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What can be said about the series $\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]$ This is a sequel to this question.
I recently was browsing through Hansen's "A Table of Series and Products", and I miraculously found the sum that I was looking for:
$$
\sum_{n=1}^\infty K_{0}\left( n z \right) \ = \ \frac{\pi}{2 z} + \frac{1}{2} \log\left( \frac{z}{4 \pi} \right) + \frac{\gamma}{2} - \frac{1}{2} \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + \frac{z^2}{4\pi^2} }} \right]
$$
What a beautiful sum (the entire book is full of such amazing results). Here, $\gamma$ is the Euler-Mascheroni constant and $K$ is the modified Bessel function of the second kind (of order 0).
I'm of course curious about the latter sum now:
$$
F(x) := \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]
$$
I'm assuming this sum cannot be evaluated exactly (although, the book is old from 1975, maybe it's out of date?)...
I am most curious, how does this function $F(x)$ look like in the limit $x \to 0$? Does it diverge? My guess would be yes since $x$ is in a denominator. My only thought of how to attack this series in the limit $x\to 0$ is to take the following series expansion in this limit:
$$
\frac{1}{\sqrt{n^2 + x^2}} = \frac{1}{n} + \frac{x^2}{2n^3} + \mathcal{O}\left( x^4 \right)
$$
Then in this limit we have something like;
$$
F[x] \approx \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{n} - \frac{x^2}{2 n^3} \right] = \frac{1}{2} x^2 \zeta(3)
$$
...in terms of the Riemann-zeta function. This seems like an answer, but I am worried there is something wrong with taking an expansion in the argument of the sum. Is this okay? Is there a better approximation?
|
Yes it is correct, note that
$$\frac{1}{\sqrt{ n^2 + x^2 }}=-\frac1n\left(1+\frac{x^2}{n^2}\right)^{-\frac12}=\frac1n\left(1-\frac{x^2}{2n^2}+o\left(\frac{1}{n^2}\right)\right)=\frac1n-\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right)$$
thus
$$\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]=\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac1n+\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]=\sum_{n=1}^{\infty} \left[ \frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]$$
|
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|
An algebraic inequality involving $\sum_{cyc} \frac1{(a+2b+3c)^2}$ I was reading through the proof of an inequality posted on a different website and the following was mentioned as being easily proven by AM-GM:
Let $a,\ b,\ c>0$, then $$\frac{1}{(a+2b+3c)^2} + \frac{1}{(b+2c+3a)^2} + \frac{1}{(c+2a+3b)^2} \le \frac{1}{4(ab + bc+ca)}$$
I checked that the inequality holds numerically, but I can't find a solution (definitely nothing obvious using AM-GM). I played around with a simpler version of the inequality:
$$\frac{1}{(a+2b)^2} + \frac{1}{(b+2a)^2} \le \frac{2}{9}\frac{1}{ab}$$
which I can prove by brute force algebra and a little AM-GM. However, brute force seems impractical for the three variable inequality. Any suggestions?
|
A full expanding gives
$$\sum_{cyc}(36a^6+80a^5b+104a^5c+21a^4b^2+189a^4c^2+58a^3b^3+82a^4bc-266a^3b^2c-122a^3c^2b-182a^2b^2c^2)\geq0,$$
which is true by AM-GM.
Also, SOS helps.
Let $2a+b=3x$, $2b+c=3y$ and $2c+a=3z$.
Thus, $a=\frac{x-2z+4y}{3}$, $b=\frac{y-2x+4z}{3}$, $c=\frac{z-2y+4x}{3}$,
$$ab+ac+bc=\frac{1}{9}\sum_{cyc}(x-2z+4y)(y-2x+4z)=\frac{1}{3}\sum_{cyc}(5xy-2x^2)$$ and we need to prove that
$$\sum_{cyc}\frac{1}{(x+y)^2}\leq\frac{27}{4\sum\limits_{cyc}(5xy-2x^2)}$$ or
$$\sum_{cyc}\left(\frac{9}{4\sum\limits_{cyc}(5xy-2x^2)}-\frac{1}{(x+y)^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{17x^2+17y^2+8z^2-2xy-20xz-20yz}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}\frac{(y-z)(-x+17y-4z)-(z-x)(-y+17x-4z)}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{-z+17x-4y}{(x+z)^2}-\frac{-z+17y-4x}{(y+z)^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-y)^2(4x^2+4y^2+23z^2-13xy+9xz+9yz)}{(x+z)^2(y+z)^2}\geq0$$ or
$$\sum_{cyc}(x^2-y^2)^2(4x^2+4y^2+23z^2-13xy+9xz+9yz)\geq0$$ or
$$\sum_{cyc}(x^2-y^2)^2(4x^2+4y^2-4xy+23z^2+9xz+9yz-9xy)\geq0,$$ for which it's enough to prove that
$$\sum_{cyc}(x^2-y^2)^2(xz+yz-xy)\geq0.$$
Indeed, since the last inequality is symmetric, we can assume $x\geq y\geq z$.
Thus, $$\sum_{cyc}(x^2-y^2)^2(xz+yz-xy)\geq(x^2-y^2)^2(xz+yz-xy)+(x^2-z^2)^2(xy+yz-xz)\geq$$
$$\geq(x^2-y^2)^2(xz+yz-xy)+(x^2-y^2)^2(xy+yz-xz)=2yz(x^2-y^2)^2\geq0$$ and we are done!
|
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|
Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural.
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$.
Now, I know I need to show that $P(a)=0$, but I do not know if it is the right path since I have not found any way to do it.
|
Let us restate the value of $P$ first.
$P = (X+2)^m+(X+2)^{2m+3}+1^{2m+3} = (X+2)^m+(X+2)^{2m+3}+1 =$
$(X+2)^m+(X+2)^{2m+3}+(X+2)^0 = (X+2)^{3m+3} = (X+2)^{(m+1)^{3}}$
And, Let us restate the value of $Q$.
$Q = X^2+4X+4+X+3 = X^2+4X+4+X+2+1 = (X+2)^2+(X+2)^1+(X+2)^0$
Do you notice the relationship between $P$ and $Q$?
|
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|
How to find the following limit algebraically? I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.
$$
\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}
$$
If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figured that the denominator can be rewritten as $ (x^2-4)(x+1) $. And then I tried to factor something in the numerator but couldn't see anything interesting. How do I find the limit algebraically?
I know that the answer is supposed to be 1, but I don't know how they got there.
Thanks in advance!
|
General tip (update) : When you can see that the denominator is equal to zero for a value $x=a$ which is the $x\to a$ of the limit, then you should try factoring on both the numerator and the denominator the factor $(x-a)$, such as you can get rid of the $\frac{0}{0}$ issue. This particular example though can also be done by following a row factorization.
Factor the expression at the numerator by taking out $x^2$ and then forming a quadratic factorization inside, as :
$$x^4+5x^3+6x^2 = x^2(x^2+5x+6) =x^2(x^2+2x+3x+6) = x^2(x+2)(x+3)$$
Then, the given limit is :
$$\lim_{x\rightarrow -2} \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)} = \lim_{x\to -2} \frac{x^2(x^2 + 5x + 6)}{(x^2-4)(x+1)} = \lim_{x \to -2} \frac{x^2(x+3)(x+2)}{(x-2)(x+2)(x+1)} $$
$$=$$
$$\lim_{x\to -2} \frac{x^2(x+3)}{(x-2)(x+1)} = 1$$
|
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|
How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$.
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
|
Compare:
$$\color{red}{1+2+\cdots +n}=\frac{n(n+1)}{2}$$
and
$$\color{red}{1+2+\cdots+n}+\color{blue}{(n+1)+(n+2)+\cdots+2n}=$$
$$\color{red}{1+2+\cdots+n}+\color{blue}{(1+2+\cdots+n)+n\cdot n}=$$
$$\color{red}{\frac{n(n+1)}{2}}+\color{blue}{\frac{n(n+1)}{2}+n^2}=\frac{2n(2n+1)}{2}$$
|
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|
If $a_{n+1}=\frac{10}{a_n}-3$ and $a_1=10$, find $\lim_{n \to \infty} a_n$
Let $a_{n+1}=\dfrac{10}{a_n}-3$, $a_1=10$ then find the limit $\lim\limits_{n \to \infty} a_n$
My Try :
$$a_2=-2 \ \ ,a_3=-8 \ \, a_4=-4.25 \ \ a_n <0$$
thus visthe monotone convergence theorem $$\lim\limits_{n \to \infty}=l$$
so: $$l=\dfrac{10}{l}-3 \to l^2+3l=10 \to l=2 , -5 $$
it is right ?
|
When you have a sequence of the form $a_{n+1}=f(a_n)$ that apparently does not lead to a closed formula for $a_n$, then you have to study the function $f(x)$.
When you graph the curve $y=f(x)=\dfrac{10}x-3$ in blue and $y=x$ in red, you notice there are two intersection points.
These are called fixed points of $f$ since $f(x)=x$. Once solved this gives $x=2$ or $x=-5$.
If the sequence would converge to $\ell$, the continuity of $a_{n+1}=f(a_n)$ will lead to $f(\ell)=\ell$ so $\ell$ will be one of the two fixed points.
On the graph we can see that $2$ is a repulsive point, and $-5$ an attractive point.
Since we are not required to do the full study for all initial seeds fo the sequence, but only for $a_1=10$, we will focus on showing it converges to $-5$.
We can see that the convergence is not a staircase (monotonic convergence) but a spiral. This means we have to show that $a_{2n}$ and $a_{2n+1}$ are both monotonic but of opposite direction.
To prove this we have to:
*
*study the sign of $a_{n+2}-a_n$, this is equivalent of studying the sign of $f(f(x))-x$.
*show that $-5$ is squeezed between $a_n$ and $a_{n+1}$, this is equivalent of studying the sign of $f(x)+5$.
First notice that $x<0\implies f(x)<0$ so as soon as $a_{n_0}<0$ then all subsequent $a_n$ with $n\ge n_0$ are also negative.
Since $a_2<0$ we will select $n_0=2$.
$f(x)+5=\dfrac {10}x-3+5=\dfrac{2(x+5)}x\quad\begin{cases} > 0 & x\in]-\infty,-5[\\<0 & x\in]-5,0[\end{cases}$
So if $a_n<-5$ then $a_{n+1}>-5$ and vice-versa and $-5$ is squeezed between $a_n$ and $a_{n+1}$ for $n\ge 2$
$f(f(x))-x=\dfrac{10}{\frac{10}{x-3}}-3-x=\dfrac{3(x+5)(x-2)}{10-3x}\quad\begin{cases} > 0 & x\in]-\infty,-5[\\<0 & x\in]-5,0[\end{cases}$
So $a_{n+2}>a_{n}$ for $a_n<-5$ and $a_{n+2}<a_n$ for $a_n>-5$.
Since $a_2>-5$ then $\begin{cases}a_{2n}>-5 & a_{2n}\searrow\\a_{2n+1}<-5 & a_{2n+1}\nearrow\end{cases}$
Now we can apply monotonicity theorem and $a_{2n}\to -5$ and $a_{2n+1}\to -5$.
This means that $a_n\to -5$.
|
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|
Solve $\int(1+\frac{1}{x})^{2}\frac{dx}{x^2}$ This seems to be a straightforward integral:
$$\int(1+\frac{1}{x})^{2}\frac{dx}{x^2}$$
By substitution substitution, $u = 1+\frac{1}{x}, du = -\frac{1}{x^2}$
$$-\int u^2= -\frac{u^3}{3}+C = -\frac{(1+\frac{1}{x})^3}{3}$$
Then I took the derivative:
$$\frac{d}{dx}-\frac{(1+\frac{1}{x})^3}{3} = -\frac{1}{3}\frac{d}{dx}(1+\frac{1}{x})^3$$
If $u = 1+ \frac{1}{x}$ then $$-\frac{1}{3}\frac{d}{du}u^3 = -u^2 = -(1+\frac{1}{x})^2$$
What went wrong with my substitution?
|
You assumed $u=1+\dfrac 1x$. So far so good.
But you didn't replace $dx$ by $du$ correctly.
You must not forget that $du= -\dfrac 1{x^2} dx$
So,
$$\frac{d(1+\dfrac 1x)^3}{dx} = \frac{d(u^3)}{du} \color{red}{\cdot \frac{du}{dx}}$$
Always remember that
$$\color{blue}{\frac{d}{dx} f(x)= \frac{d}{du} f(x) \cdot \frac {du}{dx}}$$
|
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|
Factor of a Mersenne number Why is it true that if 7 divides 91 then $(2^7-1) $ divides $(2^{91}-1)$?
1) $2^{91}-1$
$7|91 \implies (2^7-1)|(2^{91}-1)$
$\implies 2^7-1$ is factor
2) $2^{1001}-1$
$7|1001 \implies (2^7-1)|(2^{1001}-1)$
$\implies 2^7-1$ is factor
|
It may be illustrative to write the numbers out in binary. I'll use $2^{21} - 1 = (2^7)^3 - 1$ instead of $2^{91} - 1$, since it's shorter:
$$\begin{aligned}
2^{21} - 1
&= \underbrace{111111111111111111111}_{21\text{ digits}}\,\vphantom1_2 \\
&= \underbrace{1111111}_{7\text{ digits}}\,\underbrace{1111111}_{7\text{ digits}}\,\underbrace{1111111}_{7\text{ digits}}\,\vphantom1_2 \\
&= 1111111_2 \times 100000010000001_2 \\
&= (2^7 - 1) \times (2^{14} + 2^7 + 1).
\end{aligned}$$
|
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Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root
set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begin{aligned}
\alpha^6&= \alpha^3 \alpha^3
\\&=(3\alpha-1) (3\alpha-1)
\\&= 3\alpha(3\alpha-1)-1(3\alpha-1)
\\&= 9\alpha^2-3\alpha-3\alpha+1
\\&=9 \alpha^2 -6 \alpha+1
\end{aligned} $$
Now looking at $\alpha^4$
$$
\begin{aligned}
\alpha^4= \alpha^3 \alpha^1
&=(3\alpha-1) \alpha
&=3 \alpha^2-\alpha
\end{aligned}
$$
Let us go back
$$ \begin{aligned}
&\alpha^6-9\alpha^4+26 \alpha^2 -24
\\
&=( 9 \alpha^2-6 \alpha+1 )
-9(3\alpha^2-\alpha)
+26 \alpha^2 -24
\\ &=9\alpha^2-6\alpha+1-27\alpha^2+9 \alpha+26 \alpha^2 -24
\\&=18 \alpha^2 + 3\alpha -23
\\& = \vdots?
\\&=0
\end{aligned}$$
|
A possible shortcut to the problem
Notice that $$\alpha^3-3 \alpha+1=0\implies \alpha^2=3-\frac1\alpha\implies\alpha^2-2=1-\frac1\alpha$$ Now to check whether this is a root, $$\begin{align}\left(1-\frac1\alpha\right)^3-3\left(1-\frac1\alpha\right)+1&=1-\frac3\alpha+\frac3{\alpha^2}-\frac1{\alpha^3}-3+\frac3\alpha+1\\&=-1+\frac3{\alpha^2}-\frac1{\alpha^3}\\&=-\frac1{\alpha^3}(\alpha^3-3\alpha+1)=0\end{align}$$ so $\alpha^2-2$ is indeed a root of the polynomial $X^3-3X+1$.
|
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|
Manipulating a functional equation
Consider a function $f$ such that $$f(x)f(y)=f(xy)+f\left (\frac {x}{y}\right )$$ then find $$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$$
And the options are as follows
A) $2f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$
B) $2f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )$
C) $f\left(\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right )- f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )$
D) $f(3)+f\left (\frac {1}{3}\right )$
Through some algebraic manipulations I have reached until the following equations
1) $f(y)\,[ f(y^2)- 1]\, =f(y^3)$
2) $f(1)=0$ or $f(1)=2$
3) $ (f(y)\,)^2- f(1)= f(y^2)$
Also the question can be simplified as to finding $f((\sqrt 2-1)^3)-f((\sqrt 2+1)^3)$
|
Presumably, it's given that $f$ is non-constant, else we can have $f = 0$ or $f=2$.
So suppose $f$ is non-constant.
Choosing $x$ such that $f(x) \ne 0$, and letting $y=1$, we get $f(1) = 2$.
Letting $x=1$, we get $f(y) = f\left(\frac{1}{y}\right)$.
Noting that
$$
\left(\frac{3-2\sqrt 2}{\sqrt 2 + 1}\right )\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right) = 1
$$
it follows that
$$f\left (\frac {3-2\sqrt 2}{\sqrt 2 + 1}\right )- f\left (\frac {3+2\sqrt 2}{\sqrt 2 - 1}\right ) = 0$$
By the same reasoning, choice $(C)$ is correct.
|
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|
Proving that $\binom{2n}{n} > n2^n , \forall n \ge 4 $ I'm trying to prove the following statement: $\binom{2n}{n} > n2^n , \forall n \ge 4 $
This is my attempt at an inductive proof:
Let $P(n)$ be the following proposition: "$\binom{2n}{n} > n2^n , \forall n \ge 4 $"
Base case:
$\binom{2*4}{4} = 70 > 4*2^4 = 64$ so $P(4)$ is true.
Inductive step:
(I assume as an inductive hypothesis that P(n) is true and try to show that P(n+1) is true)
$\binom{2(n+1)}{n+1} = \binom{2n+2}{n+1} = \binom{2n+1}{n+1} + \binom{2n+1}{n}$
$= \binom{2n}{n} + \binom{2n}{n+1} + \binom{2n}{n} + \binom{2n}{n-1}$
$= 2 \binom{2n}{n} + \binom{2n}{n+1} + \binom{2n}{n-1}$
$= 2 \binom{2n}{n} + \frac{(2n)!}{(n+1)!(2n-(n+1))!} + \frac{(2n)!}{(n-1)!(2n-(n-1))!}$
$= 2 \binom{2n}{n} +2 \frac{(2n)!}{(n+1)!(n-1)!} $
By the inductive hypothesis we have:
$ 2 \binom{2n}{n} +2 \frac{(2n)!}{(n+1)!(n-1)!} > 2n2^n +2 \frac{(2n)!}{(n+1)!(n-1)!} = (2^{n+1}) n + 2 \frac{(2n)!}{(n+1)!(n-1)!} $
And here I'm stuck.
If I could prove that $\frac{(2n)!}{(n+1)!(n-1)!} \ge 2^n , \forall n \ge 4 $ the last step would show that P(n+1) is true. But I haven't been able to prove that. Can anyone help me ?
|
$$\frac{(2n)!}{(n+1)!(n-1)!} =\frac n{n+1}\cdot{2n\choose n}\ge \frac{n^2}{n+1}2^n$$
A more direct proof for a better claim:
$2n\choose n$ is the largest of $2n+1$ summands in the expansion of $(1+1)^{2n}$, hence
$$ {2n\choose n}\ge\frac1{2n+1}\cdot 4^n.$$
|
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Limit of a standard dissection I am looking for the limit of the following expression $$\lim_{k\rightarrow\infty}\frac{(\frac{1}{k}+\frac{\epsilon}{j})^j(\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}$$
with $j<k$ and some $\epsilon>0$.
What I have done so far:
We can rewrite it as $$\lim_{k\to \infty}{(1+\frac{k\epsilon}{j})^j(1-\frac{k\epsilon}{k-j})^{k-j}}$$. I have no idea how to compute the limit of the above expression as the first term goes to infinity and the latter one to zero.(I have thought about using L'Hospital and derive the expression $j$ times until the first term has power 0, i.e. 1) I have noticed that for different value of $\epsilon$, this sequence might converge to 0 or diverge. How to find the criteria of $\epsilon$, such that this sequence converges?
|
Note that for $k\to +\infty$
*
*$(\frac{1}{k}+\frac{\epsilon}{j})^j =\left(\frac{1}{k}\right)^{j}(1+\frac{\epsilon k}{j})^j $
*$(\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}=\left(\frac{1}{k}\right)^{k-j}(1-\frac{\epsilon k}{k-j})^{k-j}$
thus
$$\frac{(\frac{1}{k}+\frac{\epsilon}{j})^j (\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}
=\frac{\left(\frac{1}{k}\right)^{j}(1+\frac{\epsilon k}{j})^j \left(\frac{1}{k}\right)^{k-j}(1-\frac{\epsilon k}{k-j})^{k-j}}{(\frac{1}{k})^k}
=\left(1+\frac{\epsilon k}{j}\right)^j \left(1-\frac{\epsilon k}{k-j}\right)^{k-j}=e^{j\log\left(1+\frac{\epsilon k}{j}\right)+(k-j)\log{\left(1-\frac{\epsilon k}{k-j}\right)}}$$
and for $0<\epsilon<1$
$$j\log\left(1+\frac{\epsilon k}{j}\right)+(k-j)\log{\left(1-\frac{\epsilon k}{k-j}\right)}=k\left[ j\frac{\log\left(1+\frac{\epsilon k}{j}\right)}{k}+\frac{(k-j)}{k}\log{\left(1-\frac{\epsilon k}{k-j}\right)}\right]\to +\infty[0+1\cdot\log(1-\epsilon)]=-\infty$$
For $\epsilon>1$
$$\left(1-\frac{\epsilon k}{k-j}\right)>-1$$
thus
$$\left(1+\frac{\epsilon k}{j}\right)^j \left(1-\frac{\epsilon k}{k-j}\right)^{k-j}\to +\infty\cdot \pm\infty=\pm\infty$$
For $\epsilon=1$
$$\left(1+\frac{k}{j}\right)^j \left(1-\frac{k}{k-j}\right)^{k-j}=\left(1+\frac{k}{j}\right)^j \left(\frac{-j}{k-j}\right)^{k-j}=\left(\frac{j+k}{j}\right)^j \left(\frac{j-k}{j}\right)^{j} \left(\frac{j}{j-k}\right)^{k}=\frac{1}{j^{2j}}(j^2-k^2)^j\left(\frac{j}{j-k}\right)^{k}$$
by ratio test of the absolute value we obtain
$$\frac{((k+1)^2-j^2)^j\left(\frac{j}{k+1-j}\right)^{k+1}}{(k^2-j^2)^j\left(\frac{j}{k-j}\right)^{k}}
=\frac{((k+1)^2-j^2)^j}{(k^2-j^2)^j} \left(\frac{j}{k+1-j}\right) {\left(\frac{k-j}{k+1-j}\right)^{k}}
=\frac{((k+1)^2-j^2)^j}{(k^2-j^2)^j} \left(\frac{j}{k+1-j}\right) {\left(1-\frac{1}{k+1-j}\right)^{k}}
=\frac{((k+1)^2-j^2)^j}{(k^2-j^2)^j} \left(\frac{j}{k+1-j}\right) {\left(1-\frac{1}{k+1-j}\right)^{j-1}} {\left(1-\frac{1}{k+1-j}\right)^{k+1-j}}\\\to 1\cdot 0\cdot 1 \cdot \frac1e=0$$
Therefore
$$\lim_{k\rightarrow\infty} \frac{(\frac{1}{k}+\frac{\epsilon}{j})^j (\frac{1}{k}-\frac{\epsilon}{k-j})^{k-j}}{(\frac{1}{k})^k}=
\begin{cases}
0\quad \text{for} \quad 0<\epsilon\le1\\
\\
\pm\infty \quad \text{for} \quad \epsilon>1\end{cases}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which of the two sums gives a better approximation to $\pi^2/6$? Which of the following two sums
$$\sum_{n=1}^{1000000}\frac1{n^2} \quad \text{or} \quad 1+\sum_{n=1}^{1000}\frac1{n^2(n+1)}$$
gives a better approximation to $\pi^2/6?$
I tested this on MATLAB and surprisingly obtained as a result the second sum. However, I do not know how to prove this rigorously.
|
$$\eqalign{\sum_{k+1}^\infty\frac{1}{n^2(n+1)}&<\sum_{k+1}^\infty\frac{1}{(n-1)n(n+1)}\cr
&=\frac12\sum_{k+1}^\infty\left(\frac{1}{(n-1)n}-\frac{1}{n(n+1)}\right)\cr
&=\frac{1}{2k(k+1)}}$$
$$\eqalign{\sum_{k^2+1}^\infty\frac{1}{n^2}&>\sum_{k^2+1}^\infty\frac{1}{n(n+1)}\cr
&=\sum_{k^2+1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)\cr
&=\frac{1}{k^2+1}}$$
So, clearly the error in the second is smaller than half the error in the first, that is
$$\frac{\pi^2}{6}-\left(1+\sum_{n=1}^k\frac{1}{n^2(n+1)}\right)<\frac{1}{2}\left(\frac{\pi^2}{6}-\sum_{n=1}^{k^2}\frac{1}{n^2}\right)$$
The problem corresponds to $k=1000$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta)-2\sin^3(\theta) $$ simply making the substitutions $$\sin(\theta)=\frac{y}{r},~\cos(\theta)=\frac{x}{r}$$ noting also that I can square both sides of the above to be substitutions we then can write down $$r=\dfrac{2yx}{r}\cdot \dfrac{x^2}{r^2}+\dfrac{y}{r}-\dfrac{2y^3}{r^3}$$ then multiplying through by $r^3$ we obtain $$r^4=2yx^3+yr^2-2y^3$$ then replacing $r^2$ with $x^2+y^2$ we get $$(x^2+y^2)^2=2xy^2+yx^2+y^3-2y^3 \\(x^2+y^2)^2=y(3x^2-y^2)$$
However I am unsure of where I have made a mistake as the true answer is$$(x^2+y^2)^2=4x^2y-(x^2+y^2)y$$ working backwards I've so far gotten to the point of asking how I would rearrange$$r=\sin (3\theta) \Rightarrow r=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ so that $$\sin(3\theta)=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ which I am afraid I'll have to ask help for the next steps. Thanks.
|
or you use $$\sin(3x)=3\sin(x)-4\sin(x)^3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=\frac{12+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-4}{5}$$
So far,
$$\sqrt{41}=6+\frac{1}{2+\frac{\sqrt{41}-4}{5}}=6+\frac{1}{2+\frac{1}{\frac{5}{\sqrt{41}-4}}}$$
But, $$\frac{5}{\sqrt{41}-4}=\frac{5(\sqrt{41}+4)}{41-16}=\frac{\sqrt{41}+4}{5}=\frac{6+\sqrt{41}-2}{5}=\color{red}{1}+\frac{\sqrt{41}-1}{5}$$
It suppose to be $2$ and not $1$.
Where is the mistake? (I triple-checked and it seems fine to me)
|
[This old question popped up in the feed today (years later), and I can see that there's an accepted answer. Case closed. Though, reading the question and its answers, I can't help but feeling that the problem could (should?) have been tackled differently. I'm leaving this answer in the hope it might inspire and help some future readers looking for solutions to this type of problems.]
Given that the problem statement gives the result upfront and only asks for a proof, I would have just gone the other way around. Starting from the continued fraction $X = [6;\overline {2,2,12}]$ (we remark here that it's a positive number), one can see without too much difficulty that
$$ X+6=12+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{X+6}}} $$
And after a few mildly tedious but very easy steps, we derive
$$
\iff X+6=12+\cfrac{1}{2+\cfrac{X+6}{2X+13}} \\
\iff X+6=12+\cfrac{2X+13}{5X+32} \\
\iff X+6=\cfrac{62X+397}{5X+32} \\
\iff 5X^2+62X+192=62X+397 \\
\iff 5X^2-205=0 \\
\iff X^2-41=0 \\
$$
Since we know $X$ is positive, the result $X=\sqrt{41}$ follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution
$$y=\frac{1-x}{1+x} \Longleftrightarrow x=\frac{1-y}{1+y}$$
we have
$$ I(k) = \int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}$$
Now we can decompose
$$\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}= \frac{1}{2k(1+k)}\frac{1}{y+a}- \frac{1}{2k(1-k)}\frac{1}{y+a^{-1}}$$
with $a=\frac{1-k}{1+k}$ runs from $0$ to $1$.
Therefore we can write
$$I(k) = \frac{1}{2k(1+k)}\color{blue}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a}} - \frac{1}{2k(1-k)}\color{red}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a^{-1}}}$$
So we have only to evaluate $\color{blue}{I_1}$ and $\color{red}{I_2}$.
Now we consider the following integral
$$ J(\sigma) = \int_0^1 \mathrm{d}x \frac{\log x}{\sqrt{x}}\frac{1}{x-\sigma^2}$$
so that $\color{blue}{I_1}=J(i\sqrt{a})$ and $\color{red}{I_2}=J(i/\sqrt{a})$.
By considering the map $x\mapsto x^2$ we can write
$$ J(\sigma)=4\int_0^1\mathrm{d}x \frac{\log x}{x^2-\sigma^2}= \frac{2}{\sigma}\left[\color{green}{\int_0^1 \frac{\log x}{x-\sigma}}-\color{green}{\int_0^1 \frac{\log x}{x+\sigma}}\right]$$
The problem then reduces to evaluate the $\color{green}{\text{green}}$ integrals. At this point I'm stuck. I think that it needs to be solved by using polylogarithms, but I don't really know how to use these functions.
Mathematica 11.0 says
$$J(\sigma)=4 \left(\frac{\Phi \left(\frac{1}{\sigma ^2},2,\frac{3}{2}\right)}{4 \sigma ^4}+\frac{1}{\sigma ^2}\right)$$
where $\Phi$ is the Lerch transcendent. I don't know if this result is true (numerical integration is somewhat problematic). However, if it is true, I don't know what to do next.
Any hint on how to proceed with the evaluation?
Thanks in advance!
|
I have completed the computation of ASM.
For $0<k<1$,
$\displaystyle I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\,dx$
Let $\rho=\dfrac{1-k}{1+k}$
Observe that, since $0<k<1$ then $0<\rho<1$.
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}I(k)&=\int_0^1 \frac{(y-1)\ln y}{\sqrt{y}\big(\left(k-1\right)y-k-1\big)\big(\left(k+1\right)y-k+1\big)}\,dy\\
&=\frac{1}{(k^2-1)}\int_0^1 \frac{(y-1)\ln y}{\sqrt{y}\left(y+\frac{1}{\rho}\right)\left(y+\rho\right)}\,dy
\end{align}$
Perform the change of variable $\displaystyle t=\sqrt{y}$,
$\begin{align}I(k)&=\frac{4}{(k^2-1)}\int_0^1 \frac{(t^2-1)\ln t}{\left(t^2+\frac{1}{\rho}\right)\left(t^2+\rho\right)}\,dt\\
&=\frac{4\rho}{(\rho-1)(k^2-1)}\left(\int_0^1 \frac{\ln t}{t^2+\rho}\,dt-\int_0^1 \frac{\ln t}{\rho t^2+1}\,dt\right)\\
\end{align}$
In the latter integral perform the change of variable $\displaystyle u=\frac{1}{t}$,
$\begin{align}I(k)&=\frac{4\rho}{(\rho-1)(k^2-1)}\left(\int_0^1 \frac{\ln t}{t^2+\rho}\,dt+\int_1^{\infty} \frac{\ln u}{u^2+\rho}\,du\right)\\
&=\frac{4\rho}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln t}{t^2+\rho}\,dt\\
\end{align}$
Since $\displaystyle \rho>0$,
$\begin{align}I(k)&=\frac{4}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln t}{\left(\frac{t}{\sqrt{\rho}}\right)^2+1}\,dt\end{align}$
Perform the change of variable $v=\dfrac{t}{\sqrt{\rho}}$,
$\begin{align}I(k)&=\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\sqrt{\rho}\right)}{v^2+1}\,dv\\
&=\frac{2\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{1}{v^2+1}\,dv+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\
&=\frac{2\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\Big[\arctan v\Big]_0^{\infty}+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\
&=\frac{\pi\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\
\end{align}$
But,
Consider the integral,
$\displaystyle J=\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv$
Perform the change of variable $w=\dfrac{1}{v}$,
$\displaystyle J=-J$.
Therefore $\displaystyle J=0$,
and,
$\begin{align}J(k)&=\frac{\pi\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\\
&=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{(\frac{1-k}{1+k}-1)(k^2-1)}\\
&=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{(\frac{-2k}{1+k})(k^2-1)}\\
&=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{2k(1-k)}\\
&=\boxed{\frac{\pi\ln\left(\frac{1-k}{1+k}\right)}{2k\sqrt{1-k^2}}}
\end{align}$
|
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|
Constructing a Hypergeometric Function I am asked to find values for $a,b$ and $c$ such that $$ \frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = 2\alpha x\ _2F_1(a,b;c;x^2)$$
I have attempted the following:
$$\frac{1}{2} ((1+x)^{2\alpha}-(1-x)^{2\alpha}) = \frac{1}{2}\sum^{\infty}_{k=0}\binom{2\alpha}{k}x^k-\sum^{\infty}_{k=0}\binom{2\alpha}{k}(-x)^k \\ = \frac{1}{2} \sum^{\infty}_{k=0}\binom{2\alpha}{k}(x^k-(-x)^k). $$
My lecturer then advised that I should try to rewrite the binomial as Pochammer Symbols which led me to:
$$\frac{1}{2}\sum^{\infty}_{k=0}\frac{(2\alpha -k +1)_k}{k!}(x^k-(-x)^k)$$
We can also see here that for values $k= 2n$ we will recieve a term equal to 0, hence we can rewrite our equation as:
$$\frac{1}{2}\sum^{\infty}_{n=0}\frac{(2\alpha -(2n+1) +1)_{(2n+1)}}{(2n+1)!}(x^{(2n+1)}-(-x)^{(2n+1)})$$
We simpify to:
$$x\frac{1}{2}\sum^{\infty}_{n=0}\frac{(2\alpha-2n)_{(2n+1)}}{(2n+1)!}(x^{2n}-(-x)^{2n})$$
Now at this point I'm not really sure where to go... I feel like I'm really close and not seeing something or maybe really far away and not aware!
Either way any help is greatly appreciated!
|
In the expansion
\begin{align}
f(x)&=\frac{1}{2}\sum_{k=0}^\infty \binom{2\alpha}{k}\left( x^k-(-x)^k \right)\\
&=x\sum_{n=0}^\infty \binom{2\alpha}{2n+1}x^{2n}\\
&=x\sum_{n=0}^\infty c_nX^{n}
\end{align}
with $X=x^2$, the ratio of two successive terms of the series is
\begin{align}
\frac{c_{n+1}}{c_n}\frac{X^{n+1}}{X^n}&=\frac{\binom{2\alpha}{2n+3}}{\binom{2\alpha}{2n+1}}X\\
&=\frac{(n-\alpha+1/2)(n-\alpha+1)}{(n+3/2)}\frac{X}{n+1}
\end{align}
then, with $a_1=\frac{1}{2}-\alpha$, $ a_2=1-\alpha$ and $ b_1=\frac{3}{2}$ and with $c_0=\binom{2\alpha}{1}=2\alpha$, it gives
\begin{align}
f(x)&=2\alpha x\sum_{n=0}^\infty \frac{(a_1)_n(a_2)_n}{(b_1)_n}\frac{X^n}{n!} \\
&=2\alpha x \ _2F_1\left(\frac{1}{2}-\alpha,1-\alpha;\frac{3}{2};x^2 \right)
\end{align}
|
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|
Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the taylor polynomial of $\log(1+x^2)$, which is $x^2-\frac{1}{2}x^4+$... to see if I could bound the absolute value of the fraction above with something like $|\frac{x^2-x^2}{x^2\sin^2x}|$ and apply the comparison lemma, but it looks like that does not work too...
|
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{\dfrac 1{x^2}-\dfrac{\log(1+x^2)}{x^4}}{\dfrac{\sin^2x}{x^2}}$$
Taking $z=x^2$, note that the numerator becomes $$\dfrac 1z-\dfrac{\log(1+z)}{z^2}=\dfrac{z-\log(1+z)}{z^2}\to\dfrac{1-\frac 1{1+z}}{2z}\to\dfrac{1}{2(z+1)^2}\to\frac 12$$
as $z=x^2\to 0$ as $x\to 0$ where the final arrowed steps use L'Hopital
The denominator goes to $1$ since $\dfrac{\sin x}x\to 1$ as $x\to 0$
Thus, the limit is $1/2$ by a final application of L'Hopital's rule.
|
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|
prove the following algebraically $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$ I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?
$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$$
using $\left( \begin{array}{c} n \\ 2\ \end{array} \right) = \dfrac{n(n-1)}{2}$?
$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = \frac{2n(2n-1)}{2} = 2n(n-1) = 2 \frac{n(n-1)}{2} + n^2 = 2\left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2 $$
Is this proof correct? if not how do I fix it?
|
It is not correct, and it should go like this:
$$\binom{2n}{2}=\frac{2n(2n-1)}{2}=n(2n-1)=2n^2-n=n(n-1)+n^2=2\frac{n(n-1)}{2}+n^2=2\binom{n}{2}+n^2$$
It looks that your work starts well and ends well but there is a bit of a muddle in the middle.
|
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|
Positive Integer Divison Proof Prove that $6|n(n + 1)(n + 2)$ for any integer $n ≥ 1$
I have attempted to do this but never seem to prove it.
|
If $n$ is even: $n = 2k \implies n(n+1)(n+2) = 2k(2k+1)(2k+2) = 2k(4k^2+6k+2)= 8k^3+12k^2+4k = 6k^3+12k^2 + 6k+2k^3-2k$. Observe that if $3 \nmid k\implies k^2-k = 0\pmod 3$ by Fermat Little's theorem. Thus $2k^3 - 2k = 0\pmod 6$. Thus $6 \mid n(n+1)(n+2)$. If $n$ is odd, then $n = 2k+1\implies n(n+1)(n+2) = (2k+1)(2k+2)(2k+3)= (m-1)m(m+1)$ with $2 \mid m = 2k+2$. So if $3 \nmid m\implies m^2-m = 0\pmod 3$ again by Fermat Little's theorem. Thus $6 \mid n(n+1)(n+2), \forall n \ge 1$.
|
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"url": "https://math.stackexchange.com/questions/2641108",
"timestamp": "2023-03-29T00:00:00",
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|
Formula for a sequence defined on $K_1(x,y) := y+0$ if $x \geq y$ and $y-1$ otherwise Define $K_1:[0,1]^2\rightarrow\mathbb{R}$ as
$$K_1(x,y) := x - \frac{1}{2} - \begin{cases}
\ +(x - y - \frac{1}{2}) & \text{if $x \geq y$},\\
\ -(y - x - \frac{1}{2}) & \text{otherwise}
\end{cases}$$
then with
$$K_n(x,y) := \int_0^1K_1(x,u)K_{n-1}(u,y)\textrm{d}u$$
cf. equation $(35)$ part $3$
show that for $n\geq 1$
$$
n!K_n(x,y) = B_n(x) - \begin{cases} B_n(x-y) ~~\textrm{ if } x\geq y\\
~\\
(-1)^nB_n(y-x)~~\textrm{ otherwise }\end{cases}
$$
Here $B_n$ are the Bernoulli Polynomials.
Verify that
\begin{align}
+\sin(2\pi k x)=(2\pi k)^1\int_0^1K_1(x,u)\cos(2\pi k u)\textrm{d}u\\
-\cos(2\pi k x)=(2\pi k)^2\int_0^1K_2(x,u)\cos(2\pi k u)\textrm{d}u\\
-\sin(2\pi k x)=(2\pi k)^3\int_0^1K_3(x,u)\cos(2\pi k u)\textrm{d}u\\
+\cos(2\pi k x)=(2\pi k)^4\int_0^1K_4(x,u)\cos(2\pi k u)\textrm{d}u
\end{align}
for all $x\in [0,1]$ and $k\in \mathbb{Z}$, $k\neq 0$,
as well as
$$
B_{n+m}(x)=\frac{(n+m)!}{m!}\int_0^1K_n(x,u)B_m(u)\textrm{d}u\\
$$
|
Functions $K_n(x, y), n\in\mathbb N,$ are considered in the area
$$\mathbb S = \{(x, y) \in [0,1]^2\}.\tag1$$
Besides this, can be used step function
$$h(x) =
\begin{cases}
1, \text{ if }x \in (0, 1],\\
0, \text{ otherwize}
\end{cases}\tag2$$
for brief notation of 2D intervals method.
In particular, we can present the issue condition in the forms of
$$K_1(x, y) =
\begin{cases}
y\text{ if } x \ge y\\
y-1\text{ if } x < y
\end{cases}
= yh(x - y) + (y - 1)h(y - x). \tag3$$
In this way, taking in account identity
$$(-1)^nB(x) = B(1-x),\quad n\in\mathbb N\tag4$$
easy to see the replacement
$$B_n(x) -
\begin{cases}
B_n(x - y)\text{ if } x\ge y\\
(-1)^nB_n(y - x)\text{ if } x < y
\end{cases}=
\begin{cases}
B_n(x) - B_n(x - y)\text{ if } x\ge y\\
B_n(x) - B_n(x + 1 - y)\text{ if } x < y.
\end{cases}.
$$
So we can prove the identity
$$n!K_n(x, y) = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y).\quad k \ge 1\tag5$$
Let us prove it by induction.
At first, for $B_1(x) = x - \frac12$
$$B_1(x) - B_1(y) = x - y\tag6,$$
and one can write
$$B_1(x) - h(x - y)B_1(x - y) - h(y - x)B_1(x + 1 - y) = h(x - y) (B_1(x) - B_1(x - y) ) + h(y - x)(B_1(x) - B_1(x + 1 - y)) = yh(x - y) + (y - 1)h(y - x) = K_1(x)),$$
so identity $(5)$ is satisfied for $k = 1$.
Then, let identity $(5)$ is satisfied for the case $n-1,$
$$(n-1)!K_{n - 1}(x, y) = B_{n - 1}(x) - h(x - y)B_{n - 1}(x - y) - h(y - x)B_{n - 1}(x + 1 - y).$$
Taking in account $(1), (6)$ and using different splittings of the integrals in the cases $x \ge y$ and $x < y,$ one can get
\begin{aligned}
(n-1)!\int_0^1\,K_1(x, u)K_{n - 1}(u,y)\,du = \int_0^1(uh(x - u) + (u - 1)h(u - x))\\
\times(B_{n - 1}(u) - h(u - y)B_{n - 1}(u - y) - h(y - u)B_{n - 1}(u + 1 - y))du\\
\end{aligned}
\begin{aligned}
= \int_0^xuB_{n - 1}(u)\,du + \int_x^1(u - 1)B_{n - 1}(u)\,du\\
\end{aligned}
\begin{aligned}
- h(x - y)\left(\int_0^yuB_{n - 1}(1 + u - y)\,du + \int_y^xuB_{n - 1}(u - y)\,du\\
+ \int_x^1(u - 1)B_{n - 1}(u - y)\,du\right)\\
\end{aligned}
\begin{aligned}
- h(y - x)\left(\int_0^xuB_{n - 1}(1 + u - y)\,du + \int_x^y(u - 1)B_{n - 1}(1 + u - y)\,du\\
+ \int_y^1(u - 1)B_{n - 1}(u - y)\,du\right).
\end{aligned}
Standartization of $B_{n-1}$ arguments can be achieved by linear substitutions, and that allows to reconstruct the issue integral to the forms of
\begin{aligned}
(n-1)!\int_0^1\,K_1(x, u)K_{n - 1}(u,y)\,du
= \int_0^xB_{n - 1}(u)\,du + \int_0^1(u-1)B_{n - 1}(u)\,du \\
\end{aligned}
\begin{aligned}
- h(x - y)\left(\int_{1 - y}^1(u + y - 1)B_{n - 1}(u)\,du + \int_0^{x - y}(u + y)B_{n - 1}(u)\,du\\
+ \int_{x-y}^{1-y}(u + y - 1)B_{n - 1}(u)\,du\right)\\
\end{aligned}
\begin{aligned}
- h(y - x)\left(\int_{1-y}^{x + 1 - y}(u + y - 1)B_{n - 1}(u)\,du + \int_{x + 1 - y}^1(u + y - 2)B_{n - 1}(u)\,du\\
+ \int_0^{1-y}(u + y - 1)B_{n - 1}(u)\,du\right).
\end{aligned}
\begin{aligned}
& = \int_0^xB_{n - 1}(u)\,du + \int_0^1(u-1)B_{n - 1}(u)\,du - \int_0^1(u + y - 1)B_{n - 1}(u)\,du \\
& - h(x - y)\int_0^{x - y}B_{n - 1}(u)\,du + h(y - x)\int_{1 - y + x}^1B_{n - 1}(u)\,du
\end{aligned}
\begin{aligned}
& = \int_0^xB_{n - 1}(u)\,du - y\int_0^1B_{n - 1}(u)\,du \\
& - h(x - y)\int_0^{x - y}B_{n - 1}(u)\,du + h(y - x)\int_{1 - y + x}^1B_{n - 1}(u)\,du
\end{aligned}
And now it has become possible to use the identities
$$\int_a^x\,B_{n - 1}(t)\,dt = \frac1n(B_n(x) - B_n(a)),\tag7$$
$$B_n(1) - B_n(0) = 0,\quad n > 1,\tag8$$
with the result
\begin{aligned}
&n!\int_0^1\,K_1(x, u)K_{n - 1}(u,y)\,du = B_n(x) - B_n(0 ) - y(B_n(1) - B_n(0))\\
& - h(x - y)(B_n(x - y) - B_n(0)) + h(y - x)(B_n(1) - B_n(x + 1 - y))\\
& = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y) - (y - h(y - x))(B_n(1) - B_n(0))\\
& = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y),
\end{aligned}
$$\boxed{n!K_n(x, y) = B_n(x) - h(x - y)B_n(x - y) - h(y - x)B_n(x + 1 - y).}$$
So identity $(5)$ is satisfied for arbitrary $n\in\mathbb N.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2641922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
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|
Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$: $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$
(3) Prove that this is true for n = k + 1:
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
$$ = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$= ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
|
Because $$2^{5n+1}+5^{n+2}=2\cdot32^n-2\cdot5^n+27\cdot5^n=2(32-5)(32^{n-1}+...+5^{n-1})+27\cdot5^n,$$
which is divided by $27.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
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|
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
|
While others have pointed towards factorizing the expression, I would like to say that it isn't always easy for one to notice that the expression can be factorized. Also, the series has some property - the coefficients are gradually increasing or decreasing.
Let's try to find the sum of the series because that would surely help in simplification.
$$Let, \; S = 1 + 2x + 3x^2 + 4x^3+5x^4+6x^5+5x^6+4x^7+3x^8+2x^9+x^{10} \,$$
$$ x . S = x + 2x^2 + 3x^3 + 4x^4 + 5x^5+6x^6+5x^7+4x^8+3x^9+2x^{10}+x^{11}$$
From here, we can get,
$$S (x-1) = -(1+x+x^2+x^3+x^4+x^5) + (x^6+x^7+x^8+x^9+x^{10}) + x^{11}$$
Now use the formula of a geometric progression in the first two series to get:
$$ S (x-1) = -\frac{(x^6-1)}{x-1} + \frac{x^6(x^5 - 1)}{x-1} + x^{11}$$
$$ S = \frac{(1-x^6) + x^6(x^5 - 1) + x^{11}(x-1)}{(x-1)^2}$$
$$ S = \frac{x^{12} - 2x^6 + 1}{(x-1)^2}$$
$$ S = \frac{(x^6 - 1)^2}{(x-1)^2}$$
Now it is indeed a perfect whole square. You can simply divide $x^6-1$ by $x-1$ to get your desired result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 4
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|
How can I solve this system of equations?? I have this system of equations and I need values for $c_1, c_2, c_3, x_1, x_2, x_3$.
\begin{align*}
c_1 + c_2 + c_3 &= 2\\
c_1x_1 + c_2x_2+ c_3x_3 &= 0\\
c_1x_1^2 + c_2x_2^2+c_3x_3^2 &= \frac{2}{3}\\
c_1x_1^3 + c_2x_2^3+c_3x_3^3 &= 0\\
c_1x_1^4 + c_2x_2^4 + c_3x_3^4 &= \frac{2}{5}\\
c_1x_1^5 + c_2x_2^5+c_3x_3^5 &= 0
\end{align*}
I have attempted using matrix but was unsuccessful. How can I obtain values for $x_i$ and $ c_i$?
|
Rewriting the system as
\begin{align}
\sum_{j=1}^3 c_j\,x_j^i&=v_i,\quad i=0,\dots,5
\tag{1}\label{1}
,
\end{align}
we can apply Prony's method
as follows.
Solve the linear system
\begin{align}
\left[\begin{matrix}
v_0 & v_1 & v_2 \\
v_1 & v_2 & v_3 \\
v_2 & v_3 & v_4
\end{matrix}\right]
\cdot
\left[\begin{matrix}
a_0 \\ a_1 \\ a_2
\end{matrix}\right]
&=
\left[\begin{matrix}
v_3 \\ v_4 \\ v_5
\end{matrix}\right]
\end{align}
for $a_0,a_1,a_2$.
The roots of polynomial
\begin{align}
x^3-a_2\,x^2-a_1\,x-a_0
\end{align}
would be the triple $x_1,x_2,x_3$.
Given that, the solution of another linear system
\begin{align}
\left[\begin{matrix}
1&1&1 \\
x_1&x_2&x_3 \\
x_1^2&x_2^2&x_3^2
\end{matrix}\right]
\cdot
\left[\begin{matrix}
c_1 \\ c_2 \\ c_3
\end{matrix}\right]
&=
\left[\begin{matrix}
v_0 \\ v_1 \\ v_2
\end{matrix}\right]
\end{align}
for $c_1,c_2,c_3$ completes the answer.
In numbers we have
\begin{align}
\left[\begin{matrix}
2 & 0 & \tfrac23 \\
0 & \tfrac23 & 0 \\
\tfrac23 & 0 & \tfrac25
\end{matrix}\right]
\cdot
\left[\begin{matrix}
a_0 \\ a_1 \\ a_2
\end{matrix}\right]
&=
\left[\begin{matrix}
0 \\ \tfrac25 \\ 0
\end{matrix}\right]
,\\
\end{align}
\begin{align}
a_0&=0,\quad a_1=\tfrac35.\quad a_2=0
,\\
x^3-\tfrac35\,x&=0
,\\
x_1&=0,\quad x_2=\tfrac{\sqrt{15}}5
,\quad x_3=-\tfrac{\sqrt{15}}5
.
\end{align}
The system for $c_j$:
\begin{align}
\left[\begin{matrix}
1&1&\phantom{-}1 \\
0& \tfrac{\sqrt{15}}5 & -\tfrac{\sqrt{15}}5 \\
0& \tfrac35 & \phantom{-}\tfrac35
\end{matrix}\right]
\cdot
\left[\begin{matrix}
c_1 \\ c_2 \\ c_3
\end{matrix}\right]
&=
\left[\begin{matrix}
2 \\ 0 \\ \tfrac23
\end{matrix}\right]
,\\
\end{align}
which results in
\begin{align}
c_1&=\tfrac89,\quad
c_2= c_3=\tfrac59
.
\end{align}
|
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"url": "https://math.stackexchange.com/questions/2644060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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|
How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$? How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$
|
By making use of the integral
$$\int_{0}^{1} \frac{(1-x)^2}{1-x^3} \, dx = \frac{1}{2} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right)$$
one can take the following path.
\begin{align}
S &= \sum_{k=0}^{\infty} \frac{1}{(3k+1)(3k+2)(3k+3)} \\
&= \sum_{k=0}^{\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+4)} = \frac{1}{2} \, \sum_{k=0}^{\infty} B(3, 3k+1),
\end{align}
where $B(n,m)$ is the Beta function, which leads to
\begin{align}
S &= \frac{1}{2} \, \sum_{k=0}^{\infty} \, \int_{0}^{1} t^{2} \, (1-t)^{3k} \, dt \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{t^{2} \, dt}{1- (1-t)^{3}} \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2} \, dx}{1- x^3} \hspace{15mm} x = 1 - t \\
&= \frac{1}{4} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right).
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/2644864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
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|
Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}{2}$, because $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ is decreasing and convergent to $e$, while $1+\frac{1}{1!}+\dots+\frac{1}{n!}$ is increasing and convergent to $e$, so $$\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}>1+\frac{1}{1!}+\dots+\frac{1}{n!}$$
which means that $\frac{1}{2}>x_n$. I also know that for $a \in [0,\frac{1}{2}), \left(1+\frac{1}{n}\right)^{n+a}$ is eventually increasing, but I don't know how to get a lower bound for $x_n$ which goes to $\frac{1}{2}$
|
If $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
then we have
$$\begin{align}x_n&=\frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\ln\left( 1+\frac{1}{n}\right)}-n\\&\sim \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{\left(\frac{1}{n}-\frac{1}{2n^2}\right)}-n
\\&= n\left( \frac{\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)}{1-\frac{1}{2n}}-1\right)\\&\sim
n\left( \ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)-1+\color{blue}{\frac{1}{2n}}\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)\right)&\to \color{red}{\frac12} \end{align}$$
Given that $$\ln\left(1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}\right)\to 1$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2646289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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|
Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem is can we prove the statement
"for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $"
only using induction ?(without using the fact that convergence of the above series)
any ideas, thanks!
|
This is unlikely. The bare induction step would be
$$S_n\le\frac{\pi^2}6\implies S_n+\frac1{n^2}\le\frac{\pi^2}6,$$ which obviously doesn't hold. There is not enough information in the inductive hypothesis.
Any information on the asymptotics of the series will be of the form $\dfrac{\pi^2}6-\epsilon(n)$, which contains "the answer" (i.e. a hint on convergence).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2650348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
Determinate all the positive integers $x$ such that $x^4+6x^3+11x^2+3x+11$ is a perfect square. Determinate all the positive integers $x$ such that $x^4+6x^3+11x^2+3x+11$ is a perfect square.
My try
With pure intuition I can say that there is no $ x $ that meets this condition, i tried by brute force some numbers and none worked, but I would like to know if what I'm guessing is right or not, and what would be a way to prove it, because i don't see the way to prove this.
|
Suppose that exist $n\in\Bbb N$ such that $$x^4+6x^3+11x^2+3x+11=n^2$$
See $\text{mod } 3$ the equation and you can prove that $n^2\not\equiv 2 \text{ mod } 3$ for all $n\in\Bbb N$. In fact, if $n=3k+a$ with $a=0,1,2$ then $n^2=9k^2+6ka+a^2\equiv a^2 \text{ mod } 3$ and this values are $0,1,1$ respectively.
Now,
If $x\equiv 0 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 0^4+6.0^3+11.0^2+3.0+11\equiv 11\equiv 2 \text{ mod } 3$
If $x\equiv 1 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 1^4+6.1^3+11.1^2+3.1+11\equiv 1+6+11+3+11\equiv 2 \text{ mod } 3$
If $x\equiv 2 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 2^4+6.2^3+11.2^2+3.2+11\equiv 16+0+44+0+11\equiv 2 \text{ mod } 3$.
This means that no have solutions
|
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|
Determining $A+B$, given $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A - \cos B = \sqrt{\frac12}$. Different approaches give different answers. The question:
Determine $A + B$ if $A$ and $B$ are acute angles such that:
$$\sin A + \sin B = \sqrt{\frac{3}{2}}$$
$$\cos A - \cos B = \sqrt{\frac{1}{2}} $$
Here are the two solutions that I found:
I think the problem with the second solution may have to do with the assumption that:
$$\cos\left(A + \frac{\pi}{6}\right) = \cos\left(B - \frac{\pi}{6}\right)$$
is equivalent to
$$A + \frac{\pi}{6} = -B + \frac{\pi}{6}$$
But can't you say that $\cos N = \cos M$ is equivalent to $\pm N = \pm M$ for all values of $N$ and $M$ (because cosine is an even function)?
|
$$\sqrt3\cos A-\sin A=\sqrt3\cos B+\sin B$$
$$\implies\cos(A+30^\circ)=\cos(B-30^\circ)$$
$$A+30^\circ=360^\circ n\pm(B-30^\circ) $$
$$-\implies A+B=360^\circ n$$ which is impossible as $0<A+B<180^\circ$
$+\implies A-B=360^\circ n-60^\circ$
As $-90<A-B<90,n=0$
Now use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html in anyone of the given relation
|
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"timestamp": "2023-03-29T00:00:00",
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|
Radical with pattern Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.
Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.
Without using a calculator, evaluate $\sqrt{a − b}$.
|
$a= \overbrace{11\ldots11}^{2018} = \dfrac{10^{2018}-1}{9}$
$b= \overbrace{22\ldots22}^{1009} = 2\cdot\dfrac{10^{1009}-1}{9}$
$\Rightarrow a-b = \dfrac{10^{2018}-1}{9}-2\cdot\dfrac{10^{1009}-1}{9} = \dfrac{10^{2018}-2\cdot10^{1009}+1}{9} = \left(\dfrac{10^{1009}-1}{3} \right)^2$
$\Rightarrow \sqrt{a-b} = \dfrac{10^{1009}-1}{3} = \overbrace{33\ldots33}^{1009}$
|
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|
Probability of 3 of a kind - Probability of Pair Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?
Here is my thinking. There are $6$ possible values for the pair. (e.g $1$,$1$,$2$,$2$...$6$,$6$). The other four dice can be any number other than the number of the pair. Numerator = $6\cdot5\cdot5\cdot4\cdot4$. Denominator = $5^6$. So I thought that the answer was $\frac{2400}{46656}$. Can this be verified? Help is greatly appreciated.
|
There are a total of $6^5=7776$ possible sets of dice rolls. To get a pair without a three-of-a-kind, we can either have one pair and the other three dice all showing different numbers, or we have two pairs and the die showing something different.
In the first case, there are $6$ ways to pick which number makes a pair and $\binom{5}{2}=10$ ways to pick which $2$ of the $5$ dice show that number. Out of the other three dice, there are $5$ ways to pick a value for the first die so that that die doesn't match the pair, $4$ ways to pick a value for the second one so it doesn't match that die or the pair, and $3$ ways to pick a value for the last die so that it doesn't match any of the others. So there are $$6\cdot 10\cdot 5 \cdot 4 \cdot 3 = 6^2 \cdot 100$$ ways to roll this case.
In the second case, to form two pairs and one die not part of those pairs, there are $\binom{6}{2}=15$ ways to pick which two numbers make the pairs, then $4$ ways to pick a value for the last die so that it doesn't match either of those pairs. There are $$\frac{5!}{2!\cdot 2!\cdot 1!}=30$$ ways order the five dice (equal to the number of ways to order XXYYZ), so that makes a total of $$15\cdot 4 \cdot 30 = 6^2\cdot 50$$ ways to roll this case.
This makes a total of $$6^2 \cdot 100 + 6^2 \cdot 50 = 6^2 \cdot 150 = 6^3 \cdot 25$$ ways to roll a pair without rolling a three-of-a-kind. So, the probability is $$\frac{\text{successful outcomes}}{\text{total outcomes}}=\frac{6^3 \cdot 25}{6^5}=\frac{25}{6^2}=\boxed{\frac{25}{36}}.$$
|
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|
"Chain Rule" for Finite Difference Operator I am trying to prove using difference calculus the following formula;
$$\Delta(a+bx)^{(n)}=bn(a+bx)^{(n-1)}$$
which is akin to the chain rule for continuous Calculus. Here we have that
$$x^{(n)}=x(x-1)...(x-n+1)$$
and
$$\Delta f(x)=f(x+1)-f(x)$$
It seemed daunting at first so I tried small values; $a=3, b=2$. Looking at the relationship with actual values would help generate an idea how the general proof would look. So I started;
$$\begin{eqnarray*}\Delta(3+2x)^{(n)}&=&[3+2(x+1)]^{(n)}-(3+2x)^{(n)}\\&=&(5+2x)^{(n)}-(3+2x)^{(n)}\\&=&(2x+5)(2x+4)(2x+3)^{(n-2)}-(2x+3)^{(n-2)}(2x+5-n)(2x+4-n)\\&=&(2x+3)^{(n-2)}[(2x+5)(2x+4)-(2x+5-n)(2x+4-n)]\\&=&(2x+3)^{(n-2)}[n(2x+5)+n(2x+4)-n^2]\\&=&n(2x+3)^{(n-2)}[(2x+5)+(2x+4)-n]\\&=&n(2x+3)^{(n-2)}(4x+9-n)
\end{eqnarray*}$$
This last couple of steps is escaping me. I can't really factor a 2 without ending up with a $1/2$n. I know that if I can manipulate the last term into something close to $2(2x+3-(n-2))$ then this should give me what I need, but this expanded is $$(4x+6-2n+4)=4x+10-2n=4x+9-n+(1-n)$$ and so I'm missing a $(1-n)$.
DId I screw up some algebra?
EDIT; @ancientmathematician, you were right, and my definition of $p(x)^{(n)}$ was incorrect, and is defined as
$$p(x)^{(n)}=p(x)p(x-1)...p(x-n+1)$$
which I'm sure will change things around. I will go through the algebra shortly and If I solve it i will answer my own question, or someone else can write it up and I'll upvote it.
|
As you mentioned, the algebra works out when using the definition you gave in the edit.
Let $h(x) = f(x)^{(n)} = f(x)f(x-1)\cdots f(x-n+1)$. Then
\begin{align}
\Delta h(x) &= f(x+1) f(x) f(x-1) \cdots f(x-n+2) - f(x)f(x-1)\cdots f(x-n+1) \\
&= f(x) f(x-1) \cdots f(x-n+2)(f(x+1) - f(x-n+1)) \\
&= (f(x+1) - f(x-n+1)) f(x)^{(n-1)}.
\end{align}
In the special case that $f(x) = a + bx$, we have
\begin{align}
f(x+1) - f(x-n+1) &= a + bx + b - (a + bx - (n-1) b) \\
&= nb.
\end{align}
So we find that
$$
\Delta h(x) = nb (a + bx)^{(n-1)}.
$$
|
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|
Inequality with hypergeometric function I'm trying to prove the following inequality:
$$_1F_2\left(\frac{a}{2},\frac{3}{2},\frac{a}{2}+1;-\frac{\pi^2}{4}\right)\ge \frac{a+6}{(a+2)(a+3)}$$
where $a$ is a positive real number.
I wrote Euler's Integral Transform for Hypergeometric Functions, but it gaves me only more complicated formula to prove.
|
Start with
$$f(a)=
{_1F_2}\left( \frac a 2; \frac 3 2, \frac a 2 +1; -\frac {\pi^2} 4 \right) =
\frac a \pi \int_0^1 t^{a-2} \sin \pi t \,dt,$$
which is a special case of
$${_1F_2}(a; b_1, b_2; z) =
\frac {\Gamma(b_2)} {\Gamma(a) \Gamma(b_2-a)}
\int_0^1 t^{a-1} (1-t)^{b_2-a-1} {_0F_1}(; b_1; z\,t) dt.$$
Expanding $\sin \pi t$ around $t=1$, the integral of $t^{a-2} (1-t)^k$ will be of order $a^{-k-1}$ for large $a$, yielding an expansion of $f(a)$ for large $a$. We have
$$f(a) = \frac 1 a + \frac 1 {a^2} - \frac {\pi^2-1} {a^3} +
O\left( \frac 1 {a^4} \right),$$
while
$$g(a) = \frac {a+6} {(a+2)(a+3)} =
\frac 1 a + \frac 1 {a^2} - \frac {11} {a^3} +
O\left( \frac 1 {a^4} \right).$$
Thus the inequality holds for large $a$.
Writing out the expansion was not really necessary for the proof, but it shows that if we want to construct a lower bound for $f(a)$, it will need to have the correct asymptotic behavior for large $a$. Writing
$$\left| \sin \pi t - \pi(1-t) \right| \leq \frac {\pi^3} 6 (1-t)^3,$$
we obtain for $a>1$
$$\left| f(a) - \frac 1 {a-1} \right| \leq \frac {\pi^2} {(a^2-1)(a+2)}.$$
And for $a > 3(\pi^2-4)/(12-\pi^2)$, $g(a)$ lies below the lower bound.
It remains to check the inequality for smaller $a$. This can be done by expanding $f(a)$ into a continued fraction:
$$f(a) = 1 - \frac {\pi^2 a} {6(a+2)}
\left( 1 + \mathop K_{k=2}^\infty \frac {A_k} {1-A_k} \right)^{-1},\\
A_k = \frac {\pi^2 (2k+a-2)} {2 k (2k+1) (2k+a)}.$$
The convergents will not be sufficient to separate $f(a)$ from $g(a)$ for large $a$, but the fifth convergent is larger than $g(a)$ for $a<23$. Since the odd convergents are smaller than $f(a)$ (not the even ones, because $-\pi^2 a/(6(a+2))$ is negative), that shows that the inequality holds for $a<23$, concluding the proof.
|
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|
Prove $10^{n+1} - 9^{n+1} = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$ In this video, James Grime shows that the number of $\mathbb{N}$ less than $10^{n+1}$ that have at least one $3$ among their digits is given by this recurrence relation:
$$T_{n+1} = 9T_{n} + 10^{n}\; where\; T_0=1$$
But later in the video, he says it can also be written as:
$$T_n = 10^{n+1} - 9^{n+1}$$
So, I set out to prove they are the same algebraically:
Solving the recurrence relation, I got:
$$T_n = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$$
So, can somebody show me a proof to show that:
$$10^{n+1} - 9^{n+1} = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$$
|
With $a^0=1$, $b^0=1$ for $a, b\ne 0$:
$(a-b)(a^{n}b^{0}+a^{n-1}b^{1}+a^{n-2}b^2+...+a^{1}b^{n-1}+a^{0}b^{n})$
$=a^{n+1}b^{0}+a^{n}b^{1}+a^{n-1}b^2+...+a^{2}b^{n-1}+a^{1}b^{n}-(a^{n}b^{1}+a^{n-1}b^{2}+...+a^{1}b^{n}+a^{0}b^{n+1})$
$=a^{n+1}b^{0}+a^{n}b^{1}+a^{n-1}b^2+...+a^{2}b^{n-1}+a^{1}b^{n}-a^{n}b^{1}-a^{n-1}b^{2}-...-a^{1}b^{n}-a^{0}b^{n+1}$
$=a^{n+1}b^{0}+a^{n}b^{1}-a^{n}b^{1}+a^{n-1}b^{2}-a^{n-1}b^{2}+...+a^{1}b^{n}-a^{1}b^{n}-a^{0}b^{n+1}$
$=a^{n+1}-b^{n+1}$
This will apply for your problem, because $a-b=10-9=1.$
|
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|
Integral calculating issue If
$$ \int \frac{1}{x+1}\,\mathrm{d}x = \ln|x+1|+C, $$
then why
$$ \int \frac{1}{x+\frac{1}{3}}\,\mathrm{d}x \not= \ln\left|x+\frac{1}{3}\right|+C? $$
|
$$\int\frac{f'}{f}=\ln(f)$$
So both of your answers are correct. But you can do the second this way too:
$$\frac{1}{x+1/3}=\frac{1}{x+1/3}\frac{3}{3}=\frac{3}{3x+1}$$
And it's integral can be written as
$$\ln(3x+1)$$
But the $2$ results are the 'same' (Their difference is only a constant):
$$\ln(3x+1)-\ln(x+1/3)=\ln\left(\frac{3x+1}{x+1/3}\right)=\ln\left(\frac{3x+1}{x+1/3}\frac{3}{3}\right)=\ln\left(3\frac{3x+1}{3x+1}\right)=\ln(3)$$
|
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|
Eigenvalue and Eigenvector of $\small\pmatrix{0 & 0 \\ 0 & -7}$ I need help working out the eigenvectors for this matrix.
$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $
The original matrix is $ \begin {pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix} $ , eigenvalues are 5,-2,
but I am not sure how to about the eigenvectors,
as for 5
$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $ $ \begin{pmatrix} x \\ y \end{pmatrix}$ = $ \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
from the first equation, $x$ and $y$ are both zero, but from the second equation $y = 0$, so what is the eigenvector?
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No, from the first equation, $x$ and $y$ are free. From the second equation, $y=0$. So your eigenvector is
$$
\begin{bmatrix}1\\0 \end{bmatrix}
$$
as you can check, the equation is satisfied.
|
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|
Geometric images of complex numbers $z$ such that triangle with vertices $z, z^2,z^3$ is a right angled triangle.
Find the geometric images of complex numbers $z$ such that triangle with vertices $z, z^2,z^3$ is a right angled triangle.
My try:
Let the complex number $z=x+yi$
Hence we need to find the locus of the point $(x,y)$ in the argand plane such that the points $(x,y)$, $(x^2-y^2, 2xy)$, $(x^3-3y^2x, 3x^2y-y^3)$
Now here I could apply the formula that the product of the gradient of two sides of triangle formed by these points is $-1$ but the method is a bit cumbersome. Is there any other geometrical interpretation of it or any algebraic simplification further?
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Let the squared-sides of the triangle be
$$\begin{align}
a^2 &\;:=\; |z^1-z^2|^2 \;=\; |z|^2 |1-z|^2 \\
b^2 &\;:=\; |z^2-z^3|^2 \;=\; |z|^4 |1-z|^2 \\
c^2 &\;:=\; |z^3-z^1|^2 \;=\; |z|^2 |1-z|^2 |1+z|^2
\end{align}$$
Trivially, we have a Pythagorean triple if $z=0$ or $z=1$. Otherwise, we can divide each of $a^2$, $b^2$, $c^2$ by $|z|^2 |1-z|^2$ to get, for $z = r(\cos\theta + i \sin\theta)$ with $r\neq0$ and $\theta\neq 0$,
$$a_\star^2 = 1 \qquad b_\star^2 = |z|^2 = r^2 \qquad c_\star^2 = |1+z|^2 =1+r^2+2r\cos\theta$$
so that $(a, b, c)$ is a (non-trivial) Pythagorean triple if and only if $(a_\star, b_\star, c_\star)$ is. Now, let's consider cases:
$$c_\star^2 = a_\star^2 + b_\star^2 \implies 1+r^2+2r\cos\theta = 1 + r^2 \implies \cos\theta = 0 \implies z = \pm p i \quad(p>0)$$
The diagram shows $z_1 = pi$ and $z_2 = -q i$. One may recognize that this is effectively the geometric mean configuration, which makes sense, since $|z_i^2|^2 = |z_i||z_i^3|$.
$$\begin{align}
a_\star^2 &= b_\star^2 + c_\star^2 \implies 1=r^2+1+r^2+2r\cos\theta \implies 2r\left(r+\phantom{r}\cos\theta\right) = 0 \implies r = -\cos\theta \\
b_\star^2 &= a_\star^2 + c_\star^2 \implies r^2 = 1 + 1 + r^2+2 r \cos\theta \implies 2\;\left(1+r\cos\theta\right) = 0 \implies r = -\sec\theta
\end{align}$$
Since $r$ must be non-negative, we restrict $z$ to Quadrants II and III. The diagram below shows typical solutions $z_3$ and $z_4$.
|
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If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$
Here's how I tried solving the problem:
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
\begin{align}
&= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\
&= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\
&= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C}
\end{align}
I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.
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You solved it all right , just took cosA.cos C in denominator instead of sinA.SinC
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
$= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A)$
$= \sin(B+A) \sin(B-A)[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}] - [\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}]\sin(C+A) \sin(C-A)$
$= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\sinA \sinc C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\sinA \sin C}$
$= \frac {\sin(B-A) \sin (A-C)} {\sin A} [\frac {\sin(A+B)} {\sin C} - \frac {\sin(A+C)} {\sin B}]$
$= \frac {\sin(B-A) \sin (A-C)} {\sin A} [\frac {\sin C} {\sin C} - \frac {\sin B} {\sin B}]$
$= \frac {\sin(B-A) \sin (A-C) \sin (A)} [\{0}]$
|
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|
Intersection of two paraboloids Consider two paraboloids. The first one is given by $x^2 + y^2 = z+5$. So, it intersects the x-y plane in the circle $x^2+y^2=5$. The second paraboloid is exactly the same as the first one, only shifted in the x-y plane. It's equation becomes $(x-1)^2+(y-1)^2=z+5$. From the figure below, it seems clear that the two should intersect in a parabola.
However, when we actually try and solve the two equations simultaneously, we get from the second equation
$$x^2 + y^2 -2x - 2y + 2 = z+5.$$
And substituting $x^2+y^2=z+5$ into this we get
$$2x+2y=2.$$
Now this is a linear function. However, the picture seems to suggest that it should be a parabola which is non-linear. What am I missing here?
|
Alternatively, exploiting symmetry profitably, you can simply express the equations as
$$
\eqalign{
& x^{\,2} + y^{\,2} = \left( {x - 1} \right)^{\,2} + \left( {y - 1} \right)^{\,2} = z + 5 \cr
& \left( {x - 1/2 + 1/2} \right)^{\,2} + \left( {y - 1/2 + 1/2} \right)^{\,2} = \left( {x - 1/2 - 1/2} \right)^{\,2} + \left( {y - 1/2 - 1/2} \right)^{\,2} = z + 5 \cr
& \left\{ \matrix{
\left( {x - 1/2} \right)^{\,2} + 1/4 + \left( {x - 1/2} \right) + \left( {y - 1/2} \right)^{\,2} + 1/4 + \left( {y - 1/2} \right) = z + 5 \hfill \cr
\left( {x - 1/2} \right)^{\,2} + 1/4 - \left( {x - 1/2} \right) + \left( {y - 1/2} \right)^{\,2} + 1/4 - \left( {y - 1/2} \right) = z + 5 \hfill \cr} \right. \cr
& \left\{ \matrix{
\left( {x - 1/2} \right)^{\,2} + \left( {y - 1/2} \right)^{\,2} = z + 9/2 \hfill \cr
\left( {x - 1/2} \right) + \left( {y - 1/2} \right) = 0 \hfill \cr} \right. \cr}
$$
That is, a paraboloid with vertex in $(1/2,\, 1/2,\, -9/2)$ cut
by the diametral plane through the vertex.
|
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|
Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$
Find the Cartesian equation of the locus described by $|z+2-7i| =
2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.
This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If anyone could point me in the right direction I'd very much appreciate that.
Here was my working:
Let $z = x+iy$
$|z|$ = $\sqrt{x^2+y^2}$
$$|z+2-7i| = 2|z-10+2i| \\
\implies |(x+iy)+2-7i| = 2|(x+iy)-10+2i| \\
\implies |(x+2)(y-7y)i| = 2|(x-10)+(y+2)i| \\
\implies \sqrt{(x+2)^2+(y-7)^2i^2} = 2\sqrt{(x-10)^2+(y+2)^2i^2} \\
\implies (x+2)^2-((y-7^2) = 2((x-10)^2-(y+2)^2)$$
$$(x^2+4x+4)-(y^2-14y+49) = 2((x^2-20x+100)-(y^2+2y+4)) \\
\implies x^2+4x+4-y^2+14y-49 = 2x^2-40x+200-2y^2-4y-8 \\
\implies -237 = x^2-44x-y^2-18y\\
\implies x^2-44x-y^2-18 = -237$$
|
HINT
Note that in general the equation
$$|z+a+bi|=k|z+c+di|\iff\frac{|z+a+bi|}{|z+c+di|}=\frac{d_1}{d_2}=k$$
describes a circle known as Circle of Apollonius.
Thus as an alternative you could find C and D by the given condition and then the center $O$ and radius $R$ of the circle
$$O=\frac{C+D}{2} \quad R=OD=OC$$
|
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|
Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2} = 3\cdot a+\frac{b}{2}$
Is this correct? How to prove it using resiudes?
Thank You.
|
If $\gcd(n,6)=1$, then $n$ is odd and $n^2 \equiv 1 \bmod 8$.
If $\gcd(n,6)=1$, then $\gcd(n,3)=1$ and $n^2 \equiv 1 \bmod 3$.
Thus $24 = lcm(8,3)$ divides $n^2-1$.
|
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|
Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
|
Hint: Alternately, by AM-GM, note $x^4+\frac14x^2\geqslant x^3$ and $\frac12x^2+\frac12\geqslant x$, so $x^4-x^3+x^2-x+1 \geqslant \frac14x^2+\frac12>\frac12$ as equality is not possible.
|
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|
Show that $f(x)=x^{5/3}-kx^{4/3}+k^2x$ is increasing for $k\neq0$ So to show the function is increasing/decreasing we differentiate and show it is more than zero/less than zero:
We have
$$f(x)=x^{5/3}-kx^{4/3}+k^2x$$
Hence,
$$f'(x)=\frac{5}{3}x^{2/3}-\frac{4k}{3}x^{1/3}+k^2$$
But how do I show
$$\frac{5}{3}x^{\frac{2}{3}}-\frac{4k}{3}x^{\frac{1}{3}}+k^2>0$$
|
Notice that you now have a new second order polynomial equation with $k$ as a variable instead of x.
You can compute the discriminant $\Delta$:
$$\Delta = (\frac{4x^{\frac{1}{3}}}{3})^2 - 4\frac{5x^\frac{2}{3}}{3} $$
$$\Delta = \frac{16x^{\frac{2}{3}}}{9} - \frac{60x^\frac{2}{3}}{9} = -\frac{44x^\frac{2}{3}}{9} < 0$$
Therefore $f'$ has no root and has the same sign as $k^2$
|
{
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"url": "https://math.stackexchange.com/questions/2671083",
"timestamp": "2023-03-29T00:00:00",
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|
Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
|
Note that:
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $
For proof, see this.
Therefore,
$\sin 4\theta = \dbinom{4}{1}\cos^3\theta \sin \theta- \dbinom{4}{3}\cos\theta \sin^3\theta = 4\cos^3 \theta \sin\theta - 4 \cos\theta \sin^3\theta$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $x=\frac{-1}{2} , y=\frac{3}{4} , z=\frac{-3}{2}$ . Find $x^3 \div y^2 z^2$ If I have Question like that :
If $ x=\frac{-1}{2} \ , \ y=\frac{3}{4} \ , \ z=\frac{-3}{2} \ $
Find a numerical Value for $$x^3 \div y^2z^2$$
First If we divide $x^3$ over $y^2$ and then multiplying the result we get $\frac{-1}{2}$
, but if I think we should first Multiply $y^2z^2$ and then we divide $x^3$ over the result , and the answer in that case is $\frac{-8}{81}$
First We notice that $$z=-2y \ \ , x=\frac{-2}{3}y$$
So $$x^3 \div y^2z^2 = (\frac{-2}{3})^3 \times y^3 \div 4\ y^4 = \frac{-8}{81} $$
For example $$6 \div 9 = \frac{2}{3}$$
And Notice if we wrote $$6 \div 9 = 6 \div 3 \times 3 = 6 \div 3^2$$
Notice $6 \div 9 = \frac{2}{3}$ but $6 \div 3 \times 3 = 6$
The write way is $$6 \div 9 =6 \div (3 \times 3) , 6 \div 3 \times 3 \neq 6 \div 9$$
And $$6\div 3^2 \neq 6 \div 3\times 3 \ \text{but} \ 6 \div 3^2 = 6 \div (3 \times 3)$$
|
Find a numerical Value for $$x^3 \div y^2z^2$$
I find this notation a bit ambiguous, although I would assume the intended interpretation is:
$$\frac{x^3}{y^2z^2} \tag{$*$}$$
The division operator $\div$ isn't used much in mathematics (at least not 'later on') and if you would write something like $x^3/y^2z^2$, the strict interpretation would be:
$$\frac{x^3}{y^2}z^2 \tag{$\star$}$$
because division and multiplication are at the same level and you apply them as they appear from left to right. Almost all (mathematical) software will interpret $x^3/y^2z^2$ in this way, although when written by humans, some will mean $(*)$ but this is to be avoided.
When you can use fractions, such as in $(*)$ and $(\star)$, no confusion is possible. If you're forced to write 'in line', then parentheses can be used to make sure $x^3/(y^2z^2)$ gives $(*)$, at least if that's what you mean. If you actually mean $(\star)$, then although $x^3/y^2z^2$ is technically correct, writing it as $x^3z^2/y^2$ or as $x^3z^2 \div y^2$ is a lot clearer and avoids this ambiguity and possible confusion.
|
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|
Why does Jacobi method only converge for one of these two equivalent ways of stating a problem? Consider the system \begin{align}a=b-2\\ b=2a+14\end{align} Implementing a simple Jacobi method and initializing at $(a,b)=(0,0)$, updating by $a^{k+1}=b^{k}-2$, $b^{k+1}=2a^k+14$, we get: $(-2,14),(12,10),(8,38),(36,30),(28,86)\cdots$, and the resulting sequence does not converge.
By contrast, consider the equivalent description \begin{align}a=\frac b2 - 7\\ b=a+2\end{align} and the corresponding update rules $a^{k+1}=\frac {b^k} 2-7$, $b^{k+1}=a^k+2$. Here, the Jacobi method quickly converges to the truth, $a=-12$, $b=-10$.
What's with the difference between the two? Why does Jacobi work on the second version and not on the equivalent first version? I get that the matrix describing this system is not strictly diagonally dominant, and that therefore Jacobi is not guaranteed to converge, but I still find it surprising that it converges for only one of the above two statements of the problem.
|
The matrix that defines the "original" linear equations is $$A_1 = \begin{pmatrix}1 & -1 \\ -2 & 1 \end{pmatrix}$$ for unknowns $\begin{pmatrix} a \\ b \end{pmatrix} $ and right-hand-side $b_1 = \begin{pmatrix} -2 \\ 14 \end{pmatrix} $. The error propagation matrix $C$ such that
$$e^{k+1} = C e^k$$ for the Jacobi method is given by $$C = I - D^{-1} A \Rightarrow C_1 = I - IA_1 = \begin{pmatrix}0 & 1 \\ 2 & 0 \end{pmatrix}$$ where $D$ is the diagonal matrix consisting of the diagonal elements of $A$. The spectral radius of $C_1$ is $\rho(C_1) = \sqrt{2} > 1$, thus convergence is not guaranteed (as observed). Now consider the second formulation with $$A_2 = \begin{pmatrix}1 & -0.5 \\ -1 & 1 \end{pmatrix}$$ with right-hand-side $b_2 = \begin{pmatrix} -7 \\ 2 \end{pmatrix} $. Here, $$C_2 = \begin{pmatrix}0 & 0.5 \\ 1 & 0 \end{pmatrix} $$ and the spectral radius is $\rho(C_2) = 1/\sqrt{2} < 1$ thus convergence is guaranteed for every starting vector.
|
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|
How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
|
$$ \ln \left( \frac{99!!}{100!!} \right) = \ln \left(\frac{99}{100} \cdot \frac{97}{98} \cdot \ldots \cdot \frac{1}{2}\right) = \sum_{k=1}^{50} (\ln(2k-1)- \ln(2k)) $$
with $f(k) = \ln(2k-1)-\ln(2k)$ a concave function of $k$ for $k \ge 1$.
Thus
$$\eqalign{ \ln \left( \frac{99!!}{100!!} \right) &= \frac{f(1)}{2} + \frac{f(50)}{2} + \sum_{k=1}^{49} \frac{f(k)+f(k+1)}{2}\cr
&\le \frac{f(1)}{2} + \frac{f(50)}{2} + \int_1^{50} f(t)\; dt\cr
&= -199 \frac{\ln(2)}{2}-100 \ln(5)+99 \ln(3)+99 \frac{\ln(11)}{2}+\frac{\ln(99)}{2}-\ln(10)\cr &< \ln(1/10)} $$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Binomial Questions
The Spring test will contain $20$ multiple choice questions, each with five responses, and only one response is correct for each question. No marks are deducted for incorrect answers.
*
*If a student randomly guesses the answer to every question, calculate the
probability that he fails the exam (i.e. scores below $40\%$).
*If a student is confident on $5$ questions and guesses all of the remaining
questions, what is the probability that he passes?
*If a student is confident on $18$ questions and guesses the remaining questions, what is the probability that he scores $100\%$?
I just want to double check to see if my thinking is correct:
*
*$20×40\% = 8$.
$$P(\text{score below}~40\%) = ~^{20}C_0\cdot\left(\frac{1}{5}\right)^0\cdot\left(\frac{4}{5}\right)^{20} + ~^{20}C_1 \cdots + ~^{20}C_8\cdot\left(\frac{1}{5}\right)^8\cdot\left(\frac{4}{5}\right)^{12}$$
*$20-5 = 15$.
$$P(\text{passing the exam})= ~^{15}C_1\cdot\left(\frac{1}{5}\right)^1\cdot\left(\frac{4}{5}\right)^{14} + ~^{15}C_2 \cdots + ~^{15}C_3\cdot\left(\frac{1}{5}\right)^3\left(\frac{4}{5}\right)^{12}.$$
*$20 - 18 = 2$.
$$P(\text{scores}~100\%)= ~^2C_2\cdot\left(\frac{1}{5}\right)^2\left(\frac{4}{5}\right)^0.$$
Many thanks!
|
$(i)$
If the student gets $8$ questions correct, then they score a $40$%, but the question asks for the probability that the student scores $below$ $40$%. Thus we have
$$\sum_{k=0}^7 {20 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{20-k}\approx0.968$$
$(ii)$
You tried to calculate the probability that the student $fails$, but the question asks for the probability of $passing$.
The student must get $3$ or more of the remaining $15$ questions correct to pass, which is the compliment of getting $0,1,$ or $2$ correct. Thus we have
$$1-\sum_{k=0}^2 {15 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{15-k}\approx0.602$$
Alternatively, you could have done
$$\sum_{k=3}^{15} {15 \choose k}\left(\frac{1}{5}\right)^k\left(\frac{4}{5}\right)^{15-k}\approx0.602$$
$(iii)$
The student must get the two remaining questions correct with probability $$0.2^2=0.04$$
which you correctly obtained.
|
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|
Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
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We have to show that for $x\ge 7$, the number $2^x-1$ has a prime divisor different from $3$ and $7$.
If $x$ has a prime factor $p\ge 5$, then $2^p-1$ divides $2^x-1$ and all prime factors of $2^p-1$ must be of the form $2kp+1$, hence there must be a prime factor greater than $7$.
Otherwise $x$ must be divisible by $4$ or by $9$ , hence $2^4-1$ or $2^9-1$ must divide $2^x-1$. This implies that $5$ or $73$ must be a prime factor. Since $x<6$ gives no solution, in fact $x=6$ is the only one.
|
{
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|
Calculating $\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3}$. I have a difficulty in calculating this limit:
$$\lim_{x \rightarrow 0} \frac{\tan x - \sin x}{x^3},$$
I have tried $\tan x = \frac{\sin x}{\cos x}$, then I unified the denominator of the numerator of the given limit problem finally I got $$\lim_{x \rightarrow 0} \frac{\sin x}{x^{3} \cos x} - \lim_{x \rightarrow 0} \frac{ \sin x}{x^3},$$
Then I got stucked, could anyone help me in solving it?
|
If we let $A_n$ be the up/down numbers we have:
$$\tan x = \sum_{n=0}^\infty\frac{A_{2n+1}}{(2n+1)!}x^{2n+1}$$
$$\sin x = \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$
Thus the first few coefficients of $\tan x - \sin x$ are:
$$0 + \frac{1}{2}x^3 + \frac{1}{8}x^5 + \cdots$$
Thus if we divide by $x^3$ we get a constant coefficient of $\dfrac{1}{2}$ and everything else vanishes as $x \to 0$.
|
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|
On finding $\sup\left\{3(-1)^n-\frac{1}{n^2+1}\right\}$ How does one find $\sup(A)$ where $A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$?
I've tried as follows, but I'm not so sure.
$\displaystyle 3(-1)^n-\frac{1}{n^2+1} \le 3-\frac{1}{n^2+1} \le 3. $ So $3$ is an upper bound for $A$.
Let $\epsilon >0$. We wish to prove that $3-\epsilon$ is not an upper bound for $A$.
For $N = 2n_0$ where $\displaystyle n_0 > \frac{1}{\sqrt{2\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} = 3-\frac{1}{N^2+1} \ge 3-\frac{1}{2N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$.
EDIT:
For $N = 2n_0$ where $\displaystyle n_0 > \frac{1}{\sqrt{\epsilon}} $ we have
$$\displaystyle 3(-1)^N-\frac{1}{N^2+1} = 3-\frac{1}{N^2+1} \ge 3-\frac{1}{N^2} > 3-\epsilon. $$
Hence $3-\epsilon$ is not an upper bound. Hence $3= \sup(A)$.
|
This sequence $$ A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$$
has a subsequence $$ \left\{3-\frac{1}{4n^2+1}: n \in \mathbb{N}\right\}$$ which is monotonically increases to $x=3$
Since x=3 is an upper bound for the set $$ A = \left\{3(-1)^n-\frac{1}{n^2+1}: n \in \mathbb{N}\right\}$$
We have $$ sup (A) = 3
|
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|
Studying the extrema of $f(x,y) = x^4 + y^4 -2(x-y)^2$
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x,y) = x^4 + y^4 - 2(x-y)^2$. Study its extrema.
So here was my approach.
We have $$\frac{\partial f}{\partial x}(x,y) = 4(x^3 -x + y),\frac{\partial f}{\partial y}(x,y)= 4(y^3 -y + x) $$
I have to find $(x_0,y_0) \in \mathbb{R}^2$ such that $ \frac{\partial f}{\partial x}(x_0,y_0)= \frac{\partial f}{\partial y}(x_0,y_0) = 0 $
So we have:
$$\left\{\begin{matrix}
x(x^2-1)+ y =0\\
y(y^2 -1) + x =0
\end{matrix}\right. $$
Thus we have $x-x^3 = y$, and by replacing $y$ with $x-x^3$ in the second line, we get:
$$ y(y^2 -1) + x =0 = (x-x^3)((x-x^3)^2 -1)+x =x^5(-x^4 -x^2 -3) = 0 $$ And the only solution for this is $x=0$. As $y = x^3 -x$ we immediately have $y=0$.
So the only extremum possible is at $(0,0)$. Now, I need to study its Hessian matrix.
We have: $$\frac{\partial ^2f}{\partial x^2}(x,y) = 12x^2 -4 , \frac{\partial ^2f}{\partial y^2}(x,y) = 12y^2 -4, \frac{\partial^2 f}{\partial x \partial y}(x,y) = \frac{\partial^2 f}{\partial y \partial x}(x,y) = 4$$
Thus $$H(x,y) \begin{bmatrix}
12x^2 -4&4 \\
4& 12y^2 -4
\end{bmatrix} $$
At $(x,y)=(0,0)$ we have
$$H(0,0) \begin{bmatrix}
-4&4 \\
4& -4
\end{bmatrix} $$
As we have $\text{det}(H(0,0)) = 0$ and $\text{Tr}(H(0,0)) = -8$ the eigenvalues are $0$ and $-8$. As it has $0$ as eigenvalue, I need to study the differential at a higher order. But here I lack understanding. What exactly should I do? Should I compute the third order partial differentials and then restudy their Hessian Matrix?
|
To solve the system you can add:
$$\begin{cases} x^3-x+y=0 \\ y^3-y+x=0 \end{cases} \Rightarrow x^3+y^3=0 \Rightarrow x=-y \Rightarrow \\ x^3-x-x=0 \Rightarrow x(x^2-2)=0 \Rightarrow x=0;\pm \sqrt{2}.$$
|
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|
Fourier series applied I have a question concerning the Fourier series:
I started with the following:
$$\cos (\alpha x)= \frac{1}{2}a_0+ \sum_{k=1}^{\infty}a_k\cos(kx).$$
I proved that this series with Fourier coefficients is equal to:
$$\cos \alpha x= \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos kx}{k^2-\alpha^2},$$
and $\alpha$ is NOT an integer.
But now I am stuck:
Suppose that $F_N(x)$ is the above Fourier series truncated at the $N$-th term. Can we show that:
$$F_N(\pi) = \cos \alpha \pi+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=N+1}^{\infty}\frac{1}{k^2-\alpha^2},$$
and hence: $$F_N(\pi) \simeq \cos \alpha \pi + \frac{2\alpha}{N \pi} \sin \alpha \pi?$$
I started by writing:
$$\cos \alpha \pi = \frac {\sin \alpha \pi}{\alpha \pi}+ \frac{2 \alpha}{\pi}\sin \alpha \pi \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\cos k\pi}{k^2-\alpha^2}$$
and $\cos k \pi = (-1)^k$. But now I am completely lost in what to do.. Can someone help me?
|
Because the series $\displaystyle \sum\limits_{k = 1}^\infty (-1)^{k - 1} \frac{\cos kπ}{k^2 - α^2} = -\sum\limits_{k = 1}^\infty \frac{1}{k^2 - α^2}$ converges, then\begin{align*}
\cos απ &= \frac{\sin απ}{απ} + \frac{2α}{π} \sin απ \sum_{k = 1}^\infty (-1)^{k - 1} \frac{\cos kπ}{k^2 - α^2}\\
&= \frac{\sin απ}{απ} - \frac{2α}{π} \sin απ \sum_{k = 1}^\infty \frac{1}{k^2 - α^2}\\
&= \frac{\sin απ}{απ} - \frac{2α}{π} \sin απ \sum_{k = 1}^N \frac{1}{k^2 - α^2} - \frac{2α}{π} \sin απ \sum_{k = N + 1}^\infty \frac{1}{k^2 - α^2}\\
&= F_N(π) - \frac{2α}{π} \sin απ \sum_{k = N + 1}^\infty \frac{1}{k^2 - α^2},
\end{align*}
which implies\begin{align*}
F_N(π) &= \cos απ + \frac{2α}{π} \sin απ \sum_{k = N + 1}^\infty \frac{1}{k^2 - α^2}\\
&\approx \cos απ + \frac{2α}{π} \sin απ \sum_{k = N + 1}^\infty \frac{1}{k^2 - k}\\
&= \cos απ + \frac{2α}{π} \sin απ \sum_{k = N + 1}^\infty \left( \frac{1}{k - 1} - \frac{1}{k} \right)\\
&= \cos απ + \frac{2α}{π} \sin απ · \frac{1}{N} = \cos απ + \frac{2α}{Nπ} \sin απ.
\end{align*}
|
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|
Calculate $f(x)$ at a specific point Question:
Calculate $f(x) = \frac{49}{x^2} + x^2$ at points for which $\frac{7}{x}+x =3$
My attempt:-
I tried to find the value of $x$ and insert in $f(x)$
$$\frac{7}{x}+x =3$$
$$7+x^2 =3x$$
$$x^2 -3x + 7=0$$
$$x = \frac{3+\sqrt{9-7*4}}{2}$$
Now $x$ is coming out to be irrational and things get a lot more difficult from there. What should I do?
|
Make use of
$$
\left(\frac{7}{x}+x \right)^2=\frac{49}{x^2}+x^2+14
$$
i.e.
$$
3^2-14=\frac{49}{x^2}+x^2=-5
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \sin \frac{x}{2} - \cos \frac{x}{2}$
So, $\frac{dy}{dx} = \frac{1}{2} \cdot (\cos\frac{x}{2} + \sin\frac{x}{2})$.
But apparently this is wrong. The correct solution is:
$\frac{dy}{dx} = -\frac{1}{2}\cdot(\cos\frac{x}{2} + \sin\frac{x}{2})$.
So I want to know what I have done wrongly here. Why is my answer not right?
|
$$y=\sqrt {1-\sin x}$$
$$\ln y= \frac 12 \cdot \ln (1-\sin x)$$
$$\Rightarrow \frac 1y \frac {dy}{dx}=\frac 12\left (\frac {-\cos x}{1-\sin x}\right) $$
$$\Rightarrow \frac {dy}{dx}=\frac {-\cos x}{2\sqrt {1-\sin x}}$$
On rationalisation of denominator this turns out to be $$-\frac 12 (\sin x/2 +\cos x/2)$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Validity of proving identities by showing LHS-RHS =0 or using reversible steps? When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:
Method 1:
Showing that $\mathrm{LHS}-\mathrm{RHS}=0$
For instance, if we are required to prove
$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$
We instead show that $\mathrm{LHS}-\mathrm{RHS}=0$ as follows:
$\mathrm{LHS} - \mathrm{RHS} = x^2\cos^2(x)+\sin^2(x) - x^2 + x^2\sin^2(x) - 1 + \cos^2(x)$
$ = x^2(\cos^2(x) + \sin^2(x)) - x^2 + (\sin^2(x) + \cos^2(x)) - 1$
$= x^2 - x^2 + 1 - 1 = 0$
Method 2: Showing that $\mathrm{LHS}=\mathrm{RHS}$ is equivalent with another equation which is true (taking care that we can always reverse the steps and showing it by putting $\iff$):
$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$
$ \iff x^2\cos^2(x)+x^2\sin^2(x) - x^2 = 1 - \sin^2(x) - \cos^2(x)$
$ \iff x^2 - x^2 = 1 - 1$
$ \iff 0 = 0$
Are these two methods of proving valid? Are there any cases where we can make fallacious argument by using these methods? Are any one of them better than the other?
|
One of the related questions here, seems to be more suitable for method $2$:
$$\frac{1+\sin x}{\cos x} = \frac{1+\sin x+\cos x}{1-\sin x+\cos x} (*)$$
Let $a =\sin x$, $b=\cos x$, then:
$(*)\Leftrightarrow \frac{1+a}{b}=\frac{1+a+b}{1-a+b}$
$\Leftrightarrow (1+a)(1-a+b) = b(1+a+b)$
$\Leftrightarrow 1-a^2+b(a+1) = b^2 + b(a+1)$
$\Leftrightarrow a^2+b^2=1$, which is provable.
For method $1$, I think you can use it to prove this simple expression:
$$a^2-b^2=(a-b)(a+b)$$
For this problem, using method $1$ or $2$ is actually the same, no method is better than the other.
For some problems when either side is hard to be made "equal" to the other side, or if you are stuck with this method because of that annoying "$\Rightarrow$", using substraction is recommended.
|
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|
Differentiate $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}
Find $\dfrac{\mathrm dy}{\mathrm dx}$ if
$y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
I can solve it as follows:
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\frac{d}{dx}\Big(2x\sqrt{1-x^2}\Big)\\&=\frac{1}{\sqrt{1-4x^2+4x^4}}\bigg(2x\frac{-2x}{2\sqrt{1-x^2}}+2\sqrt{1-x^2}\bigg) \\
&=\frac{1}{\sqrt{(1-2x^2)^2}}\bigg(\frac{-2x^2}{\sqrt{1-x^2}}+2\sqrt{1-x^2}\bigg)\\
&=\frac{1}{|1-2x^2|}\frac{-2x^2+2-2x^2}{\sqrt{1-x^2}}\\
&=\frac{2(1-2x^2)}{|1-2x^2|\sqrt{1-x^2}}\\
\end{align}
$$
As $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\implies|x|<\frac{1}{\sqrt{2}}\implies 0<x^2<\frac{1}{2}\implies 0<2x^2<1\\\implies-1<-2x^2<0\implies 0<1-2x^2<1$
Thus, $|1-2x^2|=1-2x^2$
$$
y'=\frac{2(1-2x^2)}{(1-2x^2)\sqrt{1-x^2}}=\frac{2}{\sqrt{1-x^2}}
$$
My Attempt
But if I try to solve it by substituting $x=\sin\alpha\implies\alpha=\sin^{-1}x$
$$
y=\sin^{-1}\bigg(2\sin\alpha\sqrt{\cos^2\alpha}\bigg)=\sin^{-1}\bigg(2\sin\alpha|\cos\alpha|\bigg)
$$
Here, $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\implies \frac{-\pi}{4}<\sin^{-1}x=\alpha<\frac{\pi}{4}\implies \cos\alpha>0\implies|\cos\alpha|=\cos\alpha$
$$
\begin{align}
y&=\sin^{-1}\bigg(2\sin\alpha\cos\alpha\bigg)=\sin^{-1}\bigg(\sin2\alpha\bigg)\\&\implies\sin y=\sin2\alpha=\sin\Big(2\sin^{-1}x\Big)\\
&\implies y=n\pi+(-1)^n(2\sin^{-1}x)=\begin{cases}n\pi+2\sin^{-1}x,\quad \text{n even}\\
n\pi-2\sin^{-1}x,\quad \text{n odd}
\end{cases}
\end{align}
$$
Thus,
$$
y'=\begin{cases}\frac{d}{dx}\Big(n\pi+2\sin^{-1}x\Big)=\frac{2}{\sqrt{1-x^2}},\quad \text{n even}\\
\frac{d}{dx}\Big(n\pi-2\sin^{-1}x\Big)=\frac{-2}{\sqrt{1-x^2}},\quad \text{n odd}
\end{cases}
$$
How do I eliminate the case $y'=\frac{-2}{\sqrt{1-x^2}}$ ?
|
At one point you have $y=\sin^{-1}(\sin 2\alpha)$. It means that $-\pi/2\le y\le \pi/2$, so your only allowed solution is $n=0$.
|
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|
Prove $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ Just for fun:
How can we prove (calculate) that $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ ? Can we use (9) from:
http://mathworld.wolfram.com/DigammaFunction.html
(9): $\sum_{k=0}^\infty \frac{(-1)^k}{3k+1} = \frac{1}{9}\left(\sqrt{3}\pi+3\ln\,2\right)$
?
Thx!
|
We can evaluate the series of interest without appealing to the Digamma Function.
Note that we can simply write
$$\begin{align}
\sum_{n=0}^{2N}\frac{(-1)^{n}}{3n+2}&=\sum_{n=0}^N\left(\frac{1}{6n+2}-\frac{1}{6n+5}\right)\\\\
&=\sum_{n=0}^N\int_0^1\left( x^{6n+1}-x^{6n+6}\right)\,dx\\\\
&=\int_0^1 x\left(\frac{1-x^{6N+6}}{1+x^3}\right)\,dx\\\\
&=\int_0^1 \left(\frac{x}{1+x^3}\right)\,dx -\int_0^1 x^{6N+7} \left(\frac{1}{1+x^3}\right)\,dx
\end{align}$$
Applying the Dominated Convergence Theorem (or alternatively, integrate by parts and observe that the second integral is $ O(N^{-1})$), we find that
$$\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+2}=\int_0^1 \frac{x}{1+x^3}\,dx$$
Can you finish now using partial fraction expansion for example?
|
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|
Given $2$ points $A$ and $B$, and a point $Q$ on the circle. Find the minimized value of $\frac{2}{3}\overline{QA} + \overline{QB}$. Problem
Given $2$ points $A(0,3)$ and $B(4,0)$ on the plane, and a point $Q$ on the circle $x^2+y^2=4$.
Find the minimum value of $\frac{2}{3}\overline{QA} + \overline{QB}$.
My Idea
Let $Q$ be $(2\cos{t},2\sin{t})$. Then
\begin{align*}
\frac{2}{3}\overline{QA} + \overline{QB}
&= \frac{2}{3}(\sqrt{(2\cos{t}-0)^2+(2\sin{t}-3)^2})+\sqrt{(2\cos{t}-4)^2+(2\sin{t}-0)^2} \\
&= \frac{2}{3} \sqrt{13-12 \sin{t}} + 2 \sqrt{5-4 \cos{t}}
\end{align*}
Taking the derivative to find the mimimum:
$$\frac{-4 \cos{t}}{\sqrt{13 - 12 \sin{t}}} + \frac{4 \sin{t}}{\sqrt{5 - 4 \cos{t}}} = 0$$
However the computation is complicated. Are there another ways to deal with this problem?
|
You were on the right track. It is geometrically pretty evident that the minimizing $Q$ lies in the first quadrant, so you just have to solve
$$ \frac{\sin t}{\sqrt{5-4\cos t}}=\frac{\cos t}{\sqrt{13-12\sin t}}\tag{1}$$
for $t\in\left(0,\frac{\pi}{2}\right)$. With such constraint $(1)$ boils down to
$$ \sin^2(t)(13-12\sin t) = \cos^2(t)(5-4\cos t)\tag{2} $$
which can be factored as
$$ (3\sin(t)+\cos(t)-2)(3\sin(t)-\cos(t)+2\cos(2t))=0.\tag{3}$$
The term $3\sin(t)-\cos(t)+2\cos(2t)$ is strictly positive for $t\in\left(0,\frac{\pi}{2}\right)$, hence the solution is provided by the solution of $3\sin(t)+\cos(t)=2$, i.e. by
$$ t=\arccos\left(\frac{2+3\sqrt{6}}{10}\right)\approx 20^\circ 47'48''.\tag{4}$$
The minimum value of $\frac{2}{3}QA+QB$ turns out to be $\frac{4}{3}\sqrt{10}$.
|
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|
Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s
|
The expansion of what is under square root gives
$$A=9a^4+6a^3-6a^3-4a^2+6a^2+4a=$$
$$a (9a^3+2a+4)=a (9 (-a-1)+2a+4)=$$
$$-a (7a+5)=$$
this must be equal to
$$B=(1-a^2)^3=1-3a^2+3a^4-a^6=$$
$$$$
the difference is
$$B-A=1+4a^2+5a+3a^4-(a+1)^2=$$
$$3a^2+3a+3a^4=3a (a^3+a+1)=0$$
It is correct.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Can you explain the absolute value sign when integrating $\int_{-1}^{1}dx \int_{0}^{x^2} \sqrt{x^2-y} dy$? I'm doing homework on the following integral:
$$\int_{-1}^{1}dx \int_{0}^{x^2} \sqrt{x^2-y} dy$$
And here is the answer:
I try to think about it, but I couldn't answer why there is an absolute value sign in the third expression at $x^3$. Can somebody explain it for me why? Thank you! :D
|
Let's compute $\int_0^{x^2} \sqrt{x^2 - y} \, dy$. Applying the substitution $u = x^2 - y$, we have $du = -dy$ and hence
$$ \int_0^{x^2} \sqrt{x^2 - y} \, dy = -\int_{x^2}^0 \sqrt{u} \, du = \int_0^{x^2} \sqrt{u} \, du = \left[ \frac{2}{3} u^{\frac{3}{2}} \right]^{u=x^2}_{u=0} = \frac{2}{3} \left( (x^2)^{\frac{1}{2}} \right)^3 = \frac{2}{3} |x|^3. $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve $a=x \lfloor x \rfloor$ How can I solve for $x$ given $a=x \lfloor x \rfloor$
Where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$, and where $a$ is a rational number.
What I've done
\begin{align}
\frac{a}{x} &= \lfloor x \rfloor \\
\implies \frac{a}{x} & \le x < \frac{a}{x}+1 \\
\end{align}
which yields two cases
\begin{align}
\sqrt{a} &\le x \quad (1) \\\\
\left(x+\frac{1+\sqrt{1+4a}}{2}\right) \left(x+\frac{1-\sqrt{1+4a}}{2}\right) &< 0 \quad (2)
\end{align}
|
If $\lfloor x \rfloor = n$, you want $x = a/n$ and $n \le a/n < n+1$. Thus (assuming $n > 0$)
$n^2 \le a < n^2 + n$. Now $(n+1)^2 = n^2 + 2 n + 1 > n^2 + n$. So:
Given $a \ge 1$,
take $n = \lfloor \sqrt{a} \rfloor$. If $a \ge n^2 + n$ there is no solution.
Otherwise, $x = a/n$.
|
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|
Find the minimum of a three variate function We consider the function
$$f(x,y,z)=(7(x^2+y^2+z^2)+6(xy+yz+zx))(x^2y^2+y^2z^2+z^2x^2)$$
Find
$$m=\min\{f(x,y,z):xyz=1\}$$
Using the AM-GM inequality it is clear that
$$m_+=\min\{f(x,y,z):xyz=1,x,y,z>0\}=13\times 9=117.$$
But $f(-1,-1,1)=45$ so clearly $m<m_+$. Numerically, it seems that $m\approx 42.0$.
What is the exact value of $m$?
|
Since $\sum\limits_{cyc}(7x^2+6xy)>0$ and $\sum\limits_{cyc}x^2y^2>0$, we see that the minimal value is non-negative.
Let $m$ be a minimal value.
Thus, $$\sum_{cyc}(7x^2+6xy)\sum_{cyc}x^2y^2\geq mx^2y^2z^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.
Hence, $$(7(9u^2-6v^2)+18v^2)(9v^4-6uw^3)-mw^6\geq0$$ or $g(w^3)\geq0,$ where $$g(w^3)=(7(9u^2-6v^2)+18v^2)(9v^4-6uw^3)-mw^6.$$
We see that $g$ is a concave function.
But the concave function gets a minimal value for an extreme value of $w^3$, which happens for an equality case of two variables.
Since $g(w^3)$ is a homogeneous, it's enough to assume $y=z=1$, which gives
$$m=\min_{x\in\mathbb R}\frac{(7x^2+12x+20)(2x^2+1)}{x^2}.$$
We obtain $$\left(\frac{(7x^2+12x+20)(2x^2+1)}{x^2}\right)'=\frac{4(x-1)(7x^3+13x^2+13x+10)}{x^3},$$
which gives that the minimum occurs when $x$ equal to the real root of the equation $$7x^3+13x^2+13x+10=0,$$
which we can get by the Cardano's formula and we can get an exact minimal value.
I got
$$m=\tfrac{2062+\sqrt[3]{4420439038+12661425\sqrt{120585}}+\sqrt[3]{4420439038-12661425\sqrt{120585}}}{105}=42.04956...$$
|
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|
Differentiate $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
Find $\frac{dy}{dx}$ if $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
The solution is given as $y'=0$ in my reference. But that doesn't seem to be a complete solution as the graph of the function is:
My Attempt
Let $x=\sin\alpha\implies \alpha=\sin^{-1}x$
$$
y=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{1-\sin^2\alpha}=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{\cos^2\alpha}\\
=\sin^{-1}(\sin\alpha)+\sin^{-1}\sqrt{\sin^2(\tfrac{\pi}{2}-\alpha)}=\sin^{-1}(\sin\alpha)+\sin^{-1}|\sin(\tfrac{\pi}{2}-\alpha)|\\
=n\pi+(-1)^n(\alpha)+
$$
How do I proceed further and find the derivative ?
|
$y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$, $-1\leq x\leq1$
Let, $x=\sin\alpha\implies \alpha=\sin^{-1}x$, We have $-\pi/2\leq\alpha\leq\pi/2\implies|\cos\alpha|=\cos\alpha$
$$
\begin{align}
y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(|\cos\alpha|)=\sin^{-1}(\sin\alpha)+\sin^{-1}(\cos\alpha)\\&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))
\end{align}
$$
Here,
$$
\tfrac{-\pi}{2}\leq\alpha\leq\tfrac{\pi}{2}\implies\sin^{-1}(\sin\alpha)=\alpha\\
0\leq\tfrac{\pi}{2}-\alpha\leq{\pi}\implies\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))=\begin{cases}\frac{\pi}{2}-\alpha,\text{ if }0\leq\tfrac{\pi}{2}-\alpha\leq\tfrac{\pi}{2}\\
\pi-(\frac{\pi}{2}-\alpha),\text{ if }\tfrac{\pi}{2}<\tfrac{\pi}{2}-\alpha\leq\pi\end{cases}
$$
Therefore,
$$
\begin{align}
y&=\sin^{-1}(\sin\alpha)+\sin^{-1}(\sin(\frac{\pi}{2}-\alpha))\\
&=\begin{cases}\alpha+\frac{\pi}{2}-\alpha=\frac{\pi}{2}\quad\quad\quad\quad\;\;\text{ if }\quad\: 0\leq\tfrac{\pi}{2}-\alpha\leq\tfrac{\pi}{2}\\
\alpha+\pi-\frac{\pi}{2}+\alpha=\frac{\pi}{2}+2\alpha\quad\text{ if }\quad\tfrac{\pi}{2}<\tfrac{\pi}{2}-\alpha\leq\pi
\end{cases}\\
&=\begin{cases}\tfrac{\pi}{2}\quad\quad\quad\text{ if }\quad0\leq\alpha\leq\tfrac{\pi}{2}\\
\frac{\pi}{2}+2\alpha\quad\text{ if }\quad \frac{-\pi}{2}\leq\alpha<0
\end{cases}\\
&=\begin{cases}\tfrac{\pi}{2}\quad\quad\quad\quad\quad\text{ if }\quad\quad 0\leq x\leq 1\\
\frac{\pi}{2}+2\sin^{-1}x\quad\text{ if }\quad -1\leq x<0
\end{cases}
\end{align}
$$
$$
\color{red}{
\frac{dy}{dx}=\begin{cases}0\quad\quad\quad\text{ if }\;\quad 0\leq x\leq 1\\
\frac{2}{\sqrt{1-x^2}}\quad\;\text{ if }\; -1\leq x<0
\end{cases}}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Russian MO 2004 $\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$ I have a doubt on a proof included in "Secrets in Inequalities" by Pham Kim Hung. The exercise is to prove $$\sqrt{a} + \sqrt{b} + \sqrt{c} \geq ab + bc + ca$$ for a, b, c whose sum is 3.
His approach is the following
He observes that:
$$2(ab + bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2)$$
And he says that the equation above is equivalent to the inequality below(which is what gives me doubt):
$$\sum_{cyc} a^2 + 2\sum_{cyc} \sqrt{a} \geq 9$$
How does he get to $2\sum_{cyc} \sqrt{a} $? Does he get this out of the blue? Or is there some logic behind this?
|
By AM-GM
$$\sum_{cyc}(a^2+2\sqrt{a})\geq3\sum_{cyc}\sqrt[3]{a^2\cdot(\sqrt{a})^2}=$$
$$=3\sum_{cyc}a=9=(a+b+c)^2=\sum_{cyc}(a^2+2ab),$$
which gives $$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+ac+bc.$$
|
{
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|
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)^n(a-b)=n\pi+(-1)^n(\sin^{-1}x-\sin^{-1}\sqrt{x})
$$
Thus,
$$
y'=\frac{d}{dx}\big[n\pi+(-1)^n(a-b)\big]=\begin{cases}\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\text{ if }n\text{ is even}\\
-\bigg[\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\bigg]\text{ if }n\text{ is odd}
\end{cases}
$$
Is it the right way to solve this problem and how do I check the solution is correct ?
Note: I think there got to be two cases for the derivative as the graph of the function is
|
$$F(x)=\sin^{-1}x-\sin^{-1}\sqrt x=\sin^{-1}x+\sin^{-1}(-\sqrt x)$$
Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$\displaystyle F(x) =\begin{cases}
\arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x \le 1 \;\text{ or }\; x^2+x > 1, -x\sqrt x< 0\iff x>0\\
\pi - \arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x > 1, 0< x,-\sqrt x \le 1\text{ which is untenable}\\
-\pi - \arcsin( x\sqrt{1-x} -\sqrt x\sqrt{1-x^2}) \;\;;x^2+x > 1, -1< x,-\sqrt x \le 0
\end{cases}$$
|
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|
Integration by Substitution of Fraction involving e
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$
The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e^{2x}+4}$
let $u = e^{2x} +4$
$\frac{dy}{dx}=2e^{2x}$
$dx=\frac{1}{2}e^{-2x}du$
$u = e^{2x} +4$
$e^{2x} = u-4$
$e^{-2x} = \frac{1}{u-4}$
$dx=\frac{1}{2}(\frac{1}{u-4})du$
Hence the integral is:
$$
\begin{aligned}
&\int\frac{2}{e^{2x}+4} \\
=& 2\int\frac{1}{u}\frac{1}{2}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u^2-4u}du\\
=&\int u^{-2}-\frac{1}{4}u^{-1}du\\
=&\frac{u^{-1}}{-1}-\frac{1}{4}\ln(u)+c\\
=&-\frac{1}{e^{2x}+4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
=&-e^{-2x}-\frac{1}{4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
\ne& \frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c\\
\end{aligned}
$$
???? help
|
You have made a basic, but common algebraic error:
$$\int\frac{1}{u^2-4u}du \neq\int u^{-2}-\frac{1}{4}u^{-1}du$$
You cannot split denominator like that. To continue your method, use partial fractions.
|
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|
Getting the volume of a sphere from an infinite product I read an article about squeezing pi from a menger sponge. You can see the original here: Squeezing pi from a menger sponge
The author finds an infinite product which, he says, approximates the volume of a sphere with radius 1. The author mentions that the infinite product he found can be written in terms of the Wallis product. The equation given is:
$$8\prod_{n=1}^\infty \frac{(2n+1)^3-3(2n+1)+2}{(2n+1)^3}=\frac{4\pi}{3}$$
I do not know how to get the left hand side of the above equation in terms of the Wallis product. Hence, I do not know how to verify the equation. Please help.
For reference, the Wallis product is given by:
$$\prod_{n=1}^\infty \frac{2n}{2n-1}\cdot \frac{2n}{2n+1}=\frac{\pi}{2}$$
Suggestions?
|
You can derive this result as follows:
First, notice that:
$$(2n+1)^3-3(2n+1)+2 = 12n^2 + 8n^3 = 4n^2(2n+3)$$
Then, the product in your expression can be written as:
$$
\begin{align*}
\prod_{n=1}^\infty \frac{(2n+1)^3-3(2n+1)+2}{(2n+1)^3} &= \prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n+1)(2n+1)(2n+1)} \overset{1}{=} \prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n-1)(2n+1)(2n+1)} \\ &\overset{2}{=} \frac{1}{3}\prod_{n=1}^\infty \frac{4n^2(2n+3)}{(2n-1)(2n+1)(2n+3)} = \frac{1}{3}\prod_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1} = \frac{\pi}{6}
\end{align*}
$$
In (1) I just included 2*1 - 1 = 1 in the product of the first term, effectively shifting the starting place once to the left. Similarly, in (2) I shifted the starting place of the product for the third term once to the right, putting the remaining 3 outside.
Multiplying this by 8 gives you the result you are looking for!
|
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|
With an appropiate substitution of the Taylor polynomial around $0$ give an estimation of $sin 1$ with the accuracy of $10^{-4}$. I know that the reminder is $R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$ and that the Taylor polynom of the sinus is $x-\dfrac {x^{3}} {3!}+\dfrac {x^{5}} {5!}-\ldots +\left( -1\right) ^{n-1}\dfrac {x^{2n-1}} {(2n-1)!}+R_{2n+1}$
$f'(x)= cos (x)$ $f'(0)=1$
$f''(x) = -sin (x)$ $f''(x) = -sin (0)=0$
$f'''(x)=-cos(x)$ $f'''(x)=-cos(0)=-1$
$f^{iv}(x)=sin(x)$ $f^{iv}(x)=sin(0)=0$
$f^{v}(x)=cos(x)=f'(x)$
$T_{5}(x)=0 + x + \frac{0}{2!}x^2+-\frac{1}{3!}x^3+\frac{0}{4!}x^4+\frac{1}{5!}x^5$
how can I get the value of $sin 1$?
|
Let $A=1-1/3!+1/5!-1/7!.$ $$\text {Then }\quad A-\sin 1=1/9!-(1/10!-1/11!)-(1/12!-1/13!)-...<1/9!$$ $$ \text {and }\quad A-\sin 1=(1/9!-1/10!)+(1/11!-1/12!)+... >0.$$ So $0<A-\sin 1<1/9!=1/362,880.$
Or we could say that with $x=1,$ $a=0$ and $f(x)=\sin x,$ we have $x-a=1, $ so for some $c\in (a,x)=(0,1)$ we have $$A-\sin 1=R_7=\frac {f^{(8)}(c)}{8!} =\frac {\sin c}{8!}.$$ And since $0<\sin c<1$ for $c\in (0,1)$ we have $0<R_7<1/8!=1/40,320.$
Remark: From the first part we see that the $c$ in the second part satisfies $\sin c<1/9.$ And we have $R_7=R_8$ because $f^{(8)}(a) =f^{(8)}(0)=\sin 0=0.$
|
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|
Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4x(2/3(900 -x^2)^1/2 = (8/3)x(900 -x^2)^1/2$
Taking the derivative
$A'(x) = 8/3(900 -x^2)^{1/2} + (8/3)x(1/2)(900 -x^2)^{-1/2}(-2x)
= (2400 - (16/3)x^2)/(\sqrt{900-x^2})$
Set the $A' = 0$
$2400 - (16/3)x^2 = 0$
$(16/3)x^2 = 2400$
$(16/3)x = \sqrt{2400} = (20\sqrt{6}) / 3$
$x = (5\sqrt{6})/12$
Then i put the value of x in the equation and get
$A = (8/3)((5 \sqrt 2)/12)(900 - ((5\sqrt 2)/12)^2)^{1/2} = 81.6...$
Is this right or?
|
Alternative way by Lagrange's multipliers
*
*$4y=8\lambda x$
*$4x=18\lambda y$
since $\lambda=0$, $y=0$, $x=0$ don't lead to any solution we can divide and obtain
*
*$\frac y x = \frac 49\frac x y \implies 9y^2=4x^2 \\\implies 8x^2=3600 \implies x^2=450\implies x=\pm15\sqrt 2\implies y=\pm 10 \sqrt 2$
and then
$$A=4|x||y|=1200$$
|
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|
Evaluate $\int_0^\infty \frac{x^2}{x^4 + 6x^2 + 13}dx$ In the context of the residue theorem, I have this integral to evaluate. The function is even, and $|\int_0^\pi\frac{R^2e^{2i\theta}iRe^{i\theta}}{R^4e^{4i\theta}+6R^2e^{2i\theta} + 13}d\theta| \leq \int_0^\pi2\frac{R^3}{R^4}d\theta \to 0$, so the problem is to find the residues in the upper halfplane.
$\int_0^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \frac12\int_{-\infty}^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \pi i\sum_{\{\Im z > 0\}}$res$(\frac{x^2}{x^4 + 6x^2 + 13})$
There are two residues to calculate:
*
*$z = \sqrt{-3 + 2i}$: $\frac{\sqrt{-3 + 2i}}{4(-3+ 2i) + 12} = -\frac i8\sqrt{-3 + 2i}$
*$z = \sqrt{-3 - 2i}: \frac i8\sqrt{-3 - 2i}$
(Wolfram Alpha if you don't want to trust me)
Giving me overall for the integral:
$\frac\pi8 (\sqrt{-3 + 2i} - \sqrt{-3 - 2i}) = $1.427346... i
But the answer is clearly not meant to be imaginary.
|
Let us try to avoid useless computations: $x^4+6x^2+13=(x^2+\alpha)(x^2+\beta)$ for a couple of conjugated complex numbers $\alpha,\beta$ with positive real part and such that $\alpha\beta=13$ and $\alpha+\beta=6$. By partial fraction decomposition we have
$$ \int_{0}^{+\infty}\frac{x^2}{(x^2+\alpha)(x^2+\beta)}\,dx = \frac{1}{\beta-\alpha}\int_{0}^{+\infty}\left(\frac{\beta}{x^2+\beta}-\frac{\alpha}{x^2+\alpha}\right)\,dx = \frac{\pi}{2\left(\sqrt{\beta}+\sqrt{\alpha}\right)}$$
and
$$\left(\sqrt{\alpha}+\sqrt{\beta}\right)^2 = \alpha+\beta+2\sqrt{\alpha\beta} = 6+2\sqrt{13} $$
hence the wanted integral equals $\frac{\pi}{2\sqrt{6+2\sqrt{13}}}$. Similarly
$$ \int_{0}^{+\infty}\frac{x^2\,dx}{x^4+Ax^2+B} = \frac{\pi}{2\sqrt{A+2\sqrt{B}}} $$
for any $A,B>0$. Lazy is good.
|
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|
Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be something easy I just can't see.
Thanks in advance.
|
From
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
one can factor an $x$ from each term as follows:
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{x \left(1 + \sqrt{1 + \frac{a^{2}}{x^2}} \right)}{x \left( 1 + \sqrt{1 + \frac{b^2}{x^2}} \right)} =
\frac{1 + \sqrt{1 + \frac{a^{2}}{x^2}} }{ 1 + \sqrt{1 + \frac{b^2}{x^2}} }.
\end{align}
From here the limit can be taken or one can expand the one more time. Using
\begin{align}
\sqrt{1 + t} = 1 + \frac{t}{2} - \frac{t^2}{8} + \mathcal{O}(t^3)
\end{align}
then
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{2 + \frac{a^2}{2 x^2} - \frac{a^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)}{2 + \frac{b^2}{2 x^2} - \frac{b^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)} = 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right).
\end{align}
Upon taking the limit the result becomes
\begin{align}
\lim_{x \to \pm \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} = \lim_{x \to \pm \infty} 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right) = 1.
\end{align}
|
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|
Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$ $I=\displaystyle\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$
Attempt:
$\dfrac{ A_0 }{ x }+\dfrac{ A_1 }{ x +1 }+\dfrac{ A_2 }{ x + 2 }...+\dfrac{ A_m }{ x +m } =\dfrac{1}{x(x+1)(x+2)(x+3)...(x+m)}$
But things got very messy.
I also thought that 1)applying integral by parts, or 2) taking terms one from left head, one from right hand and use some kind of a symmetry, or 3) using trigonometric identites etc.
I cannot see the solution, any hint, help would be perfect. Thank you in advanced.
|
Maybe the following reduction will be helpful:
$$x(x+1)(x+2)...(x+(m-2))(x+(m-1))(x+m)=x(x+m)(x+1)(x+(m-1))(x+2)(x+m(m-1))...=(x^2+mx)(x^2+mx+1(m-1))(x^2+mx+2(m-2))$$
|
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|
Interior Area of Circle of Circles
I am looking for the area of the white region interior to a set of circles with radius A, oriented on the edge of a larger circle with radius B, spaced apart from each other with distance C. You can assume that C is less than 2 times A, so that each smaller circle overlaps with its neighbors. To clarify, I am looking for the area of the circle with radius b, not covered by small circles.
|
It's more convenient
to use the number of circles $n$
instead of the distance between their centers.
Then
\begin{align}
\phi&=\tfrac\pi{n}
,
\end{align}
the distance between the centers $|C_iC_{i+1}|=2\,R\sin\tfrac\phi2$
and
the total area $S$ of the interior region
consists of $n$ areas $S_p$ of petals $P_1OP_2$.
\begin{align}
S&=n\,S_p
,\\
S_p
&=2\,S_{\triangle OC_0P_1}-S_{\mathrm{seg}\,C_0P_1P_2}
\\
&=|OC_0|\cdot|P_1F|-\tfrac\theta2\,|C_0P_1|^2
\\
&=R\cdot r_a\sin\tfrac\phi2-\tfrac\theta2\,r^2
\tag{1}\label{1}
.
\end{align}
In $\triangle P_1OF$,
$\triangle P_1FC_0$,
$\triangle P_1C_0G$
\begin{align}
\angle C_0GO&=\tfrac\pi2
,\\
\angle GOC_0&=\tfrac\phi2=\tfrac\pi{n}
,\\
\angle FP_1O&=\tfrac\pi2-\tfrac\phi2
,\\
\angle FC_0P_1&=\tfrac\theta2
,\\
\angle C_0P_1F&=\tfrac\pi2-\tfrac\theta2
,\\
\angle GP_1C_0&
=\tfrac\phi2+\tfrac\theta2
=\tfrac\pi{n}+\tfrac\theta2
,\\
|C_0G|&=R\sin\tfrac\phi2=R\sin\tfrac\pi{n}
,
\end{align}
\begin{align}
\sin\angle GP_1C_0&=
\sin(\tfrac\pi{n}+\tfrac\theta2)
=\tfrac{|C_0G|}{r}
=\tfrac{R}r\,\sin\tfrac\pi{n}
,\\
\tfrac\theta2&=
\arcsin\left(\tfrac{R}r\,\sin\tfrac\pi{n}\right)-\tfrac\pi{n}
,\\
\sin\tfrac\theta2&=
\tfrac{R}{r}\,\sin\tfrac\pi{n}\,
\left(
\cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}}
\right)
,\\
r_a&=\frac{r}{\sin\tfrac\pi{n}}\,\sin\tfrac\theta2
\\
&=
R\,
\left(
\cos\tfrac\pi{n}-\sqrt{\tfrac{r^2}{R^2}-\sin^2\tfrac\pi{n}}
\right)
.
\end{align}
Finally,
\begin{align}
S(n,R,r)&=
\pi\,r^2+
n\cdot \left(
R^2\sin(\tfrac\pi{n})\,
\left(
\cos(\tfrac\pi{n})-\sqrt{\tfrac{r^2}{R^2}-\sin^2(\tfrac\pi{n})}
\right)
-r^2\,
\arcsin\left(\tfrac{R}r\,\sin(\tfrac\pi{n})\right)
\right)
.
\end{align}
Edit
An example with (more-or-less) nice expression for the area.
Let $n=6$, $R=\sqrt3$, $r=1$.
Then
\begin{align}
r_a&=\sqrt3\left(\tfrac{\sqrt3}2-\sqrt{\tfrac13-\tfrac14} \right)
=\tfrac32-\sqrt{1-\tfrac34}=1
,\\
\tfrac\theta2&=\arcsin(\tfrac{\sqrt3}2)-\tfrac\pi6=\tfrac\pi6
,\\
S(6,\sqrt3,1)
&=
\pi+6\cdot
\left(
\tfrac32\,
\left(
\tfrac{\sqrt3}2
-\sqrt{\tfrac{1}{3}-\tfrac{1}{4}}
\right)
-1\cdot
\arcsin\tfrac{\sqrt3}2
\right)
\\
&=
\pi+
9\,
\left(
\tfrac{\sqrt3}2
-\tfrac{\sqrt3}6
\right)
-6\cdot
\tfrac\pi3
\\
&=
3\sqrt3-\pi\approx 2.05
.
\end{align}
In the image below four grid cells represent one square unit:
|
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|
How and the best way to resolve this matrix equation I have the information that: distribution of probabilities is
$$ P_{X} =\{\frac{11}{24},\frac{7}{24},\frac{1}{8},\frac{1}{8}\}$$
And the matrix has the form;
$$A= \left[
\begin{array}{cccc}
p_{1}&p_{2}&p_{2}&p_{4}\\
p_{2}&p_{1}&p_{2}&p_{5}\\
p_{3}&p_{3}&p_{2}&p_{5}\\
p_{3}&p_{3}&p_{2}&p_{5}\\
\end{array}
\right] $$
I have resolve:
$$A\cdot P_{X}=P_{X}$$
I can see that the sums of the columns is $1$
$$4p_{2}=1 \rightarrow p_{2}=\frac{1}{4}$$
Also we can see row 4 and row 5 are equal
so
$$ \left[
\begin{array}{cccc}
p_{1}&\frac{1}{4}&\frac{1}{4}&p_{4}\\
\frac{1}{4}&p_{1}&\frac{1}{4}&p_{5}\\
p_{3}&p_{3}&\frac{1}{4}&p_{5}\\
p_{3}&p_{3}&\frac{1}{4}&p_{5}\\
\end{array}
\right]
\cdot
\left[
\begin{array}{c}
\frac{11}{24}\\
\frac{7}{24}\\
\frac{1}{8}\\
\frac{1}{8}\\
\end{array}
\right]=\left[
\begin{array}{c}
\frac{11}{24}\\
\frac{7}{24}\\
\frac{1}{8}\\
\frac{1}{8}\\
\end{array}
\right]
$$
multiply except the last row(and add the last equation $=1$):
$$\frac{11}{24}p_{1}+\frac{1}{8}p_{4}=\frac{17}{48}\\
\frac{7}{24}p_{1}+\frac{1}{8}p_{5}=\frac{7}{48}\\
\frac{3}{4}p_{3}+\frac{1}{8}p_{5}=\frac{3}{32}\\
p_{1}+\frac{1}{4}+p_{3}+p_{4}+p_{5}=1$$
How can I continue and solve this?
Thank you very much
|
Hints: write the system as a matrix equation, say $Ax=b$, where $x=(p_1,\cdots,p_5)^t$ is a column vector of variables, $b$ is a constant column vector, consider the argumented matrix $B=[A \ b].$\begin{bmatrix}
1&0&1&1&1&\frac{3}{4}\\
\frac{11}{24}&0&0&\frac{1}{8}&0&\frac{17}{48}\\
\frac{7}{24}&0&0&0&\frac{1}{8}&\frac{3}{32}\\
0&0&\frac{3}{4}&0&\frac{1}{8}&\frac{3}{4}
\end{bmatrix}
For above matrix $B$, by a successive elementary transformation of rows , simplifying $B$ to get an upper-triangular matrix, then solving the last equation to get solutions and iterating step by step
|
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|
Find the condition that one of the lines $ax^2+2hxy+by^2=0$ Find the condition that one of the lines $ax^2+2hxy+by^2=0$ may coincide with one of the lines $a_1x^2+2h_1xy+b_1y^2=0$.
My Attempt:
Here,
$$ax^2+2hxy+by^2=0$$
$$(\dfrac {y}{x})^{2}+\dfrac {2h}{b}.(\dfrac {y}{x})+\dfrac {a}{b}$$
Let $y=mx$ be a line represented by above equation:
Also,
$$a_1x^2+2h_1xy+b_1y^2=0$$
$$(\dfrac {y}{x})^2 + \dfrac {2h_1}{b_1} (\dfrac {y}{x})+\dfrac {a_1}{b_1}=0$$
And, let $y=m_1x$ be a line represented by the above equation.
How do I move further?
|
Given lines are $ax^2+2hxy+by^2=0$ and $a_1x^2+2h_1xy+b_1y^2=0$
Let $y=mx$ and then we get,
First let us consider the equation $ax^2+2hxy+by^2=0$
$$ax^2+2hx(mx)+bm^2x^2=0$$
$$x^2(a+2hm+m^2b)=0$$
$$m^2b+2hm+a=0........(1)$$
Now consider the equation $a_1x^2+2h_1xy+b_1y^2=0$
$$a_1x^2+2h_1x(mx)+b_1m^2x^2=0$$
$$x^2(m^2b_1+2h_1m+a_1)=0$$
$$m^2b_1+2h_1m+a_1=0........(2)$$
Now from $(1)$ and $(2)$ we get
$$\frac{m^2}{2ha_1-2ah_1}=\frac{m}{ab_1-a_1b}=\frac{1}{2bh_1-2hb_1}$$
$$m^2=\frac{ha_1-h_1a}{bh_1-hb_1}\mbox{ and }m=\frac{ab_1-a_1b}{2bh_1-2hb_1}$$
$$\left[\frac{ab_1-a_1b}{2bh_1-2hb_1}\right]^2=\frac{ha_1-h_1a}{bh_1-hb_1}$$
$$\frac{(ab_1-a_1b)^2}{4(bh_1-hb_1)^2}=\frac{ha_1-h_1a}{bh_1-hb_1}$$
$$(ab_1-a_1b)^2=4(ha_1-h_1a)(bh_1-hb_1)$$
|
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If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that
$$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$
My attempt:
$a$, $b$, $c$ are in arithmetic progression, so
$$\begin{align}
2b&=a+c \\[4pt]
2\sin B &= \sin A+ \sin C \\[4pt]
2\sin(A+C) &=2\sin\frac {A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\sin\frac{A+C}{2}\;\cos\frac{A+C}{2}&=\sin\frac{A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\cos\frac{A+C}{2}&=\cos\frac{A-C}{2}
\end{align}$$
|
Hint:
$$\dfrac21=\dfrac{\cos\dfrac{A-C}2}{\cos\dfrac{A+C}2}$$
Apply Componendo and Dividendo
$$\dfrac{2+1}{2-1}=?$$
|
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|
How can I do the Euclidian's algorithm and the Extended Euclidean algorithm at the same time? This is what my lecture notes have but I cannot find anything like it online and there is no explanation in the notes. The example given is for 903 and 444.
Thank you.
|
I like to write these as (simple) continued fractions.
$$ \gcd( 903, 444 ) = ??? $$
$$ \frac{ 903 }{ 444 } = 2 + \frac{ 15 }{ 444 } $$
$$ \frac{ 444 }{ 15 } = 29 + \frac{ 9 }{ 15 } $$
$$ \frac{ 15 }{ 9 } = 1 + \frac{ 6 }{ 9 } $$
$$ \frac{ 9 }{ 6 } = 1 + \frac{ 3 }{ 6 } $$
$$ \frac{ 6 }{ 3 } = 2 + \frac{ 0 }{ 3 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccc}
& & 2 & & 29 & & 1 & & 1 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 2 }{ 1 } & & \frac{ 59 }{ 29 } & & \frac{ 61 }{ 30 } & & \frac{ 120 }{ 59 } & & \frac{ 301 }{ 148 }
\end{array}
$$
$$ $$
$$ 301 \cdot 59 - 148 \cdot 120 = -1 $$
$$ \gcd( 903, 444 ) = 3 $$
$$ 903 \cdot 59 - 444 \cdot 120 = -3 $$
|
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|
How can I evaluate $\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$? I'm here wondering if this integral that our math teacher gave us (students) is even possible to evaluate? I just started to study real analysis so I find this very disturbing. Here you go, and if anyone has any idea I will be very grateful.
$$\lim_{n \rightarrow \infty} \int_n^\infty \frac{n^2 \arctan {\frac{1}{x}}}{x^2+n^2}\ dx$$
|
Sorry for overlooking some constant, the following is for the integral $\displaystyle\int_{n}^{\infty}\dfrac{n\tan^{-1}(1/x)}{x^{2}+n^{2}}dx$, the integral in question will be addressed in the second part.
By change of variable $u=x/n$, then the integral is
\begin{align*}
\int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}du.
\end{align*}
But
\begin{align*}
\int_{1}^{\infty}\dfrac{1}{1+u^{2}}du<\infty
\end{align*}
and
\begin{align*}
\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}\leq\dfrac{\tan^{-1}1}{1+u^{2}}
\end{align*}
for all $u\geq 1$, $n\geq 1$, use Lebesgue Dominated Convergence Theorem to conclude that the limit is zero.
It seems that there is an elementary answer.
\begin{align*}
&\lim_{n\rightarrow\infty}\int_{1}^{\infty}\dfrac{\tan^{-1}(1/(nu))}{1+u^{2}}du\\
&=\lim_{n\rightarrow\infty}\tan^{-1}u\tan^{-1}(1/(nu))\bigg|_{u=1}^{u=\infty}+\lim_{n\rightarrow\infty}\int_{1}^{\infty}\dfrac{\tan^{-1}u}{1+(1/(nu))^{2}}\dfrac{1}{nu^{2}}du\\
&=\lim_{n\rightarrow\infty}n\int_{1}^{\infty}\dfrac{\tan^{-1}u}{n^{2}u^{2}+1}du,
\end{align*}
but
\begin{align*}
n\int_{1}^{\infty}\dfrac{1}{n^{2}u^{2}+1}du&=\int_{n}^{\infty}\dfrac{1}{v^{2}+1}dv\\
&=\dfrac{\pi}{2}-\tan^{-1}n\\
&\rightarrow 0.
\end{align*}
However,
\begin{align*}
n\int_{1}^{\infty}\dfrac{\tan^{-1}u}{n^{2}u^{2}+1}du\leq\dfrac{\pi}{2}\cdot n\int_{1}^{\infty}\dfrac{1}{n^{2}u^{2}+1}du.
\end{align*}
For the integral in question:
So the integral is $\displaystyle\int_{1}^{\infty}\dfrac{n\tan^{-1}(1/(nu))}{1+u^{2}}$, and we have $n\tan^{-1}(1/(nu))\rightarrow(1/u)$ by L'Hopital rule or some sort, and note that $\tan^{-1}v$ behaves like $v$ for small $|v|$.
And in fact, one deduces using Taylor formula or some sort that $n\tan^{-1}(1/(nu))\leq 1/u$, and that
\begin{align*}
\int_{1}^{\infty}\dfrac{1}{u(1+u^{2})}\leq\int_{1}^{\infty}\dfrac{1}{1+u^{2}}du<\infty,
\end{align*}
so Lebesgue Dominated Convergence Theorem implies that the limit in question is then
\begin{align*}
\int_{1}^{\infty}\dfrac{1}{u(1+u^{2})}du&=\int_{1}^{\infty}\left(\dfrac{1}{u}-\dfrac{1}{1+u^{2}}\right)du\\
&=\lim_{M\rightarrow\infty}\left(\log u\bigg|_{u=1}^{u=M}-\dfrac{1}{2}\log(1+u^{2})\bigg|_{u=1}^{u=M}\right)\\
&=\lim_{M\rightarrow\infty}\left(\dfrac{1}{2}\log 2+\log\dfrac{M}{\sqrt{1+M^{2}}}\right)\\
&=\dfrac{1}{2}\log 2+\log 1\\
&=\dfrac{1}{2}\log 2.
\end{align*}
|
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|
Sum of the first n terms of series $\frac{x^{3n}}{3n(3n-1)(3n-2)}$ I need to find sum of first n terms of series $\sum_{1}^{\infty} {\frac{x^{3n}}{3n(3n-1)(3n-2)}}$. I tried but I just don't know how to transform it into any known form of power series.
EDIT: I tried partial fraction decomposition and telescoping, something cancels but not for every $n$.
|
$$ \frac{1}{3n(3n-1)(3n-2)} = \frac{1}{2}\left[\frac{1}{3n-2}-\frac{2}{3n-1}+\frac{1}{3n}\right] = \frac{1}{2}\int_{0}^{1}\left(z^{3n-3}-2 z^{3n-2}+z^{3n-1}\right)\,dz $$
hence
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)} = \frac{1}{2}\int_{0}^{1}\sum_{n\geq 1}x^{3n} z^{3n-3}(1-z)^2\,dz=\frac{1}{2}\int_{0}^{1}\frac{x^3(1-z)^2}{1-x^3 z^3}\,dz. $$
On the other hand $\sum_{n\geq 1}\frac{x^n}{n}=-\log(1-x)$, hence
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n} = -\frac{1}{3}\log(1-x^3) $$
and by letting $\omega=\exp\left(\frac{2\pi i}{3}\right)$,
$$ \sum_{n\equiv 1\!\!\pmod{3}}\frac{x^n}{n}=-\frac{1}{3}\left[\log(1-x)+\omega^2 \log(1-\omega x)+\omega \log(1-\omega^2 x)\right]$$
$$ \sum_{n\equiv 2\!\!\pmod{3}}\frac{x^n}{n}=-\frac{1}{3}\left[\log(1-x)+\omega \log(1-\omega x)+\omega^2 \log(1-\omega^2 x)\right]$$
due to the discrete Fourier transform. In particular
$$ \sum_{n\geq 1}\frac{x^{3n}}{3n(3n-1)(3n-2)}=-\frac{1}{6}\left[(x^2-2x+1)\log(1-x)+(\omega^2 x^2-2\omega x)\log(1-\omega x)+(\omega x^2-2\omega^2 x) \log(1-\omega^2 x)\right]. $$
|
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|
$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ is convergent? Is $\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....\forall x \in \mathbb{R}$ convergent? If it is convergent find the sum of the series.
Gives series $$\sin x-\frac{1}{2}\sin ^2x+\frac{1}{4}\sin ^3x-\frac{1}{8}\sin ^4x....=\sum _{n=1}^\infty \frac{(-1)^{n+1}}{2^{n-1}}\sin^n x$$
$$\left|\sum _{n=1}^\infty \frac{(-1)^{n+1}}{2^{n-1}}\sin^n x\right|\le \sum _{n=1}^\infty \frac{1}{2^{n-1}}=2$$ so is it convergent?
|
It converges to the limit $\frac{\sin x}{1+\frac{1}{2}\sin x}$ ;$a=\sin x$ and $r$(common ratio)$=-\frac{1}{2}\sin x.$ Since $|a|<1$.
|
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|
How to find $y^{(y^2-6)}$? $$\frac{3}{1-3^{x-2}} + \frac{3}{1-3^{2-x}} = y$$ $$x≠2$$ $$y^{(y^2-6)} = ?$$What is $y^{(y^2-6)}$? Could you please explain to me how to solve this question step by step?
|
Calling $z = 3^{x-2}$ we have
$$
3\left(\frac{1}{1-z}+\frac{1}{1-z^{-1}}\right) = y \Rightarrow y = 3
$$
so finally
$$y^{y^2-6} = 3^{9-6} = 27$$
|
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|
Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in advanced.
|
Hint:
$$\frac{(5+6)+5^2}{3^1}=12$$
$$\frac{(5+6)(5^2+6^2)+5^4}{3^2}=144=12^2$$
$$\frac{(5+6)(5^2+6^2)(5^4+6^4)+5^8}{3^4}=20736=12^4$$
There is a pattern there. See if you can prove that the pattern continues.
One way is to generalize how something like $(5+6)(5^2+6^2)+5^4$ squares to become equal to $(5+6)(5^2+6^2)(5^4+6^4)+5^8$. [Because then just square both sides of one of the above equations to get to the next equation.]
As in:
$$
\begin{align}
\left((5+6)(5^2+6^2)+5^4\right)^2&=(5+6)^2(5^2+6^2)^2+2(5+6)(5^2+6^2)5^4+5^8\\
&=(5+6)(5^2+6^2)\left[(5+6)(5^2+6^2)+2(5^4)\right]+5^8
\end{align}
$$
So now it is about generalizing that something like $(5+6)(5^2+6^2)+2(5^4)$ is the same as $(5^4+6^4)$. I think writing all the $6$'s as $5+1$ and using the binomial theorem might get you there.
|
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|
Why two different results to surface areal calculation using Guldin? Guldin's first rule, also known as Pappus first centroid theorem states that a linear figure creates an area when rotated that is the product of the distance the centroid is moved and the length of the figure.
I should therefore get the same resulting area when I rotate a triangle, as when I rotate the three segments making up the sides of the triangle.
Thus, I must do something wrong when I do not get that. Here are my calculations, provided triangle $ABC$, where $A=(1,1)$, $B=(1,4)$ og $C=(4,3)$.
In procedure 1, I shall calculate the contribution from each segment, then add.
In procedure 2, I shal find the centroid and the perimeter and "directly" calculate the surface area of the volume of rotation.
Procedure 1:
$r_{AB}=1,L_{AB}=3,A_{AB}=2\cdot\pi\cdot r_{AB}\cdot L_{AB}=6\pi$.
$r_{BC}=\frac{5}{2},L_{BC}=\sqrt{(4-1)^2+(3-4)^2}=\sqrt{10},A_{BC}=2\cdot\pi\cdot r_{BC}\cdot L_{BC}=5\pi\sqrt{10}$.
$r_{AC}=\frac{5}{2},L_{AC}=\sqrt{(4-1)^2+(3-1)^2}=\sqrt{13},A_{AC}=2\cdot\pi\cdot r_{AC}\cdot L_{AC}=5\pi\sqrt{13}$.
Consequently, the total surface area amounts to $A_1=A_{AB}+A_{BC}+A_{AC}=\pi\left(6+5\sqrt{10}+5\sqrt{13}\right)$.
Procedure 2:
Triangle $ABC$ perimeter $L=L_{AB}+L_{BC}+L_{AC}=3+\sqrt{10}+\sqrt{13}$.
Radius of rotation for triangle $ABC$: $r=2$.
Total surface area: $A_2=2\pi r L=4\pi\left(3+\sqrt{10}+\sqrt{13}\right)$.
I expected $A_1=A_2$, but I get $A_1=A_2+\pi\left(-6+\sqrt{10}+\sqrt{13}\right)$.
|
The two plane figures below are not the same:
In partricular, their centroids are not in the same spot.
The centroid of the filled-in triangle at right is at (3, 3) as you found, but the centroid of the empty triangle at left is not.
The centroid of the empty triangle is at the weighted mean of the midpoints of the line segments.
Midpoints: $(1, 2.5), (2.5, 3.5), (2.5, 2)$
Weights: $3, \sqrt{10}, \sqrt{13}$
We only care about the $x$-coordinate of the centroid, though:
$\overline{x} = \frac{1 \cdot 3 + 2.5 \cdot \sqrt{10} + 2.5 \cdot \sqrt{13}}{3 + \sqrt{10} + \sqrt{13}}$
Multiply numerator and denominator by 2:
$\overline{x} = \frac{6 + 5 \sqrt{10} + 5 \sqrt{13}}{2\left(3 + \sqrt{10} + \sqrt{13}\right)}$
So the surface area calculated this way is $A_2 = 2 \pi r L = 2 \pi \overline{x} L$, which is
$A_2 = 2\pi\overline{x}L = 2\pi \frac{6 + 5 \sqrt{10} + 5 \sqrt{13}}{2\left(3 + \sqrt{10} + \sqrt{13}\right)} \left(3 + \sqrt{10} + \sqrt{13}\right)$
$A_2 = \pi \left(6 + 5 \sqrt{10} + 5 \sqrt{13}\right)$
, as desired.
|
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|
How do I factorise $r^4+r^2+1$? How do I factorise $r^4+r^2+1$ ?
$(r^2+r+1)(r^2-r+1)$ gives $r^4+r^2+1$
But how to split it into these factors?
I generally find roots and then write the factors, but $r^4+r^2+1$ seems to have no real root.
Thanks!
|
$$x^4+x^2+1=\frac{x^6-1}{x^2-1}=\frac{(x^3-1)(x^3+1)}{(x-1)(x+1)}
=(x^2+x+1)(x^2-x+1).$$
|
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|
An inequality involving three consecutive primes Can you provide a proof or a counterexample to the following claim :
Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .
I have tested this claim up to $10^{10}$ .
For $p>5$ we get $\pi(2p)-\pi(p) \ge 2$ , a result by Ramanujan .
This means that $q<2p$ and $r<2p$ , so $\frac{1}{2p}<\frac{1}{q}$ and $\frac{1}{2p}<\frac{1}{r}$ which implies $\frac{1}{p} < \frac{1}{q} + \frac{1}{r}$ . If we square both sides of inequality we get $\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}{qr} + \frac{1}{r^2}$ . Now , I don't know how to rule out term $\frac{2}{qr}$ .
|
The inequality holds for all $p$ large enough. Let $a>1$ be such that $a^{-2}+a^{-4}=1$ and $p_n$ be the $n$-th prime. By the Prime Number Theorem there is an $N$ such that $p_{n+1}<a\,p_n$ for all $n\ge N$.If $p\ge p_N$, then $q<a\,p$ and $r<a\,q<a^2\,p$ and
$$
\frac{1}{q^2}+\frac{1}{r^2}>\frac{1}{a^2\,p^2}+\frac{1}{a^4\,p^2}=\frac{1}{p^2}.
$$
|
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|
If $x$, $y$, and $z$ are real numbers such that $x+y+z=8$ and $x^2+y^2+z^2=32$, what is the largest possible value of $z$? I tried swapping $z$ from the first equation to the second, and got $$x^2 + x y - 8 x + y^2 - 8 y + 16=0$$ Not sure where to go from there, and if I'm on the right track at all.
|
$$\max z$$
subject to $$x+y+z=8$$
$$x^2+y^2+z^2=32$$
The Lagrangian is $$z-\lambda(8-x-y-z)-\mu(32-x^2-y^2-z^2)$$
Differentiating wrt x:
$$8\lambda+2\mu x=0\tag{1}$$
Differentiating wrt $y$:
$$8\lambda+2\mu y=0\tag{2}$$
Differentiating wrt $z$:
$$1+8\lambda+2\mu z=0\tag{3}$$
If $\mu=0$, equation $(1)$ and $(2)$ gives us $\lambda=0$ and we get a contradiction in equation $(3)$.
Hence $\mu \ne 0$. From equation $(1)$ and $(2)$, we get $x=y$.
$$2x+z=8$$
$$2x^2+z^2=32$$
$$2\left( \frac{8-z}2\right)^2+z^2=32$$
$$\left( 8-z\right)^2+2z^2=64$$
$$3z^2-16z=0$$
$$z(3z-16)=0$$
Hence maximum $z$ occurs at $\frac{16}3$
|
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|
Maximum number of circles tangent to two concentric circles In a recent math contest, the following question arose:
Two concentric circles of radii 1 and 9 make a ring. In the interior of this ring $n$ circles are drawn without overlapping, each being tangent to both of the circles of the ring. What is the largest possible value for $n$?
I solved it like this:
The radius of each of the small circles must be $(9-1)/2=4$. I connected the centres of two of the small circles together, and also to the centre of the large circle. Let the central angle be $\theta$. The triangle formed has side lengths $4+1=5$, $4+1=5$, and $4+4=8$. This can be split in two to get two 3-4-5 triangles. Since these triangles are right, we can solve for $\theta$:
$$
\begin{align}
\sin\frac{\theta}{2}&=\frac{4}{5}\\
\frac{\theta}{2}&=\arcsin{\frac{4}{5}}\\
\theta&=2\arcsin\frac{4}{5}.
\end{align}
$$
Now the answer to the question is simply $\left\lfloor\frac{2\pi}{\theta}\right\rfloor=3$.
However, this contest was a no calculator contest, and thus I was not able to compute $\arcsin(4/5)$ for the answer. How do you solve this question without a calculator?
|
Note that $\displaystyle \frac{\sqrt{2}}{2}<\frac{4}{5}<\frac{\sqrt{3}}{2}$. We have $\displaystyle \sin45^\circ<\sin\frac{\theta}{2}<\sin60^\circ$.
$90^\circ<\theta<120^\circ$.
Hence $\displaystyle 4>\frac{360^\circ}{\theta}>3$.
|
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|
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ to $A(y+1)^2$, $B(y+1)^1$ and $C$.
My question is why do we have to substitute $x-1=y$ and why can't equate coefficients of $x^2$, $x^1$ and $x^0$ to $A$, $B$ and $C$ without substituting? Thanks in advance.
|
Let the quotient be $Q(x)$ and the remainder be $ax^2+bx+c$. Then
$$(x+1)^n=(x-1)^3Q(x)+ax^2+bx+c$$
If we put $y=x-1$, then we have
$$(y+2)^n=y^nQ(y+1)+a(y+1)^2+b(y+1)+c$$
We have
$$(y+2)^n=2^n+\binom{n}{1}2^{n-1}y+\binom{n}{2}2^{n-2}y^2+\textrm{terms involving higher powers of }y$$
So, we have
$$a(y+1)^2+b(y+1)+c=2^n+\binom{n}{1}2^{n-1}y+\binom{n}{2}2^{n-2}y^2=2^n+2^{n-1}ny+2^{n-3}n(n-1)y^2$$
$a$, $b$ and $c$ can be found by comparing the coefficients.
|
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|
Show that the solutions from one quadratic equation are reciprocal to the solutions of another quadratic equation From Sullivan's Algebra & Trigonometry book: Chapter 1.2; Exercise 116:
Show that the real solutions of the equation $ax^2+bx+c=0$ are the
reciprocals of the real solutions of the equation $cx^2+bx+a=0$.
Assume that $b^2-4ac\geq0$.
The most I reached in my attempt to solve the problem was to prove that, as $ax^2+bx+c=0$ can be expressed as $a^2x^2+abx+b^2$, and that $cx^2+bx+a=0$ can be expressed as $b^2x^2+abx+a^2$; thus, the solutions from the one will be the reciprocal of the another...
$a^2x^2+abx+b^2=(ax+b)^2$
$(ax+b)^2= 0$
$x_1 = -\frac{b}{a}$ (root of multiplicity two)
$b^2x^2+abx+a^2=(bx+a)^2$
$(bx+a)^2=0$
$x_2 = -\frac{a}{b}$ (root of multiplicity two)
$x_1x_2=0$
The problem here is that the solution only works for perfect squares; I want to know, and that's what the problem is asking for, to prove it for the $ax^2+bx+c$ form.
Another attempts to solve the problem...
I've tried to solve it using the quadratic formula, I started from the equation $\frac{2a}{-b+\sqrt{b^2-4ac}}=0$ and then try to get to the expression $\frac{-b+\sqrt{b^2-4ac}}{2c}=0$, but I wasn't cappable, don't even know it it's possible to do or correct to try.
I also tried the same from above with the reciprocal of the actual equation: $\frac{1}{ax^2+bx+c}=0$. But for me it was also impossible to do anything with that expression.
|
To ensure that the equations are quadratic, we need the assumption that $a,c\ne0$.
If $\alpha$ is a root of $ax^2+bx+c=0$, then $a\alpha^2+b\alpha+c=0$.
Note that $\alpha\ne0$, otherwise, $c=0$.
So $\displaystyle c\left(\frac{1}{\alpha}\right)^2+b\left(\frac{1}{\alpha}\right)+a=\frac{a\alpha^2+b\alpha+c}{\alpha^2}=0$.
$\displaystyle \frac{1}{\alpha}$ is a root of $cx^2+bx+a=0$.
|
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|
Find greatest value of $a^2+b^2$ If $f(x)=x^3+3x^2+4x+ a \sin x + b\cos x ~ \forall x \in \mathbb{R}$ is an injection then the greatest value of $a^2+b^2$ is _______?
To ensure injection, we must ensure that there is no maxima/minima in any interval which is equivalent to $f'(x)\neq 0$.
Note that $f'(x)=3x^2+6x+4+a \cos x - b \sin x \neq 0$. It can be observed that $3x^2+6x+4>0$ with its minimum value being $1$.
So, our condition can be reduced to $a \cos x - b \sin x > -1$.
Again, $a \cos x - b \sin x +1$ can be written as $\frac{a}{\sqrt {a^2+b^2}} \cos (-x) - \frac{b}{\sqrt{a^2+b^2}} \sin (-x) +1$ such that $\sin (\theta -x) + \frac{1}{\sqrt {a^2+b^2}} >0$.
I couldn't proceed anymore!
|
Introducing $t=x+1$, rewrite the requirement on $f'(x)$ as:
$$
3t^2+1+A\cos(t+\phi) \ge 0.
$$
To maximize the value of $A=\sqrt{a^2+b^2}$ the expression should attend its minimal value as far from $t=0$ as possible, which corresponds to the choice $\phi=0$.
In this case for sufficiently large $A$ two symmetric minima on both sides of $t=0$ will appear. If $A$ is increased further the minima become negative.
The critical values of the parameters $A$ and $t$ correspond to the case when the function value in the minima is 0:
$$\begin{cases}
3t^2+1+A\cos t&=0\\
6t-A\sin t&=0
\end{cases}.
$$
For the critical value of $t$ one obtains then the equation:
$$
(3t^2+1)\sin t+6t\cos t=0,
$$
which minimal positive solution is:
$$
t_*\approx2.49037.
$$
The corresponding value of $A$ is:
$$
A_*=\frac{6t_*}{\sin t_*}\approx24.6508.
$$
Its square gives the greatest possible value of $a^2+b^2$.
|
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|
$f(x) = \frac{x^3}{6}+\frac{1}{2x}$, $\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$ $$f(x) = \frac{x^3}{6}+\frac{1}{2x}$$
$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$
Let's start by deriving the function, we have
$$f'(x) = \dfrac{x^4-1}{2x^2}$$
Hence we get
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = ?$$
Am I right?
UPDATE:
If we have a definite integral:
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = \dfrac{x^4-3}{6x}+C$$
Then
$$\dfrac{x^4-3}{6x}+C = \boxed{\frac{14}{3}}$$
|
Hint: Note that $$1+\left(\frac{x^4-1}{2x^2}\right)^2=\frac{(x^4+1)^2}{4x^4}$$
|
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|
Proving inequalities (1)
Let $a, b, c > 0$ $;abc=8$ ,
$$\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}
+ \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}
+ \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \ge \frac43
$$
(ref: original image)
I tried using AM-GM to eliminate the square root on the bottom but I am stuck, what is the general strategy here and maybe I can use its symmetry?
|
By AM-GM $$1+a^3=(1+a)(1-a+a^2)\leq\left(\frac{1+a+1-a+a^2}{2}\right)^2=\frac{(2+a^2)^2}{4}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{a^2}{(2+a^2)(2+b^2)}\geq\frac{1}{3}$$ or
$$3\sum_{cyc}a^2(2+c^2)\geq\prod_{cyc}(2+a^2)$$ or
$$\sum_{cyc}(a^2b^2+2a^2)\geq72,$$ which is true by AM-GM:
$$\sum_{cyc}(a^2b^2+2a^2)\geq3\sqrt[3]{a^4b^4c^4}+6\sqrt[3]{a^2b^2c^2}=3\cdot16+6\cdot4=72.$$
Done!
|
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|
Proof of Motzkin numbers recurrence I know the Motzkin numbers are given as $$M_n=\sum_{k=0}^{\lfloor n/2\rfloor}{n \choose 2k}C_k,$$ where $$C_k=\frac{1}{1+k}{2k\choose k}.$$
The recurrence relation is given as $M_n=M_{n-1}+\sum_{k=0}^{n-2}M_kM_{n-2-k}$. Is there a direct way to prove this recurrence using either Pascal's identity or summation by parts?
|
One possible approach starts from
$$M_n = \sum_{k=0}^{\lfloor n/2\rfloor} {n\choose 2k} C_k.$$
This is
$$\sum_{k=0}^{\lfloor n/2\rfloor} {n\choose n-2k} C_k
\\ = \sum_{k=0}^{\lfloor n/2\rfloor} C_k [z^{n-2k}] (1+z)^n C_k
= [z^n] (1+z)^n \sum_{k=0}^{\lfloor n/2\rfloor} C_k z^{2k}.$$
Now when $2k\gt n$ there is no contribution to the coefficient
extractor and hence we may extend the sum to infinity:
$$[z^n] (1+z)^n \sum_{k\ge 0} C_k z^{2k}.$$
Using the generating function of the Catalan numbers this becomes
$$[z^n] (1+z)^n \frac{1-\sqrt{1-4z^2}}{2z^2}$$
which is
$$\mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
(1+z)^n \frac{1-\sqrt{1-4z^2}}{2z^2}.$$
With $z/(1+z) = w$ we have $z = w/(1-w)$ and $dz = 1/(1-w)^2 \; dw$
and obtain
$$\mathrm{Res}_{w=0}
\frac{1}{w^n} \frac{(1-w)^3}{2w^3} (1-\sqrt{1-4w^2/(1-w)^2})
\frac{1}{(1-w)^2}
\\ = \mathrm{Res}_{w=0}
\frac{1}{w^{n+1}} \frac{1-w-\sqrt{1-2w-3w^2}}{2w^2}.$$
This shows that
$$M_n = [w^n] \frac{1-w-\sqrt{1-2w-3w^2}}{2w^2},$$
which is also the LHS of the recurrence. We get for the RHS
$$[w^{n-1}] \frac{1-w-\sqrt{1-2w-3w^2}}{2w^2}
+ [w^{n-2}] \frac{(1-w-\sqrt{1-2w-3w^2})^2}{4w^4}
\\ = [w^{n}] \frac{2w-2w^2-2w\sqrt{1-2w-3w^2}}{4w^2}
+ [w^{n}] \frac{(1-w-\sqrt{1-2w-3w^2})^2}{4w^2}.$$
The numerator is
$$2w-2w^2-2w\sqrt{1-2w-3w^2} + 2 - 4w - 2w^2
- 2(1-w)\sqrt{1-2w-3w^2}
\\ = 2 - 2w - 4w^2 - 2 \sqrt{1-2w-3w^2}.$$
We therefore conclude that the RHS is
$$[w^n] \left(-1 + \frac{1-w-\sqrt{1-2w-3w^2}}{2w^2}\right),$$
which is the same as the LHS when $n\ge 2,$ as claimed.
|
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|
Order of accuracy for trapezoidal integration I have a question regarding the order of accuracy for certain function using the trapezoidal formula.
I know that from theory the formula is second order accurate, but when working with matlab I get different answers.
I calculated the order of accuracy by plotting the logarithm of the error vs the logarithm of the step size $H$.
When using functions which are infinitely differentiable, for example $\sin(x)$ or $e^x$, I get a slope of $-2$, as expected.
But when using the formula $\arctan(\sqrt{x})$ I get a slope of about -1.5.
Why does the order of accuracy depend on the function that is integrated?
|
You can compute the error in the trapeze method by using integration by parts, let $f:(0,1) \to \Bbb R$.
Then the error is the sum of local errors:
$$E=\sum_{k=0}^{n-1} \left(\int_\frac{k}{n}^\frac{k+1}{n}\left(f(t)-\frac{f\left(\frac{k}{n}\right)+f\left(\frac{k+1}{n}\right)}{2} \right) dt \right)$$
integrating by parts:
$$\int_\frac{k}{n}^\frac{k+1}{n}\left(f(t)-\frac{f\left(\frac{k}{n}\right)+f\left(\frac{k+1}{n}\right)}{2} \right) =0-\int_\frac{k}{n}^\frac{k+1}{n}f'(t) \left(t-\frac{k+\frac{1}{2}}{n} \right) dt$$
and integrating by parts again:
$$\int_\frac{k}{n}^\frac{k+1}{n}\left(f(t)-\frac{f\left(\frac{k}{n}\right)+f\left(\frac{k+1}{n}\right)}{2} \right) =0+\int_\frac{k}{n}^\frac{k+1}{n}f''(t) \left(t-\frac{k}{n} \right)\left(t-\frac{k+1}{n} \right) dt$$
If $f''$ is bounded the idea to recover the standard bound is to write:
$$|E|=\left|\sum_{k=0}^{n-1}\int_\frac{k}{n}^\frac{k+1}{n}f''(t) \left(t-\frac{k}{n} \right)\left(t-\frac{k+1}{n} \right) dt \right|\leq \|f''\|_{\infty} \left|\sum_{k=0}^{n-1}\int_\frac{k}{n}^\frac{k+1}{n} \left(t-\frac{k}{n} \right)\left(t-\frac{k+1}{n} \right) dt \right|=\|f''\|_{\infty} n \cdot \frac{1}{6 n^3}$$
Here we can write, as $f''$ is not bounded:
$$E=\int_0^\frac{1}{n}f''(t)\left(t-0\right)\left(t-\frac{1}{n} \right)dt+\sum_{k=1}^{n-1}\int_\frac{k}{n}^\frac{k+1}{n}f''(t) \left(t-\frac{k}{n} \right)\left(t-\frac{k+1}{n} \right) dt$$
so for $f(t)=\sqrt{t}$ (in fact or any function that looks like $\sqrt{t}$ near $0$ and is smooth far from $0$):
$$|E|\leq\int_0^\frac{1}{n} \left|\frac{1}{4t^\frac{3}{2}}t\left(\frac{1}{n} -t\right)\right|dt+\sum_{k=1}^{n-1} \frac{1}{4n^2}\int_\frac{k}{n}^\frac{k+1}{n}|f''(t)| dt$$
so as $f''$ is of constant sign:
$$|E|\leq\frac{1}{3n^\frac{3}{2}}+\frac{1}{4n^2}\int_\frac{1}{n}^1 |f''(t) |dt=\frac{1}{3n^\frac{3}{2}}+\frac{1}{4n^2}\left|f'(1)-f'\left(\frac{1}{n} \right) \right|\leq\frac{1}{3n^\frac{3}{2}}+\frac{1}{4n^2}+\frac{1}{2n^\frac{3}{2}}$$
so the $\frac{3}{2}$ rate of convergence. Note that the problem lies near $t=0$, as for example the part between $0$ and $\frac{1}{n}$ contribute with a $n^\frac{-3}{2}$ in itself.
|
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|
Quick way to determine if a piecewise-defined function is injective/surjective. Let $f(x): \mathbb{R} \to \mathbb{R}$ be the map determined by:
$f(x)=\begin{cases} x &x \ge 2\\
\frac{x^3}{4} &-1\le x < 2\\
x &x < -1\end{cases}$
Is there an easy way to determine if this function is injective and surjective?
Injectivity means that $f(x) = f(y)$, hence for both $x>2$ and $x< -1$, $x = y$. Is the same true for $x^3\over4$?
As for surjectivity, I couldn't figure it out.
|
injectivity: If $x \geq 2$, then $f(x) = x \geq 2$. Moreover, if $f(x_1) = f(x_2)$, then $f(x_1) = x_1 = x_2 = f(x_2)$.
If $x < - 1$, then $f(x) = x < -1$. Moreover, if $f(x_1) = f(x_2)$, then $f(x_1) = x_1 = x_2 = f(x_2)$.
It remains to establish that if $-1 \leq x < 2$, then $-1 \leq x < 2$ and that, if this is the case, then $f(x_1) = f(x_2) \implies x_1 = x_2$.
We will show that $f$ is strictly increasing on this interval. Suppose $x_1, x_2 \in [-1, 2)$ and $x_1 > x_2$. Then
\begin{align*}
f(x_1) - f(x_2) & = \frac{x_1^3}{4} - \frac{x_2^3}{4}\\
& = \frac{1}{4}(x_1^3 - x_2^3)\\
& = \frac{1}{4}(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2)\\
& = \frac{1}{4}(x_1 - x_2)\left(x_1^2 + x_1x_2 + \frac{1}{4}x_2^2 + \frac{3}{4}x_2^2\right)\\
& = \frac{1}{4}(x_1 - x_2)\left[\left(x_1 + \frac{x_2}{2}\right)^2 + \frac{3}{4}x_2^2\right]\\
& > 0
\end{align*}
Thus, $f$ is strictly increasing on $[-1, 2)$. Moreover, if $x \in [-1, 2)$, then
$$f(-1) = \frac{(-1)^3}{4} = -\frac{1}{4} \leq x < \frac{2^3}{4} = 2$$
so, if $x \in [-1, 2)$, then $-1 < f(x) < 2$.
Consequently, $f$ is injective.
surjectivity: Since $f: \mathbb{R} \to \mathbb{R}$, for the function to be surjective, we require that every real number be in the range. We have shown that
\begin{align*}
x \geq 2 & \implies f(x) \geq 2\\
-1 \leq x < 2 & \implies -\frac{1}{4} \leq x < 2\\
x < -1 & \implies f(x) < -1
\end{align*}
Thus, $f(x) \notin [-1, -1/4)$. Hence, $f$ is not surjective.
The results above can be obtained more easily by sketching the function's graph. Since a horizontal line crosses the graph at most once, the function is injective. Since the horizontal line $y = -1/2$ does not cross the graph, the function is not surjective.
|
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|
Integral: $\int\frac{x^2}{\sqrt{4-x^2}} dx$ So, I am really just having one issue with this integral, but I will go through the steps I have taken.
Consider $$\int\frac{x^2}{\sqrt{4-x^2}} dx$$
First, I set $x = 2\sin\theta$, found $dx = 2\cos\theta\space d\theta$ and plugged this back in, making $$4\int \frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}} \cos\theta \space d\theta$$
Using $\sin^2\theta + \cos^2\theta = 1$, I canceled $\sqrt{1-\sin^2\theta}$ with $\cos\theta$, leaving $$4\int\sin^2\theta\space d\theta$$
Using $\sin^2\theta = \frac{1}{2}(1 -\cos2\theta)$, I simplified, ending up with $$4(\frac{\theta}{2} - \frac{\sin 2\theta}{4})$$
When trying to put the solution back in terms of $x$, $\sin 2\theta$ is giving me trouble. Using the relationship from before, I got that $\theta = \sin^{-1}(\frac{x}{2})$. I am able to obviously substitute this back in for $\theta$, but again I am lost what to do with $\sin 2\theta$.
|
Hint
Just use that $$\sin 2\theta=2\sin \theta\cos\theta=2\left(\frac x2\right)\left(\sqrt {1-\frac {x^2}{4}}\right) =\frac {x\sqrt {4-x^2}}{2}$$
|
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|
Infinite sum converging to 2 How do I compute
$$\sum_{r=1}^{\infty} \frac{8r}{4r^4 +1}$$
Calculating first few terms tells me that the sum converges to 2.
I have also tried squeezing the term.
|
Partial fraction expansion gives us$$\begin{align*} & \sum\limits_{r=1}^n\frac {8r}{4r^4+1}=\sum\limits_{r=1}^n\frac 2{2r^2-2r+1}-\sum\limits_{r=1}^n\frac 2{2r^2+2r+1}\\ & =\left[2+\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2-2n+1}\right]-\left[\frac 2{5}+\frac 2{13}+\cdots+\frac 2{2n^2+2n+1}\right]\end{align*}$$Notice how all but the last fraction in the second sum cancels out with the fractions in the first sum. Continuing on indefinitely until the end gives us$$\sum\limits_{r=1}^n\frac {8r}{4r^4+1}=2-\frac 2{2n^2+2n+1}$$As $n\to\infty$, the fraction tends to zero, so your sum equals$$\sum\limits_{r\geq1}\frac {8r}{4r^4+1}=2$$
EDIT: To find the partial fraction decomposition, we first factor the denominator as a product of two quadratics. This can be done by adding and subtracting $4r^2$ so the quartic factors as$$4r^4+1=(2r^2+2r+1)(2r^2-2r+1)$$Now, the decomposition is set up as$$\frac {8r}{(2r^2+2r+1)(2r^2-2r+1)}=\frac {Ar+B}{2r^2-2r+1}+\frac {Cr+D}{2r^2+2r+1}$$
|
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|
Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist? My solution is the following:
approaching by the y-axis:
$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$
approaching by $y=x$
$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$
So I think,that this limit exists.
is it correct in this form?
|
HINT:
$$
\frac{x^3+y^3}{x^2+y^2}=x\frac{x^2}{x^2+y^2}+y\frac{y^2}{x^2+y^2}
$$
But your method doesn't answer the question.
|
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|
Curve of degree four touching the line Let the curve $y=x^4+Ax^3+Bx^2+Cx+D$ touches the line $y=px+q$ at x=2 and x=3 where A, B, C, D, p, q $\in$R. If the area bounded by the curve and the line is $\frac{1}{\alpha}$ then the number of divisors of $\alpha$ are.
My approach was that I used $\frac{dy}{dx}=m=p$
Then I substituted $p=4x^3+3Ax^2+2Bx+C$ at x=2 and x=3
As it touches the line at x=2 and 3 I used the following substitution
$px+q=x^4+Ax^3+Bx^2+Cx+D$ by putting x=2 and x=3 but getting more and more complicated.
Please suggest some short cut
|
\begin{align}
f_1(x)&=x^4+Ax^3+Bx^2+Cx+D
,\\
f_2(x)&=px+q
.
\end{align}
We have a system of four equations
\begin{align}
f_1(2)&=f_2(2)
,\\
f_1(3)&=f_2(3)
,\\
f'_1(2)&=f'_2(2)
,\\
f'_1(3)&=f'_2(3)
,
\end{align}
which allows to express $A,B,C,D$
in terms of given numbers $p,q$
as
\begin{align}
A &= -10,\quad
B = 37,\quad
C = p-60,\quad
D = 36+q.
\end{align}
The area in question is
\begin{align}
\int_2^3(f_1(x)-f_2(x))\,dx
&=\frac{65}4\,A+\frac{19}3\,B
+\frac52\,C+D+\frac{211}5-\frac52\,p-q
\\
&=\frac1{30}
,
\end{align}
hence, $\alpha=30$.
|
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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.