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Prove line connecting intersection of tangents and opposite vertex bisects segment containing intersection of tangents and a vertex Let $\triangle ABC$ be an isosceles triangle with $AB=BC$. Let $\Gamma$ be the circumcircle of $\triangle ABC$. Let the tangents at $A$ and $B$ intersect at $D$, and let $DC\cap\Gamma=E\neq C$. Prove that $AE$ bisects segment $BD$.
This is my progress:
I let $\Gamma_1$ be the circumcircle of $\triangle ADE$, so $AE$ becomes the radical axis of $\Gamma_1,\Gamma$. Now, if I can prove that $\Gamma_1$ is tangent to $BD$ at $D$, I would be done by applying the following well known fact: The radical axis of two intersecting circles (the common chord of the two circles) bisects the common external tangent of the two circles.
However I'm having some trouble proving that the circle is tangent at $D$. I would appreciate any ideas on how to do this.
|
Let $|BA|=|BC|=a$, $|AC|=b$,
$\angle ABC=\beta$,
$\angle CAB=\angle ACB=\alpha=\tfrac\pi2-\tfrac\beta2$.
Since $\triangle AEC \cong\triangle FED$,
let's find the scaling factor,
\begin{align}
\frac{|EC|}{|DE|}
&=
\frac{|DC|-|DE|}{|DE|}
=
\frac{|DC|}{|DE|}-1.
\end{align}
From $\triangle DBC$:
\begin{align}
|DC|^2
&=
|BD|^2+a^2-2a|BD|\cos(\tfrac\pi2+\tfrac\beta2)
=
|BD|^2+a^2+2a|BD|\sin\tfrac\beta2
\end{align}
Also, using
the tangent-secant theorem,
\begin{align}
|DE| \cdot|DC|
&=
|AD|^2=|BD|^2,
\end{align}
hence
\begin{align}
\frac{|DC|}{|DE|}
&=
1+
\left(\frac{2a}{|BD|}\right)^2
+
\frac{a}{|BD|}\sin\tfrac\beta2
\\
\frac{|EC|}{|DE|}
&=
\left(\frac{2a}{|BD|}\right)^2
+
\frac{a}{|BD|}\sin\tfrac\beta2
\quad (1)
\end{align}
Substitution
\begin{align}
\frac{a}{|BD|}
&=
2\sin\tfrac\beta2
\end{align}
into (1) gives
\begin{align}
\frac{|EC|}{|DE|}
&=
8\sin^2\tfrac\beta2
\end{align}
Then
\begin{align}
|FD|
&=
\frac{|DE|}{|EC|}\cdot b
=
\frac{b}{8\sin^2\tfrac\beta2}
\end{align}
Finally
\begin{align}
\frac{|BD|}{|FD|}
&=
\frac{8\sin^2\tfrac\beta2}{b}
\cdot
\frac{a}{2\sin\tfrac\beta2}
=
\frac{2\sin\tfrac\beta2}{
\tfrac{b}2/a
}
=2.
\end{align}
|
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"url": "https://math.stackexchange.com/questions/1380515",
"timestamp": "2023-03-29T00:00:00",
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|
Help with C is Euler's constant and $\Gamma(0)=\infty$ in paper I am referring to a paper by S. Nadarajah & S. Kotz.
The notation is simple enough to understand, however i having trouble with $C$ is Euler’s constant and $\Gamma(0)=\infty$
by equation (2.3) I have
$$F(z)= \displaystyle\lambda\int_{0}^{\infty}y^{-1-1}exp\left(-\frac{\lambda}{y}\right)erfc\left(-\frac{z}{\sqrt{2}\sigma} y\right)~dy - 1$$
by lemma 1(defined $\displaystyle p>0 , \alpha < 0 , \arrowvert \text{arg}(c) \arrowvert < \frac{\pi}{4}$) , then
\begin{align}
\displaystyle F(z) &= \lambda\left[\frac{1}{\lambda} + \frac{2z}{\sqrt{2\pi}\sigma}\Gamma(0)G\left(\frac{1}{2};\frac{3}{2},1,\frac{1}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) + \frac{z}{\sqrt{2\pi}\sigma}\Gamma(0)G\left(\frac{1}{2};\frac{3}{2},\frac{1}{2},1;-\frac{\lambda^2z^2}{8\sigma^2}\right) \right. \\
& \hspace{10mm} \left. + \frac{z^2 \lambda}{4\sqrt{\pi}\sigma^2}\Gamma \left(-\frac{1}{2}\right) G\left(1;2,\frac{3}{2},\frac{3}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) \right]-1
\end{align}
note that $\Gamma(-\frac{1}{2})=-2\sqrt{\pi}$ and $\Gamma(0)=\infty$
so $~~~~~~~~\displaystyle F(z) = \frac{\lambda z}{\sqrt{2}\sigma}\left[\frac{3\Gamma(0)}{\sqrt{\pi}}G\left(\frac{1}{2};\frac{3}{2},1,\frac{1}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right) - \frac{\lambda z}{\sqrt{2}\sigma}G\left(1;2,
\frac{3}{2},\frac{3}{2};-\frac{\lambda^2z^2}{8\sigma^2}\right)\right] $
but the equation is not equal to equation(2.1)
I would be really happy if someone could help me.help please
|
Look at this part of the formula given in Lemma 1:
$$
\begin{multline}
-\frac{2cp^{\alpha+1}}{\sqrt{\pi}}\Gamma(-\alpha-1)G\left(\frac{1}{2},\frac{3}{2},\frac{3+\alpha}{2},1+\frac{\alpha}{2}; -\frac{c^2p^2}{4}\right) \\
+ \frac{1}{c^\alpha\sqrt{\pi}\alpha}\Gamma\left(\frac{\alpha+1}{2}\right) G\left(-\frac{\alpha}{2},1-\frac{\alpha}{2},\frac{1}{2},\frac{1-\alpha}{2}; -\frac{c^2p^2}{4}\right)
\end{multline}
$$
We cannot substitute $\alpha$ for $-1$ in this expression because it would yield two occurences of $\Gamma(0)$. But this gives $-\infty +\infty$ and there possibly is a limit here.
Below is an heuristic derivation of the limit at $\alpha=-1$. Substitute $\alpha$ for $-1$ except in the $\Gamma(\cdot)$:
$$
\begin{multline}
-\frac{2c}{\sqrt{\pi}}\Gamma(-\alpha-1)G\left(\frac{1}{2},\frac{3}{2},1,\frac{1}{2}; -\frac{c^2p^2}{4}\right) \\ -
\frac{c}{\sqrt{\pi}}\Gamma\left(\frac{\alpha+1}{2}\right) G\left(\frac{1}{2},\frac{3}{2},\frac{1}{2},1; -\frac{c^2p^2}{4}\right) \\
= -\frac{c}{\sqrt{\pi}}\left(2\Gamma(-\alpha-1)+\Gamma\left(\frac{\alpha+1}{2}\right) \right)G\left(\frac{1}{2},\frac{3}{2},1,\frac{1}{2}; -\frac{c^2p^2}{4}\right)
\end{multline}
$$
Now, I have not tried to know why, but it seems that $$2\Gamma(-\alpha-1)+\Gamma\left(\frac{\alpha+1}{2}\right) \to -3C \quad \text{when $\alpha \to -1$:}$$
> f <- function(x) 2*gamma(-x-1) + gamma((1+x)/2)
> f(-0.999999999)/3
[1] -0.5772157
> digamma(1) # this is the Euler constant
[1] -0.5772157
|
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|
Find min & max of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$ Problem: Find the maximum and minimal value of $f(x,y) = x + y + x^2 + y^2$ when $x^2 + y^2 = 1$.
Since $x^2 > x$ (edit $x^2 \geq x$) for all $x \in \mathbb{R}$, $f$ is bowl-ish with a minimal value in the bottom.
This is a critical point which means that we can set partial derivatives of $f$ equal to $0$ and try to solve for $x$ and $y$
$\nabla f = (1+2x, 1+2y) = (0,0) \implies (x,y) = (\frac{-1}{2}, \frac{-1}{2})$
So we get the minimal value $f(\frac{-1}{2}, \frac{-1}{2}) = \frac{-1}{2} + \frac{-1}{2} + (\frac{-1}{2})^2 \frac{-1}{2})^2 = -\frac{1}{2}$
But how about the maximal value? How does $x^2 + y^2 = 1$ restrict $f$?
|
Your method to get the minimum gets in the entire plane, but you want the minimum on the unit circle $x^2+y^2=1$.
You can use Lagrange multipliers. You could also parameterize the unit circle on one variable then find the extrema on that variable. Two parametrizations are
$$x=\cos\theta,\quad y=\sin\theta,\quad 0\le\theta<2\pi$$
or
$$y=\pm\sqrt{1-x^2},\quad -1\le x\le 1$$
In your problem it looks like the cosine/sine parametrization would be easier.
|
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"timestamp": "2023-03-29T00:00:00",
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|
10th derivative of a function I want to find $f^{(10)}(0)$ where $f(x)=\ln(2+x^2)$.
I know that it can be done "by hand", but I believe there is a smarter way.
I think I should use Taylor series and the fact that $f^{(n)}(0)=a_n*n!$ , but I'm not sure how.
|
\begin{align}
f(x) &= \ln(2 + x^{2}) = \ln 2 + \ln\left( 1 + \frac{x^{2}}{2}\right) \\
&= \ln 2 + \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{2k}}{2^{k} \, k} \\
&= \ln 2 + \frac{x^{2}}{2} - \frac{x^{4}}{8} + \frac{x^{6}}{24} - \frac{x^{8}}{64} + \frac{x^{10}}{160} - \cdots \\
f(x) &= \ln 2 + \frac{x^{2}}{2!} - 3 \, \frac{x^{4}}{4!} + 30 \, \frac{x^{6}}{6!} - 630 \, \frac{x^{8}}{8!} + 22680 \, \frac{x^{10}}{(10)!} - 1247400 \, \frac{x^{12}}{(12)!} + \cdots
\end{align}
From this series expansion it can be determined that
\begin{align}
f^{(2n+1)}(0) &= 0 \hspace{10mm} \text{for} \quad n \geq 0 \\
f^{(2n)}(0) &= \frac{(-1)^{n-1} \, (2n)!}{2^{n} \, n} = \frac{(-1)^{n-1} \, (2n-1)!}{2^{n-1}} \hspace{10mm} \text{for} \quad n \geq 1 \\
f(0) &= \ln 2.
\end{align}
|
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|
Why $\int \frac{dx}{\sqrt{9-x^2}} = sin^{-1}\frac{x}{3}$? I don't understand how the two formulas are equal since the right side involves trigonometric sine that the left is devoid of.
|
You can use the following substitution
$$ (*) \qquad x=\dfrac{3}{2}\sin t \implies \text{ d}x = \dfrac{3}{2} \cos t \text{ d}t $$
This'll yield the following equality
$$ \underbrace{\int \dfrac{1}{\sqrt{9-4x^2}} \text{ d}x \ \overset{(*)}= \ \dfrac{1}{3} \int \dfrac{1}{|\cos t|} \dfrac{3}{2} \cos t \text{ d}t}_{\because \ \sqrt{9-4\left(\frac{9}{4}\sin (t)\right)^2} \ = \ 3 |\cos (t)|} \ = \underbrace{\dfrac{1}{2} \int \mathrm{sgn}\left(\cos(t)\right) \text{ d}t}_{\because \ \mathrm{sgn}\left(\cos(t)\right) \ \cdot \ |\cos(t)| \ =: \ \cos(t) } = \dfrac{1}{2} \ t \ \mathrm{sgn}\left(\cos(t)\right) + \mathcal{C} $$
Finally, reverting our substitution back, we have that
$$ t = \arcsin \left( \dfrac{2x}{3} \right) $$
and that
$$ \mathrm{sgn}\left(\cos(t)\right) = \mathrm{sgn}\left( \sqrt{1-\left(\dfrac{2x}{3}\right)^2} \right) = +1 $$
so we finally have that
$$ \int \dfrac{1}{\sqrt{9-4x^2}} \text{ d}x \ = \ \dfrac{1}{2} \arcsin \left( \dfrac{2x}{3} \right) + \mathcal{C} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1385533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
how to solve $3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$ $$A = 3 - \cfrac{2}{3 - \cfrac {2}{3 - \cfrac {2}{3 - \cfrac {2}{...}}}}$$
My answer is:
$$\begin{align}
&A = 3 - \frac {2}{A}\\
\implies &\frac {A^2-3A+2}{A}=0\\
\implies &A^2-3A+2=0\\
\implies &(A-1)\cdot(A-2)=0\\
\implies &A=1\;\text{ or }\; A=2
\end{align}$$
I should note that I'm not sure if the above answer is true. Because I expected just one answer for A (A is a numeric expression), but I found two, $1$ and $2$. This seems to be a paradox.
|
This continued fraction is the limit of
the sequence $a_n=3-2/a_{n-1}$. Computing the first few terms shows that $2$ is the correct limit; if our initial term $a_1=3$ were different then the limit could be $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1386966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Solve this trigonometric equation $ \sin2x-\sqrt3\cos2x=2$ Solve equation: $$ \sin2x-\sqrt3\cos2x=2$$
I tried dividing both sides with $\cos2x$ but then I win $\frac{2}{\cos2x}$.
|
$\bf{My\; Solution::}$ Given $\displaystyle \sin 2x -\sqrt{3}\cos 2x = 2$
We can write it as $$\displaystyle \sin 2x \cdot \frac{1}{2}-\cos 2x\cdot \frac{\sqrt{3}}{2} = 1\Rightarrow \sin \left(2x-\frac{\pi}{3}\right) =1=\sin \frac{\pi}{2}$$
Above we have used the formula $$ \sin \alpha\cdot \cos \beta-\cos \alpha\cdot \sin \beta=\sin (\alpha-\beta).$$
So $$\displaystyle 2x-\frac{\pi}{3} = n\pi+(-1)^n\cdot \frac{\pi}{2}\Rightarrow x=\frac{n\pi}{2}+(-1)^n\cdot \frac{\pi}{4}+\frac{\pi}{6}\;,$$ Where $n\in \mathbb{Z}$
|
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|
Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understand?
$$\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$$
How do I prove the above equation for all integers where $n\geq1$?
|
Hint:
$$\frac { 1 }{ 2 } \left( \frac { 1 }{ n } -\frac { 1 }{ n+2 } \right) =\frac { 1 }{ n\left( n+2 \right) } $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $ Prove that:
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $
What I've tried:
I solved the improper integral: $\int_0^\infty \frac{1}{1+x^2} = \lim_{b\to \infty} \arctan b -\arctan 0 = \frac{\pi}2 $.
Now, everything in the sum (and in the integral) is positive so the sum must be lower than the integral (it has more members), so the claim is wrong (originally it was a "Prove/Disprove" problem...). However, if we only add the first 3 members we get 1.7 which is bigger than $\frac{\pi}2 $.
Also, I checked with wolfram and the sum is somewhere around 2 so the claim is true. I have no idea how to prove it.
Someone suggested I would read Basel's problem ,in which the integral is also smaller than the sum (starting from 1) but it led me nowhere and I don't think that this kind of complex solution is needed here.
|
The key result to use is that if $f(x)$ is a continuous strictly decreasing function, then $f(n+1) < \int_{n}^{n+1} f(x)\,dx < f(n)$.
If we take the sum over all values of $n$ starting at $n=0$ using the right inequality, we get that $$\frac{\pi}{2} = \int_{0}^\infty \frac{dx}{1+x^2} < \sum_{n=0}^\infty \frac{1}{1+n^2},$$
and if we sum over all values of $n$ starting at $n=1$ and use the left inequality, we get
$$ \sum_{n=2}^\infty \frac{1}{1+n^2} < \int_{1}^\infty \frac{dx}{1+x^2} = \frac{\pi}{4}.$$
We then get the desired inequality as follows:
$$ \sum_{n=0}^\infty \frac{1}{1+n^2} = 1 + \frac{1}{2} + \sum_{n=2}^\infty \frac{1}{1+n^2} < \frac{3}{2} + \frac{\pi}{4} < \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}.$$ (This elaboration is necessary since $\frac{1}{1+x^2}$ is increasing on $[-1,0]$, as noted in comments.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^3+y^3=72$ and $xy=8$ then find the value of $x-y$. I recently came across a question,
If $x^3+y^3=72$ and $xy=8$ then find the value of $(x-y)$.
By trial and error I found that $x=4$ and $y=2$ satisfies both the conditions. But in general how can I solve it analytically? I tried using $a^3+b^3=(a+b)(a^2+b^2-ab)$ and also $a^3+b^3=(a+b)^3-3ab(a+b)$. But both ways aren't working.
Please explain how do I solve these types of questions analytically.
|
You may solve this by brute force. Write $x=8/y$. The first equation now reads
$x^3+\frac{512}{x^3}=72$.
Multiply both sides by $x^3$, and you have
$x^6+512-72x^3=0$.
Define $u\equiv x^3$, and the equation becomes
$u^2-72u+512=0$.
Use the quadratic formula to solve for $u$, solve back for x, and solve for y.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\int \sec^3x\,dx$ I've tried substitution, replacing $\sec^2x$ with $\tan^2x + 1$, and parts and I just hit dead ends every time... Do you need knowledge of higher-level calculus to solve this?
|
$\bf{My\; Solution::}$ Let $\displaystyle I = \int \sec^3 x dx = \int \frac{1}{\cos^3 x} dx = \int\frac{1}{\sin^3\left(\frac{\pi}{2}-x\right)}dx$
Now Let $\displaystyle \left(\frac{\pi}{2}-x\right) = t\;,$ Then $dx = -dt$
So Integral $\displaystyle I = -\int\frac{1}{\sin^3 t} dt =-\frac{1}{8}\int \frac{1}{\sin^3 \frac{t}{2}\cdot \cos^3 \frac{t}{2}}dt = -\frac{1}{8}\int\frac{\sec^6 \frac{t}{2}}{\tan^3 \frac{t}{2}}dt = \frac{1}{8}\int\frac{\sec^4 \frac{t}{2}\cdot \sec^2 \frac{t}{2}}{\tan^3 \frac{t}{2}}dt$
Now Let $\displaystyle \tan \frac{t}{2} = u\;,$ Then $\displaystyle \sec^2 \frac{t}{2}dt = 2du$
So Integral $\displaystyle I = -\frac{1}{4}\int\frac{(1+u^2)^2}{u^3}du = -\frac{1}{4}\int\frac{1+u^4+2u^2}{u^3}du = -\frac{1}{4}\int \left[u^{-3}+u+2u^{-1}\right]du$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Triangle Area problem I've been trying to solve the following:
Let $ABC$ be a triangle with sides $a, b $ and $ c$, inradius $r$ and exradii $r_a, r_b$ and $r_c$. If $A'B'C'$ is another triangle with sides $\sqrt{a}, \sqrt{b}$ and $\sqrt{c}$ show that $Area(A'B'C')=\frac {\sqrt{r(r_a+r_b+r_c)}} {2}$.
I tried to combine various formulas and theorems involving exradii and inradii but it got me nowhere so far.. maybe there's a better way to approach the problem! Any advice or help will be appreciated!!
|
Brute force method:
Firstly, Heron's formula can be simplified in this form:
The area $\Delta$ of a triangle with sides $a,b,c$ is given by $16\Delta^2 $ $= (a + b + c)(b + c - a) (c + a - b)(a + b - c) \\= 2(a^2 b^2 + b^2 c^2 + c^2 a^2) - (a^4 + b^4 + c^4)$
Therefore area of triangle with sides as $\sqrt a, \sqrt b, \sqrt c$ is $\frac 1 4 \sqrt{2 (ab + bc + ca) - (a^2 + b^2 + c^2)}$.
Also we have $r_a + r_b + r_c = 4R + r$. (Steiner's formula)
So it is enough to show that
$ \frac{\sqrt{r(r_a+r_b+r_c)}}{2} = \frac{1}4\sqrt{2 (ab + bc + ca) - (a^2 + b^2 + c^2)}$
which is equivalent to
$4r (4R + r) = 2 (ab + bc + ca) - (a^2 + b^2 + c^2)$.
Also $\Delta = \frac 12 (a + b + c)\cdot r = \frac{abc}{4R}$, so
$4r (4R + r) = 2 (ab + bc + ca) - (a^2 + b^2 + c^2)$
$\Leftrightarrow 4 \frac{2 \Delta}{a + b + c} \left(\frac{abc}{\Delta} + \frac{2 \Delta}{a + b + c} \right) = 2 (ab + bc + ca) - (a^2 + b^2 + c^2)$
(Note that here $\Delta$ is the area of the triangle with sides $a,b,c$)
$\Leftrightarrow 8 (abc (a + b + c) + 2 \Delta^2) = (a + b + c)^2(2 (ab + bc + ca) - (a^2 + b^2 + c^2))$
$\Leftrightarrow 8 abc (a + b + c) + (a + b + c)(b + c - a) (c + a - b)( a + b - c) = (a + b + c)^2(2(ab + bc + ca) - (a^2 + b^2 + c^2))$
$\Leftrightarrow 8 abc + (b + c - a)(c + a - b)(a + b - c) = (a + b + c)(2 (ab + bc + ca) - (a^2 + b^2 + c^2))$
$\Leftrightarrow 8 abc - 2 abc + a^2 (b + c) + b^2 (c + a) + c^2 (a + b) - (a^3 + b^3 + c^3)
= 6 abc + a^2 (b + c) + b^2 (c + a) + c^2 (a + b) - (a^3 + b^3 + c^3) $
which is true.
|
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|
Why is this sum zero? I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$ factors and is
$$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\dots \left(1-\frac{i-1}{n}\right)\left(1-\frac{i^2}{n}\right).$$
It seems, amazingly, that the answer is 0. How can one show this?
|
The following holds true for $n\geq 1$
\begin{align*}
\sum_{k=1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)=0\tag{1}
\end{align*}
Note, the empty product is set equal to $1$. We start by transforming the product into a somewhat more convenient form .
\begin{align*}
\sum_{k=1}^{\infty}&\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\\
&=\sum_{k=1}^{n}\left(1-\frac{k^2}{n}\right)\frac{1}{n^{k-1}}\prod_{j=1}^{k-1}(n-j)\tag{2}\\
&=n!\sum_{k=1}^{n}\left(1-\frac{k^2}{n}\right)\frac{1}{n^k}\frac{1}{(n-k)!}\\
&=\frac{n!}{n^n}\sum_{k=0}^{n-1}\left(1-\frac{(n-k)^2}{n}\right)\frac{n^k}{k!}\tag{3}\\
\end{align*}
In (2) we set the upper limit of the sum to $n$ since for values greater than $n$ the product contains a factor zero. In the last step (3) we changed the index summation $k\rightarrow n-k$. In the following we skip the factor $\frac{n!}{n^n}$ and show the sum is equal to zero.
\begin{align*}
\sum_{k=0}&\left(1-\frac{(n-k)^2}{n}\right)\frac{n^k}{k!}\\
&=\frac{1}{n}\sum_{k=0}^{n-1}(n-n^2+2kn-k^2)\frac{n^k}{k!}\\
&=(1-n)\sum_{k=0}^{n-1}\frac{n^k}{k!}+2\sum_{k=1}^{n-1}k\frac{n^k}{k!}
-\frac{1}{n}\sum_{k=2}^{n-1}k(k-1)\frac{n^k}{k!}-\frac{1}{n}\sum_{k=1}^{n-1}k\frac{n^k}{k!}\tag{4}\\
&=(1-n)\sum_{k=0}^{n-1}\frac{n^k}{k!}+2n\sum_{k=0}^{n-2}\frac{n^k}{k!}
-n\sum_{k=0}^{n-3}\frac{n^k}{k!}-\sum_{k=0}^{n-2}\frac{n^k}{k!}\tag{5}\\
&=\frac{n^{n-1}}{(n-1)!}-n\frac{n^{n-1}}{(n-1)!}+n\frac{n^{n-2}}{(n-2)!}\\
&=\frac{n^{n-1}}{(n-1)!}\left(1-n+(n-1)\right)\\
&=0\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
Comment:
*
*In (4) we set the lower bounds of the index accordingly and use $k^2=k(k-1)+k$ to be able to cancel out factors of the factorial conveniently.
*In (5) we adjust the index accordingly to obtain equal summands for telescoping in the next step.
Added 2015-08-13: A note to the elegant approach of @BarryCipra due to a comment of @hypergeometric.
Let $R$ be a positive integer less than $n$. We obtain from (1)
\begin{align*}
\sum_{k=1}^{\infty}&\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\\
&=\sum_{k=1}^{R}\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)+
\sum_{k=R+1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\\
&=\prod_{r=1}^R\left(1-\frac{r}{n}\right)\left(\sum_{k=1}^{R}\left(1-\frac{k^2}{n}\right)\prod_{j=k}^R\left(1-\frac{j}{n}\right)^{-1}\right.\\
&\qquad\qquad\qquad\qquad+\left.\sum_{k=R+1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=R+1}^{k-1}\left(1-\frac{j}{n}\right)\right)\tag{6}\\
&=\prod_{r=1}^R\left(1-\frac{r}{n}\right)\left(R+\sum_{k=R+1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=R+1}^{k-1}\left(1-\frac{j}{n}\right)\right)\tag{7}
\end{align*}
Comment:
*
*Observe the last line of @BarryCipras representation corresponds to (7) with $R=3$.
*Since we know that (7) is valid, we proceed from (6) to (7) by claiming the validity of the identity
\begin{align*}
\sum_{k=1}^{R}\left(1-\frac{k^2}{n}\right)\prod_{j=k}^R\left(1-\frac{j}{n}\right)^{-1}=R\qquad\qquad 1\leq R < n
\end{align*}
This identity is a nice by-product.
*
*When setting $R\geq n$ in (7) we observe, that the leftmost product contains a factor zero. But we can't use this argument in this derivation, since we have to consider in (6) the factors $\left(1-\frac{j}{n}\right)^{-1}$ which are undefined in case $j=n$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof of the identity: $c\sin \frac{A-B}{2} \equiv (a-b) \cos \frac{C}{2}$ Trigs is not my strongest apparently...
I need to prove $c\sin \frac{A-B}{2} = (a-b) \cos \frac{C}{2}$ for a general triangle $ABC$.
Here is what I do, or rather, here is how I fail at proving it:
$\cos \frac{C}{2} \equiv \sin \frac{A+B}{2}$, so $\displaystyle{\frac{\sin \frac{A-B}{2}}{\sin \frac{A+B}{2}} \equiv \displaystyle{\frac{a-b}{c}}}$.
This implies: $\displaystyle{\frac{\tan \frac{A}{2}-\tan \frac{B}{2}}{\tan \frac{A}{2}+\tan \frac{B}{2}} \equiv \frac{a-b}{c}}$.
Now, imagine graphing an angle bisector from angle $A$ and then from angle $B$, the point where they intersect, let's call it $K$. From that point drop a perpendicular on $AB$, let's call that point $L$. Hence, $\tan \frac{A}{2} = \frac{KL}{AL}$ and $\tan \frac{B}{2} = \frac{KL}{LB}$. Plugging those in, gives us: $$\frac{LB-AL}{c} \equiv \frac{a-b}{c}$$ And now I have no clue how to show that $LB-AL = a-b$.
If you could let me know how to show that and/or you know a better way of proving the identity, please share.
|
Here's a trigonograph:
(This space intentionally left blank.)
|
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|
How to prove the trigonometric identity $\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$ I am doing some practice questions for a Math class and I was told that similar questions would be in the exam. So I need to learn this but I have no idea where to even start with this question:
$$\frac{\cot x}{1- \tan x} + \frac{\tan x}{1 - \cot x} - 1 = \sec x \csc x$$
Hint: use standard factorization for the difference of 2 cubes, e.g. $$a^3-b^3 = (a-b)(a^2 + ab + b^2)$$
Help please I find that looking at the working when someone does these questions allows me to learn the method. Thanks in advance.
|
Given $$\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} = \frac{1}{\sin x}\cdot \left(\frac{\cos^2 x}{\cos x-\sin x}\right) - \frac{1}{\cos x}\cdot \left(\frac{\sin^2 x}{\cos x-\sin x}\right)$$
So $$\displaystyle = \frac{\cos^3 x-\sin^3 x}{\sin x\cdot \cos x\cdot (\cos x-\sin x)} = \frac{(\cos x-\sin x)\cdot (\cos^2 x+\sin^2 x+\sin x\cdot \cos x)}{\sin x\cdot \cos x \cdot (\cos x-\sin x)}$$
So $$\displaystyle = \frac{1+\sin x\cdot \cos x}{\sin x\cdot \cos x} = \sec x\cdot \csc x+1$$
So $$\displaystyle \frac{\cot x}{1-\tan x}+\frac{\tan x}{1-\cot x} = 1+\sec x\cdot \csc x$$
|
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|
Family of Lines Problem:
Consider a family of straight lines $(x+y)+\lambda(2x-y+1)=0$. Find the equation of the straight line belonging to this family which is farthest from $(1,-3)$.
$$$$
Any help with this problem would be really appreciated!
|
We can write the equation as $\displaystyle (2\lambda+1)x+(1-\lambda)y+\lambda+1 =0$
Now Perpendicular Distance of Line from Point $\bf{P(1,-3)}$ is $ = PQ=\displaystyle \left|\frac{2\lambda+1+3\lambda-3+\lambda+1}{\sqrt{(2\lambda+1)^2+(1-\lambda)^2}}\right|$
Now $\displaystyle PQ = \left|\frac{6\lambda-1}{\sqrt{5\lambda^2+2\lambda+2}}\right|\;,$ Then we have to $\bf{Maximize}$ the value of $$\displaystyle \frac{6\lambda-1}{\sqrt{5\lambda^2+2\lambda+2}}$$
Now Let $$\displaystyle f(\lambda) = \frac{(6\lambda-1)^2}{5\lambda^2+2\lambda+2}$$.(Bcz If $f(\lambda)$ is $\bf{Max.}\;,$ Then $\left[f(\lambda)\right]^2$ is $\bf{Max.}$)
Now Let $$\displaystyle z = \frac{36\lambda^2+1-12\lambda}{5\lambda^2+2\lambda+2}\Rightarrow 5\lambda^2\cdot z+2\lambda \cdot z+2z = 36\lambda^2-12\lambda+1$$
So we get $$\displaystyle \left(5z-36\right)\lambda^2+2(z+6)\lambda+(2z-1) =0\;,$$ For real roots, Then $\bf{Discriminant\geq 0}$
So $$\displaystyle 4\left(z+6\right)^2-4\cdot (5z-36)\cdot (2z-1)\geq 0$$
So $(z+6)^2-(5z-36)\cdot (2z-1)\geq 0$
So $$\displaystyle z^2+36+12z-10z^2+77z-36\geq 0$$
So $$\displaystyle -9z^2+89z\geq 0\Rightarrow 9z^2-89z\leq 0$$
So we get $$\displaystyle z\leq \frac{89}{9}$$
So $$\displaystyle \bf{Max.(z) = \frac{89}{9}}$$
Now Put $\displaystyle z = \frac{89}{9}$ in $$\displaystyle z = \frac{(6\lambda-1)^2}{5\lambda^2+2\lambda+2}\;,$$ We Get $\displaystyle \lambda = -\frac{13}{11}$
|
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|
find the limit of $ \lim_{(x,y) \rightarrow (0,0)}\frac{2xy^3+x^2y^3}{x^4+2y^4}$ find the limit of $$ \lim_{(x,y) \rightarrow (0,0)}\frac{2xy^3+x^2y^3}{x^4+2y^4}$$
I have absolutely no idea how to proceed with that. I would prefer a solution that would involve use of squeeze theorem
|
Let $y=kx$. So,
$$
\frac{2xy^3+x^2y^3}{x^4 + y^4} = \frac{k^3x^3(2x+x^2)}{x^4(1+k^4)} = \frac{k^3}{1+k^4}(x+2)
$$
and limit depends on $k$; so, it doesn't exists.
|
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|
Finding the derivative of an absolute value This one I just don't know how to derive.
$\ln\|x^4\cos x\|$
I know the derivative of $\ln\ x$, is just $\frac{1}{x}$ . It is the absolute value that throws me off. My question is, does the absolute value stay as is or does it disappear?
*
*$\frac{1}{|x^4\cos x|}$
*$\frac{1}{x^4\cos x}$
Or is there an extra step that I should preform?
|
HINT:
$$\frac{d|x|}{dx}=\text{sgn}(x)$$
for $x\ne0$. Now, use the chain rule, but make sure that $x^4 \cos x\ne0$
SPOILER ALERT: SCROLL OVER SHADED AREA TO REVEAL ANSWER
Let $y=x^4\cos x$. Then, we have $$\frac{d}{dx}\log|x^4\cos x|=\frac{d\log |y|}{d|y|}\text{sgn(y)}\frac{dy}{dx}=\frac{1}{|x^4\cos x|}\text{sgn}(x^4\cos x)(4x^3\cos x-x^4\sin x)=\frac{1}{x^4\cos x}(4x^3\cos x-x^4\sin x)$$Another way to approach the problem is to split it into two parts, (i) $x^4\cos x>0$, and (ii) $x^4\cos x<0$. On Part (i), we have $$\frac{d}{dx}\log|x^4\cos x|=\frac{d}{dx}\log(x^4\cos x)=\frac{1}{x^4\cos x}(4x^3\cos x-x^4\sin x)$$On Part (ii), we have $$\frac{d}{dx}\log|x^4\cos x|=\frac{d}{dx}\log(-x^4\cos x)=\frac{1}{-x^4\cos x}(-4x^3\cos x+x^4\sin x)=\frac{1}{x^4\cos x}(4x^3\cos x-x^4\sin x)$$as expected!
|
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|
$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}$ $\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}$
I tried:
$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}=\int \frac{dx}{\sin x \sqrt{\sin2x\cos \alpha+\cos 2x\sin \alpha}}$,then i could not solve and changed the method.
Let $\sqrt{\sin(2x+\alpha)}=u\Rightarrow dx=\frac{u du}{\cos(2x+\alpha)}$
$\int \frac{dx}{\sin x \sqrt{\sin(2x+\alpha)}}=$ But this way is also not working.Please help me suggesting the proper method.Thanks..
|
Given $\displaystyle \int\frac{1}{\sin x\sqrt{\sin (2x+a)}}dx = \int\frac{1}{\sin x\sqrt{\sin 2x\cdot \cos a+\cos 2x\cdot \sin a}}dx$
So we get $\displaystyle = \int\frac{1}{\sin x\sqrt{2\sin x\cdot \cos x\cdot \cos a+(\cos^2 x-\sin^2 x)\cdot \sin a}}$
$\displaystyle = \int\frac{1}{\sin^2 x\sqrt{2\cot x\cdot \cos a+(\cot^2 x-1)\cdot \sin a}}dx $
$\displaystyle = \int\frac{\csc^2 x}{\sqrt{2\cot x\cdot \cos a+(\cot^2 x-1)\cdot \sin a}}dx$
Now Let $\displaystyle \cot x= t\;,$ Then $\csc^2 xdx = -dt$
So Integral $\displaystyle = -\int\frac{1}{\sqrt{2t\cdot \cos a+(t^2-1)\cdot \sin a}}dt = -\frac{1}{\sqrt{\sin a}}\int\frac{1}{\sqrt{t^2+2t\cdot \cot a-1}}dt$
So we get $\displaystyle = -\frac{1}{\sqrt{\sin a}}\int\frac{1}{\sqrt{(t+\cot a)^2-(\csc a)^2}}dt$
So we get $\displaystyle = \frac{1}{\sqrt{\sin a}}\cdot \ln\left|t+\cot a+\sqrt{(t+\cot a)^2-\csc^2 a}\right|+\mathcal{C}$
|
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|
Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together. Let a complex number $\alpha,\alpha\neq1$,be a root of the equation $z^{p+q}-z^p-z^q+1=0$,where $p$ and $q$ are distinct primes.Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together.
Since $\alpha$ is a root of $z^{p+q}-z^p-z^q+1=0$.Therefore,$\alpha^{p+q}-\alpha^p-\alpha^q+1=0$ factorised to $(\alpha^p-1)(\alpha^q-1)=0$,but since $\alpha\neq1$
Therefore,either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$ or both equal to zero.Because we have studied that if $ab=0$,then either $a=0$ or $b=0$ or both equal to zero.But this is in contradiction to what is to be proved.My query is why cannot both be zero?
|
If $a$ is a root of $1+z+z^2+\cdots+z^{m-1}=0\ \ \ \ (1),a\ne1$
and is a root of $\dfrac{z^m-1}{z-1}=0$
If $a$ is a root of $1+z+z^2+\cdots+z^{n-1}=0\ \ \ \ (2),a$ is a root of is a root of $\dfrac{z^n-1}{z-1}=0$
Now Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ says $(a^m-1,a^n-1)=a^{(m,n)}-1$
$$\implies\left(\dfrac{a^m-1}{a-1},\dfrac{a^n-1}{a-1}\right)=\dfrac{a^{(m,n)}-1}{a-1}$$
If $(m,n)=1,$
$$\left(\dfrac{a^m-1}{a-1},\dfrac{a^n-1}{a-1}\right)=1$$
$\implies(1),(2)$ can not have any root in common
|
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|
Find inverse of 15 modulo 88. Here the question: Find an inverse $a$ for $15$ modulo $88$ so that $0 \le a \le 87$; that is, find an integer $a \in \{0, 1, ..., 87\}$ so that $15a \equiv1$ (mod 88).
Here is my attempt to answer:
Find using the Euclidean Algorithm, we need to find $\gcd(88, 15)$, that must equal to $1$ to be possible to find an inverse of $15 \pmod{88}$.
\begin{align*}
88 & = 5 \times 15 + 13\\
15 & = 1 \times 13 + 2\\
13 & = 6 \times 2 + 1\\
2 & = 2 \times 1 + 0
\end{align*}
So,
$$\gcd(88, 15) = 1$$
Now, we need to write this into the form:
$$\gcd(88, 15) = 88x + 15y.$$
And find $x$ and $y$.
\begin{align*}
1 & = 13(1) + 2(-6)\\
& = 13(7) + 15(-6)\\
& = 88(7) + 15(-41)
\end{align*}
So, $x = 7$ and $y = -41$.
So, an inverse of $15 \pmod{88} = -41$. Now, I need to find an inverse that is between $0$ and $87$. What is a good easy approach to find other inverses? Any ideas please?
|
For a systemetic way, you have to use the Extended Euclidean Algorithm: if you have a Bézout's relation:
$$u\times 88+v\times 15=1$$
the you know the inverse of $15$ is $v\bmod88$.
The following shows how to display the computation, taking into account that, in Euclid's algorithm, the remainder at each step is a linear combination of $88$ and $15$:
\begin{array}[t]{lrrrr}
\text{Successive Divisions}& r_i & u_i & v_i & q_i\\
\hline
& 88 & 1 & 0 & \\
88 = {\color{red}5} \times 15 +\color{blue}{13} & 15 &0 & 1 & \color{red}5 \\
15 = {\color{red}1} \times 13 + \color{blue}{2} & \color{blue}{13} & 1 & -5 & \color{red}1 \\
13 = {\color{red}6} \times 2 + \color{blue}{1} & \color{blue}{2} & -1 & 6 & \color{red}6 \\
& \color{blue}{1} & 7 & -41\\[2ex]
\hline
\end{array}
Thus the inverse of $15$ modulo $88$ is $\;-41\equiv 47\mod 88$.
|
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|
Evaluation of $\sum_{r=1}^{n}r\cdot (r-1)\cdot \binom{n}{r} = $ Evaluation of $\mathop{\displaystyle \sum_{r=1}^{n}r\cdot (r-1)\cdot \binom{n}{r} = }$
$\bf{My\; Try::}$ Given $$\displaystyle \sum_{r=1}^{n}r\cdot (r-1)\cdot \binom{n}{r}\;,$$ Now Using the formula $$\displaystyle \binom{n}{r} = \frac{n}{r}\cdot \binom{n-1}{r-1}.$$
$$\mathop{\displaystyle =\sum_{r=1}^{n}r\cdot (r-1)\cdot \frac{n}{r}\cdot \binom{n-1}{r-1}}$$
$$\mathop{\displaystyle = n\sum_{r=2}^{n}(r-1)\cdot \frac{n-1}{r-1}\cdot \binom{n-2}{r-2} = n\cdot (n-1)\sum_{r=2}^{n}\binom{n-2}{r-2}}$$
$$\mathop{\displaystyle = n\cdot (n-1)\cdot \left[\binom{n-2}{0}+\binom{n-2}{1}+..........+\binom{n-2}{n-2}\right] = n(n-1)\cdot 2^{n-2}}$$
My question is can we solve it using combinatorial argument, If yes then plz explain here
Thanks
|
$\sum_{r=0}^{n} r(r-1)\binom{n}{r} = 2 \sum_{r=0}^n \binom{n}{r} \binom{r}{2}$. The summand is the number of ways of choosing from $n$ objects an $r$-tuple and a 2-subtuple, so the right-hand sum is just the number of ways of choosing a 2-tuple multiplied by the number of possible tuples the 2-subtuple could have been drawn from. That is, the number of 2-tuples multiplied by $2^{n-2}$, or $\binom{n}{2} \times 2^{n-2}$.
Therefore, the left-hand sum is $n(n-1) \times 2^{n-2}$.
To expand on the $2^{n-2}$: the set of tuples from which a given 2-subtuple $(a,b)$ has been drawn, bijects with the set of tuples drawn from a set of size $n-2$, by the bijection "remove $a$ and $b$ from the tuple".
|
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Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$ Find:
$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$
I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What method to use?
Result should be $\frac{1}{4}$
|
$$\lim_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=L$$
Using L'hopital:
$$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{2\sin x \cos x}=L$$
Reordering the denominator:
$$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{\sin 2x }=L$$
Using L'hopital Again
$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x }=L$$
You cannot use l'hopital again because $\lim_{x\to 0} \cos 2x = 1 > 0$
so replacing:
$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x } = \dfrac{\frac{0-\frac{1}{2}}{2}+\frac{1}{4}}{2\cdot 1} = 0$$
|
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|
Let $X$ denote the number of tosses required to get the 5th head and $Y$ the number between the 6th and 7th heads. Are $X$ and $Y$ independent?
Let $X$ denote the number of tosses required to get the 5th head and $Y$ the number between the 6th and 7th heads. Are $X$ and $Y$ independent?
Y will always depend on X . NO ?
i know geometric distribution has lack of memory property . but in this the underlying distribution is i think negative binomial.
i think 'no, they are not independent ' . help please.
|
What you want to show is that
$$
P(X=x \wedge Y=y) = P(X=x)P(Y=y)
$$
for all values $x$ and $y$ of $X$ and $Y$.
Let $Z$ be the number of tosses between the 5th and 6th head. Then
\begin{align*}
P(X=x \wedge Y=y)
&= \sum_{z=1}^\infty \underbrace{\binom{x-1}{4}p^5q^{x-5}}_{X=x} \underbrace{\vphantom{\binom{x-1}{4}}q^{z-1}p}_{Z=z} \underbrace{\vphantom{\binom{x-1}{4}}q^{y-1}p}_{Y=y} \\
&= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p \cdot \sum_{z=1}^\infty q^{z-1}p \\
&= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p \cdot \frac{p}{1-q} \\
&= \binom{x-1}{4}p^5q^{x-5} \cdot q^{y-1}p = P(X=x)P(Y=y)
\end{align*}
Once you separate the events like this it boils down to the tosses being independent of each other.
|
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Evaluation of trignometric limit I want to find the following limit without L'Hospital.
$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$
Maybe I should try to get rid of the radical.
|
let us make a change of variable $x = 3\pi/4 + h.$ then we have
$$\begin{align}\tan(x) &= \tan(3\pi/4 + h) = \frac{-1 + \tan h}{1 + \tan h} \\
&= (\sin h - \cos h)(\sin h + \cos h \cdots)^{-1} \\
&=(-1 + h \cdots)(1+h +\cdots)^{-1} \\
&=(-1 + h \cdots)(1-h +\cdots) \\
&=-1+2h+\cdots\\
\tan^{1/3} x &=-1 + \frac23 h+\cdots\\
1+\tan^{1/3} x &= \frac23 h+\cdots \tag 1\\
\sqrt 2\cos x &= \sqrt 2\cos(3\pi/4+h)= -\cos h - \sin h\\
&=-1 - h + h^2/2+\cdots\\
2\cos^2 x &= 1+2h+\cdots\\
1-2\cos^2 x &= -2h+\cdots \tag 2
\end{align}$$
from $(1)$ and $(2),$ $$\lim_{x \to 3\pi/4} \frac{1+\tan^{1/3} x}{1-2\cos^2 x }=\frac{2h/3+\cdots}{-2h + \cdots} = -\frac 13. $$
|
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|
If $\frac{dy}{dx}\frac{dx}{dy} = 1$, does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? I know $\frac{dy}{dx}\frac{dx}{dy} = 1$ because the chain rule says $1 = \frac{dy}{dy} = \frac{dy}{dx}\frac{dx}{dy}$. But does $\frac{d^2 y}{dx^2} \frac{d^2 x}{dy^2} = 1$? Or would that be too good to be true?
|
If you want an example where the derivatives don't vanish, consider $y=x^3$ (so $x=y^{1/3}$). Then:
$$
\frac{dy}{dx}=3x^2
$$
and
$$
\frac{dx}{dy}=\frac{1}{3}y^{-2/3}=\frac{1}{3}x^{-2}
$$
and
$$
\frac{dy}{dx}\frac{dx}{dy}=3x^2\cdot\frac{1}{3x^2}=1.
$$
On the other hand,
$$
\frac{d^2y}{dx^2}=6x
$$
and
$$
\frac{d^2x}{dy^2}=-\frac{2}{9}y^{-5/3}=-\frac{2}{9x^5}.
$$
However,
$$
\frac{d^2y}{dx^2}\frac{d^2x}{dy^2}=6x\cdot\left(-\frac{2}{9x^5}\right)=-\frac{4}{3x^4}\not=1.
$$
|
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|
Arithmetic Derivatives: Arithmetic Logarithmic Derivative Problem In Calculus, whenever we see a constant and want to take the derivative of it, it always is 0. However in Number Theory, we have something called the arithmetic derivative in which we can differentiate to get some nonzero term.
So we can denote the arithmetic derivative the same way as in calculus, say for some $x$, we can say $x'$ to be the arithmetic derivative. Some properties of arithmetic derivatives are that:
*
*For all primes, the arithmetic derivative is $1$.
*Product Rule: $(xy)'=x'y+xy'$
*$0'=1'=0$
Now, there is also some lesser-known sub-part to the arithmetic derivative called the Arithmetic Logarithmic Derivative, $L(n)$, which is equal to $\frac{n'}{n}$. My question is that less than $100$, how many pairs of distinct positive integers (call them $a$ and $b$) does $L(a)=L(b)$?
|
Since $a=b$ is a trivial solution. For $a \neq b$:
$$L(a)=L(b) \\ \frac{a'}{a}=\frac{b'}{b} \\ a'b-b'a=0 \\ \frac{a'b-b'a}{b^2}=0 \\ \left(\frac{a}{b}\right)'=0$$
Now what is $\displaystyle\left(\frac{a}{b}\right)'$?
Lemma 1:
Let prime factorization of $n$ be $\prod_{i=1}^{k} p_i^{a_i}$, where $p_i$'s are distinct prime numbers and write $n=p_1\frac{n}{p_1}$, It follows that:
$$n'=p_1' \frac{n}{p_1}+p_1 \left(\frac{n}{p_1} \right)'=\frac{n}{p_1}+p_1 \left(\frac{n}{p_1} \right)'=\frac{n}{p_1}+p_1 \left(p_1 \frac{n}{p_1^2} \right)' \\=\frac{n}{p_1}+p_1 \left(p_1' \frac{n}{p_1^2} +p_1 \left(\frac{n}{p_1^2} \right)' \right)=2\frac{n}{p_1}+p_1^2 \left(\frac{n}{p_1^2} \right)' $$
Continuing this procedure we get:
$$n'=a_1 \frac{n}{p_1}+p_1^{a_1} \left(\frac{n}{p_1^{a_1}} \right)' \ \ \ \star$$
Now $\frac{n}{p_1^{a_1}}$ is $p_1$ free and if you do the same thing for decreasing power of $p_2$ in $\frac{n}{p_1^{a_1}}$ you will get:
$$\left(\frac{n}{p_1^{a_1}} \right)'=a_2 \frac{n}{p_1^{a_1} p_2}+p_2^{a_2} \left(\frac{n}{p_1^{a_1} p_2^{a_2}} \right)'$$
Put this back in the $\star$:
$$n'=a_1 \frac{n}{p_1}+a_2 \frac{n}{p_2}+p_1^{a_1} p_2^{a_2} \left(\frac{n}{p_1^{a_1} p_1^{a_1}} \right)'$$
Which leads us to:
$$n'=a_1 \frac{n}{p_1}+a_2 \frac{n}{p_2}+... +a_k \frac{n}{p_k}+ p_1^{a_1} p_2^{a_2} ... p_k^{a_k} \left(\frac{n}{p_1^{a_1} p_1^{a_1} ... p_k^{a_k}} \right)' \\ n'=a_1 \frac{n}{p_1}+a_2 \frac{n}{p_2}+... +a_k \frac{n}{p_k} $$
Here, the essence of this proof is that for a natural number $a$ with prime factorization, say $\prod_{i=1}^{k} p_i^{m_i}$, then we have:
$$a'=a \sum_{i=1}^{k} \frac{m_i}{p_i}$$
Lemma 2:
Let us first verify that the Quotient Rule applies to Arithmetic Derivatives:
Note that:
$$a'=\left(b\cdot\frac{a}{b}\right)'=b\cdot\left(\frac{a}{b}\right)'+b'\cdot\frac{a}{b}$$
This is by Leibniz Rule.
So,$$b\cdot\left(\frac{a}{b}\right)'=a'-b'\cdot\frac{a}{b}=\frac{a'b-b'a}{b}$$
$$\therefore \left(\frac{a}{b} \right)'=\frac{a'b -b'a}{b^2}$$
Lemma 3:
If $a=\prod_{i=1}^{k} p_i^{m_i}$ and $b=\prod_{j=1}^{l} q_j^{n_j}$
$$\left(\frac{a}{b} \right)'=\frac{a'b -b'a}{b^2} \\ =\frac{ba\sum_{i=1}^{k} \frac{m_i}{p_i} -ab \sum_{j=1}^{l} \frac{n_j}{q_j}}{b^2} \\ =\frac{a}{b} \left(\sum_{i=1}^{k} \frac{m_i}{p_i} - \sum_{j=1}^{l} \frac{n_j}{q_j} \right)$$
Therefore, if $\prod_{i=1}^{k} p_i^{x_i}$ is a factorization of a rational number $x$ in prime powers, (where some $x_i$ may be negative) then we can write:
$$x'=x \sum_{i=1}^{k} \frac{x_i}{p_i}$$
Here, the essence of this proof is to show that if we have $\frac{a}{b}=\prod_{i=1}^{k} p_i^{x_i}$ then:
$$\left(\frac{a}{b}\right)'=\frac{a}{b} \sum_{i=1}^{k} \frac{x_i}{p_i}$$
Now for this problem:
$$\left(\frac{a}{b}\right)'=\frac{a}{b} \sum_{i=1}^{k} \frac{x_i}{p_i}=0 \\ \sum_{i=1}^{k} \frac{x_i}{p_i}=0 \\ x_1 p_2 p_3 ... p_k+x_2 p_1 p_3 ... p_k+ ... + x_k p_1 p_2 ... p_k=0 \ \ \ \ \ (1) $$
Here we can conclude that $x_i=c_i p_i$ and equation $(1)$ becomes:
$$c_1+c_2+...+c_k=0$$ Therefore the solutions to the equation $\left(\frac{a}{b}\right)'=0 \ \ \ \ (2)$ are of the form:
$$\frac{a}{b}=\prod_{i=1}^{k} (p_i^{p_i})^{c_i}$$
where $\sum c_i=0$.
For this problem since $a$ and $b$ are less than $100$ only possible values for $\frac{a}{b}$ are:
$$\frac{a}{b}=\frac{2^2}{3^3}, \frac{3^3}{2^2}$$
Now if $\frac{a}{b}$ is a solution to the equation $(2)$ then $\frac{ma}{mb}$ is a solution too. So the solutions are:
$$(2^2, 3^3), (2^3, 2\times 3^3), (3\times 2^2, 3^4), (3^3, 2^2), (2\times 3^3, 2^3), (3^4, 3\times 2^2) \\ = (4,27),(8,54),(12,81),(27,4),(54,8),(81,12)$$
|
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|
Inequality between altitude and sides in triangle Let $a,b,c$ be the side lengths and $h_a,h_b,h_c$ the altitudes each connect a vertex to the opposite side and are perpendicular to that side. Then we need to prove $h_a^2+h_b^2+h_c^2\leq\dfrac14(a+b+c)^2$.
I know the inequality $h_a^2+h_b^2+h_c^2\leq\dfrac34(a^2+b^2+c^2)$ by using Cauchy inequality. But I could not extend the proof for the above inequality. Does one help me to prove this?
|
Following ASCII advocate's idea, the problem boils down to proving that:
$$ 4\Delta^2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq (a+b+c)^2 \tag{1}$$
or, using Heron's formula,
$$ (a+b-c)(a-b+c)(-a+b+c)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\leq (a+b+c). \tag{2}$$
Through Ravi substitution that turns out to be equivalent to:
$$ 8xyz\left(\frac{1}{(x+y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}\right)\leq 2(x+y+z)\tag{3} $$
or to:
$$ \sum_{cyc}\frac{1}{(x+y)^2}\leq\sum_{cyc}\frac{1}{4xy}\tag{4}$$
that is trivial as a consequence of the AM-GM inequality.
|
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|
Solve $\cos \frac{4x}{3}=\cos x+1$
Solve the equation \begin{equation} \cos \frac{4x}{3}=\cos x+1\tag 1\end{equation}
I had tried by taking $\cos\dfrac x3=t$ and from this we have $\displaystyle\cos\frac{4x}3=2\left(2t^2-1\right)^2-1; \cos x=4t^3-3t$
$(1) \iff t\left(8t^3-4t^2-8t+3\right)=0$
But I can't solve $8t^3-4t^2-8t+3=0$ because it gives me the approximate roots when I need exact roots
|
There is a trigonometric method for solving cubic equations. Not very pleasant but doable. Applying the method to the case of $$8t^3-4t^2-8t+3=0$$ Using $A=-\frac{1}{2}$, $B=-1$, $C=\frac{3}{8}$, $Q=-\frac{13}{36}$, $R= -\frac{43}{432}$, $D=-\frac{257}{6912}$, $$\theta= \cos ^{-1}\left(-\frac{43}{26 \sqrt{13}}\right)$$ the solutions are then $$t_1=\frac{1}{6}+\frac{\sqrt{13}}{3} \cos \left(\frac{\theta }{3}\right)$$ $$t_2=\frac{1}{6}+\frac{\sqrt{13}}{3} \cos \left(\frac{\theta +2\pi}{3}\right)$$ $$t_3=\frac{1}{6}+\frac{\sqrt{13}}{3} \cos \left(\frac{\theta +4\pi}{3}\right)$$As Jack D'Aurizio pointed it out, $t_1 \gt 1$ must be discarded.
|
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|
Discrete mathematics question $$(n+1)^2+(n+2)^2+(n+3)^2+\dots+(2n)^2=\frac{n(2n+1)(7n+1)}{6}$$
Prove the statement using mathematical induction.
|
Let $s(n)=1^2+2^2+...+n^2=\sum_{j=1}^nj^2.$. You want calculate $s(2n)-s(n)$. We know that $s(n)=\frac{n(n+1)(2n+1)}{6}$, then $s(2n)=\frac{2n(2n+1)(4n+1)}{6}$. It follows that
$$s(2n)-s(n)=\frac{2n+1}{6}(8n^2+2n-n^2-n)=\frac{2n+1}{6}(7n^2+n)=\frac{n(2n+1)(7n+1)}{6}.$$
|
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If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$ If $x^4+7x^2y^2+9y^4=24xy^3$,show that $\frac{dy}{dx}=\frac{y}{x}$
I tried to solve it.But i got stuck after some steps.
$x^4+7x^2y^2+9y^4=24xy^3$
$4x^3+7x^2.2y\frac{dy}{dx}+7y^2.2x+36y^3.\frac{dy}{dx}=24x.3y^2\frac{dy}{dx}+24y^3$
$\dfrac{dy}{dx}=\dfrac{24y^3-4x^3-14y^2x}{14x^2y+36y^3-72xy^2}$
How to move ahead?Or there is some other elegant way to solve it.
|
Notice, we have $$x^4+7x^2y^2+9y^4=24xy^3\tag 1$$
Now, differentiating both the sides w.r.t. $x$ as follows $$\frac{d}{dx}(x^4+7x^2y^2+9y^4)=\frac{d}{dx}(24xy^3)$$
$$4x^3+14x^2y\frac{dy}{dx}+14xy^2+36y^3\frac{dy}{dx}=72xy^2\frac{dy}{dx}+24y^3$$
$$\frac{dy}{dx}(14x^2y+36y^3-72xy^2)=24y^3-4x^3-14xy^2$$
$$\frac{dy}{dx}=\frac{24y^3-4x^3-14xy^2}{14x^2y+36y^3-72xy^2}$$
$$=\frac{y}{x}\cdot \frac{x}{y}\left(\frac{24y^3-4x^3-14xy^2}{14x^2y+36y^3-72xy^2}\right)$$
$$=\frac{y}{x}\left(\frac{24xy^3-4x^4-14x^2y^2}{14x^2y^2+36y^4-72xy^3}\right)$$
Now, substituting the value of $x^4$ from (1), we get
$$\frac{dy}{dx}=\frac{y}{x}\left(\frac{24xy^3-4(24xy^3-7x^2y^2-9y^4)-14x^2y^2}{14x^2y^2+36y^4-72xy^3}\right)$$
$$=\frac{y}{x}\left(\frac{14x^2y^2+36y^4-72xy^3}{14x^2y^2+36y^4-72xy^3}\right)=\frac{y}{x}$$ Hence, we get
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\frac{dy}{dx}=\frac{y}{x}}}$$
|
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|
Correctness of the definite integral Consider the integral
\begin{eqnarray*}
I & = & \int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}\\
& = & \int_{-1}^{0}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}+\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}
\end{eqnarray*}
Use the substitutions $u=-\sqrt{1-x^{2}}$ and $v=\sqrt{1-x^{2}}$
into first and second integrals respectively.
\begin{eqnarray*}
= & -{\displaystyle \int_{-1}^{0}}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}} & +{\displaystyle \int}_{0}^{1}\frac{dv}{(1+v)^{\frac{3}{2}}(1-v)^{\frac{1}{2}}}
\end{eqnarray*}
Noticing that
\begin{eqnarray*}
-\int_{-1}^{0}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}}=-\int_{-1}^{1}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}}+\int_{0}^{1}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}}
\end{eqnarray*}
and
\begin{eqnarray*}
\int_{0}^{1}\frac{dv}{(1+v)^{\frac{3}{2}}(1-v)^{\frac{v}{2}}}= & {\displaystyle \int_{-1}^{1}}\frac{dv}{(1+v)^{\frac{3}{2}}(1-v)^{\frac{1}{2}}} & -\int_{0}^{1}\frac{dv}{(1+v)^{\frac{1}{2}}(1-v)^{\frac{3}{2}}}
\end{eqnarray*}
Finally leading to the result
\begin{eqnarray}
\int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})} & ={\displaystyle \int}_{-1}^{1}\frac{dx}{(1+x)^{\frac{3}{2}}(1-x)^{\frac{1}{2}}} & -\int_{-1}^{1}\frac{dx}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{3}{2}}}
\end{eqnarray}
The value of the integral is 2. But the R.H.S. seem to be difference of two divergent integrals. Can someone suggest whether the derivation is still consistent and valid?
|
Let $$\displaystyle I = \int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}dx= \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}dx$$
Above we Used $\displaystyle \bullet \int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx\;,$ If $f(-x) = f(x)$
Now Put $x=\sin \phi\;,$ Then $dx = \cos\phi d\phi$ and Changing limit, We get
So $$\displaystyle I = 2\int_{0}^{\frac{\pi}{2}}\frac{\cos \phi}{\cos \phi\cdot (1+\cos \phi)}d\phi = \int_{0}^{\frac{\pi}{2}}\frac{1-\cos \phi}{\sin^2 \phi}d\phi$$
So $$\displaystyle I = 2\int_{0}^{\frac{\pi}{2}}\csc^2 \phi-\int_{0}^{\frac{\pi}{2}}\csc\phi\cdot \cot \phi d\phi = -2\left[\cot \phi\right]_{0}^{\frac{\pi}{2}}+2\left[\csc \phi \right]_{0}^{\frac{\pi}{2}}$$
So $$\displaystyle I = 2\cdot \lim_{a\rightarrow 0}\cot (a)+2-2\cdot \lim_{b\rightarrow 0}\csc (b)$$
Yes the given Integral is Divergent.
By applying L'Hospital's rule , we evaluate the limit $$\lim_{a \rightarrow 0 } \left( \cot a - \csc a \right) = 0 $$,
So we get
$$ I = 2 $$
|
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|
Understanding how to use $\epsilon-\delta$ definition of a limit I finally understand the geometric intuition behind the $\epsilon-\delta$ definition of a limit, which is actually quite neat:
But I'm having trouble actually using the definition to come to a conclusion.
For (a solved) example, to prove that $\lim_{n\rightarrow \infty} \frac{3n+5}{2n+7} = \frac{3}{2}$, the following solution is given:
Proof:
Let $a_n = \frac{3n+5}{2n+7}$. Then, $\left | a_n-\frac{3}{2} \right |=\frac{11}{2(2n+7)}<\frac{3}{n}$. Given $\epsilon > 0$, choose $n_0 \in \mathbb{N}$ such that $n_0 > \frac{3}{\epsilon}$. Then, for all $n \ge n_0, \left | a_n-\frac{3}{2} \right | < \epsilon $. Therefore, $\lim a_n = \frac{3}{2}$.
I'm not quite sure I understand this proof. Namely:
*
*Where did $\frac{3}{n}$ come from?
*I guess we just replaced n with $\epsilon$ in the third sentence.
*How did the fourth sentence follow from the third, and why does that show that 3/2 is the limit?
So I tried it myself on another problem, where $a_n = \frac{2n+5}{6n-3}$, and the limit being $\frac{1}{3}$:
Let $a_n = \frac{2n+5}{6n-3}$. Then, $\left | a_n-\frac{1}{3} \right |=\left | \frac{6n-6n+15-3}{3(6n-3)} \right |= \left | \frac{12}{3(6n-3)} \right |= 4\left | \frac{1}{6n-3} \right | < \epsilon$. Dividing by 4, $\left | \frac{1}{6n-3} \right | < \frac{\epsilon}{4}$.
And that's as far as I logically get... I assume the next step is to choose an $n_0 \in \mathbb{N}$ such that $n_0 > \frac{\epsilon}{4}$, and isolating epsilon, $\epsilon < 4n_0$.
but as you can tell, I'm really lost.
|
HINT:
Note that
$$\left|a_n-\frac{1}{3}\right|=\left|\frac{2n+5}{6n-3}-\frac{1}{3}\cdot\frac{2n-1}{2n-1}\right|=\left|\frac{2n+5-(2n-1)}{6n-3}\right|=\left|\frac{6}{6n-3}\right|=\left|\frac{2}{2n-1}\right|$$
Taking $n_0\ge\max\left(\frac{1}{\epsilon}+1,2\right)$ we have the inequality, since
$$n\ge n_0\quad\Longrightarrow\quad\left|a_n-\frac{1}{3}\right|=\left|\frac{2}{2n-1}\right|<\left|\frac{2}{2n-2}\right|=\frac{1}{n-1}<\frac{1}{1/\epsilon}=\epsilon$$
|
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|
Number of Arithmetic Operations in Gaussian-elimination/Gauss-Jordan Hybrid Method for Solving Linear Systems I am stucked at this problem from the book Numerical Analysis 8-th Edition (Burden) (Exercise 6.1.16) :
Consider the following Gaussian-elimination/Gauss-Jordan hybrid method for solving linear systems:
First apply the Gaussian-elimination technique to reduce the system to triangular form. Then use the $n$-th equation to eliminate the coefficients of $x_{n}$ in each of the first $n-1$ rows. After this is completed use the $(n-1)$-st equation to eliminate the coefficients of $x_{n-1}$ in the first $n-2$ rows, ans so on. The system will eventually appear as the reduced system we get by applying Gauss-Jordan method to the original system.
Show that this method requires $\frac{n^3}{3}+\frac{3}{2}n^2-\frac{5}{6}n$ multiplication/divisions and $\frac{n^3}{3}+\frac{n^2}{2}-\frac{5}{6}n$ additions/subtractions.
But when I tried to calculate the total number of multiplication/divisions I got $\frac{n^3}{3}+\frac{3}{2}n^2-\frac{11}{6}n$, Here's what I've done (Sorry for being a bit long):
The augmented matrix for the system has the form
$$ \left[
\begin{array}{cccc|c}
a_{1,1} & a_{1,2} & \cdots & a_{1,n} & a_{1,n+1} \\
a_{2,1} & a_{2,2} & \cdots & a_{2,n} & a_{2,n+1} \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
a_{n,1} & a_{n,2} & \cdots & a_{n,n} & a_{n,n+1}
\end{array}
\right] $$
after applying Gaussian-elimination (without backward substitution) to this system we get
$$ \left[
\begin{array}{cccc|c}
\hat{a}_{1,1} & \hat{a}_{1,2} & \cdots & \hat{a}_{1,n} & \hat{a}_{1,n+1} \\
0 & \hat{a}_{2,2} & \cdots & \hat{a}_{2,n} & \hat{a}_{2,n+1} \\
\vdots & \ddots & \ddots & \vdots & \vdots \\
0 & \cdots & 0 & \hat{a}_{n,n} & \hat{a}_{n,n+1}
\end{array}
\right] $$
We know that the total number of multiplications/divisions and additions/subtractions of the Gaussian-elimination technique is $\frac{n^3}{3}+\frac{1}{2}n^2-\frac{5}{6}n$ and $\frac{1}{3}n^3-\frac{1}{3}n$ respectively.
Now we will use the $i$-th row to eliminate the coefficients of $x_i$ in each of the first $i-1$ rows for each $i=n,n-1,...,2$ ($i$ starts at $n$ and goes down to $2$), and we will get:
$$ \left[
\begin{array}{cccc|c}
\hat{\hat{a}}_{1,1} & 0 & \cdots & 0 & \hat{\hat{a}}_{1,n+1} \\
0 & \hat{\hat{a}}_{2,2} & \cdots & 0 & \hat{\hat{a}}_{2,n+1} \\
\vdots & \ddots & \ddots & \vdots & \vdots \\
0 & \cdots & 0 & \hat{\hat{a}}_{n,n} & \hat{\hat{a}}_{n,n+1}
\end{array}
\right] $$
Since for each $i$ we do $i - 1$ divisions (inorder to calculate $\frac{\hat{a}_1,i}{\hat{a}_{i,i}}$ , $\frac{\hat{a}_2,i}{\hat{a}_{i,i}}$ , ... , $\frac{\hat{a}_{i-1},i}{\hat{a}_{i,i}}$), After that we subtract each of the elements in the row $j=1,2,...,i-1$, $\frac{\hat{a}_j,i}{\hat{a}_{i,i}}$ times the $i$-th row, and so additional $i-1$ multiplications are carried out, and we get that the total number of multiplications/divisions in this step is $\Sigma_{i=2}^n2(i-1)=n^2-n$.
And so we get that the total number of multiplications/divisions required to apply the hybrid method is $\frac{n^3}{3}+\frac{1}{2}n^2-\frac{5}{6}n + \frac{1}{2}n^2-\frac{1}{2}n = \frac{1}{3}n^3+\frac{3}{2}n^2-\frac{11}{6}n$ which is different from $\frac{n^3}{3}+\frac{3}{2}n^2-\frac{5}{6}n$.
Thanks for any help.
|
*
*Gaussian elimination :
(n+1)(n-1)+(n)(n-2)+...+(3)(1)=sigma(k=1 to n-1)(k+2)(k)=n^3/3+n^2/2-5n/6
2.Backward gaussian method :
2(n-1)+2(n-2)+...+2(1)=2*sigma(k=1 to n-1)(k)=n^2-n
3.Find the solution :
n
Sum of multi./div. = sum of 1,2,3 = n^3/3+3n^2/2-5n/6
I tried to upload my picture to explain it in detail but i dont know how to do it...
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
How to solve $ \int \limits _0 ^{2\pi} \dfrac{dx}{(\alpha +\beta\cos x)^2} $ I am trying to solve this integral, I think that it could be solve using the complex. $$ \int \limits _0 ^{2\pi} \dfrac{dx}{(\alpha +\beta\cos x)^2} $$
|
Let $$\displaystyle I = \int \frac{1}{(\alpha+\beta \cos x)^2}dx\;,$$ Now Let $$\displaystyle t = \frac{\beta+\alpha\cos x}{\alpha+\beta \cos x}$$
So $$\displaystyle \frac{dt}{dx} = \frac{\left(\alpha+\beta \cos x\right)\cdot -(\alpha \sin x)-(\beta+\alpha \cos x)\cdot (-\beta \sin x)}{(\alpha+\beta\cos x)^2} = \frac{(\beta^2 -\alpha^2)\sin x}{(\alpha+\beta\cos x)^2}$$
So we get $$\displaystyle \frac{dt}{(\beta^2 -\alpha^2)\sin x} = \frac{dx}{(\alpha+\beta \cos x)^2}$$
So Integral $$\displaystyle I = \frac{1}{(\beta^2-\alpha^2)}\int\frac{1}{\sin x}dt$$
Now above we take $$\displaystyle t = \frac{\beta+\alpha\cos x}{\alpha+\beta \cos x}\Rightarrow \cos x = \frac{\alpha t-\beta}{\alpha-\beta t}$$
So Using $$\displaystyle \sin x= \sqrt{1-\cos^2 x} = \frac{\sqrt{\alpha^2 -\beta^2}\cdot \sqrt{1-t^2}}{(\alpha-\beta t)}$$
So Integral $$\displaystyle I = -\frac{1}{(\alpha^2-\beta^2)^\frac{3}{2}}\int \frac{(\alpha-\beta t)}{\sqrt{1-t^2}}dt $$
So $$\displaystyle I= \frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\int \frac{t}{\sqrt{1-t^2}}dt-\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\int\frac{1}{\sqrt{1-t^2}}dt$$
Now Let Put $(1-t^2) = u^2\;,$ Then $-tdt = udu$ in first Integral ,
and $t=\sin \phi\;,$ Then $dt = \cos \phi d\phi$ in Second Integral, We get
$$\displaystyle I = \frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\int \frac{-u}{u}du-\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\int\frac{\cos \phi}{\cos \phi}d\phi$$
So we get $$\displaystyle I = -\frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sqrt{1-t^2}-\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sin^{-1}(t)+\mathcal{C}$$
So we get $$\displaystyle I = -\frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sqrt{(\beta^2-\alpha^2)}\cdot \sin x -\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sin^{-1}\left(\frac{\alpha+\beta \cos x}{\beta+\alpha \cos x}\right)+\mathcal{C}$$
Now Given $$\displaystyle\int_{0}^{2\pi}\frac{1}{(\alpha+\beta \cdot \cos x)^2}dx = 2\int_{0}^{\pi}\frac{1}{(\alpha+\beta \cos x)^2}dx$$
Above we have used $$\displaystyle \bullet\; \int_{0}^{2a}f(x)dx = 2\int_{0}^{a}f(x)dx\;,$$ If $f(2a-x) = f(x)$
So $$\displaystyle 2\int_{0}^{\pi}\frac{1}{(\alpha+\beta \cos x)^2}dx = 2\left[-\frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sqrt{(\beta^2-\alpha^2)}\cdot \sin x -\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sin^{-1}\left(\frac{\alpha+\beta \cos x}{\beta+\alpha \cos x}\right)\right]_{0}^{\pi}$$
So we get $$\displaystyle \int_{0}^{2\pi}\frac{1}{(\alpha+\beta \cdot \cos x)^2}dx = \frac{2\pi\cdot \alpha}{(\alpha^2-\beta^2)\cdot \sqrt{\alpha^2-\beta^2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1412037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
cyclic three variable inequality
Let $a,b,c$ be nonnegative real numbers and $a+b+c=3$. Prove the inequality
$$
\sqrt{24a^2b+25}+\sqrt{24b^2c+25}+\sqrt{24c^2a+25}\le 21
$$
I have tried to find the solution using classical inequalities, but failed. Any idea?
|
By C-S
$$\left(\sum_{cyc}\sqrt{24a^2b+25}\right)^2\leq\sum_{cyc}(24a^2b+25)(a+3b+5c)\sum_{cyc}\frac{1}{a+3b+5c}$$
Thus, it remains to prove that
$$\sum_{cyc}(24a^2b+25)(a+3b+5c)\sum_{cyc}\frac{1}{a+3b+5c}\leq441$$ or
$$\sum_{cyc}(648a^2b+25(a+b+c)^3)(a+3b+5c)\sum_{cyc}\frac{1}{a+3b+5c}\leq441(a+b+c)^3$$ or
$$\sum_{cyc}(20a^6+47a^5b+352a^5c-209a^4b^2+509a^4c^2-275a^3b^3+$$
$$+690a^4bc-556a^3b^2c-143a^3c^2b-435a^2b^2c^2)\geq0.$$
Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that
$$4779(u^2-uv+v^2)a^4+27(263u^3-211u^2v+130uv^2+263v^3)a^3+$$
$$+9(365u^4-58u^3v-453u^2v^2+965uv^3+365v^4)a^2+$$
$$+3(173u^5+169u^4v-739u^3v^2+356u^2v^3+1156uv^4+173v^5)a+$$
$$+(u-2v)^2(20u^4+127u^3v+219u^2v^2+93uv^3+5v^4)\geq0,$$
which is obvious.
Done!
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
The differential equation $\frac{dy}{dx} = \frac{y}{x} - \frac{1}{y}\;$ I am learning differential equations and can do the basic examples. However, how can you solve the differential equation
$$\frac{dy}{dx} = \frac{y}{x} - \frac{1}{y}\;?$$
|
You already have nice solutions with substitutions. If one cannot come up with the right substitution, maybe this is a solution that works better:
Multiplying the differential equation with $y$ gives
$$
yy'=\frac{y^2}{x}-1,
$$
or
$$
\frac{1}{2}\bigl(y^2\bigr)'=\frac{y^2}{x}-1.
$$
This is a linear and first order differential equation for $y^2$, we write it as
$$
\bigl(y^2\bigr)'-\frac{2}{x}y^2=-2
$$
The integrating factor is $1/x^2$, so
$$
\Bigl(\frac{1}{x^2}y^2\Bigr)'=-\frac{2}{x^2}.
$$
Integrating,
$$
\frac{1}{x^2}y^2=\frac{2}{x}+C,
$$
where $C$ is an arbitrary constant. We can write this as
$$
y^2=2x+Cx^2\quad \text{or}\quad y=\pm\sqrt{2x+Cx^2}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1414760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Bound on maximum angle between vectors I have two vectors $\mathbf v_1$ and $\mathbf v_2$:
$$\mathbf v_1 = \begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix},
\mathbf v_2 = \begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}$$
The components of these vectors each differ by at most a constant $\epsilon$:
\begin{align}
|x_2 - x_1| &\le \epsilon\\
|y_2 - y_1| &\le \epsilon\\
|z_2 - z_1| &\le \epsilon\\
\end{align}
The angle $\theta$ between these vectors is given by:
\begin{align}
\theta = \arccos \frac{\mathbf v_1 \cdot \mathbf v_2}{\left\| \mathbf v_1 \right\| \, \left\| \mathbf v_2 \right\|}
\end{align}
How can I derive a bound on the maximum on the angle between vectors $\mathbf v_1$ and $\mathbf v_2$ only based on the constant $\epsilon$?
|
Assume that $\mathbf{v}_1$ is fixed and $\mathbf{v}_2$ can be varied within the constraints. Furthermore, assume none of $x_1,y_1,z_1$ are zero, and $\epsilon \ll \min(|x_1|,|y_1|,|z_1|)$.
WLOG consider only $\mathbf{v}_1$ in the positive octant, i.e. $x_1,y_1,z_1>0$ and with $x_1\ge y_1\ge z_1$. The angle between $\mathbf{v}_1$ and $\mathbf{v}_2$ is maximised when the endpoint of $\mathbf{v}_2$ is at one of the vertices of a cube of edge length $2\epsilon$ centered at $(x_1,y_1,z_1)$ and with faces parallel to the coordinate planes. It can be shown that the particular vertex at which the angle is maximum is $(x_1-\epsilon,y_1+\epsilon,z_1+\epsilon)$ if $x\ge(y+z)$; else $(x_1+\epsilon,y_1-\epsilon,z_1-\epsilon)$ if $x\le(y+z)$. In either case, the approximation for the upper bound on angle will be the same.
Then we have for the choice of $(x_1-\epsilon,y_1+\epsilon,z_1+\epsilon)$:
$$\begin{align}
\cos\theta&=\frac{\mathbf{v}_1 \cdot \mathbf{v}_2}{||\mathbf{v}_1||\,||\mathbf{v}_2||} \\[1em]
&=\frac{(x_1-\epsilon,y_1+\epsilon,z_1+\epsilon)\cdot(x_1,y_1,z_1)}{||(x_1-\epsilon,y_1+\epsilon,z_1+\epsilon)||\,||(x_1,y_1,z_1)||} \\[1em]
&=\frac{x_1(x_1-\epsilon)+y_1(y_1+\epsilon)+z_1(z_1+\epsilon)}{\sqrt{x_1^2+y_1^2+z_1^2}\cdot\sqrt{(x_1-\epsilon)^2+(y_1+\epsilon)^2+(z_1+\epsilon)^2}} \tag{1}
\end{align}$$
Now let
$$s_1=x_1-y_1-z_1 \tag{2}$$ and $$s_2=x_1^2+y_1^2+z_1^2 \tag{3}$$
which are fixed by our choice of $\mathbf{v}_1$. Then (1) reduces to
$$\cos\theta=\frac{s_2-s_1\epsilon}{\sqrt{s_2}\cdot\sqrt{s_2-2s_1\epsilon+3\epsilon^2}}$$
so
$$\begin{align}
\sec\theta &= \frac{\sqrt{1-2\dfrac{s_1}{s_2}+3\dfrac{\epsilon^2}{s_2}}}{1-\dfrac{s_1}{s_2}\epsilon} = \frac{\sqrt{1-2\dfrac{s_1}{s_2}+\dfrac{s_1^2}{s_2^2}\epsilon^2+\dfrac{(3s_2-s_1^2)}{s_2^2}\epsilon^2}}{\sqrt{1-2\dfrac{s_1}{s_2}\epsilon+\dfrac{s_1^2}{s_2^2}\epsilon^2}} \\[1em]
&= \sqrt{1+\frac{\dfrac{(3s_2-s_1^2)}{s_2^2}\epsilon^2}{\left(1-\dfrac{s_1}{s_2}\epsilon\right)^2}} \\[1em]
&\approx 1+\frac{(3s_2-s_1^2)}{2s_2^2}\epsilon^2 \tag{4}
\end{align}$$
because $\sqrt{1+x}=1+\frac{1}{2}x+\frac{(\frac{1}{2})(-\frac{1}{2})}{2}x^2+\ldots \approx 1+\frac{1}{2}x$ by the binomial series, and in the denominator $1-\frac{s_1}{s_2}\epsilon \to 1$ as $\epsilon \to 0$.
But the Taylor Series for $\sec\theta$ is:
$$\sec\theta=1+\frac{1}{2}\theta^2+\ldots \approx 1+\frac{1}{2}\theta^2 \tag{5}$$
So equating the approximations in (4) and (5):
$$\theta_{max} \approx \frac{\sqrt{3s_2-{s_1}^2}}{s_2}\,\epsilon \tag{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Trying to show $|\overrightarrow{a}\times\overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}⋅\overrightarrow{b})^2$
If $\overrightarrow{a} = \langle a_1, a_2, a_3 \rangle$ and $\overrightarrow{b} = \langle b_1, b_2, b_3 \rangle$, then the cross product of $\overrightarrow{a}$ and $\overrightarrow{b}$ is the vector
$$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$$
Using the above definition show algebraically that
$$|\overrightarrow{a} \times \overrightarrow{b}|^2=|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2$$
I'm not sure if I'm on the right track with this problem, but I've started with the given expression
$\overrightarrow{a} \times \overrightarrow{b} =(a_2 b_3 - a_3 b_2)\overrightarrow{i} + (a_3 b_1 - a_1 b_3)\overrightarrow{j} + (a_1 b_2 - a_2 b_1)\overrightarrow{k}$
with the goal of algebraically manipulating it into the form
$|\overrightarrow{a}|^2|\overrightarrow{b}|^2-(\overrightarrow{a}\cdot\overrightarrow{b})^2$
Consider the steps:
$|\overrightarrow{a} \times \overrightarrow{b}|^2$
$|\langle a_2 b_3 - a_3 b_2, a_3 b_1 - a_1 b_3, a_1 b_2 - a_2 b_1 \rangle|^2$
$\sqrt{(a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2}^2$
$(a_2 b_3 - a_3 b_2)^2 + (a_3 b_1 - a_1 b_3)^2 + (a_1 b_2 - a_2 b_1)^2$
which expands to:
$(a_3^2 b_2^2 - 2 a_2 a_3 b_3 b_2 + a_2^2 b_3^2) + (a_3^2 b_1^2 - 2 a_1 a_3 b_3 b_1 + a_1^2 b_3^2) + (a_2^2 b_1^2 - 2 a_1 a_2 b_2 b_1 + a_1^2 b_2^2)$
but I'm not really sure where to go from here. I could group the squared terms together:
$a_3^2 b_2^2 + a_2^2 b_1^2 + a_2^2 b_3^2 + a_3^2 b_1^2 + a_1^2 b_3^2 + a_1^2 b_2^2 - 2 a_2 a_3 b_3 b_2 - 2 a_1 a_3 b_3 b_1 - 2 a_1 a_2 b_2 b_1$
but no recognizable forms are really achieved (like the expanded expression for a dot product, $\overrightarrow{a} \cdot \overrightarrow{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$).
How I algebraically prove this? Thanks for your help!
|
Let me try.
You already have: $$\begin{eqnarray}LHS &=& a_1^2b_2^2 + a_1^2b_3^2 + a_2^2b_1^2 + a_2^2b_3^2 + a_3^2b_1^2 + a_3^2b_2^2 - 2 a_1a_2b_1b_2 - 2 a_2a_3b_2b_3 - 2a_3a_1b_3b_1\\ & =& a_1^2b_2^2 + a_1^2b_3^2 + a_2^2b_1^2 + a_2^2b_3^2 + a_3^2b_1^2 + a_3^2b_2^2 + a_1^2b_1^2 + a_2^2b_2^2 + a3^2b_3^2 -(a_1^2b_1^2 + a_2^2b_2^2 + a3^2b_3^2 + 2 a_1a_2b_1b_2 + 2 a_2a_3b_2b_3 + 2a_3a_1b_3b_1) \\
&=& (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2) - (a_1b_1+a_2b_2+a_3b_3)^2 \\
&=& RHS\end{eqnarray}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimum value of $\cos x+\cos y+\cos(x-y)$ What is the minimum value of $$ \cos x+\cos y+\cos(x-y). $$ Here $x,y$ are arbitrary real numbers. Mathematica gives (with NMinimize) $-3/2$. But I don't know if this is correct and if so, how to prove it.
|
The local max or min of a 2-variable function comes where both partial derivatives are 0. So if we say that
$$
z = \cos x+\cos y+\cos(x-y)
$$
then
$$
\frac{\partial z}{\partial x} = \sin x + \sin(x-y) = 0
$$
and
$$
\frac{\partial z}{\partial y} = \sin y - \sin(x-y) = 0
$$
adding the two yields
$$
\begin{align}
\sin (x) + \sin(y) &= 0
\\ \rightarrow
\sin (x) &= -\sin(y)
\\ \rightarrow
x &= -y
\end{align}
$$
plugging that into the first partial yields
$$
\begin{align}
\sin x + \sin(x+x) &= 0
\\ \rightarrow
\sin x &= - \sin(2x)
\\ \rightarrow
x &= -2x + 2\pi\text{*}
\\ \rightarrow
3x &= 2\pi
\\ \rightarrow
x &= \frac{2\pi}{3} \approx 2.1
\end{align}
$$
and
$$
y = -\frac{2\pi}{3} \approx -2.1
$$
plugging those values into the original equation yields
$$
\begin{align}
\mathcal z &= \cos (\frac{2\pi}{3})+\cos (-\frac{2\pi}{3})+\cos(\frac{4\pi}{3})\\
&= -\frac{1}{2} -\frac{1}{2} -\frac{1}{2}\\
&= -\frac{3}{2}
\end{align}
$$
*x=0 would fit as well but that would be a local maximum. The proof of that is left for the student.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Confusing probability problems based on product rule and combinations I am going thru probability exercise. Faced first problem:
Book Q1. Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. What is the probability the largest number appearing on the selected tickets is 7?
My logic: if one of six tickets should be 7, the $\color{red}{\text{remaining 5}}$ can be any of 1 to 7, so it should be $7^5$.
But turns out that the given solution is $\frac{7^6-6^6}{10^6}$.
My Q1. Though I understood the logic behind $\frac{7^6-6^6}{10^6}$, I was wondering what is exact logical mistake with $7^5$? I guessed that $7^5$ completely ignores what should be 6th ticket, it only puts restriction on 5 tickets. Is it like that?
Then I came across similar but more involved problem, with significant difference from above one that it performs action without replacement:
Book Q2. Three numbers are chosen at random without replacement from (1,2,3,...,10). What is the probability that the minimum number is 3 or the maximum number is 7?
My logic: Noticing that this is without replacement, I guessed the solution should be
$$
=
\begin{pmatrix}
\text{selections with}\\
\text{minimum}\\
\text{number is 3}
\end{pmatrix}
+
\begin{pmatrix}
\text{selections with}\\
\text{maximum}\\
\text{number is 7}
\end{pmatrix}
-
\begin{pmatrix}
\text{selections with}\\
\text{maximum}\\
\text{number is 7}\\
\text{and minimum}\\
\text{number is 3}
\end{pmatrix}
$$
$$=
\frac{
\overbrace{(\binom{8}{3}-\binom{7}{3})}^{\text{#selections with min 3}}
+
\overbrace{(\binom{7}{3}-\binom{6}{3})}^{\text{#selections with max 7}}
-
\overbrace{3\times {^3P_3}}^{\text{#selections with max 7 and min 3}}
}
{\binom{10}{3}
}
$$
But the book solutions says:
P(minimum 3) or P(maximum 7)
P(minimum 3) $=\frac{\binom{7}{2}}{\binom{10}{3}}=\frac{21}{120}$
P(max 7) $=\frac{\binom{6}{2}}{\binom{10}{2}}=\frac{15}{120}$
Thus the solution is $\frac{11}{40}$
My Q2. How even by books logic the solution $\frac{11}{40}$ is achieved. I am unable to understand it as I find the explanation insufficient.
My Q3. If book Q2 answer is correct then why for book Q1 solution is not $7^5$ which is what I initially guessed (because the only difference being with / without replacement, the logic of getting $\color{red}{\text{remaining m}}$ stuffs out n should remain same)?
My Q4. If we make first question without replacement, will the solution be $\frac{\binom{7}{6}-\binom{6}{6}}{\binom{10}{6}}$?
My Q5. What will be the solution if we make book Q2 with replacement?
My Q6. Where my logic for solution to Book's Q2 is wrong?
|
Addressing your Q1, the difference between the answer you gave and the answer that the book gives is that your solution assumes that a specific selection is 7. That is, $\frac{7^5}{10^6}$ is the probability that a fixed ticket is 7 and the other five take values from 1 to 7 while the books solution just assumes that at least one of the tickets is 7.
Q2 The second problem is much different as there is no order on the sets of chosen numbers. Note that $\binom{8}{3}-\binom{7}{3}=\binom{7}{2}$ and $\binom{7}{3}-\binom{6}{3}=\binom{7}{2}$, so that part of your answer is consistent with the book solution--though the logic that the book uses might be a tad bit more straightforward. In each of the above cases, the book solution fixes the given number and then chooses two other numbers at random from the allowed set (either $\{$greater than 3$\}$ or $\{$less than 7$\}$). Using this type of reasoning we can quickly enumerate the number of choices with max 7 and min 3: two of the choices are fixed and there are only three possibilities for the third choice. Hence the solution is
$$
\frac{\binom{7}{2}+\binom{6}{2}-3}{\binom{10}{3}}=\frac{11}{40}.
$$
For your Q3, again, the implied order in the first problem makes it easy to accidentally specify an order during the solution process (and as a result get the wrong answer). Note that your solution for the second problem was not necessarily wrong (I don't know what $3\times^3P_3$ means), you just used a different method than the book.
For your Q4 you are correct. This is the same thing as $$\frac{\binom{7}{5}}{\binom{10}{6}}.$$
Q5 The answer would be
$$
\frac{(8^3-7^3)+(7^3-6^3)-5}{10^3}.
$$
Let me know if you have any questions about how I arrived at the above answer.
|
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|
Solving the infinite series $1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$ I have the following question:
Evaluate the infinite series:
$$S=1-\frac{2^3}{1!}+\frac{3^3}{2!}-\frac{4^3}{3!}+\cdots$$
(a) $\displaystyle\frac1e$ (b) $\displaystyle\frac{-1}e$ (c) $\displaystyle\frac{2}e$ (d) $\displaystyle\frac{-2}e$
Now in the book they have given a strange explanation using $$(n+1)^3=[n(n-1)(n-2)+6n(n-1)+7n+1].$$ I don't understand how did they get this. So I have two doubts:
*
*How did they get this "trick"?
*If you would have got this question in a competitive exam, how would you have solved it (in the minimum amount of time), either directly or with numerical methods?
|
Since $(n+1)^3 = \color{red}{1}\cdot n(n-1)(n-2)+\color{red}{6}\cdot n(n-1) +\color{red}{7}\cdot n+\color{red}{1}$ we have:
$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{(-1)^n (n+1)^3}{n!} &=& \frac{13}{2}+\sum_{n\geq 3}\frac{(-1)^n (n+1)^3}{n!}\\&=&\frac{13}{2}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-3)!}+\color{red}{6}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-2)!}+\color{red}{7}\cdot\sum_{n\geq 3}\frac{(-1)^n}{(n-1)!}+\color{red}{1}\cdot\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\frac{13}{2}-\sum_{n\geq 0}\frac{(-1)^n}{n!}+6\cdot\sum_{n\geq 1}\frac{(-1)^n}{n!}-7\cdot\sum_{n\geq 2}\frac{(-1)^n}{n!}+\sum_{n\geq 3}\frac{(-1)^n}{n!}\\&=&\sum_{n\geq 0}(-1+6-7+1)\frac{(-1)^n}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n}{n!}=\color{red}{-\frac{1}{e}}.\end{eqnarray*}$$
We simply get the initial decomposition by induction. Obviously $(n+1)^3-n(n-1)(n-2)$ is a quadratic polynomial in $n$, whose leading term is $6n^2$. So we substract $6n(n-1)$ and we are left with a linear polynomial and so on. Another chance is to tackle our problem this way:
$$\begin{eqnarray*} \sum_{n\geq 0}\frac{(-1)^n(n+1)^3}{n!}&=&\sum_{n\geq 1}\frac{(-1)^n n(n+1)^2}{n!}+\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}\\&=&\sum_{n\geq 0}\frac{(-1)^n(n+1)^2}{n!}-\sum_{n\geq 0}\frac{(-1)^n(n+2)^2}{n!}\\&=&-\sum_{n\geq 0}\frac{(-1)^n(2n+3)}{n!}\\&=&-3\sum_{n\geq 0}\frac{(-1)^n}{n!}+2\sum_{n\geq 0}\frac{(-1)^n}{n!}=\frac{2-3}{e}.\end{eqnarray*}$$
|
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|
Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$
Find the area of the circle that falls between the circle $x^2+y^2=5$ and the lines $x^2-4y^2+6x+9=0$.
I tried to solve this question. The lines are $x-2y+3=0$ and $x+2y+3=0$ which intersect at $(-3,0)$ and the circle has its center at $(0,0)$ and radius $\sqrt5$. The figure was asymmetric and calculations were difficult. So I shifted the origin to $(-3,0)$. In the new system the equations become $X+2Y=0$ and $X-2Y=0$ and the circle equation becomes $(X-3)^2+Y^2=5$. Now I found the point of intersection of lines and the circle which comes out to be $(\frac{4}{5},\frac{2}{5})$
Now when I integrate and required area is 4 times the integral $\int_{0}^{4/5}\frac{x}{2}dx+\int_{4/5}^{\sqrt5}\sqrt{5-(x-3)^2}dx$
but the answer $5(\pi-\arcsin\frac{4}{5})+\frac{24}{5}$ is still elusive and the calculations have not simplified even after origin shifting. Is the origin shifting futile. What is the correct way to solve it? Please help me.
|
If I may refer to sketch of Omran Kouba, you need even not rotate each pink segment by $\pm \tan ^{-1} \frac12$, you can straightaway use the standard formula of area of circular segment ( subtract sector and triangle areas)..
|
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|
Evaluating the limit $\lim_{x\rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}$ The question is :
$$\lim_{x\rightarrow 1} \frac{\sqrt{x+3}-2}{x-1}$$
I know I probably have to do some sort of factorisation of the numerator in order to cancel the denominator, but the surd has me stumped I'm afraid.
|
If you multiply both the numerator and the denominator by the conjugate of the expression, you'll get:
\begin{align*}
\lim_{x \to 1}\frac{\sqrt{x + 3} -2}{x - 1} \cdot \frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} & = \lim_{x \to 1} \frac{x + 3 - 4}{(x-1)(\sqrt{x + 3} + 2)}\\
& = \lim_{x \to 1} \frac{x-1}{(x-1)(\sqrt{x + 3} + 2)}\\
& = \lim_{x \to 1} \frac{1}{\sqrt{x + 3} + 2}\\
& = \frac{1}{4}
\end{align*}
Like pointed out by André Nicolas, this trick (multiplying by the conjugate) is quite classic.
|
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|
Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$
Find roots of equation $(x^2+1)\cdot \arccos\left(\frac{2x}{1+x^2}\right)+2x\cdot \mathrm{sgn}(x^2-1)=0$
One root is $x=1$ (checking functions $\arccos$ and $\mathrm{sgn}$).
Second root is $x=0.442$.
How to find the second root?
How to check if this equation only has two roots?
|
We can start by simplifying the sign function. We know that $x^2 - 1$ is negative between $-1 < x < 1$ and positive or zero otherwise (I'm assuming that's the definition you're using for it), so we have the following two equations:
$$(x^2 + 1) \cdot \arccos{\left(\frac{2x}{1 + x^2}\right)} - 2x = 0, -1 < x < 1$$
$$(x^2 + 1) \cdot \arccos{\left(\frac{2x}{1 + x^2}\right)} + 2x = 0, x < -1, x > 1$$
The first equation simplifies to :
$$ \arccos{\left(\frac{2x}{1 + x^2}\right)} = \frac{2x}{x^2 + 1}$$
We can do the division since $x^2 + 1$ is always positive. Now, let's use the substitution $u = \frac{2x}{x^2 + 1}$:
$$\arccos{(u)} = u$$
Which has the sole solution $u \approx 0.739085$. Back substituting u and solving for x gives $0.739085 = \frac{2x}{x^2 + 1}$ which leads to $x = 0.442$ (the one you mentioned) as well as $x = 2.264$.
Now, noting that the second equation reduces to:
$$ \arccos{\left(\frac{2x}{1 + x^2}\right)} = \frac{-2x}{x^2 + 1}$$
Can you use the same substitution and continue the process to find the roots?
|
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|
If $(x+1)^4+(x+3)^4=4$ then how to find the sum of non-real solutions of the given equation? If $(x+1)^4+(x+3)^4=4$ then how to find the sum of non-real solutions of the given equation?
I took $x+2=t$ and got $t^2=-6+\sqrt(40)/2$.How to proceed?
|
Put $t=x+2$ then you obtain $$(t-1)^4 +(t+1)^4 =4 .$$
hence $$((t-1)^2)^2 -2(t^2-1 )^2 +((t+1)^2)^2 =4 -2(t^2-1 )^2$$ hence $$16t^2 =4 -2(t^2-1 )^2$$ and let $s=t^2 -1 $ we obtain $$16 s +12 +2s^2 =0$$ therefore $$s^2 +8s +6 =0$$ thus $$x=-2\pm i\sqrt{3+\sqrt{10}}$$ or $$x=-2 \pm \sqrt{\sqrt{10}- 3}$$
|
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|
Finding the exact value of $\tan(\cos^{-1} x)$
Find the exact values of:
a) $\tan(\cos^{-1} x)$
b) $\cos(\tan^{-1} x)$
c) $\sec(\sin^{-1} x)$
My work is the following.
a) $\tan x = \sin x / \cos x$ so
$$\tan(\cos^{-1} x) = \sin(\cos^{-1} x) / \cos(\cos^{-1} x) = \sin(\cos^{-1}x) / x$$
b) $\cos(\tan^{-1} x) = \cos(\sin^{-1} x / \cos^{-1} x)$
c) $\sec(\sin^{-1} x)= 1/\cos(\sin^{-1} x)$
Is my answer correct?
|
Useful things to note
$$
\tan^2 x + 1 = \sec^2 x \implies \tan x = \pm\sqrt{\frac{1}{\cos^2 x}-1}
$$
And
$$
\cos^2 x+\sin^2 x = 1\implies \cos x =\pm\sqrt{1-\sin^2 x}
$$
Can you use these to your advantage?
|
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|
What is the sum $\sum_{k=0}^{n}k^2\binom{n}{k}$? What should be the strategy to find
$$\sum_{k=0}^{n}k^2\binom{n}{k}$$
Can this be done by making a series of $x$ and integrating?
|
Let $$\displaystyle S=\sum^{n}_{k=0}k^2 \binom{n}{k} \;,$$ Now using $\displaystyle \bullet \; \binom{n}{k} = \frac{n}{k}\cdot \binom{n-1}{k-1}$
So we get $$\displaystyle S = \sum^{n}_{k=0}k^2 \cdot \frac{n}{k}\binom{n-1}{k-1}\;,$$ Again using $\displaystyle \bullet \; \binom{n-1}{k-1} = \frac{n-1}{k-1}\cdot \binom{n-2}{k-2}$
So $$\displaystyle S = n\sum^{n}_{k=1}k\binom{n-1}{k-1} = n\sum^{n}_{k=0}[(k-1)+1]\binom{n-1}{k-1}$$
So we get $$\displaystyle S=n\sum^{n}_{k=1}(k-1)\binom{n-1}{k-1}+n\sum^{n}_{k=1} \binom{n-1}{k-1}$$
So we get $$\displaystyle S=n\sum^{n}_{k=2}(k-1)\cdot \frac{n-1}{k-1}\cdot \binom{n-2}{k-2}+n\sum^{n}_{k=1}\binom{n-1}{k-1}$$
So we get $$\displaystyle S=n(n-1)\sum^{n}_{k=0}\binom{n-2}{k-2}+n\sum^{n}_{k=0}\binom{n-1}{k-1}$$
Now Using $\displaystyle \bullet\; \sum^{n}_{k=0}\binom{n}{k} = 2^{k}$
So we get $$\displaystyle S = n(n-1)\cdot 2^{n-2}+n\cdot 2^{n-1}$$
|
{
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|
Solving an integral with substitution method I want a hint for the following integral:
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}}$$
where $a$ is a real constant.
In fact,
$$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}} = -\frac{\sqrt{1- a^{2}+x^{2}}}{x(1- a^{2})}$$
why?
|
Assume a$\ne $$1 or-1$,Put $x$=$\sqrt {1-a^2}$$tan$$\theta$ and then simplify..
$\int \frac{dx}{x^{2}\sqrt{1- a^{2}+x^{2}}}$=$\int \frac{sec^2\theta d\theta}{({1-a^2})tan^2\theta sec\theta}$=$\int \frac{cos\theta d\theta}{({1-a^2})sin^2\theta}$=
$\int \frac{d(sin\theta)}{({1-a^2})sin^2\theta}$=$\frac{-1}{({1-a^2})sin\theta}$+c
=-$\frac{\sqrt{1-a^2+x^2}}{x(1-a^2)}$+c
|
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|
Minimum of $\frac{1}{x+y+z}+\frac{1}{x+y+w}+\frac{1}{x+z+w}-\frac{2}{x+y+z+w}$ Let $x,y,z,w\geq 0$ and $0\leq x+y,y+z,z+x,x+w,y+w,z+w\leq 1$. What is the minimum of $$F(x,y,z,w)=\frac{1}{x+y+z}+\frac{1}{x+y+w}+\frac{1}{x+z+w}-\frac{2}{x+y+z+w}?$$
We have $F(1/2,1/2,1/2,1/2)=F(1,0,0,0)=1$, so the minimum is at most $1$. Since the constraints are on $x+y,y+z$, etc. instead of $x,y,z,w$, taking partial derivative with respect to $x,y,z,w$ doesn't help much. We cannot change one variable without affecting the others. Moreover, it is not easy to create new variables $a=x+y,b=y+z,\ldots$ and write $F$ nicely in terms of $a,b,\ldots$
|
Not a very easy problem, but the following solution is nice enough ;)
Denote $a=y+z,b=z+w,c=y+w$ then $y=\frac{a+c-b}{2},z=\frac{a+b-c}{2},w=\frac{b+c-a}{2}$.
The constraints become
\begin{align}
x\ge 0 \\
a+b-c\ge 0 \\
b+c-a\ge 0\\
c+a-b\ge 0 \\
0\le a,b,c \le 1 \\
x+ \frac{a+b-c}{2} \le 1 \\
x+ \frac{b+c-a}{2} \le 1 \\
x+ \frac{c+a-b}{2} \le 1
\end{align}
and the function $F$ becomes
$$F = \frac{1}{a+x} + \frac{1}{b+x} + \frac{1}{c+x} - \frac{2}{x+\frac{a+b+c}{2}}.$$
WLOG, suppose that $c=\min(a,b,c)$.
We will show that $F\ge 1$.
From the well-known inequality $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \ge \frac{9}{p+q+r} \quad \forall p,q,r > 0$ with equality iff $p=q=r$ (which follows from AM-GM inequality: $(p+q+r)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right) \ge 3\sqrt[3]{pqr}\cdot 3 \sqrt[3]{\frac{1}{pqr}}=9$), we have
$$F\ge \frac{9}{(a+x)+(b+x)+(c+x)} - \frac{4}{2x+a+b+c},$$
or $F\ge f(x)$ where $$f(x) = \frac{9}{3x+s} - \frac{4}{2x+s}$$ and $s=a+b+c$.
It is easy to show that $f(x)$ is decreasing on $[0,+\infty)$ by taking its derivative, or by just re-writing it as $$f(x) = \frac{2s}{(3x+s)(2x+s)} + \frac{3}{3x+s}.$$
Thus, since $x\le 1-\frac{a+b-c}{2}$ we have
\begin{align}
f(x) \ge f\left(1-\frac{a+b-c}{2}\right) &= \frac{9}{3\left(1-\frac{a+b-c}{2}\right) + a+b+c} - \frac{4}{2\left(1-\frac{a+b-c}{2}\right) + a+b+c} \\
&=\frac{18}{6+5c-a-b} - \frac{2}{c+1} \\
&\ge \frac{18}{6+3c} - \frac{2}{c+1} \quad (\text{since }a+b\ge 2c) \\
&= 1 + \frac{c(1-c)}{(c+1)(c+2)} \\
&\ge 1.
\end{align}
Equality occurs if and only if $a=b=c$ and $x=1-\frac{a+b-c}{2}$ and $(c=0 \text{ or } c=1)$, or equivalently, $(x,a,b,c)=(1,0,0,0)$ or $(x,a,b,c)=(1/2,1,1,1)$, i.e., $(x,y,z,w)=(1,0,0,0)$ or $(x,y,z,w)=(1/2,1/2,1/2,1/2)$.
We are done.
|
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|
Coefficient of the generating function $G(z)=\frac{1}{1-z-z^2-z^3-z^4}$ I am seeking the coefficient $a_n$ of the generating function
$$G(z)=\sum_{k\geq 0} a_k z^k = \frac{1}{1-z-z^2-z^3-z^4}$$
The combinatorial background of this question is to solve the recurrence
$$a_n=a_{n-1}+a_{n-2}+a_{n-3}+a_{n-4},\qquad (a_0,a_1,a_2,a_3)=(1,1,2,4).$$
My first idea was to use that $\frac{1}{1-x}=1+x+x^2+...$ which after some expansion leads to
$$a_n = \sum_{k_1+2k_2+3k_3+4k_4 = n}\binom{k_1+k_2+k_3+k_4}{k_1}\binom{k_2+k_3+k_4}{k_2}\binom{k_3+k_4}{k_3}$$
At this point I have no idea how to continue. Looking for a recurrence for $a_n$ would mean to run in circles. I feel that another combinatorial technique is needed, which I dont know. Any ideas?
|
Here is another variant to extract the coefficients $a_n$ resulting in a different formula.
We obtain
\begin{align*}
\frac{1}{1-z-z^2-z^3-z^4}&=\frac{1}{1-z(1+z+z^2+z^3)}\\
&=\frac{1}{1-z\frac{1-z^4}{1-z}}\tag{1}\\
&=\frac{1-z}{1-2z+z^5}\\
&=(1-z)\sum_{k=0}^\infty z^k(2-z^4)^k\tag{2}\\
&=(1-z)\sum_{k=0}^\infty z^k\sum_{l=0}^k\binom{l}{k}(-1)^lz^{4l}2^{k-l}\tag{3}
\end{align*}
Comment:
*
*In (1) we use the formula for the finite geometric series.
*In (2) we apply the geometric series expansion.
*In (3) we apply the binomial theorem.
Denoting with $[z^n]$ the coefficient of $z^n$ of a series we obtain from (3)
\begin{align*}
[z^n]\sum_{k=0}^\infty &z^k\sum_{l=0}^k\binom{k}{l}(-1)^lz^{4l}2^{k-l}\\
&=\sum_{k=0}^n [z^{n-k}]\sum_{l=0}^k\binom{k}{l}(-1)^lz^{4l}2^{k-l}\tag{4}\\
&=\sum_{k=0}^n[z^k]\sum_{l=0}^{n-k}\binom{n-k}{l}(-1)^lz^{4l}2^{n-k-l}\tag{5}\\
&=\sum_{k=0}^{\lfloor{n/4}\rfloor}[z^{4k}]\sum_{l=0}^{n-4k}\binom{n-4k}{l}(-1)^lz^{4l}2^{n-4k-l}\tag{6}\\
&=\sum_{k=0}^{\lfloor{n/4}\rfloor}\binom{n-4k}{k}(-1)^k2^{n-5k}\tag{7}\\
\end{align*}
Comment:
*
*In (4) we apply the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$ and set the upper limit of the outer sum to $n$ since the power of $z^{n-k}$ is non-negative.
*In (5) we change the order of summation $k\rightarrow n-k$.
*In (6) we observe that due to the factor $z^{4l}$ only multiples of $4$ contribute to the coefficient of $z^n$.
*In (7) we select the coefficient with $k=l$ accordingly.
Respecting the factor $(1-z)$ in (3) we conclude from (3) and (7) the coefficients $a_n$ of the series $\frac{1}{1-z-z^2-z^3-z^4}$
are given by
\begin{align*}
\qquad\qquad\color{blue}{a_n=\sum_{k=0}^{\lfloor{n/4}\rfloor}\left[\binom{n-4k}{k}-\frac{1}{2}\binom{n-4k-1}{k}\right](-1)^k2^{n-5k}}\qquad\qquad
\color{blue}{n\geq 0}
\end{align*}
In general we have
\begin{align*}
\qquad\qquad&[z^n]\frac{1}{1-z-z^2-\cdots-z^p}=[z^n]\frac{1-z}{1-2z+z^{p+1}}\\
&\qquad=\sum_{k=0}^{\lfloor{n/p}\rfloor}\left[\binom{n-pk}{k}-\frac{1}{2}\binom{n-pk-1}{k}\right](-1)^k2^{n-(p+1)k}\qquad\ \ n\geq 0
\end{align*}
|
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|
Determinant of remainder of a primitive matrix modulo 2 I'm trying to prove the following relation for a matrix $A\in \mathbb{Z}^{m\times m} $, $m\geq 2$. It is assumed that the characteristic polynomial of $A$ is primitive modulo $2$:
If $C$ is defined to be the remainder of $A^{2^m-1}\pmod{4}$ , i.e.
$$A^{2^m-1}\equiv I+2C \pmod{4}$$
Then prove
$$\textrm{det}(C)\equiv \textrm{det}(C+I)\equiv 1 \pmod{2}$$
My try:
$$2C\equiv A^{2^m-1}-I\pmod{4}\Rightarrow C\equiv \frac{1}{2}(A^{2^m-1}-I)\pmod{2}$$
$$ \textrm{det}(C)\equiv \textrm{det}(\frac{1}{2}(A^{2^m-1}-I)) \pmod{2}$$
Then for $C+I$
$$C+I\equiv \frac{1}{2}(A^{2^m-1}-I)+I \equiv \frac{1}{2}(A^{2^m-1}+I)\pmod{2}$$
Therefore to prove the statement we must have:
$$ \textrm{det}(\frac{1}{2}(A^{2^m-1}-I))\equiv \textrm{det}(\frac{1}{2}(A^{2^m-1}+I)) \equiv 1 \pmod{2} $$
Or $$\frac{1}{2^m}\textrm{det}(A^{2^m-1}-I)\equiv \frac{1}{2^m}\textrm{det}(A^{2^m-1}+I) \equiv 1 \pmod{2}$$
$$\Rightarrow\textrm{det}(A^{2^m-1}-I)\equiv \textrm{det}(A^{2^m-1}+I) \equiv 2^{m} \pmod{2^{m+1}}$$
If we call $P_{A^{2^m-1}}(\lambda)$ the characteristic polynomial of $A^{2^m-1}$ then I know that $\textrm{det}(A^{2^m-1}-\lambda I)\equiv P_{A^{2^m-1}}(\lambda)$, and therefore
$$\textrm{det}(A^{2^m-1}-I)= P_{A^{2^m-1}}(1)\,,\textrm{det}(A^{2^m-1}+I)= P_{A^{2^m-1}}(-1)$$
EDIT: I've found out here that if eigenvalues of $A$ are $y_i$ then eigenvalues of $A^n$ will be $y_i^n$, however $y_i$ might be complex. Therefore:
$$P_{A}(\lambda)=\prod_{i=0}^m(\lambda-y_i)\Rightarrow P_{A^{2^m-1}}(\lambda)=\prod_{i=0}^m(\lambda-y_i^{2^m-1})$$
Then I tried to compute each coefficient of $P_{A^{2^m-1}}$ in terms of coefficients of $P_{A}$, which leads to some kind of generalization of Newton identities I asked it here, however it seems it is not possible to compute all of them (Question has received no general answer yet).
I couldn't continue it further.
Notes:
I've tested a few matrices (of degrees 2,3,4,5) with computer and the statement was true for them.
Primitive here means that the characteristic polynomial of $A$, $q(t)=\textrm{det}(A-tI)=t^m+a_1t^{m-1}+\ldots+a_m$, be primitive (irreducible) modulo 2.
$$A=\left(\begin{array}{cccc}
a_1 & \ldots & a_{m-1}& a_m\\
1 & \ldots & 0 & 0\\
\vdots & \ddots & \vdots &\vdots\\
0 & \ldots &1 & 0
\end{array}\right)$$
Important Edit:
An additional assumption is required for $P_A(x)$ modulo 4. Which is:
If $P_A(x)\equiv f(x^2)+xg(x^2)\pmod 2$ then the above claim is not true only if:
$P_A(x)\equiv f(x)^2+xg(x)^2\pmod 4$
|
The claim is not true for $m=2$: Take
$$
A=\pmatrix{1& 3\\1& 0}, \quad p_A = x^2-x-3
$$
$$
A^3 =\pmatrix{7 & 12 \\ 4 & 3} \equiv \pmatrix{3&0\\0&3} \equiv I + 2I \pmod 4
$$
hence $2C \equiv 2I \pmod 4$, $C\equiv I \pmod 2$, and $C+I\equiv 0\pmod2$, in contradiction to the claim $det(C+I)\equiv 1\pmod 2$.
Here is another counterexample for $m=3$:
$$
A =\pmatrix{0&0&1\\ 1&0&3\\0&1&2},
\quad
A^7 \equiv I \pmod 4,
$$
hence $C=0$. The characteristic polynomial is $p_A=x^3+x+1$ modulo $2$, which is irreducible.
In the constructions above, I exploited the following. Denote by $p_{A,2}$ and $p_{A,4}$ the characteristic polynomials of $A$ modulo $2$ and $4$. The difference between them, $p_{A,4} - p_{A,2}$ was chosen to have degree $m-1$.
This difference has even coefficients.
Then using the factorization of $A^{2^m-1}-I=(A-I)(A^{2^m-2}+\dots + I)$, using $p_{A,4}(A)=0$ to reduce the degree of the polynomial, one gets an expression for $C$, which depends on $A$ and $(p_{A,4}-p_{A,2})/2$. Chosing the latter to get $\det(C)=0$ was then the last step.
In the case $m=2$, the claim is true if and only if $p_{A,4}$ has at least one coefficient from $\{2,3\}$. Here, $p_{A,2}=x^2+x+1$ anyway.
It holds
$$
A^3-1 = (A-I)(A^2+A+I).
$$
If $p_{A,4} = x^2+x+1$, then $A^3-I\equiv 0 \pmod 4$, and $C=0$.
If not, then $p_{A,4} = x^2+x+1 + 2q$, where $q\in \{1,x,1+x\}$. Then
$$
A^3-1 \equiv (A-I)(2M) \pmod 4,
$$
where $M\in \{I,A,I+A\}$. Hence
$$
C\equiv (A-I)M \equiv (A+I)M\pmod 2,
$$
and $\det(C)\equiv 1\pmod 2$, since
$$
\det(A) \equiv p_{A,2}(0) = 1, \ \det(I+A)\equiv p_{A,2}(1) = 1 \pmod 2.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Does the series $\sum\frac{(-1)^n cos(3n)}{n^2+n}$ converge absolutely? $\sum|\frac{(-1)^n \cos(3n)}{n^2+n}| \le \sum\frac{1}{n^2+n}$
since $-1 \le cos(3n) \le 1$
$\sum\frac{1}{n^2+n} = \sum\frac{1}{n(n+1)} = \sum(\frac{1}{n} - \frac{1}{n+1})$
$\int(\frac{1}{x} - \frac{1}{x+1}) dx = \log|x| + \log|x+1| + C$
which diverges.
Therefore by the integral test, the series diverges.
However, the answer is that the series converges absolutely. Where have I gone wrong?
|
Of course the mistake lies in the integral test:
$$
\frac{d}{dx} \left( \log |x| + \log |x+1| \right) = \frac{1}{x}+\frac{1}{x+1},
$$
so your antiderivative is wrong.
|
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|
Find $n$, where its factorial is a product of factorials I need to solve $3! \cdot 5! \cdot 7! = n!$ for $n$.
I have tried simplifying as follows:
$$\begin{array}{}
3! \cdot 5 \cdot 4 \cdot 3! \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3! &= n! \\
(3!)^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^3 \cdot 5^2 \cdot 4^2 \cdot 7 \cdot 6 &= n! \\
6^4 \cdot 5^2 \cdot 4^2 \cdot 7 &= n! \\
\end{array}$$
I really didn't see this helping me.
I then tried $6 \cdot 6 \cdot 6 \cdot 6 \cdot 25 \cdot 16 \cdot 7$, but $25$ only has $5$ as a double factor.
Any ideas?
|
We have
$$3!\cdot 5!\cdot 7!=(1\cdot 2\cdot 3)\cdot (1 \cdot 2\cdot 3\cdot 4\cdot 5)\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7,$$
and combining some of those gives
$$1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot \underbrace{(2\cdot 4)}_8\cdot \underbrace{(3\cdot 3)}_9\cdot \underbrace{(2\cdot 5)}_{10}=10!$$
|
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|
Evaluate the limit of ratio of sums of sines (without L'Hopital): $\lim_{x\to0} \frac{\sin x+\sin3x+\sin5x}{\sin2x+\sin4x+\sin6x}$ Limit to evaluate:
$$\lim_{x \rightarrow 0} \cfrac{\sin{(x)}+\sin{(3x)}+\sin{(5x)}}{\sin{(2x)}+\sin{(4x)}+\sin{(6x)}}$$
Proposed solution:
$$
\cfrac{\sin(x)+\sin(3x)+\sin(5x)}{\sin(2x)+\sin(4x)+\sin(6x)}
\Bigg/ \cdot\ \cfrac{1/x}{1/x}\Bigg/=
\frac{\cfrac{\sin(x)}x + \cfrac{\sin(3x)}{3x} \cdot 3 + \cfrac{\sin(5x)}{5x} \cdot 5}
{\cfrac{\sin(2x)}{2x} \cdot 2 + \cfrac{\sin(4x)}{4x} \cdot 4 + \cfrac{\sin(6x)}{6x} \cdot 6}
$$
Using $\lim_{x \rightarrow 0} \frac{\sin x}x=1$, we get
$$\frac{1+1\cdot 3+1\cdot 5}{1\cdot 2+1\cdot 4+1\cdot 6} = \frac 9{12} = \frac 3 4$$
Please tell me if I am correct.
|
HINT:
$$\sin (ax) =(ax)+O(x^3)$$
as $x\to 0$.
|
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|
$e^{\cot^2\theta}+\sin^2\theta-2\cos^22\theta+4=4\sin\theta$ in $[0,10\pi]$ Number of solution of the equation $e^{\cot^2\theta}+\sin^2\theta-2\cos^22\theta+4=4\sin\theta$ in $[0,10\pi]$ is $(A)2\hspace{1cm}(B)3\hspace{1cm}(C)4\hspace{1cm}(D)5$
In this question,$e^{\cot^2\theta}$ varies from $1$ to infinity,whereas $RHS$ varies from $-4$ to $4$.How can i find the number of solutions without using graphing calculator.Please help me.
|
We rewrite the equation as $e^{\cot^2\theta}=2\cos^22\theta-\sin^2\theta+4\sin\theta-4$. The equation has a period of 2π. L.H.S.≥1, then R.H.S.≥1, i.e. $2\cos^22\theta-\sin^2\theta+4\sin\theta-4≥1$, then we have $\cos4\theta≥(sin\theta-2)^2$. Since $\cos4\theta$∈[-1,1] and $(sin\theta-2)^2$∈[1,9] for θ∈[0,2π], θ must be π/2. Therefore we can see that the equation has solutions {π/2,5π/2,9π/2,13π/2,17π/2} for θ∈[0,10π].
|
{
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|
How to find this indefinite integral? Find $\displaystyle\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}$
I think I should change dx and let x = something. What is your suggestion?
|
Let me try to use $x=\sinh^2\theta$, then $dx=2\sinh\theta\cosh\theta d\theta=\sinh2\theta d\theta$
$$\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}$$
$$=\int \frac{\sinh2\theta d\theta}{2\sinh\theta+\cosh\theta+1}$$
$$=\int \frac{(e^{2\theta}-e^{-2\theta})/2}{e^\theta+(e^\theta+e^{-\theta})/2+1}d\theta$$
$$=\int \frac{e^{4\theta}-1}{e^\theta(3e^{2\theta}+2e^\theta-1)}d\theta$$
let $y=e^\theta$, $dy=e^\theta d\theta, i.e. d\theta=\frac{1}{y}dy$, the integral becomes
$$\int \frac{y^4-1}{y^2(3y^2+2y-1)}dy$$
$$=\int \frac{(y^2+1)(y-1)}{y^2(3y-1)}dy$$
$$=\int \frac{y^3-y^2+y-1}{y^2(3y-1)}dy$$
$$=\frac13\int [1+\frac{-2y^2+3y-3}{y^2(3y-1)}]dy$$
$$=\frac13\int [1-\frac2{9}\frac{9y^2-2y}{y^2(3y-1)}+\frac1{9}\frac{23y-27}{y^2(3y-1)}]dy$$
$$=\frac y3-\frac2{27}\ln|3y^3-y^2|+\frac1{27}\int[\frac{27}{y^2}-\frac{174}{3y-1}+\frac{58}y]dy$$
$$=\frac y3-\frac2{27}\ln|3y^3-y^2|-\frac1y-\frac{58}{27}\ln|3y-1|+\frac{58}{27}\ln|y|+C$$
$$=\frac y3-\frac1y-\frac{20}{9}\ln|3y-1|+2\ln|y|+C$$
Therefore $\displaystyle\int \frac{dx}{2\sqrt x+\sqrt{x+1}+1}=\frac y3-\frac1y-\frac{20}{9}\ln|3y-1|+2\ln|y|+C$, where $y=e^\theta$, $x=\sinh^2\theta$ and $C$ is constant. Also I find that $\theta\in[1+\sqrt2,+\infty)$. Did I write something wrong?
|
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|
Deriving an equation for the Sum of Alternating Combinations Consider the equation $$A(n,m) = \sum_{i = 0}^m (-1)^i {n \choose i}$$
To get a feel for it, I let $n=5$ and $m \in \{1,2,...,5\}$
for $m=1$ $$A(5,1) = (-1)^0 {5 \choose 0} - {5 \choose 1} = 1 - 5 = -4$$
for $m=2$, $$A(5,2) = A(5,1) + {5 \choose 2} = -4 + 10 = 6$$
for $m=3$, $$A(5,3) = A(5,2) - {5 \choose 3} = 6 - 10 = -4$$
for $m=4$, $$A(5,4) = A(5,3) + {5 \choose 4} = -4 + 5 = 1$$
for $m=5$, $$A(5,5) = A(5,4) + {5 \choose 5} = 1 - 1 = 0$$
Now This pattern also suggests that for any $k \geq 1$ then we have $A(k,k)= 0$
Now I am required to find a simpler equation for $A(n,m)$, But I am not able to come up with it , Any suggestions ?
|
Using the binomial identities: Negating the upper index $\binom{n}{i}=(-)^i\binom{i-n-1}{i}$ and parallel sumation $\sum_{k=0}^{n}\binom{m+k}{k}=\binom{n+m+1}{n}$ we get :
$$
\eqalign{
\sum_{i=0}^{m}(-)^i\binom{n}{i} &= \sum_{i=0}^{m}(-)^i(-)^i\binom{i-n-1}{i} \cr
&= \sum_{i=0}^{m}\binom{i-n-1}{i} = \binom{m-n}{m} \cr
&= (-)^m\binom{m-m+n-1}{m} = (-)^m\binom{n-1}{m} \cr
}
$$
|
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|
$[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function. Solve the equation in interval $[0,\pi]:[\cos x+\sin x]=[\cos x]+[\sin x]$,where [.] is the greatest integer function.
How should i start this question,breaking it into intervals is difficult.Please guide me.
|
Here $x\in \left[0,\pi\right]\;,$ Then $0\leq \sin x\leq 1$. So we get $\displaystyle \lfloor \sin x\rfloor = 0,1$
Similarly $x\in \left[0,\pi\right]\;,$ Then $-1 \leq \cos x\leq 1$. So we get $\displaystyle \lfloor \cos x\rfloor = -1, 0,1$
Similarly $x\in \left[0,\pi\right]\;,$ Then $-1 \leq \sin x+ \cos x\leq \sqrt{2}$. So we get $\displaystyle \lfloor \sin x+\cos x\rfloor =-1, 0,1$
$\bullet \; $ If $\displaystyle \lfloor \sin x+\cos x\rfloor=0\;,$ Then $\lfloor \sin x \rfloor =0 $ and $\lfloor \cos x \rfloor =0$
$\bullet \; $ If $\displaystyle \lfloor \sin x+\cos x\rfloor=0\;,$ Then $\lfloor \sin x \rfloor =1 $ and $\lfloor \cos x \rfloor =-1$
$\bullet \; $ If $\displaystyle \lfloor \sin x+\cos x\rfloor=1\;,$ Then $\lfloor \sin x \rfloor =0 $ and $\lfloor \cos x \rfloor =1$
$\bullet \; $ If $\displaystyle \lfloor \sin x+\cos x\rfloor=1\;,$ Then $\lfloor \sin x \rfloor =1$ and $\lfloor \cos x \rfloor =0$
$\bullet \; $ If $\displaystyle \lfloor \sin x+\cos x\rfloor=-1\;,$ Then $\lfloor \sin x \rfloor =0$ and $\lfloor \cos x \rfloor =-1$
|
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|
Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:dt$ Hence
$$I_1=\int_{1}^{0}\frac{-3t^3}{7}(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3}{7}\int_{0}^{1}(1-t^3-1)(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3I_2}{7}+\frac{3A}{7}$$
where $A=\int_{0}^{1}(1-t^3)^{\frac{-6}{7}}dt$
Similarly using substitution $x^3=1-t^7$ for $I_2$ and proceeding as above we get
$$I_2=\frac{-7I_1}{3}+\frac{7B}{3}$$ where $B=\int_{0}^{1}(1-t^7)^{\frac{-2}{3}}dt$
But we need to find $I_1-I_2$ so got stuck up here
|
Consider the area in the first quadrant bounded by the curve $x^7+y^3 = 1$.
The curve in the first quadrant can be written as $y = (1-x^7)^{1/3}$ for $0 \le x \le 1$. Hence, the area bounded by the curve is given by $\displaystyle\int_{0}^{1}(1-x^7)^{1/3}\,dx$.
We can also write the curve in the first quadrant as $x = (1-y^3)^{1/7}$ for $0 \le y \le 1$. So, the area bounded by the curve is also $\displaystyle\int_{0}^{1}(1-y^3)^{1/7}\,dy$.
Therefore, $\displaystyle\int_{0}^{1}(1-x^7)^{1/3}\,dx = \int_{0}^{1}(1-y^3)^{1/7}\,dy$, and thus, the difference is $\displaystyle\int_{0}^{1}\left[(1-x^7)^{1/3} - (1-x^3)^{1/7}\right]\,dx = 0 $.
|
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|
Cauchy–Schwarz type Inequality for 4 variables
For $x,y,z,t \geqslant 0$, prove that
$$ \tag{1} (x+2y+3z+4t)(x^2+y^2+z^2+t^2) \geqslant \frac{35-\sqrt{10}}{54}(x+y+z+t)^3$$
Observations
*
*This inequality is not symmetric nor cyclic. We cannot order the variables.
*This has 4 variables. My first attempt is to assume that $x+y+z+t=1$, then plug in $t=1-x-y-z$ to $(1)$ and try to reduce number of variables. But there are so many cases to consider.
*I think Cauchy–Schwarz is a good one. I just don't know how to get the cosntant $\frac{35-\sqrt{10}}{54}$ .
|
This is a typical problem that can be solved by brute force with homogeneization techniques and Lagrange multipliers. Since the inequality is homogeneous, it is not restrictive to assume $x^2+y^2+z^2+t^2=1$, then compute
$$ \max_{x^2+y^2+z^2+t^2=1} f(x,y,z,t)=\max_{x^2+y^2+z^2+t^2=1}\frac{x+2y+3z+4t}{(x+y+z+t)^3} $$
by imposing $2x = \lambda\frac{\partial f}{\partial x}$ and so on. Lagrange equations then give that the stationary points are eigenvectors for the rank-$2$ matrix
$$ M=\begin{pmatrix}2&5&8&11\\ 1&4&7&10 \\ 0&3&6&9 \\ -1&2&5&8 \end{pmatrix}$$
that has a unique eigenvector with positive coordinates. It follows that the maximum is achieved at
$$ (x,y,z,t) = \mu\left(\sqrt{10}-1,\frac{2}{3}\sqrt{10}-\frac{1}{3},\frac{1}{3}\sqrt{10}+\frac{1}{3},1\right) $$
and $\mu$ can be computed by imposing that $x^2+y^2+z^2+t^2=1$. That also proves
$$\max_{x^2+y^2+z^2+t^2=1} f(x,y,z,t)=\frac{35-\sqrt{10}}{54}$$
as wanted.
|
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|
Uniformly distributed variables: what does the sum reveal. Say $U_1, U_2 \sim U(1,0)$ are independent uniformly randomly distributed variables on $[0, 1]$.
What lower bound $C$ should I enforce on their sum in order to believe (with a probability $p$) that at least one of the variables themselves is equal to at least $p$? In other words, what $C=C(p)$ solves the following
$$
P(U_1 \ge p \; \vee \; U_2 \ge p \mid U_1 + U_2 \ge C) = p.
$$
Geometrically this brings me to two distinct cases, namely:
*
*Case $C<p$ and the relation
$$
p = \frac{1-p^2}{1-C^2/2}.
$$
*Case $C\ge p$ and the relation
$$
p = \frac{1-p^2 - (C-p)^2}{1-C^2/2}.
$$
But what does it imply? How to interpret this intuitively? And where are the trivialities which undoubtedly arise when $C \notin [0,2]$?
|
Denote $A = \max \{ U_1, U_2\}$ and $B=U_1 + U_2$. And I suppose you want to find the smallest such $C$ for any $p$, thus the least information about $U_1 + U_2$.
First, suppose $C \le 0$, then we have $$\mathbb{P}(A \ge p \mid B \ge C) = \mathbb{P}(A \ge p) = 1- p^2.$$
If we want $\mathbb{P}(A \ge p)=p$, we must solve $1-p^2-p=0$, which yields $p = \frac{1}{2} ( \sqrt{5}-1)$. If we want $\mathbb{P}(A\ge p) \ge p$, then we have $p \le \frac{1}{2} ( \sqrt{5}-1)$.
So, if you have no information about the sum of $U_1 + U_2$, you can up to $p = \frac{1}{2}(\sqrt{5}-1)$ say that the maximum of $U_1$ and $U_2$ is at least $p$ with probability greater than or equal to $p$.
Now, if we take $0 \le C < p$, we have $$\mathbb{P}(A\ge p \mid B \ge C) \ge \mathbb{P}( A \ge p ).$$
So this is only interesting when $p \ge \frac{1}{2}( \sqrt{5} -1)$. This gives $$\mathbb{P}( A \ge p \mid B \ge C ) = \frac{ 1 -p^2}{1 - \frac{1}{2}C^2} \ge p$$
Solving yields
\begin{eqnarray}
1 - p^2 &\ge& p - \frac{p}{2}C^2 \\
\frac{p}{2}C^2 &\ge& p + p^2 -1 \\
C^2 &\ge& -\frac{2}{p} + 2 + 2p
\end{eqnarray}
Since $C \le p$, we find that this can only work for approximately $p\le 0.68889$, which is bigger than $\frac{1}{2}(\sqrt{5}-1)$.
So, knowing that the sum greater than a constant, but not knowing if the sum is bigger than $p$, you can say that the maximum of $U_1$ and $U_2$ is at least $p$ with probability at least $p$.
So now the case $C \ge p$ remains, which has to be split into $1 \ge C \ge p$ and $2p \ge C \ge 1$, as $\mathbb{P}(B \ge C) = 2 - 2C + \frac{1}{2} C^2$ for $C \ge 1$, and when $C \ge 2p$, we have $\mathbb{P}(A \ge p \mid B \ge C) = 1$.
For $1 \ge C \ge p$, we find $C^2(1- \frac{1}{2}p) - 2pC -1+p +2p^2 \le 0$, plotting this numerical, we find that this holds for all $p \le 0.75$ approximately. ($x=C$ and $y=p$).
So the final case is $1 \ge C\ge 2p$, which a numerical approach yields that this doesn't hold.
|
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|
Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital. I am trying to solve the limit
$$\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$
Without using L'Hopital.
Evaluating yields $\frac{0}{0}$. When I am presented with roots, I usually do this:
$$\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} \cdot \frac{3+\sqrt{5+x}}{3+\sqrt{5+x}}$$
And end up with
$$\frac{9 - (5+x)}{(1-\sqrt{5-x}) \cdot (3+\sqrt{5+x})}$$
Then
$$\frac{9 - (5+x)}{3+\sqrt{5+x}-3\sqrt{5-x}-(\sqrt{5-x}\cdot\sqrt{5+x})}$$
And that's
$$\frac{0}{3+\sqrt{9}-3\sqrt{1}-(\sqrt{1}\cdot\sqrt{9})}= \frac{0}{3+3-3-3-3} = \frac{0}{-3} = -\frac{0}{3}$$
Edit: Ok, the arithmetic was wrong. That's $\frac{0}{0}$ again.
But the correct answer is
$$-\frac{1}{3}$$
But I don't see where does that $1$ come from.
|
For every $x\in [-5,5]\setminus\{4\}$ we have:
\begin{eqnarray}
\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}&=&\frac{(3-\sqrt{5+x})\color{green}{(3+\sqrt{5+x})}\color{blue}{(1+\sqrt{5-x})}}{(1-\sqrt{5-x})\color{blue}{(1+\sqrt{5-x})}\color{green}{(3+\sqrt{5+x})}}=\frac{[9-(5+x)](1+\sqrt{5-x})}{[1-(5-x)](3+\sqrt{5+x})}\\
&=&\frac{-\color{red}{(x-4)}(1+\sqrt{5-x})}{\color{red}{(x-4)}(3+\sqrt{5+x})}=-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}},
\end{eqnarray}
it follows that
$$
\lim_{x\to4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}=\lim_{x\to4}\left(-\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}\right)=-\frac{1+1}{3+3}=-\frac13.
$$
|
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|
Prove $u^2_n-u_{n-1}u_{n+1}=(-1)^n$ if $u_1=1,u_2=2,u_n=u_{n-1}+u_{n-2},n>2$ Prove $u^2_n-u_{n-1}u_{n+1}=(-1)^n$ if $u_1=1,u_2=2,u_n=u_{n-1}+u_{n-2},n>2$
If $n=3$ equality is true.
How to prove this by induction?
|
We may use the following famous trick: Notice that
$$ \begin{pmatrix}u_{n+2} \\ u_{n+1}\end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} u_{n+1} \\ u_n \end{pmatrix}. $$
Recursively using this relation, we have
$$ \begin{pmatrix}u_{n+1} & u_n \\ u_{n} & u_{n-1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n-1} \begin{pmatrix} u_2 & u_1 \\ u_1 & u_0 \end{pmatrix}. $$
Here, we extrapolate the sequence $(u_n)$ using the recurrence relation so that $u_0 = 1$. Now taking the determinant gives the desied result.
|
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|
What are the methods of dividing numbers to get weird values like $16\over 17$ without a calculator? I tried estimating it to somewhere near $16\over 20$, but it's a far stretch from getting the actual $16\over 17$. How can one do so? Conventionally, I think for numbers such as $50\over 17$, or for any large numbers, we have methods to do division and we can get $2+{16\over 17}$, but we're still left with $16\over 17$, which I have no idea how to find (or at least estimate till a good accuracy).
More interestingly, how would computers or calculators even find these values?
|
Long division:
$$
\begin{array}{cccccccccc}
& & & 0 & . & 9 & 4 & 1 & 1 & 7 & 6 \\ \\
17 & ) & 1 & 6 & . & 0 & 0 & 0 & 0 & 0 & 0 \\
& & 1 & 5 & & 3 \\ \\
& & & & & 7 & 0 \\
& & & & & 6 & 8 \\ \\
& & & & & & 2 & 0 \\
& & & & & & 1 & 7 \\ \\
& & & & & & & 3 & 0 \\
& & & & & & & 1 & 7 \\ \\
& & & & & & & 1 & 3 & 0 \\
& & & & & & & 1 & 1 & 9 \\ \\
& & & & & & & & 1 & 1 & 0 \\
& & & & & & & & 1 & 0 & 2 \\
& & & & & & & & & & \text{et} & \text{cetera}
\end{array}
$$
|
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|
Probability that the sum of 6 dice rolls is even Question:
6 unbiased dice are tossed together. What is the probability that the sum of all the dice is an even number?
I think the answer would be 50%, purely by intuition. However, not sure if this is correct. How should I go about solving such a problem?
|
The generating function approach:
$$P(x)=(1+x+x^2+x^3+x^4+x^5)^6=\sum a_i x^i$$
Then $a_i$ counts the number of ways of getting a total of $i+6$ from $6$ dice.
Now, to find the even terms, you can compute $$\frac{P(1)+P(-1)}{2}=\sum_i a_{2i}.$$
But $P(1)=6^6$ and $P(-1)=0$. So $$\frac{P(1)+P(-1)}{2}=\frac{6^6}{2},$$ or exactly half, as you conjectured.
For another example, let $N_{i}$ be the number of ways to roll $6$ dice and getting a value $\equiv i\pmod{5}$. Then it turns out that if $z$ is a primitive $5$th root of unity, then the value can be counted by defining:
$$Q_i(x)=x^{6-i}(1+x+x^2+x^3+x^4+x^5)^6$$
then computing $$N_i=\frac{Q_i(1)+Q_i(z)+Q_i(z^2)+Q_i(z^3)+Q_i(z^4)}{5}$$
This gives the result:
$$N_i =\begin{cases}\frac{6^6+4}{5}&i\equiv 1\pmod 5\\
\frac{6^6-1}{5}&\text{otherwise}
\end{cases}$$
More generally, if $N_{n,i}$ is the number of ways to get $\equiv i\pmod 5$ when $n$ dice are rolled, you get:
$$N_{n,i} =\begin{cases}\frac{6^n+4}{5}&i\equiv n\pmod 5\\
\frac{6^n-1}{5}&\text{otherwise}
\end{cases}$$
It's this simple because of the fact that $6=5+1$.
If each die has $d$ sides, and you ask what are the number of ways to get a total $\equiv i\pmod {d-1}$, then you get:
$$N_{d,n,i} =\begin{cases}\frac{d^n+{d-2}}{d-1}=\frac{d^n-1}{d-1}+1&i\equiv n\pmod {d-1}\\
\frac{d^n-1}{d-1}&\text{otherwise}
\end{cases}$$
|
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|
Evalutating $\lim_{x\to0}\frac{1-\cos x}{x^2}$ $$\lim_{x\to0}\frac{1-\cos x}{x^2}$$
I know there are many ways to calculate this. Like L'Hopital. But for learning purposes I am not supposed to do that. Instead, I decided to do it this way:
Consider that $\cos x = 1- \sin^2 \frac{x}{2}$ (from the doulbe-angle formulas here). Therefore:
$$\frac{1-(1- \sin^2 \frac{x}{2})}{x^2} = \frac{\sin^2 \frac{x}{2}}{x^2}$$
Let us split this:
$$\frac{\sin \frac{x}{2}}{x} \cdot \frac{\sin \frac{x}{2}}{x}$$
I will just work with the left one for now. At the end, I will just multiply it by itself.
$$\frac{\sin \frac{x}{2}}{x}$$
Hm. Remember that $\frac{\sin x}{x} = 1$ when $x\to0$. If only the denominator were an $\frac{x}{2}$ instead of $x$ I could do this... Well, I can! Let's just
$$\frac{\sin \frac{x}{2}}{x}\cdot\frac{\frac{1}{2}}{\frac{1}{2}}$$
Looks a bit dumb but it kind of makes sense. This will yield
$$\frac{\color{red}{\sin \frac{x}{2}}\cdot \frac{1}{2}}{\color{red}{\frac{x}{2}}}$$
Hurray, we can now apply the formula thingy and end up with
$$\frac{\color{red}1\cdot\frac{1}{2}}{\color{red}1} = \frac{1}{2}$$
Now I need to multiply this thing by itself.
$$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$
And that is the answer.
... Which is wrong. The correct answer is
$$\frac{1}{2}$$
What did I do wrong in these steps? I know many of you are able to solve this with a variety of different methods, but I'm particularly interested in this specific procedure I just tried to use. What was my mistake?
|
$$\begin{align}
\lim_{x\to 0}\frac{1-\cos x}{x^2}\frac{1+\cos x}{1+\cos x} & = \lim_{x\to 0}\frac{\sin^2x}{x^2(1+\cos x)}\\
& = \lim_{x\to 0}\frac{\sin^2x}{x^2}\lim_{x\to 0}\frac 1{1+\cos x}\\
& = 1\cdot\frac 12=\frac 12.
\end{align}$$
|
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|
Correct understanding of Manly's $S$ statistic formula I am working on social associations of bats and have a problem with correct understanding of formula from:
Bejder, Lars, D Fletcher, and S BrÄger. 1998. “A Method for Testing
Association Patterns of Social Animals.” Animal Behaviour 56 (3):
719–25. doi:10.1006/anbe.1998.0802.
$$S =\sum_{i=1}^3\sum_{j=1}^3(o_{ij} - e_{ij})^2/3^2$$
Is this matrix a correct representation of formula above?
$$
\begin{matrix}
& i=1 & i=2 & i=3\\
j=1 & (o_{11} - e_{11})^2/3^2 & (o_{21} - e_{21})^2/3^2 & (o_{31} - e_{31})^2/3^2 \\
j=2 & (o_{12} - e_{12})^2/3^2 & (o_{22} - e_{22})^2/3^2 & (o_{32} - e_{32})^2/3^2 \\
j=3 & (o_{13} - e_{13})^2/3^2 & (o_{23} - e_{23})^2/3^2 & (o_{33} - e_{33})^2/3^2 \\
\end{matrix}
$$
|
Hint: Let your matrix be M. Then
$$\begin{pmatrix} 1&1&1\end{pmatrix} \times M \times \begin{pmatrix} 1\\1\\1\end{pmatrix}=S$$
|
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|
How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$
Thanks in advance
Update: I actually solved it, and this is the way that I wanted to:
$\lim \:_{x\to \:\infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)\:=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x+\sqrt{1+\sqrt{x}}}+\sqrt{x}}\right)=\:\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{1+\sqrt{x}}}{\sqrt{x^2\left(1+\frac{\sqrt{1+\sqrt{x}}}{x}\right)+\sqrt{x}}}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x\left(\frac{1}{x}+\frac{1}{\sqrt{x}}\right)}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{x}\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{x}\left(\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{x}}}+1\right)}\right)=\lim \:\:_{x\to \:\:\infty \:\:}\left(\frac{\sqrt{\frac{1}{x}+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x^2}+\frac{1}{\sqrt{x}}}}+1}\right)=\frac{0}{2}=0$
Thanks for your two answers guys. I really appreciate it ! This community is awesome !
|
For clarity, I will use $y=\sqrt{x}$ The limit becomes
$$
\lim_{y\to \infty}( \sqrt{y^2+\sqrt{1+y}}-y)= \lim_{y\to \infty}( \sqrt{y^2+\sqrt{1+y}}-y)\frac{\sqrt{y^2+\sqrt{1+y}}+y}{\sqrt{y^2+\sqrt{1+y}}+y}
=\lim_{y\to \infty} \frac{\sqrt{1+y}}{\sqrt{y^2+\sqrt{1+y}}+y} \\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{\sqrt{y^2+y^2\sqrt{1/y^4+1/y^3}}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{\sqrt{y^2(1+\sqrt{1/y^4+1/y^3})}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{|y|\sqrt{1+\sqrt{1/y^4+1/y^3}}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{y\sqrt{1+\sqrt{1/y^4+1/y^3}}+y}\\
=\lim_{y\to \infty}\frac{\sqrt{y}\sqrt{1/y+1}}{y\left(\sqrt{1+\sqrt{1/y^4+1/y^3}}+1\right)}\\
=\lim_{y\to \infty}\frac{\sqrt{1/y+1}}{\sqrt{y}\left(\sqrt{1+\sqrt{1/y^4+1/y^3}}+1\right)}\\
\to \frac{1}{\infty \cdot 2}=0
$$
|
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|
Where is this converging to on $y=x$? Well I was playing with graphs and I started plotting equations as the following:
$$\underbrace{x+y}_{degree=1}=1 \tag{1}$$
$$\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{2}$$
$$\underbrace{x^3+y^3+x^2y+xy^2}_{degree=3}+\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{3}$$
$$\underbrace{x^4+x^3y+x^2y^2+xy^3+y^4}_{degree=4}+\underbrace{x^3+y^3+x^2y+xy^2}_{degree=3}+\underbrace{x^2+y^2+xy}_{degree=2}+\underbrace{x+y}_{degree=1}=1 \tag{4}$$
and so on ...
And here are the plots : (Click on them to get a better pic)
Zoom:1
Zoom:2
Zoom:3
It seems like these graphs are converging to some value on the (red dashed line) $y=x$.
What is this value?
My Attempt:
Since the graphs are converging to $y=x$, hence we solve the two equations as :
(Let's take degree 2)
$$\Rightarrow x^2+y^2+xy+x+y=1 \space\space\space and \space\space\space y=x$$
$$\Rightarrow x^2+x^2+x^2+x+x=1$$
$$\Rightarrow 3x^2+2x=1$$
Hence if we take $degree=n$:
The equation becomes : $\sum\limits_{n=1}^{\infty} (n+1)x^n=1$
*
*Am I right?
*How do you solve this?
*Any more comments on this question?
*How to represent equations $(1),(2),(3),(4),\cdots$ in a more general way?
WolframAlpha showed that when solved (IDK how?) $x=1\pm \frac{1}{\sqrt{2}}$ and via plotting it looks like $x=1 - \frac{1}{\sqrt{2}}$.
How?
Thanks!!
|
If you think of the both sides of that equation as functions of $x$, you're looking for a value $x$ with
$$
f(x) = g(x)
$$
where $f$ is a sum and $g$ is a constant function.
The sum happens to also be
$$
F'(x)
$$
where
\begin{align}
F(x)
&= \sum_{n=1}^\infty x^{n+1}\\
&= \sum_{n=2}^\infty x^{n}\\
&= -1-x + \sum_{n=0}^\infty x^{n}\\
&= -1-x + \frac{1}{1-x}
\end{align}
where this last step is from the geometric series formula.
Hence
$$
F'(x) = -1 + \frac{1}{(1-x)^2}
$$
Setting this to $1$, we get
\begin{align}
1 &= -1 + \frac{1}{(1-x)^2}\\
2 &= \frac{1}{(1-x)^2}\\
2(1-x)^2 &= 1
(1-x)^2 &= \frac{1}{2}\\
1-x &= \pm\frac{1}{\sqrt{2}}\\
-x &= -1 + \pm\frac{1}{\sqrt{2}}\\
x &= 1 + \pm\frac{1}{\sqrt{2}}
\end{align}
Pretty nifty!
A partial answer to question 4:
Suppose you add a new variable, $z$, and include in your formulas enough copies of $z$ in each term to make the terms all have the same degree. So you get
\begin{align}
F_1 (x, y, z) &= x + y\\
F_2(x,y, z) &= x^2 + xy + y^2 + xz + yz \\
F_3(x, y, z) &= x^3 + x^2y + xy^2 + y^3 + x^2z + xyz + y^2 z + xz^2 + yz^2
\ldots
\end{align}
Then each $F_k$ is almost a sum of every $k$th power monomial in the three variables:
\begin{align}
G_1 (x, y, z) &= x + y + z\\
G_2(x,y, z) &= x^2 + xy + y^2 + xz + yz + z^2 \\
G_3(x, y, z) &= x^3 + x^2y + xy^2 + y^3 + x^2z + xyz + y^2 z + xz^2 + yz^2 + z^3
\ldots
\end{align}
the only difference being the addition of the $z, z^2, z^3$ at the end of each one.
The equations you're solving are
\begin{align}
F_1 (x, y, z) &= 1\\
F_2(x,y, z) &= 1 \\
F_3(x, y, z) &= 1\\
\ldots
\end{align}
which I propose to rewrite as
\begin{align}
G_1 (x, y, z) &= 0 \text{ and } z = 1\\
G_2(x,y, z) &= 0 \text{ and } z = 1\\
G_3(x, y, z) &= 0 \text{ and } z = 1\\
\ldots
\end{align}
which shows that the things you've drawn are the $z = 1$ slices of certain very symmetric surfaces in 3-space, ones defined (for $k$) as the 0-level sets of the sum of all homogeneous symmetric monomials of total degree $k$.
Is that the kind of "general" statement you were looking for?
BTW: cool question!
|
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|
How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \cos^2x)(1) \longrightarrow\,
= \sin^2x - \cos^2x
\end{align}
Is that true?
|
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&=1-\frac{1}{2}\sin^2 (2x)\\
&=1-\frac{1}{2}\left(\frac{1-\cos 4x}{2}\right)\\
&=\frac{3}{4}+\frac{1}{4}\cos 4x
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Li Shanlan's combinatorial identities I am struggling to prove the following combinatorial identities:
$$(1)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+r}{m+n} = \binom{p}{m}\binom{p}{n},\quad \forall n\in\mathbb N,p\ge m,n$$
$$(2)\quad\sum_{r=0}^m \binom{m}{r}\binom{n}{r}\binom{p+m+n-r}{m+n} = \binom{p+m}{m}\binom{p+n}{n},\quad \forall p\in\mathbb N$$
The book I found them in says they were discovered and proved by Chinese mathematician Li Shanlan in the 19th century but I have failed to find any of his works translated into English in the Internet.
I am looking for an either combinatorial or algebraic solution.
|
Very striking identities. Hard to guess how Li Shanlan discovered them...
Concerning (1), multiply by $x^n y^m z^p$ and add up.
In the rhs we have
$$
\sum_{p\ge 0}z^p\sum_{n=0}^p {p \choose n} x^n \sum_{m=0}^p {p \choose m} y^m = \sum_{p\ge 0}(1+x)^p(1+y)^p z^p = \frac{1}{1-z(1+x)(1+y)}.
$$
The lhs is tougher. Denoting $t = \frac{z}{1-z}$,
$$
\sum_{r\ge 0}\sum_{n\ge r}{n\choose r}x^n \sum_{m\ge r}{m\choose r}y^m \sum_{p\ge m+n-r} {p+r\choose m+n}z^p \\
= \sum_{r\ge 0}\sum_{n\ge r}{n\choose r}x^n \sum_{m\ge r}{m\choose r}y^m\frac{z^{m+n-r}}{(1-z)^{m+n+1}} = \sum_{r\ge 0}\frac1{z^r(1-z)}\sum_{n\ge r}{n\choose r}(tx)^n \sum_{m\ge r}{m\choose r}(ty)^m\\
= \sum_{r\ge 0}\frac1{z^r(1-z)}\frac{(tx)^r}{(1-tx)^{r+1}}\frac{(ty)^r}{(1-ty)^{r+1}} = \frac{1}{(1-z)(1-tx)(1-ty)}\frac{1}{1-\frac{1}{z}\frac{tx}{1-tx}\frac{ty}{1-ty}}\\
= \frac{1}{(1-tx)(1-ty)\left(1-z-\frac{txy}{(1-tx)(1-ty)}\right)} = \frac{1}{(1-z)(1-tx)(1-ty) - txy}\\
= \frac{1}{(1-z-xz)(1-ty) - txy} = \frac{1}{1-xz -yz - txy + txyz} \\
= \frac{1}{1-z-xz -yz - xyz} = \frac{1}{1-z(1+x)(1+y)}.
$$
|
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|
Prove using factor theorem. Using factor theorem, show that $a+b$,$b+c$ and $c+a$ are factors of
$(a + b + c)^3$ - $(a^3 + b^3 + c^3)$
How do we go about solving this ?
Thanks in advance !
|
To show, for example that $a+b$ is a factor of $p(a,b,c) = (a+b+c)^3 -(a^3 + b^3 + c^3)$, we consider $p$ as a polynomial in $a$, with constants $b$ and $c$, let's write $$ p_{b,c}(a) = p(a,b,c)$$
Now $a - (-b)$ is a factor of $p_{b,c}$ iff $-b$ is a root of $p_{b,c}$. We have
\begin{align*}
p_{b,c}(-b) &= (-b + b + c)^3 -\bigl( (-b)^3 + b^3 + c^3 \bigr) \\
&= c^3 - c^3\\
&= 0
\end{align*}
So $$ a+b \mid p_{b,c}(a) $$
|
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|
Proving differentiability I just had a question on proving differentiability by showing that the difference quotient exists. I understand in the case of a function like $f(x)=x^2$, where you end up with $((x+h)^2 - x^2)/h = 2x + h = 2x$ as h goes to infinity, but in the case of a function such as $1/x^n$, how do you address the "n" portion since you cannot expand and divide out the $h$ in the denominator?
|
METHOD 1: BRUTE FORCE USE OF THE DEFINITION OF THE DERIVATIVE
We can actually approach this with brute force using the binomial expansion. We can write
$$(x+h)^n=x^n\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)$$
Therefore,
$$\begin{align}
\frac{1}{(x+h)^n}-\frac{1}{x^n}&=\frac{1}{x^n}\left(\frac{1-\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}{\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}\right)\\\\
&=\frac{1}{x^n}\left(\frac{-\left(n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}{\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}\right)\\\\
&=\frac{-\left(n\frac h{x^{n+1}}+\frac{n(n-1)}{2!}\frac{h^2}{x^{n+2}}+\cdots+\frac{h^n}{x^{2n}}\right)}{\left(1+n\frac hx+\frac{n(n-1)}{2!}\frac{h^2}{x^2}+\cdots+\frac{h^n}{x^n}\right)}\tag 1
\end{align}$$
Dividing $(1)$ by $h$ and letting $h\to 0$, we obtain the result $-\frac{n}{x^{n+1}}$ as expected.
METHOD 2: USE OF THEOREMS
Theorem 1: The product rule. $(fg)'=fg'+f'g$.
Theorem 2: The quotient rule. $(f/g)'=-fg'/g^2+f'/g$.
Theorem 3: The derivative of $x$ is $1$.
We use induction to show that $(x^n)'=nx^{n-1}$ using Theorem $2$ and Theorem $3$.
First observe that Theorem $3$ forms a base case.
Then, we postulate that for some $k$, $(x^k)'=kx^{k-1}$ and examine $(x^{k+1})'$.
Note that $(x^{k+1})'=(x\times x^{k})'$. By Theorem 2, the product rule, we have $(x\times x^{k})'=x(x^k)'+x^k(x)'=x(kx^{k-1})+x^k=(k+1)x^k$.
Therefore, by induction we have proven that $(x^n)'=nx^{n-1}$.
Next, we analyze $(1/x^n)'$. Using Theorem 1 with $f=1$ and $g=x^n$ yields, $(1/x^n)'=-nx^{n-1}/x^{2n}+0=-n/x^{n+1}$ and we are done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$ What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$
The lines $x+y=2$ and $x-y=2$ are perpendicular to each other and the circle is touching both the lines,these lines are tangents to the circle.Let points of tangency be $P$ and $Q$,let the center of the circle be $O$ and let the point where the lines $x\pm y=2$ meet be $R$.$OP$ is perpendicular to $PR$ and $OQ$ is perpendicular to $QR$,therefore $OPRQ$ is a square and the point $(-1,1)$ does not lie on any of the lines $x\pm y=2$.
But now i am stuck,how to solve further.Please help me.
|
Here is another approach. The general equation for a circle is
$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {R^2}$$
with the center at $(a,b)$ and the radius $R$. As the line $x+y=2$ is tangent to your circle, hence the line and the circle just have one intersection that you called point $P$, so the coordinates of $P$ is the solution to the nonlinear algebraic system
$$\left\{ \matrix{
{\left( {{x_P} - a} \right)^2} + {\left( {{y_P} - b} \right)^2} = {R^2} \hfill \cr
{x_p} + {y_p} = 2 \hfill \cr} \right.$$
by assumption this system has exactly one solution. Considering ${y_p} = 2 - {x_p}$ and putting it into the first equation you may obtain a quadratic equation in terms of ${x_p}$. In order to force this quadratic equation have just one real root, it's discriminant must be zero. Carrying out the computations gives
$$\left\{ \matrix{
{\Delta _P} = 8{R^2} - 4{\left( {a + b - 2} \right)^2} = 0 \hfill \cr
{x_p} = {a \over 2} - {b \over 2} + 1 \hfill \cr
{y_p} = - {a \over 2} + {b \over 2} + 1 \hfill \cr} \right.$$
where the ${\Delta _P}$ was the discriminant of the aforementioned quadratic equation. You can repeat the same process for the point $Q$, the only intersection of the circle and the line $x-y=2$, which results in
$$\left\{ \matrix{
{\Delta _Q} = 8{R^2} - 4{\left( {a - b - 2} \right)^2} = 0 \hfill \cr
{x_Q} = {a \over 2} + {b \over 2} + 1 \hfill \cr
{y_Q} = {a \over 2} + {b \over 2} - 1 \hfill \cr} \right.$$
Now using the ${\Delta _P}$ and ${\Delta _Q}$ equations, you obtain
$$\left\{ \matrix{
\left| {a - b - 2} \right| = \left| {a + b - 2} \right| \hfill \cr
R = {1 \over {\sqrt 2 }}\left| {a - b - 2} \right| = {1 \over {\sqrt 2 }}\left| {a + b - 2} \right| \hfill \cr} \right.$$
Then two cases are possible according to the first equation above.
Case 1. $b=0$
In this case the radius $R$ , the equation of circles, and coordinates of intersection points becomes
$$\left\{ \matrix{
R = {1 \over {\sqrt 2 }}\left| {a - 2} \right| \hfill \cr
{\left( {x - a} \right)^2} + {y^2} = {1 \over 2}{\left( {a - 2} \right)^2} \hfill \cr
\left\{ \matrix{
{x_p} = {a \over 2} + 1 \hfill \cr
{y_p} = - {a \over 2} + 1 \hfill \cr} \right.\,\,\,,\left\{ \matrix{
{x_Q} = {a \over 2} + 1 \hfill \cr
{y_Q} = {a \over 2} - 1 \hfill \cr} \right. \hfill \cr} \right.$$
Case2. $a=2$
In this case the radius $R$ , the equation of circles, and coordinates of intersection points becomes
$$\left\{ \matrix{
R = {1 \over {\sqrt 2 }}\left| b \right| \hfill \cr
{\left( {x - 2} \right)^2} + {\left( {y - b} \right)^2} = {1 \over 2}{b^2} \hfill \cr
\left\{ \matrix{
{x_p} = - {b \over 2} + 2 \hfill \cr
{y_p} = {b \over 2} \hfill \cr} \right.\,\,\,,\left\{ \matrix{
{x_Q} = {b \over 2} + 2 \hfill \cr
{y_Q} = {b \over 2} \hfill \cr} \right. \hfill \cr} \right.$$
So there are two family of circles, one having the center on the x-axis and one having the center on the line $x=2$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Find the equation of the locus of the intersection of the lines below Find the equation of the locus of the intersection of the lines below
$y=mx+\sqrt{m^2+2}$
$y=-\frac{ 1 }{ m }x+\sqrt{\frac{ 1 }{ m^2 +2}}$
By graphing, I have got an ellipse as locus : $x^2+\dfrac{y^2}{2}=1$.
The given lines form tangent and normal to above ellipse.
Is there any other nice way to eliminate $m$ w/o using graphing ?
I have tried eliminating $m$ by solving the intersection point, but it's looking very messy. Thanks!
|
Notice,
Solving the given equations of the straight lines: $y=mx+\sqrt{m^2+2}$ & $y=-\frac{1}{m}x+\sqrt{\frac{1}{m^2+2}}$,
as follows
$$mx+\sqrt{m^2+2}=-\frac{1}{m}x+\sqrt{\frac{1}{m^2+2}}$$
$$mx+\frac{x}{m}=\frac{1}{\sqrt{m^2+2}}-\sqrt{m^2+2}$$
$$\frac{m^2+1}{m}x=\frac{1-m^2-2}{\sqrt{m^2+2}}\implies x=\frac{-m}{\sqrt{m^2+2}}$$
substituting value of $x$ in first equation we get
$$y=m\frac{-m}{\sqrt{m^2+2}}+\sqrt{m^2+2}\implies y=\frac{2}{\sqrt{m^2+2}}$$
If the intersection point is $(h, k)$ then we have
$$h=\frac{-m}{\sqrt{m^2+2}}\tag 1$$$$ \ k=\frac{2}{\sqrt{m^2+2}}\tag 2$$
dividing (1) by (2), we get
$$\frac{h}{k}=-\frac{m}{2}\implies m=-2\frac{h}{k}$$
Now, squaring (1), we get
$$h^2=\frac{m^2}{m^2+2}$$ setting the value of $m$ we get
$$h^2=\frac{\left(-\frac{2h}{k}\right)^2}{\left(-\frac{2h}{k}\right)^2+2}$$
$$4\frac{h^2}{k^2}+2=\frac{4}{k^2}$$
$$h^2+\frac{k^2}{2}=1$$
Substituting $h=x$ & $k=y$, the locus of the intersection point is given as
$$\color{red}{x^2+\frac{y^2}{2}=1}$$
Above equation represents an ellipse.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of $\frac{\frac{1}{e}(1+x)^{1/x}-1+\frac{x}{2}}{x^2}$ when $x\to0$
Find the limit of $\dfrac{\frac{1}{e}(1+x)^{1/x}-1+\frac{x}{2}}{x^2}$ when $x\to0$.
I tried applying L'Hospital rule, but it is not working here. How should I solve this?
|
With the definition of $a^b$ we have for the numerator
\begin{align*}
(1+x)^\frac{1}{x} \cdot \exp(-1) -1+\frac{x}{2}&= \exp\left(\frac{1}{x} \cdot \ln(1+x) \right) \cdot \exp(-1) -1+\frac{x}{2}\\
&=1-\frac{x}{2}+\frac{11 x^2}{24}-1+\frac{x}{2} +\mathcal{O}(x^3)\\
&= \frac{11x^2}{24} +\mathcal{O}(x^3)
\end{align*}
Now the limit is equal to
$$\lim_{x\to 0} \frac{\frac{11}{24} x^2}{x^2}=\frac{11}{24}$$
Maybe some words how to derive they taylor series without calculating to much.
We have
$$\log(1+x)= \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} = x-\frac{x^2}{2} \pm \dots$$
and
$$\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}+\dots$$.
So we have
\begin{align*}
\exp( \log(1+x)\cdot x^{-1} -1)&= 1+ \log(1+x)\cdot x^{-1} -1 + \frac{(\log(1+x)\cdot x^{-1} -1)^2}{2} +\mathcal{O}((\log(1+x)\cdot x^{-1} -1)^3)\\
&= 1-\frac{x}{2} +\frac{x^2}{3} + \left(\frac{-x}{2}+\frac{x^2}{3}\right)^2 \cdot \frac{1}{2}+ \mathcal{O}(x^3)
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/1466855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
find the sum of $ \sum_{n=1}^{\infty} \frac{x^{2n}(-1)^n}{n} $ $ \sum_{n=1}^{\infty} \frac{x^{2n}(-1)^n}{n} $
compute the sum of this
$\frac{1}{1+x^2}$ = $ \sum_{n=1}^{\infty} x^{2n}(-1)^n $
and then integrate both sides
$ \sum_{n=1}^{\infty} \frac{x^{2n+1}(-1)^n}{2n+1} $
but how to get n from there
|
You know that:
$\begin{align}
\frac{1}{1 - z}
&= \sum_{n \ge 0} z^n \\
\int_0^z \frac{\mathrm{d} t}{1 - t}
&= - \ln (1 - z) \\
&= \sum_{n \ge 0} \frac{z^{n + 1}}{n + 1}
\end{align}$
so that:
$\begin{align}
\sum_{n \ge 1} \frac{(-1)^n x^{2 n}}{n}
&= - \ln (1 + x^2)
\end{align}$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$. I have a homework as follow:
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$.
Please help to prove it.
EDIT: MY ATTEMPT
Suppose that $5\mid a^2-2b^2$, then $a^2-2b^2=5n$,where $n\in Z$,
then $a^2-2b^2=(a+\sqrt2b)(a-\sqrt2b)=5n$,
Since 5 is a prime number, we get that $5\mid (a+\sqrt2b)$or $5\mid (a-\sqrt2b)$,
If $5\mid (a+\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
If $5\mid (a-\sqrt2b)$, then $5\mid a$ and $5\mid b$, contradiction.
Thus $5\nmid a^2-2b^2$.
|
Hint$\ \ (ab^{-1})^2\equiv 2\,\Rightarrow\,(ab^{-1})^4\equiv 4\,$ contra little Fermat
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$? How to find the absolute maximum of $f(x) = (\sin 2\theta)^2 (1+\cos 2\theta)$ for $0 \le \theta \le \frac{\pi}2$?
I found the derivative to be $f'(x)= (2\sin 4\theta)(1+\cos 2\theta) + (\sin 2\theta)^2 (-2\sin 2\theta)$.
Now I know that I have to set this equal to zero and solve for $\theta$ but that is where I get stuck, can I solve for $\theta$ or is there a different way that is easier?
|
In order to find the stationary points of $g^2$ it is enough to find the roots of $g$ and the stationary points of $g$. Since:
$$ \sin^2(2\theta)(1+\cos(2\theta)) = 2\sin^2(2\theta)\cos^2(\theta) $$
we just need to find the stationary points of:
$$ 2\sin(2\theta)\cos(\theta) = \sin(3\theta)+\sin(\theta) $$
that occur for $\cos\theta = 0$ or $3\cos^2\theta-2=0$. If $\cos^2\theta=\frac{2}{3}$, $\sin^2\theta=\frac{1}{3}$. That gives:
$$ \sin^2(2\theta)(1+\cos(2\theta))\leq 8\cdot\frac{1}{3}\cdot\frac{4}{9} = \frac{32}{27}.$$
That also follows from the AM-GM inequality.
|
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"timestamp": "2023-03-29T00:00:00",
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|
On evaluating the Riemann zeta function, including that $\zeta(2)\gt \varphi$ where $\varphi$ is the golden ratio A week ago, I got the following :
For a positive integer $k$, using Cauchy–Schwarz inequality,
$$\left(\sum_{n=1}^{\infty}\frac{1}{n^k(n+1)^k}\right)^2\lt \left(\sum_{n=1}^{\infty}\frac{1}{n^{2k}}\right)\left(\sum_{n=1}^{\infty}\frac{1}{(n+1)^{2k}}\right)=\zeta(2k)(\zeta(2k)-1),$$
i.e.
$$\zeta(2k)^2-\zeta(2k)-\left(\sum_{n=1}^{\infty}\frac{1}{n^k(n+1)^k}\right)^2\gt 0.$$
So,
$$\zeta(2k)\gt \dfrac{1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}}{2}.$$
From this, we can have the followings :
$$\zeta(2)\gt \varphi,\qquad\zeta(4)\gt \dfrac{1+\sqrt{1+4\left(\dfrac{\pi^2}{3}-3\right)^2}}{2},\qquad \zeta(6)\gt \dfrac{1+\sqrt{1+4\left(10-\pi^2\right)^2}}{2}$$where $\varphi=\frac{1+\sqrt 5}{2}$ is the golden ratio.
Now let us define a sequence $\{a_k\}$ as
$$a_k=\zeta(2k)-\dfrac{1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}}{2}$$
Then, it seems that $\{a_k\}$ is decreasing :
$$a_1\approx 0.0269,\quad a_2\approx 0.0044,\quad a_3\approx 0.0006.$$
But I've been facing difficulty in proving that.
Question : Is $\{a_k\}$ decreasing? If so, how can we prove that?
|
Let's look at the
first few terms
of each side of
$$\zeta(2k)
\gt \dfrac{1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}}{2}.
$$
$\zeta(2k)
\approx 1+\frac1{4^k}+\frac1{9^k}
$
and
$\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}
\approx \frac1{2^k}+\frac1{6^k}
$
so,
since
$\sqrt{1+x}
\approx 1+\frac{x}{2}-\frac{x^2}{8}
$,
$\begin{array}\\
1+\sqrt{1+4\left(\sum_{n=1}^{\infty}\dfrac{1}{n^k(n+1)^k}\right)^2}
&\approx 1+\sqrt{1+4\left(\frac1{2^k}+\frac1{6^k}\right)^2}\\
&\approx 1+\left(1+2\left(\frac1{2^k}+\frac1{6^k}\right)^2 -2\left(\frac1{2^k}+\frac1{6^k}\right)^4\right)\\
&\approx 2+\left(2\left(\frac1{4^k}+\frac{2}{12^k}\right) -2\frac1{16^k}\left(1+\frac1{3^k}\right)^4\right)\\
&\approx 2+2\left(\frac1{4^k}+\frac{1}{12^k}-\frac1{16^k}\right)\\
\end{array}
$
so the inequality becomes,
approximately,
$1+\frac1{4^k}+\frac1{9^k}
> 1+\frac1{4^k}+\frac1{12^k}-\frac1{16^k}
$
so the difference of the two sides
is approximately
$\frac1{9^k}-\frac1{12^k}+\frac1{16^k}
$,
and this is decreasing.
Computationally,
using just these first few terms
is not to accurate for small $k$.
This difference is,
for $k=1,2,3$,
${0.0902778, 0.00930748, 0.00103718}
$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How should I calculate $\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$ How should I calculate the below limit
$$\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$$
I have no idea where to start from.
|
First we use an observation by @Stan in the comment. Note that as $(1 +\frac{x}{n})^n$ is increasing in $n$ whenever $|x|<n$,
$$ \left(\frac{k}{n}\right)^n = \left(1 + \frac{k-n}{n}\right)^n \le e^{k-n}, $$
(here we assume that $x:= k-n$ is fixed and varies the remaining two $n$'s. This sequence is increasing and tends to $e^{k-n}$, as $|x| = |k-n| < n$. See here). Then we have
$$\begin{split}
\frac{1^n + 2^n + \cdots + n^n}{n^n} &= \sum_{k=1} ^n \left(\frac{k}{n}\right)^n \\
&\le \sum_{k=1}^n e^{k-n} \\
&= 1 + e^{-1} + e^{-2} + \cdots e^{1-n} \\
&\le \frac{1}{1-e^{-1}} = \frac{e}{e-1}.
\end{split}
$$
This implies
$$\limsup_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \le \frac{e}{e-1}.$$
On the other hand, fix $k$. Then for all $n >k$, we have
$$\begin{split}
\frac{1^n + 2^n + \cdots + n^n}{n^n} &\ge \frac{(n-k)^n + (n-k+1)^n + \cdots + n^n} {n^n}\\
&= \left( 1 - \frac kn\right)^n + \left( 1 - \frac {k-1}n\right)^n + \cdots +1
\end{split}$$
Then for all $\epsilon >0$, there is $N\in \mathbb N$ so that
$$ \left| \left( 1 - \frac {j-1}n\right)^n - e^{-(j-1)} \right| < \epsilon$$
whenever $n \ge N$ and for all $j = 1, 2 , \cdots, k+1$ (Note $k$ is fixed, so this $N$ can be found)
In particular, this implies
$$
\frac{1^n + 2^n + \cdots + n^n}{n^n} \ge e^{-k} + e^{-(k-1)} + \cdots + 1 - (k+1) \epsilon.
$$
Thus
$$
\liminf_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \ge e^{-k} + e^{-(k-1)} + \cdots + 1 - (k+1) \epsilon.
$$
Now let $\epsilon \to 0$ and then $k \to \infty$, we have
$$
\liminf_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} \ge \frac{1}{1-e^{-1}} = \frac{e}{e-1}.
$$
This implies
$$\lim_{n\to \infty} \frac{1^n + 2^n + \cdots + n^n}{n^n} = \frac{e}{e-1}.$$
|
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|
Calculate $\sum\limits_{n=1}^{\infty} \frac{a\cos(nx)}{a^2+n^2}$
I have to calculate $\sum\limits_{n=1}^{\infty} \frac{a\cos(nx)}{a^2+n^2}$ for $x\in(0,2\pi)$.
I have used the function $f(x)=e^{ax}$ and I have calculated the Fourier coefficients which are:
$$a_0=\dfrac 1{2a} \dfrac {e^{2\pi a}-1}{\pi}$$
$$a_n=\dfrac {e^{2\pi a}-1}{\pi} \dfrac{a}{a^2+n^2}$$
$$b_n=\dfrac {e^{2\pi a}-1}{\pi} \dfrac{-n}{a^2+n^2}$$
In the end, when it is written as Fourier series:
$$e^{ax}=\frac{e^{2\pi a}-1}{\pi}\left(\frac{1}{2a}+\sum^\infty_{n=1}\frac{a\cos(nx)-n\sin(nx)}{a^2+n^2}\right),\text{ for }x\in(0,2\pi).$$
My question is how can I use all these facts to calculate $\sum\limits_{n=1}^{\infty} \frac{a\cos(nx)}{a^2+n^2}$?
|
Or equivallently, we can choose : $f(x)=\cosh ax$, then
$$f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos nx + b_n \sin nx$$
where, since $f(x)$ is symmetric, $b_n =0$ and also
$$a_n=\frac{2}{\pi}\int_0^\pi\cosh ax \cos nx \;\mathrm{d}x=\frac{2a}{\pi}\frac{\sinh\pi a}{a^2+n^2}\,\cos{\pi n}$$
hence
$$\cosh ax= \frac{2}{\pi}\frac{\sinh\pi a}{2a}+a\sinh\pi a\sum_{n=1}^\infty\frac{2}{\pi}\frac{\cos{\pi n}\cos n x}{a^2+n^2}$$
For $x=\pi-y$ we get :
$$\cosh a(\pi-y)= \frac{2}{\pi}\frac{\sinh\pi a}{2a}+a\sinh\pi a\sum_{n=1}^\infty\frac{2}{\pi}\frac{\cos n y}{a^2+n^2}$$
Rearanging and renaming $y \rightarrow x$ :
$$\sum_{n=1}^\infty\frac{a \cos n x}{a^2+n^2}= \frac{\pi\cosh a(\pi-x)}{2\sinh \pi a}-\frac{1}{2a} \quad\quad ;\text{for}\quad x\in [0,2\pi]$$
Addendum :
$$\sum_{n=-\infty}^\infty\frac{a \cos n x}{a^2+n^2}= \frac{\pi\cosh a(\pi-x)}{\sinh \pi a}\quad\quad ;\text{for}\quad x\in [0,2\pi]$$
Substituing $x=0$
$$\sum_{n=-\infty}^\infty\frac{a}{a^2+n^2}= \pi\coth \pi a$$
and $x=\pi$
$$\sum_{n=-\infty}^\infty\frac{a(-1)^n}{a^2+n^2}= \pi\operatorname{csch} \pi a$$
which coincides with @Lucian's commentary.
|
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|
A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
|
Factorise this into the quadratic:
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
$$(3 \sin \theta - \cos \theta )(sin \theta + 2\cos \theta) = 0$$
This gives 2 equations.
$$3 \sin \theta - \cos \theta = 0$$
$$\sin \theta + 2\cos \theta = 0$$
You should be able to solve it by dividing both by cos. And then solve for tan.
(Make sure you check the cases if $$\cos \theta = 0$$ separately afterwards to see if you missed any solutions when dividing by cos)
|
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|
How to prove this inequality about $xyz=1$ Let $x,y,z>0$,and such $xyz=1$, show that
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3$$
I tried use $AM-GM$ inequality
$$\dfrac{x+y}{x^3+x}+\dfrac{y+z}{y^3+y}+\dfrac{z+x}{z^3+z}\ge 3\sqrt{\dfrac{(x+y)(y+z)(z+x)}{(x^3+x)(y^3+y)(z^3+z)}}$$
which shows that $$(x+y)(y+z)(z+x)\ge (x^3+x)(y^3+y)(z^3+z)$$
Am I on the right lines?but I don't have any idea how to start proving it
|
Since $\frac{x+y}{x+x^3}=\frac{(x+y)yz}{(x+x^3)yz}=\frac{1+y^2z}{1+x^2}$ and similar equalities hold, we may rewrite the inequality as:
$$\sum \frac{1+y^2z}{1+x^2}\ge 3$$.
According to Cauchy-Schwarz:
$$\sum \frac{1+y^2z}{1+x^2}=\sum \frac{(1+y^2z)^2}{(1+x^2)(1+y^2z)}\ge \frac{(3+\sum y^2z)^2}{\sum (1+x^2)(1+y^2z)}$$
Therefore it suffices to show:
$$9+6\sum y^2z+\left(\sum y^2z\right)^2\ge \sum (1+xy+x^2+y^2z)$$
$$\iff 3\left[\sum y^2z-\sum xy\right]+\left[\sum y^4z^2+2\sum y^2z^3x-3\sum x^2\right]\ge 0 $$
According to AM-GM, we have:
$$\sum y^2z-\sum xy=\frac{1}{3}\sum [2x^2y+y^2z-3x^{\frac{4}{3}}y^\frac{4}{3}z^\frac{1}{3}]\ge 0$$
and
$$\sum y^4z^2+2\sum y^2z^3x-3\sum x^2=\sum [y^4z^2+2y^3zx^2-3y^{\frac{10}{3}}z^{\frac{4}{3}}x^{\frac{4}{3}}]\ge 0$$
and so we are done. Equality occurs at $(x y, z)=(1, 1, 1)$.
Note: $\sum$ denotes the cyclic sum.
|
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|
Finding all positive integer solutions for $x+y=xyz-1$ How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2, z=1$
Any hints would be appreciated.
|
Note that for $x,y$ positive integers $$xy\ge x+y+1$$ for $x>2$ and $y\ge 3$ (or vice verse). Therefore you can control only $y=1$(or vice verse) and $y=x=2$
If $x=1$ the equation becomes:
$$y(z-1)=1$$ and the solutions are $z=2$ $y=2$ and $z=3$, $y=1$. The same thing for $y=1$. While for $y=x=2$ there aren't solutions.
|
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|
How to calculate the line integral and substitute $dx\,\ dy$ in a question on Green's theorem The question states that:
Verify Green's Theorem on the plane for $\oint_C (2x-y^3)dx-xydy$ where C is the boundary of the region enclosed by the circles $x^2+y^2=1$ and $x^2+y^2=9$.
My attempt:
*
*While verifying the L.H.S. I divided the line integral into 2 parts for the 2 circles and substituted $x=r\cos \theta$ and $y=r\sin \theta$ where the limits for $r$ will be different for the 2 circles. I then put $dx=dr\cos \theta -r\sin \theta d\theta$ and $dy=dr\sin \theta +r\cos \theta d\theta$. The integral became very complex and after computing both of them and adding them, I got $60\pi +\frac{273}{12}$. But the correct answer should be just $60\pi$.
*For the R.H.S. I don't know what to substitute $dx\,\ dy$ with and hence cannot proceed.
|
On a circle the radius $r$ is a constant, so we have:
$$
x=r\cos \theta \quad \Rightarrow \quad dx=-r\sin \theta d\theta
$$
and
$$
y=r\sin \theta \quad \Rightarrow \quad dy=r \cos \theta d \theta
$$
(this is the mistake in your solution, I suppose).
So, for a circle of radius $r$ we have:
$$
\oint_C (2x-y^3)dx-xydy=
\oint_C\left[(2r\cos \theta-r^3\sin^3 \theta)(-r\sin \theta)-r^3\cos^2 \theta \sin \theta \right]d\theta=
$$
$$
=\oint_C\left[-2r^2\cos \theta\sin \theta+r^4\sin^4 \theta-r^3\cos^2 \theta \sin \theta \right]d\theta=
$$
$$
=\int_0^{2\pi}-r^2\sin 2\theta d\theta + \int_0^{2\pi}r^4\sin^4 \theta d \theta -\int_0^{2\pi} r^3\cos^2 \theta \sin \theta d\theta
$$
I suppose that you can integrate these functions and find the result $\frac{3}{4}r^4 \pi$, that, for $r=1$ is $\frac{3}{4} \pi$ and for $r=3$ is $\frac{3}{4}(81) \pi$, so the difference (the boundary of the region enclosed) is $60 \pi$.
For the question 2. In the OP I don't see an RHS. (?)
If the RHS is from Green's theorem, than it is:
$$
\int \int _D\left(\dfrac{\partial}{\partial y}(2x-y^3)-\dfrac{\partial}{\partial x}xy \right) da
$$
where $da$ is the area element. So, using the same substitution and $da= rdrd\theta$ in polar coordinates, we have:
$$
\int \int _D\left(-y+3y^2 \right)dx dy=
$$
$$
\int_0^{2\pi} \int_1^3 \left(-r\sin \theta+3r^2 \sin^2 \theta \right)r dr d\theta= \int_0^{2\pi}\left(\frac{-26}{3}\sin \theta+60\sin^2 \theta \right)d \theta = 60 \pi
$$
|
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|
If $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $ then find $a+b$ $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}+ax+b=1 $
if I use this $$\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right| $$
I find $a=1,b=4$ but if I try to multiple by its conjugate
$$\lim _{x\rightarrow -\infty }\dfrac {x^{2}+6x+3-\left( ax+b\right) ^{2}} {\left| x\right| \sqrt {1+\dfrac {6} {x}+\dfrac {3} {x^{2}}}-ax-b}=\lim _{x\rightarrow -\infty } \dfrac {x\left( x+6+\dfrac {3} {x}-a^{2}x-2ab-\dfrac {b^{2}} {x}\right) } {x\left( -\sqrt {1+\dfrac {6} {x}+\dfrac { {3}} {x^{2}}}-a-\dfrac {b} {x}\right) }$$
in order to lim of this equal to $1$ $$x\left(\underbrace { 1-a^{2}}_0\right) +6-2ab=-1-a$$
if $a=1$ then $ b=4$
if $a=-1$ so $b=-3$ but wolfram says $ \lim _{x\rightarrow -\infty }\sqrt {x^{2}+6x+3}-x-3=\infty$
can u tell me where is my mistake?
|
Note that
$$
\lim _{x\rightarrow -\infty }\sqrt {ax^{2}+bx+c}=\left| x+\dfrac {b} {2a}\right|
$$
is meaningless, because the limit cannot depend on $x$.
One way can be doing a substitution, $x=-1/t$, so the limit becomes
$$
\lim_{t\to0^+}\sqrt{\frac{1}{t^2}-\frac{6}{t}+3}-\frac{a}{t}+b=
\lim_{t\to0^+}\frac{\sqrt{1-6t+3t^2}-a+bt}{t}
$$
In order that the limit is finite, we must have that the numerator has limit $0$; since the numerator is continuous, this means $a=1$. Now the limit is just the derivative at $0$ of the function
$$
t\mapsto \sqrt{1-6t+3t^2}-1+bt
$$
so compute at $0$ the expression
$$
\frac{-6+6t}{2\sqrt{1-6t+3t^2}}+b
$$
which gives
$$
-3+b=1
$$
Thus $a=1$ and $b=4$.
|
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|
A limit problem about $a_{n+1}=a_n+\frac{n}{a_n}$ Let $a_{n+1}=a_n+\frac{n}{a_n}$ and $a_1>0$. Prove $\lim\limits_{n\to \infty} n(a_n-n)$ exists.
In my view, maybe we can use
$${a_{n + 1}} = {a_n} + \frac{n}{{{a_n}}} \Rightarrow {a_{n + 1}} - \left( {n + 1} \right) = \left( {{a_n} - n} \right)\left( {1 - \frac{1}{{{a_n}}}} \right).$$
And then
$${a_n} - n = \left( {{a_1} - 1} \right)\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} .$$
By Stolz formula, we have
\begin{align*}
&\mathop {\lim }\limits_{n \to \infty } n\left( {{a_n} - n} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{\frac{1}{{{a_n} - n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{\frac{1}{{{a_{n + 1}} - \left( {n + 1} \right)}} - \frac{1}{{{a_n} - n}}}}\\
= &\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{{a_n} - {a_{n + 1}} + 1}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right]\left( {{a_n} - n} \right)}}{{ - \frac{n}{{{a_n}}} + 1}}\\
= &\mathop {\lim }\limits_{n \to \infty } {a_n}\left[ {{a_{n + 1}} - \left( {n + 1} \right)} \right].
\end{align*}
And how can we continue?
|
Let's start with some observations:
1) You can clearly see from what you have developed that $a_n\geq n$ (equality for $a_1=1$. For the rest we assume $a_1=1+\alpha>1$).
2) It is also easy to see that $a_{n}-n$ is decreasing. So if $a_1=\alpha+1$ then $a_n-n<\alpha$.
3) Consider the product $\prod\limits_{k = 1}^{n - 1} {\left( {1 - \frac{1}{{{a_k}}}} \right)} $, we can show it converges to zero using following inequalities:
$$
\sum\limits_{k = 1}^{n - 1}\log {\left( {1 - \frac{1}{{{a_k}}}} \right)} \leq \sum\limits_{k = 1}^{n - 1}({ - \frac{1}{{{a_k}}}}) \leq \sum\limits_{k = 1}^{n - 1} \frac{-1}{k+\alpha}\to-\infty
$$
Hence $a_n-n\to 0$.
4) Define $c_n=a_n-n$ and $b_n=nc_n$; we have:
$$
b_{n+1}-b_n=c_n(\frac{c_n}{n+c_n}-\frac{1}{n+c_n}).
$$
The term on the right goes to zero and therefore $b_{n+1}-b_n\to 0$.
|
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|
If $1+2^n+4^n$ is prime number then prove that $n=3^k$ for some $k\in \mathbb{N}$. If $1+2^n+4^n$ is prime number then prove that $n=3^k$ for some $k\in \mathbb{N}$.
I've looked at https://math.stackexchange.com/a/186723/283318 for an inspiration. In base 2, $1+2^n+4^n=10\ldots010\ldots01$ and I am completely stuck.
|
Let's solve a slightly more general problem and let's do it a novel way: Consider the function, for fixed $a$:
$$f_a(n)=1+a^n+a^{2n}.$$
We are here wondering about the case $a=2$. We will prove the following statement:
$f_a(1)$ divides $f_a(3n+1)$ and $f_a(3n+2)$ for any non-negative integer $n$.
This proves to be remarkably easy; we can notice that $f_a$ satisfies the following recurrence (whose derivation is shown below; it is easy to verify algebraically, but looks like magic if I just plop it in):
$$f_a(n+3)=(1+a+a^2)f_a(n+2)-(a+a^2+a^3)f_a(n+1)+a^3f(n)$$
We take this mod $f_a(1)=1+a+a^2$ since we only care about whether $f_a(1)$ divides $f_a(n)$. Notice that $a^3\equiv 1\pmod{f_a(1)}$ as it can be written $(a-1)f_a(1)+1$. The other coefficients are obviously $0$ mod $f_a(1)$ so we get
$$f_a(n+3)\equiv f_a(n)\pmod{f_a(1)}$$
which tells us that if $f_a(1)$ divides $f_a(n)$ it divides $f_a(n+3)$ as well. Obviously $f_a(1)$ divides itself, so it divides $f_a(3n+1)$ for non-negative integer $n$. Moreover, one may check the following identity (found by polynomial division):
$$1+a^2+a^4=(1-a+a^2)(1+a+a^2)$$
which tells us that $f_a(1)$ divides $f_a(2)$ too.
From here, it is easy: Notice that for $f_a(n)$ to be prime, it must be that $n$ is either $1$ or a multiple of $3$ since otherwise $f_a(1)$ is a proper factor. However, if we factor $n=3^k\cdot m$ for $m$ coprime to $3$. Then $f_a(3^k\cdot m)=f_{a^{3^k}}(m)$ which may only be a prime if $m$ is $1$ (since $m$ is not a multiple of $3$). Thus, $n=3^k$ if $f_a(n)$ is prime.
To find the recurrence, I noted that
$$\frac{1}{1-cx}=1+cx+c^2x^2+c^3x^3+\ldots$$
so the series $f(n)$ has generating function
$$\frac{1}{1-x}+\frac{1}{1-ax}+\frac{1}{1-a^2x}=f(0)+f(1)x+f(2)x^2+\ldots$$
which, putting the fraction together gives
$$\frac{\text{something awful}}{(1-x)(1-ax)(1-a^2x)}=f(0)+f(1)x+f(2)x^2+\ldots$$
or expanding the denominator
$$\frac{\text{something awful}}{1-(1+a+a^2)x+(a+a^2+a^3)x^2-a^3x^3}=f(0)+f(1)x+f(2)x^2+\ldots$$
The awful stuff on top is a quadratic polynomial. Thus, if we multiply through by the denominator, the coefficients of $x^3$ and higher on the left hand must vanish in order to equal the left side; the coefficient of $x^{n+3}$ is $$f(n+3)-(1+a+a^2)f(n+2)+(a+a^2+a^3)f(n+1)-a^3f(n)=0$$
and then we just isolate $f(n+3)$ to get the recurrence.
|
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|
Recurrence Relation $T_n=\sum_{r=0}^{n-1} T_r+2^n$ From a recent solution I posted here, working from an alternative path would have led to the following recurrence relation which involves a summation term:
$$T_n=\sum_{r=0}^{n-1}T_r+2^n; \qquad T_0=1\qquad (n>1)$$
How can this be solved?
Edit 3 (replaces previous edits)
Initial condition is $T_0=1$. For the problem as stated, $T_1=3$ (not $4$) and does not need to be specified as an initial condition.
This is actually different from the previous problem, which would have resulted in $T_n=\sum_{r=0}^{n-1}+2^{n+1}-1$.
For clarification
For the previous problem that I referred to here, the recurrence relatioships, the first two terms, and the closed form solution are as follows:
$$T_n=2T_{n-1}+\color{blue}{2^n}\\
T_n=\sum_{r=0}^{n-1}T_r+\color{blue}{2^{n+1}-1}\\
T_0=1, T_1=\color{blue}4\\
T_n=(\color{blue}n+1)2^n$$
For the present problem as posted here, the recurrence relatioships, the first two terms, and the closed form solution are as follows:
$$T_n=2T_{n-1}+\color{red}{2^{n-1}}\\
T_n=\sum_{r=0}^{n-1}T_r+\color{red}{2^n}\\
T_0=1, T_1=\color{red}3\\
T_n=(\color{red}{\frac n2}+1)2^n$$
|
Use generating functions. Define $T(z) = \sum_{n \ge 0} T_n z^n$, adjust indices in the recurrence:
$\begin{align}
T_{n + 1} = \sum_{0 \le r \le n} T_r + 2 \cdot 2^r
\end{align}$
Multiply by $z^n$ and sum over $n \ge 0$, recognize some sums in the result:
$\begin{align}
\frac{T(z) - T_0}{z}
= \frac{T(z)}{1 - z} + \frac{2}{1 - 2 z}
\end{align}$
Plug in $T_0 = 1$, solve for $T(z)$ as partial fractions:
$\begin{align}
T(z)
&= \frac{1 - z}{1 - 4 z + 4 z^2} \\
&= \frac{1 - z}{(1 - 2 z)^2} \\
&= \frac{1}{2 (1 - 2 z)^2} + \frac{1}{2 (1 - 2 z)}
\end{align}$
Using the binomial theorem and a geometric series:
$\begin{align}
T_n
&= \frac{1}{2} \cdot (-1)^n \binom{-2}{n} \cdot 2^n
+ \frac{1}{2} \cdot 2^n \\
&= \frac{1}{2} \binom{n + 2 - 1}{2 - 1} \cdot 2^n
+ \frac{1}{2} \cdot 2^n \\
&= (n + 2) \cdot 2^{n - 1}
\end{align}$
|
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|
Prove that $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ I've seen this identity on examsolutions, but I'm unsure on how to prove it.
$$\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$$
|
Substitute the following: $v := \frac{x+y}{2}$ and $w := \frac{x-y}{2}$
Hence: $$\cos(x) + \cos(y) = \cos(v + w) + \cos(v - w)$$
Using the addition theorem of the cosinus yields:
$$= \cos(v) \cdot \cos(w) - \sin(v) \cdot \sin(w) + \cos(v) \cdot \cos(w) + \sin(v) \cdot \sin(w)$$
$$= 2 \cdot \cos(v) \cdot \cos(w) = 2 \cdot \cos(\frac{x+y}{2}) \cdot \cos(\frac{x-y}{2})$$
|
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|
How to evaluate the definite integral by the limit definition $\int_{-1}^1 x^3 dx$? Solve the definite integral by the limit definition:
$$\int_{-1}^1 x^3 dx$$
The formula:
$$\int_a^bf(x)dx= \lim_{n\rightarrow \infty} \sum_{i=1}^n f(c_i)\Delta x_i$$
Get the variables:
$$\Delta x_i : \frac{b-a}{n} = \frac{1-(-1)}{n} = \frac{2}{n}$$
$$f(c_i) : a + i(\Delta x_i) = \left(-1+\frac{2i}{n}\right)$$
Now plug them into the formula I get:
$$\lim_{n\rightarrow \infty} \sum_{i=1}^n \left(-1+\frac{2i}{n}\right)^3\left(\frac{2}{n}\right)$$
Take out the delta and distribute the cube:
$$\lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \sum_{i=1}^n \left(-1+\frac{2i}{n}\right)^3 = \lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \sum_{i=1}^n \left(-1+\frac{8i^3}{n^3}\right) $$
Expand the summation and take out constants, and properties of Simga:
$$\lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \left[- \sum_{i=1}^n 1+ \frac{8}{n^3} \sum_{i=1}^n i^3\right] = \lim_{n\rightarrow \infty} \left(\frac{2}{n}\right) \left[-n + \frac{8}{n^3}\left( \frac{n^2(n+1)^2}{4}\right)\right]$$
Distribute the $\frac{2}{n}$ and simplify:
$$\lim_{n\rightarrow \infty}\left[-2 + \frac{16}{n^4}\left( \frac{n^2(n+1)^2}{4}\right)\right] = \lim_{n\rightarrow \infty}\left[-2 + \frac{4}{n^2}\left( \frac{n^2+2n+1}{1}\right)\right]$$
Distribute and simplify/cancel:
$$\lim_{n\rightarrow \infty}\left[-2 + \frac{4n^2}{n^2} + \frac{8n}{n^2} + \frac{4}{n^4}\right] = \lim_{n\rightarrow \infty}\left[-2 + 4 + \frac{8}{n} + \frac{4}{n^4}\right] $$
I keep getting the limit is -2 (-2+4), but the book says it's 0. Where did I go wrong?
|
You made an "algebra capital sin" when you wrote
$$
\left(-1+\frac{2i}n\right)^3=-1+\frac{8i^3}{n^3}.
$$
|
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|
Diophantine equation: $y^2=1+12x+16x^2$ The diophantine equation $$y^2=1+12x+16x^2$$ only has solutions $x=0, y=\pm1$ according to wolfram alpha. How would I go about proving these are the only solutions?
Similarly the equation $$y^2=5+12x+16x^2$$ has solutions $x=-1, y=\pm3$.
Is there a general method with regards to these types of equations? Thanks.
|
$1+12x+16x^2$ is closest to $(4x+1)^2$ and $(4x+2)^2$ regardless of $x\geq0$ or $x<0$, as it is easy to see the distance from $(4x)^2$ and $(4x+3)^2$ to $1+12x+16x^2$ are strictly larger. You need to check both equality and finds out it has integer solution only when $x=0$
Similarly,
$5+12x+16x^2$ is closest to $(4x+1)^2$ and $(4x+2)^2$ as well, the equality holds only when $5+12x+16x^2=(4x+1)^2$ and $x=-1$.
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Show that $a = b = c = 0$ for $a\sqrt{2} + b = c\sqrt{3}$ is This is the following question:
Suppose that $a, b, c$ are integers such that
$a\sqrt{2} + b = c\sqrt{3}$
(i) By squaring both sides of the equation, show that $a = b = c = 0$
The answer says that you put the equation into the following form:
if $ab \neq 0$
$\sqrt{2} = \frac{3c^2 − 2a^2 − b^2}{2ab}$
is rational — a contradiction and so a = 0 or b = 0.
Why would $a$ or $b$ be 0? (I get that you cannot express an irrational number as the quotient of two rational numbers).
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If $a,b \in \mathbb{Z}$, then
$$
2a^{2} + b^{2} + 2ab\sqrt{2} = 3c^{2},
$$
and then
\begin{align}
(*)\ \ \ \ ab\sqrt{2} = \frac{3c^{2}-2a^{2}-b^{2}}{2} .
\end{align}
The number $\sqrt{2}$ is irrational;
so $ab \neq 0$ leads to a contradiction,
and hence $ab = 0$.
We claim that $a=b=c = 0$; without loss of generality,
let $a = 0$ and let $b \neq 0$.
Then from $(*)$ we have
$|b| = \sqrt{3}|c|$,
which,
by the fact that $\sqrt{3}$ is irrational,
shows that $bc \neq 0$ is absurd;
hence $bc = 0$.
But either $b \neq 0$ or $c \neq 0$ also gives absurdity,
so $b=c=0$.
|
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|
Find the complex number $z$ such that it satisfies: 1.$|z+\frac{1}{z}|=\frac{\sqrt{13}}{2} $, 2.$[Im (z)]^2+ [Re(z)]^2=2$ Find the complex number $z$ such that it satisfies: $$\left|z+\frac{1}{z}\right|=\frac{\sqrt{13}}{2} $$$$\Im (z)^2+ \Re(z)^2=2$$$$\frac{\pi}{2}<\arg(z)<\pi$$ then find $$z^{1991}$$
Know I was thinking that this second condition might mean whole part to the power of two, but am unsure about that, just wanted some input on how to do these types of assignments because we having done anything like this before..
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Hint
Let $z=a+bi$ then
$$a^2+b^2=2,\dfrac{1}{z}=\dfrac{a-bi}{a^2+b^2}$$
and
$$|z+\dfrac{1}{z}|^2=\left(a+\dfrac{a}{a^2+b^2}\right)^2
+\left(b-\dfrac{b}{a^2+b^2}\right)^2=\dfrac{13}{4}$$
so
$$(a+\frac{a}{2})^2+(b-\frac{b}{2})^2=\frac{13}{4}$$
so
$$9a^2+b^2=13$$
then
$$a^2=\dfrac{11}{8},b^2=\dfrac{5}{8}$$
so
$$z=\sqrt{a^2+b^2}e^{ix}=\sqrt{2}e^{ix}$$ then you can do it
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|
If the coefficient of $x^{50}$ in the expansion of $(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998}$........ Problem :
If the coefficient of $x^{50}$ in the expansion of $(1+x)^{1000}+2x(1+x)^{999}+3x^2(1+x)^{998} +\cdots +1001x^{1000}$ is $\lambda$ then the value of $\frac{1952! 50!}{1001!}\lambda$
Please guide how to find the value of $\lambda$ in this will be of great help as I am not getting any clue how to proceed further in this problem , Thanks
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Let $$\displaystyle S = (1+x)^{1000}+2x(1+x)^{999}+..1000x^{999}(1+x)+1001x^{1000}........(1)\times \frac{x}{(1+x)}$$
So we get $$\frac{x\cdot S}{(1+x)} = x(1+x)^{999}+2x^2(1+x)^{998}+.........+1000x^{1000}+\frac{1001x^{1001}}{1+x}....(2)$$
So we get $$S\left[1-\frac{x}{1+x}\right] = (1+x)^{1000}+\underbrace{\left[x(1+x)^{999}+x^2(1+x)^{998}+......+x^{1000}\right]}_{S'}-\frac{1001x^{1001}}{1+x}$$
Now $$S'=x(1+x)^{999}+x^2(1+x)^{998}+.........+x^{1000}........(1)\times \frac{x}{1+x}$$
So $$\frac{S'\cdot x}{1+x} = x^2(1+x)^{998}+........+\frac{x^{1001}}{1+x}.......(2)$$
So we get $$\frac{S'}{1+x} = x(1+x)^{999}-\frac{x^{1001}}{1+x}\Rightarrow S'= x(1+x)^{1000}-x^{1001}$$
So we get $$\frac{S}{1+x}= (1+x)^{1000}+x(1+x)^{1000}-x^{1001}-\frac{1001x^{1001}}{1+x}$$
So we get $$S=(1+x)^{1002}-x^{1001}(1+x)-1001x^{1001}$$
Now Coefficient of $x^{50}$ in $$(1+x)^{1002}-x^{1001}(1+x)-1001x^{1001}$$ is $$= \binom{1002}{50}$$
|
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|
Infinitely many numbers that are the sum of two squares and the sum of two cubes, but...
I have to show that there are infinitely many natural numbers which
are simultaneously a sum of two squares and a sum of two cubes but
which are not a sum of two 6-th powers.
How to approach this question? Is there some simple condition that gives that a number cannot be the sum of two sixth powers?
|
If we start with $n^3+5^3$ we have for sure a sum of two cubes. Since:
$$ n^3+5^3 = (n+5)((n-3)^2+(n+16)) $$
if we set $n=m^2-16$ we have:
$$ (m^2-16)^3 + 5^3 = (m^2-11)((m^2-19)^2+m^2) $$
and if we set $m=l+6$ we have:
$$ (l^2+12 l +20)^3 + 5^3 = ((l+5)^2+2l)((l^2+12 l+17)^2+(l+6)^2) $$
and if $l$ is twice a square, the LHS is a sum of two squares, since the set of numbers that are the sum of two squares is a semigroup due to Lagrange's identity. By setting $l=2q^2$, we have:
$$ (4q^4+24q^2+20)^3+5^3 = ((2q^2+5)^2+(2q)^2)((4q^4+24q^2+17)^2+(2q^2+6)^2)$$
but if we assume $q\equiv 0\pmod{7}$, the LHS is $\equiv 5\pmod{7}$, hence it cannot be the sum of two sixth powers, since $n^6\in\{0,1\}\pmod{7}$.
A more sophisticated, non constructive argument may be the following one. In the range $[1,N]$, with $N$ being a huge number, there are at least $\frac{1}{4} N^{2/3}$ integers that are the sum of two positive cubes. Among these numbers, at least $\frac{1}{12\log^3 N} N^{2/3}$ are the sum of two squares (we may apply a rough sieve with respect to the primes $\equiv 3\pmod{4}$). However, in the interval $[1,N]$ there are no more than $N^{1/3}$ numbers that are the sum of two sixth powers, hence infinitely many numbers are the sum of two squares and the sum of two cubes, but not the sum of two sixth powers.
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Let $p> 7$, prove that $\left(\frac{2}{q}\right) = (-1)^{\frac{q^2-1}{8}}$ with $q$ an odd prime Let $p> 7$, prove that $\left(\frac{2}{q}\right) = (-1)^{\frac{q^2-1}{8}}$. with $q$ an odd prime. We can by using the following verifications:
$$\left(\frac{2}{p}\right) = \left(\frac{8-p}{p}\right) = \left(\frac{p}{p-8}\right) = \left(\frac{2}{p-8}\right)$$
I don't see how to use these verifications for the proof to be honest and how to prove these verifications.
Suppose $p \equiv 1,3 \pmod 4$.
If $p \equiv 1 \pmod {4}$ we get that:
$p-8 \equiv 1 \pmod {8}$ or $p-8 \equiv 5 \pmod {8}$.
if $p \equiv 3 \pmod {4}$, we get that:
$p-8 \equiv 3 \pmod {8}$ or $p-8 \equiv -1 \pmod {8}$.
So we see that the second equation works only if $p-8 \equiv p \equiv 1 \pmod{4}$. But what about the other case. I am quite confused here, can i get a hint in the right direction please?
Kees
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Hint : For a prime $p>2$, we have $$(\frac{2}{p})=1$$, if and only if $$p\equiv \pm1\ (\ mod\ 8)$$
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|
Dominated Convergence Is there a function that dominates
$$f_n(x) = \frac{1}{(1+\frac{x}{n})^nx^{\frac{1}{n}}}$$
on $(1,\infty)$ for all $n$? I need to apply DCT to get $e^{-x}$.Obviously MCT doesn't apply since $x^{\frac{1}{n}}$ is decreasing but $(1+\frac{x}{n})^n$ is increasing to $e^x$. The best I got was $\frac{1}{1+x}$ but it's not in $L^1$.
|
First, we have for $x\in[1,\infty)$
$$x^{-1/n}\le 1 \tag 1$$
Next, we show that $\left(1+\frac xn\right)^n$ is increasing function of $n$ for $x>0$. To that end, we analyze the ratio
$$\begin{align}
\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}&=\left(\frac{n(n+1+x)}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac xn\right)\\\\
&=\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac xn\right) \tag 2\\\\
&\ge \left(1-\frac{x}{n+x}\right)\left(1+\frac xn\right) \tag 3\\\\
&=1
\end{align}$$
where we used Bernoulli's Inequality to go from $(2)$ to $(3)$. We have shown that $\left(1+\frac xn\right)^n$ is increasing in $n$. Therefore, we can assert for $n\ge 2$
$$\begin{align}
\frac{1}{\left(1+\frac xn\right)^n}&\le \frac{1}{\left(1+\frac x2\right)^2}\\\\
&=\frac{1}{1+x+\frac14x^2}\tag 4
\end{align}$$
Putting all of this together, we have from $(1)$ and $(4)$
$$\frac{1}{\left(1+\frac xn\right)^nx^{1/n}}\le \frac{1}{1+x+\frac14 x^2}$$
Inasmuch as the integral of the dominating function satisfies the condition
$$\int_1^{\infty}\frac{1}{1+x+\frac14 x^2}\,dx<\infty$$
then by the Dominated Convergence Theorem, we have
$$\begin{align}
\lim_{n\to \infty}\int_1^{\infty}\frac{1}{\left(1+\frac xn\right)^nx^{1/n}}\,dx&=\int_1^{\infty}\lim_{n\to \infty}\left(\frac{1}{\left(1+\frac xn\right)^nx^{1/n}}\right)\,dx\\\\
&=\int_1^\infty e^{-x}\,dx\\\\
&=1
\end{align}$$
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