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Solve the equation $9^x+3^{2x-1}=2^{x+3.5}+2^{x+0.5}$ Solve the equation $$9^x+3^{2x-1}=2^{x+3.5}+2^{x+0.5}$$
The equation is equivalent to $$3^{2x}+3^{2x}\cdot3^{-1}=2^x\cdot2^{3.5}+2^x\cdot2^{0.5}$$ which is $$3^{2x}\left(1+\dfrac{1}{3}\right)=2^x\left(2^\frac72+2^\frac12\right)$$ We can write the last equation as $$\dfrac433^{2x}=9\sqrt2\cdot2^x,$$ or $$\dfrac{3^{2x}}{2^x}=\dfrac{9\sqrt2}{\frac43}\iff\left(\dfrac{9}{2}\right)^x=\dfrac{3^3}{2^{2-\frac12}}=\dfrac{3^3}{2^\frac32}$$ I don't see where my solution goes wrong as the answer is $x=\dfrac{3}{2}=1.5$. Any thoughts on that?
| you just need to simplify the last expression as
$\displaystyle \left(\frac{9}{2}\right)^{x} \ =\ \frac{27\sqrt{2}}{4} \ =\ \left(\frac{3^{3}}{2^{( 3/2)}}\right) \ =\ \left(\frac{9^{1.5}}{2^{1.5}}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Composing function The function is $f: \Bbb{R}\rightarrow\Bbb{R}$ defined as $f(x)= 2/ (x -3).$
I need to find $(f o f)(1).$
I would like to ask which of the following answers are the right one for writing this function.
$( f o f) ( 1 ) = ( f ( f ( 1 ) ) )= ( f ( 2/ 1 - 3) = ( 2 / 1 - 3 - 2 ) = -1/2 $
or
$( f o f ) ( 1 ) = ( f ( f (1) ) )= f( 2/ 1 - 3 ) = 2 / ( 2/ 1 - 3 ) - 3 = 2 / ( - 2 ) - 3 = -4$
| At first note that we should consider a different domain of the function $f$:
\begin{align*}
&f:\mathbb{R}\setminus\{3\}\to\mathbb{R}\\
&f(x)=\frac{2}{x-3}
\end{align*}
since $f$ is not defined at $x=3$.
We can calculate $(f\circ f)(1)=f(f(1))$ as follows:
\begin{align*}
\color{blue}{f(f(1))}=
\begin{cases}
\frac{2}{f(1)-3}=\frac{2}{\frac{2}{1-3}-3}=\frac{2}{-1-3}\color{blue}{=-\frac{1}{2}}\\
f\left(\frac{2}{1-3}\right)=f(-1)=\frac{2}{-1-3}\color{blue}{=-\frac{1}{2}}
\end{cases}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Parametrizations of $ x^4+y^4+z^4=9t^2 $ integer solutions I would like to derive all the parametrizations for the nontrivial solutions of this Diophantine equation:
$ x^4+y^4+z^4=9t^2 $
I already know that with the Fauquembergue's parametrization I can find infinite solutions (essentially multiplying by 3 the terms in the parametrization, or in the Pythagorean triple), like this one:
$ 60^4 + 45^4 + 36^4 = 9(1443)^2 $
$ 108^4+135^4+180^4 = 9(12987)^2 $
But I even know that there are more solutions (I found the last two from $ x^4+y^4+z^4=t^4 $):
$(155, 260, 296, 37747)$
$(95800, 217519, 414560, 59496731787)$
$ (2682440, 15365639, 18796760, 141668657747643) $
and they don't seem to come out from the result I found on here (none of the fourth powers is divisible by 3).
So which kind of parametrization will give those solutions?
| The Diophantine equation,
$$x^4+y^4+z^4 = nt^2$$
has quite an interesting history, being studied by Fauquembergue, Proth, Ramanujan, etc.
I. Case n = 1
As alluded to by the OP, Fauquembergue found that if $a^2+b^2 = c^2$, then,
$$(ab)^4+(ac)^4+(bc)^4 = (a^4+a^2b^2+b^4)^2$$
So using the simplest Pythagorean triple $(3,4,5)$ and multiplying both sides by $3^4$, we get the first solution of the OP,
$$ 60^4 + 45^4 + 36^4 = 9(1443)^2 $$
II. Case n = 2
Even simpler, Proth found that if $a+b+c = 0$, then,
$$a^4+b^4+c^4 = 2(ab+ac+bc)^2$$
extended by Ramanujan to,
$$a^4(b-c)^4 + b^4(a-c)^4 + c^4(a-b)^4 = 2(ab+ac+bc)^4$$
and so on for all $2(ab+ac+bc)^k$ with even $k.$
III. Case n = 9
We can use Fauquembergue's identity and multiply both sides by $3^4$ to get $n=9$. But if we wish that both sides have no common factor, we can employ a particular case of the result by Demjanenko-Elkies,
$$(85v^2 + 484v - 313)^4 + (68v^2 - 586v + 10)^4 + (2u)^4 = 9t^2$$
where $t = 3(119v^2 - 68v + 121)^2$ and,
$$22030 + 28849v - 56158v^2 + 36941v^3 - 31790v^4 = u^2$$
This is an elliptic curve, with one initial rational point $v = -31/467$ and an infinite more. The first point gives,
$$2682440^4+15365639^4+18796760^4 = 9\times141668657747643^2$$
noticed by the OP.
IV. Remark
The small co-prime solution found by the OP,
$$155^4+260^4+296^4 = 9\times37747^2$$
can also be treated as the initial point of an elliptic curve, and it would be nice to know if it satisfies a simpler polynomial identity than the Demjanenko-Elkies.
| {
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"url": "https://math.stackexchange.com/questions/4621481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $\frac{x^2}{x^2 + y^2}$ is irreducible in $\Bbb{Q}[x,y,\frac{1}{x^2+y^2}]_0$ I was doing Vakil's FOAG, in exercise 5.4 N needs to investigate an important example that $\Bbb{Q}[x,y,\frac{1}{x^2+y^2}]_0$ is not UFD (where we take localization of $\Bbb{Q}[x,y]$ at $x^2+y^2$ and take the degree zero part of it).
Ravi gives an example :
$$(\frac{x^2}{x^2+y^2})(\frac{y^2}{x^2+y^2}) = (\frac{xy}{x^2+y^2})^2$$
However I have some difficulty in showing $\frac{x^2}{x^2+y^2}$ is irreducible, notice that is not irreducible in $\Bbb{Q}[x,y,\frac{1}{x^2+y^2}]$ but the example indicate that it is irreducible in the degree zero part of $\Bbb{Q}[x,y,\frac{1}{x^2+y^2}]$. in this case we may have to prove it by definition?
| We have $p / (x^2 + y^2)^m * q / (x^2 + y^2)^n = x^2 / (x^2 + y^2)$. Rearranging, we get $pq = x^2 (x^2 + y^2)^{m+n-1}$. Assume that neither $p$ nor $q$ are divisible by $x^2+y^2$. Note that as $p / (x^2 + y^2)^m $ has degree 0 and $deg(p) \geq 0$, $m$ cannot be negative, and similarly for $n$.
Now, $\mathbb{Q}[x,y]$ is a UFD and $x^2 + y^2$ is irreducible in it. Since $x^2 + y^2$ does not divide $pq$, it cannot divide the right side either, so $m+n-1 = 0$. WLOG let $m = 1, n = 0$, so $p$ has degree 2 and $q$ has degree 0.
This implies $q$ is invertible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What is the probability that the sum of 6 four-sided dice is less than or equal to 14? The solution given is shown below.
My question is how did they count the numerator like that?
What is the explanation for it please?
$$\begin{align}
\frac{C_6^{14}-C_1^6\times C_6^{10}+C_2^6 \times C_6^6}{4^6}&=\frac{3003-6\times 210+15\times 1}{4^6}\\
&= \frac{1758}{4^6}\\
&= \frac{879}{2048}
\end{align}$$
I understand the denominator namely because each of the four sided dice has four choices and six of them so, all possible outcomes will be $4^6$.
I believe that the 4-sided dice here has 1, 2, 3, 4 printed on its faces.
Any help is appreciated.
| (I'm assuming you're familiar with Stars-and-Bars. If not, see true blue anil's solution which deals with that part of it.)
Consider finding the sum of 6 positive integers that is at most $14$, which is equivalent to 7 non-negative integers that sum to exactly $14-6=8$ (where the last variable is the dummy variable for the remainder). By Stars-and-bars, there are ${14 \choose 6 }$ ways of doing so.
Let $A_i$ be the event that the $i$th integer is 5 or more.
We're interested in $ | \cup A_i |$, which is the complement to "each is at most 4".
Notice that since $ 4 + 4 + 4 + 1 + 1 + 1 = 15 > 14$, hence at most 2 dice can show 5 or more. So $|A_i \cap A_j \cap A_k| = 0 $.
Thus, by PIE, $ | \cup A_i | = \sum |A_i| - \sum |A_i \cap A_j|$ (since the larger terms are all 0).
By Stars-and-Bars again, $|A_i| = {14 - 4 \choose 6 } $ and $ |A_i \cap A_j | = { 14 - 4 - 4 \choose 6}$.
There are clearly $ 6 \choose 1 $ events $ A_i$ and $ {6 \choose 2 }$ events $ A_i \cap A_j$.
So $ | \cup A_i |= \sum |A_i| - \sum |A_i \cap A_j| = { 6 \choose 1 } { 10 \choose 6} - { 6 \choose 2 } { 6 \choose 6}$
Hence, the number of ways is $ {14 \choose 6} - { 6 \choose 1 } { 10 \choose 6} + { 6 \choose 2 } { 6 \choose 6}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving a first order ordinary differential equation with no initial conditions
Solve the following differential equation. $$ \dfrac{dy}{dx} = \dfrac{3y - 3x}{2x+3y} $$
Answer:
Let $y = xv$. Then we have:
\begin{align*}
\dfrac{dy}{dx} &= x \dfrac{dv}{dx} + v \\
v &= \dfrac{ y}{x} \\
\dfrac{dy}{dx} &=
\dfrac{ 3\left( \dfrac{y}{x} \right) - 3} {2+3\left( \dfrac{y}{x}\right) } \\
x \dfrac{dv}{dx} + v &= \dfrac{3v-3}{2+3v} \\
x \dfrac{dv}{dx} &= \dfrac{3v-3 - v(2+3v)}{2+3v} \\
x \dfrac{dv}{dx} &= \dfrac{3v-3 -2v - 3v^2}{2+3v} \\
x \dfrac{dv}{dx} &= \dfrac{-3v^2+v-3}{3v+2} \\
\dfrac{dx}{x} &= \dfrac{ 3v + 2}{-3v^2 + v -3 } \,\,\, dv
\end{align*}
Using an online integral calculator, I find:
$$ \int \dfrac{ 3x+2 }{ -3x^2 + x - 3} \,\, dv =
-\dfrac{ \ln( 3x^2 - x + 3 ) }{2} +
\dfrac{4 \arctan\left( \dfrac{ 6x - 1 }{ \sqrt{35} } \right) } { \sqrt{35} }.
+ C$$
Now we have:
\begin{align*}
\ln{x} &=
-\dfrac{ \ln( 3v^2 - v + 3 ) }{2} +
\dfrac{4 \arctan\left( \dfrac{ 6v - 1 }{ \sqrt{35} } \right) } { \sqrt{35} }.
+ C
\end{align*}
According to ChatGPT, the correct solution is:
$$ y = (2x+3y)(x+C) $$
Where did I go wrong?
| Notice that
$$y=(2x+3y)(x+c)\iff y=\frac{2x(x+c)}{-3(x+c)+1},$$when $(x+c)\not=\frac{1}{3}$. Setting $c=0$ and find $dy/dx$ we have $$\frac{dy}{dx}=\frac{2(2-3x)x}{(3x-1)^2}\not=\frac{3(\frac{2x^2}{1-3x})-3x}{2x+3(\frac{2x^2}{1-3x})}.$$
So your function $y$ is not solution of the ode.
Instead, your first way is correct. Setting the substitution $y/x:=v$ we need to solve the ode
$$x\frac{dv}{dx}+v=\frac{3(v-1)}{3v+2}$$
Thus, we get by variable separation
$$-\sqrt{\frac{5}{7}}\arctan(\frac{6v-1}{\sqrt{35}})-\frac{1}{2}\ln|3v^2-v+3|=\ln|x|+c,$$with $c$ be an arbitrary constant, then substitution back give the implicit solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4627965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the total number of objects(stones). I have $N$ stones. Then the stones are arranged in ascending order of weights. If I remove three stones that are heaviest, then the total weight of the stones decreases by $35$%. Now if I remove the three lightest stones, the total weight of the stones further decreases by a factor of $\frac{5}{13}$.
Find the value of $N$.
What is tried:Let the total weight be $S$.
After removing three stones that are heaviest the total weight is $0.65 S$ and after further removing three stones that are lightest the total weight is $0.40 S$.
Now using average weights we have:
$$\frac{0.65}{N-3} < \frac1N \text{ and } \frac{0.65}{N-3} < \frac{0.4}{N-6}$$
from which we can conculde that $N>8$ and $N< 11$. So we are having two possible values of $N$ which are $9$ or $10$.
But how to narrow it down between $9$ and $10$ which one is correct?
Source: Homework Question.
| For another approach: The three heaviest stones weigh $\frac{7}{20} S$. The three lightest stones weigh $\frac{5}{20} S$. The $N-6$ other stones weigh $\frac{8}{20} S$. The averages of these three groups must be in order:
$$ \frac{5 S}{3 \cdot 20} < \frac{8 S}{(N-6)20} < \frac{7 S}{3 \cdot 20} $$
$$ \frac{24}{5} > N-6 > \frac{24}{7} $$
There's only one integer in that range. $N=10$.
| {
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"url": "https://math.stackexchange.com/questions/4629953",
"timestamp": "2023-03-29T00:00:00",
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Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.
My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdots)x + (1 + 2 + 3 + 5 +...)\\&= (S_n - 2)x + (S_n - 1)\\& = (a_{n+2} - 3)x + (a_{n+2} - 2)\end{aligned}$$
where $S_n$ is sum of first $n$ terms of Fibonacci series and $a_n$ is its $n$th term.
I'm not getting any idea how to solve it further.
| $x^3 = 2x + 1$ has an evident rational solution, that being $-1.$ Next, $x+1$ divides $x^3 - 2x - 1,$ the quotient is $x^2 - x-1$ for
$$ x^3 - 2x - 1 = (x+1) (x^2 - x - 1) $$
However $-1$ does not satisfy $x^4 = 3x + 2.$
We are left with those numbers $x$ with $x^2 = x+1.$ We can find the values of $x^n$ by a simple rule: if $x^n = ax + b,$ then
$$x^{n+1} = a x^2 + bx = a (x+1) + bx = (a+b)x + a. $$
The coefficient pairs for $x^n$ come out
$$
\begin{array}{ccc}
2: & 1 & 1 \\
3: & 2 & 1 \\
4 : & 3 & 2 \\
5 : & 5 & 3 \\
\end{array}
$$
and so on. Given a row $(a,b) $ the next row is what happens when we multiply it on the right side by
$$ M=
\left(
\begin{array}{cc}
1 & 1 \\
1 & 0 \\
\end{array}
\right)
$$
All of the pairs are of the form $(1 \hspace{2mm} 1) \; M^j . $ Actually, now that I think of it, all of the pairs are of the form $(1 \hspace{2mm} 0) \; M^k $ and are just the top row of that $M^k. \;$ These are, of course, consecutive fibonacci numbers
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Check an $\epsilon-\delta$ proof of $\lim_{ (x,y) \to (3,1)} \frac{x}{y} = 3$ I am going to prove
$$\lim_{ (x,y) \to (3,1)} \frac{x}{y} = 3$$
by using $\epsilon-\delta$ argument.
My Attempts
Preliminary Analysis
For every $\epsilon >0$, we need to find a suitable $\delta$ such that:
If $\Vert (x,y)-(3,1)\Vert _2 < \delta$ then $\left| \frac{x}{y} - 3\right| < \epsilon.$
First note that
$$\left| \frac{x}{y} - 3\right| = \left| \frac{x}{y} - 3y + 3y -3\right| \le \left|\frac{1}{y}\right||x-3| + 3|y-1|.$$
It is obvious that
$$|x-3| \le \Vert (x,y)-(3,1)\Vert_2\; \text{and} \; |y-1| \le \Vert (x,y)-(3,1)\Vert_2.$$
Suppose $\Vert (x,y)-(3,1)\Vert_2 < \frac{1}{2}$ then $|y-1| <\frac{1}{2}.$ As a result we have
$$\frac{1}{2} < y < \frac{3}{2} \Longrightarrow \frac{1}{|y|} < 2.$$
Hence,
$$\left| \frac{x}{y} - 3\right| \le \left|\frac{1}{y}\right||x-3| + 3|y-1|< 2\delta+3\delta = 5\delta.$$
Formal Proof
For every $\epsilon > 0$, we must
choose
$$\delta= \min\left\{\frac{1}{2}, \frac{\epsilon}{5}\right\}.$$
If $\Vert (x,y)-(3,1)\Vert _2 < \delta$ then $\left| \frac{x}{y} - 3\right| < \epsilon.$
Is my proof correct?
| Your proof is not correct because
$\left|\dfrac xy-3\right|=\left|\dfrac xy-3y+3y-3\right|\leqslant\left|\dfrac1y\right||x-3|+3|y-1|$
is wrong (see my comment).
So you could proceed as follows:
$\begin{align}\left|\dfrac xy-3\right|&=\left|\dfrac 1y\right|\,\big|x-3y\big|=\left|\dfrac 1y\right|\,\big|x-3-3\big(y-1\big)\big|\leqslant\\&\leqslant \left|\dfrac1y\right|\bigg(\big|x-3\big|+3\big|y-1\big|\bigg).\end{align}$
For any $\,\varepsilon>0\,,\,$ we choose $\,\delta=\min\left\{\dfrac12,\dfrac{\varepsilon}8\right\}>0\,.$
If $\;\big\Vert(x,y)-(3,1)\big\Vert_2<\delta\,,\;$ then
$\big|x-3\big|\leqslant\big\Vert(x,y)-(3,1)\big\Vert_2<\delta\;\;,$
$\big|y-1\big|\leqslant\big\Vert(x,y)-(3,1)\big\Vert_2<\delta\;\;,$
$\left|\dfrac1y\right|=\dfrac1{\big|1+(y\!-\!1)\big|}\leqslant\dfrac1{1-\big|y\!-\!1\big|}<\dfrac1{1\!-\!\delta}\leqslant\dfrac1{1\!-\!\frac12}=2\;.$
Consequently,
$\left|\dfrac xy-3\right|\leqslant \left|\dfrac1y\right|\bigg(\big|x-3\big|+3\big|y-1\big|\bigg)<2\big(\delta+3\delta\big)=8\delta<\varepsilon\,.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Divergence theorem example I have this problem:
Verify the divergence theorem $$
\oint_{S} \mathbf{A} \cdot d \mathbf{S}=\int_{v} \nabla \cdot \mathbf{A} d v$$ for the following case: $\mathbf{A}=2 \rho z \mathbf{a}_{\rho}+3 z \sin \phi \mathbf{a}_{\phi}-4 \rho \cos \phi \mathbf{a}_{z}$ and $S$ is the surface of the wedge $0<\rho<2$, $0<\phi<45^{\circ}=\pi/4, 0<z<5$
So, I have solved both sides of the equation:
\begin{align*}
\oint_{S} \mathbf{A} \cdot d \mathbf{S} &=\oint_{S} \left(2 \rho z ,+3 z \sin \phi ,-4 \rho \cos \phi\right)\cdot \left(\rho d\phi dz,d\rho dz, \rho d\phi d\rho\right)\\
&= \oint_{S} \left(2 \rho z\rho d\phi dz +3 z \sin \phi d\rho dz -4 \rho \cos \phi \rho d\phi d\rho\right)\\
&= \left(\iint_{\rho=0}+\iint_{\rho=2}+\iint_{\phi=0}+\iint_{\phi=\pi/4}+\iint_{z=0}+\iint_{z=5}\right)\\
& \ \ \ \ \left(2 \rho^2 z d\phi dz + 3 z \sin \phi d\rho dz -4 \rho^2 \cos \phi d\phi d\rho\right)\\
&=\iint_{\rho=2}2 \rho^2 z d\phi dz +\iint_{\phi=\pi/4} 3 z \sin \phi d\rho dz+\\
& \ \ \ \ +\iint_{z=0}-4 \rho^2 \cos \phi d\phi d\rho+\iint_{z=5}-4 \rho^2 \cos \phi d\phi d\rho\\
&= 8\int_{0}^{\pi/4}d\phi \int_0^5 zdz+\frac{3}{\sqrt{2}}\int_{0}^{5}zdz \int_{0}^{2}d\rho -8\int_0^2 \rho^2 d\rho \int_{0}^{\pi/4}\cos \phi d\phi\\
&= 8\left(\frac{\pi}{4}\right)\left(\frac{5^2}{2}\right)+\frac{3}{\sqrt{2}}\left(\frac{5^2}{2}\right)\left(2\right)-8\left(\frac{2^3}{3}\right)\left(\sin\left(\frac{\pi}{4}\right)-\sin 0\right)\\
&=25\pi +\frac{75}{\sqrt{2}}-\frac{64}{3\sqrt{2}}\\
&= 25\pi +\frac{1}{\sqrt{2}}\left(\frac{75*3-64}{3}\right)\\
&= 25\pi +\frac{1}{\sqrt{2}}\left(\frac{161}{3}\right)\\
\end{align*}
and
\begin{align*}
\int_{v} \nabla \cdot \mathbf{A} d v &=\int_v \left(\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(\rho A_\rho\right)+\frac{1}{\rho} \frac{\partial A_\phi}{\partial \phi}+\frac{\partial A_z}{\partial z}\right)\rho d\rho d\phi dz\\
&=\int_v \left(\frac{1}{\rho} \frac{\partial}{\partial \rho}\left(2\rho^2z\right)+\frac{1}{\rho} \frac{\partial (3z\sin \phi)}{\partial \phi}+\frac{\partial (-4\rho \cos\phi)}{\partial z}\right)\rho d\rho d\phi dz\\
&= \int_v \left(4z+\frac{3z\cos\phi}{\rho}+0\right)\rho d\rho d\phi dz\\
&= \int_v \left(4\rho z+3z\cos\phi\right) d\rho d\phi dz\\
&= 4\int_{0}^{2}\rho d\rho\int_{0}^{\pi/4}d\phi \int_0^5 zdz +3\int_{0}^{2}d\rho \int_{0}^{\pi/4}\cos\phi d\phi \int_0^5 zdz\\
&= 4\left(\frac{2^2}{2}\right)\left(\frac{\pi}{4}\right)\left(\frac{5^2}{2}\right)+3\left(2\right)\left(\sin\left(
\frac{\pi}{4}
\right)\right)\left(\frac{5^2}{2}\right)\\
&= 25\pi +\frac{150}{2\sqrt{2}}
\end{align*}
However, I didn't get same results, do you know where is the error and why? Maybe I have missed some basic property.
I think is something related to
$$\iint_{z=0}-4 \rho^2 \cos \phi d\phi d\rho+\iint_{z=5}-4 \rho^2 \cos \phi d\phi d\rho$$
but how am I justify those expressions cancel out each other?
| On the $z = 0$ surface, the normal vector is $-\hat{z}$, not $\hat{z}$. That means that you should have
\begin{eqnarray}
&=&\iint_{\rho=2}2 \rho^2 z d\phi dz +\iint_{\phi=\pi/4} 3 z \sin \phi d\rho dz\\
& &\ \ \ \ \color{red}{-} \iint_{z=0}-4 \rho^2 \cos \phi d\phi d\rho+\iint_{z=5}-4 \rho^2 \cos \phi d\phi d\rho\\
&=& 8\int_{0}^{\pi/4}d\phi \int_0^5 zdz+\frac{3}{\sqrt{2}}\int_{0}^{5}zdz \int_{0}^{2}d\rho - \color{red}{0}
\end{eqnarray}
and then it all works out.
Note that this also applies to the $\phi = 0$ surface. The only reason you didn't run into that problem with it is that the integrand is identically 0 there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Number of distinct arrangement of $(a,b,c,d,e)$
If $a<b<c<d<e $ be
positive integer such that
$a+b+c+d+e=20$.
Then number of distinct
arrangement of $(a,b,c,d,e)$ is
Here the largest value of $e$ is $10$
like $a\ b\ c\ d\ e$
as $ \ \ 1\ 2\ 3\ 4\ 10$
And least value is $6$
like $ a\ b\ c\ d\ e$
as $\ \ 2\ 3\ 4\ 5\ 6$
Now after that solution given in book as
Total number of ways
$ \displaystyle =\binom{4}{0}+\frac{\binom{4}{1}}{4}+\frac{\binom{4}{2}}{3}+\frac{\binom{4}{3}}{2}+\frac{\binom{4}{4}}{1}$
$\displaystyle = 1+1+2+2+1=7$
I did not understand last $2$ line
i e solution given in book
Please have a look on that part
| Equivalent problem; find non-negative integers $p\leq q\leq r\leq s\leq t$ such that $p+q+r+s+t=5$. The 5 numbers you want will be given by
$$
\begin{aligned}
a &= 1+p \\
b &= 2+q \\
c &= 3+r \\
d &= 4+s \\
e &= 5+t
\end{aligned}
$$
Repeated stars and bars and division by number of permutations will give you that expression
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Create unique table rotation for an event I am planning to host an event with 14 people participating in it. There will be five tables in total. (4 tables with 3 people each and 1 with 2).
I want to rotate the tables 5-6 times and to ensure that everyone gets to meet as many people as possible and minimize (if not completely avoid) that some people see each other twice.
I figured out how to do tables of 2, but I struggle with tables of 3.
I would very much appreciate your help!
| The best that I could come up with was this seven round schedule. Out of the $14\times 13/2=91$ pairs of people, this schedule causes all but six pairs of people to meet each other, with only six pairs of people meeting twice.
Table 1
Table 2
Table 3
Table 4
Table 5
Round 1
$4, 6, 7$
$1, 2, 5$
$9, 12, 14$
$3, 10, 13$
$8, 11$
Round 2
$1, 6, 11$
$8, 9, 10$
$7, 12, 13$
$2, 3, 14$
$4, 5$
Round 3
$3, 4, 8$
$5, 10, 12$
$1, 7, 14$
$2, 11, 13$
$6, 9$
Round 4
$4, 13, 14$
$5, 9, 11$
$3, 6, 12$
$2, 7, 10$
$1, 8$
Round 5
$5, 6, 13$
$8, 12, 14$
$2, 4, 9$
$3, 7, 11$
$1, 10$
Round 6
$6, 10, 14$
$3, 9, 13$
$5, 7, 8$
$1, 4, 12$
$2, 11$
Round 7
$3, 5, 14$
$4, 10, 11$
$1, 7, 9$
$2, 6, 12$
$8, 13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given a triangle $\triangle ABC$, with an internal point $K$, find $\angle AKC$. As title suggests, the question is to find the measure of $\angle AKC$ in the triangle $\triangle ABC$, with an internal point $K$ and some given angles.
This is a pretty challenging problem, I'm going to post my solution below as an answer, but I'm not sure if my answer is correct and if my method is correct. Please share your own answers and approaches. I want to see if my answer is correct and if there are any other ways to arrive at an answer.
| We have $\angle ABC = 66^\circ$ and $\angle AKC = 150^\circ$. Assume $BC = 1$, then $AC = \frac{\sin 48^\circ}{\sin 66^\circ} = \frac{\sin 48^\circ}{\cos 24^\circ} = 2 \sin 24^\circ$ and $KC = \frac{\sin 12^\circ}{\sin 150^\circ} = 2\sin 12^\circ$.
Now
$$AK^2 = AC^2 + KC^2 - 2AC \cdot KC \cdot \cos \angle ACK = 4\sin^2 24^\circ + 4\sin^2 12^\circ - 8 \sin 12^\circ \sin 24^\circ \cos 48^\circ.$$
Note that $\sin^2 12^\circ = \frac{(1 - \cos 24^\circ)}{2}$ and $\sin 12^\circ \cos 48^\circ = \frac{(\sin 60^\circ - \sin 36^\circ)}{2}$. Now everything is in $24^\circ$ and $36^\circ$ and the values can be calculated. The result is $\frac{(3 - \sqrt{5})}{2}$. So $AK = \frac{(\sqrt{5} - 1)}{2}$.
$$\cos X = \frac{AK^2 + KC^2 - AC^2}{2AK \cdot KC}$$
Calculate again and get $\cos X = -\frac{\sqrt{7 - \sqrt{5} - \sqrt{30 - 6\sqrt{5}}}}{4}$. Compare this with a cos table and get $X = 102^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4637245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Mysterious sum equal to $\frac{7(p^2-1)}{24}$ where $p \equiv 1 \pmod{4}$ Consider a prime number $p \equiv 1 \pmod{4}$ and $n_p$ denotes the remainder of $n$ upon division by $p$. Let $A_p=\{ a \in [[0,p]] \mid {(a+1)^2}_p<{a^2}_p\}$.
I Conjecture
$$\sum_{a \in A_p } a=\dfrac{7(p^2-1)}{24}$$
**Addition, a simpler way to prove equality after have answer **.
We denote the sequence $s_p=\sum_{a=1}^{p-1} a({(a+1)^2}_p-{a^2}_p)$.
By expanding, we verify that $s_p=-\sum_{a=1}^{p-1} {a^2}_p$.
We can also verify that if $a\in A_p$, then
${(a+1)^2}_p-{a^2}_p =({2a+1})_p-p$
and if $a\notin A_p$, then
${(a+1)^2}_p-{a^2}_p =({2a+1})_p.$
We deduce $p\sum_{p \in A_p } a =\sum_{a=1}^{p-1} a ({2a+1})_p-s_p=\sum_{a=1}^{p-1} a ({2a+1})_p + \sum_{a=1}^{p-1} {a^2}_p$.
to finish, it is easy to calculate $\sum_{a=1}^{p-1} a ({2a+1})_p\quad \forall p\in\mathbb N$ and $\sum_{a=1}^{p-1} {a^2}_p\quad \forall p\equiv 1\mod 4$
| I will leave several details for you to check, ask me if there are any issues with verifying them.
Let $$A_p=\{ a \in [[0,p-1]] \mid {(a+1)^2}_p<{a^2}_p\},$$
$$B_p=\{ a \in [[0,p-1]] \mid {(a+1)^2}_p>{a^2}_p\},$$
$$C_p=\{ a \in [[0,p-1]] \mid {(a+1)^2}_p={a^2}_p\}.$$
Then $A_p \cup B_p \cup C_p = [[0,p-1]]$. Check that $C_p$ is the singleton set $\{(p-1)/2 \}$.
Now, note that $a \in A_p \iff (p-1-a) \in B_p$ (Verify!).
Let $A_p' = A_p \cap [[0,(p-1)/2]]$, $B_p' = B_p \cap [[0,(p-1)/2]]$.
Then $A_p = A'_p \cup A''_p$ where $A_p'' = \{ (p-1)-b | b\in B_p' \}$.
Final thing to note is that $A_p' = \{ \lfloor \sqrt{jp} \rfloor | 1 \leq j \leq (p-5)/4 \}$, so $|A_p'| = (p-5)/4$ and $|B_p'| = (p+3)/4$.
Now, we just do a computation:
\begin{align}
\sum_{a \in A_p'} a + \sum_{a \in A_p''} a &= \sum_{a \in A_p'} a + \sum_{a \in B_p'} [(p-1) - a] \\ &= 2 \sum_{a \in A_p'} a + \sum_{a \in B_p'} (p-1) - \sum_{a=0}^{(p-3)/2} a \\ &= 2 \sum_{j=0}^{(p-5)/4} \lfloor \sqrt{jp} \rfloor + \frac{(p-1)(p+3)}4 - \frac{(p-3)(p-1)}8
\end{align}
Now, consider the subset of $\mathbb Z \times \mathbb Z$
$$\eqalign{
&\left\{(t,j)\ \bigg|\ 1 \leq j \leq \frac{p-5}{4}\ \hbox{and}\
1 \leq t \leq \sqrt{jp} \right\}\cr
&\qquad\qquad= \left\{(t,j)\ \bigg|\ 1 \leq t \leq \frac{p-3}{2}\ \hbox{and}\
\frac{t^2}{p} \leq j \leq \frac{p-5}{4} \right\}\cr}
$$
Comparing both cardinalities we get (q.r. = quadratic residues)
\begin{align}
\sum_{j=0}^{(p-5)/4} \lfloor \sqrt{jp} \rfloor &= \sum_{t=1}^{(p-3)/2} \left \{ \frac {(p-5)} 4 - \left \lfloor \frac{t^2} p \right \rfloor \right \} \\ &= \frac{(p-3)(p-5)}{8} - \sum_{t=1}^{(p-3)/2} \frac{t^2} p + \sum_{t=1}^{(p-3)/2} \frac{t^2 \pmod p} p \\ &= \frac{(p-3)(p-5)}{8} - \frac{(p-1)(p-2)(p-3)}{24p} \\ &\ \ \ + \frac{\text{Sum of q.r. mod p}- ((p-1)/2)^2 \pmod p}p
\end{align}
Now, if $p \equiv 1 \pmod 4$, then $((p-1)/2)^2 \equiv (3p+1)/4 \pmod p$ and sum of quadratic residues is $p(p-1)/4$.
Thus,
\begin{align}
\sum_{j=0}^{(p-5)/4} \lfloor \sqrt{jp} \rfloor &= \frac{(p-3)(p-5)}{8} - \frac{(p-1)(p-2)(p-3)}{24p} + \frac{p(p-1)- (3p+1)}{4p} \\ &= \frac{(p-1)(p-5)}{12}.
\end{align}
Now, the final sum is
\begin{align}
\sum_{a \in A_p} a &= \frac{(p-1)(p-5)}{6} + \frac{(p-1)(p+3)}4 - \frac{(p-3)(p-1)}8 \\ &= \frac{7(p^2-1)}{24}.
\end{align}
Here are two problems employing similar techniques. (That is where I learnt this method from)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 1,
"answer_id": 0
} |
Find limit of sequence $(x_n)$ Find limit of sequence $(x_n)$:
$$x_1 = a >0$$
$$x_{n+1} = \frac{n}{2n-1}\frac{x_n^2+2}{x_n}, n \in Z^+$$
I think I can prove $(x_n)$ is low bounded (which is obvious that $x_n>0$) and decreasing sequence. Then I can calculate the limit of sequence is $\sqrt{2}$
All my attempts to prove it's a decreasing sequence have been unsuccessful.
My attemps: Try to prove $x_{n+1}-x_{n} <0$ from a number $N_0$ large enough. It lead to I have to prove $x_n \ge \sqrt{\frac{2n}{n-1}}$ and I stuck.
Does anyone have an idea?
| A recursion shows that the $x_n$ for $n \ge 2$ are well-defined and bounded below by $\sqrt{2}$, since $x^2+2-2\sqrt{x} = (x-\sqrt{2})^2$ is always positive. Set $y_n = x_n-\sqrt{2}$. Then for $n \ge 2$,
\begin{eqnarray*}
y_{n+1}
&=& \frac{n(x_n^2+2)-\sqrt{2}(2n-1)x_n}{(2n-1)x_n} \\
&=& \frac{n(x_n-\sqrt{2})^2+\sqrt{2}x_n}{(2n-1)x_n} \\
&=& \frac{ny_n^2}{(2n-1)x_n} + \frac{\sqrt{2}}{2n-1} \\
&\le& \frac{n}{2n-1}y_n+ \frac{\sqrt{2}}{2n-1},
\end{eqnarray*} since $y_n \le x_n$.
For $n \ge 2$, $n/(2n-1) \le 2/3$, so
$$y_{n+1} \le \frac{2}{3}y_n+\frac{1}{3}z_n, \text{ where } z_n = \frac{3\sqrt{2}}{2n-1}$$
Given $\epsilon>0$, one can find an integer $N$ such that for every $n \ge N$, $z_n \le\epsilon$ so $$y_{n+1}-\epsilon \le \frac{2}{3}(y_n-\epsilon),$$
so by recursion
$$y_n-\epsilon \le \Big(\frac{2}{3}\Big)^{n-N}(y_N-\epsilon).$$
As a result, $\limsup_{n \to +\infty} y_n \le \epsilon$. Since this is true for every $\epsilon>0$, we get the desired conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Limit of $(|x| + |y|)\ln(x^2 + y^4)$ at $(0,0)$ I want to show that $$\lim\limits_{(x,y) \to (0,0)} (\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4) = 0$$
First I let $\lVert (x,y) \rVert = \lvert x \rvert + \lvert y \rvert\ < \delta$, and assume that $x,y < 1$ so $x^2 + y^4 < \lvert x \rvert + \lvert y \rvert$. Then $\ln(x^2 + y^4) < \ln(\lvert x \rvert + \lvert y \rvert)$. However, $\lvert \ln(x^2 + y^4)\rvert > \lvert \ln(\lvert x \rvert + \lvert y \rvert)\rvert$, which is where I am stuck because I wanted to show that $\lvert(\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4)\rvert < \lvert x \rvert + \lvert y \rvert < \delta $. It does not seem like this approach will work & I am not sure what else I can try.
| We have that by generalized mean inequality
$$\frac{\lvert x \rvert + \lvert y \rvert}2 \le \sqrt[4]{\frac{x^4 +y^4}2} \iff \lvert x \rvert + \lvert y \rvert\le\frac 2{\sqrt[4]2}\sqrt[4]{x^4 +y^4}$$
then assuming wlog $x^2 + y^4<1$
$$(\lvert x \rvert + \lvert y \rvert)\left|\ln(x^2 + y^4)\right|
\le \frac 2{\sqrt[4]2}\sqrt[4]{x^4 +y^4}\left|\ln\left(x^4 +y^4\right)\right|= \\
=\frac 8{\sqrt[4]2}\sqrt[4]{x^4 +y^4}\left|\ln\left(\sqrt[4]{x^4 +y^4}\right)\right|
\to 0$$
since for $t\to 0^+$ we have $t\ln t \to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve the differential equation: $(x^2-y^2)dx+2xydy=0$. Given $(x^2-y^2)dx+2xydy=0$
My solution-
Divide the differential equation by $dx$
$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$
$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$
Divide both sides by $2xy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$
This is a homogenous differential equation. Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}[v-\frac{1}{v}]$
$\Rightarrow x\frac{dv}{dx}=-\frac{v^2+1}{2v}$
$\Rightarrow -\frac{2v}{v^2+1}dv=\frac{dx}{x}$
Integrating both sides
$\Rightarrow -\log|v^2+1|=\log x+\log c$
$\Rightarrow -\log|\frac{y^2}{x^2}+1|=\log xc$
$\Rightarrow -\log|\frac{x^2+y^2}{x^2}|=\log xc$
$\Rightarrow \frac{x^2}{x^2+y^2}= xc$
$\Rightarrow x= c(x^2+y^2)$
$\Rightarrow y=\pm \sqrt{xc-x^2}$
Kindly review my solution and let me know if there are other methods of solving such problems.
| Another approach let's call $u=x^2+y^2$.
Then $xu'=2x^2+2xyy'=2x^2+(y^2-x^2)=x^2+y^2=u$
Which solves to $u=cx\iff x^2+y^2=cx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluate: $\int \frac{2}{(1-x)(1+x^2)}dx$ Given $\int \frac{2}{(1-x)(1+x^2)}dx$
The most obvious approach is to use Partial fractions
Let $\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}$
$\Rightarrow \frac{2}{(1-x)(1+x^2)}=\frac{A+Ax^2+Bx-Bx^2+C-Cx}{(1-x)(1+x^2)}$
We get $A=1, B=1, C=1$
The integral now becomes
$\int[\frac{1}{1-x}+\frac{x+1}{1+x^2}]dx$
$\Rightarrow \int[\frac{1}{1-x}+\frac{x}{1+x^2}+\frac{1}{1+x^2}]dx$
Now, we can simply integrate term by term
$\Rightarrow -\log|1-x|+\frac{\log(1+x^2)}{2}+\tan^{-1}x+c $
$\Rightarrow \log |\frac{\sqrt{1+x^2}}{1-x}|+\tan^{-1}x+c $
Please review my solution and if you have any other way of integrating then please share your solution.
| Here is another way to integrate
\begin{align}
&\int \frac{2}{(1-x)(1+x^2)}dx\\
=& \int \frac{2(1+x)}{1-x^4}dx= \int \frac{2x}{1-x^4}+ \frac{2}{1-x^4} \ dx\\
=& \int \frac{1}{1-x^4} \ d(x^2)+\int \frac{1}{1-x^2}+\frac{1}{1+x^2}\ dx\\
= &\ \frac12\ln|\frac{1+x^2}{1-x^2}|+ \frac12\ln|\frac{1+x}{1-x}|+\tan^{-1}x+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show the value of $\mathbb{E}(X^2Y^2)$ Suppose that $(X,Y)$ is distributed ranodm 2-vectors having the normal distribution with $\mathbb{E}X=\mathbb{E}Y=0,\mathbb{Var}(X)=\mathbb{Var}(Y)=1,$ and $\text{Cov}(X,Y)=\theta\in(-1,1).$ Show the value of $\mathbb{E}(X^2Y^2).$
The joint pdf of $(X,Y)$ is $$p(x,y)=\frac{1}{2\pi\sqrt{1-\theta^2}}\exp\left\{-\frac{1}{2(1-\theta^2)} \left [ x^2-2\theta xy+y^2\right ]\right\}.$$
Then $$\mathbb{E}(X^2Y^2)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x^2y^2\cdot p(x,y)\mathrm{d}x\mathrm{d}y.$$
Let $$\begin{cases}
u=\frac{x-y}{\sqrt{2}} \\
v=\frac{x+y}{\sqrt{2}}
\end{cases}\Rightarrow \mathbb{E}(X^2Y^2)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{(u^2-v^2)^2}{2}\cdot p\left (\frac{u+v}{\sqrt{2}},\frac{u-v}{\sqrt{2}}\right )\mathrm{d}u\mathrm{d}v.$$
But I don't know how to deal with this integral
$$\begin{align*}
&\Delta:=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{(u^2-v^2)^2}{2}\cdot \exp\left \{-\frac{1}{2(1-\theta^2)}\left [ (1-\theta)u^2+(1+\theta)v^2\right ]\right \}\mathrm{d}u\mathrm{d}v\\
&\quad=4\int_{0}^{\infty}\int_{0}^{\infty}\frac{(u^2-v^2)^2}{2}\cdot \exp\left \{-\frac{1}{2(1-\theta^2)}\left [ (1-\theta)u^2+(1+\theta)v^2\right ]\right \}\mathrm{d}u\mathrm{d}v,\quad \theta\in(-1,1).
\end{align*}$$
| $$\begin{align*}
\Bbb E\left[X^2Y^2\right] &= \frac1{2\pi \sqrt{1-\theta^2}} \int\limits_0^\infty \int\limits_0^\infty \left(u^2-v^2\right)^2 e^{\Large-\frac{(1-\theta) u^2+(1+\theta)v^2}{2(1-\theta^2)}} \, du \, dv \\
\tag{1} &= \frac4\pi \int\limits_0^\infty \int\limits_0^\infty \left((1+\theta)s^2-(1-\theta)t^2\right)^2 e^{-(s^2+t^2)} \, ds \, dt \\[1ex]
\tag{2} &= \frac4\pi \int\limits_0^{\pi/2} \int\limits_0^\infty \bigg((1+\theta)^2 r^4\cos^4(\phi) - 2(1-\theta^2)r^4\cos^2(\phi)\sin^2(\phi) \\
&\qquad \qquad \qquad \qquad + (1-\theta)^2 r^4 \sin^4(\phi)\bigg) re^{-r^2} \, dr \, d\phi \\[1ex]
\tag{3} &= \frac4\pi \int\limits_0^{\pi/2} \bigg((1+\theta)^2 \cos^4(\phi) - 2(1-\theta^2) \cos^2(\phi) \sin^2(\phi) \\
&\qquad \qquad \qquad \qquad + (1-\theta)^2 \sin^4(\phi)\bigg) \, d\phi \\[1ex]
\tag{4} &= \frac4\pi \left(\frac{3\pi}{16} (1+\theta)^2 - \frac\pi8(1-\theta^2) + \frac{3\pi}{16} (1-\theta)^2\right) \\[1ex]
&= 1+2\theta^2
\end{align*}$$
*
*$(1)$ : substitute $(u,v)=\left(\sqrt{2(1+\theta)}\,s,\sqrt{2(1-\theta)}\,t\right)$
*$(2)$ : substitute $(s,t)=(r\cos(\phi),r\sin(\phi))$
*$(3)$ : integrate by parts;
$$\int\limits_0^\infty r^5 e^{-r^2} \, dr = 1$$
*$(4)$ : reduce powers via $\cos^2(\alpha)=\dfrac{1+\cos(2\alpha)}2$ and $\sin^2(\alpha)=\dfrac{1-\cos(2\alpha)}2$ and integrate
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Linear Regression of Quadratic Points I was casually experimenting with Desmos's linear regression and noticed some patterns in the best fit line for points which fit y = x^2.
For (0,0) and (1,1) the best fit line is y = 1x+0 and r = 1 (the correlation coefficient).
For (0,0) , (1,1) , (2,4) , ... , (10,100) the best fit line is y = 10x-15 and r = .963.
And for (0,0) , (1,1) , (2,4) , ... , (20,400) the best fit line is y = 20x-63.333 and r = .965.
My questions are:
*
*Why is the best fit slope always equal to the # of points - 1?
*Does the y-intercept follow the curve -$1/6 x^2 + 1/2 x - 1/3$ where x is the # of points?
*Does the correlation coefficient approach $1$, and why does it even increase as more points are added?
| What is the formula for the line of best fit here?
In this case, because the points we take follow an algebraic pattern, we can calculate a linear regression on the points $\{(0,0), (1,1), (2,4), \dots, (n,n^2)\}$ in terms of $n$, and see what happens.
To do this, we start with the overconstrained system of equations that a line $y=ax+b$ passes through all $n+1$ points: $0 = a\cdot0+b$, $1=a\cdot1+b$, and so on through $n^2 = a\cdot n+b$. In matrix form, $$\begin{bmatrix}1 & 0 \\ 1 & 1 \\ 1 & 1 \\ \vdots & \vdots \\ 1 & n\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix} = \begin{bmatrix}0^2 \\ 1^2 \\ 2^2 \\ \vdots \\ n^2\end{bmatrix}.$$ There is no such line; to find the least-squares solution, we multiply by the transpose of the coefficient matrix: $$\begin{bmatrix}1 & 1 & 1 & \cdots & 1\\ 0 & 1 & 2 & \cdots & n\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 1 \\ 1 & 1 \\ \vdots & \vdots \\ 1 & n\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix} = \begin{bmatrix}1 & 1 & 1 & \cdots & 1\\ 0 & 1 & 2 & \cdots & n\end{bmatrix}\begin{bmatrix}0^2 \\ 1^2 \\ 2^2 \\ \vdots \\ n^2\end{bmatrix}.$$ This simplifies to a $2\times 2$ system $$\begin{bmatrix}n+1 & \frac{n(n+1)}{2} \\ \frac{n(n+1)}{2} & \frac{n(n+1)(2n+1)}{6}\end{bmatrix} \begin{bmatrix}a \\ b\end{bmatrix} = \begin{bmatrix}\frac{n(n+1)(2n+1)}{6} \\ \frac{n^2(n+1)^2}{4}\end{bmatrix}$$ where the increasingly complicated formulas in the cells are the sum of our $x$-coordinates, the sum of their squares, and the sum of their cubes. Solving this system gets us a general formula for $a$ and $b$: the line of best fit is $$y = nx -\frac{n(n-1)}{6}.$$
(This does fit the conjecture formula in the question.)
Why?
It's hard to answer "why" questions when the real "why" is "because the formula said so".
It is reasonable to expect a slope approximately equal to $n$, because this is the slope from the first to the last point: $\frac{n^2-0}{n-0}$. This doesn't tell us that the slope must be exactly $n$, but sometimes math does nice things for us.
We can also get some insight into the problem by rescaling. Divide all $x$-coordinates by $n$ and all $y$-coordinates by $n^2$. Our $n+1$ points are now $(0,0), (\frac1n, \frac1{n^2}), (\frac2n, \frac4{n^2}), \dots, (1,1)$: they are $n+1$ evenly-spaced points on the graph of $y=x^2$ for $x$ between $0$ and $1$. As $n$ increases, our line of best fit should get closer and closer to some line best approximating that graph...
...and it does; rescaling our line of best fit gives us $y = x - \frac16 + \frac1{6n}$, and this approaches $y = x - \frac16$ as $n$ increases.
What about the correlation coefficient?
The most convenient formula to use here is
$$r_{xy} = \frac{n\sum x_i y_i - \sum x_i\sum y_i}
{\sqrt{n\sum x_i^2-\left(\sum x_i\right)^2}~\sqrt{n\sum y_i^2-\left(\sum y_i\right)^2}}$$
Here, each sum is one of the familiar expressions $ \frac{n(n+1)}{2}$, $ \frac{n(n+1)(2n+1)}{6}$, or $\frac{n^2(n+1)^2}{4}$, except that $\sum y_i^2 = \sum_{i=0}^n i^4$, which has a more complicated formula I'm scared to write down.
To avoid getting into those weeds, we approximate $\sum x_i \approx n^2/2$, $\sum x_i^2 = \sum y_i \approx n^3/3$, $\sum x_iy_i \approx n^4/4$, and $\sum y_i^2 \approx n^5/5$. (The general rule for the approximation is that when we have a sum of $r^{\text{th}}$ powers, $\sum_{i=0}^n i^r$, it is approximately $n^{r+1}/(r+1)$. One way to get there is to replace the sum by an integral.)
This approximation simplifies the numerator to about $n^5/12$, the first square root to $\sqrt{n^4/12}$, and the second square root to $\sqrt{4n^6/45}$. Overall, the factors of $n$ cancel, and we get $$r_{xy} \approx \frac{\sqrt{15}}{4} \approx 0.968246.$$
Because we took our approximations, this only gets us the limiting behavior: if you like, the correlation coefficient for $y = x^2$ when $0 \le x \le 1$, with line of best fit $y = x - \frac16$. The correlation coefficient for small $n$ will only approach this value. (That may mean it sometimes increases as we increase $n$ - but it won't increase all the way to $1$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$ Background
I recently found out about Lobachevsky's integral formula, so I tried to create a problem on my own for which I'd be able to apply this formula. The problem is presented below.
Problem
Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$
Attempt
If we define $f(x) := \frac{1+\cos(x)}{2+\cos(x)}$, which most definitely is $\pi$ - periodic.
The integral is, using our notation above, on the form
$$I = \int_{-\infty}^\infty \frac{\sin(x)}{x}f(x) dx.$$
The integrand is even, so we might as well compute
$$ I = 2 \int_{0}^\infty \frac{\sin(x)}{x}f(x) dx.$$
We will now have to make use of a theorem.
Lobachevsky's integral formula states that if $f(x)$ is a continous $\pi$ - periodic function then we have that $$ \int_0^\infty \frac{\sin(x)}{x}f(x) dx= \int_0^{\pi/2} f(x) dx.$$
Substituing our $f(x)$ yields us
$$ \int_0^{\pi/2} \frac{1+\cos(x)}{2+\cos(x)} dx = \pi/2 - \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx $$
where
$$I_2 = \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx = \int_0^{\pi/2}\frac{\sec^2(x/2)}{3+\tan^2(x/2)}dx.$$
Letting $ u = \tan(x/2)/\sqrt{3}$, for which $du = \sec^2(x/2)/(2\sqrt{3})dx$, therefore gives us:
$$ I_2 = \int_0^{1/\sqrt{3}}\frac{2\sqrt{3}}{3u^2+3} = \frac{\pi}{3\sqrt{3}}.$$
Finally we can compute $I$ to
$$I = 2\left(\frac{\pi}{2} - \frac{\pi}{3\sqrt{3}}\right) = \frac{\pi(3\sqrt{3}-2)}{3\sqrt{3}}.$$
I've tried calculating this integral in Desmos where it gives me $0$ when I calculate the integrand on the interval $(-\infty, \infty)$, and something negative for $(0,\infty)$. This contradicts my answer.
I also tried typing it into Wolfram, without success. Can anyone confirm the validity of my result?
| Another approach:
Using the Fourier series $$1 + 2 \sum_{k=1}^{\infty} (-1)^{n} a^{n}\cos(nx) = \frac{1-a^{2}}{1+2a \cos (x) +a^{2}}, \quad |a| <1, $$ we have
$$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\frac{1-a^{2}}{1+2a \cos (x) +a^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x}\left(1+2 \sum_{n=1}^{\infty} (-1)^{n} a^{n} \cos(nx) \right) \, \mathrm dx \\ &= \pi + 2 \sum_{n=1}^{\infty} (-1)^{n}a^{n}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \cos(nx) \, \mathrm dx \\ &= \pi - 2a \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \cos (x) \, \mathrm dx + 2 \sum_{n=2}^{\infty} (-1)^{n} a^{n} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \, \cos(nx) \mathrm dx \\ &= \pi - 2a \left(\frac{\pi}{2} \right)+2\sum_{n=2}^{\infty} (-1)^{n}a^{n}(0) \\ &= \pi \left(1-a \right) . \end{align}$$
Rewriting the integral as
$$\frac{1}{1+a^{2}}\int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1-a^{2}}{1+\frac{2a}{1+a^{2}}\cos(x)} \, \mathrm dx,$$ and letting $a= 2-\sqrt{3}$, we get $$\sqrt{3} \int_{-\infty}^{\infty}\frac{\sin (x)}{x} \frac{1}{2+\cos(x)} \, \mathrm dx = \pi \left(\sqrt{3}-1 \right). $$
Therefore, $$ \begin{align} \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1+\cos (x)}{2+ \cos (x)} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \, \mathrm dx - \int_{-\infty}^{\infty} \frac{\sin (x)}{x} \frac{1}{2+ \cos (x)} \, \mathrm dx \\ &= \pi - \frac{\pi}{\sqrt{3}} \left(\sqrt{3} - 1 \right) \\ &= \frac{\pi}{\sqrt{3}}. \end{align} $$
The one issue with this approach is justification for switching the order of integration and summation. Fubini's theorem is not satisfied.
Fortunately, we can use Sangchul Lee's result from the addendum of this answer to justify the switching.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Proof by induction of inequality. For $n \geq 2, 3^n \gt 2^n+n^2$ I'm trying to understand proof by induction on inequalities.
One problem I have is:
Prove that for $n \geq 2$ $3^n \gt 2^n+n^2$
I can prove the base case as $n=2$:
$3^2 \gt 2^2+2^2$
$9 \gt 8$
What I don't understand is the inductive step. How do I show for some $k$ where $k \geq 2$, $3^{k+1} \gt 2^{k+1} + (k+1)^2$.
| Suppose $3^k > 2^k + k^2$ you have to show $3^{k+1} > 2^{k+1} + (k+1)^2$. From the inductive step, $3^{k+1} = 3\cdot 3^k > 3(2^k+k^2)= 2^{k+1}+2^k+k^2+2k^2>2^{k+1}+k^2+2k+1=2^{k+1}+(k+1)^2$. This is true because $2^k \ge 2k$, and $2k^2 > 1$. Thus by induction, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what $c \in \mathbb{Q}$ is $\sqrt{2}+c\sqrt{3}$ a primitive element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$? I am trying to find the $c\in\mathbb{Q}$ so that $\sqrt{2}+c\sqrt{3}$ is a primitive element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, i.e. $\mathbb{Q}(\sqrt{2}+c\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$.
In class we have already shown that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, so I know that c=1 is possible.
If $c=0\in\mathbb{Q}$, then we would have $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2})$, which is not possible since $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, I am not sure if I have to prove this further.
So know considering $c\in\mathbb{Q}\setminus\{0\}$, I followed the prove for $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$:
$"\supseteq"$: Since $\sqrt{2},\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, then $\sqrt{2}+c\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, by the closure of $\mathbb{Q}(\sqrt{2},\sqrt{3})$.
$"\subseteq"$: Since $(\sqrt{2}+c\sqrt{3})^{-1}=\frac{1}{\sqrt{2}+c\sqrt{3}}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$ and $\frac{\sqrt{2}-c\sqrt{3}}{\sqrt{2}-c\sqrt{3}}=1\in\mathbb{Q}$, then $\frac{1}{\sqrt{2}+c\sqrt{3}}\frac{\sqrt{2}-c\sqrt{3}}{\sqrt{2}-c\sqrt{3}}=c\sqrt{3}-\sqrt{2}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. Therefore, $\sqrt{2}+c\sqrt{3}+(c\sqrt{3}-\sqrt{2})=2c\sqrt{3}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. And since $\frac{1}{2c}\in\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2}+c\sqrt{3})$, $\sqrt{3}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. Using the same procedure we can show that $\sqrt{2}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$.
Is this correct?
| We need to decide whether $\theta=\sqrt{2}+c\sqrt{3}$ has degree $4$ over $\mathbb Q$.
This reduces to deciding whether $1,\theta,\theta^2,\theta^3$ are linearly independent over $\mathbb Q$.
We have
$$
\pmatrix{1 \\ \theta \\ \theta^2 \\ \theta^3}
=
\pmatrix{
1 & 0 & 0 & 0 \\
0 & 1 & c & 0 \\
2+3c^2 & 0 & 0 & 2c \\
0 & 2+9c^2 & 3+6c & 0 \\
}
\pmatrix{1 \\ \sqrt2 \\ \sqrt3 \\ \sqrt6}
$$
The determinant of the matrix is $2 c (9 c^3 - 4 c - 3)$ and so is zero for rational $c$ iff $c=0$.
This argument relies on knowing that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $4$ over $\mathbb Q$ and that $1,\sqrt2,\sqrt3,\sqrt6$ is a basis.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Solving inequality $\frac{7x+12}{x} \geq 3$ I was sitting with my friend in a park, when he thought of and asked for solving this inequality:
$$
\frac{7x+12}{x} \geq 3
$$
I was confident enough that I could solve this example as continued:
$$
7x + 12 \geq 3x
$$
$$
4x \geq -12
$$
$$
x \geq -3
$$
He simply told me to put $x = -4$, and in the equation, I applied:
$$
\frac{-28+12}{-4}
$$
It got $\frac{-16}{-4}$ which is 4, so it is false that $x \geq -3$. Where I am doing a mistake, and what would be the correct solution.
| You made a mistake when you multiplied both sides of the inequality by $x$. Since multiplying an inequality by a negative number reverses the direction of the inequality, you have to consider cases when you multiply both sides of the inequality by $x$.
Method 1
Case 1: $x > 0$
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
7x + 12 & \geq 3x\\
4x + 12 & \geq 0\\
4x & \geq -12\\
x & \geq -3
\end{align*}
Since $x > 0$ and $x \geq -3$, $x > 0$.
Case 2: $x < 0$
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
7x + 12 & \leq 3x\\
4x + 12 & \leq 0\\
4x & \leq -12\\
x & \leq -3
\end{align*}
Since $x < 0$ and $x \leq -3$, $x \leq -3$.
Since the two cases are mutually exclusive and exhaustive, $x > 0$ or $x \leq -3$. Therefore, the solution set is
$$S = (-\infty, -3] \cup (0, \infty) = ]-\infty, -3] \cup ]0, \infty[$$
Method 2:
We can avoid cases if we first subtract $3$ from each side of the inequality.
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
\frac{7x + 12}{x} - 3 & \geq 0\\
\frac{7x + 12}{x} - \frac{3x}{x} & \geq 0\\
\frac{4x + 12}{x} & \geq 0\\
\frac{x + 3}{x} & \geq 0
\end{align*}
Equality holds when $x = -3$. The strict inequality holds when $x + 3$ and $x$ have the same sign. $x + 3 > 0$ and $x > 0 \implies x > 0$. $x + 3 < 0$ and $x < 0 \implies x < -3$. Thus, the solution set is
$$S = \{-3\} \cup (-\infty, -3) \cup (0, \infty) = (-\infty, -3] \cup (0, \infty)$$
Method 3
Since the direction of the inequality is preserved if we multiply both sides of the inequality by a positive number, we multiply both sides of the inequality by $x^2 > 0$.
\begin{align*}
\frac{7x + 12}{x} & \geq 3\\
x(7x + 12) & \geq 3x^2\\
7x^2 + 12x & \geq 3x^2\\
4x^2 + 12x & \geq 0\\
x^2 + 3x & \geq 0\\
x(x + 3) & \geq 0
\end{align*}
Equality holds if $x = 0$ or $x = -3$. However, the original expression is not defined when $x = 0$, so equality holds if and only if $x = -3$. The strict inequality holds when $x + 3$ and $x$ have the same sign. $x + 3 > 0$ and $x > 0 \implies x > 0$. $x + 3 < 0$ and $x < 0 \implies x < -3$. Thus, the solution set is
$$S = \{-3\} \cup (-\infty, -3) \cup (0, \infty) = (-\infty, -3] \cup (0, \infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(\frac{1}{x}+x\right)^2} d x\tag*{} $
Averaging them gives the exact value of the integral
$\displaystyle \begin{aligned}I & =\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2} d x \\& =\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+4} d x \\& =\frac{1}{4}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{2}\right)\right]_0^{\infty} \\& =\frac{1}{4}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\& =\frac{\pi}{4}\end{aligned}\tag*{} $
I guess that we can similarly evaluate the general integral
$$
I_n=\int_0^{\infty} \frac{d x}{\left(x+\frac{1}{x}\right)^{2 n}}
$$
by mapping $x\mapsto \frac{1}{x}$ and then averaging.
$$
I_n=\frac{1}{2} \int_0^{\infty} \frac{1+\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^{2 n}} d x=\frac{1}{2} \int_0^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left[\left(x-\frac{1}{x}\right)^2+4\right]^n}
$$
Letting $x-\frac{1}{x}=\tan \theta$ yields
$$
\begin{aligned}
I_n & =\frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2 \sec ^2 \theta d \theta}{4^n \sec ^{2 n} \theta} =\frac{1}{4^n} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta= \boxed{\frac{\pi(2 n-3) ! !}{4^n(2 n-2) ! !}}
\end{aligned}
$$
where the last answer comes from the Wallis cosine formula.
My question: How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$?
| $$
\begin{align}
\int_{0}^{\infty}\frac{1}{\left(x+\frac{1}{x}\right)^{2n}}dx &= \int_{0}^{\infty}\frac{x^{2n}}{\left(1+x^{2}\right)^{2n}}dx \\
&= \frac{1}{2}\int_{-\infty}^{\infty}\frac{x^{2n}}{\left(1+x^{2}\right)^{2n}}dx \\
&= \frac{1}{2}\int_{-\infty}^{\infty}\frac{x^{2n}}{\left(x-i\right)^{2n}\left(x+i\right)^{2n}}dx \\
&= \frac{1}{2}\cdot2\pi i\operatorname{Res}\left(\frac{z^{2n}}{\left(z-i\right)^{2n}\left(z+i\right)^{2n}},\ z=i\right) \tag{1}\\
&= \frac{\pi i}{\left(2n-1\right)!} \lim_{z\to i}\frac{d^{2n-1}}{dz^{2n-1}}\frac{\left(z-i\right)^{2n}z^{2n}}{\left(z-i\right)^{2n}\left(z+i\right)^{2n}} \\
&= \frac{\pi i}{\left(2n-1\right)!} \lim_{z\to i}\frac{d^{2n-1}}{dz^{2n-1}}\left(1-\frac{i}{z+i}\right)^{2n} \\
&= \frac{\pi i}{\left(2n-1\right)!}\lim_{z\to i}\frac{d^{2n-1}}{dz^{2n-1}}\sum_{j=0}^{2n} \binom{2n}{j}\left(-1\right)^{j}\frac{i^{j}}{\left(z+i\right)^{j}} \\
&= \frac{\pi i}{\left(2n-1\right)!}\lim_{z\to i}\sum_{j=1}^{2n}\binom{2n}{j}\left(-1\right)^{j}\frac{d^{2n-1}}{dz^{2n-1}}\frac{i^{j}}{\left(z+i\right)^{j}} \\
&= \frac{\pi i}{\left(2n-1\right)!}\lim_{z\to i}\left(-1\right)^{2n-1}\sum_{j=1}^{2n}\binom{2n}{j}\left(-1\right)^{j}\frac{\left(j+2n-2\right)!}{\left(j-1\right)!\left(z+i\right)^{j+2n-1}} \\
&= \frac{\pi i}{\left(2n-1\right)!}\left(-1\right)^{2n-1}\sum_{j=1}^{2n}\binom{2n}{j}\left(-i\right)^{j}\frac{\left(j+2n-2\right)!}{\left(j-1\right)!\left(2i\right)^{j+2n-1}} \\
&= \frac{\sqrt{\pi}Γ\left(n-\frac{1}{2}\right)}{4^{n}Γ\left(n\right)} \tag{2}\\
\end{align}
$$
For $(1)$ we define a contour $C := [-R,R] \cup \Gamma$ (such that $\Gamma$ is a semicircle of radius $R$) and $(2)$ do a lot of algebra and simplifying.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Define $f\left(x\right) = \frac{x}{x + \frac{x}{x + \frac{x}{x + \frac{x}{x + \vdots}}}}$. What is $f'\left(x\right)$, the derivative? Since $f\left(x\right)$ is indeed an infinitely deep continued fraction, I have seen that $f\left(x\right) = \frac{x}{x + f\left(x\right)}$, but taking the derivative of both sides from there has not worked. I have tried to roughly figure out what $f\left(x\right)$ is in terms of $x$, and the work is down here (except I have also run into a dead end with this one once I then took the derivative in the end).
$f\left(x\right) = \frac{x}{x + f\left(x\right)}$
$f\left(x\right)\left(x + f\left(x\right)\right) = x$
$x \cdot f\left(x\right) + {\left(f\left(x\right)\right)}^{2} = x$
${\left(f\left(x\right)\right)}^{2} = x - x \cdot f\left(x\right)$
${\left(f\left(x\right)\right)}^{2} = x\left(1-f\left(x\right)\right)$
$\frac{\left(f\left(x\right)\right)^{2}}{1-f\left(x\right)} = x$
I don't know what to do from there. Maybe I'm just complicating things more than they should be. :]
| We have $f^2(x) + xf(x) - x = 0$ equation, there are two functions satisfying this equation:
$$f_+(x) = \frac{-x+\sqrt{x^2+4x}}{2}$$ and
$$f_-(x) = \frac{-x - \sqrt{x^2+4x}}{2}$$
both functions are undefined for $ -4 < x < 0$. There are also the following properties for these functions:
$$x > 0 \implies f_+(x) > 0, f_-(x) < 0$$
and
$$x<-4 \implies f_+(x) > 2, f_-(x) < 2$$
Also
$$f(x) = \lim_{n \to \infty}f_n(x), f_0(x) = 1, f_{n+1}(x) = \frac{x}{x+f_n(x)}$$
There following is true:
$$x < -4,f_n(x) < 2 \implies f_{n+1}(x) < 2$$
and
$$ x > 0, f_n(x) > 0 \implies f_{n+1}(x) > 0$$
Since $0 < f_0(x) < 2 $, $x < -4 \implies f(x) \le 2$ and $x > 0 \implies f(x) \ge 0$.
This means that
$f(x) = f_-(x)$ for $x < -4$
and
$f(x) = f_+(x)$
for $x > 0$.
So,
$$f_-'(x) = \frac{-\frac{x+2}{\sqrt{x^2+4x}}-1}{2}$$
and
$$f_+'(x) = \frac{\frac{x+2}{\sqrt{x^2+4x}}-1}{2}$$
and $f'(x)$ is not defined for $-4 \le x \le 0$.
Bonus: let $g(x) = f'(x)$.
$$ \forall x, x < -4 \lor x>0: (g(x)-f'_-(x))(g(x)-f'_+(x)) = 0$$
$$ g^2(x) -(f'_-(x)+f'_+(x))g(x) + f'_-(x)f'_+(x) = 0$$
$$ g^2(x) + g(x) + \frac{1-\frac{(x+2)^2}{x^2+4x}}{4} = 0$$
$$ g^2(x) + g(x) - \frac{1}{x^2+4x} = 0$$
$$ g(x)(g(x)+1) = \frac{1}{x^2+4x}$$
Let $d(x) = x^2+4x$.
$$g(x) = \frac{1}{d(x)+d(x) g(x)}$$
$$g(x) = \frac{1}{d(x)+d(x) \frac{1}{d(x) + d(x)\frac{1}{d(x) +...}}}$$
Notice also that $f'(x) > 0$, while second solution of quadratic equation for $g(x)$ takes negative values. Since $d(x) > 0$ for those $x$ for which $f'(x)$ is defined, recurring fraction for $g(x)$ takes positive values. Therefore:
$$f'(x) = \frac{1}{d(x)+d(x) \frac{1}{d(x) + d(x)\frac{1}{d(x) +...}}}$$
$$f'(x) = \frac{1}{d(x) + \frac{1}{1+\frac{1}{d(x)+\frac{1}{1+...}}}}$$
| {
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"url": "https://math.stackexchange.com/questions/4647660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that $n^3+2n$ is divisible by $3$ I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!
Problem:
For any natural number $n , n^3 + 2n$ is divisible by $3.$
This makes sense
Proof:
Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$
$2 \times 0 = 0.$ So it is divisible by $3.$
Induction: Assume that for an arbitrary natural number $n$,
$n^3+ 2n$ is divisible by $3.$
Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use
the induction hypothesis. Got it
$$( n + 1 )^3+ 2( n + 1 ) = ( n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$
$$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying
and regrouping}\}$$
$$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out
the 3}\}$$
which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$
by the induction hypothesis. What?
Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
| $$
\begin{align}
n^3+2n &= n(n^2+2) \\
&= n(n^2−1+3)\\
&= n((n^2−1^2)+3)\\
&= n((n−1)(n+1)+3)\\
&= n(n−1)(n+1)+3n
\end{align}
$$
The key is to identify the factor $(n^2-1^2)=(n-1)(n+1)$. Looking at the result you can tell that it must be divisable by 3 because one of $n$, $(n-1)$ or $(n+1)$ is a multiple of 3. Ben Albert gives a similar solution but I believe $3n^2$ should be $3n$ instead.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| Gathering as many proofs as we can? Write the series recursively:
$$S(n) = S(n - 1) + n \tag{1}$$
Substitute $n \to n+1$ :
$$S(n + 1) = S(n) + n + 1\tag{2}$$
Equation (2) subtract Equation (1):
$$S(n+1) - S(n) = S(n) + 1 - S(n - 1) \tag{3}$$
And write it up:
$$\begin{cases}
S(n+1) &= 2S(n) -S(n-1) + 1 \\
S(n) &= S(n) \end{cases} \tag{4}$$
Which can now be written in matrix form:
$$
\begin{bmatrix} S(n+1) \\ S(n) \end{bmatrix} =
\begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix}
\begin{bmatrix} S(n) \\ S(n-1) \end{bmatrix} +
\begin{bmatrix} 1 \\ 0 \end{bmatrix}
\tag{5}$$
And then converting the affine equation (5) to the linear equation (6):
$$
\begin{bmatrix} S(n+1) \\ S(n+0) \\ 1\end{bmatrix} =
\begin{bmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} S(n) \\ S(n-1) \\ 1\end{bmatrix}
\tag{6}$$
And closing the equation:
$$
\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} =
\begin{bmatrix} 2 & -1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^n
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\tag{6}$$
Then finding the Jordan form of the 3x3 matrix:
$$\begin{align}
\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} &=
\left(\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^{-1}\right)^n
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\\ &=
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{bmatrix}^n
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}^{-1}
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\\ &=
\begin{bmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 1 & \binom{n}{1} & \binom{n-1}{2} \\ 0 & 1 & \binom{n}{1} \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} 0 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}
\begin{bmatrix} S(1) \\ S(0) \\ 1\end{bmatrix}
\end{align}
\tag{7}$$
And multiplying the matrices out:
$$\begin{bmatrix} S(n+1) \\ S(n) \\ 1\end{bmatrix} =
\begin{bmatrix}
\frac{ (2n+2)S(1) - 2nS(0) + {n}^{2} + n}{2} \\
\frac{ 2nS(1) + (2 - 2n)S(0)+{n}^{2}-n}{2} \\
1
\end{bmatrix} \tag{8}$$
And given that $S(0) = 0$ and $S(1) = 1$, we get that:
$$S(n) = \frac{n^2 + n}{2} \tag{9}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that this sum $\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$ How does one prove that for a prime $p \geq 5$ the sum : $$\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.
Since each term of $\displaystyle \sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 < k < \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.
How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$
| Since we are working in the field $\mathbb{F}_p$ we can write
$$\frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdots k}$$ as
$$\frac{(-1)(-2) \cdots (-(k-1))}{1 \cdot2 \cdots k}$$
= $$\frac{(-1)^{k-1}}{k}$$
Let $N = [\frac{2p}{3}]$ and $M = [\frac{N}{2}]$
Thus what we need is
$$ \sum_{k=1}^{N} \frac{(-1)^{k-1}}{k}$$ $$ = \sum_{k=1}^{N} \frac{1}{k} - 2 \sum_{k=1}^{M}\frac{1}{2k}$$ $$ =\sum_{k=M+1}^{N}\frac{1}{k}$$
Now $N+M+1 =p$ so we can rewrite as
$$\frac{1}{N} + \frac{1}{M+1} + \frac{1}{N-1} + \frac{1}{M+2} + \cdots = $$
$$\frac{p}{N(M+1)} + \frac{p}{(N-1)(M+2)} + \cdots $$
which is $0$.
There are $N-M$ terms, which is even, so each term gets paired off.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a binomial sum identity $\sum _{k=0}^n \binom nk \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}$ Mathematica tells me that
$$\sum _{k=0}^n { n \choose k} \frac{(-1)^k}{2k+1} = \frac{(2n)!!}{(2n+1)!!}.$$
Although I have not been able to come up with a proof.
Proofs, hints, or references are all welcome.
| In trying to prove
$$S_n = \sum_{k=0}^n \frac{(-1)^k}{2k+1} {n\choose k}
= \frac{(2n)!!}{(2n+1)!!}$$
we introduce
$$f(z) = n! (-1)^n \frac{1}{2z+1} \prod_{q=0}^n \frac{1}{z-q}.$$
This has the property that for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z)
= n! (-1)^n \frac{1}{2k+1}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = n! (-1)^n \frac{1}{2k+1}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= \frac{(-1)^k}{2k+1} {n\choose k}.$$
Note that the residue of $f(z)$ at infinity is zero by inspection and
residues sum to zero so that we have
$$S_n = \sum_{k=0}^n \mathrm{Res}_{z=k} f(z)
= - \mathrm{Res}_{z=-1/2} f(z)
\\ = - \frac{1}{2} \mathrm{Res}_{z=-1/2} \; n! (-1)^n
\frac{1}{z+1/2} \prod_{q=0}^n \frac{1}{z-q}
= - \frac{1}{2} n! (-1)^n
\prod_{q=0}^n \frac{1}{-1/2-q}
\\ = \frac{1}{2} n! \prod_{q=0}^n \frac{1}{1/2+q}
= 2^n n! \prod_{q=0}^n \frac{1}{2q+1}
= \frac{(2n)!!}{(2n+1)!!}.$$
This is the claim.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
| Notice that $$\sin 3\theta=3\sin \theta-4\sin^3 \theta$$ Now substitute $\theta=\frac{\pi}{10}$ in the above equation, we get
$$\sin \frac{3\pi}{10}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies \cos\left( \frac{\pi}{2}-\frac{3\pi}{10}\right)=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies \cos\frac{\pi}{5}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies 1-2\sin^2\frac{\pi}{10}=3\sin \frac{\pi}{10}-4\sin^3 \frac{\pi}{10}$$ $$\implies 4\sin^3\frac{\pi}{10}-2\sin^2 \frac{\pi}{10}-3\sin \frac{\pi}{10}+1=0$$ It is obvious that the sum of all the coefficients of above cubic equation is $0$ hence above cubic equation has one real root $1$. Now it can be easily factorized as follows $$\left(\sin \frac{\pi}{10}-1\right)\left(4\sin^2 \frac{\pi}{10}+2\sin \frac{\pi}{10}-1\right)=0$$
$$\color{red}{ \text{if}\quad \sin\frac{\pi}{10}-1=0 \implies \sin\frac{\pi}{10}=1}$$ but $\frac{\pi}{10}<\frac{\pi}{2} \implies \sin\frac{\pi}{10}<1$ Hence, above value is unacceptable
$$\color{blue}{ \text{if}\quad 4\sin^2\frac{\pi}{10}+2\sin\frac{\pi}{10}-1=0 \implies \sin\frac{\pi}{10}=\frac{-2\pm \sqrt{(-2)^2-4(4)(-1)}}{2(4)}}$$ $$\sin\frac{\pi}{10}=\frac{-1\pm \sqrt{5}}{4}$$ but $0<\frac{\pi}{10}<\frac{\pi}{2}\implies 0<\sin\frac{\pi}{10}<1$ Thus we get
$$\sin\frac{\pi}{10}=\frac{-1+\sqrt{5}}{4}$$
$$\implies \cos\left(\frac{\pi}{2}-\frac{\pi}{10}\right)=\frac{-1+\sqrt{5}}{4}$$
$$\implies \color{blue}{\cos\frac{2\pi}{5}=\frac{-1+\sqrt{5}}{4}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it with us?
| Following proof rely on this integral identity :
$$\int_{a}^{1}\frac{\arccos x}{\sqrt{x^2-a^2}}\mathrm{d}x=-\frac{\pi}{2}\ln a\qquad ;\,a\in(0,1]$$
We will prove it later on. Now, let's make a power series :
$$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\int_0^1\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^n}{n}\,\mathrm{d}x=-\int_0^1\frac{\ln(1-x)}{x}\,\mathrm{d}x=-\int_0^1\frac{\ln x}{1-x}\,\mathrm{d}x$$
Inserting the formula above we get :
$$\zeta(2)=\frac{2}{\pi}\int_0^1\int_{x}^{1}\frac{\arccos y}{(1-x)\sqrt{y^2-x^2}}\,\mathrm{d}y\,\mathrm{d}x$$
Interchanging the order of integration :
$$\zeta(2)=\frac{2}{\pi}\int_0^1\int_{0}^{y}\frac{\arccos y}{(1-x)\sqrt{y^2-x^2}}\,\mathrm{d}x\,\mathrm{d}y\tag{A}$$
But, with help of substitution $x=y \cos{\theta}$ and universal $t=\tan\frac{\theta}{2}$ : $$\int_{0}^{y}\frac{\mathrm{d}x}{(1-x)\sqrt{y^2-x^2}}=\int_{0}^{\frac{\pi}{2}}\frac{\mathrm{d}\theta}{1-y \cos{\theta}}=\int_{0}^{1}\frac{\frac{2\mathrm{d}t}{1+t^2}}{1-y\frac{1-t^2}{1+t^2}}=\frac{\pi-\arccos{y}}{\sqrt{1-y^2}}$$
Plugging this to $(A)$ we get :
$$ \begin{align*}&\zeta(2)=\frac{2}{\pi}\int_{0}^{1}\frac{\pi\arccos{y}-\arccos^2 y}{\sqrt{1-y^2}}\,\mathrm{d}y=\frac{2}{\pi}\left(\frac{\pi}{2}\arccos^2 y- \frac{1}{3}\arccos^3 y\right)\bigg{|}_{1}^{0}= \\ \\
&\frac{2}{\pi}\left(\frac{\pi}{2}\left(\frac{\pi}{2}\right)^2-\frac{1}{3}\left(\frac{\pi}{2}\right)^3\right) = \frac{2}{\pi}\left(\frac{\pi}{2}\right)^3 \left(1-\frac{1}{3}\right) =\frac{\pi^2}{6} \end{align*}$$
ADDENDUM : Proof of the apriori integral :
$$\begin{align*}&\int_{a}^{1}\frac{\arccos x}{\sqrt{x^2-a^2}}\mathrm{d}x=\int_{a}^{1}\frac{\arccos\left(\frac{x}{y}\right)}{\sqrt{x^2-a^2}}\bigg{|}_{y=x}^{y=1}\mathrm{d}x=\int_{a}^{1}\int_{x}^{1}\frac{x}{y}\frac{\mathrm{d}y\,\mathrm{d}x}{\sqrt{x^2-a^2}\sqrt{y^2-x^2}} = \\ \\
& \int_{a}^{1}\int_{a}^{y}\frac{x}{y}\frac{\mathrm{d}x\,\mathrm{d}y}{\sqrt{x^2-a^2}\sqrt{y^2-x^2}} = \frac{\pi}{2}\int_{a}^{1}\frac{\mathrm{d}y}{y} = -\frac{\pi}{2}\ln a
\end{align*}$$
Where the inner integral was computed via substitution $x^2=a^2\cos^2\theta+y^2\sin^2\theta$ it is clear, taking differential, that $2x\;\mathrm{d}x=2\left(y^2-a^2\right)\sin\theta\cos\theta\;\mathrm{d}\theta$, then :
$$(x^2-a^2)(y^2-x^2)=(a^2\cos^2\theta+y^2\sin^2\theta-a^2)(y^2-a^2\cos^2\theta-y^2\sin^2\theta)=(y^2\sin^2\theta-a^2\sin^2\theta)(y^2\cos^2\theta-a^2\cos^2\theta)=(y^2-a^2)^2\sin^2\theta\cos^2\theta$$
Or $$\sqrt{x^2-a^2}\sqrt{y^2-x^2}= \left(y^2-a^2\right)\sin\theta\cos\theta\ = x\,\mathrm{d}x$$
Therefore :
$$\int_{a}^{y}\frac{x\,\mathrm{d}x}{\sqrt{x^2-a^2}\sqrt{y^2-x^2}}=\int_{0}^{\frac{\pi}{2}}\frac{x\,\mathrm{d}x}{x\,\mathrm{d}x}=\frac{\pi}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving: $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $ $$\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $$
I am very much inquisitive to see how this trigonometrical identity can be proved.
PS:I am not much of interested about an inductive proof.
| You can prove it by induction since
$\sin t = 2 \cos \dfrac{t}{2}\sin \dfrac{t}{2}$
$\sin t = 2^2 \cos \dfrac{t}{2}\cos \dfrac{t}{4}\sin \dfrac{t}{4}$
$\sin t = {2^3}\cos \dfrac{t}{2}\cos \dfrac{t}{4}\cos \dfrac{t}{8}\sin \dfrac{t}{8}$
So we conjecture:
$$\sin t = 2^n\sin\dfrac{t}{2^n} \prod_{k=1}^{n} \cos\dfrac{t}{2^k} $$
It is true for $n=1$
$$\sin t = 2^1\sin\dfrac{t}{2^1} \prod_{k=1}^{1} \cos\dfrac{t}{2^1} = 2 \cos \dfrac{t}{2}\sin \dfrac{t}{2} $$
But then for $n+1$ we get
$$\sin t = 2^{n+1}\sin\dfrac{t}{2^{n+1}} \prod_{k=1}^{n+1} \cos\dfrac{t}{2^{k}} $$
$$\sin t = {2^n}2\sin \frac{t}{{{2^{n + 1}}}}\cos \frac{t}{{{2^{n + 1}}}}\prod\limits_{k = 1}^n {\cos } \frac{t}{{{2^k}}}$$
$$\sin t = {2^n}\sin \frac{t}{{{2^n}}}\prod\limits_{k = 1}^n {\cos } \frac{t}{{{2^k}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
| $$\displaystyle \sin(60^{\circ}-20^{\circ}) = \sin 40^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$
$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ} = 2 \sin 20^{\circ} \cos 20^{\circ}$$
$$\displaystyle \frac{\sqrt{3}}{2} \cos 20^{\circ} - 2 \sin 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin 20^{\circ}$$
Multiply by $\displaystyle \frac{2}{\sin 20^{\circ}}$
$$\displaystyle \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ} = 1$$
| {
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"url": "https://math.stackexchange.com/questions/10661",
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How to solve this differential equation? $y''-\frac{y'}{x}=4x^{2}y$ How to solve $y''-\displaystyle\frac{y'}{x}=4x^{2}y$ ?
I know that the solution of this equation is: $y = e^{x^{2}}$, but I cannot resolve.
First I thought that $z=y'$ could be, but was not.
| Note that we have $xy'' - y' = 4x^3y$. The LHS in some sense is dimensional consistent and looks something similar to a quotient rule provided we divide by $x^2$. So dividing by $x^2$ and doing algebraic manipulations we get $(\frac{1}{x}y')' = 4xy$. Now this looks familiar to some extent.
Rewriting, we get $(\frac{y'}{2x})' = 2xy$.
Now let $\frac{y'}{2x} = z(x)$. Plug this in and simplify to get $z' = \frac{y'}{z}y$
(Replace $2x$ by $\frac{y'}{z}$).
So we have $z^2 = y^2 + c$.
Thus we have now converted a second order differential equation in terms of first order differential equation, viz,
$\frac{1}{2x}\frac{dy}{dx} = \pm \sqrt{y^2 + c}$.
where $c$ is a constant.
(You could plug this in and check that this satisfies the second order differential equation.)
We now need other conditions (boundary/ initial conditions) to completely solve the problem i.e. to eliminate $c$ and other constant which will arise after solving the first order differential equation to get $y(x) = \exp(x^2)$.
(Note that taking $c =0 $ we get a simple ode and the solution to which is $y(x) = y(0) \exp(\pm x^2)$).
$\textbf{EDIT:}$
The first order ODE can be solved as follows:
$\textbf{CASE 1:}$
Let $c > 0$, then let $c = a^2$
Rearranging, we get
$\frac{dy}{\sqrt{y^2 + a^2}} = \pm d(x^2)$
$y = a \tan(\theta)$, we get $dy = a \sec^2(\theta) d\theta$.
Hence, the ode now becomes,
$\sec(\theta) d\theta = \pm d(x^2)$
$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.
$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$
$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$
Substitute for $\theta$ in terms of $y$ to get,
$\frac{y}{a} \pm \sqrt{1+(\frac{y}{a})^2} = K \exp(\pm x^2)$
$\textbf{CASE 2:}$
Let $c > 0$, then let $c = -a^2$.
Rearranging, we get
$\frac{dy}{\sqrt{y^2 - a^2}} = \pm d(x^2)$
$y = a \sec(\theta)$, we get $dy = a \sec(\theta) \tan(\theta) d\theta$.
Hence, the ode now becomes,
$\sec(\theta) d\theta = \pm d(x^2)$
$d(log(\sec(\theta) + \tan(\theta))) = \pm d(x^2)$.
$log(\sec(\theta) + \tan(\theta)) = \pm (x^2 + k)$
$\sec(\theta) + \tan(\theta) = \exp(\pm (x^2 + k))$
Substitute for $\theta$ in terms of $y$ to get,
$\frac{y}{a} \pm \sqrt{(\frac{y}{a})^2 - 1} = K \exp(\pm x^2)$
$\textbf{CASE 3:}$
Let $c = 0$.
The equation, we have now is $\frac{dy}{dx} = \pm 2xy$.
Solving, we get $y(x) = K \exp(\pm x^2)$.
| {
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How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$
how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$
Thank you
| As noted above, observe that
\begin{eqnarray}
\sum_{r = 1}^{d} x^{r} = \frac{x(x^{d} - 1)}{x - 1}.
\end{eqnarray}
Differentiating both sides and multiplying by $x$, we find
\begin{eqnarray}
\sum_{r = 1}^{d} r x^{r} = \frac{dx^{d + 2} - x^{d+1}(d+1) + x}{(x - 1)^{2}}.
\end{eqnarray}
Substituting $x = 2$,
\begin{eqnarray}
\sum_{r = 1}^{d} r 2^{r} = d2^{d + 2} - (d+1) 2^{d+1} + 2 = (d - 1) 2^{d+1} + 2.
\end{eqnarray}
| {
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Minimum and maximum There are five real numbers $a,b,c,d,e$ such that
$$a + b + c + d + e = 7$$$$a^2+b^2+c^2+d^2+e^2 = 10$$
How can we find the maximum and minimum possible values of any one of the numbers ?
Source
| Maximum value that can be taken by one of the variables is $\frac{9}{5}$ and the minimum value is $1$.
This can be argued based on symmetry as follows (and I hope I have not overlooked anything in my symmetry argument).
Say we want to find the extreme values of $a$. By symmetry when $a$ reaches the extremum the remaining variables have to be equal. So when $a$ takes the extremum $b=c=d=e$.
Hence, $a + 4b = 7$ and $a^2 + 4b^2 = 10$ i.e. $a^2 + 4 (\frac{7-a}{4})^2 = 10$
i.e. $a^2 + \frac{(a-7)^2}{4} = 10$ i.e. $5a^2 -14a + 49 = 40 $ i.e. $5a^2 - 14a + 9 =0$.
Hence, $(5a-9)(a-1)=0$.
So, minimum is $a=1$ and maximum is $a = \frac{9}{5}$.
The round about way would be to use Lagrange multiplier technique and set up 7 equations in 7 unknowns and solve it.
| {
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How to prove a formula for the sum of powers of $2$ by induction? How do I prove this by induction?
Prove that for every natural number n, $ 2^0 + 2^1 + ... + 2^n = 2^{n+1}-1$
Here is my attempt.
Base Case: let $ n = 0$ Then, $2^{0+1} - 1 = 1$ Which is true.
Inductive Step to prove is: $ 2^{n+1} = 2^{n+2} - 1$
Our hypothesis is: $2^n = 2^{n+1} -1$
Here is where I'm getting off track. Lets look at the right side of the last equation: $2^{n+1} -1$ I can rewrite this as the following.
$2^1(2^n) - 1$ But, from our hypothesis $2^n = 2^{n+1} - 1$ Thus:
$2^1(2^{n+1} -1) -1$ This is where I get lost. Because when I distribute through I get this.
$2^{n+2} -2 -1$ This is wrong is it not? Am I not applying the rules of exponents correctly here? I have the solution so I know what I'm doing is wrong. Here is the correct proof.
| let $S = 2^0 + 2^1 + 2^2 + .... + 2^{n-1}$
so $2S = 2^1 + 2^2 + 2^3 + 2^n$
then $2S - S = S = (2^1 - 2^0) + (2^2 - 2^1) + ... + (2^{n-1} - 2^{n-1}) + 2^n = 2^n - 1$
and we got $2^0 + 2^1 + 2^2 + .... + 2^{n-1} = 2^n - 1$
| {
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Probability of 3 of a kind with 7 dice Similar questions:
Chance of 7 of a kind with 10 dice
Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$
Probability was never my thing, so please bear with me.
I've reviewed the threads above to the best of my ability, but I still wonder how to go about finding a match of 3 from 7 dice.
At least three match, but no more (two sets of three is okay, a set of three and a set of four is not):
(a) : $ \frac{6 \binom{7}{3} 5^4}{6^7} $
In the other discussions, this wasn't desired since it would allow for a second triple to occur, or even a quadruple.
Odds of a quadruple with the remaining 4 dice:
(b) : $(1/5)^4 $
Then, the probability that from rolling 7 dice that there is at least three that match, and no more than three, would be:
(c) : $ \frac{6 \binom{7}{3} * 5^4}{6^7}- (1/5)^4 $
Exactly two sets of three:
(d) : $ \frac{6 \binom{7}{3} \binom{4}{3} \binom{1}{1}}{ 6^7} $
Maybe? My thought process was that if $\binom{7}{3}$ will give me a set of three, then with the remaining 4, I could pick 3 yielding $\binom{4}{3}$ with 1 leftover. I realize this is probably wrong. Why? What would be the proper way to go about this?
Exactly one set of three:
Then to find the probability that there is one and only one set of three from 7 dice, we could take the probability of one or more sets of three (c) and subtract the probability of exactly two sets (d), for:
$ \frac{6 \binom{7}{3} 5^4}{ 6^7} - (1/5)^4 - \frac{6 \binom{7}{3} \binom{4}{3} \binom{1}{1} }{ 6^7} $
(e) : $ \left(\frac{6 \binom{7}{3}}{ 6^7}\right) \left( 5^4 - \binom{4}{3} \binom{1}{1} \right) - (1/5)^4 $
Is this at all on the right path?
Thank you!
PS.
Sorry about the syntax, but I couldn't figure out how to make the standard nCr() symbol with MathJaX.
| You are on the right track with $6^7 = 279936$ as the denominator. To find how many cases have three but no more matching, I would start by looking at the four partitions of 7 into up to 6 parts where the largest is 3: 3+3+1, 3+2+2, 3+2+1+1, 3+1+1+1+1. You can then work out each systematically, taking account both the numbers that appear and the order they appear in.
I think the first (which has two sets of three) is $$\frac{6!}{2!\;1!\;3!} \times \frac{7!}{3!\;3!\;1!} = 8400$$ which is ten times what you have in (c). The others (just one three-of-a-kind) are
$$\frac{6!}{1!\;2!\;3!} \times \frac{7!}{3!\;2!\;2!} + \frac{6!}{1!\;1!\;2!\;2!} \times \frac{7!}{3!\;2!\;1!\;1!} + \frac{6!}{1!\;4!\;1!} \times \frac{7!}{3!\;1!\;1!\;1!\;1!} = 113400$$.
So I get about 0.405 for the probability of exactly one three-of-a-kind (but no four or more) and about 0.435 for the probability of the one or more threes-of-a-kind (but no four or more)
| {
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Identity for $\sum\limits_{j = a}^{N} \binom{N}{j} \binom{j}{a} d^{-j}$? I have run across the following multinomial series:
$$ \sum_{j = a}^{N}
\binom{N}{j} \binom{j}{a} d^{-j}
$$
Here, $d>1$. This seems like a formula which has either a well-known identity, or which has no further closed form or simplification. Can anyone shed any light on this formula, or possibly point me to a standard reference?
| Suppose we seek to evaluate
$$\sum_{j=a}^N {N\choose j}{j\choose a} d^{-j}.$$
Start from
$${j\choose a}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{a+1}} (1+z)^j \; dz.$$
Note that with $j$ and $a$ integers when $j\lt a$ then $j-a \lt 0$ and
$j-(a+1) \lt -1$ so we may extend the sum to start at $j=0$ because
the extra contribution is zero.
This gives for the sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{a+1}}
\sum_{j=0}^N {n\choose j} \times (1+z)^j \times d^{-j} \; dz.$$
This simplifies to
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{a+1}}
\left(\frac{1+z}{d}+1\right)^N \; dz$$
which is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{a+1}}
\left(\frac{1+d+z}{d}\right)^N \; dz.$$
Hence the closed form is
$$[z^a] \left(\frac{1+d+z}{d}\right)^N
= \frac{1}{d^N} [z^a] (1+d+z)^N
= \frac{1}{d^N} {N\choose a} (1+d)^{N-a}$$
This finally yields
$$\frac{1}{d^a} \frac{1}{d^{N-a}} {N\choose a} (1+d)^{N-a}
= \frac{1}{d^a} {N\choose a}
\left(1 + \frac{1}{d}\right)^{N-a}.$$
A trace as to when this method appeared on MSE and by whom starts at this
MSE link.
| {
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Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions
Prove that $x^{2} \equiv 1 \pmod{2^k}$ has exactly four incongruent solutions.
My attempt:
We have,
$x^2 - 1 = (x - 1) \times (x + 1)$, then
$(x - 1)(x + 1) \equiv 0 \pmod{2^k}$
which implies,
$2^k|(x - 1)$ or $2^k|(x + 1) \implies x \equiv \pm 1 \pmod{2^k} (1)$
Furthermore, $2^{k-1} \equiv 0 \pmod{2^k} \Leftrightarrow 2^{k-1} + 1 \equiv 1 \pmod{2^k}$.
Multiply both sides by $-1$, we have another congruent namely $-(2^{k-1} + 1) \equiv -1 \pmod{2^k}$
Hence, $x \equiv \pm(1 + 2^{k-1}) \pmod{2^k} (2)$
From $(1)$ and $(2)$, we can conclude that $x^{2} \equiv 1 \pmod{2^k}$ have four incongruent solutions.
Am I in the right track?
Thanks,
| Note that $(x - 1)(x + 1) \equiv 0 \mod{2^k}$ will imply that $x$ must be an odd integer. So we may assume that $x = 2m + 1$. Putting value of $x$ in the equation we have $4m(m+1) \equiv 0 \mod{2^k}$. This means that $2^{k-2}$ divides $m(m+1)$ if I assume $k>2$. Note that if $k \leq 2$ then there is no condition on $m$. So all residue classes of odd integer satisfy the above equation. So now assume that $k > 2$. if $m$ is even then $m$ is divisible by $2^{k-2}$ and $x = 2^{k -1}t +1$ $t \in \mathbb{Z}$. But if $m$ is odd, then $m+1$ is divisible by $2^{k - 2}$ and in this case $x= 2(m +1)-1 = 2^{k-1}t - 1$ for $t \in \mathbb{Z}$. In the first we shall have only two non congruent solution namely $1$,$2^{k -1} +1$ whereas in the second case the incongruent solutions are $-1$ and $2^{k-1} - 1$.
| {
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Proving $2 ( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} )$ is a root of$ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$ How can one show that the number
$2 \left( \cos \frac{4\pi}{19} + \cos \frac{6\pi}{19}+\cos \frac{10\pi}{19} \right)$
is a root of the equation
$\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x$?
| Let $$f(x) = 2(\cos \frac{4\pi x}{19} + \cos \frac{6\pi x}{19} + \cos \frac{10\pi x}{19})$$
Then we claim that the following are true:
$$f(1) + f(4)^2 = 4$$
$$f(4) + f(16)^2 = 4$$
$$f(16) + f(64)^2 = 4$$
Noting that $\displaystyle f(4) \lt 0$, $\displaystyle f(16) \lt 0$ and $\displaystyle f(64) = f(1) \gt 0$, we can show the identity about $\displaystyle f(1)$.
In order to prove the above three identities,
We use the fact that if
$\displaystyle P(z) = z^4 + z^{-4} + z^6 + z^{-6} + z^{10} + z^{-10}$
and if $\displaystyle w$ is a root of $\displaystyle z^{19} = -1$ then
$$P(w) + P^2(w^4) = 4$$
After some (slightly tedious, but easy) algebraic manipulation, the above identity involving $\displaystyle P(w)$ basically boils down to proving that
$$w^{18} - w^{17} + w^{16} - w^{15} + \dots + w^{2} - w = -1$$
which follows easily from the fact that
$$ 0 = w^{19} + 1 = (w+1)(w^{18} - w^{17} + w^{16} - w^{15} + \dots + w^{2} - w + 1)$$
Now since $\displaystyle P(z) = P(-z)$, the identity is also true of $\displaystyle w$ is a root of $\displaystyle z^{19} = 1$
Thus applying the identity with $\displaystyle P(w)$ three times to $\displaystyle c, c^4, c^{16}$, where $\displaystyle c = \cos \frac{\pi}{19} + i \sin \frac{\pi}{19}$ we get the above three identities involving $\displaystyle f(1), f(4), f(16)$ and thus we are done.
| {
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If $x^2 \equiv a \pmod{p}$, can we use Jacobi symbol to determine the equation has solution? To determine a quadratic congruence equation has any solution, we have to evaluate
$$\bigg( \dfrac{a}{p} \bigg)$$, So can we apply the algorithm of Jacobi symbol to evaluate this?
If yes, what's the easiest way to evaluate this symbol? Let's not take efficiency into account because I don't have a computer in my exam, so I just want to learn the simplest way which I will be able recall it easily during the exam.
For example, evaluate $$\bigg( \dfrac{2819}{4177} \bigg)$$
$$(\dfrac{2819}{4177}) = (\dfrac{4177}{2819}) \text{ because } \frac{(2819-1)(4177-1)}{4} = \text{ even } $$
$$= (\dfrac{1358}{2819}) = (\dfrac{2}{2819}) \cdot (\dfrac{679}{2819})$$
$$= (-1) \cdot (\dfrac{679}{2819}) \text{ because } \frac{2819^2 - 1}{8} = \text{ odd }$$
$$= (-1) \cdot (-1) \cdot (\dfrac{2819}{679}) = (\dfrac{2819}{679}) \text{ because } \frac{(m-1)(n-1)}{4} = \text{ odd }$$
$$= (\dfrac{2819}{679}) = (\dfrac{103}{679}) = (-1) \cdot (\dfrac{679}{103}) \text{ because } \frac{(103-1)(679-1)}{4} = \text{ odd }$$
$$= -(\dfrac{61}{103}) = -(\dfrac{103}{61}) \text{ because } \frac{(61-1)(103-1)}{4} = \text{ even }$$
$$= -(\dfrac{42}{61}) = -(\dfrac{2}{61}) \cdot (\dfrac{21}{61}) = -(\dfrac{21}{61}) \text{ because } \frac{61^2 - 1}{8} = \text{ even }$$
$$= -(\dfrac{61}{21}) \text{ because } \frac{(21-1)(61-1)}{4} = \text{ even }$$
$$= -(\dfrac{19}{21}) = -(\dfrac{21}{19}) \text{ because } \frac{(19-1)(21-1)}{4} = \text{ even }$$
$$= -(\dfrac{2}{19}) = (-1) \cdot (-1) = 1 \text{ because } \frac{19^2-1}{8} = \text{ odd }$$.
Since in the book example, the author often skipped many steps, so I just want to make sure that I understand it correctly.
Any suggestion and idea would be greatly appreciated.
Thank you,
| It looks to me like what you have done is fine. Did you check to see whether 4177 is prime? If it isn't, then Jacobi symbol 1 doesn't guarantee quadratic congruence solvable, e.g., $x^2\equiv2\pmod{15}$ has no solution even though the symbol is 1.
| {
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Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is
Write each of the following as a
decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%
For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{6} = 0.\bar{6}$ and a fraction $66\frac{2}{3}\% = \frac{66*3 + 2}{3*100}\% = \frac{200}{3}\times\frac{1}{100} =\frac{2}{3}$.
For (iv), I follow a similar path to establish that the fraction is $\frac{1}{6}$ and the decimal is $0.1\bar{6}$.
What I don't understand is a part of the model answer for this exercise. They say
$66\frac{2}{3}\% = 66.\bar{6}\% = 0.\bar{6} = \frac{6}{9} = \frac{2}{3}$
and
$16\frac{2}{3}\% = 16.\bar{6}\% = 0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$
I did not do much work with recurring decimals before and I don't know how to justify the 9 and 90 in the denominator.
Could you please explain?
| Slightly informally, you just multiply to make the repeating decimals subtract out.
$\begin{align} 10x &=6.\bar{6} \\x &=0.\bar{6} \\9x &=6 \\ x&=\frac{6}{9} \end{align}$
The case for $0.1\bar{6}$ is similar:
$\begin{align} 100x &=16.\bar{6} \\ 10x &=1.\bar{6} \\ 90x &=15 \\ x&=\frac{15}{90} \end{align}$
| {
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Formal Power Series Say I differentiate this twice:
$$\dfrac{1}{1+3x} = 1 - 3x + 9x^2 -\cdots+ (-3)^n x^n+\cdots $$
I got
$$\dfrac{18}{(1+3x)^3} = 18 - 162x + \cdots + n\cdot(n-1)(-3)^nx^{n-2}+\cdots$$
If I wanted to get $$\dfrac{1}{(1+3x)^3} = \cdots$$do I just move the 18 over ? Would that work?
$$\frac1{(1+3x)^3} = 1 - 9x + \cdots + \dfrac{n(n-1)}{18}(-3)^nx^{n-2}+\cdots $$
| Yes. Note that you can write your series as
$$1/(1+3x)^3 = 1 - 9x + \cdots + \dfrac{n(n-1)}{3\cdot3\cdot2}(-3)^nx^{n-2}+\cdots $$
$$1/(1+3x)^3 = 1 - 9x + \cdots + \dfrac{n(n-1)}{ 2}(-3x)^{n-2}+\cdots $$
$$1/(1+3x)^3 = 1 - 9x + \cdots +{n \choose2}(-3x)^{n-2}+\cdots $$
or changing the index
$$1/(1+3x)^3 = 1 - 9x + \cdots +{n+2 \choose2}(-3x)^{n}+\cdots $$
In general you can write
$$\dfrac{1}{(1-x)^{k+1}}=\sum_{n=0}^{\infty}{n+k\choose k}x^n$$
Where $$\displaystyle {n+k\choose k}$$ means $$\dfrac{(n+1)(n+2)\cdots(n+k)}{k!}$$
| {
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periodicity of function If $f(x+1) + f(x-1) = \sqrt3f(x)$, then what is the period of $f(x)$?
| $$ f(x+1)+f(x-1) = \sqrt{3} f(x) $$
$$ \sqrt{3}f(x+1) + \sqrt{3}f(x-1)= 3f(x) $$
$$ f(x)+f(x+2)+f(x-2)+f(x)= 3 f(x)$$
$$ f(x+2)+f(x-2)=f(x) $$
$$ f(x+4)+f(x) = f(x+2) $$
Adding last two equations give ,
$$ f(x+4)+f(x-2)= 0$$
$$ f(x+10)+f(x+4)= 0 \implies f(x+10) = f(x-2) \implies f(x+12)=f(x) $$
Thus period = 12 $\Box$
| {
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Trig identities I need to perform the indicated operation and simplify $(1+\sin t)^{2} + \cos^{2} t$
The book is telling me that it turns into $1 + 2\sin^2t + \cos^2t$, how is is possible? Basic math tells me that 2(3) is equal to six and that $3^2 = 9$ so there is no way that $\sin^2$ can be turned into $2\sin$
| I don't understand what your book is suggesting.
If you first expand the squared binomial, remembering that $(a+b)^2 = a^2 + 2ab + b^2$, we have
$$(1+\sin t)^2 = 1^2 + 2\times 1 \times \sin t + \sin^2 t = 1 + 2\sin t + \sin^2 t.$$
Then, use the fact that $\sin^2 t + \cos^2 t = 1$. So we have:
$$\begin{align*}
(1+\sin t)^2 + \cos^2 t &= \Bigl( 1 + 2\sin t + \sin^2 t\Bigr) + \cos^2 t\\
&= 1 + 2\sin t + \Bigl( \sin^2 t + \cos^2 t\Bigr)\\
&= 1 + 2\sin t + 1\\
& = 2 + 2\sin t\\
& = 2 (1 + \sin t).
\end{align*}$$
| {
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Seriously: What is the inverse of this 2-by-2 matrix? Okay, I have a hangover and it must be a stupid error, but I just don't get it:
The inverse of a 2-by-2 matrix $A=\left( \matrix{a & b \\ c& d} \right )$ is $\frac{1}{det A}\left( \matrix{d & -b \\ -c& a} \right ) = \frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right )$. Now when I apply this recipe to the matrix
$$G:=\frac{1}{1+x^2+y^2}\left( \matrix{1+y^2 & -xy \\ -xy & 1+x^2} \right ) = \left( \matrix{\frac{1+y^2}{1+x^2+y^2} & \frac{-xy}{1+x^2+y^2} \\ \frac{-xy}{1+x^2+y^2} & \frac{1+x^2}{1+x^2+y^2}} \right )$$ I get
$$det G = \frac{1+x^2+y^2+x^2y^2 -x^2y^2}{(1+x^2+y^2)^2} = \frac{1}{1+x^2+y^2}$$
and hence
$$G^{-1} = (1+x^2+y^2)\left( \matrix{1+x^2 & xy \\ xy & 1+y^2} \right )$$
But direct calculation shows that the inverse of $G$ is $\left( \matrix{1+x^2 & xy \\ xy & 1+y^2} \right )$, without the factor in front. Am I mad or still drunk? Can someone help my addled brain? Thanks - seriously!
| You just made a mistake in "... and hence $G^{-1}=$...". Your $\det G$ is correct.
In your problem $\frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right )=(1+x^2+y^2)\left( \matrix{d & -b \\ -c& a} \right )$. You should know what $a,b,c,d$ are now.
| {
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A hyperbola as a constant difference of distances I understand that a hyperbola can be defined as the locus of all points on a plane such that the absolute value of the difference between the distance to the foci is $2a$, the distance between the two vertices.
In the simple case of a horizontal hyperbola centred on the origin, we have the following:
*
*$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
*$c = \sqrt{a^2 + b^2} = a\varepsilon = a\sqrt{1 + \frac{b^2}{a^2}}$
The foci lie at $(\pm c, 0)$.
Now, if I'm not wrong about that, then this should be pretty basic algebra, but I can't see how to get from the above to an equation given a point $(x,y)$ describing the difference in distances to the foci as being $2a$. While I actually do care about the final result, how to get there is more important.
Why do I want to know this? Well, I'd like to attempt trilateration based off differences in distance rather than fixed radii.
| We will use a little trick to avoid work. We want to have
$$\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2+y^2}=\pm 2a.\qquad\text{(Equation 1)}$$
Rationalize the numerator, by multiplying "top" and "bottom" by
$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}.$
After the (not very dense) smoke clears, we get
$$\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}}=\pm 2a.$$
Flip it over, do some easy algebra. We get
$$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}=\pm \frac{2cx}{a}.\qquad\text{(Equation 2)}$$
From Equations 1 and 2, by adding, we get
$$2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right).$$
Cancel the $2$'s, square. We get
$$x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}.$$
Now it's basically over, the $2cx$ terms cancel. Multiply through by $a^2$, put $c^2=a^2+b^2$, and rearrange.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find $\sin \theta$ and $\cos \theta$ given $\tan 2\theta$ Can you guys help with verifying my work for this problem. My answers don't match the given answers.
Given $\tan 2\theta = -\dfrac{-24}{7}$, where $\theta$ is an acute angle, find $\sin \theta$ and $\cos \theta$
I used the identity, $\tan 2\theta = \dfrac{2\tan \theta}{1 - tan^2 \theta}$ to try and get an equation in $\tan \theta$.
$$
\begin{align}
-\dfrac{24}{7} &= \dfrac{2\tan \theta}{1 - \tan^2 \theta} \\
-24 + 24\tan^2 \theta &= 14 \tan \theta \\
24tan^2 \theta - 14\tan \theta - 24 &= 0 \\
12tan^2 \theta - 7\tan \theta - 12 &= 0 \\
\end{align}
$$
Solving this quadratic I got,
$$ \tan \theta = \dfrac{3}{2} \text{ or } \tan \theta = -\dfrac{3}{4}$$
$$\therefore \sin \theta = \pm \dfrac{3}{\sqrt{13}} \text{ and } \cos \theta = \pm \dfrac{2}{\sqrt{13}}$$
Or,
$$\therefore \sin \theta = \pm \dfrac{3}{5} \text{ and } \cos \theta = \mp \dfrac{4}{5}$$
The given answer is,
$$\sin \theta = \dfrac{4}{5} \text{ and } \cos \theta = \dfrac{3}{5}$$
I thought I needed to discard the negative solution assuming $\theta$ is acute. But they haven't indicated a quadrant. Do I assume the quadrant is I only? What am i missing? Thanks again for your help.
| Given $ \tan(2\theta)= -\frac{24}{7}$ From the relation between $\sin(\theta)$, $\cos(\theta)$ and $\tan(\theta)$, we get $$ \frac{\sin(2\theta)}{\cos(2\theta)}= -\frac{24}{7} \implies \sin(2\theta)= -\frac{24}{7} \cos(2\theta)$$and $$ \sin(2\theta)^2 + \cos(2\theta)^2=1$$ $$\cos(2\theta) = \pm \frac{7}{25} \implies 2 \cos^2(\theta)-1 = \pm \frac{7}{25}$$ Case 1: Rational number on the right is positive, $$\cos^2(\theta)=\frac{16}{25} \implies \cos(\theta) = \pm \frac{4}{5} $$ Solution to case 1: $$\cos(\theta)=\frac{4}{5}$$$$ \sin(\theta)=\frac{3}{5}.$$ Both sine and cosine functions are positive, for $\theta$ being acute. Case 2:Rational number on the right is negative $$\cos^2(\theta)=\frac{9}{25} \implies \cos(\theta) = \pm \frac{3}{5} $$Solution to case 2: $$\cos(\theta)=\frac{3}{5} $$$$ \sin(\theta)=\frac{4}{5}.$$ Both sine and cosine functions are positive, for $\theta$ being acute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.
Prove that,
$$
\dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2}
$$
| $\textbf{Method 1.}$ You have
\begin{align*}
\frac{1+ \sin{A} - \cos{A}}{1+\sin{A} + \cos{A}} &= \frac{ \cos^{2}\frac{A}{2} + \sin^{2}\frac{A}{2} + 2\cdot \sin\frac{A}{2}\cdot\cos\frac{A}{2} - \Bigl(\cos^{2}\frac{A}{2} - \sin^{2}\frac{A}{2}\Bigr)}{\cos^{2}\frac{A}{2} + \sin^{2}\frac{A}{2} + 2\cdot \sin\frac{A}{2}\cdot\cos\frac{A}{2} + \Bigl(\cos^{2}\frac{A}{2} -\sin^{2}\frac{A}{2}\Bigr)} \\ &=\frac{ \Bigl(\cos\frac{A}{2} + \sin\frac{A}{2}\Bigr)^{2} - \Bigl(\cos\frac{A}{2}+\sin\frac{A}{2}\Bigr) \cdot \Bigl(\cos\frac{A}{2} - \sin\frac{A}{2}\Bigr)}{\Bigl(\cos\frac{A}{2} + \sin\frac{A}{2}\Bigr)^{2} + \Bigl(\cos\frac{A}{2}+\sin\frac{A}{2}\Bigr) \cdot \Bigl(\cos\frac{A}{2} - \sin\frac{A}{2}\Bigr)} \\ &= \frac{ \cos\frac{A}{2} + \sin\frac{A}{2} - \cos\frac{A}{2} + \sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2} + \cos\frac{A}{2} -\sin\frac{A}{2}} \qquad \Bigl[ \text{cancelling}\ \Bigl(\small\cos\frac{A}{2} +\sin\frac{A}{2}\Bigr) \ \Bigr] \\ &= \tan\frac{A}{2}
\end{align*}
$\textbf{Method 2.}$ Here is another way of seeing this. By using something called Componendo and Dividendo.
*
*http://en.wikipedia.org/wiki/Componendo_and_dividendo
Let $$\frac{1+\sin{A} -\cos{A}}{1+\sin{A} + \cos{A}} = \frac{k}{1}$$ Now applying componendo and dividendo we get $$\frac{1+ \sin{A}}{-\cos{A}} = \frac{k+1}{k-1} \Longrightarrow \frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{-\cos\frac{A}{2}+\sin\frac{A}{2}} = \frac{k+1}{k-1} \qquad (1)$$
Again using componendo dividendo on $(1)$ we get $$\frac{\cos\frac{A}{2}+\sin\frac{A}{2} -\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2}+\sin\frac{A}{2} +\cos\frac{A}{2}-\sin\frac{A}{2}} = \frac{k+1 +k-1}{k+1-k+1} \Longrightarrow k=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}=\tan\frac{A}{2}$$
$\textbf{Method 3.}$ Another way of looking into this would be,
\begin{align*}
\frac{1+\sin{A}-\cos{A}}{1+\sin{A}+\cos{A}} =\frac{(1-\cos{A})+\sin{A}}{(1+\cos{A}) + \sin{A}} &= \frac{ 2\:\sin^{2}\frac{A}{2} + 2\cdot\sin\frac{A}{2}\cdot\cos\frac{A}{2}}{2\:\cos^{2}\frac{A}{2} + 2 \cdot\sin\frac{A}{2}\cdot \cos\frac{A}{2}} \\ &= \frac{2\:\sin\frac{A}{2}}{2\:\cos\frac{A}{2}} \cdot \Biggl(\frac{\cos\frac{A}{2}+\sin\frac{A}{2}}{\cos\frac{A}{2} + \sin\frac{A}{2}}\Biggr) \\ &= \tan\frac{A}{2}
\end{align*}
$\textbf{Method 4.}$ My younger brother mentioned about this method. Multiplying numerator and deominator by $(1+\sin{A}-\cos{A})$ we get
\begin{align*}
\frac{1+\sin{A}-\cos{A}}{1+\sin{A}+\cos{A}} \times \frac{1+\sin{A}-\cos{A}}{1+\sin{A}-\cos{A}} &= \frac{(1+\sin{A}-\cos{A})^{2}}{(1+\sin{A})^{2}-\cos^{2}{A}} \\ &= \frac{2 + 2\: \sin{A} -2\cdot\sin{A}\cdot \cos{A} -2\cos{A}}{2\: \sin{A} + 2\sin^{2}{A}} \\ &= \frac{(2-2\cos{A}) \cdot (1+\sin{A})}{2\:\sin{A} \cdot (1+\sin{A})} \\ &=\frac{1-\cos{A}}{\sin{A}} = \frac{2\: \sin^{2}\frac{A}{2}}{2\cdot \sin\frac{A}{2} \cdot \cos\frac{A}{2}} \\ &=\tan\frac{A}{2}
\end{align*}
Oh once again when I looked at the question I realize that you attempted this method. Hopefully now it's clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/50093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 5
} |
Analysis of Algorithms: Solving Recursion equations-$T(n)=3T(n-2)+9$ I need your help with solving this recursion equation: $T(n)=3T(n-2)+9$
with the initial condition: $T(1)=T(2)=1.$
I need to find $T(n),$ the complexity of the algorithm which works that way.
I tried to do that with induction.
Edit: The answer should depend on $n.$
Thank you.
| Induction is a way to prove things; it's not generally helpful in solving computational problems, except perhaps in proving that a guessed solution is correct.
Note that the values of $T(n)$ for odd $n$ are completely independent of the values for even $n$, so you can look at it as essentially two separate instances of the recursion $f(n+1) = 3f(n) + 9$ with initial condition $f(1) = 1$. There are lots of ways to solve that recurrence. One is to 'unwind' it, which is a bit like doing an induction: $f(n) = 3f(n-1)+9 = 3(3f(n-2)+9)+9$ = $3^2f(n-2)+3$ and so on. If you continue in this fashion, you'll find that after $k$ steps you have $f(n)$ = $3^kf(n-k) + 9 \sum_{i=0}^{k-1}3^i$; in particular, when $k=n-1$ you have $f(n) = 3^{n-1}f(1)+9\sum_{i=0}^{n-2}3^i$ = $3^{n-1} + 9 \cdot \frac{3^{n-1}-1}{2} = \frac{1}{2}\left(11 \cdot 3^{n-1} - 9\right)$.
Now $T(2n) = f(n) = \frac{1}{2}\left(11 \cdot 3^{n-1} - 9\right)$, and $T(2n-1) = f(n) = \frac{1}{2}\left(11 \cdot 3^{n-1} - 9\right)$.
A neater way to solve the $f$ recurrence is to let $g(n) = f(n) - d$ for some constant $d$ yet to be determined. Then $f(n) = g(n) + d$, so the recurrence can be written $g(n+1)+d$ = $3 \left(g(n)+d \right) + 9$ = $3g(n) + 3d + 9$, or $g(n+1)=3g(n) + 2d + 9$. Setting $d=-9/2$ makes this $g(n+1)=3g(n)$ with initial condition $g(1)=f(1)+9/2=11/2$. This is simple exponential growth: $g(n) = 3^{n-1}g(1)= \frac{11}{2} \cdot 3^{n-1}$. Hence $f(n) = g(n)-9/2 = \frac{1}{2}\left(11 \cdot 3^{n-1} - 9\right)$, of course giving the same solution for the original problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/50153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
A fun Pascal-like triangle Inspired by Pascal, I put on some shackles and a thorny belt. Inspiration came pouring in, and I thought of the following triangle:
$$
\begin{array}{rcccccccccc}
& & & & & 1\\\
& & & & 1 & & 1\\\
& & & 1 & & \frac{1}{2} & & 1\\\
& & 1 & & \frac{2}{3} & & \frac{2}{3} & & 1\\\
& 1 & & \frac{3}{5} & & \frac{3}{4} & & \frac{3}{5} & & 1\\\
1 & & \frac{5}{8} & & \frac{20}{27} & & \frac{20}{27} & & \frac{5}{8} & & 1\\\
& ... & & & &... & & & & ... &
\end{array}
$$
Let's call the corresponding entry ${n \choose k}$ because there is clearly no danger of confusion. The construction rule is very simple. Instead of having
$${n+1 \choose k} = {n \choose k-1} + {n \choose k},$$
as usual, we have
$${n+1 \choose k} = \frac{1}{{n \choose k-1} + {n \choose k}}.$$
It's easy to see by induction that all entries of the triangle lie between $1/2$ and $1$. Also, for fixed $k$, ${n \choose k}$ converges to a limit $C_k$. For example, $C_1=1/\phi$, where $\phi$ is the golden ratio (this should be obvious! think of Fibonacci numbers...). We can determine easily $C_{k+1}$ in terms of $C_k$ by taking the limit in the construction rule, which yields $C_k=(C_{k-1}+C_k)^{-1}$. In particular, all of the $C_k$'s are algebraic numbers.
I would like to know if anybody here can prove interesting properties of this triangle, or of the numbers $C_k$.
Enjoy! I'm going to take off the shackles and belt now.
| The sequence $C_k$ is not monotonic as @JohnM originally claimed. Here's a quick but not so elegant proof of convergence of $C_k$. As a side effect, we'll also know that the sequence goes alternately above and below its limit, namely $1/\sqrt{2}$.
Solving for $C_{k+1}$ in terms of $C_k$, we get $C_{k+1} = \frac{\sqrt{C_k^2 + 4} - C_k}{2}$. I'll directly show that the difference sequence $|C_{k}-\frac{1}{\sqrt{2}}|$ is decreasing exponentially fast, which establishes the required convergence. I'll abbreviate $C_k$ by $c$ for convenience. We have:
$$
\frac{C_{k+1} - \frac{1}{\sqrt{2}}}{C_{k} - \frac{1}{\sqrt{2}}} = \frac{\sqrt{c^2+4} - (c + \sqrt{2})}{2 (c - \frac{1}{\sqrt{2}})}
= \frac{-\sqrt{2}}{\sqrt{c^2+4} + c + \sqrt{2}},
$$
after some straightforward rearrangement. (Notice the minus sign.) Finally, notice that the denominator is at least $2+\sqrt{2} \geq 2\sqrt{2}$ for $c \geq 0$. Hence the ratio is at most $1/2$ in magnitude. In particular, we have $|C_k - \frac{1}{\sqrt{2}}| \leq A 2^{-k}$ for some constant $A$, and we are done. The negative sign shows that the sequence is alternately above and below the limit. $\Box$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 1,
"answer_id": 0
} |
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
| Since people (not from this site, but still...) keep bugging me, and I am unable to edit my previous answer, here's Mathematica code for plotting this monster:
Plot[{With[{w = 3 Sqrt[1 - (x/7)^2],
l = 6/7 Sqrt[10] + (3 + x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x + 1)^2],
h = (3 (Abs[x - 1/2] + Abs[x + 1/2] + 6) -
11 (Abs[x - 3/4] + Abs[x + 3/4]))/2,
r = 6/7 Sqrt[10] + (3 - x)/2 - 3/7 Sqrt[10] Sqrt[4 - (x - 1)^2]},
w + (l - w) UnitStep[x + 3] + (h - l) UnitStep[x + 1] +
(r - h) UnitStep[x - 1] + (w - r) UnitStep[x - 3]],
1/2 (3 Sqrt[1 - (x/7)^2] + Sqrt[1 - (Abs[Abs[x] - 2] - 1)^2] + Abs[x/2] -
((3 Sqrt[33] - 7)/112) x^2 - 3) (Sign[x + 4] - Sign[x - 4]) - 3*Sqrt[1 - (x/7)^2]},
{x, -7, 7}, AspectRatio -> Automatic, Axes -> None, Frame -> True,
PlotStyle -> Black]
This should work even for versions that do not have the Piecewise[] construct. Enjoy. :P
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
"answer_count": 10,
"answer_id": 2
} |
Inequality involving sides of a triangle How can one show that for triangles of sides $a,b,c$ that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2$$
My proof is long winded, which is why I am posting the problem here.
Step 1: let $a=x+y$, $b=y+z$, $c=x+z$, and let $x+y+z=1$ to get
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}<2$
Step 2: consider the function $f(x)=\frac{1-x}{1+x}$, and note that it is convex on the interval (0,1), so the minimum of
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}$
is reached when the function takes the extreme points. i.e. $x=y=0, z=1$.
But going back to the fact that this is a triangle, we note that $x=y=0 \implies a=0$ which is not possible, so the inequality is strict.
| $\displaystyle\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} <
\frac{a+a}{b+c+a} + \frac{b+b}{c+a+b} + \frac{c+c}{a+b+c} = 2$
since $a<b+c$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to prove the following identities?
Prove:
\begin{align}
\tan(A) + \cot(A) & = 2 \text{cosec}(2A)\\
\tan(45^{\circ}+A^{\circ}) - \tan(45^{\circ}-A^{\circ}) & = 2 \tan(2A^{\circ})\\
\text{cosec}(2A) + \cot(2A) & = \cot(A)
\end{align}
I have got all the formulas that I need but I just couldn't solve these. Some help, please?
| EDITED. In general there are two different techniques we can use to prove a trigonometric identity $A=B$. One is to transform one side into the other:
$$A=A_1=A_2=\dots =A_n=B.$$
The other is to look at the identity $A=B$ as a whole and convert it into an equivalent one and repeat the process until one known identity is found:
$$A=B\Leftrightarrow A'=B'\Leftrightarrow A''=B''\Leftrightarrow\dots\Leftrightarrow A^{*}=B^{*}.$$
The following hints are intended for proving your $3^{rd}$ identity by this second technique. Use
$$\frac{\cos 2A}{\sin 2A}=\frac{2\cos ^{2}A-1}{2\sin A\cos A}=\frac{%
\cos A}{\sin A}-\frac{1}{2\sin A\cos A}$$
to obtain
$$\csc 2A+\cot 2A=\cot A\Leftrightarrow \frac{1}{2\cos A}+\cos A-\frac{1}{%
2\cos A}=\cos A.$$
Added. Proof:
$$\csc 2A+\cot 2A=\cot A\tag{1}$$
$$\begin{eqnarray*}
&\Leftrightarrow &\frac{1}{\sin 2A}+\frac{\cos 2A}{\sin 2A}=\frac{\cos A}{%
\sin A} \\
&\Leftrightarrow &\frac{\sin A}{\sin 2A}+\frac{\cos 2A}{\sin 2A}\sin A=\cos A
\\
&\Leftrightarrow &\frac{1}{2\cos A}+\left( \frac{\cos A}{\sin A}-\frac{1}{%
2\sin A\cos A}\right) \sin A=\cos A \\
&\Leftrightarrow &\frac{1}{2\cos A}+\left( \cos A-\frac{1}{2\cos A}\right)
=\cos A \\
\end{eqnarray*}$$
$$\Leftrightarrow \cos A=\cos A\tag{2},$$
which is an identity. Thus $(1)$ is also an identity.
Your $1^{st}$ identity can be proved by the first technique:
$$\begin{eqnarray*}
\tan A+\cot A &=&\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin
^{2}A+\cos ^{2}A}{\cos A\sin A} \\
&=&\frac{1}{\cos A\sin A}=\dfrac{1}{\dfrac{\sin (2A)}{2}}=\cdots
\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/56122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Elementary central binomial coefficient estimates
*
*How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
*Does anyone know any better elementary estimates?
Attempt. We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$
Then we have
$$\left(1+\frac{k}{2(n-k)}\right)>\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$
So maybe, for the lower bound, we have
$$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}>\frac{2^n}{\sqrt{4n}}.$$
By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes
$$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$
but this isn't $>\frac{2^n}{\sqrt{4n}}$.
| For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\le\frac{n+\frac13}{n+\frac43}\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\le4\sqrt{\frac{n+\frac13}{n+\frac43}}\tag{2}
\end{align}
$$
Inequality $(2)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac13}}{4^n}\text{ is decreasing}}}\tag{3}
$$
For $n\ge0$, we have (by cross-multiplication)
$$
\begin{align}
\left(\frac{n+\frac12}{n+1}\right)^2
&=\frac{n^2+n+\frac14}{n^2+2n+1}\\
&\ge\frac{n+\frac14}{n+\frac54}\tag{4}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}
&=4\frac{n+\frac12}{n+1}\\
&\ge4\sqrt{\frac{n+\frac14}{n+\frac54}}\tag{5}
\end{align}
$$
Inequality $(5)$ implies that
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\frac{\sqrt{n+\frac14}}{4^n}\text{ is increasing}}}\tag{6}
$$
Note that the formula in $(3)$, which is decreasing, is bigger than the formula in $(6)$, which is increasing. Their ratio tends to $1$; therefore, they tend to a common limit, $L$.
Theorem $1$ from this answer says
$$
\lim_{n\to\infty}\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n}=1\tag{7}
$$
which means that
$$
\begin{align}
L
&=\lim_{n\to\infty}\frac{\sqrt{n}}{4^n}\binom{2n}{n}\\
&=\frac1{\sqrt\pi}\tag8
\end{align}
$$
Combining $(3)$, $(6)$, and $(8)$, we get
$$
\boxed{\bbox[5pt]{\displaystyle\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag9
$$
Assymptotic Observation
If we know that $n\ge n_0$, we can replace $\frac13$ in $(1)-(3)$ and $(9)$ by $\frac14+\frac1{16n_0+12}$. Thus, asymptotically,
$$
\boxed{\bbox[5pt]{\displaystyle\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}}}\tag{10}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "39",
"answer_count": 7,
"answer_id": 2
} |
Integral question: $\int \frac{x^{5}+x+3}{x{^3}-5x^{2}}\mathrm dx$ i have integral $\int \frac{x^{5}+x+3}{x{^3}-5x^{2}}\mathrm dx$,
so first step is to divide polynoms, and i get:
$\frac{x^{5}+x+3}{x{^3}-5x^{2}}$ = $x^{2} + 5x + 25$ with remainder $125x^2 + x + 3$
is it corrent to divide this polynomial division into two integrals:
int 1: $\int x^{2}+5x+25 \mathrm dx$
int 2: $\int \frac{125x^{2}+x+3}{x{^2}(x-5)}\mathrm dx$
first integral solve directly from tables, and make partial fractions from second integral, then just merge solutions into one expression.
| Yes, it is all correct.
$$
x^5 + x^2 + 3 = 125 x^2 + x + 3 + (x^3-5 x^2)(x^2 + 5 x + 25)
$$
Hence
$$
\begin{eqnarray}
\frac{x^5 + x + 3 }{x^3-5 x^2} &=& x^2 + 5 x + 25 + \frac{125 x^2 +x+ 3}{x^3-5 x} \\ &=&
x^2 + 5 x + 25 -\frac{3}{5 \, x^2} - \frac{8}{25 x} + \frac{3133}{25} \frac{x - 5}{x^2-5} \end{eqnarray}
$$
Therefore
$$
\int \frac{x^5 + x + 3 }{x^3-5 x^2} \mathrm{d} x =
\left(\frac{x^3}{3} + \frac{5}{2} x^2 + 25 x\right) + \left( \frac{3}{5} \frac{1}{x} - \frac{8}{25} \log x + \frac{3133}{25} \log(x-5) \right) + C
$$
| {
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Simplifying the expression $\sqrt[4]{\frac{g^3h^4}{4r^{14}}}$? How would I simplify this expression?
$$\sqrt[4]{\frac{g^3h^4}{4r^{14}}}\ ?$$
I did this
$$\begin{align*}
\sqrt[4]{g^3h^3h^4}\\
h\sqrt[4]{g^3h^3}\\
\sqrt[4]{4r^{14}}\\
\sqrt[4]{2r^2r^{12}}\\
r^3\sqrt[4]{2r^2}\\
\end{align*}$$
But I am stuck?
Yes that is correct
| Why did $h^4$ become $h^3h^4$? Why did $4$ become $2$?
In general, remember that for $a$ and $b$ positive, $\sqrt[4]{ab} = \sqrt[4]{a}\sqrt[4]{b}$, and that $(a^r)^{s} = a^{rs}$. Added: If $g$, $h$, and $r$ are positive, then you can rewrite what you have as:
$$\sqrt[4]{\frac{g^3h^4}{4r^{14}}} = \left( g^3\times h^4 \times 4^{-1} \times r^{-14}\right)^{1/4},$$
and then using the laws of exponents you get
$$\begin{align*}
\left(g^3\times h^4\times 4^{-1} \times r^{-14}\right)^{1/4} &= g^{3/4}\times h^{4/4} \times 4^{-1/4}\times r^{-14/4}\\
&= g^{3/4}\times h \times 4^{-1/4}\times r^{-3}\times r^{-2/4} \\
&= \frac{h\sqrt[4]{g^3}}{r^34^{1/4}r^{1/2}}\\
&= \frac{h\sqrt[4]{g^3}}{r^3(2^2)^{1/4}r^{1/2}}\\
&=\frac{h\sqrt[4]{g^3}}{r^3 2^{1/2}r^{1/2}}\\
&= \frac{h\sqrt[4]{g^3}}{r^3\sqrt{2r}}.
\end{align*}$$
If $r$ is not known to be positive, then you shoudl replace the $r^3$ and the $r$ in the last step with $|r|^3$ and $|r|$. $g$ must be positive for the original expression to be sensible; if $h$ is not known to be positive, then you should replace the $h$ at the end with $|h|$.
| {
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} |
Evaluate $\int \sqrt{1+x^{\frac{3}{2}}} \operatorname d x$ I can't figure out how to integrate
$$\int \sqrt{1+x^{\frac{3}{2}}} \operatorname d x$$
I've tried substitution by letting $u = x^3$, but it didn't go anywhere. I also tried to integrate using a trigonometric substitution, but that also got me nowhere. Then I tried Wolfram Alpha, and it got me even nowhere-er!
If you could give me a hint as to where to go, I'll try to answer this question at a later time. Thanks!
| Let $I = \int \sqrt{1 + x^{3/2}} \, \mathrm{d} x$. Integrating by parts:
$$
\begin{eqnarray}
I &=& x \sqrt{1 + x^{3/2}} - \int x \frac{3}{4} \, \frac{\sqrt{x}}{\sqrt{1+x^{3/2}}} \, \mathrm{d} x = x \sqrt{1 + x^{3/2}} - \frac{3}{4} \int \sqrt{1+x^{3/2}} \, \mathrm{d} x + \frac{3}{4} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}} \\
&=& -\frac{3}{4} I + x \sqrt{1 + x^{3/2}} + \frac{3}{4} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}}
\end{eqnarray}
$$
Thus
$$
I = \frac{4}{7} x \sqrt{1 + x^{3/2}} + \frac{3}{7} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}}
$$
The latter integral is not elementary and can be evaluated in terms of Gauss hypergeometric function:
$$
I = \frac{4}{7} x \sqrt{1 + x^{3/2}} + \frac{3 x}{7} \, {}_2F_1\left(\frac{1}{2}, \frac{2}{3} ; \frac{5}{3} ; -x^{3/2} \right)
$$
| {
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How to prove by induction that $a^{2^{k-2}} \equiv 1\pmod {2^k}$ for odd $a$? Let $a \in $ Z be odd. How to prove by induction that $a^{2^{k-2}} \equiv 1\pmod {2^k}$ for all $k \geq 3$.
My attempt: let $k=3$
$$\begin{align*}
a^{2^{k-2}}-1&= a^{2^{3-2}}-1=a^{2}-1\\
& \stackrel{\text{a is odd}}{=} (2l+1)^2-1=4l^2+4l=4l(l+1)\\ &\stackrel{\text{choose } l=2}{=} 8 \cdot 3 = 8q
\end{align*}$$
or $2^3|a^2-1 \Longleftrightarrow a^2 \equiv 1\pmod {2^3}$.
Induction assumption:
Let us assume that claim holds when $k=n$ or
$2^n|a^{2^{n-2}}-1$ or $a^{2^{n+2}} =q'2^n+1 $.
Look at the statement $a^{2^{k-2}}-1$ with the value $k=n+1$.
Then
$$\begin{align*}
a^{2^{n+1-2}}-1 &= a^{2^{n-2}\cdot 2}-1=(a^{2^{n-2}})^2-1\\
&\stackrel{\text{assumption}}{=} (q')^2(2^n)^2+2q'\cdot 2^n +1 -1\\
&= 2^n((q')^22^n+2q')=2^nq''\\
&\Longrightarrow 2^n|a^{2^{n+1-2}}-1.
\end{align*}$$
Claim holds, i.e. $a^{2^{k-2}} \equiv 1\pmod {2^k}$.
| For the case $k=3$, you cannot choose $l$. You must argue that $4l(l+1)$ is a multiple of $8$, which can be done by arguing that $l(l+1)$ must be even.
Your induction step seems ok, but it could be written more simply as
$$
a^{2^{n+1-2}}-1 = a^{2^{n-2}\cdot 2}-1=(a^{2^{n-2}}-1)(a^{2^{n-2}}+1)
$$
and use that $a^{2^{n-2}}+1$ is even.
| {
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} |
Help with volume integration I need help solving this integral ($\hat{z}$ denotes the polar axis):$$\int_V\dfrac{\vec{r}\cdot(\vec{r}-c\hat{z})}{|\vec{r}|^3|\vec{r}-c\hat{z}|^3} dV$$ Where $V$ denotes all space.
Attempt: $$2\pi \int_0^\infty \int_0^{\pi}\dfrac{\vec{r}\cdot(\vec{r}-c\hat{z})}{|\vec{r}|^3|\vec{r}-c\hat{z}|^3} \sin \theta d\theta \;|\vec{r}|^2dr $$
let $r = |\vec{r}|$, as $\hat{r}\cdot \hat{z} = \cos \theta $
$$2\pi \int_0^\infty \int_0^{\pi} \dfrac{r - c\cos \theta}{|\vec{r}-c\hat{z}|^3} \sin \theta d\theta dr$$
$$2\pi \int_0^\infty \int_0^{\pi} \dfrac{r - c\cos \theta}{(r^2+c^2-2 rc \cos \theta)^{3/2}}\sin \theta d\theta dr$$
Let $x = \cos \theta$
$$2\pi \int_0^\infty \int_{-1}^{1} \dfrac{r-cx}{(r^2+c^2 - 2 r c x)^{3/2}}dx dr$$
How do I reduce this further?
| This answers the original version of the question:
Use $ \vert \vec{r} - c \hat{z} \vert^2 = r^2 + c^2 - 2 r c \cos \theta$. Now integration with respect to $\theta$ can be carried out, change $t = \cos\theta$.
$$
\begin{eqnarray}
2 \pi \int_0^ \pi \frac{r(1-\cos \theta)}{ \left( r^2 + c^2 - 2 \, c \cdot r \cdot \cos \theta \right)^{3/2}} \sin \theta \, \mathrm{d} \theta &=&
2 \pi \int_{-1}^1 \frac{r (1 -t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t \\ &=& 2 \pi \frac{ c^2 + r^2 - \vert c^2 - r^2 \vert }{c^2 r ( c+ r)}
\end{eqnarray}
$$
Integration with respect to $r$ is now trivial:
$$
\begin{eqnarray}
\int_0^\infty 2 \pi \frac{ c^2 + r^2 - \vert c^2 - r^2 \vert }{c^2 r ( c+ r)} \mathrm{d} r &=& \int_0^c 2 \pi \frac{ 2 r^2 }{c^2 r ( c+ r)} \mathrm{d} r + \int_c^\infty 2 \pi \frac{ 2 c^2 }{c^2 r ( c+ r)} \mathrm{d} r \\
& = & 2 \pi \frac{2 - 2 \log 2}{c} + 2 \pi \frac{2 \log 2}{c} = \frac{4 \pi}{c}
\end{eqnarray}
$$
It now remains to show steps to integrate with respect to $t$:
$$
\begin{eqnarray}
\frac{r (1 -t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } &=& \frac{1}{c} \frac{ (r^2 + c^2 - 2 r c t) - (r-c)^2 }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \\ &=&
\frac{1}{c} \frac{1}{\left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{1/2}} - \frac{1}{c} \frac{(r-c)^2 }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} }
\end{eqnarray}
$$
Now integration with respect to $t$ can be done trivially using $\int ( a-b t)^{-c} \mathrm{d} t =\frac{ (a - b t)^{1-c}}{b (c-1)}$.
Modified version
$$
\begin{eqnarray}
2 \pi \int_0^ \pi \frac{(r-c \cos \theta)}{ \left( r^2 + c^2 - 2 \, c \cdot r \cdot \cos \theta \right)^{3/2}} \sin \theta \, \mathrm{d} \theta &=&
2 \pi \int_{-1}^1 \frac{(r -c t) }{ \left(r^2 + c^2 - 2 \, c \cdot r \cdot t \right)^{3/2} } \mathrm{d} t \\ &=& 2 \pi \frac{ 1 - \operatorname{sign}(c - r ) }{r^2}
\end{eqnarray}
$$
Integrating with respect to $r$:
$$
\int_0^\infty 2 \pi \frac{ 1 - \operatorname{sign}(c - r ) }{r^2} \mathrm{d} r =
\int_c^\infty 2 \pi \frac{ 2 }{r^2} \mathrm{d} r = \frac{4\pi}{c}
$$
The logic of carrying out integration with respect to $t$ is the same.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
\int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C
\end{align*}$$
So
$$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^2}\right)=\pi/\sqrt3+2\ln2.$$
However the result seems to be $ \pi/\sqrt3 $ only.
Why is there this $ 2\ln2 $?
Detail:
$$ \begin{align*}
t \arcsin\left(\frac{2t}{1+t^2}\right)&- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\
&= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt
\end{align*} $$
| The Weierstrass substitution in a slightly different form from that in which I'm accustomed to seeing it will do it.
We have
$$
\begin{align}
t & = \tan\frac x2 \\ \\
\frac{2\;dt}{1+t^2} & = dx \\ \\
\frac{2t}{1+t^2} & = \sin x \\ \\
\frac{1-t^2}{1+t^2} & = \cos x
\end{align}
$$
That's the usual Weierstrass substitution. Now, differentiate the first line above to get
$$
dt = \frac12\sec^2\frac x2\;dx
$$
so this is
$$
\frac{dx}{2\cos^2\frac x2}
$$
and by the cosine half-angle formula, this is
$$
\frac{dx}{1+\cos x}.
$$
By the third line above, we have
$$
\arcsin\left(\frac{2t}{1+t^2}\right) = \arcsin \sin x = x
$$
(if $0\le x\le \pi/2$). Therefore the desired integral becomes
$$
\int \frac{x\;dx}{1+\cos x} = \int x\;dt.
$$
Integrating by parts, we get
$$
xt - \int t\;dx = x\tan\frac x2 - \int \tan \frac x2 \; dx = x\tan\frac x2 - 2\log\cos\frac x2 + C.
$$
As $t$ goes from $0$ to $\sqrt{3}$, $x$ goes from $0$ to $\pi/3$, and there you have it.
Correction: As $t$ goes from $0$ to $\sqrt{3}$, the function $t\mapsto2t/(1+t^2)$ goes from $0$ up to $1$ and then starts going down again. It reaches its maximum at $t=1$. So $\sin x$ goes from $0$ up to $1$ and then starts going down again. Thus $x$ goes from $0$ to $2\pi/3$.
This creates problems when one says $\arcsin\sin x = x$, since that applies when $x$ is between $0$ and $\pi/2$. For $x$ between $\pi/2$ and $2\pi/3$, we'd have $\arcsin\sin x = \pi-x$ and we need to examine that interval separately.
| {
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equation of the tangent to the following curves? example:
$f(x)=x^2 (2,4)$
$= x^2$
at x= 2
$f(x) 2^2=4$
$y=mx+c$
$y=4x+c$
$(2,4)$
$4=4\cdot2+c$
$4-8=c$
$c=-4$
$y=4x-4$
but below question
I don't know how to solve?
$F(x)=x^2-2x+5$ at $x= -1$
the answer is $y=-4x+4$
thanks in advance! :)
| Added: concerning the first part of your question, you should edit it. As a minimum, something like the following.
Example: $f(x)=x^2$. At $x=2$, $f(x)=2^2=4$. The equation of the tangent at $(2,4)$ is
$$y=mx+c,$$
where $m$ is $f'(2)=4$. And so, $y=4x+c$, and $$4=4\cdot 2+c\Leftrightarrow 4-8=c\Leftrightarrow c=-4.$$ Therefore, $y=4x-4$.
Using your notation, the equation of a straight line is
$$y=mx+c\tag{1},$$
where $m$ is its slope. If this line passes through the point $P(a,b)
$, then the coordinates $a,b$ must satisfy the equation
$b=ma+c\tag{2}.$
Subtracting $(2)$ from $(1)$, we get
$$y-b=m(x-a),\tag{3}$$
which is equivalent to
$$y=mx-ma+b.\tag{3'}$$
The slope of the tangent line to the graph of the function $f(x)$ at the point $
(a,b)=(a,f(a))$ is equal to the derivative of the function at $x=a$, i.e. $m=f^{\prime }(a)$ (Wikipedia link). Therefore, the equation of the
tangent is given by (see sketch)
$$y=f^{\prime }(a)x-f^{\prime }(a)a+f(a).\tag{4}$$
For the second function $F(x)=x^{2}-2x+5$, the equation of the tangent at $(-1,F(-1))=(-1,(-1)^{2}-2(-1)+5)=(-1,8)$ is
$$y=F^{\prime }(-1)x-F^{\prime }(-1)\left( -1\right) +8.$$
The derivative $F^{\prime }(x)=2x-2$ and $F^{\prime }(-1)=2\left( -1\right)
-2=-4$. Consequently, the equation of the tangent is
$$y=-4x-(-4)\left( -1\right) +8,$$
which is equivalent to
$$y=-4x+4.\tag{5}$$
| {
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challenging alternating infinite series involving $\ln$ I ran across an infinite series that is allegedly from a Chinese math contest.
Evaluate:
$\displaystyle\sum_{n=2}^{\infty}(-1)^{n}\ln\left(1-\frac{1}{n(n-1)}\right).$
I thought perhaps this telescoped in some fashion. So, I wrote out
$\ln(1/2)-\ln(5/6)+\ln(11/12)-\ln(19/20)+\ln(29/30)-..............$
Separated the positive and negative using log properties:
$\ln(1/2)+\ln(11/12)+\ln(29/30)+......=\ln(\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot)$
$-(\ln(5/6)+\ln(19/20)+\ln(41/42)+......=-\ln(\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot) $
$\ln(\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot)-\ln(\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot)$
$=\displaystyle \ln\left(\frac{\frac{1}{2}\cdot \frac{11}{12}\cdot \frac{29}{30}\cdot\cdot\cdot}{\frac{5}{6}\cdot \frac{19}{20}\cdot \frac{41}{42}\cdot\cdot\cdot}\right)$
Maybe come up with a general term at the end of the partial sum? The terms in the numerator are $n=2,4,6,....$ and those in the denominator are $n=3,5,7,.....$
$\frac{N(N-1)-1}{N(N-1)}$. But, I always end up with a limit of 1. This then gives $\ln(1)=0$.
The series does converge. I managed to do some cancellations, but failed to wrap it up.
I thought maybe I was onto something. I suppose I am and not seeing it. What would be a good plan of attack for this one? Since it was in a contest, I assume it can be done. Any thoughts?
Thanks very much.
| Use $ \log\left(1 - \frac{1}{n(n-1)}\right) = \int_0^1 \frac{\mathrm{d} t}{n(1-n)+t}$.
Then
$$ \begin{eqnarray}
\sum_{n=2}^\infty \frac{(-1)^n}{n - n^2 + t} &=& \sum_{n=2}^\infty \frac{2 (-1)^n }{\sqrt{4 t+1}} \left(\frac{1}{2 n+\sqrt{4 t+1}-1}-\frac{1}{2 n-\sqrt{4
t+1}-1}\right) \\&=&
\frac{1}{2 \sqrt{4 t+1}} \left( \psi ^{(0)}\left(-\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)-\psi
^{(0)}\left(\frac{1}{4}-\frac{1}{4} \sqrt{4 t+1}\right) \right) \\
&+& \frac{1}{2 \sqrt{4 t+1}} \left(\psi
^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}+\frac{1}{4}\right) -\psi
^{(0)}\left(\frac{1}{4} \sqrt{4 t+1}-\frac{1}{4}\right)\right)
\end{eqnarray}
$$
The latter comes about from $\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+a}\right) = \gamma + \psi^{(0)}(a+1)$, and the summation above was split into summation over even and odd integers.
Integrating this expression out produces:
$$
\text{log$\Gamma $}\left(\frac{1}{2}-\frac{\phi}{2} \right)-\text{log$\Gamma
$}\left(-\frac{1}{2}+\frac{\phi }{2}\right)-\text{log$\Gamma $}\left(-\frac{\phi
}{2}\right)+\text{log$\Gamma $}\left(\frac{\phi }{2}\right)+\log (2)
$$
where $\phi$ is Golden ratio.
Integration is trivial as $\frac{\mathrm{d} t}{\sqrt{4 t+1}} = \mathrm{d}\left(\frac{\sqrt{4 t+1}}{2}\right)$, and $\int \psi^{(0)}(u) \mathrm{d} u = \log\Gamma(u) + C$.
Numerical check in Mathematica:
In[85]:= N[
Log[2] - LogGamma[-GoldenRatio/2] +
LogGamma[-(GoldenRatio/2) + 1/2] - LogGamma[GoldenRatio/2 - 1/2] +
LogGamma[GoldenRatio/2], 20]
Out[85]= -0.56655310975860303045 + 0.*10^-21 I
In[84]:= NSum[(-1)^n Log[1 - 1/(n (n - 1))], {n, 2, \[Infinity]},
WorkingPrecision -> 20]
Out[84]= -0.566553109758603
| {
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How to convert a fraction into ternary? In a ternary system how is $\dfrac{1}{2}=0.\bar1$ ,$\dfrac{1}{3}=0.1=0.0\bar2$??, etc. In general, how does one write the ternary expression for a given fraction?
| I assume you know how to convert integers from base 10 to base 3, so this answer will address fractions between $0$ and $1$.
To say, for example, that $\frac{5}{8} = 0.625$ means that $\frac{5}{8} = 6 \frac{1}{10} + 2 \frac{1}{100} + 5 \frac{1}{1000}$. So to convert a fraction $\frac{a}{b}$ to ternary means we want to find coefficients $c_1, c_2, c_3, \ldots$ such that $$\frac{a}{b} = c_1 \frac{1}{3} + c_2 \frac{1}{3^2} + c_3 \frac{1}{3^3} + \cdots.$$
Finding these coefficients can be automated. In fact, what follows is exactly the long division algorithm taught in elementary school for converting fractions to decimals, adapted to base 3. (You could also simply convert everything to base 3 first and then do long division, as is mentioned in the comments. For the algorithm described below, though, all the calculations can be done in base 10, which is nice since we're used to working in base 10.)
To find $c_1$, multiply the equation above by $3$ to obtain $$\frac{3a}{b} = c_1 + c_2 \frac{1}{3} + c_3 \frac{1}{3^2} + \cdots .$$ Divide $b$ into $3a$ to get $\frac{3a}{b} = \frac{qb + r}{b}$, with $q$ the quotient and $r$ the remainder. The quotient $\frac{qb}{b} = q$ will equal $c_1$, the integer part of the right-hand side, and $\frac{r}{b}$ will be the fractional part; i.e., $$\frac{r}{b} = c_2 \frac{1}{3} + c_3 \frac{1}{3^2} + \cdots.$$ Then multiply by $3$ again, and repeat the procedure until it terminates or starts repeating itself.
In tabular form and applied to $\frac{5}{8}$, this process looks like the following, where the numerator of the fraction in each line comes from $3r$ in the previous line.
$$\begin{matrix} \text{current fraction } & \text{quotient } q & \text{remainder } r & 3r \\
\frac{5}{8} & 0 & 5 & 15 \\
\frac{15}{8} &1 & 7 & 21\\
\frac{21}{8} &2 & 5 & 15\\
\frac{15}{8} &1 & 7 & 21\\
\frac{21}{8} &2 & 5 & 15\\
\vdots & \vdots & \vdots & \vdots \\
\end{matrix}$$
The base 3 representation comes from the quotients, so $\frac{5}{8}$ in ternary must be $0.\overline{12}_3$.
So why is this equivalent to long division? With long division, after finding a quotient and a remainder in a particular step, you then "carry down the $0$," which entails concatenating a $0$ on the end of the remainder. Mathematically, adding a $0$ on the end of a number in base 10 means that you are multiplying it by $10$. Since we're working in base 3 here, we want to multiply the remainder by $3$. That's the only difference between what I've said here and the long division algorithm taught in elementary school.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Why do definitions of distinct conic sections produce a single equation? I understand how to get from the definitions of a hyperbola — as the set of all points on a plane such that the absolute value of the difference between the distances to two foci at $(-c,0)$ and $(c,0)$ is constant, $2a$ — and an ellipse — as the set of all points for which the sum of these distances is constant — to the equation $$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1,\qquad\text{(A)}$$ and I also understand if $a>c$ we can define $b^2=a^2-c^2$, yielding $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad\text{(E)}$$ and if $a<c$ we can define $b^2=c^2-a^2$, yielding $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.\qquad\text{(H)}$$ I also understand (by simply "looking" at the graphs of these two cases) that the former, (E), corresponds to an ellipse, and the latter, (H), to a hyperbola.
However, it seems that the general equation, (A), obscures a distinction that was specified in the definitions: if two different definitions produce the same equation, hasn't something been lost in the process? At some point the derivations must have taken a step that wiped out the some feature of the equations — (B) and (C) below — that distinguishes the definitions. I see that one can "restore" a distinction by considering the relationship between $a$ and $b$, as above, but how that distinction maps back to the distinction between the definitions is obscure to me.
What steps in the derivations of (A) from the respective definitions, (B) and (C), is obscuring information that distinguishes those definitions? Is something going on here that can be generalized?
The derivations I'm referring to are pretty standard, they appear in many texts and also in several places on this site, but are repeated here for reference.
From Spivak's Calculus (p. 66): a point $(x,y)$ is on an ellipse if and only if
$$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}= 2a\qquad\text{(B)}$$
$$\sqrt{(x+c)^2+y^2}= 2a-\sqrt{(x-c)^2+y^2}$$
$$x^2+2cx+c^2+y^2=4a^2-4a\sqrt{(x-c)^2+y^2}+x^2-2cx+c^2+y^2$$
$$4(cx-a^2)=-4a\sqrt{(x-c)^2+y^2}$$
$$c^2x^2-2cxa^2+a^4=a^2(x^2-2cx+c^2+y^2)$$
$$(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$$
$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$$
From a related post: a point $(x,y)$ is on a hyperbola if and only if
$$\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2-y^2}=\pm 2a\qquad\text{(C)}$$
$$\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}}=\pm 2a$$
$$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2-y^2}=\pm \frac{2cx}{a}$$
$$2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right)$$
$$x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}$$
$$\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\qquad\text{(A)}$$
| Notice that E and H have two constants, A has three. So A forms a superset of E and H and you should see graph of A to study how E and H are fitted together to form A in a certain composite geometrical pattern/relationship.
Because a certain manipulation on A can bring it into form of E or H, we can see that a three parameter plot contains subset graphs of E and H. In fact A is a set of orthogonally intersecting confocal E and H.
| {
"language": "en",
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"source": "stackexchange",
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} |
Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$
| The fraction $x = \frac{a}{b}$ is the solution to $b \cdot x = a$. Similarly $y = \frac{c}{d}$ solves $d \cdot y = c$. Now, multiply the left-hand-side with $b \cdot x$ and the right hand-size with $a$ which preserves the equality since $a = b \cdot x$. We get
$$
\begin{eqnarray}
(b \cdot x) \cdot ( d \cdot y) &=& c \cdot a \\
(b \cdot d) \cdot (x \cdot y) & = & (c \cdot a)
\end{eqnarray}
$$
Thus $ ( x \cdot y) $ is the fraction $\frac{a \cdot c}{b \cdot d}$.
| {
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"source": "stackexchange",
"question_score": "4",
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} |
How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$? How do I prove that $\cos{\frac{2\pi}{7}}\notin\mathbb{Q}$?
Should I use some geometrical approach or apagoge?
| Since Chebyshev polynomials have the property $T_n\left( \cos \theta \right) = \cos(n \theta)$,
we can use this property with $n=7$ and $\theta = \frac{2 \pi}{7}$.
Let $z = \cos\left( \frac{2\pi}{7} \right)$. Then $T_7(z) = 1$
Let $p(x) = T_7(x)-1 = 64 x^7 - 112 x^5 +56 x^3 - 7x - 1 = (x-1)( 8 x^3 + 4 x^2 - 4 x -1 )^2.$
So $z$ must be a root of $q(x) = (2 x)^3 + (2 x)^2 - 2 (2 x) - 1$. Observe that $2z$ is a root of a monic polynomial, and thus is an algebraic integer. If $z$ were rational, then $2z$ must be an integer. Since $-2 \le 2 \cos \theta \le 2$, there are only 5 possibilities to consider, which are ruled out because $q(1) = 7$, $q(-1) = q(0) = q(1/2) = -1$, $q(-1/2) = 1$.
| {
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"source": "stackexchange",
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} |
Partial Fractions of form $\frac{1}{(ax+b)(cx+d)^2}$ When asked to convert something like $\frac{1}{(ax+b)(cx+d)}$ to partial fractions, I can say
$$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$
Then why can't I split $(cx+d)^2$ into $(cx+d)(cx+d)$ then do
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{cx+d}$$
The correct way is
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$$
| Because
$$\frac{B}{cx+d}+ \frac{C}{cx+d} = \frac{B+C}{cx+d}\ne\frac{1}{(cx+d)^2}$$ for any $B$ and $C$.
| {
"language": "en",
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"source": "stackexchange",
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$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
Sol:
$P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
$P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.
$P(1): 3$.
Hence it is true for $n=n_0 = 1$.
Let it be true for $n=k$.
$P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.
We have to prove
$P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.
Taking LHS:
$$\begin{align*}
1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right].
\end{align*}$$
| As I understand it you are trying to show that for $n\geq 1$:
$$
1\cdot 3+2\cdot 4+\cdots +n(n+2)=\frac{1}{6}n(n+1)(2n+7).
$$
You first showed it for $n=1$:
$$
3=1\cdot 3=\frac{1}{6}(1)(2)(9)=3
$$
Now we assume that the formula holds for some $n\geq 1$ and we have the statement for $n+1$
$$
3\cdot 1+\cdots +n(n+1)+(n+1)(n+3).
$$
By induction hypothesis the sum of the first $n$ terms is $\frac{1}{6}n(n+1)(2n+7)$. So
$$
3\cdot 1+\cdots +n(n+1)+(n+1)(n+2)=\frac{1}{6}n(n+1)(2n+7)+(n+1)(n+3).
$$
We may factor out an $n+1$ to get
$$
\frac{1}{6}n(n+1)(2n+7)+(n+1)(n+3)=(n+1)\left(\frac{1}{6}n(2n+7)+n+3\right).
$$
Then factor out a $\frac{1}{6}$ to get
$$
\frac{1}{6}(n+1)(n(2n+7)+6n+18)=\frac{1}{6}(n+1)(2n^2+7n+6n+18).
$$
Finally,
$$
\frac{1}{6}(n+1)(2n^2+7n+6n+18)=\frac{1}{6}(n+1)(n+2)(2(n+1)+7).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/72660",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ What is the method to calculate the Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ ?
| The formula for the cosine of a difference yields
$$
\begin{align}
\cos(\pi/4-y)
&= \frac{1}{\sqrt{2}}\cos(y)+\frac{1}{\sqrt{2}}\sin(y)\\
&=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}(\sin(y)+\cos(y)-1)\\
&=\frac{1}{\sqrt{2}}+x\tag{1}
\end{align}
$$
Noting that $x=\frac{1}{\sqrt{2}}(\sin(y)+\cos(y)-1)$, it is easy to show that
$$
2\sqrt{2}x+2x^2=\sin(2y)\tag{2}
$$
Now the series for $\sin^{-1}(x)$ can be gotten by integrating the series for $\dfrac{1}{\sqrt{1-x^2}}$. Using the binomial theorem, we get
$$
(1-x^2)^{-\frac{1}{2}}=\sum_{k=0}^\infty\binom{2k}{k}\frac{x^{2k}}{4^k}\tag{3}
$$
Integrating $(3)$, we get
$$
\sin^{-1}(x)=\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}\frac{x^{2k+1}}{4^k}\tag{4}
$$
Combining $(1)$, $(2)$, and $(4)$, we get that
$$
\begin{align}
\cos^{-1}\left(\frac{1}{\sqrt{2}}+x\right)
&=\frac{\pi}{4}-y\\
&=\frac{\pi}{4}-\frac{1}{2}\sin^{-1}(2\sqrt{2}x+2x^2)\\
&=\frac{\pi}{4}-\frac{1}{2}\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}\frac{(2\sqrt{2}x+2x^2)^{2k+1}}{4^k}\\
&=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(\sqrt{2}x+x^2)^{2k+1}\tag{5}
\end{align}
$$
To get $2n$ terms of the Taylor series for $\cos^{-1}\left(\frac{1}{\sqrt{2}}+x\right)$, you only need $n$ terms of $(5)$.
Afterthought:
A nicer series, that doesn't involve all the $\sqrt{2}$s would be
$$
\cos^{-1}\left(\frac{1+x}{\sqrt{2}}\right)=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(x+\tfrac{1}{2}x^2)^{2k+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/75481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Evaluating an expression using snake oil I have to evaluate this expression: $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}$,
(In the original question we had $\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{k}$)
this is what I have done:
$$\begin{aligned}
\sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n
& = \sum_k\sum_n\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n \\
& = \sum_k\binom{2k}{k}(-2)^{-k}\sum_n\binom{n}{k}x^n \\
& = \sum_k\binom{2k}{k}(-2)^{-k}\frac{x^k}{(1-x)^{k+1}} \\
& = \frac{1}{1-x}\sum_k\binom{2k}{k}(\frac{x}{2x-2})^k
\end{aligned}$$
now we know that $\sum_k\binom{2k}{k}x^k=\frac{1}{\sqrt{1-4x}}$ and so we get
$$ \sum_n\sum_k\binom{n}{k}\binom{2k}{k}(-2)^{-k}x^n
= \frac{1}{\sqrt{1-x^2}}
$$
So the sum that I'm looking for is equal to $\sum_{k=0}^n\binom{-1/2}{k}(-1)^k\binom{-1/2}{n-k}$
there is a way to express this expression without using sums or menus signs?
| Added: This post answers the original question, which is to compute $\sum_{k=0}^\infty \binom{n}{k} \binom{2k}{k} (-2)^k$.
Assuming the typo is corrected you correctly arrived at the result:
$$
\sum_{n=0}^\infty c_n x^n = \frac{1}{\sqrt{(1-x)(1+7 x)}}
$$
Where $c_n = \sum_{k=0}^\infty \binom{n}{k} \binom{2k}{k} (-2)^k$. Using
$$
\frac{1}{\sqrt{1-x}} = \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+1/2)}{n!} x^n
\qquad
\frac{1}{\sqrt{1+7x}} = \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+1/2)}{n!} (-7 x)^n
$$
Thus
$$
\frac{1}{\sqrt{(1-x)(1+7x)}} = \sum_{n=0}^\infty x^n \sum_{m=0}^n \frac{1}{\pi} (-7)^m \frac{\Gamma(m+1/2) \Gamma(n-m+1/2)}{m! (n-m)!}
$$
The sum over $m$ is the closed form which can, alternatively, be represented as a hypergeometric polynomial, giving
$$ \begin{eqnarray}
c_n &=& \sum_{m=0}^n \frac{1}{\pi} (-7)^m \frac{\Gamma(m+1/2) \Gamma(n-m+1/2)}{m! (n-m)!} \\ &=&
\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right) n!} {}_2 F_1\left( \frac{1}{2}, -n; \frac{1}{2}-n ; -7 \right) \\ &=&
{}_2 F_1 \left( \frac{1}{2}, -n ; 1 ; 8 \right)
\end{eqnarray}
$$
Added: To address OP's skepticism that the above is the correct solution to the question posed, here is some numerical verification using Mathematica:
In[73]:= Table[{Sum[
Binomial[n, k] Binomial[2 k, k] (-2)^k, {k, 0, n}],
Hypergeometric2F1[1/2, -n, 1, 8], (
Gamma[1/2 (1 + 2 n)] Hypergeometric2F1[1/2, -n, 1/2 - n, -7])/(
Sqrt[\[Pi]] Gamma[1 + n])}, {n, 1, 6}]
Out[73]= {{-3, -3, -3}, {17, 17, 17}, {-99, -99, -99}, {609, 609,
609}, {-3843, -3843, -3843}, {24689, 24689, 24689}}
| {
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"timestamp": "2023-03-29T00:00:00",
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The limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0? Is there any elementary proof that the limit of infinite product $\prod_{k=1}^{\infty}(1-\frac{1}{2^k})$ is not 0?
| This is similar to the beginning of Alex's answer, however it proceeds differently.
We express the log of the product as a sum of logarithms
$$\small L=-\log(\prod_{k=1}^\infty (1-{1 \over 2^k})) = -\sum_{k=1}^\infty \log(1-{1 \over 2^k})$$
Next we expand each of the logs according to its power series and write this as a double sum; then we change order of summation and sum up columnwise; this is allowed because both directions provide convergent sums. So
$$\small \begin{array} {lllllll}
-\log(1-1/2)=&+1/2 & +1/2 \cdot 1/2^2 & +1/3 \cdot 1/2^3 & +1/4 \cdot 1/2^4 & \ldots \\
-\log(1-1/4)=&+1/4 & +1/2 \cdot 1/4^2 & +1/3 \cdot 1/4^3 & +1/4 \cdot 1/4^4 & \ldots \\
-\log(1-1/8)=&+1/8 & +1/2 \cdot 1/8^2 & +1/3 \cdot 1/8^3 & +1/4 \cdot 1/8^4 & \ldots \\
-\log(1-1/16)=&+1/16 & +1/2 \cdot 1/16^2 & +1/3 \cdot 1/16^3 & +1/4 \cdot 1/16^4 & \ldots \\
\cdots & & \cdots &\cdots &\cdots & \\
\hline \\
L = & 1 & + 1/2 \cdot 1/3 & + 1/3 \cdot 1/7 & + 1/4 \cdot 1/15 & \ldots \\
\hline \\
\zeta(2)=&1&+ 1/2\cdot 1/2& + 1/3 \cdot 1/3 & + 1/4 \cdot 1/4 & \dots \\
\end{array}
$$
We compare L with $\small \zeta(2) $ and observe, that the absolute value each term in L is smaller than the corresponding term in $\small \zeta(2)$ thus L is a finite/converging sum $\small 0 \lt L \lt \zeta(2) \lt \infty $ and thus $\small \exp(-L) \ne 0 $ . QED .
| {
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"source": "stackexchange",
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Finding pairs of integers such that $x^2+3y$ and $y^2+3x$ are both perfect squares Can we find pairs $(x,y)$ of positive integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares? Thanks a lot in advance. My progress is minimal.
| $x^2+3y=a^2$
$y^2+3x=b^2$
$a,b \in \mathbb{N}$
assume wlog $x\ge y$
$a^2> x^2$
$a > x$ but since $x\ge y$ we have $x+2>a$
so $a=x+1 $
$3y=2x+1$
$x=3k+1 ,y=2k+1$
$4k^2+4k+1+9k+3=b^2$
solving for k we get
$16b^2+105=t^2$ is a perfect square
$b=1,2,4,13$
so $b=2,4,13$ and $x=y=1$ or $x=16 $, $y=11 $ and if $y\ge x$ we can get $x=11 $, $y=16$
Done!!!!
| {
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Is this proof about the form $2^n \pm a$ correct? I want to prove following statement :
For prime numbers $p$ greater than $3$, it is true that:
$a)$ if $p=2^n-a$ and $a=6k+1$, then $n$ is an odd number.
$b)$ if $p=2^n+a$ and $a=6k-1$, then $n$ is an odd number.
$c)$ if $p=2^n-a$ and $a=6k-1$, then $n$ is an even number.
$d)$ if $p=2^n+a$ and $a=6k+1$, then $n$ is an even number.
where $n \in \mathbf{Z}^+, k\in \mathbf{Z}^\ast$.
Proof :
$a)$ Lemma $1$ : $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+....+b^{n-1})$
$ p=2^n-(6k+1)=2(2^{n-1}+1)-6k-3=2(2^{n-1}+1)-3(2k+1)$
Let's suppose that $n$ is even then $n-1$ is odd and by the Lemma $1$:
$ (2+1) \mid (2^{n-1}+1) $ and since $(2+1) \mid 3(2k+1) $ we may conclude that $p$ is composite number.
So, we have contradiction, therefore $n$ must be odd number.
$b)$ Similarly as case $a)$
$c)$ $p=2^n-(6k-1)=2(2^{n-1}-1)-6k+3=2(2^{n-1}-1)-3(2k-1)$
let's suppose that $n$ is odd then $n-1$ is even ,therefore
$2^{n-1}-1=2^{2t}-1=(2^t-1)(2^t+1)$ , so :
$3 \mid (2^{n-1}-1)$ and since $ 3 \mid 3(2k-1)$ we may conclude that $p$ is composite number.
So,we have contradiction,therefore $n$ must be even number.
$d)$ Similarly as case $c)$
Is this an acceptable proof?
| Yes, the proofs are fine. You did not do part (d), but it requires nothing new, so it is quite reasonable to omit it. In the formal statement of the lemma, the conditions on the parameters should have been made explicit.
You chose to avoid congruence notation. I would probably instead observe first that $2^n \equiv 1 \pmod{3}$ if $n$ is even, and $2^{n}\equiv -1\pmod{3}$ if $n$ is odd. This can be proved in the style that you used, or more neatly by using properties of congruences: since $2\equiv -1\pmod{3}$, it follows that $2^n \equiv (-1)^n \pmod{3}$.
Then for example for (b), if $n$ is odd, and $a\equiv -1\pmod{6}$, then $2^n+a\equiv 1+(-1)=0\pmod{3}$. If we wish, we can even gather all four results together into one.
For divisibility of $2^n \pm 1$ by $3$, congruence language has no advantage over the techniques that you used. For more complicated situations, congruence language becomes more and more essential.
| {
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Proving that a polynomial is positive A Finnish mathematics competition asked to prove that for all $x$ we have $x^8-x^7+2x^6-2x^5+3x^4-3x^3+4x^2-4x+\frac{5}{2}\geq 0$ for all real $x$. I heard that it follows from Hilbert's problem that one can prove this by writing the polynomial as sum of squares. How can I find such a representation? I managed to prove the inequality by considering cases $x\leq 0$, $0<x<1$ and $x\geq 1$ separately but I was unable to find a solution based on sum of squares.
| I'm not sure if this is what you're looking for, but it works and is quite elementary.
We can take the highest three powers and do this:
$$
x^8-x^7+x^6 = x^6(x^2-x+1) = x^6\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]
= x^6 \left(x-\dfrac{1}{2}\right)^2 + \dfrac{3}{4} x^6,
$$
which gives us the expression
$$
x^6 \left(x-\dfrac{1}{2}\right)^2 + \dfrac{3}{4} x^6 + −2x^5+3x^4−3x^3+4x^2−4x + \dfrac{5}{2}.
$$
Doing this three more times results in
$$
x^6 \left(x-\dfrac{1}{2}\right)^2 + \dfrac{3}{4}x^4\left(x-\dfrac{4}{3}\right)^2 + \dfrac{5}{3}x^2\left(x-\dfrac{9}{10}\right)^2 + \dfrac{53}{20}\left(x-\dfrac{40}{53}\right)^2 + \dfrac{105}{106} \geq 0.
$$
| {
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variance of a sum We draws 10 from a finite population of 30, without replacement! 15 members out of 30 have the number 1, 10 members have the number 2 and 5 members have the number 3.
The Expected value is
$$\mathbb E(X)= 1 \cdot \frac12+2 \cdot \frac13+3 \cdot \frac 16 = \frac 53 $$
Variance
$$V(X)= \left( 1 -\frac53 \right)^2 \cdot \frac12+ \left( 2-\frac53 \right)^2 \cdot \frac13+ \left( 3-\frac53 \right)^2 \cdot \frac16 = \frac59 $$
I need the Variance of $$X_{1}+...+X_{10}$$ and $$Var( \frac1{10}( X_{1}+...+X_{10}))$$
I found this formula $$V(X_{1}+...+X_{10})= \sum\limits_{i=1}^{10} \operatorname{Var}(X_i)+\sum\limits_{i\neq j}^{} \operatorname{Cov}(X_i,X_j)$$
But how i can solve this?
With Wolframalpha?
thx
| It is simpler to work it out directly. Since the total of the sample $Z = X_1+\ldots+X_{10}$ does depend on the order of sampling of each individual components, the distribution of $Z$ would be the same if, instead of sampling without replacement, we sampled 10 elements at once.
Suppose a sample contains $k_1$ balls with number 1, $k_2$ balls with number 2, and $k_3$ balls with number 3. The probability of obtaining such a configuration is:
$$
p(k_1, k_2, k_3) = \mathbb{P}(K_1=k1,K_2=k_2,K_3=k_3) = \frac{ \binom{15}{k_1} \binom{10}{k_2} \binom{5}{k_3}}{ \binom{30}{10} } \mathsf{1}_{k_1+k_2+k_3=10}
$$
The triple $(K_1,K_2,K_3)$, thus follows the multi-variate hypergeometric distribution.
The mean is
$$
\mathbb{E}(Z) = \mathbb{E}(K_1 + 2K_2+3K_3) = \mathbb{E}(K_1) + 2 \mathbb{E}(K_2) + 3 \mathbb{E}(K_3) = 1 \cdot \frac{10}{30} 15 + 2 \cdot \frac{10}{30} 10 + 3 \cdot \frac{10}{30} 5 = \frac{50}{3}
$$
The variance then
$$ \begin{eqnarray}
\mathbb{Var}(Z) &=& \mathbb{Var}(K_1) + 4 \mathbb{Var}(K_2) + 9 \mathbb{Var}(K_3) + 4 \mathbb{Cov}(K_1,K_2) + 6 \mathbb{Cov}(K_1,K_3) + 12 \mathbb{Cov}(K_2,K_3) \\
&=& \frac{50}{29} + 4 \cdot \frac{400}{261} + 9 \cdot \frac{250}{261} + 4 \cdot \left( -\frac{100}{87}\right) + 6 \cdot \left( - \frac{50}{87} \right) + 12 \cdot \left( -\frac{100}{261} \right) = \frac{1000}{261}
\end{eqnarray}
$$
| {
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Proving $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$ Could anyone help me prove that $2,3,1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducible in $\mathbb{Z}[\sqrt{-5}]$?
As $6=2*3=(1+\sqrt{-5})(1-\sqrt{-5})$ so $\mathbb{Z}[\sqrt{-5}]$ is not a UFD. Therefore is not a PID or euclidean domain
| Since $\sqrt{-5} \cdot \sqrt{-5}=-5 \in \mathbb{Z}$ holds, we have $\mathbb{Z}[\sqrt{-5}]=\{a+b \sqrt{-5} \mid a, b \in \mathbb{Z}\}.$
We use
$$
(a+b \sqrt{-5})(c+d \sqrt{-5})=a c-5 b d+\left(a d+bc\right) \sqrt{5}
$$
and $|a+b \sqrt{-5}|^2=a^2+5 b^2$.
$1+\sqrt{-5}$ is not a unit: it is valid that
$$
(1+\sqrt{-5})(a+b \sqrt{-5})=a-5 b+(a+b) \sqrt{5} \neq 1,
$$
since otherwise $a=-b$ and $-6b=0$.
If $(1+\sqrt{-5})=(a+b \sqrt{-5})(c+d \sqrt{-5})$, then
$$
6=1+5=|1+\sqrt{-5}|^2=\left(a^2+5 b^2\right)\left(c^2+5 d^2\right)
$$
holds.
There is no $a, b \in \mathbb{Z}$ with $a^2+5 b^2=2$ and $a^2+5 b^2=3$, since $a^2 \in\overline{0}, \overline{1}, \overline{4}\rangle \pmod{5}$
and if $a^2+5 b^2=1$, then $a+b \sqrt{-5}$ would be invertible with
$$
(a-b \sqrt{-5})^{-1}=(a-b \sqrt{-5}),
$$
since
$$
(a+b \sqrt{-5})(a-b \sqrt{-5})=a^2+5 b^2. \quad\square
$$
| {
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How to find $n$'th term of the sequence $3, 7, 12, 18, 25, \ldots$? $$3, 7, 12, 18, 25, \ldots$$
This sequence appears in my son's math homework. The question is to find the $n$'th term. What is the formula and how do you derive it?
| for the series: 3, 7, 12, 18, 25, …
$3=0+3\\7=3+4\\12=7+5\\18=12+6\\25=18+7$
We can find that: each term equals previous term plus n+2
$t_0=0$
$t_1=t_0+(1+2)=(1+2)$
$t_2=t_1+(2+2)=(1+2)+(2+2)$
$t_3=t_2+(3+2)=(1+2)+(2+2)+(3+2)$
...
$t_n=t_{n-1}+(n+2)$
$=(\color{red}1+\color{green}2)+(\color{red}2+\color{green}2)+(\color{red}3+\color{green}2)+...+(\color{red}n+\color{green}2)$
$=\color{red}{(1+...+n)}+\color{green}{(2n)}$
$=\frac{n(n+1)}{2} + 2n$
$=\frac{(n^2+5n)}{2}$
| {
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How to compute the characteristic polynomial of $A$ The matrix associated with $f$ is:
$$ \left(\begin{array}{rrr}
3 & -1 & -1 \\
-1 & 3 & -1 \\
-1 & -1 & 3
\end{array}\right) .
$$
First, I am going to find the characteristic equation of $\det(A- \lambda I)$. Please correct me if I'm wrong.
$$= (3-\lambda)(3-\lambda)(3-\lambda)-1-1-(3-\lambda)-(3-\lambda)-(3-\lambda)
=-\lambda^3+9\lambda^2-24\lambda+16 .$$
How to factor this? I know the $\lambda$'s should be $1$, $4$ and $4$. But how am I supposed to find these values?
| You may also use Cardano's formula to find the roots $x_1,x_2,x_3$ of a cubic polynomial $ax^3+bx^2+cx+d$.
$$
\begin{align}
p &= 2 b^3-9 a b c+27 a^2 d,\\
q &= \sqrt{p^2-4 \left(b^2-3 a c\right)^3},\\
x_1 &= -\frac{b}{3 a}
-\frac{1}{3 a} \sqrt[3]{\frac{p+q}{2}}
-\frac{1}{3 a} \sqrt[3]{\frac{p-q}{2}},\\
x_2 &= -\frac{b}{3 a}
+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\frac{p+q}{2}}
+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\frac{p-q}{2}},\\
x_3 &= -\frac{b}{3 a}
+\frac{1-i \sqrt{3}}{6 a} \sqrt[3]{\frac{p+q}{2}}
+\frac{1+i \sqrt{3}}{6 a} \sqrt[3]{\frac{p-q}{2}}.
\end{align}
$$
(There are some subtleties in handling $q$ when the operand inside the square root is negative. See the Wikipedia entry for details.) In your case we have $a=-1, b=9, c=-24, d=16$. Hence $p=-54, q=0$ and $x_1=1, x_2=x_3=4$.
| {
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$\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt $$
\begin{align}
\int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx
\end{align}
$$
Then, setting
$$\begin{array}{l l}
u=\cos^{-1} x & v=x \\
u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\
\end{array}$$
Then by the IBP technique, we have:
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &=\cos^{-1} (x) \cdot x - \int x \cdot -\frac{1}{\sqrt{1-x^2}} \; dx\\
&= x \cos^{-1} (x) - \int -\frac{x}{\sqrt{1-x^2}} \; dx\\
\end{array}$$
Now at this point suppose I have overlooked the possibility of using integration by substitution (setting $u=1-x^2$) to simplify the second integral. Instead, I attempt to reapply IBP to the second integral $\int -\frac{x}{\sqrt{1-x^2}} \; dx$.
I let
$$\begin{array}{l l}
u= -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-\frac{1}{2}} \qquad & v= \frac{x^2}{2} \\
u' = - \left( -\frac{1}{2} \right) (1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}} \qquad & v'=x\\
\end{array}$$
Then by IBP again,
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &= x \cos^{-1} (x) - \left( -\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx \right) \\
\end{array}$$
At this stage, I can see no way to proceed. Can anyone see a reasonable way to salvage this solution, continuing along this line of reasoning? Or was approaching the second integral by IBP doomed to fail?
| Put $x = \cos{\nu}$. Then you have $dx = -\sin\nu \ d\nu$. So your integral is now $$I =\int \nu \cdot \sin\nu \ d\nu$$
This is easy to evaluate by parts. Take $u = \nu$ and $dv = \sin\nu \ d\nu$. Then you have
\begin{align*}
I &= uv - \int v \ du \\ &= \Bigl[ -\nu \cdot \cos\nu \Bigr] + \int \cos\nu \ d\nu
\end{align*}
| {
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Is this a proper use of induction? ($(n^2+5)n$ is divisible by 6) Just want to get input on my use of induction in this problem:
Question. Use mathematical induction to prove that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
Proof by mathematical induction.
(1) show base case ($n=1$) is true:
$$
((1)^2 + 5) (1) = 6
$$
$6$ is indeed divisible by $6$, so base case ($n=1$) is true
(2a) Assume case $n$ is true: $(n^2+5)n$ is divisible by $6$.
(2b) Show that case $n$ $\implies$ case $(n+1)$
$$
\begin{align*}
((n+1)^2+5)(n+1)
&\rightarrow
((n^2+2n+1)+5)(n+1)
\\ &\rightarrow
[(n^2+5)+(2n+1)](n+1)
\\ &\rightarrow
(n^2+5)n + (n^2+5)+(2n+1)n+ (2n+1)
\\ &\rightarrow
(n^2+5)n + [(n^2+5)+(2n^2+n)+ (2n+1)]
\\ &\rightarrow
(n^2+5)n + [(3n^2+3n)+6]
\end{align*}
$$
Now we can see case $(n+1)$ $= (n^2+5)n + (3n^2+3n)+6$.
We know $6$ is divisible by $6$ and we are assuming $(n^2+5)n$ is divisible by $6$ already, so all we need to do is show $(3n^2+3n)$ is divisible by $6$:
Letting $n=1$ for $(3n^2+3n)$ gives: $(3(1)^2+3(1)) = 6$
Thus, it has been demonstrated that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
I'm not sure if letting $n=1$ for that last expression is enough to prove it is divisible by $6$
| The statement that
$3n^2+3n$ is divisible by $6$ for all $n$.
and the statement that
$3(1)^2+3(1)=6$ is divisible by $6$.
are clearly not the same :) You can prove the former statement by showing that $n^2+n=n(n+1)$ is always even (use that either $n$ is even, or $n+1$ is).
By the way, use equals signs ($=$) instead of arrows ($\rightarrow$).
Finally: for any number $k$, "case $k$" refers to the statement
$(k^2+5)k$ is divisible by $6$.
As a statement, it is true, or possibly false. Your goal is to prove that "case $n$" is true for every number $n$. Therefore, you should not think things like
"case $(n+1)$ = $(n^2+5)n+3n^2+3n+6$"
which doesn't really make sense.
| {
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Right triangle where the perimeter = area*k I was doodling on some piece of paper a problem that sprung into my mind. After a few minutes of resultless tries, I advanced to try to solve the problem using computer based means.
The problem stated is
Does a right angle triangle with integer sides such that
$$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ exist?
(Here P is the perimeter of the triangle. Sum of the sides. And A is the area of the triangle. a*b/2 in a triangle with a b c and c is the hypotenuse)
Obviously for the simple cases such triangles exists. For an example
when
$ A=2P \qquad $ the triangle $12,16,20$ works
$ A=P \qquad \; \; $ the triangle $6,8,10$ works
$2A=P \qquad $ the triangle $3,4,5$ works
I tried solving this by hand, first for the special case where $2P=A$. This ended up giving me
$$ \frac{a \cdot b}{2} = A $$
and
$$ P = \frac{A}{2} = \frac{a \cdot b}{4} \qquad \text{also} \qquad P = a + b + c $$
So
$$ \frac{ab}{4} = a + b + c $$
By knowing that this is a right angle this leads to the equation (Using the Pythagorean theorem)
$$ a^2 + b^2 = c^2 $$
Now we have two equations and three unknowns, which also needs to be integers! Sadly I was not able to continue from here. I have only learned how to solve linear Diophantine equations. Not a system of nonlinear Diophantine equations.
Just to restate my question below =)
Is there a right angle triangle with integer sides such that
$$P = A\cdot k \quad \quad \text{or} \quad \quad k\cdot P = A$$ where $(P , \ A , \ k )\in \mathbb{N}$ ?
Regards, Werner
| If you choose positive integers $a$ and $b$, you can make a right triangle with sides
$a^2-b^2\text{, } 2ab \text{ and } a^2+b^2,$
and all right triangles can be found this way by varying the values of $a$ and $b$.
If we have such a triangle, its area is
$$\frac{1}{2}(a^2-b^2)\times 2ab = ab(a^2-b^2)$$
and its perimeter is
$$a^2-b^2 + 2ab + a^2+b^2 = 2a(a+b).$$
For the first part of your question we want to set
$$ab(a^2-b^2)\times k = 2a(a+b).$$
We can write this as
$$ab(a-b)(a+b) = 2a(a+b)$$
and then cancel the $a$ and the $(a+b)$ on each side (neither are zero so it's ok) and get
$$b(a-b)k = 2\quad\text{, and finally}$$
$$k = \frac{2}{b(a-b)}.$$
We want $k$ to be an integer, so it can only be 1 or 2.
If $k=1$ , $b(a-b)$ must equal $2$ and this can happen in two ways: when $a=3\text{ and }b=2$, which produces the 5-12-13 triangle, and when $a=3$ and $b=1$, which produces the 6-8-10 triangle.
If $k=2$ , $b(a-b)$ must equal $1$ and the only way this can happen is when $a=2\text{ and }b=1$. These values of $a$ and $b$ produce the 3-4-5 triangle.
For the second part of your question we want to set
$$ab(a^2-b^2) = 2a(a+b)\times k.$$
We can reduce this as before and arrive at
$$k = \frac{b(a-b)}{2}.$$
There are lots of different values of $a$ and $b$ that make $k$ an integer.
| {
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Trigonometric integral I would like to know if there is some way to calculate exactly
$$\int_{0}^{\pi/2} \frac{\cos^3 x}{\sqrt {\sin x}(1+\sin^3 x)}\,dx$$
The numerical value is 1,52069
| From Dilip Sarwate's suggestion on:
$$2\int_0^1\frac{1-y^4}{1+y^6}\mathrm dy = 2\int_0^1\frac{(1-y^2)(1+y^2)}{(1+y^2)(y^4-y^2+1)}\mathrm dy = 2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy \; .$$
The denominator can be split further into
$$2\int_0^1\frac{(1-y^2)}{(y^4-y^2+1)}\mathrm dy = 2\int_0^1\frac{(1-y^2)}{(y^2+\sqrt{3}y+1)(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$
Using partial fractions this becomes
$$2\int_0^1\frac{(1-y^2)}{(y^2-\sqrt{3}y+1)(y^2+\sqrt{3}y+1)}\mathrm dy = 2\int_0^1\frac{\frac{y}{\sqrt{3}}+\frac{1}{2}}{(y^2+\sqrt{3}y+1)}-\frac{\frac{y}{\sqrt{3}}-\frac{1}{2}}{(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$
The last part can be rewritten as
$$\frac{1}{\sqrt{3}}\int_0^1\frac{2y+\sqrt{3}}{(y^2+\sqrt{3}y+1)}-\frac{2y-\sqrt{3}}{(y^2-\sqrt{3}y+1)}\mathrm dy \; .$$
And can easily be solved by substitution to give as Zev Chonoles already mentioned in the commentary: $\frac{\ln(7+4\sqrt{3})}{\sqrt{3}}$ .
| {
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Converting to partial fractions I would like to see a step by step conversion of this fraction to two partial fractions.
$\frac{x^2}{1-x^2}$
| As André said, because the degree of the numerator is greater than or equal to that of the denominator, the first step is to divide out to get polynomial quotient and a rational function remainder whose numerator has smaller degree than its denominator:
$$\frac{x^2}{1-x^2}=-1+\frac1{1-x^2}\;.$$
Now expand the second term. The denominator factors as $(1-x)(1+x)$, with no repeated factors, so you expect to be able to decompose the fraction as
$$\frac1{1-x^2}=\frac{A}{1-x}+\frac{B}{1+x}\tag{1}$$
for some constants $A$ and $B$. Now recombine the righthand side of $(1)$:
$$\begin{align*}
\frac1{1-x^2}&=\frac{A}{1-x}+\frac{B}{1+x}\\
&=\frac{A(1+x)+B(1-x)}{1-x^2}\\
&=\frac{(A-B)x+(A+B)}{1-x^2}\;.
\end{align*}\tag{2}$$
The left and right sides of $(2)$ are fractions with the same denominator, and they’re supposed to be the same function, so we must choose $A$ and $B$ in such a way that they have the same numerator:
$$1=(A-B)x+(A+B)\;.\tag{3}$$
Two polynomials are identically equal (i.e., equal for all values of the variable) if and only if they have identical coefficients, so the next step is to equate coefficients. On the lefthand side of $(3)$ the coefficient of $x$ is $0$, and the constant term is $1$; on the righthand side the coefficient of $x$ is $A-B$, and the constant term is $A+B$. Thus, we want $A$ and $B$ to satisfy the system
$$\left\{\begin{align*}
A-B&=0\\
A+B&=1
\end{align*}\right.$$
This system is easily solved to yield $A=B=1/2$, which we can insert into $(1)$ to get $$\frac1{1-x^2}=\frac12\left(\frac1{1-x}+\frac1{1+x}\right)$$ and, finally, $$\frac1{1-x^2}=-1+\frac12\left(\frac1{1-x}+\frac1{1+x}\right)\;.$$
Several shortcuts are possible, but this is the procedure that they’re cutting short.
| {
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The inequality $b^n - a^n < (b - a)nb^{n-1}$ I'm trying to figure out why $b^n - a^n < (b - a)nb^{n-1}$.
Using just algebra, we can calculate
$ (b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) $
$ = (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) - (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1}) $
$ = b^n - a^n, $
but why is it necessarily true that $(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}$?
Note: I am interested in an answer to that last question, rather than in another way to prove the general inequality in the title...
| This inequality can fail if $a=b$ or $a$ and $b$ have differing signs: e.g. $b=1$, $a=-3$, and $n=3$. So let's assume that $a\not=b$ and $a,b\ge0$. Division yields
$$
\frac{b^n-a^n}{b-a}=\sum_{k=1}^nb^{n-k}a^{k-1}\tag{1}
$$
If $a<b$ then obviously, $\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}< nb^{n-1}$, thus $\dfrac{b^n-a^n}{b-a}< nb^{n-1}$ and because $b-a>0$,
$$
b^n-a^n<(b-a)nb^{n-1}\tag{2}
$$
If $a>b$ then obviously, $\sum\limits_{k=1}^nb^{n-k}a^{k-1}=b^{n-1}\sum\limits_{k=1}^n\left(\frac{a}{b}\right)^{k-1}> nb^{n-1}$, thus $\dfrac{b^n-a^n}{b-a}> nb^{n-1}$ and because $b-a<0$,
$$
b^n-a^n<(b-a)nb^{n-1}\tag{3}
$$
If both $a$ and $b$ are negative, the inequality holds for even $n$, and is reversed for odd $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric equality $\cos x + \cos 3x - 1 - \cos 2x = 0$ In my text book I have this equation:
\begin{equation}
\cos x + \cos 3x - 1 - \cos 2x = 0
\end{equation}
I tried to solve it for $x$, but I didn't succeed.
This is what I tried:
\begin{align}
\cos x + \cos 3x - 1 - \cos 2x &= 0 \\
2\cos 2x \cdot \cos x - 1 - \cos 2x &= 0 \\
\cos 2x \cdot (2\cos x - 1) &= 1
\end{align}
So I clearly didn't choose the right path, since this will only be useful if I become something like $a \cdot b = 0$.
All tips will be greatly appreciated.
Solution (Addition to the accepted answer):
The problem was in writing $\cos 3x$ in therms of $cos x$. Anon pointed out that it was equal to $4\cos^3 x - 3\cos x$ but I had to work may way thru it to actually prove that. So I write it down here, maybe it's of use to anybody else.
\begin{align}
\cos 3x &= \cos(2x + x)\\
&= \cos(2x)\cdot \cos x - \sin(2x)\sin x\\
&= (\cos^2 x - \sin^2 x) \cdot \cos x - 2\sin^2 x \cdot \cos^2 x\\
&= \cos^3 x - \sin^2x\cdot \cos x - 2\sin^2x\cdot \cos^2 x\\
&= \cos^3 x - (1 - \cos^2 x)\cos x - 2(1-\cos^2 x)\cos x\\
&= cos^3 x - \cos x + \cos^3 x - 2\cos x + 2\cos^3 x\\
&= 4\cos^3 x - 3\cos x
\end{align}
EDIT: Apparently you can write all of the formulas $\cos(n\cdot x)$ with $n \in \{1, 2, 3, …\}$ in terms of $\cos x$. Why didn't my teacher tell me that! I don't have time to proof it myself now, but I'll definitely adapt my answer tomorrow (or any time soon)!
| You can also use the identity:
$$\cos(A)+\cos(B) = 2 \cos(\frac{A+B}{2}) \cos (\frac{A-B}{2})$$
Thus
\begin{equation}
\begin{split}
\cos x + \cos 3x - 1 - \cos 2x &= \cos x + \cos 3x - \cos(0) - \cos 2x \
&= 2\cos(\frac{x+3x}{2})\cos(\frac{x-3x}{2})- 2\cos(\frac{0+2x}{2})\cos(\frac{0-2x}{2})
\end{split}
\end{equation}
Then $2 \cos(x)$ becomes a common factor, and you can use the corresponding formula for difference this time to factor the rest....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
Simplifying $1 - x + x^2 - x^3 + ... + x^{98} - x^{99}$ to an equivalent expression. I am doing an exercise to see the error when solving this polynomial for $x = 1.00001$ using nested multiplication.
I believe the correct way to achieve this simplification (based on a lecture) is to multiply the polynomial by $\frac{1+x}{1+x}$; however, my algebra skills are not really up to par so I am failing to see the purpose of this. I assume it is to cancel out most of the terms in the polynomial - but which ones are being canceled?
| Here's the beginning:
$$\begin{align}
&\ \ \ 1-x+x^2-x^3+\cdots+ x^{98}-x^{99}\\
&= \left(1-x+x^2-x^3+\cdots+ x^{98}-x^{99}\right)\frac{1+x}{1+x}\\
&= \frac{1(1+x)-x(1+x)+x^2(1+x)+\cdots +x^{98}(1+x)-x^{99}(1+x)}{1+x}\\
&=\frac{1+x-x-x^2+x^2+x^3-x^3\cdots +x^{98}+ x^{99}-x^{99} -x^{100}}{1+x}\\
\end{align}$$
Can you see where this goes?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/104889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Diophantine equation $x^3+z x^2-z y^2=0$ I'm not familiar with diophantine equations. At most my approaches doesn't give results. I need to solve the following equation
$$
x^3+zx^2-zy^2=0
$$
where $x,y,z\in\mathbb{Z}$
| We have $x^2 (x+z) = z y^2.$ If $x=0$ things simplify, so let us assume that $x,y,z \neq 0$ until such time as we can work those in.
$$ x+z = z \; \frac{y^2}{x^2}.$$
Define a rational number $r = \pm y / x,$ so
$$ x + z = r^2 z. $$ Then
$$ x = (r^2 -1)z, $$ and
$$ y = \pm x = \pm r (r^2 -1)z. $$
ORIGINAL: All rational solutions are given by a rational parameter $r$ and any value of $z,$ with
$$ x = (r^2 - 1) z, \; \; y = \pm r (r^2 -1) z = \pm r x. $$
Then
$$ x + z = r^2 z, $$
$$ x^2 = (r^2 -1)^2 z^2, $$
$$ y^2 = r^2 (r^2 -1)^2 z^2, $$
$$ x^2 (x + z) = (r^2-1)^2 z^2 \cdot r^2 z = r^2 (r^2-1)^2 z^3, $$
$$ y^2 z = r^2 (r^2 -1)^2 z^2 \cdot z = r^2 (r^2 -1)^2 z^3. $$
So
$$ 0 = x^2 (x + z) - z y^2 = x^3 + z x^2 - z y^2. $$
Next, how do we get this to come out as integers, without searching for a common denominator? taking the parameter $r = p/q$ with $\gcd(p,q) = 1,$ we get some integer $s$ with
$$ x = (p^2 - q^2 )q s, \; y = \pm (p^2 - q^2 )p s, \; z = q^3 s. $$
Which says that we can multiply any time we like by some $s,$ and for the moment we might as well take it to be 1, resulting in a primitive two-parameter solution
$$ x = (p^2 - q^2 )q , \; y = \pm (p^2 - q^2 )p , \; z = q^3. $$
Looking again, we can accomplish $\pm y$ by simply negating $p,$ so we get a prettier answer with
$$ x = (p^2 - q^2 ) \;q , \; \; \; y = (p^2 - q^2 ) \; p , \; \; \; z = q^3. $$
This takes care of nonzero $z.$ If $z=0,$ then $x=0$ but $y$ can be anything. However, having erased the factor $s,$ we are looking at only primitive solutions, so that this would force $y = \pm 1$ if $y$ is nonzero. As this situation is covered by $p = -1,0,1, \; \; q = 0,$ I would say we are in good shape.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/105860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
System $a+b+c=4$, $a^2+b^2+c^2=8$. find all possible values for $c$. $$a+b+c=4$$$$a^2+b^2+c^2=8$$
I'm not sure if my solution is good, since I don't have answers for this problem. Any directions, comments and/or corrections would be appreciated.
It's obvious that $\{a,b,c\}\in[-\sqrt8,\sqrt8]$. Since two irrational numbers always give irrational number when added, if we assume that one of $a,b $ or $c$ is irrational, for example $a$, then one more has to be irrational, for example $b$, such that $a=-b$. That leaves $c=4$ in order to satisfy the first equation, but that makes the second one incorrect. That's why all of $a,b,c$ have to be rational numbers. $(*)$
I squared the first equation and got $ab+bc+ca=4$. Then, since $a=4-(b+c)$, I got quadratic equation $b^2+(c4)b+(c-2)^2=0$. Its' discriminant has to be positive and perfect square to satisfy $(*)$.
$$ D= -c(3c-8) $$
From this, we see that $c$ has to be between $0$ and $8/3$ in order to satisfy definition of square root. Specially, for $c=0$ we get $D=0$ and solution for equation $b=\frac{-c+4}{2}=2$. Since the system is symmetric, we also get $c=2$. Similar, for $c=8/3$ we get $b=2$.
In order for $D$ to be perfect square, one of the following has to be true:
$$ -c=3c-8 $$
$$ c=n^2 \land 3c-8=1 $$
$$ 3c-8=n^2 \land c=1 $$
However, only the first one is possible, so $c=2$, and for that we get $b=\frac{(-2+4\pm2)}{2} \Rightarrow b=2 \lor b=0$.
So, all possible values for $c$ are $c\in\{0, 2, \frac{8}3\}$.
EDIT: For $a=b$, there is one more solution: $c=2/3$. Why couldn't I find it with method described above?
| All the points are given by
$$ \left( \frac{4}{3}, \; \frac{4}{3}, \; \frac{4}{3} \; \right) + \frac{\sqrt{12}}{3} \; \left(1, \; -1, \; 0 \; \right) \; \cos t + \frac{2}{3} \; \left(1, \; 1, \; -2 \; \right) \; \sin t $$ where the points with a zero and a pair of 2's occur at $t = \frac{\pi}{2},\frac{7 \pi}{6},\frac{11 \pi}{6}, $ and the third component is given by
$$ \frac{4}{3} (1 - \sin t) $$ which varies between $0$ and $\frac{8}{3},$ the latter happening at $t = \frac{3\pi}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/106530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Manipulating complex numbers Express the complex number $(-1-i)^{48}(-3+i\sqrt{3})^{36}$ in the form a + bi?
I got $ (2^{24})(12^{18}) + 0i $?
Did I get it right?
| Note: $\:x\: =\: -1\:-\:i\: \Rightarrow\:x^4 = (2i)^2 =\: -4.\ $ Recall $\:\zeta^3 = 1\:$ for $\:\zeta = (-1 - \sqrt{-3})/2\:$
thus $\:y\: =\: \sqrt{-3} - 3\ \Rightarrow\: y^6 \: =\: \left(\sqrt{-3}\: (1+\sqrt{-3})\right)^6\ =\ (\sqrt{-3}\:(-2\:\zeta))^6 \:=\: - 2^6\cdot 3^3$
Therefore $\:x^{48} y^{36} = (x^8 y^6)^6 = (2^4\: (-2^6\cdot 3^3))^6 =\: 2^{60} 3^{18}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$ I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:
In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
coefficients of $x^3$?
| $$
\begin{align}
\left(\frac1{\sqrt{x}}+1-\frac12x\right)^8
&=\sum_{k=0}^8\binom{8}{k}\left(\frac1{\sqrt{x}}\right)^k\left(1-\frac12x\right)^{8-k}\\
&=\color{#00A000}{\sum_{k=0}^4\binom{8}{2k}\left(\frac1x\right)^k\left(1-\frac12x\right)^{8-2k}}\\
&+\color{#C00000}{\frac1{\sqrt{x}}\sum_{k=0}^3\binom{8}{2k+1}\left(\frac1x\right)^k\left(1-\frac12x\right)^{7-2k}}\tag{1}
\end{align}
$$
The terms from the red part of $(1)$ will all have fractional powers of $x$ in them, so we only need to consider the green part of $(1)$. Furthermore, for each $k$, the highest power of $x$ in the green part is $8-3k$. Therefore, we only need to consider $k=0$ and $k=1$.
For $k=0$, the term in the green sum is $\left(1-\frac12x\right)^8$ and the coefficient of $x^3$ there is $\binom{8}{3}\left(-\frac12\right)^3$.
For $k=1$, the term in the green sum is $\binom{8}{2}\frac1x\left(1-\frac12x\right)^6$ and the coefficient of $x^3$ there is $\binom{8}{2}$ times the coefficient of $x^4$ in $\left(1-\frac12x\right)^6$, which is $\binom{8}{2}\binom{6}{4}\left(-\frac12\right)^4$.
Therefore, the coefficient of $x^3$ in $\left(\frac1{\sqrt{x}}+1-\frac12x\right)^8$ is $\binom{8}{3}\left(-\frac12\right)^3+\binom{8}{2}\binom{6}{4}\left(-\frac12\right)^4=\frac{77}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/109748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Possibility to simplify $\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{\pi }{{\sin \pi a}}} $ Is there any way to show that
$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}} = \frac{1}{a} + \sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)}^k}\left( {\frac{1}{{a - k}} + \frac{1}{{a + k}}} \right)}=\frac{\pi }{{\sin \pi a}}} $$
Where $0 < a = \dfrac{n+1}{m} < 1$
The infinite series is equal to
$$\int\limits_{ - \infty }^\infty {\frac{{{e^{at}}}}{{{e^t} + 1}}dt} $$
To get to the result, I split the integral at $x=0$ and use the convergent series in $(0,\infty)$ and $(-\infty,0)$ respectively:
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{ - \left( {k + 1} \right)t}}} $$
$$\frac{1}{{1 + {e^t}}} = \sum\limits_{k = 0}^\infty {{{\left( { - 1} \right)}^k}{e^{kt}}} $$
Since $0 < a < 1$
$$\eqalign{
& \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to - \infty } \frac{{{e^{\left( {k + a} \right)t}}}}{{k + a}} = \frac{1}{{k + a}} \cr
& \mathop {\lim }\limits_{t \to \infty } \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} - \mathop {\lim }\limits_{t \to 0} \frac{{{e^{\left( {a - k - 1} \right)t}}}}{{k + a}} = - \frac{1}{{a - \left( {k + 1} \right)}} \cr} $$
A change in the indices will give the desired series.
Although I don't mind direct solutions from tables and other sources, I prefer an elaborated answer.
Here's the solution in terms of $\psi(x)$. By separating even and odd indices we can get
$$\eqalign{
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a + 2k + 1}}} \cr
& \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k}}} - \sum\limits_{k = 0}^\infty {\frac{1}{{a - 2k - 1}}} \cr} $$
which gives
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)$$
$$\sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} = \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) + \frac{1}{a}$$
Then
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} + \sum\limits_{k = 0}^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a - k}}} - \frac{1}{a} = \cr
& = \left\{ {\frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right)} \right\} - \left\{ {\frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right)} \right\} \cr} $$
But using the reflection formula one has
$$\eqalign{
& \frac{1}{2}\psi \left( {1 - \frac{a}{2}} \right) - \frac{1}{2}\psi \left( {\frac{a}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi a}}{2} \cr
& \frac{1}{2}\psi \left( {1 - \frac{{a + 1}}{2}} \right) - \frac{1}{2}\psi \left( {\frac{{a + 1}}{2}} \right) = \frac{\pi }{2}\cot \frac{{\pi \left( {a + 1} \right)}}{2} = - \frac{\pi }{2}\tan \frac{{\pi a}}{2} \cr} $$
So the series become
$$\eqalign{
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \frac{\pi }{2}\left\{ {\cot \frac{{\pi a}}{2} + \tan \frac{{\pi a}}{2}} \right\} \cr
& \sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} = \pi \csc \pi a \cr} $$
The last being an application of a trigonometric identity.
| This is a very elegant and quick way to evaluate this sum with complex analysis. Consider
$$g(z) = \pi \csc (\pi z)f(z)$$
$\csc$ has poles at $2 \pi n$ and $2 \pi n + \pi$ for $n \in \mathbb Z$. Assuming $f(z)$ has no poles at any integer, the residue of $g(z)$ at $2\pi n$ is
$$\operatorname*{Res}_{z = 2 n} g(z) = \lim_{z\to 2 n}(z-2 n)\pi \csc (\pi z)f(z) = \lim_{z\to 2 n}\pi \left(\frac{z-2 n}{\sin (\pi z)}\right)f(z) = f(n)$$
and at $2 \pi n + \pi$:
$$\operatorname*{Res}_{z = 2 n + 1} g(z) = \lim_{z\to 2 n + 1}(z-(2 n + 1))\pi \csc (\pi z)f(z) = \lim_{z\to 2 n + 1}\pi \left(\frac{z-2 n - 1}{\sin (\pi z)}\right)f(z) = -f(n)$$
Let $C_N$ be the square contour with the verticies $\left(N+\frac{1}{2}\right)(1+i)$, $\left(N+\frac{1}{2}\right)(-1+i)$, $\left(N+\frac{1}{2}\right)(-1-i)$ and $\left(N+\frac{1}{2}\right)(1-i)$.
By residue theorem, we have
$$\int_{C_N}g(z)\,dz = \sum_{n=-N}^N (-1)^n f(n) + S$$
where $S$ is the sum of the residues of the poles of $f$. Now, seeing that the left side vanishes as $N \to \infty$ (see here), we have
$$\sum_{k=-\infty}^\infty (-1)^k f(k)=-\sum \text{Residues of }\pi \csc (\pi z)f(z)$$
Clearly the only singularity of $f(z)=\frac{1}{a+z}$ is at $z_0=-a$. Thus
$$\operatorname*{Res}_{z=z_0} \,(\pi \csc (\pi z)f(z))=\lim_{z \to z_0} (z-z_0)\pi \csc (\pi z)f(z)=\lim_{z \to -a} \pi \csc (\pi z)\frac{z+a}{z+a}=-\pi \csc (\pi a)$$
Thus
$$\sum_{k=-\infty}^\infty (-1)^k f(k)=-\operatorname*{Res}_{z=z_0}\,(\pi \csc (\pi z)f(z))=-(-\pi \csc (\pi a))=\frac{\pi}{\sin (\pi a)}$$
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 4,
"answer_id": 3
} |
smallest ODD number n whose sum of divisors (including n) is at least 3n As usual, for a positive integer $n$, let $\sigma(n)$ denote the sum of all positive divisors of $n$ (including 1 and $n$).
What is the smallest ODD number such that $\sigma(n) \ge 3n$?
For comparison, the answers to some related questions are:
Smallest $n$ of any parity satisifying $\sigma(n) \ge 3n$ is $n=120=2^3\cdot3\cdot5$. We have $\sigma(120)=360$.
Smallest $n$ of any parity satisifying $\sigma(n) > 3n$ is $n=180=2^2\cdot3^2\cdot5$. We have $\sigma(180)=546$.
Smallest ODD $n$ satisifying $\sigma(n) \ge 2n$ is $n=945=3^3\cdot5\cdot7$. We have $\sigma(945)=1920$.
I suspect that the answer to my above question is
$$
1310112879075 = 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29
$$
but I can't quite prove it.
I can prove that the answer must be divisible by each of the primes from 3 to 23, and then I can use trial and error to compare various candidates (e.g. compare multiplication by 29 with multiplication by $3^3$, and such "playing around").
| Nope, the smallest is
$$
n=1018976683725=3^3 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29,
$$
which has $\frac{\sigma(n)}{n}=\frac{34283520}{11350339}\approx 3.02048423399513$.
Just for the record, I found this by using your observation that any candidate must be a multiple of $m:=3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23$. Your answer is $9765m$, and by running a for loop over all smaller (odd) multiples, I came upon the (unique) smaller answer of $n=9135m$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/111720",
"timestamp": "2023-03-29T00:00:00",
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Showing $ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}}$ I would like to show that:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\frac{\pi}{3\sqrt{3}} $$
We have:
$$ \sum_{n=0}^{\infty} \frac{1}{(3n+1)(3n+2)}=\sum_{n=0}^{\infty} \frac{1}{3n+1}-\frac{1}{3n+2} $$
I wanted to use the fact that $$\arctan(\sqrt{3})=\frac{\pi}{3} $$ but $\arctan(x)$ can only be written as a power series when $ -1\leq x \leq1$...
| An important trick here is that sigma and integral signs can be changed around.
$$\int \sum^b_{n=a} f\left(n,x\right)\, dx = \sum^b_{n=a} \int f\left(n,x\right) \,dx$$
And this is because
$$\int \sum^b_{n=a} f(n,x)\, dx$$
$$\int f\left(a,x\right) + f((a+1),x) + f((a+2),x) + \dots +f\left((b-1),x\right) + f(b,x) $$
$$ = \int f(a,x)\,dx + \int f((a+1),x) \,dx + \dots + \int f((b-1),x)\, dx + \int f(b,x)\, dx$$
Therefore
$$\begin{align*}
\sum_{n=0}^m \frac{1}{(3n+1)(3n+2)}
=& \sum_{n=0}^m \left( \frac{1}{3n+1} - \frac{1}{3n+2} \right) \\
=& \sum_{n=0}^m \int_0^1 \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
=& \int_0^1 \sum_{n=0}^m \left( x^{3n} - x^{3n+1} \right) \mathrm{d} x \\
=& \int_0^1 \left( \frac{(1-x^{3m+3}) (1-x)}{1-x^3} \right) \mathrm{d} x =
\int_0^1 \frac{1-x^{3m+3}}{1+x + x^2} \mathrm{d} x
\end{align*}
$$
Also because
$$ \sum_{n=0}^m \left( x^{3n} - x^{3n+1} \right) = \frac{(1-x^{3(m+1)})(1-x)}{1-x^3} $$
Now let us see how the final integral
$$\sum_{n=0}^\infty \frac{1}{(3n+1)(3n+2)} = \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} $$
is evaluated.
$$ x^2+x+1 = \left(x+\frac{1}{2}\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2$$
therefore if you skip two steps of substitution and do it once
$$ x+\frac{1}{2} = \frac{\sqrt{3}}{2} tan \theta$$
$$ dx = \frac{\sqrt{3}}{2} sec^{2} \theta$$
$$
\begin{eqnarray}
\int \frac{dx}{1+x+x^2} = \int \frac{ \frac{\sqrt{3}}{2} sec^{2} \theta}{\frac{3}{4} sec^{2} \theta} {\mathrm{d} \theta}
&=& \frac{2}{\sqrt{3}} \theta \\
&=& \frac{2}{\sqrt{3}} tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right)
\end{eqnarray}
$$
$$ \Rightarrow \int_0^1 \frac{\mathrm{d} x}{1+x+x^2} = \frac{2\sqrt{3}}{3} \left( tan^{-1} ( \frac{3}{\sqrt{3}} ) - tan^{-1} ( \frac{1}{\sqrt{3}} ) \right)$$
$$ = \frac{2\sqrt{3}}{3} \left( \frac{\pi}{3} - \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 6,
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} |
Finding a limit of a series How would you calculate this limit it just blew me off on my midterms i seem to have calculated the limit correctly but my process is bougus < what my friend said.
$$
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2}
$$
How i calculated the limit:
$$
\lim_{n\to\infty}\frac{n \sqrt{n^2 \frac{n}{n^2}{}} +n}{\sqrt{n^4 n^{-1}}+2} =
\lim_{n\to\infty}\frac{n^2 \sqrt{\frac{n}{n^2}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2} =
\lim_{n\to\infty}\frac{n^2 \sqrt{\frac{n}{n^2}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2}\\
\frac{n^2 \sqrt{\frac{1}{n}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2}=
\frac{\frac{n^2\sqrt{0}}{n^2\sqrt{0}}+\frac{n}{n^2\sqrt{0}}}{\frac{n^2\sqrt{0}}{n^2\sqrt{0}}+\frac{2}{n^2\sqrt{0}}}= \frac{1}{1}=1
$$
I followed a book where an example was given where it said you can transform
a expression like so:
$$
\frac{1}{n} \sqrt{n^2 + 2} = \sqrt{\frac{1}{n^2} (n^2+2)}
$$
or
$$
\sqrt{n^2+1} = n \sqrt{1+\frac{1}{n^2}}
$$
| I didn't get how it appeared $\sqrt{n^2+\frac1n}$ already in the first numerator - but the answer you got is correct. Nevertheless, for such limits with $n\to\infty$ and powers both in numerator and denominator, you should always divide both numerator and denominator by the highest power of $n$. Here the highest power is $3/2$ so you get
$$
\lim\limits_n\frac{n\sqrt{n}+n}{\sqrt{n^3}+2} = \lim\limits_n\frac{1+\frac{1}{\sqrt{n}}}{{1+\frac{2}{n^{3/2}}}} = 1
$$
since both limits $\lim\limits_n\left(1+\frac{1}{\sqrt{n}}\right)$ and $\lim\limits_n\left(1+\frac{2}{n^{3/2}}\right)$ exist, finite and equal to the same number: $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Gaussian curvature of an ellipsoid proportional to fourth power of the distance of the tangent plane from the center? Is it true that the Gaussian curvature of an ellipsoid is proportional to fourth power of the distance of the tangent plane from the center? I can verify that it holds at the places where the major axes intersect the surface. (Mathworld has an equation for the Gaussian curvature, which simplifies at those points.) But verifying that it holds elsewhere seems like it would get ugly.
The motivation for this question is that Lord Kelvin proved that the charge density on a conducting ellipsoid is proportional to the distance of the tangent plane from the center, while McAllister (I W McAllister 1990 J. Phys. D: Appl. Phys. 23 359 doi:10.1088/0022-3727/23/3/016) finds that under certain assumptions, the charge density on a conducting surface is proportional to the fourth root of the absolute value of the Gaussian curvature. However, I think the assumptions of McAllister's result fail for the ellipsoid (actually I only have access to the abstract, so I'm not sure), so it would be nontrivial to learn that this held for the ellipsoid. (The proportionality is definitely not universal. For a pair of conducting spheres that are far apart and connected by a wire, the exponent is not 1/4. For a deep concavity, all of this definitely fails -- you get a a Faraday cage, which excludes the electric field almost completely.)
| I did the same thing for the "elliptic hyperboloids," of one sheet or two sheets, see
http://mathworld.wolfram.com/EllipticHyperboloid.html
where they have some pictures and formulas. There is enough detail, it is not necessary to use the parametrizations.
I am going to use a fixed number $$ \delta = \pm 1$$ and surface
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \delta = \frac{z^2}{c^2}, $$
or
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = - \delta. $$
It is a bit of a mess, but after you put back in all three variables you get
$$ K = \frac{\delta a^6 b^6 c^6}{ \left( b^4 c^4 x^2 + c^4 a^4 y^2 + a^4 b^4 z^2 \right)^2} $$
If we eliminate $z$ as on the website, we get
$$ K = \frac{\delta a^6 b^6 c^6}{ \left(\delta a^4 b^4 c^2 + b^4 c^2 (c^2 + a^2) x^2 + c^2 a^4 (b^2 + c^2) y^2 \right)^2} $$
which uses, and I kept missing factors in this one,
$$ a^4 b^4 z^2 = \delta a^4 b^4 c^2 + a^2 b^4 c^2 x^2 + a^4 b^2 c^2 y^2. $$
For the distance, fix some point $(x_0, y_0, z_0)$ on the hyperboloid, find the normal vector $ \left(\frac{x_0 }{a^2}, \; \frac{y_0 }{b^2}, \; - \frac{z_0 }{c^2} \right)$ using the gradient of the defining equation, write the tangent plane
$$ \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} - \frac{z_0 z}{c^2} = - \delta. $$
Find the multiple of the normal vector (leaving the origin) lying in the tangent plane. Oh, the multiplier is
$$ t = \frac{-\delta a^4 b^4 c^4}{ b^4 c^4 x_0^2 + c^4 a^4 y_0^2 + a^4 b^4 z_0^2 } $$
So the closest point to the origin is that $t$ times the given normal vector, and the distance of that point from the origin is
$$ d = \frac{ a^2 b^2 c^2}{ \sqrt{ b^4 c^4 x_0^2 + c^4 a^4 y_0^2 + a^4 b^4 z_0^2} } $$
where at one point we use $|\delta| = 1.$
Alright, dropping the $0$ subscripts, we get
$$ d^4 = \frac{ a^8 b^8 c^8}{ \left( b^4 c^4 x^2 + c^4 a^4 y^2 + a^4 b^4 z^2 \right)^2} $$
and
$$ K = \frac{ \delta \; d^4}{a^2 b^2 c^2} $$
A few features to notice: the hyperboloid of one sheet has negative Gauss curvature. The hyperboloid of two sheets has positive Gauss curvature again. The tangent plane, in either case, never passes through the origin itself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is $\lim_{n\rightarrow \infty} \frac{1}{n^2}(\ln(\frac{2^n}{3^n})+\ln(\frac{5^n}{4^n})+\cdots+\ln(\frac{(3n-1)^n}{(n+2)^n}))$? Per the title of this question, how does one go about calculating $$\lim_{n\rightarrow \infty} \frac{1}{n^2}\left(\ln\left(\frac{2^n}{3^n}\right)+\ln\left(\frac{5^n}{4^n}\right)+\cdots+\ln\left(\frac{(3n-1)^n}{(n+2)^n}\right)\right)\ ?$$
Thanks!
| For a mechanical* way to do this:
Hint:
$$\log \frac{(3k-1)}{k+2} = \log 3 + \log (1- \frac{1}{3k}) - \log (1 + \frac{2}{k})$$
and
$$\log (1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots $$
for $|x| \lt 1$.
More details:
Using the above hint, the $k^{th}$ term is
$$ n \log \frac{(3k-1)}{k+2} = n \log 3 + \frac{5n}{3k} + \mathcal{O}(\frac{n}{k^2})$$
and so your sum is
$$ \frac{1}{n^2} \sum_{k=1}^{n} (n \log 3 + \frac{5n}{3k} + \mathcal{O}(\frac{n}{k^2}))$$
$$ = \frac{1}{n^2}(n^2\log 3 + \frac{5n\log n}{3} + \mathcal{O}(n))$$
Here we used the estimate $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \log n + \mathcal{O}(1)$ and $\displaystyle \sum_{k=1}^{n} \frac{1}{k^2} = \mathcal{O}(1)$
Thus the sum is
$$ = \log 3 + \frac{5\log n}{3n} + \mathcal{O}(\frac{1}{n})$$
and so your limit is
$$\log 3$$
Note that we don't really need the estimate $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = \log n + \mathcal{O}(1)$
All we need to show is that $\displaystyle \sum_{k=1}^{n} \frac{1}{k} = o(n)$ and this easily follows from the following classic theorem:
If $\displaystyle a_n \to 0$, then $\displaystyle \frac{1}{n} \sum_{k=1}^{n} a_k \to 0$
*As Didier points out (see comments below), this last theorem can be used to skip all the mechanical calculations done above by applying it directly to the terms of your sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum_{n=0}^\infty \frac1{n^3+1}$ if it can express in terms of elementary functions. Evaluate $\sum_{n=0}^\infty \frac1{n^3+1}$ if it can express in terms of elementary functions .
You may refer to this and this articles.
| Use $n^3+1 = (n+1)(n^2-n+1) = (n+1)(n-\omega)(n - \bar{\omega})$, where $\omega = \mathrm{e}^{i \pi/3}$. Then
$$
\frac{1}{n^3+1} = \frac{1}{3} \frac{1}{n+1} - \frac{1}{3} \frac{\omega}{n -\omega} - \frac{1}{3} \frac{\bar{\omega}}{n -\bar{\omega}} = \frac{1}{3} \frac{\omega +\bar{\omega}}{n+1} - \frac{1}{3} \frac{\omega}{n -\omega} - \frac{1}{3} \frac{\bar{\omega}}{n -\bar{\omega}}
$$
Therefore
$$
\sum_{n=0}^m \frac{1}{n^3+1} = \frac{\omega}{3} \sum_{n=0}^m \left(\frac{1}{n+1} - \frac{1}{n-\omega} \right) + \frac{\bar{\omega}}{3} \sum_{n=0}^m \left(\frac{1}{n+1} - \frac{1}{n-\bar{\omega}} \right)
$$
Thus
$$
\begin{eqnarray}
\sum_{n=0}^\infty \frac{1}{n^3+1} &=& \frac{\omega}{3} \sum_{n=0}^\infty \left(\frac{1}{n+1} - \frac{1}{n-\omega} \right) + \frac{\bar{\omega}}{3} \sum_{n=0}^\infty \left(\frac{1}{n+1} - \frac{1}{n-\bar{\omega}} \right) \\
&=& \frac{\omega}{3} \left( \gamma + \psi(-\omega) \right) + \frac{\bar{\omega}}{3} \left( \gamma + \psi(-\bar{\omega}) \right)
\end{eqnarray}
$$
where $\psi(x)$ is the digamma function, and $\gamma$ is the Euler-Mascheroni constant.
Thus the sum is not elementary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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