$ to $(5,5)$ I'm looking at paths starting from $(0,0)$ with the following allowable steps :
1) from $(x,y)$ to $(x,y+1)$
2) from $(x,y)$ to $(x+1,y)$
3) from $(x,y)$ to $(x+2,y+1)$
how can I determine the total number of paths from $(0,0)$ to $(5,5)$?
and then generally from $(0,0)$ to $(n,n)$ using the same allowable steps. Is it possible to do this using binomial theorem?
|
We algebraically encode the base steps
\begin{align*}
&(1,0)\qquad\to\qquad x\\
&(0,1)\qquad\to\qquad y\\
&(2,1)\qquad\to\qquad x^2y
\end{align*}
so that each step can be encoded as $(x+y+x^2y)$. Denoting the coefficient of $x^n$ with the coefficient of operator $[x^n]$ we want to find
\begin{align*}
[x^5y^5]\sum_{q=0}^\infty(x+y+x^2y)^q\tag{1}
\end{align*}
which means we consider paths of length $q\geq 0$ and for each path we get the number of possibilities from $(0,0)$ to $(5,5)$ and sum up all these numbers.
We obtain from (1)
\begin{align*}
\color{blue}{[x^5y^5]}&\color{blue}{\sum_{q=0}^\infty\left(x+y+x^2y\right)^q}\\
&=[x^5y^5]\sum_{q=0}^\infty \sum_{{j+k+l=q}\atop{j,k,l\geq0}}\binom{q}{j,k,l}x^jy^k\left(x^2y\right)^l\tag{2}\\
&=[x^5y^5]\sum_{q=0}^\infty \sum_{{j+k+l=q}\atop{j,k,l\geq0}}\binom{q}{j,k,l}x^{j+2l}y^{k+l}\\
&=[x^5]\sum_{q=0}^\infty\sum_{k=0}^5\sum_{{j+k+(5-k)=q}\atop{j\geq 0}}\binom{q}{j,k,5-k}x^{j+10-2k}\tag{3}\\
&=[x^5]\sum_{q=0}^\infty\sum_{k=3}^5\binom{q}{q-5,k,5-k}x^{q+5-2k}\tag{4}\\
&=\binom{6}{1,3,2}+\binom{8}{3,4,1}+\binom{10}{5,5,0}\tag{5}\\
&=\frac{6!}{1!3!2!}+\frac{8!}{3!4!1!}+\frac{10!}{5!5!0!}\\
&\,\,\color{blue}{=592}
\end{align*}
Comment:
*
*In (2) we expand the trinomial using multinomial coefficients.
*In (3) we select the coefficient of $y^5$ by setting $k=0,\ldots,5$ and $l=5-k$.
*In (4) we set the lower limit of $k$ to $3$ since $x^{j+10-2k}$ has exponent less than $5$ otherwise. The index relation in (3) also shows that $j+5=q$ and we eliminate $j$.
*In (5) we select the coefficient of $x^5$.
The steps $x+y+x^2y$ indicate a recurrence relation
\begin{align*}
a_{m,n}&=a_{m-1,n}+a_{m,n-1}+a_{m-2,n-1}\qquad\ m\geq 2,n\geq 1\\
a_{m,0}&=1=a_{0,n}\qquad\qquad\qquad\qquad\qquad\quad m,n\geq 0\\
a_{1,n}&=n\qquad\qquad\qquad\qquad\qquad\qquad\qquad n\geq 1
\end{align*}
Some values of the recurrence relation are listed in the table below, which verifies the solution
\begin{align*}
\color{blue}{a_{5,5}}&=a_{4,5}+a_{5,4}+a_{3,4}=241+296+55\color{blue}{=592}
\end{align*}
|
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|
Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a_i^2\right ) \left ( \sum_{i=1}^{n}b_i^2 \right ) }$$
My attempt:
$n=3$
$a_1=\sqrt{ab}$, $a_2=\sqrt{bc}$, $a_3=\sqrt{ac}$
$b_1=\frac{\sqrt{a}}{\sqrt{b}}$, $b_2=\frac{\sqrt{b}}{\sqrt{c}}$, $b_3=\frac{\sqrt{c}}{\sqrt{a}}$
Plugging it in,
$$ab+bc+ac+\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq a^2 + b^2 + c^2$$
There are $3$ unwanted fractions. Is there any way to remove them?
|
We should set $a_1=b_3=a, a_2=b_1= b$ and $a_3=b_2=c$ in the Cauchy-Schwarz inequality, to get:
$$(ab+bc+ca)^2\leq (a^2+b^2+c^2)(b^2+c^2+a^2)=(a^2+b^2+c^2)^2$$
and therefore:
$$a^2+b^2+c^2\geq |ab+bc+ca|\geq ab+bc+ca$$
Of course, we don't need any restriction over $a,b,c$ (they don't have to be integers or ordered, they can be any real number).
|
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|
Help with trig sub integral I've gone over my work twice, and while I dropped a negative the first time, I believe that my work is sound. Webwork is not accepting it, however.
Here is the problem and my solution. Sorry for bad handwriting
|
Let $x=2\sin{t}$.
Thus, $$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{x^2-4+4}{\sqrt{4-x^2}}dx=4\int\frac{1}{\sqrt{4-x^2}}dx-\int\sqrt{4-x^2}dx=$$
$$=4\arcsin\frac{x}{2}-4\int\cos^2tdt=4\arcsin\frac{x}{2}-2\int(1+\cos2t)dt.$$
Can you end it now?
Also, let $x-2=v$.
Thus, we can use the similar way: $$\int\frac{x^2}{\sqrt{4x-x^2}}dx=\int\frac{x^2}{\sqrt{4-(x-2)^2}}dx=\int\frac{v^2+4v+4}{\sqrt{4-v^2}}dv=$$
$$=\int\frac{v^2-4+4v+8}{\sqrt{4-v^2}}dv=\int\frac{4v}{\sqrt{4-v^2}}dv-\int\sqrt{4-v^2}dv+8\int\frac{1}{\sqrt{4-v^2}}dv.$$
|
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find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(\lambda+2)^2$.
By Cayley-Hamilton Theorem, we have $Ch_{A}(A)=0\implies (A -2)^3(A+2)^2=0\implies$ Either $(A -2)^3=0$ or $(A+2)^2=0$
Considering
$$(A+2)^2=0\implies A^2+4A+4I=0$$
$$\implies I=\left(-\frac{1}{4}\right)A^2+(-A)$$
$$\implies A^{-1}=\left(-\frac{1}{4}\right)A+(-I).$$
On comparing just recent equation with $A^{-1}=\alpha A+\beta I$,we get $\alpha =-\frac{1}{4}$ and $\beta =-1$
But the correct answer is $\alpha=\frac{1}{4}$ and $\beta =0$..
I wanted to know where I'm wrong?
|
Since $A$ is diagonalizable, and the characteristic polynomial is $A$ is $c_A(x)=(x-2)^3(x+2)^2$, it follows that $A=PDP^{-1}$, where $P$ is an invertible $5\times5$ matrix, and
$$D=\begin{pmatrix}2&0&0&0&0\\0&2&0&0&0\\0&0&2&0&0\\0&0&0&-2&0\\0&0&0&0&-2\end{pmatrix}.$$
It follows that $A^{-1}=PD^{-1}P^{-1}$. Note that
$$D^{-1}=\begin{pmatrix}\frac{1}{2}&0&0&0&0\\0&\frac{1}{2}&0&0&0\\0&0&\frac{1}{2}&0&0\\0&0&0&-\frac{1}{2}&0\\0&0&0&0&-\frac{1}{2}\end{pmatrix}.$$
So we have that $4\cdot D^{-1}=D$. Hence
$$\begin{align*}
4\cdot A^{-1} &= 4\cdot PD^{-1}P^{-1}\\
&= P(4\cdot D^{-1})P^{-1}\\
&= PDP^{-1}\\
&= A\end{align*}$$
So $4\cdot A^{-1}=A$. It follows that $A^{-1}=\frac{1}{4}\cdot A$. Hence $\alpha=\frac{1}{4}$ and $\beta=0$.
As I pointed out in the comments, you made a mistake when you assumed that $(A-2I)^2(A+2I)^2=0$ implied that either $(A-2I)^2=0$ or $(A+2I)^2=0$. Also, note that the Cayley-Hamilton theorem turned out to not be necessary for this particular problem.
|
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|
Tough multivariable inequality: Minimize $a^2 + b^3 + c^4$ given $a + b^2 + c^3 = \frac{325}{9}$ Let $a,$ $b,$ and $c$ be positive real numbers such that $a + b^2 + c^3 = \frac{325}{9}.$ Find the minimum value of
$a^2 + b^3 + c^4.$
|
Note that $(a-2)^2+(c^2+2c+3)(c-3)^2+\frac{1}{27}(3b+4)(3b-8)^2\ge 0$ for all $a, b, c\ge 0$, i.e.
$a^2 + b^3 + c^4 + \frac{1093}{27} - 4(a + b^2 + c^3) \ge 0$ for all $a,b,c\ge 0$.
Thus, under the condition $a, b, c\ge 0;\ a+b^2+c^3=\frac{325}{9}$, we have $a^2 + b^3 + c^4 \ge 4\cdot \frac{325}{9} - \frac{1093}{27} = \frac{2807}{27}$.
The minimum is $\frac{2807}{27}$ at $a=2, b= 8/3, c=3$.
|
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On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$$
What I have managed to do so far is to convert $S$ to two rather difficult integrals as follows.
Starting with the result
$$\frac{H_{2n}}{2n} = -\int_0^1 x^{2n - 1} \ln (1 - x) \, dx \tag1$$
Multiplying (1) by $(-1)^n H_n/n$ then summing the result from $n = 1$ to $\infty$ gives
$$S = -2 \int_0^1 \frac{\ln (1 - x)}{x} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n}. \tag2$$
From the following generating function for the harmonic numbers
$$\sum_{n = 1}^\infty \frac{H_n x^n}{n} = \frac{1}{2} \ln^2 (1 - x) + \operatorname{Li}_2 (x),$$
replacing $x$ with $-x^2$ leads to
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n} x^{2n} = \frac{1}{2} \ln^2 (1 + x^2) + \operatorname{Li}_2 (-x^2).$$
Substituting this result into (2) yields
$$S = -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx,$$
or, after integrating the first of the integrals by parts twice
$$S = -\frac{5}{2} \zeta (4) + 4 \zeta (3) \ln 2 - 8 \int_0^1 \frac{x \operatorname{Li}_3 (x)}{1 + x^2} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx. \tag3$$
I have a slim hope the first of these integrals can be found (I cannot find it). As for the second of the integrals, it is proving to be a little difficult.
Can someone find each of the integrals appearing in (3)? Or perhaps an alternative approach to the sum will deliver the closed-form I seek, I am fine either way.
Update
Thanks to Ali Shather, the first of the integrals can be found. Here
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \ dx &=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{2n-1}\ln(1-x)\ dx\\
&= -\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n^3}\\
&=-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}\\
&=-4 \operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3}.
\end{align}
And using the result I calculated here, namely
$$\operatorname{Re} \sum_{n=1}^\infty i^n\frac{H_n}{n^3} = \frac{5}{8} \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{195}{256} \zeta (4) + \frac{5}{192} \ln^4 2 - \frac{5}{32} \zeta (2) \ln^2 2 + \frac{35}{64} \zeta (3) \ln 2,$$
gives
\begin{align}
\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx &= -\frac{5}{2} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{195}{64} \zeta (4) - \frac{5}{48} \ln^4 2\\
& \qquad + \frac{5}{8} \zeta (2) \ln^2 2 - \frac{35}{16} \zeta (3) \ln 2.
\end{align}
|
Using your integral representation, the sum equals to: $$\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}= -2 \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x^2)}{x} \, dx - \int_0^1 \frac{\ln (1 - x) \ln^2 (1 + x^2)}{x} \, dx$$ $$\small=-2 C^2+2 \pi C \log (2)-4 \pi \Im(\text{Li}_3(1+i))+3 \text{Li}_4\left(\frac{1}{2}\right)+\frac{21}{8} \zeta (3) \log (2)+\frac{487 \pi ^4}{5760}+\frac{\log ^4(2)}{8}+\frac{1}{8} \pi ^2 \log ^2(2)$$
For the second integral and its derivation, see here.
|
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|
Which one is larger? I want to prove that for positive integer $n$,
$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n}$$
but I am stuck on how to proceed. Can someone help me?
|
Consider
$$a_n=\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n=\left(1-\frac{1}{n^2}\right)^n\implies \log(a_n)=n \log\left(1-\frac{1}{n^2}\right)$$ So, by Taylor
$$\log(a_n)=n\left(-\frac{1}{n^2}-\frac{1}{2 n^4}-\frac{1}{3
n^6}+O\left(\frac{1}{n^8}\right)\right)=-\frac{1}{n}-\frac{1}{2 n^3}-\frac{1}{3
n^5}+O\left(\frac{1}{n^7}\right)$$
$$a_n=e^{\log(a_n)}=1-\frac{1}{n}+\frac{1}{2 n^2}-\frac{2}{3 n^3}+\frac{13}{24 n^4}++O\left(\frac{1}{n^5}\right)$$ and, as Ross Millikan commented, this is an alternating series and then ...
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$
Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrless \xi\gtrless x$$
Thus $$\frac{x}{(1+x)^n-(1-x)^n}=\frac{1}{2f'(\xi)}=\frac{1}{2n(1+\xi)^{n-1}}$$
Can we go on from here?
|
Well, for $0 < x < 1$, we have,
$S = \sum_{n=1}^{\infty}\frac{x}{(1+x)^n - (1-x)^n}
= \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{(1+x)^{n-1} + \cdots + (1-x)^{n-1}}
\geq (\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(1+x)^{n-1}}
= \frac{1}{1+x}\sum_{n=1}^{\infty}\frac{1}{n(1+x)^n}).$
Using power series for natural logarithm,
$S = -(1+x)\ln{\frac{x}{1+x}}$.
Now, let the required limit be L.
$L \leq (\lim_{x\to0^{+}}-((1+x)\frac{1}{2} - \frac{(1+x)\ln{1+x}}{2\ln{x}})
= -\frac{1}{2})$.
Hence, the upper bound.
|
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|
Given a solution $u = f(x-c_1t)+g(x-c_2t)$ to $Au_{tt}+2Bu_{tx}+Cu_{xx}=0$ what equation should $c_1$ and $c_2$ satisfy We find a solution to the equation $$Au_{tt}+2Bu_{tx}+Cu_{xx}=0$$ as $$u = f(x-c_1t)+g(x-c_2t)$$ with aribitrary $f,g,$ and real $c_1 < c_2.$
*
*What equation should satisfy $c_1$ and $c_2$
*When does this equation have such roots?
So far I have used the solution to rewrite the PDE as
$$f''(x-c_1t)\left[Ac_1^2 - 2Bc_1 + C\right] + g''(x-c_2t)\left[Ac_2^2 - 2Bc_2 +C\right]=0.$$ I'm stuck on what I ought to do next. I recognize that within each bracket is a quadratic in the respective $c.$ I'm just not sure where to go from here.
|
Calling
$$
M = \left(
\begin{array}{cc}
A & B \\
B & C \\
\end{array}
\right)
$$
we have
$$
\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)^{\dagger}M\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)u=0
$$
or
$$
\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)^{\dagger}T^{-1}\Lambda T\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)u=0
$$
with
$$
T = \left(
\begin{array}{cc}
\frac{A-C-\sqrt{4 B^2+(A-C)^2}}{2 B} & 1 \\
\frac{A-C+\sqrt{4 B^2+(A-C)^2}}{2 B} & 1 \\
\end{array}
\right)\\
\Lambda=\left(
\begin{array}{cc}
\frac{1}{2} \left(A+C-\sqrt{4 B^2+(A-C)^2}\right) & 0 \\
0 & \frac{1}{2} \left(A+C+\sqrt{4 B^2+(A-C)^2}\right) \\
\end{array}
\right) = \left(
\begin{array}{cc}
\lambda_1 & 0 \\
0 & \lambda_2 \\
\end{array}
\right)
$$
and now calling
$$
\left(
\begin{array}{c}
\partial \eta \\
\partial \xi \\
\end{array}
\right) = T\left(
\begin{array}{c}
\partial t \\
\partial x \\
\end{array}
\right)
$$
the PDE reduces to
$$
\lambda_1 u_{\eta\eta}+\lambda_2 u_{\xi\xi}=0
$$
and also we can verify by substitution that
$$
u(\eta,\xi) = f(\eta)+g(\xi)
$$
is a solution, or changing variables again
$$
u(t,x) = f\left(\frac{\lambda_1}{B} t + x\right) + g\left(\frac{\lambda_2}{B} t+x\right)
$$
|
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|
Evaluate of: ${\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$ How do we evaluate this infinite product with a sum within it?
$$\large{\prod_{n=1}^{\infty}\left[1+\frac{1}{\sum_{j=1}^{n}F_j^2}\right]^{(-1)^n+1}}$$
Where $F_j$ is the Fibonacci number
If I open the product, it does not help me. I am sure there must be an equivalent form of this $1+\frac{1}{\sum_{j=1}^{n}F_j^2}$ into an easier manageable form.
Due to lack of knowledge in this field, I can not do much.
We can rewrite as (due to a hint)
$${\prod_{n=1}^{\infty}\left(1+\frac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}}$$
|
We use Cassini's identity:
$$F_{n-1}F_{n+1}-F_n^2=(-1)^n\Rightarrow F_{2n-1}F_{2n+1}-F_{2n}^2=1$$
Then
$$\prod_{n=1}^\infty \left( 1+\dfrac{1}{F_nF_{n+1}}\right)^{(-1)^n+1}=\left( \prod_{n=1}^\infty \left( 1+\dfrac{1}{F_{2n}F_{2n+1}}\right) \right)^2$$
Let $P_n$ be
$$P_n=\prod_{k=1}^n \left( 1+\dfrac{1}{F_{2k}F_{2k+1}}\right),\quad P_1=1+\dfrac{1}{F_2F_3}=\dfrac{F_4}{F_3}$$
As
\begin{align*}
F_{2n}F_{2n+1}+1 &= F_{2n}F_{2n+1}+F_{2n-1}F_{2n+1}-F_{2n}^2\\
&= F_{2n+1}(F_{2n}+F_{2n-1})-F_{2n}^2\\
&= F_{2n+1}^2-F_{2n}^2=(F_{2n+1}+F_{2n})(F_{2n+1}-F_{2n})\\
&= F_{2n+2}F_{2n-1}
\end{align*}
and
$$P_2=P_1\cdot \left( 1+\dfrac{1}{F_4F_5}\right) =\dfrac{F_4}{F_3}\cdot \left( \dfrac{F_4F_5+1}{F_4F_5}\right) =\dfrac{F_4}{F_3}\cdot \dfrac{F_6\cdot F_3}{F_4F_5}=\dfrac{F_6}{F_5}$$
We suppose that $P_n=\dfrac{F_{2(n+1)}}{F_{2n+1}}$. Then,
\begin{align*}
P_{n+1} &= P_n\cdot \left( 1+\dfrac{1}{F_{2(n+1)}F_{2(n+1)+1}}\right) \\
&= \dfrac{F_{2(n+1)}}{F_{2n+1}}\left( \dfrac{F_{2(n+1)}F_{2(n+1)+1}+1}{F_{2(n+1)}F_{2(n+1)+1}}\right) \\
&= \dfrac{F_{2(n+1)}}{F_{2n+1}}\left( \dfrac{F_{2(n+2)}F_{2n+1}}{F_{2(n+1)}F_{2(n+1)+1}}\right) \\
&= \dfrac{F_{2(n+2)}}{F_{2(n+1)+1}}
\end{align*}
Finally,
$$\lim_{n\to \infty} P_n^2=\left( \lim_{n\to \infty} \dfrac{F_{2(n+1)}}{F_{2(n+1)+1}}\right)^2 =(\varphi)^2$$
|
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|
Evaluate $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$ Denote $g(n,j)=\sum _{k=1}^n \frac{k^j (-1)^{n-k} \binom{n}{k}}{\frac{1}{2} n (n+1)-k}$, then how can we show that:
*
*$g(n,1)=\frac{n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}$
*$g(n,n)=\frac{\left(\frac{1}{2} n (n+1)\right)^{n-1} n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}$
*$g(n,n+1)=\frac{\left(\frac{1}{2} n (n+1)\right)^n n!}{\prod _{k=1}^n \left(\frac{1}{2} n (n+1)-k\right)}-n!$
This post is related. Any help will be appreciated!
|
We seek to evaluate
$$G_{n,j} = \sum_{k=1}^n \frac{k^j (-1)^{n-k} {n\choose k}}
{\frac{1}{2}n(n+1)-k}.$$
With this in mind we introduce the function
$$F_n(z) = n! \frac{z^{j-1}}{\frac{1}{2}n(n+1)-z}
\prod_{q=1}^n \frac{1}{z-q}.$$
This has the property that the residue at $z=k$ where $1\le k\le n$ is
the desired sum term. We find
$$\mathrm{Res}_{z=k} F_n(z)
= n! \frac{k^{j-1}}{\frac{1}{2}n(n+1)-k}
\prod_{q=1}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^{n} \frac{1}{k-q}
\\ = n! \frac{k^{j}}{\frac{1}{2}n(n+1)-k}
\frac{1}{k} \frac{1}{(k-1)!}
\frac{(-1)^{n-k}}{(n-k)!}
\\ = \frac{k^{j}}{\frac{1}{2}n(n+1)-k}
(-1)^{n-k} {n\choose k}.$$
We will evaluate this using the fact that residues sum to zero and if
$(n+1)-(j-1) \ge 2$ or $n\ge j$ the residue at infinity is zero, so we
have in this case
$$G_{n,j} = - \mathrm{Res}_{z=\frac{1}{2} n(n+1)} F_n(z)
= n! \frac{(\frac{1}{2} n(n+1))^{j-1}}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}.$$
We thus have
$$\bbox[5px,border:2px solid #00A000]{
G_{n,1} = \frac{n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}}$$
and
$$\bbox[5px,border:2px solid #00A000]{
G_{n, n} = \frac{(\frac{1}{2} n(n+1))^{n-1} n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}.}$$
When $j\gt n$ we must use the formula
$$G_{n,j} = - \mathrm{Res}_{z=\frac{1}{2} n(n+1)} F_n(z)
- \mathrm{Res}_{z=\infty} F_n(z).$$
We have
$$- \mathrm{Res}_{z=\infty} F_n(z)
= \mathrm{Res}_{z=0} \frac{1}{z^2} F_n(1/z)
\\ = n! \times \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1}{z^{j-1}}
\frac{1}{\frac{1}{2}n(n+1)-1/z}
\prod_{q=1}^n \frac{1}{1/z-q}
\\ = n! \times \mathrm{Res}_{z=0} \frac{1}{z^{j+1}}
\frac{z}{\frac{1}{2}n(n+1)z-1}
\prod_{q=1}^n \frac{z}{1-qz}
\\ = n! \times \mathrm{Res}_{z=0} \frac{1}{z^{j-n}}
\frac{1}{\frac{1}{2}n(n+1)z-1}
\prod_{q=1}^n \frac{1}{1-qz}.$$
In particular when $j=n+1$ we just need the constant term and find
$$n! \frac{1}{\frac{1}{2}n(n+1)\times 0 -1}
\prod_{q=1}^n \frac{1}{1-q\times 0} = -n!$$
we thus have
$$\bbox[5px,border:2px solid #00A000]{
G_{n, n+1} = \frac{(\frac{1}{2} n(n+1))^{n} n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)} - n!.}$$
The general case for $j\gt n$ is
$$n! \times \mathrm{Res}_{z=0} \frac{1}{z^{j}}
\frac{1}{\frac{1}{2}n(n+1)z-1}
\prod_{q=1}^n \frac{z}{1-qz}$$
which yields
$$-n! \sum_{q=0}^{j-1} \left(\frac{1}{2}n(n+1)\right)^q
{j-1-q\brace n}$$
so that the closed form is (here we must have $j-1-q\ge n$)
$$\bbox[5px,border:2px solid #00A000]{
G_{n, j} = \frac{(\frac{1}{2} n(n+1))^{j-1} n!}
{\prod_{q=1}^n (\frac{1}{2} n(n+1)-q)}
- [[j\gt n]] n! \sum_{q=0}^{j-1-n} \left(\frac{1}{2}n(n+1)\right)^q
{j-1-q\brace n}.}$$
|
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|
Find the CDF if $P(X=1)=\frac{1}{2}$ and $P(a
The Problem: Let $X$ be a random variable that satisfies $P(X=1)=\frac{1}{2}$ and $P(a<X<b|X\ne1)=\frac{b-a}{2}$ for $0\leq a<b\leq2$. Find the cumulative distribution function of $X$.
My Thoughts: We have that
$$P(0<X<2)=P(0<X<2|X\ne1)P(X\ne1)+P(0<X<2|X=1)P(X=1)=1.$$
Therefore, $P(X\leq t)=0$ for $t\leq0$, and $P(X\leq t)=1$ for $t\geq2$. Now if $0<t<1$ then
\begin{equation}\begin{split}
P(0<X<2)
&=P(X=t)+P(0<X<t)+P(t<X<1)+P(X=1)+P(1<X<2)\\
&=P(X=t)+\frac{t}{4}+\frac{1-t}{4}+\frac{1}{2}+\frac{1}{4}\\
&=P(X=t)+1\\
&=P(X=t)+P(0<X<2),
\end{split}\end{equation}
so that $P(X=t)=0$. Similarly, $P(X=t)=0$ for $1<t<2$.
Next, we note that for $0<t<1$ we have
$$P(X\leq t)=P(0<X<t|X=1)P(X=1)+P(0<X<t|X\ne1)P(X\ne1)=\frac{t}{4}.$$
If $t=1$, then
$$P(X\leq1)=\frac{1}{2}+P(0<X<1)=\frac{1}{2}+P(0<X<1|X\ne1)P(X\ne1)=\frac{3}{4}.$$
Finally, if $1<t<2$ then
$$P(X\leq t)=P(0<X\leq1)+P(1<X\leq t)=\frac{3}{4}+P(1<X<t|X\ne1)=\frac{3}{4}+\frac{t-1}{8}.$$
Therefore, we have
$$F_X(t)=\begin{cases}
0&\text{if }t\leq0\\
\dfrac{t}{4}&\text{if }0<t<1\\
\dfrac{3}{4}+\dfrac{t-1}{4}&\text{if }1\leq t<2\\
1&\text{if }t\geq2.
\end{cases}$$
|
Your end result is okay.
Let me add a route that makes things less error prone.
Let $Y$ have uniform distribution on interval $[0,2]$ and let $Z$ be a random variable degenerated at $1$ in the sence that $P(Z=1)=1$.
Then we find:
$$F_{X}\left(x\right)=P\left(X=1\right)P\left(X\leq x\mid X=1\right)+P\left(X\neq1\right)P\left(X\leq x\mid X\neq1\right)=\frac{1}{2}F_{Z}\left(x\right)+\frac{1}{2}F_{Y}\left(x\right)$$
It can easily be checked that the RHS is the same as your final result.
Here $F_Z(x)=\mathbf{1}_{\left[1,\infty\right)}\left(x\right)$ and $F_{Y}\left(x\right)=\begin{cases}
0 & \text{if }x\leq0\\
\frac{1}{2}x & \text{if }0<x<2\\
1 & \text{otherwise}
\end{cases}$
|
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|
Which expression about power series is correct? I found something weird about power series.
In my textbook,
$$ \frac{1}{1-x} = 1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n \quad |x|<1 $$
and $ \frac{1}{2+x} $ could be expressed using above equation.
$$ \frac{1}{2+x} = \frac{1}{2 \left(1 + \frac{x}{2} \right)} = \frac{1}{2 \left[ 1- \left( - \frac{x}{2} \right) \right]} \\ = \frac{1}{2} \sum_{n=0}^\infty \left( - \frac{x}{2} \right)^n = \sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n $$
which has interval of convergence $ (-2,2) $.
And below is my solution.
$$ \frac{1}{2+x} = \frac{1}{1-(-1-x)} = 1+(-1-x)+(-1-x)^2+(-1-x)^3+\cdots \\ =\sum_{n=0}^\infty (-1-x)^n \quad |-1-x|<1 $$
which has interval of convergence $ (-2,0) $
Here are my questions.
*
*Why above two power series have different interval of convergence from one equation $ \frac{1}{2+x} $? Why does it appear?
*How do I find out right answer?
|
You are just comparing two series developed around different points:
- one around $x=0$;
- the other around $x=-1$.
and remember that the radius of convergence is the distance from that point to the next irregularity ($x=-2$ in this case).
|
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|
Show that $Cov(AX) = A^T Cov(X) A$ For simplicity, assume random variable $X$ has mean $0$. I want to prove that $$Cov(AX) = A^T Cov(X) A$$ where A is a scalar matrix.
I tried to prove it from the definition of covariance:
$$Cov(AX) = E(AX)^T(AX) = E(X^TA^TAX)$$
I am wondering how do I bring $A$ out of the expectation. Thanks in advance.
|
Are you sure the statement you’re trying to prove isn’t $ACov(X)A^T$? Also, I believe in your definition of the covariance matrix you were applying the transpose operator to the wrong parentheses.
$$Cov(AX)=E[(AX)(AX)^T]=E(AXX^TA^T)=AE(XX^T)A^T=ACov(X)A^T$$
The statement you were trying to prove is actually incorrect. Let $x_1,x_2=\{-1,0,1\}$ be uniform random variables and $X=\begin{pmatrix} x_1 \\x_2\end{pmatrix}$ and $A= \begin{pmatrix}1 && 0 \\ 1 && 1 \end{pmatrix}$ then
$$AX=\{
\begin{pmatrix} -1 \\ -2 \end{pmatrix},
\begin{pmatrix} -1 \\ -1 \end{pmatrix},
\begin{pmatrix} -1 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ -1 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \end{pmatrix},
\begin{pmatrix} 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 1 \\ 1 \end{pmatrix},
\begin{pmatrix} 1 \\ 2 \end{pmatrix}
\}$$
so $$Cov(AX)=\begin{pmatrix}2/3 && 2/3 \\ 2/3 && 4/3 \end{pmatrix}$$
but $$A^TCov(X)A=
\begin{pmatrix} 1 && 1 \\ 0 && 1 \end{pmatrix}
\begin{pmatrix} 2/3 && 0 \\ 0 && 2/3 \end{pmatrix}
\begin{pmatrix} 1 && 0 \\ 1 && 1 \end{pmatrix}=
\begin{pmatrix}4/3 && 2/3 \\ 2/3 && 2/3 \end{pmatrix}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim_{x \to \infty} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$ $$\underset{x\to \infty}{\lim} \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
My Attempt:
$$L = \underset{t\to 0}{\lim} \frac{\left( \left( \frac{t+1}{t-1} \right)^{\frac 1t} - e^2\right)} {t^2}$$
I Now have a $\frac 00$ form that I could use L'Hopital rule with, but I don't want to differentiate the ugly looking function in the numerator. Is there an easier way to solve these kinds of problems? Maybe a taylor series expansion for $(1+t)^{\frac 1t}, t \to 0$ forms would come in handy here and I could just subtract the $e^2$ from the resulting expansion.
|
$$ \left( \left( \frac{x+1}{x-1} \right)^x - e^2\right) x^2$$
Start with
$$y=\left( \frac{x+1}{x-1} \right)^x\implies \log(y)=x\log\left( \frac{x+1}{x-1} \right)=x\log\left(1+ \frac{2}{x-1} \right)$$ Now, by Taylor
$$\log(y)=x\left(\frac{2}{x}+\frac{2}{3 x^3}+\frac{2}{5 x^5}+O\left(\frac{1}{x^7}\right)\right)={2}+\frac{2}{3 x^2}+\frac{2}{5 x^4}+O\left(\frac{1}{x^6}\right)$$
$$y=e^{\log(y)}=e^2+\frac{2 e^2}{3 x^2}+\frac{28 e^2}{45
x^4}+O\left(\frac{1}{x^6}\right)$$
$$(y-e^2)x^2=\frac{2 e^2}{3}+\frac{28 e^2}{45 x^2}+O\left(\frac{1}{x^4}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$ Mathematica gives $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $~_2F_1$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summation for this has been discussed at MSE:
A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$
|
One way to look at this is to take:
$\begin{align*}
\sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
&= \sum_k \frac{(-1)^k}{k + 1} \binom{n}{k} \binom{n}{n - k} \\
&= [z^n] \left(
\sum_k \frac{1}{k + 1} \binom{n}{k} z^k
\right)
\cdot (-1)^n \left(
\sum_k (-1)^k \binom{n}{k} z^k
\right)
\end{align*}$
For the pieces, you know that:
$\begin{align*}
\sum_k \binom{n}{k} z^k
&= (1 + z)^n \\
\sum_k \frac{1}{k + 1} \binom{n}{k} z^k
&= \frac{1}{z} \sum_k \frac{1}{k + 1} \binom{n}{k} z^{k + 1} \\
&= \frac{1}{z} \int_0^z (1 + t)^n \, d t \\
&= \frac{1}{z} \frac{(1 + z)^{n + 1} - 1}{n + 1}
\end{align*}$
Thus:
$\begin{align*}
\sum_k \frac{(-1)^k}{k + 1} \binom{n}{k}^2
&= (-1)^n [z^n] \frac{(1 + z)^{n + 1} - 1}{z (n + 1)} (1 - z)^n \\
&= \frac{(-1)^n}{n + 1}
[z^{n + 1}] ((1 + z)^{n + 1} - 1) (1 - z)^n \\
&= \frac{(-1)^n}{n + 1}
[z^{n + 1}] ((1 - z^2)^n (1 + z) \\
&= \frac{(-1)^n}{n + 1}
\left(
[z^n] (1 - z^2)^n
[z^{n - 1}] (1 - z^2)^n
\right) \\
&= \frac{(-1)^n}{n + 1}
\left(
(-1)^{n} \binom{2 n}{n}
+ (-1)^{n - 1} \binom{2 n}{n - 1}
\right) \\
&= \frac{1}{n + 1} \left( \binom{2 n}{n} - \binom{2 n}{n - 1} \right) \\
&= \frac{1}{(n + 1)^2} \binom{2 n}{n}
\end{align*}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplify Geometric series Can this equation simplify to the property of a sum of a geometric series, such as $ \frac{1}{1-r} $
$$\sum_{y=1}^{\infty}y^2q^{y}p$$
I understand that
$$\sum_{y=1}^{\infty}yq^{y} = q \sum_{y=0}^{\infty}(y-1)q^{y-1} = q \frac{d}{dq} \sum_{y=0}^{\infty}q^{y} = q \frac{d}{dq}\frac{1}{1-q} $$
|
\begin{align}
\sum_y y^2 q^y p
&= p \sum_y (y(y-1)+y) q^y \\
&= p q^2 \sum_y y(y-1) q^{y-2} + p q \sum_y y q^{y-1} \\
&= p q^2 \frac{d^2}{dq^2} \sum_y q^y + p q \frac{d}{dq} \sum_y q^y \\
&= p q^2 \frac{d^2}{dq^2} \frac{1}{1-q} + p q \frac{d}{dq} \frac{1}{1-q} \\
&= p q^2 \frac{2}{(1-q)^3} + p q \frac{1}{(1-q)^2} \\
&= \frac{pq(2 q +(1-q))}{(1-q)^3} \\
&= \frac{pq(1+ q)}{(1-q)^3} \\
\end{align}
|
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|
Rewriting linear combinations of complex numbers into real numbers for probabilities Below is from Norris' Markov Chains. From the matrix decomposition of $P^n$, it is clear that $p_{11}^{(n)}=a+b(\frac{i}{2})^n+c(-\frac{i}{2})^n$ for some complex numbers $a,b,c$ that depends on the entries of $U, U^{-1}$. But how are we able to rewrite this in the form $\alpha + (\frac{1}{2})^n(\beta \cos\frac{n\pi}{2}+\gamma \sin\frac{n\pi}{2})$ for some real numbers $\alpha,\beta,\gamma$ because $p_{11}^{(n)}$ is real?
|
The disciplined approach here is to check this for 3 consecutive values, say $n= 0,1,2$. That is enough to uniquely specify and solve for $a$, $b$ and $c$ (using a Vandermonde matrix, shown at the end).
Based on limitting values you should be able to see $a =\alpha$. As far as the non-real eigenvalues and the real trig function approach, applying Euler's formula gives
$b\cdot \cos\frac{n\pi}{2}+ b\cdot i\cdot \sin\frac{n\pi}{2} + c\cdot \cos\frac{3n\pi}{2}+c\cdot i\cdot \sin\frac{3n\pi}{2} $
$= b\cdot \cos\frac{n\pi}{2}+ b\cdot i\cdot \sin\frac{n\pi}{2} + c\cdot \cos\frac{n\pi}{2}-c\cdot i\cdot \sin\frac{n\pi}{2}$
$= (b+c)\cdot \cos\frac{n\pi}{2}+ (b-c)\cdot i\cdot \sin\frac{n\pi}{2}$
where the middle equality makes use of the fact that $i$ and $-i$ (or $\lambda_2,\lambda_3$ on the unit circle) are conjugate pairs so they have the same real portions and 'opposite' imaginary portions given by cosine and sine respectively. Non-real eigenvalues necessarily come in conjugate pairs for matrices that have entirely real components.
so to make this very explicit in your particular problem, in the original formulation you have
$\left[\begin{matrix}1 & 1 & 1\\1 & \frac{i}{2} & - \frac{i}{2}\\1 & - \frac{1}{4} & - \frac{1}{4}\end{matrix}\right]\mathbf v = \mathbf e_1$
and your coefficient vector is
$\mathbf v=\left[\begin{matrix}a\\ b \\ c \end{matrix}\right] =\left[\begin{matrix}1 & 1 & 1\\1 & \frac{i}{2} & - \frac{i}{2}\\1 & - \frac{1}{4} & - \frac{1}{4}\end{matrix}\right]^{-1}\mathbf e_1 = \left[\begin{matrix}\frac{1}{5}\\\frac{2}{5} + \frac{i}{5}\\\frac{2}{5} - \frac{i}{5}\end{matrix}\right]$
and of course
$b+c = \frac{2}{5} + \frac{i}{5} + \frac{2}{5} - \frac{i}{5} = \frac{4}{5}$ and
$b-c = \frac{2}{5} + \frac{i}{5} -\big(\frac{2}{5} - \frac{i}{5}\big) = i\cdot \frac{2}{5}$
and when you plug this into
$ (b+c)\cdot \cos\frac{n\pi}{2}+ (b-c)\cdot i\cdot \sin\frac{n\pi}{2}\longrightarrow \frac{4}{5}\cdot \cos\frac{n\pi}{2}- \frac{2}{5}\cdot \sin\frac{n\pi}{2}$
|
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|
Integrate $\int \frac{1}{1+x+x^4}dx$
Find $$\int \frac{{\rm d}x}{1+x+x^4}.$$
WA gives a result, which introduces complex numbers. But according to the algebraic fundamental theorem, any rational fraction can be integrated over the real number field.
How to do this?
|
\begin{align}
\int \frac{{\rm d}x}{1+x+x^4}
\end{align}
It looks like the main problem is factorization
of the quartic, so here it is one more variant:
for
\begin{align}
a&=\sqrt{\frac1{\sqrt3}\,\cos\Big(\tfrac13\,\arctan(\tfrac19\,\sqrt{687})\Big)}
\\
&\approx 0.72713608449
,
\end{align}
\begin{align}
x^4+x+1&=
\Big(x^2-2\,a\,x+2\,a^2+\sqrt{4\,a^4-1}\Big)
\left(x^2+2\,a\,x+\frac1{2\,a^2+\sqrt{4\,a^4-1})}\right)
\\
&\approx
(x^2-1.45427216898\,x+1.40126836794)
(x^2+1.45427216898\,x+0.713639173537)
.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that the function $f(x)=\cos x+\cos\left(\frac{\sqrt{3}}{2}x\right)$ is not periodic. Question: Show that the function $f(x)=\cos x+\cos\left(\frac{\sqrt{3}}{2}x\right)$ is not periodic.
My approach: Let us assume for the sake of contradiction that $T>0$ be the period of $f$. Then we must have $\forall a\in\mathbb{R}$ $$\int_a^{a+T}f(x)dx=\int_0^Tf(x)dx \hspace{1cm}...(1)$$
Therefore by setting $a=2\pi$ in $(1)$ we have, $$2\cos\left(\sqrt{3}\pi+\frac{\sqrt{3}}{4}T\right)\sin\frac{\sqrt{3}}{4}T=2\sin\frac{\sqrt{3}}{4}T\cos\frac{\sqrt{3}}{4}T \\\implies\sin\frac{\sqrt{3}}{4}T\left(\cos\left(\sqrt{3}\pi+\frac{\sqrt{3}}{4}T\right)-\cos\frac{\sqrt{3}}{4}T\right)=0$$
$$\implies \sin\frac{\sqrt{3}}{4}T=0, \cos\left(\sqrt{3}\pi+\frac{\sqrt{3}}{4}T\right)-\cos\frac{\sqrt{3}}{4}T=0.$$ This implies that either $T=\frac{4}{\sqrt 3}n\pi$ for some $n\in\mathbb{N}$ or $T=\left(\frac{4}{\sqrt 3}m-2\right)\pi$ for some $m\in\mathbb{N}$.
How to proceed after this?
|
$$f(x)=\cos x+\cos\frac{\sqrt{3} x}{2}$$
Let $f_1(x)$ be periodic with a period of $T_1$ and $f_2(x)$ be periodic with a period of $T_2$. Then $f(x)=f_1(x)+f_2(x)$ only if $T_1/T_2$ is rational. Then the period is given by $T=LCM(T_1,T_2).$
Here $T_1=2\pi$ and $T_2=\frac{4\pi}{\sqrt{3}}$ This means $T_1/T_2=\frac{\sqrt{3}}{2}$, it being irrational the given $f(x)$ is not periodic. Equivalently, here the LCM of $T_1$ and $T_2$ does not exist.
|
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|
Find all solutions to $\tan 2x = \cos \frac x2$ on the interval $[0,2\pi]$ I tried to solve this equation for almost 24 hours, using all the trigonometry identities I can think of, to no avail.
$$\tan 2x = \cos \frac x2$$
So far I've gotten to this point,
$$\cos^2 2x (\cos x + 3) = 2$$
but I've not made much of a progress since. I know one of the solutions is $x = \pi$, but I don't know how to solve for the rest.
Can someone give me a hint on how to solve this?
|
$\tan 2x = \cos \frac{x}{2} \iff \frac {\sin 2x}{\cos 2x} = \pm \sqrt{\frac {1+\cos x}{2}} \iff \frac{\sin^2 2x}{\cos^2 2x} = \frac{1 + \cos x}{2} \iff 2\sin^2 2x=(1 + \cos x)\cos^2 2x \\ \text{condition:} \quad x \neq \frac{\pi}{4} + \frac{k\pi}{2}, k \in \mathbb{Z}$
$ \iff 8\sin^2x \cos^2x = (1 + \cos x)(\cos^2x - \sin^2x)^2$
$\iff 8\sin^2x \cos^2x = (1 + \cos x)(\cos^4x - 2\cos^2x \sin^2x + \sin^4x)$
$\iff 8\sin^2x \cos^2x = \cos^4x - 2\cos^2x \sin^2x + \sin^4x + \cos^5x - 2\cos^3x \sin^2x + \cos x \sin^4x$
$\iff 10\sin^2x \cos^2x = \cos^4x + \sin^4x + \cos^5x - 2\cos^3x \sin^2x + \cos x \sin^4x$
$\iff 10(1 - \cos^2x) \cos^2x = \cos^4x + (1 - \cos^2x)^2 + \cos^5x - 2\cos^3x (1 - \cos^2x) + \cos x (1 - \cos^2x)^2$
$\iff 10\cos^2x -10\cos^4x = \cos^4x + 1 -2\cos^2x +\cos^4x+\cos^5x -2\cos^3x+2\cos^5x+\cos x-2\cos^3x+\cos^5x$
$\iff 4cos^5x +12\cos^4x-4\cos^3x -12\cos^2x + \cos x + 1 = 0$
$\iff 4cos^3x(\cos^2x - 1) +12\cos^2x(\cos^2x - 1) + \cos x + 1 = 0$
$\iff 4cos^3x(\cos x + 1)(\cos x - 1) +12\cos^2x(\cos x + 1)(\cos x - 1) + \cos x + 1 = 0$
$\iff (\cos x + 1)(4cos^3x(\cos x - 1) +12\cos^2x(\cos x - 1) + 1) = 0$
$\iff (\cos x + 1)(4cos^4x + 8\cos^3x -12\cos^2x + 1) = 0$
$\iff \cos x + 1 = 0 \quad \text{OR} \quad 4cos^4x + 8\cos^3x -12\cos^2x + 1 = 0$
Now we can easily find the solution(s) of first equation, and for the second equation we will be able to find solution(s) using substitution $\cos x = t$...
|
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|
Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\\
&\leq \frac{1}{k}\cdot\frac{2}{k}\cdots\cdot\frac{k-2}{k} \\
&\qquad \vdots \\
&\leq \frac{1}{k}\cdot\frac{2}{k}
= \frac{2}{k^2}
\end{align*}
That is
$$\sum_{k=1}^{\infty}\frac{k!}{k^k} \leq \sum_{k=1}^{\infty}\frac{2}{k^2} $$
And since right-hand side of the inequality is finite, so is left-hand side and therefore the series is convergent.
However, I dont find this way of solving the assignment elegant and I believe there is a cleaner way. Appreciates all help I can get.
|
Here is another elementary solution. Using the inequality $x(a-x) \leq \frac{a^2}{4}$ for $0 \leq x \leq a$, we get
$$ \frac{k!}{k^k} = \frac{(k-1)!}{k^{k-1}} = \prod_{j=1}^{k-1} \frac{\sqrt{j(k-j)}}{k} \leq \frac{1}{2^{k-1}}. $$
Therefore the series converges by the Direct Comparison Test.
|
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|
For non-negative reals such that $a+b+c\geq x+y+z$, $ab+bc+ca\geq xy+yz+zx$, and $abc\geq xyz$, show $a^k+b^k+c^k\geq x^k+y^k+z^k$ for $0
Let $a$, $b$, $c$, $x$, $y$, $z$ be non-negative real numbers such that
$$a+b+c \geq x+y+z,$$
$$ab+bc+ca \geq xy+yz+zx,$$
$$ abc \geq xyz$$
Show that
$$a^k+b^k+c^k \geq x^k+y^k+z^k, \quad 0 < k < 1$$
This is case $n=3$ of a problem posted by Ji Chen in the Art of Problem Solving forums, 2008.
I have posted a partial result in an answer below.
I have a proof when $r = \frac12,$ for weaker conditons
$$a+b+c = x+y+z,$$
$$\min(x, y, z) \leqslant \min(a, b, c),$$
$$\max(a, b, c) \leqslant \max(x, y, z).$$
Indeed, if $u, v > 0$ it's easy check
$$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$
Assume $x \geqslant y \geqslant z$ and $a \geqslant b \geqslant c$ then $x \geqslant a, \; c \geqslant z.$ Therefore
$$\begin{aligned}\sqrt{x}+\sqrt{y}+\sqrt{z} - \sqrt{a} - \sqrt{b} - \sqrt{c} & \leqslant \frac{x-a}{2\sqrt{a}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& \leqslant \frac{x-a}{2\sqrt{b}}+ \frac{y-b}{2\sqrt{b}}+ \frac{z-c}{2\sqrt{c}} \\& =\frac{x+y+z-a-b-c}{2\sqrt{b}}-\frac{(c-z)(\sqrt{b}-\sqrt{c})}{2\sqrt{bc}} \leqslant 0.\end{aligned}$$
|
Let $a$, $b$ and $c$ be different numbers, $a+b+c=u$, $ab+ac+bc=v$ and $abc=w$.
Now, by $u$, $v$ and $w$ we can get any permutations of $a$, $b$ and $c$, but since $a^k+b^k+c^k$ is a symmetric expression, we can think that $a^k+b^k+c^k=f(u,v,w)$ and we need to prove that
$f$ increases as a function of $u$, as a function of $v$ and as a function of $w$.
Id est, it's enough to prove that: $\frac{\partial f}{\partial u}\geq0,$ $\frac{\partial f}{\partial v}\geq0$ and $\frac{\partial f}{\partial w}\geq0.$
Indeed, $$1=\frac{\partial(a+b+c)}{\partial u}=\frac{\partial a}{\partial u}+\frac{\partial b}{\partial u}+\frac{\partial c}{\partial u},$$
$$0=\frac{\partial(ab+ac+bc)}{\partial u}=\frac{\partial a}{\partial u}b+\frac{\partial b}{\partial u}a+\frac{\partial a}{\partial u}c+\frac{\partial c}{\partial u}a+\frac{\partial b}{\partial u}c+\frac{\partial c}{\partial u}b=$$
$$=(b+c)\frac{\partial a}{\partial u}+(a+c)\frac{\partial b}{\partial u}+(a+b)\frac{\partial c}{\partial u}.$$
Also, $$0=\frac{\partial(abc)}{\partial u}=bc\frac{\partial a}{\partial u}+ac\frac{\partial b}{\partial u}+ab\frac{\partial c}{\partial u}.$$
Now, the determinant of this system it's
$$\Delta=\left|\left(\begin{array}{cc} 1&1&1\\b+c&a+c&a+b\\bc&ac&ab\end{array}\right)\right|=\sum_{cyc}(ab(a+c)-bc(a+c))=$$
$$=\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c);$$
$$\Delta_{\frac{\partial a}{\partial u}}=\left|\left(\begin{array}{cc}1&1&1\\0&a+c&a+b\\0&ac&ab\end{array}\right)\right|=(a+c)ab-ac(a+b)=a^2(b-c),$$
which gives $$\frac{\partial a}{\partial u}=\frac{a^2(b-c)}{(a-b)(a-c)(b-c)}=\frac{a^2}{(a-b)(a-c)}.$$
Similarly, $$\frac{\partial b}{\partial u}=\frac{b^2}{(b-a)(b-c)}$$ and
$$\frac{\partial c}{\partial u}=\frac{c^2}{(c-a)(c-b)}.$$
By the same way we can get that
$$\left(\frac{\partial a}{\partial v},\frac{\partial b}{\partial v},\frac{\partial c}{\partial v}\right)=\left(-\frac{a}{(a-b)(a-c)},-\frac{b}{(b-a)(b-c)},-\frac{c}{(c-a)(c-b)}\right)$$ and
$$\left(\frac{\partial a}{\partial w},\frac{\partial b}{\partial w},\frac{\partial c}{\partial w}\right)=\left(\frac{1}{(a-b)(a-c)},\frac{1}{(b-a)(b-c)},\frac{1}{(c-a)(c-b)}\right).$$
Now, let $a>b>c$.
Thus, by the Lagrange's theorem there is $t\in[b,a],$ for which
$$\frac{a^{k+1}-b^{k+1}}{a-b}=(k+1)t^k\geq(k+1)b^k,$$ which gives
$$a^{k+1}\geq b^{k+1}+(k+1)b^k(a-b).$$
Also, there is $p\in[c,b],$ for which
$$\frac{b^{k+1}-c^{k+1}}{b-c}=(k+1)p^k\leq(k+1)b^k,$$ which says
$$c^{k+1}\geq b^{k+1}-(k+1)b^k(b-c).$$
Thus, $$\frac{\partial f}{\partial u}=\sum_{cyc}\frac{\partial f}{\partial a}\frac{\partial a}{\partial u}=\sum_{cyc}\frac{ka^{k-1}\cdot a^2}{(a-b)(a-c)}=k\sum_{cyc}\frac{a^{k+1}}{(a-b)(a-c)}\geq$$
$$\geq k\left(\frac{b^{k+1}+(k+1)b^k(a-b)}{(a-b)(a-c)}+\frac{b^{k+1}}{(b-a)(b-c)}+\frac{b^{k+1}-(k+1)b^k(b-c)}{(c-a)(c-b)}\right)=0.$$
By the same way we can prove that $\frac{\partial f}{\partial v}\geq0$ and $\frac{\partial f}{\partial w}\geq0.$
Now, since our proof is valid for $a\rightarrow b^+$ and for $b\rightarrow c^+$ and since $f$ is a continues function, we have that $f$ increases for any positive $a$, $b$ and $c$, which ends a proof.
|
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|
Conclusion from the general triangle in a rectangle to an equilateral triangle
Let for the areas
$|\Delta ABP| =: F_B$,
$|\Delta PCQ| =: F_C$,
$|\Delta AQD| =: F_D$ and
$|\Delta APQ| =: F_\Delta$.
I know $~~~ F_\Delta = \sqrt{(F_B+F_C+F_D)^2-4 F_B F_D}~~~$ (see here);
is there an easy way to show, that $$F_B+F_D=F_C,~~$$ if $\Delta APQ$ is an equilateral triangle
with that formula above for $F_\Delta$?
Just as example for an equilateral triangle:
|
To proved the equality $F_B + F_D = F_C$ when $\triangle APQ$ is equilateral, you don't need that formula of $F_\Delta$. It can be proved directly using a little bit of trigonometry.
When $\triangle APQ$ is equilateral, let $s$ be its side and let $\alpha, \beta, \gamma$ be angles illustrated below.
The three angles can be parameterized by a single $\theta \in (-\frac{\pi}{12},\frac{\pi}{12})$ as
$$(\alpha,\beta,\gamma) = \left(\frac{\pi}{12} + \theta, \frac{\pi}{12} - \theta, \frac{\pi}{4} - \theta\right)$$
In terms of these angles, the areas of the three right angled triangles are
$$\begin{align}
F_B &= \frac12 (s\cos\alpha)(s\sin\alpha) = \frac{s^2}{4}\sin(2\alpha) = \frac{s^2}{4} \sin\left(\frac{\pi}{6} + 2\theta\right)\\
F_D &= \frac12 (s\cos\beta)(s\sin\beta) = \frac{s^2}{4}\sin(2\beta) = \frac{s^2}{4} \sin\left(\frac{\pi}{6} - 2\theta\right)\\
F_C &= \frac12 (s\cos\gamma)(s\sin\gamma) = \frac{s^2}{4}\sin(2\gamma) = \frac{s^2}{4}\sin\left(\frac{\pi}{2} - 2\theta\right) = \frac{s^2}{4}\cos(2\theta)\end{align}$$
With help of the trigonometric identity $$\sin(\phi+\psi) + \sin(\phi-\psi) = 2\sin \phi\cos \psi$$
and the fact $\sin\frac{\pi}{6} = \frac12$, we find
$$F_B + F_D = \frac{s^2}{4}\times 2 \sin\frac{\pi}{6} \cos(2\theta) = \frac{s^2}{4}\cos(2\theta) = F_C$$
|
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|
Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$
Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replacing $2/5$ with $x$)
After some manipulations,
$$y+1=(1-2x)\frac{dy}{dx}$$
Integrating and substituting $x=\dfrac{2}{5}$, we get $y=\sqrt{5}-1$.
Is there any other way to solve this question?
|
The hint:
It's a Taylor expansion for $f(x)=\frac{1}{\sqrt{1-2x}}-1.$
|
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|
Verify the integral $\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$ I'm stuck solving the integral
$$\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$$
This is what I got so far
\begin{align}
\int_{1}^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}} - u} \,du &= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{u - u\cdot(1 + \sqrt{u^2 + 1})} \,du \\
&= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{-u\cdot \sqrt{u^2 + 1}} \,du = -\bigg( \underbrace{\int_1^z \frac{1}{u\sqrt{u^2 + 1}} \,du}_{(*)} + \int_1^z \frac{1}{u}\,du\bigg) \\
&= - (*) - \ln(z)
\end{align}
Calculation of $(*)$ yields
\begin{align*}
(*) &= \int_1^z \frac{1}{u \cdot \sqrt{u^2 + 1}} \,du = \int_1^z \frac{u}{u^2 \cdot \sqrt{u^2 + 1}} \,du = \int_1^z \frac{\varphi'(u)}{\varphi(u)^2 - 1} \,du = \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u^2 -1} \,du \\
&= \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u-1} \,du - \frac{1}{2} \int_{\sqrt{2}}^{\sqrt{z^2 + 1}} \frac{1}{u+1} \,du = \frac{1}{2} \bigg[\ln(u-1) - \ln(u+1)\bigg]_{\sqrt{2}}^{\sqrt{z^2 + 1}} \\
&= \frac{1}{2} \left(\ln(\frac{\sqrt{z^2 + 1} - 1}{\sqrt{z^2 + 1} + 1}) - \ln(\frac{\sqrt{2} - 1}{\sqrt{2} + 1})\right)
\end{align*}
Thus we have the result
\begin{align*}
&-\frac{1}{2} \left(\ln(\frac{\sqrt{z^2 + 1} - 1}{\sqrt{z^2 + 1} + 1}) - \ln(\frac{\sqrt{2} - 1}{\sqrt{2} + 1})\right) - \ln(z) \\
&= -\frac{1}{2} \left(\ln(\frac{(\sqrt{z^2 + 1} - 1)^2}{(\sqrt{z^2 + 1} + 1) \cdot (\sqrt{z^2 + 1} - 1)}) - \ln(\frac{(\sqrt{2} - 1)^2}{(\sqrt{2} + 1)\cdot(\sqrt{2} - 1)})\right) - \ln(z) \\
&= -\frac{1}{2} \left(\ln(\frac{z^2 + 1 - 2\sqrt{z^2 + 1} + 1}{z^2}) - \ln(\frac{2 - 2\sqrt{2} + 1}{2 - 1})\right) - \frac{1}{2}\ln(z^2) \\
&= -\frac{1}{2} \left(\ln(z^2 - 2\sqrt{z^2 + 1} + 2) - \ln(3 + 2\sqrt{2})\right) \\
&= \frac{1}{2} \ln(3 + 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)
\end{align*}
The correct solution is given by
$$\ln(\frac{\sqrt{z^2 + 1} + 1}{z^2(1 + \sqrt{2})})$$
I'm not able to simplify my solution to the given solution. Did I integrate wrong or is there a way to verify the solution?
Edit: I think, that I made an mistake somewhere, since the solution should equal zero for $z = 1$ but $1$ is not root of my solution..
Edit 2: I found the mistake. I switched "+" and "-" in the line of the second last equality. My question remains how to simplify my solution to the given solution.. My solution then looks like this
$$\frac{1}{2} \ln(3 - 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)$$
|
$$\frac{1}{2} \ln(3 - 2\sqrt{2}) - \frac{1}{2}\ln(z^2 - 2\sqrt{z^2 + 1} + 2)$$
$$=\frac{1}{2}{ln(\sqrt{2}-1)^2}-\frac{1}{2}{ln(\sqrt{z^2+1}-1)^2}$$
$$=\frac{1}{2}({ln \frac{(\sqrt{2}-1)^2}{(\sqrt{z^2+1}-1)^2})}$$
$$={ln \frac{(\sqrt{2}-1)}{(\sqrt{z^2+1}-1)}}$$
$$={ln \frac{(\sqrt{z^2+1}+1)}{z^2(\sqrt{2}+1)}}$$
|
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|
How do I solve this limit without derivatives or series expansions? I have tried most of the common tricks but I still don't see what to do
I did get it to that form since I know that limit(ln(x+1)/x)=1, but the problem is that I can't get rid of x in the denominator.
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Let $n>1$ then from binomial theorem we know that $$(1+x)^{1/n}=1+\frac{x}{n}-\frac{n-1}{2n^2}x^2+\frac{(n-1)(2n-1)}{6n^3}x^3-\dots$$ for $0<x<1$. Clearly the series on right is alternating (ignoring the first term). You can also see that the terms of the series decrease in absolute value as $$\frac{(n-1)(2n-1)\dots((m+1)n-1)x^{m+1}}{(m+1)!n^{m+1}}\cdot\frac{m!n^m}{(n-1)(2n-1)\dots(mn-1)x^m}$$ equals $$\frac {((m+1)n-1)x}{(m+1)n}$$ and is thus less than $1$.
It follows that partial sums of the series give bounds for $(1+x) ^{1/n}$ as $$1 +\frac{x} {n} - \frac{(n-1)x^2}{2n^2}<(1+x)^{1/n}<1+\frac{x}{n}-\frac{(n-1)x^2}{2n^2}+\frac {(n-1)(2n-1)x^3}{6n^3}\tag{1}$$ which means that $$x-\frac{(n-1)x^2}{2n}<n((1+x)^{1/n}-1)<x-\frac{(n-1)x^2}{2n}+\frac{(n-1)(2n-1)x^3}{6n^2}$$ Letting $n\to\infty $ we get $$x-\frac{x^2}{2}\leq \log(1+x)\leq x-\frac{x^2}{2}+\frac{x^3}{3}$$ for $0<x<1$. And this means that $$-\frac{1}{2}\leq \frac{\log(1+x)-x}{x^2}\leq - \frac{1}{2}+\frac {x}{3}$$ for $0<x<1$. Taking limits as $x\to 0^+$ we get $$\lim_{x\to 0^+}\frac{\log(1+x)-x}{x^2}=-\frac{1}{2}$$ To deal with the case when $x\to 0^-$ we can put $x=-t$ so that $t\to 0^+$ and then the expression under limit in question becomes $$\frac{\log(1-t)+t}{t^2}$$ Adding and subtracting $(\log(1+t) - t) /t^2$ in above expression we can transform it into $$\frac{\log(1-t^2)}{t^2}-\frac{\log(1+t)-t}{t^2}$$ The first fraction tends to $-1$ and second fraction as shown above tends to $-1/2$ as $t\to 0^+$. Hence the desired limit is $-1/2$.
Note: Since the inequality $(1)$ is algebraic it should be possible to prove it by algebraic means, but I have not yet looked into that part.
You can also use integrals if that is allowed. Just note that for positive $t$ we have $$1-t<\frac {1}{1+t}<1-t+t^2$$ and integrating the above with respect to $t$ on interval $[0,x]$ we get $$x-\frac{x^2}{2}<\log(1+x)<x-\frac {x^2}{2}+\frac{x^3}{3}$$ and then we can proceed exactly as before.
|
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|
Range of $f(x) = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $ Here's what I did :
$$\text{Let }\quad y = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $$
$$\text{Let } \sin^2x=t$$
$$\Rightarrow\ (y-1)t^2-(y+1)t+2y+1=0 $$
$$\text{Since } t= \sin x\text{ is real,} $$
$$\text{Discriminant} \geqslant 0 $$
$$ \Rightarrow (y+1)^2-4(y-1)(2y+1) \geqslant 0$$
$$ \Rightarrow\ -7y^2+6y+5 \geqslant 0$$
$$\Rightarrow\ 7y^2-6y-5 \leqslant 0$$
$$\Rightarrow \ y\in\left[\frac{3-2\sqrt{11}}{7},\frac{3+2\sqrt{11}}{7}\right] $$
But the correct answer (according to wolfram) is
$$y \in \left[\frac{3-2\sqrt{11}}{7},\frac{1}{2}\right] $$
Please guide me how I should proceed further .
|
Introducing $t = \sin{x}, t\in [-1, 1]$, gives a new function $y(t) = \frac{t^2+t-1}{t^2-t+2}$, and a task is to find a $\min_{t\in [-1, 1]}y(t)$ and $\max_{t\in [-1, 1]}y(t)$.
1) $y'(t) = \frac{(2t+1)(t^2-t+2) - (2t-1)(t^2-t+2)}{(t^2-t+2)^2} = \frac{-2t^2+8t}{(t^2-t+2)^2} = 0 \Leftrightarrow \left\{\begin{aligned}-2t^2 + 8t &= 0 \\t^2 -t +2 &\ne 0\\\end{aligned}\right. \Leftrightarrow t = 0 \text{ and } t = 4 ( \notin [-1, 1])$
2)
- $y(0) = -\frac{1}{2}$
*
*$y(-1) = \frac{(-1)^2+(-1)-1}{(-1)^2-(-1) + 2} = \frac{1}{4}$
*$y(1) = \frac{1^2+1-1}{1^2- 1 + 2} = \frac{1}{2}$
range of $y = \left[-\frac{1}{2}, \frac{1}{2}\right]$
|
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|
Finding Maximum area of the isosceles triangle $A = \{(x , y ,z): 1 \leq x , y, z \leq 6$ and $x , y ,z$ are the sides of a isosceles triangle$\}$
How can I find the maximum area of the triangle whose sides $a , b , c$ , $(a , b , c) \in A$?
Can anyone please give me some hint?
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Let us first find out the set of sample points $S = \{(x , y ,z) : 1 \leq x , y , z \leq 6$ and $ x , y, z $ are the sides of an isosceles triangle $\} $.\
So $S= \{(2,2,1),(2,2,3),(3,3,1),(3,3,2),(3,3,4),(3,3,5),(4,4,1),\\
(4,4,2),(4,4,3),(4,4,5),(4,4,6),(5,5,1),(5,5,2),(5,5,3)\\
(5,5,4)(5,5,6)(6,6,1)(6,6,2)(6,6,3)(6,6,4)(6,6,5)\\
(1,1,1)(2,2,2)(3,3,3)(4,4,4)(5,5,5)(6,6,6)\}$\
Total number of elements of $S$ is $27$.\
Now we know area of a triangle with sides $a , b ,c$ is\
$A = \sqrt{x(x-a)(x-b)(x-c)}$ where $x = \frac{a+b+c}{2}$\
$ = \frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$.
Now we need to find $\max _{(a, b,c)\in S}\frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$.\
Let's say $x = (b+c-a) $\ $y = (a+c-b) $\ $z = (a+b-c) $.\
Now as $a , b ,c$ are the sides of a triangle we can say $x>0$ , $y>0$ , $z>0$ .\
So we can say $\frac{x+y}{2} = c$ , $\frac{z+y}{2} = a$ , $\frac{x+z}{2} = b$\
Now we can write \
$ \sqrt xy \sqrt xz \sqrt yz \leq \frac{x+y}{2} \frac{x+z}{2} \frac{z+y}{2}$ [By applying A.M $\geq$ G>M ]\
$\Rightarrow (b+c-a) (a+c-b) (a+b-c) \leq abc$ \
$\Rightarrow \frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)} \leq \frac{1}{4} \sqrt{ (a+b+c)abc}$\
[We can do so because $(b+c-a)(a+c-b)(a+b-c) >0$ and $abc >0$ ]\
If $a = b= c$ then the equality occurs.\ .
Now we can say for a particular triangle with sides of lengths $ m , n , p$ and $ m \geq n \geq p$ ,
the area of the triangle with sides of length $m$ will always be greater than the area of the triangle with sides of lengths $ m , n, p$.\
So we can say $A$ will assume it's maximum when $a = b= c = 6$. As all points of $S$ are $(x , y , z)$ where $1 \leq x , y , z \leq 6$.
$\max _{(a, b,c)\in S}\frac{1}{4} \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)} = \frac{1}{4} \sqrt{(6+6+6)6.6.6} =9 \sqrt3$.
So we will have only one point in the set of sample points $S$ , $(6 , 6 , 6)$ ,for which the triangle will be of maximum area.\
the probability that the triangle is of maximum area given that it is an isosceles triangle, is $\frac{1}{27}$.
|
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|
Solve the differential equation $ x^4y^3=xy' + y$
Using the change of the dependent variable $z = y^{−2}$, solve the differential equation:
$$xy' + y = x^4y^3$$.
My attempt:
$$xy' + y = x^4y^3 \tag 1$$
Now dividing $(1)$ by $y^2$
$$\frac{xy'}{y^2} + \frac{1}{y} = x^4y$$
Now put $z= \frac{1}{y^2}$,
$$\frac{dz}{dx} = \frac{-2}{y^3}\frac{dy}{dx}=\frac{-2}{y^3}y'$$
$$y'=\frac{-y^3dz}{2dx}$$
$$xy' z +\frac{1}{y} = x^4y$$
After that im not able to proceed further.
|
It would have been easier to divide by $y^3$ initially instead of $y^2$. Also, instead of just replacing $\frac{1}{y^2}$ with $z$ in your first term, if you used your $y'=\frac{-y^3dz}{2dx}$ instead, you would then get
$$\begin{equation}\begin{aligned}
-\frac{xyz'}{2} +\frac{1}{y} & = x^4y \\
-\frac{xz'}{2} + \frac{1}{y^2} & = x^4 \\
-\frac{xz'}{2} + z & = x^4
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
I trust you can now proceed to solve this.
|
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|
Roots of the equation $(1-4x)^4+32x^4=\frac{1}{27}$ find all real roots of the equation $(1-4x)^4+32x^4=\dfrac{1}{27}$
i try to use binomial expansion and am-gm inequality but i don't know how to do next.
|
If $x<0$ so
$$(1-4x)^4+32x^4>1>\frac{1}{27}.$$
If $x>\frac{1}{4}$ so
$$(1-4x)^4+32x^4>32\left(\frac{1}{4}\right)^4>\frac{1}{27}.$$
Id est, it's enough to solve our equation for $x\in\left[0,\frac{1}{4}\right]$.
But in this case by Holder we obtain:
$$(1-4x)^4+32x^4=\frac{1}{27}(1+2)^3((1-4x)^4+2(2x)^4)\geq\frac{1}{27}(1-4x+2\cdot2x)^4=\frac{1}{27}.$$
But we have equality and the equality in Holder occurs for $(1,2)||((1-4x)^4,32x^4)$ only,
which gives
$$1-4x=2x$$ or $$x=\frac{1}{6}.$$
|
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|
What does this double summation with mod evaluate to? Let $X = \{1,2,3,\dots\}$ and $Y = \{0,1\}$.
Define $f:X\times Y \rightarrow \mathbb{R}$ by
$$
f(x,y) =
\begin{cases}
-2^{-x} &\text{ if } \mod(x, 2) = y \\
(y+1) 2^{-x} &\text{ if } \mod(x,2) = y+1 \\
0 &\text{ otherwise}
\end{cases}
$$
I've been able to calculate that
$$
\sum_{x}\sum_{y}f(x,y) = -\frac{1}{3}
$$
How do I compute the opposite summation? Namely
$$
\sum_{y}\sum_{x}f(x,y)
$$
|
Note you have
$$\begin{equation}\begin{aligned}
\sum_{x}f(x,0) & = 2^{-1} - 2^{-2} + 2^{-3} - 2^{-4} + \ldots \\
& = \frac{1}{2}\sum_{i=0}^{\infty}\frac{1}{2}\left(\right)^{i} \\
& = \frac{\frac{1}{2}}{1 - \left(-\frac{1}{2}\right)} \\
& = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} \\
& = \frac{1}{3}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
$$\begin{equation}\begin{aligned}
\sum_{x}f(x,1) & = -2^{-1} + 2^{-1} - 2^{-3} + 2^{-3} + \ldots \\
& = (-2^{-1} + 2^{-1}) + (-2^{-3} + 2^{-3}) + \ldots \\
& = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note I was able to do the bracketing above due to the series being absolutely convergent (as the sum of the absolute values would be that of a geometric series with $r = \frac{1}{4}$, so its sum would be $\frac{4}{3}$), as explained in the Rearrangements and unconditional convergence section of Wikipedia's "absolute convergence" article. Thus,
$$\begin{equation}\begin{aligned}
\sum_{y}\sum_{x}f(x,y) & = \sum_{y}\left(\sum_{x}f(x,y)\right) \\
& = \sum_{x}f(x,0) + \sum_{x}f(x,1) \\
& = \frac{1}{3}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
|
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|
$e^{-x^4\sin^2x}\in L_2(R)$ proving $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$ I am trying to prove that $e^{-x^4\sin^2x}\in L_2(R)$. It means that $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$. So, i tried to use some usual methods, but $\sin(x)$ has no limit of infinity and I can't pick up an integrable upper bound to use Weierstrass theorem. Please give me some hints.
|
We have
\begin{align*}
& \int_{ - \infty }^{ + \infty } {e^{ - 2x^4 \sin ^2 x} dx} = 2\int_0^{ + \infty } {e^{ - 2x^4 \sin ^2 x} dx}
\\ &
= 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx} + 2\sum\limits_{n = 1}^\infty {\int_{\left( {n - \frac{1}{2}} \right)\pi }^{\left( {n + \frac{1}{2}} \right)\pi } {e^{ - 2x^4 \sin ^2 x} dx} }
\\ &
= 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx} + 2\sum\limits_{n = 1}^\infty {\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {e^{ - 2(x + \pi n)^4 \sin ^2 x} dx} }
\\ &
< 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx} + 2\sum\limits_{n = 1}^\infty {\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {e^{ - 2(x + \pi n)^4 (2x/\pi )^2 } dx} }
\\ &
< 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx} + 2\sum\limits_{n = 1}^\infty {\int_{-\frac{\pi }{2}}^{ \frac{\pi }{2} } {e^{ - 8\pi ^2 \left( {n - \frac{1}{2}} \right)^4 x^2 } dx} }
\\ &
< 2\int_0^{\pi /2} {e^{ - 2x^4 \sin ^2 x} dx} + 2\sum\limits_{n = 1}^\infty {\int_{-\infty}^{ + \infty } {e^{ - 8\pi ^2 \left( {n - \frac{1}{2}} \right)^4 x^2 } dx} }
\\ &
< \pi + 2\sqrt {\frac{2}{\pi }} \sum\limits_{n = 1}^\infty {\frac{1}{{(2n - 1)^2 }}} = \pi + \frac{{\pi ^{3/2} }}{{2\sqrt 2 }}<+\infty.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Diophantine equation from "Solving mathematical problems" by Terence Tao
Find all integers $n$ such that the equation $\frac{1}{a} + \frac{1}{b} = \frac{n}{a+b}$ is satisfied for some non-zero values of $a$ and $b$ (with $a + b \neq 0$).
I'm reading "Solving mathematical problems" by Terence Tao and I'm a bit stuck on this particular example. It starts as follows:
First multiplying out the denominators we get
$$\frac{a+b}{ab} = \frac{n}{a+b}$$
and from here follows $$(a+b)^2 = nab.$$
Now expanding this we get
$$a^2+2ab+b^2-nab=0 \Leftrightarrow a²+ab(2-n)+b^2.$$
From here on he suggests to use the quadratic formula to get
$$a= \frac{b}{2}[(n-2) \pm \sqrt{(n-2)^2-4}]$$
which I don't quite see how he came up with...
Also he then notes that "This looks very messy, but actually we can turn this messiness to our advantage. We know that $a, b$, and $n$ are integers, but there is a square root in the formula. Now this can only work if the term inside the square root, $(n-2)^2-4$ is a perfect square."
Could someone enlighten me on the part "Now this can only work if the term inside the square root, $(n-2)^2-4$ is a perfect square." what is he stating right here?
|
$$a^2 +a\left[b(2-n)\right] + b^2 = 0$$
is a quadratic equation in $a$, so we apply the quadratic formula. \begin{align*}
a &= \frac{-\left[b(2-n)\right] \pm \sqrt{\left[b(2-n)\right]^2 - 4(1)(b^2)}}{2} \\
&= \frac{-b(2-n)}{2} \pm \frac{1}{2}\sqrt{b^2( 4 - 4n + n^2) - 4b^2} \\
&= \frac{b}{2}(n-2) \pm \frac{1}{2}\sqrt{b^2( - 4n + n^2)} \\
&= \frac{b}{2}(n-2) \pm \frac{1}{2}|b|\sqrt{ n^2 - 4n + 4 - 4} \\
&= \frac{b}{2}(n-2) \pm \frac{1}{2}|b|\sqrt{ (n-2)^2 - 4} \\
&= \frac{b}{2} \left[ (n-2) \pm \sqrt{ (n-2)^2 - 4} \right] \text{,}
\end{align*}
where regardless of the sign of $b$, $\{|b|, -|b|\} = \{b,-b\} $ (in some order), so we get the last line.
We require $a$ is an integer, so the right hand side of this is an integer. This requires $b[\dots]$ is an even integer (even, to cancel the division by $2$ at the front). Now $n$ is a known integer, so $(n-2)^2 - 4$ is an integer. Call this integer
$$ D = (n-2)^2 - 4 \text{.} $$
If $D$ is a perfect square, $D = d^2$, then
$$ \frac{b}{2} \left[ (n-2) \pm \sqrt{ D} \right] = \frac{b}{2} \left[ (n-2) \pm d \right] \text{,} $$
and, excepting the evenness requirement, everything in sight is sums differences and products of integers, so is an integer.
If $D$ is not a perfect square, $\sqrt{D}$ is not an integer; it isn't even a rational number. So $(n-2) \pm \sqrt{ D}$ is an irrational number, and multiplying by the integer $b$ yields an irrational number. Dividing by $2$ leaves the number irrational. But $a$ is rational (it's even an integer), so this cannot be what happens.
|
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|
Finding integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square I am tring to find "integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square". Make this statement into equation I have
\begin{align}
x + 1 = a^2, \qquad \frac{x}{2} + 1 = b^2
\end{align}
where $x,a,b \in \mathbb{N}$.
From knowledge of odd perfect square is of the form of $8k+1$. I noticed $x$ should be multiple of 16.
From trial and errors I found the smallest integer $x$ is $48$. i.e.,
\begin{align}
48 +1 = 7^2, \qquad 24+1 = 5^2
\end{align}
and the second one is 1680. What is the general form of $x$ and how one can find that?
|
From the second equation, we have: $$x=2b^2-2$$
Substituing, we have:
$$2(b^2-1)-(a^2-1)=0$$
And so:
$$2b^2-a^2=1 \leftrightarrow a^2-2b^2=-1$$
This is a simply Pell equation with $d=2$. The first solution is $(a_1,b_1)\rightarrow(1,1)$ and in general:
$$a_{n} = 6 a_{n-1} - a_{n-2}, \: \: b_{n} = 6 b_{n-1} - b_{n-2}, n\in N \land n\geq3$$
Substituing again, we have that $x$ is equal tro:
$$x=a_n^2-1=(6a_{n-1}^2-a_{n-2})^2-1$$
|
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|
How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2x^3y+3x^2y^2+2xy^3+y^4=(x^2+xy+y^2)^2$$ without knowing the answer.
I know about $$(a+b+c)^2$$ but then how would i chose my $a, b$ and $c$?
There are many $xy$ combinations here, with different powers, so would I chose the lowest as my $b$, and $x^2$ as my $a$ and $y^2$ as my $c$, or perhaps is there another formula that can help me.
|
Note,
$$\begin{array}
& & x^4+2x^3y+3x^2y^2+2xy^3+y^4 \\
& =( x^4+ 2x^2y^2+y^4 )+ 2x^3y+x^2y^2+2xy^3\\
&=( x^2+y^2)^2+ (2x^3y+2xy^3) + x^2y^2 \\
&=( x^2+y^2)^2+ 2xy(x^2+y^2) + (xy)^2 \\
&=(x^2+xy+y^2)^2
\end{array}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Bounding the integral of $\ln(1+e^{-x})$ from $0$ to $n$ using given double inequality Let $f(x) := \ln(1+e^{-x})$ and
$$u_n := \sum_{k=1}^n \frac{(-1)^{k-1}}{k^2}.$$
Given that for $x>0$
$$ \tag{1} \sum_{k=1}^n \frac{(-1)^{k-1}}{k}e^{-kx} - \frac{e^{-(n+1)x}}{n+1} \le f(x) \le \sum_{k=1}^n \frac{(-1)^{k-1}}{k}e^{-kx} + \frac{e^{-(n+1)x}}{n+1} $$
it is asked to prove that
$$u_n - \dfrac{1}{(n+1)^2} \le \int_0^n f(x) dx \le u_n + \frac{1}{(n+1)^2}$$
I tried to integrate the first double-inequality form $0$ to $n$ but I get extra terms of unknown sign. For the left side for instance I get
$$ \tag{2} \left( u_n - \dfrac{1}{(n+1)^2} \right) + \left(\dfrac{e^{-(n+1)n}}{(n+1)^2} -
\displaystyle\sum_{k=1}^n \dfrac{(-1)^{k-1}}{k^2} e^{-nk} \right) \le \displaystyle\int_0^n f(x) dx $$
I tried to prove that the second term is positive (sufficient condition) but in vain.
thanks for any advice.
|
It is possible to give a closed formula for the integral. We have
\begin{align}
\int_0^n \ln(1+e^{-x}) dx = \int_{-n}^0 \ln(1+e^x) dx = -\left. \mathrm{Li}_2(-e^{x}) \right|_{x=-n}^0 &= \mathrm{Li}_2(-e^{-n}) - \mathrm{Li}_2(-1) \\& = \mathrm{Li}_2(-e^{-n}) + \frac{\pi^2}{12},
\end{align}
since it is well-known that $$\mathrm{Li}_2(-1) = -\eta(2) = - \zeta(2)/2 =- \pi^2/12,$$ where $\eta$ denotes the Dirichlet eta function. Now, one can easily do some numerical checks: Looking at
$$E(n):= \mathrm{Li}_2(-e^{-n}) + \frac{\pi^2}{12} - \Big( u_n - \frac{1}{(n+1)^2} \Big)$$
one get that $E(1) < 0$ and $E(3) < 0$. Thus, at least for $n=1$ and $n=3$ the inequality is wrong.
Conjecture: For $n>3$ the inequality $$\int_0^n \ln(1+e^{-x}) \, dx \geq u_n - \frac{1}{(n+1)^2}$$ seems to be true.
However, the Taylor-type inequality (1) is not suitable for the verification of this conjecture. If we define
$$a_n:= \sum_{k=1}^n \frac{(−1)^{k}}{k^2} e^{-nk},$$
then one has the alternating series estimate
$$|a_n - \mathrm{Li}_2(-e^{-n})| \leq \frac{e^{-n(n+1)}}{(n+1)^2}.$$
Additionally, one has
$$\mathrm{Li}_2(-e^{-n}) = - \int_0^{e^{-n}} \frac{\ln(1+t)}{t} dt.$$
Because of $|t - \ln(1+t)| \leq |t|/2$ for $|t| \leq 1/2$, we see that
$$-e^{-n} \leq \mathrm{Li}_2(-e^{-n}) \leq - \frac{e^{-n}}{2}$$
for all $n \in \mathbb{N}$. Thus, we may conclude that
$$ \frac{e^{-n(n+1)}}{(n+1)^2} - \sum_{k=1}^n \frac{(-1)^{k-1}}{k^2} e^{-nk} =\frac{e^{-n(n+1)}}{(n+1)^2} +a_n \leq 2 \frac{e^{-n(n+1)}}{(n+1)^2} -\frac{e^{-n}}{2} \leq - \frac{(1-e^{-2})}{2} e^{-n}.$$
This shows that the second term in (2) is always negative!
Let us prove that $\int_0^n \ln(1+e^{-x}) \, dx \geq u_n - (n+1)^{-2}$ for all $n>3$.
Starting with the formula
$$\int_0^n \ln(1+e^{-x}) \, dx = \mathrm{Li}_2(-e^{-n}) + \frac{\pi^2}{12} = \mathrm{Li}_2(-e^{-n}) + u_n + \sum_{k=n+1}^\infty \frac{(-1)^{k-1}}{k^2}$$
we have to show that
$$\mathrm{Li}_2(-e^{-n})+ \sum_{k=n+1}^\infty \frac{(-1)^{k-1}}{k^2} \geq - \frac{1}{(n+1)^2}$$
for $n>3$. If $n$ is even, then the sum is positive and $\mathrm{Li}_2(-e^{-n}) \geq - e^{-n}$, as we have already seen in the previous step. Since $e^{-n} \geq - (n+1)^2$, if $n>3$, the inequality follows. If $n$ is odd, then
$$\mathrm{Li}_2(-e^{-n})+ \sum_{k=n+1}^\infty \frac{(-1)^{k-1}}{k^2} = - \frac{1}{(n+1)^2} + \mathrm{Li}_2(-e^{-n}) + \sum_{k=n+2}^\infty \frac{(-1)^{k-1}}{k^2}$$
and hence we need to verify that
$$\mathrm{Li}_2(-e^{-n}) + \sum_{k=n+2}^\infty \frac{(-1)^{k-1}}{k^2} \geq 0.$$
Here we may use the estimate
$$\sum_{k=n+2}^\infty \frac{(-1)^{k-1}}{k^2} \ge \frac{1}{(n+2)^2} - \frac{1}{(n+3)^2}$$
and note that
$$\mathrm{Li}_2(-e^{-n}) + \frac{1}{(n+2)^2} - \frac{1}{(n+4)^2} \geq - e^{-n} + \frac{1}{(n+2)^2} - \frac{1}{(n+3)^2} \geq 0$$
holds for $n \geq 6$. One case here is missing, namely $n=5$. Here one need to use the more precise estimate
$$\sum_{k=n+2}^\infty \frac{(-1)^{k-1}}{k^2} \ge \frac{1}{(n+2)^2} - \frac{1}{(n+3)^2} + \frac{1}{(n+4)^2} - \frac{1}{(n+5)^2}.$$
All in all, we conclude that $E(1)<0$, $E(2)>0$, $E(3)<0$ and $E(n)>0$ for $n>3$. Thus, the left-hand side of the inequality is valid with two exception, namely if $n=1$ or $n=3$.
|
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|
Find the value of $\big|\frac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\frac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\big|$ If $$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}=1,$$ where $\theta_1$ and $\theta_0$ do not differ by an odd multiple of $\pi$, then find the value of $$\left|\dfrac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|.$$
My attempt is as follows:
Attempt $1$:
$$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}$$
$$\dfrac{\sin\left(\theta_1+\theta_2\right)}{\cos\theta_2\sin\theta_2}=\dfrac{\sin\left(\theta_0+\theta_2\right)}{\cos\theta_2\sin\theta_2}$$
$$\sin\left(\theta_1+\theta_2\right)-\sin\left(\theta_0+\theta_2\right)=0$$
$$2\sin\dfrac{\left(\theta_1-\theta_0\right)}{2}\cos\dfrac{2\theta_2+\theta_1+\theta_0}{2}=0$$
either $\theta_1-\theta_0=2n\pi$ or $2\theta_2+\theta_1+\theta_0=2n\pi$
Unfortunately it is given that $\theta_1-\theta_0$ is not equal to odd multiple of $\pi$ but we are getting $\theta_1-\theta_0$ as even multiple of $\pi$ so we cannot rule out one of the factor. Due to this reason I didn't find any way ahead.
Attempt $2$:
$$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=1$$
$$\sin\left(\theta_1+\theta_2\right)=\cos\theta_2\sin\theta_2$$
$$2\sin\left(\theta_1+\theta_2\right)=\sin2\theta\tag{1}$$
In the similary way
$$2\sin\left(\theta_0+\theta_1\right)=\sin2\theta\tag{2}$$
But by doing this we are tending towards result obtained in Attempt $1$:
Attempt $3$:
$$\left|\left(1-\dfrac{\sin\theta_0}{\sin\theta_2}\right)\left(1-\dfrac{\sin\theta_1}{\sin\theta_2}\right)+\dfrac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|$$
$$\left|1-\dfrac{\sin\theta_1}{\sin\theta_2}-\dfrac{\sin\theta_0}{\sin\theta_2}+\dfrac{2\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\right|$$
Also not getting anything from here, what to do?
|
Hint:
Observe that $\theta_1,\theta_0$ are the roots of
$$\cos x\sin\theta_2+\sin x\cos\theta_2-\sin\theta_2\cos\theta_2=0$$
$$\iff\cos x\sin\theta_2=\sin\theta_2\cos\theta_2(\sin\theta_2-\sin x)$$
Now squaring both sides and replacing $\cos^2 x$ with $1-\sin^2x,$ we find
$$(1-\sin^2x)\sin^2\theta_2=\cos^2\theta_2(\sin\theta_2-\sin x)^2$$
$$\implies \sin^2x-\sin x\cdot 2\cos^2\theta_2\sin\theta_2-\sin^2\theta_2(1-\cos^2\theta_2)=0$$
So, $\sin\theta_1\sin\theta_0=-\dfrac{\sin^4\theta_2}1$
Similarly, $\cos\theta_1\cos\theta_0=?$
|
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|
In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to
My attempt is as follows:-
Squaring both equations and adding
$$9+16+24\sin(P+Q)=37$$
$$\sin(P+Q)=\dfrac{1}{2}$$
either $P+Q=\dfrac{\pi}{6}$ or $P+Q=\dfrac{5\pi}{6}$
If $P+Q=\dfrac{\pi}{6}$, then $R=\dfrac{5\pi}{6}$ otherwise $R=\dfrac{\pi}{6}$
Let's see case $1$: $P+Q=\dfrac{\pi}{6}$
$$3\sin P+4\cos\left(\dfrac{\pi}{6}-P\right)=6$$
$$3\sin P+4\left(\dfrac{\sqrt{3}}{2}\cos P+\dfrac{1}{2}\cdot\sin P\right)=6$$
$$3\sin P+2\sqrt{3}\cos P+2\sin P=6$$
$$5\sin P+2\sqrt{3}\cos P=6\tag{1}$$
$$4\left(\dfrac{1}{2}\cdot\cos P-\sin P\cdot\dfrac{\sqrt{3}}{2}\right)+3\cos P=1$$
$$-2\sqrt{3}\sin P+5\cos P=1\tag{2}$$
$$\cos P=\dfrac{12\sqrt{3}+5}{37}$$
$$\sin P=\dfrac{30-2\sqrt{3}}{37}$$
Using calculator I found $\cos P=0.69$, this means $P>\dfrac{\pi}{6}$ because $\cos \dfrac{\pi}{6}=0.866$, this mean $Q$ will be negative because $Q=\dfrac{\pi}{6}-P$. So this cannot be the case hence $P+Q$ would be $\dfrac{5\pi}{6}$ and $R$ will be $\dfrac{\pi}{6}$
This is the correct answer also , but I want to know does there exist any better way to decide on the value of $P+Q$. I am asking this because I had to use the calculator for finding the value of $\cos P$.
|
A solution using complex numbers geometry :
The 2 relationships can be grouped into a single one by adding the second one to $i$ times the first one, giving :
$$3e^{iP}+4ie^{-iQ}=1+6i \ \ \iff \ \ \underbrace{3e^{iP}}_A+\underbrace{4e^{i(\pi/2-Q)}}_B=\underbrace{1+6i}_C \tag{1}$$
This defining relationship between points ("affixes") of these complex numbers can be written under a vectorial form :
$$\vec{OA}+\vec{OB}=\vec{OC}$$
meaning that $OBCA$ is a parallelogram with prescribed lengths $OA, OB, OC$ which will not leave much degrees of freedom as we are going to see it.
Remark: the polar angles of $\vec{OA}$ and $\vec{OB}$ are $P$ and $\pi/2-Q$ resp. (the latter being in $(-\pi/2,\pi/2)$).
Therefore
$$\alpha := angle(OB,OA)=P-(\pi/2-Q)\tag{2}$$
In parallelogram $OBCA$, we have the following classical relationship between the sides and the diagonals (see here).
$$p^2+q^2=2(a^2+b^2)\tag{3}$$
With $a=OA=3, b=OB=4, p=OC=\sqrt{1^2+6^2}=\sqrt{37}$, we deduce from (3) that the second diagonal has its length $q$ given by :
$$37+q^2=2(3^2+4^2) \ \ \implies \ \ AB=q=\sqrt{13}$$
Let us apply the cosine formula to triangle $OAB$ :
$$AB^2=OA^2+OB^2-2OA.OB \cos \alpha \ \ \iff \ \ 13=3^2+4^2-2.3.4 \cos \alpha$$
giving
$$\cos(OB,OA)=\cos \alpha = \dfrac12 \ \ \implies \ \ \alpha := angle(OB,OA)=\dfrac{\pi}{3}\tag{4}$$
Identifying (4) and (2), we get :
$$P+Q=\dfrac{\pi}{2}+\dfrac{\pi}{3}=\dfrac{5\pi}{6} \ \ \implies \ \ R=\pi-(P+Q)=\dfrac{\pi}{6}$$
as awaited.
But there is a case we haven't yet considered :
It has been assumed implicitly that the polar angle of $\vec{OB}$ is less than the polar angle of $\vec{OA}$. We could have had the inverse situation, which geometrically corresponds to a symmetry of parallelogram $OBCA$ with respect to its diagonal $OC$. Fortunately, this cannot arise, because the polar angle for $\vec{OB}$ would have been outside $(-\pi/2,\pi/2)$, contradicting the remark done upwards.
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.