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System of congruent equations Given the following system, how do you prove that there only is one solution congruent 8?
$$3x+7y \equiv 2 (\text{mod } 8)$$
$$4x+5y \equiv 7 (\text{mod } 8)$$
My main idea have been to solve the equation, then to come to the solution that there only is one solution, however I get flawed results to say the least. Here is my method:
$$\left\{
\begin{array}{ll}
3x+7y \equiv 2 (\text{mod } 8)
\\4x+5y \equiv 7 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+14y \equiv 4 (\text{mod } 8)
\\8x+10y \equiv 14 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+6y \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
$$
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+18 \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
-2x\equiv -14 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
2x\equiv 6 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
Which then gives:
$$
\left\{
\begin{array}{ll}
x=4t+3
\\y=4s+3
\end{array}
\right.$$
$t$ and $s$ being arbitary whole numbers
So solution given congruent 8 is:
$$x\equiv 3 (\text{mod } 8) \text{ or } x\equiv 7 (\text{mod } 8)$$
$$y\equiv 3 (\text{mod } 8) \text{ or } y\equiv 7 (\text{mod } 8)$$
However when you put that into the original equation, the answer is wrong. So where did i do wrong...
| Write this linear system in matrix form:
$$\begin{pmatrix}3&7\\4&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}2\\7\end{pmatrix}$$
and multiply both sides by the adjugate matrix, which is invertible mod. $8$:
\begin{align}
&\begin{pmatrix}5&1\\4&3\end{pmatrix}\begin{pmatrix}3&7\\4&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}5&1\\4&3\end{pmatrix}\begin{pmatrix}2\\7\end{pmatrix}\\[1ex]
\iff &\begin{pmatrix}3&0\\0&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}1\\5\end{pmatrix}\\[1ex]
\iff 3&\begin{pmatrix}3&0\\0&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv3\begin{pmatrix}1\\5\end{pmatrix}\\[1ex]\iff & \begin{pmatrix}1&0\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\color{red}x\\\color{red}y\end{pmatrix}\equiv\begin{pmatrix}\color{red}3\\\color{red}7\end{pmatrix}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determine the value of $\cos(2)$ I'm trying to prove that the function $f(x)=\cos(x)$ has an $x$-intercept in the interval $[0,2]$. To do that I showed that $f(0)=\cos(0)=1>0$. Then I would have to show that $\cos(2)<0$, however how would I workout the value of $\cos(2)$? My professor mentioned something about using the Taylor Polynomial but I'm not sure.
| Just for the fun of it.
Using the same idea as Felix Marin, let us build the $[n,n]$ Padé approximant of $\cos \left(x+\frac{2\pi }{3}\right)$ around $x=0$ and evaluate it for $x=2-\frac{2\pi }{3}$.
For $n=1$
$$-\frac{\frac{1}{2}+\frac{7 }{4 \sqrt{3}} x}{1+\frac{1}{2 \sqrt{3}}x }\implies \cos(2)\approx -\frac{9+21 \sqrt{3}-7 \sqrt{3} \pi }{18+6 \sqrt{3}-2 \sqrt{3} \pi }\approx -0.415961$$ which corresponds to a relative error of $\color{red}{\text{0.04455 %}}$.
For $n=2$
$$-\frac{\frac{1}{2}+\frac{17 }{12 \sqrt{3}}x-\frac{19 }{72}x^2 } { 1-\frac{1}{6 \sqrt{3}}x+\frac{5 }{36}x^2}\implies \cos(2)\approx \frac{90-153 \sqrt{3}-114 \pi +51 \sqrt{3} \pi +19 \pi ^2}{252-18 \sqrt{3}-60 \pi
+6 \sqrt{3} \pi +10 \pi ^2}\approx -0.416147$$ which corresponds to a relative error of $\color{red}{\text{0.00003 %}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx $ ,then find $\sum_{n\in \mathbb{Z}}|c_n|^2$ Let $$c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx .$$ Then what will be the value of $$\sum_{n\in \mathbb{Z}}|c_n|^2.$$
My attempt:
\begin{align*}
c_n & =\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx\\
& = \frac{1}{\sqrt{2\pi}}\cdot\frac{1}{i(n-i)} \left(e^{i(n-i)x}\right) \Big|_{x=-\pi}^{x=\pi}\\
& = \frac{1}{\sqrt{2\pi}}\cdot\frac{1}{i(n-i)} \cos (n\pi) (2\sinh \pi)\\
\implies |c_n|^2 &= \frac{1}{2\pi}\cdot \frac{1}{n^2+1} \cos^2n\pi (4\sinh^2\pi)\\
&= \frac{2\sinh^2\pi}{\pi(n^2+1)}
\end{align*}
So, $$\sum_{n\in \mathbb{Z}}\frac{2\sinh^2\pi}{\pi(n^2+1)}=\frac{2\sinh^2\pi}{\pi}\sum_{n\in \mathbb{Z}}\frac{1}{n^2+1}$$
After that I stuck, what to do?
| from here David Cardon
$$\sum_{k=-\infty}^{\infty}\frac1{1+k^2}=\pi\coth (\pi)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
I don't know how to solve this integral
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$
What I've done so far is:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$
So
$$x^2+x-4 = A(x^2-4)+Bx$$
And I did this so I could figure out the value of A and B but I am having a rough time trying to calculate A because of the $x^2$.
I've also seen that the solution solves this exercise by separating the fraction's numerator like this:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{x^2}{x(x^2+4)}dx+\int \frac{x}{x(x^2+4)}dx + \int \frac{-4}{x(x^2+4)}dx $$
When do I know that I have to use this? Is what I though correct? If yes, how do I continue it?
Thank you very much. Please, let me know if something is not very clear in my question.
Agapita.
| Hint:
It must be
$$\dfrac{x^2+x-4}{x(x^2+4)}=\dfrac Ax+\frac{Bx+C}{x^2+4}$$
then
\begin{align*}
x^2+x-4&=A(x^2+4)+(Bx+C)x\\
&=(A+B)x^2+Cx+4A
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Factorising $x^2-x+1$ I attempted to factorise $x^2-x+1$, which seemed fairly straightforward:
$$
\begin{align}
x^2-x+1 &= (x+1)^2-3x \\
&= (x+1-\sqrt{3x})(x+1+\sqrt{3x}) \\
&= (x-\sqrt{3x}+1)(x+\sqrt{3x}+1)
\end{align}$$
Technically neither $(x-\sqrt{3x}+1)$ nor $(x+\sqrt{3x}+1)$ are linear functions, which is why I didn't get the "correct" answer:
$$(x-\frac{1}{2}-i\frac{\sqrt{3}}{2})(x-\frac{1}{2}+i\frac{\sqrt{3}}{2})$$
But I am curious as to why Desmos only considers the positive cases of $x$? If $(x-\sqrt{3x}+1)(x+\sqrt{3x}+1)$ does expand perfectly into $x^2-x+1$, why isn't it graphed the same way?
| The function $(x+1-\sqrt{3x})(x+1+\sqrt{3x})$ is only defined for $x \ge 0$, therefore it is not the same function as $x^2-x-1$, which is defined on $\forall x \in \mathbb{R}$. Why? Think about how the square root is defined.
For $x < 0$, you have $|x|=-x$, hence this factorization becomes
$$ x^2 - x + 1 = (x+1)^2 + 3|x| = \left(x+1-i\sqrt{3|x|}\right)\left(x+1+i\sqrt{3|x|}\right) $$
and you no longer have real values.
For the "correct" answer, you need to complete the square
$$ x^2 - x + 1 = \left(x-\frac12\right)^2 + \frac34 = \left(x-\frac12 -
i\frac{\sqrt3}{2}\right)\left(x+\frac12 + i\frac{\sqrt3}{2}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of rational expression
If $x,y,z$ are distinct real number, Then minimum value of $\displaystyle \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2$
Try: $$\left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq \frac{x^2+y^2+z^2}{(x-y)^2+(y-z)^2+(z-x)^2}$$
could some help me to solve this, thanks
| we have to prove that $$ \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq 2$$
this is true since this inequality is equivalent to $${\frac { \left( {x}^{3}-y{x}^{2}-z{x}^{2}-{y}^{2}x+3\,zyx-{z}^{2}x+{y}
^{3}-z{y}^{2}-{z}^{2}y+{z}^{3} \right) ^{2}}{ \left( y-z \right) ^{2}
\left( -z+x \right) ^{2} \left( x-y \right) ^{2}}}
\geq 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sin x - \cos x = 1$ I have been trying to solve $\sin x - \cos x = 1$ by squaring both sides but has not been able to obtain the solution. Here is what I did:
$$\begin{align}(\sin x - \cos x)^2 &=1^2\\\sin^2x-2\sin x\cos x+\cos^2x&=1\\1-\sin 2x&=1\\\sin2x&=0\end{align}$$
Obviously $x=0$ is not a solution. May I ask why this is the case or where did things go wrong?
Thank You
| $$
\begin{aligned}
\sin x-\cos x &=1 \\
-\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x &=-\frac{1}{\sqrt{2}} \\
\cos \left(x+\frac{\pi}{4}\right) &=-\frac{1}{\sqrt{2}} \\
x+\frac{\pi}{4} &=2 n \pi \pm \frac{3 \pi}{4} \\
x &=\frac{\pi}{2}(4 n+1) \text { or }(2 n-1) \pi, \text{ where }n\in Z.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
} |
Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$ Got these problems as separate sections of a question in a book's chapter on 'Divisibility & primes'.
*
*Show that if $x^2 + 5y^2 =1$, then $x = \pm 1$.
Can state it in terms of two factors as : $(x + y\sqrt{-5})(x-y\sqrt{-5}) =x^2 + 5y^2$, with
(i) $(x + y\sqrt{-5}) = 1$, (ii) $(x - y\sqrt{-5}) = 1$
Adding both (i) & (ii), get: $x = 1$, Subtracting (ii) from (i), get: $(y\sqrt{-5}) = 0$.
Unable to pursue after that, as $x=-1$ is not possible.
*Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$
The hint given is to use the unique factorization theorem for the integers, by supposing $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$. The hint asks to show that: $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. I need some more hint or help to pursue, as squaring $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$ does not lead to $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. My attempt is stated for squaring both sides below:L.H.S.: $(x + y\sqrt{-5})(x + y\sqrt{-5}) => x^2 + 2xy(-5) -5y^2$ R.H.S.: $(a + b\sqrt{-5})^2(c + d\sqrt{-5})^2 => (a + b\sqrt{-5})(a + b\sqrt{-5})(c + d\sqrt{-5})(c + d\sqrt{-5}) => (a^2 -5b^2 +2ab\sqrt{-5})(c^2 -5d^2 +2cd\sqrt{-5})$
*Find all primes less than 50 in integers that can be written in the form $x^2 + 5y^2$.
No clue except to first find the primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.
Then trying to see if the factorization works, so starting with $2 = x^2 + 5y^2$, but cannot think further. Do I need to have $y$ as a imaginary number only, as $\sqrt{-5}$, or anything will work.
| Both $x$ and $y$ are drawn from $\mathbb Z$, right?
*
*If $x = \pm 1$, then $x^2 = 1$. Then $5y^2 = 0$, so obviously $y = 0$. Okay, I kinda took that backwards. Going forward: if $y \in \mathbb R$, then $y^2 \geq 0$. Therefore, if $y \neq 0$, then $5y^2 > 4 > 1$.
*Unless there is some restriction on $x$ and/or $y$, this statement is false. Simply choose $x$ and $y$ such that $\gcd(x, y) > 1$. Then $\gcd(x, y) \mid (x + y \sqrt{-5})$. Of course, as Tony showed, $\gcd(x, y) = 1$ is no guarantee either.
*Try fixing $y = 0$, then... um, that doesn't work, squares can't also be prime. Okay, try fixing $y = 1$, then $x^2 + 5y^2$ gives the sequence 5, 6, 9, 14, 21, 30, 41, from which you cull out 5 and 41. Then fix $y = 2$, giving 20, 21, 24, 29, 36, 45, which gives just 29. No need to bother with $y = 3$ for the specified range. So we've got 5, 29, 41. Just three primes, but that's enough to find http://oeis.org/A033205 in the OEIS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
} |
Extremal problem with positive integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}-b\sqrt{7}>0$. Find the minimum value of
$$
S=(a\sqrt{3}-b\sqrt{7})(a+b).
$$
Attempt I have tried and guess that the minimum value of $S$ is $(55+36)(55\sqrt{3}-36\sqrt{7})$, where $(55,36)$ is the integer solution of the Pell equation $3a^2-7b^2=3$.
| Here is a sketch proof that the greatest lower bound for $S$ is $(3/2)(1/\sqrt3 + 1/\sqrt7) = 1.43297\ldots$. Call this $m$. Note that $(3 - \sqrt3m)/\sqrt3 = (\sqrt7m - 3)\sqrt7 = k$, say.
For integers $a,b$ such that $\sqrt3a - \sqrt7b > 0$ define
$$\delta(a,b) = (\sqrt3a - \sqrt7b)(a + b) - m.$$
As noted in the link, we can't have $3a^2 - 7b^2 = 1$ or $2$ (consider residues mod $3$ and $7$). But, as noted by Piyush Davyanakar, there are infinitely many $(a,b)$ such that $3a^2 - 7b^2 = 3$. We can get these more simply by $(a,b) = (1,0), (55, 36), (6049, 3960)$ and in general $a_{n+1} = 110a_n - a_{n-1}$, $b_{n+1} = 110b_n - b_{n-1}$.
This can be proved together with $3a_na_{n-1} - 7b_nb_{n-1} = 165$ by simultaneous induction.
Since $3a^2 - 7b^2 \ge 3$ we have
$$\delta(a,b)(\sqrt3a + \sqrt7b) \ge 3(a + b) - m(\sqrt3a + \sqrt7b) = (3 - \sqrt3m)a - (\sqrt7m - 3)b = k(\sqrt3a - \sqrt7b) > 0.$$
On the other hand,
$$\delta(a,b)(\sqrt3a + \sqrt7b)^2 = k(3a^2 - 7b^2)$$
and this equals $3k$ for infinitely many $(a,b)$. So $\delta(a,b)$ can be made as small as we like.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice factorization of $2^{3x+1}$
Any help specifically on how to treat $2^{3x+1}$ would be appreciated.
| If $x^2+x+1=0,x=w$ where $w$ is a complex cube root of unity.
$$x^{2m}+x^m+1=w^{2m}+w^m+1= \begin{cases}(w^3)^{2n}+(w^3)^n+1=3 &\mbox{if }m=3n\\
w^{2(3n+1)}+w^{3n+1}+1=w^2+w+1=0 & \mbox{if }m=3n+1\\=\cdot=0 & \mbox{if }m=3n+2 \end{cases} $$
$\implies x^2+x+1$ will divide $x^{2m}+x^m+1$ if $3\nmid m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that 2 is not a Gaussian prime? To prove an algebraic number is not a Gaussian prime, need find factors of it.
Let us assume, $(a + bi)(x +yi) = 2, \exists a,b,x,y \in \mathbb{Z}$; so $(ax -by) + i(ay + bx) =2$.
As imaginary part is null in $2$, so $ax - by =2, ay + bx =0$.
Now, how to reach from here, the fact that $a=x=1, b=i, y=-i$.
I read in here in a comment by @StevenStadnicki, that an easier way is to multiply both sides of $(a + bi)(x +yi) = 2$ by the complex conjugates, but is not clear, as my below attempt shows.
$$
\begin{align}
& ax - by =2, ay + bx =0 & \ \\
\implies & axy - by^2 =2y, axy + bx^2 =0 \text{ }{ -(i)} & \ \\
\implies & b = 0
\end{align}
$$
By the answer given below, my attempt is stated below:
Two set of values are possible, one in which $(a^2 + b^2) = 4, (x^2 + y^2) = 1$, second in which $(a^2 + b^2) = 2, (x^2 + y^2) = 2$.
Possible combinations for each of the two cases are stated below:
Case (i): $(a^2 + b^2) = 4, (x^2 + y^2) = 1$ $\implies a= \pm 2, b = 0, x = \pm 1, y = 0$, or $a= \pm 2, b = 0, x = 0, y = \pm 1$. But, as $ax -by =2, ay + bx = 0$. But being algebraic number, need have non-zero imaginary part, hence no algebraic factors of $2$.
Case (ii) $(a^2 + b^2) = 2, (x^2 + y^2) = 2$ $\implies$ possible values are : $ a= \pm 1, b = \pm 1, x = \pm 1, y = \pm 1$.
But, as : $ax -by =2, ay + bx = 0$, so the valid combinations are: 1. $a=1, b=-1, x=1, y=1$; 2. $a=1, b=1, x=1, y=-1$, 3. $a=-1, b=1, x=-1, y=-1$.
Here, combination 3 is same as 1, so can be ignored also valid.
So, the solution has values of $a =1,b=1, x = 1, y = -1$.
If there is a redundancy or unnecessary steps, please tell.
| If $(a+bi)(x+yi)=2$, $(a-bi)(x-yi)=2$ and therefore $(a^2+b^2)(x^2+y^2)=4$. In how many ways can $4$ be written as the product of two integers, each of which is a sum of $2$ squares? Well, $4=4\times1=(2^2+0^2)\times(1^2+0^2)$ and $4=2\times2=(1^2+1^2)\times(1^2+1^2)$. Therefore, there are few possibilities for you to test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\cos A } - 1$$
$$\Rightarrow \tan A + \cot A =\frac{ \sqrt 2 }{\cos A} -1 + \cot A $$
and the 2nd way as -
$$(\sin A + \cos A)^2 = 2$$
$$\sin ^2 A + \cos ^2 A + 2\sin A\cos A = 2 $$
$$\Rightarrow 2\sin A\cos A=1$$
$$\Rightarrow \sin A\cos A=\frac12$$
As we can see the first way is unable to give an answer in absolute Real Number, and the second way doesn't go even near to what is required to proof.
I know few trigonometry identities as per my textbook, those are
*
*$\sin^2 A + \cos^2 A = 1$
*$1 + \cot^2 A = \csc^2 A$
*$\tan^2A + 1 = \sec^2 A$
| Your second attempt is actually what helps here. Note:
$$\begin{align}\tan A+\cot A&=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\\&=\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}\\&=\frac{1}{\sin A\cos A}\\&=\frac{1}{1/2}\\&=2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Jordan decomposition - help with calculation of transformationmatrices? Let $$A=\begin{pmatrix}5&1&1 \\-1&3&1\\0&0&4\end{pmatrix}$$
The Jordan-decomposition is $$A=\begin{pmatrix}-1&-1&-1/2 \\1&0&0\\0&0&-1/2\end{pmatrix}\begin{pmatrix}4&1&0 \\0&4&1\\0&0&4\end{pmatrix}\begin{pmatrix}0&1&0 \\-1&-1&1\\0&0&-2\end{pmatrix}$$
And it drives me crazy that I don't get it !
The characteristic polynomial is $p(\lambda)=(\lambda-4)^3$ therefore $\lambda=4$ has algebraic multiplicity $3$
Then
$(A-4I)=\begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0\end{pmatrix} \Rightarrow \dim(\text{kernel})=2<3$
$(A-4I)^2=\begin{pmatrix}0&0&2 \\0&0&-2\\0&0&0\end{pmatrix}\Rightarrow \dim(\text{kernel})=1<3$
The matrix is not diagonalizable, so we get
$$J=\begin{pmatrix}4&1&0 \\0&4&1\\0&0&4\end{pmatrix}$$
Now I need the eigenvectors/eigenspaces/generalized eigenvetors/spaces and here is the part I have trouble with.
$(A-4I)v=\begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0\end{pmatrix}v=0 \Rightarrow v_1=-v_2 \land v_3=0 \Rightarrow v=\begin{pmatrix}1\\-1\\0\end{pmatrix} $
$(A-4I)^2u=\begin{pmatrix}0&0&2 \\0&0&-2\\0&0&0\end{pmatrix}u=0 \Rightarrow u_3=0\Rightarrow u=\begin{pmatrix}1\\0\\0\end{pmatrix}, u*=\begin{pmatrix}0\\1\\0\end{pmatrix}$
I don't understand how/what wolfram alpha did -.-
What did I wrong ? Thanks for answers.
| We are looking for P such that $P^{-1}AP=J$ thus you need to solve the following systems $AP=PJ$ that is
$Av_1=4v_1$
$Av_2=v_1+4v_2$
$Av_3=v_2+4v_3$
where $v_1$,$v_2$,$v_3$ are the columns of P.
Once we find $v_1$ from the first system, we can find $v_2$ from the second and finally $v_3$ from the third.
See also this example
Making matrix upper triangular by finding Jordan Normal Form
More in detail
$Av_1=4v_1\implies (A-4I)v_1=0\implies \begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0\end{pmatrix}v_1=0\\\implies v_1=\begin{pmatrix}-1 \\1\\0\end{pmatrix}$
$Av_2=v_1+4v_2\implies (A-4I)v_2=v_1 \implies \begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0 \end{pmatrix} v_2=\begin{pmatrix}-1 \\1&\\0\end{pmatrix}\\\implies v_2=\begin{pmatrix}-1 \\0\\0\end{pmatrix}$
$Av_3=v_2+4v_3\implies (A-4I)v_3=v_2 \implies \begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0 \end{pmatrix} v_3=\begin{pmatrix}-1 \\0&\\0\end{pmatrix}\\\implies v_3=\begin{pmatrix}-\frac12 \\0\\-\frac12\end{pmatrix}$
Thus
$$P=\begin{pmatrix}-1&-1&-1/2 \\1&0&0\\0&0&-1/2\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Smallest $n$ such that $125|2^n+3^n$
Find the smallest positive integer $n$ such that $125|2^n+3^n$
Suppose $125|2^n+3^n$. Now, $n$ must be odd because if $n$ is even then
$$0\equiv 2^n+3^n\equiv2^n+(-2)^n\equiv 2^n+2^n\equiv2^{n+1}\pmod 5,$$
which is in contrast to hypothesis. How to go on from here?
| As $3\equiv2^7\pmod{125}$
$$2^n+3^n\equiv2^n+(2^7)^n\equiv2^n(2^{6n}+1)\pmod{125}$$
So, we need $2^{6n}\equiv-1\pmod{125}$
Using Show that $r$ is a primitive root (mod $p^k$), $2$ is a primitive root $5^n$ for $n\ge1$
$$\implies6n\equiv\dfrac{\phi(125)}2\pmod{\phi(125)}$$
$$\iff3n\equiv25\pmod{50}$$
As $17\cdot3\equiv1\pmod{50}$
$$ n\equiv25\cdot17\pmod{50}\equiv25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| We can rewrite the problem as a first-order difference equation:
\begin{align*}
a_{n+1} &= \frac{1}{3} a_n + 3 \\ a_0 &= 4
\end{align*}
We want to know $a_{2018}$.
There are a number of standard techniques to solve these, that are covered in discrete math courses. One trick is temporarily ignore the $+3$ at the end and find a general solution to $a_{n+1} = \frac{1}{3} a_{n}$. It's not too hard to see that $a_{n} = \frac{1}{3^n}$ satisfies that, and so does any multiple of it.
OK, now what about the $3$? Let's look for a constant solution to $a_{n+1} = \frac{1}{3}a_n + 3$. That is, we solve $a = \frac{1}{3}a + 3 $ for $a$. We get $a = \frac{9}{2}$.
Combining the two, we have a solution $$a_n = c \cdot \left(\frac{1}{3}\right)^n + \frac{9}{2}$$
This sequence does satisfy $a_{n+1} = \frac{1}{3}a_n + 3$ for any $c$, as you can check. We now just need to fit the initial condition $a_0 = 4$. Substituting $n=0$ gives
$$
4= c \cdot 1 + \frac{9}{2} \implies c = - \frac{1}{2}
$$
So our particular solution is
$$
a_n = -\frac{1}{2} \cdot \left(\frac{1}{3}\right)^n + \frac{9}{2}
= \frac{9}{2} - \frac{1}{2\cdot 3^n}
$$
Therefore
$$
a_{2018} = \frac{9}{2} - \frac{1}{2\cdot 3^{2018}}
$$
This is very, very close to $\frac{9}{2} = 4.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Verify that for $k=3/2 ,$ $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$ This is a part of a proof I am studying on:
Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\}) $
For $k=3/2 ,$ it's shown that $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$
How is it concluded this way?
| Note that in general
$$\sin x=\sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.$$
thus
$$\sin 2x =2x-\frac{8x^3}{6}+\frac{32x^3}{120}...$$
and
$$\frac{\sin 2x}{2x} =1-\frac{4x^2}{6}+\frac{16x^2}{120}...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Problem in Triangles. $A{A_1}, B{B_1}, C{C_1}$ are the median of triangle ${ABC}$ whose centroid is ${G}$ . If the points ${A}, {C_1} ,{G}, {B_1}$ are concyclic, then show that ${2a^2 =c^2 +b^2} $ if $a,b,c$ are sides of triangle?
| Let $a$, $b$ and $c$ be sides-lengths of $\Delta ABC$ in the standard notation.
Thus, by the Ptolemy theorem we obtain
$$AC_1\cdot B_1G+AB_1\cdot C_1G= C_1B_1\cdot AG$$ or
$$\frac{c}{2}\cdot\frac{1}{6}\sqrt{2a^2+2c^2-b^2}+\frac{b}{2}\cdot\frac{1}{6}\sqrt{2a^2+2b^2-c^2}=\frac{a}{2}\cdot\frac{1}{3}\sqrt{2b^2+2c^2-a^2}$$ or after squaring of the both sides
$$2a^4+b^4+c^4-3a^2b^2-3a^2c^2-b^2c^2+bc\sqrt{(2a^2+c^2-b^2)(2a^2+2b^2-c^2)}=0,$$ which gives
$$(2a^4+b^4+c^4-3a^2b^2-3a^2c^2-b^2c^2+bc)^2-b^2c^2(2a^2+c^2-b^2)(2a^2+2b^2-c^2)=0$$ or
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)(b^2+c^2-2a^2)^2=0,$$ which gives $$b^2+c^2=2a^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring the polynomial $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
can you help me factoring the following polynomial?
$6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
I don't know where to start from.
| Note that $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$ is homogenous. So set $u=b/a$. Then $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = a^4(6 + u +5u^2 + u^3 - u^4)$.
We try integer roots of $6 + u +5u^2 + u^3 - u^4$ by considering the divisors of $6$. We find $6 + u +5u^2 + u^3 - u^4=-(u + 2) (u - 3) (u^2 + 1)$.
This gives
$
6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = -(2 a + b) (b-3 a) (a^2 + b^2)
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ Prove that $n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ is given by
$$T_n =\left[\sqrt{2n+\frac{1}{2}}\right]$$ where $[.]$ is Floor function
My Try:
Its clear that $1$st Term is $1$, $3$rd Term is $2$, $6$th term is $3$ and so on
We have Triangular numbers as $1,3,6,10,...$ whose $m$th term is given by $\frac{m(m+1)}{2}$
Hence for the given sequence
$$T_{\frac{m(m+1)}{2}}=m$$
Letting $$\frac{m(m+1)}{2}=n$$ we get Quadratic in $m$ as
$$m^2+m-2n=0$$ So $m$ takes $\frac{-1\pm \sqrt{1+8n}}{2}$ and since $m$ is a positive integer we have
$$m=\frac{-1+\sqrt{1+8n}}{2}$$
hence
$$T_n=\frac{-1+\sqrt{1+8n}}{2}$$
how to proceed further?
| Note that $$T_n=\frac{-1+\sqrt{1+8n}}{2}$$ is true only if $$n=m(m+1)/2.$$ Now you have to define $T_n$ for other values of $n$. Since terms repeat between two such triangular numbers and you know the value of $T_n$ for $n=m(m+1)/2$ the rest of the solution should be obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let $x$ and $y$ are linearly interdependent and that $x+y+z = 0$. Show that $\theta(x,y) + \theta(y,z) + \theta(z,x) = 2\pi$ In the book of Linear Algebra by Werner Greub, at page 202 Q.11.a, it is asked that
Let $x,y,z$ be three vectors of a plane such that $x$ and $y$ are
linearly interdependent and that $x+y+z = 0$.
a-) Prove that the ordered pairs $x,y; y,z; z,x$ represent the same
orientation. Then show that $$\theta(x,y) + \theta(y,z) + \theta(z,x)
= 2\pi$$ where the angles refer to the above orientation.
I can see that they represent the same orientation but couldn't show the followed result. I was hoping you can help me on that point.
| Let $X=\theta(y,z)$, $Y=\theta(z,x)$, $Z=\theta(x,y)$.
$$\cos Z=\frac{x\cdot y}{|x|\cdot|y|}$$
$$\cos X=\frac{-x\cdot y-|y|^2 }{|y|\cdot|x+y|}$$
$$\cos Y=\frac{-x\cdot y-|x|^2 }{|x|\cdot|x+y|}$$
We'll need also $\sin X$ and $\sin Y$:
$$\sin X=\sqrt{1-\frac{(x\cdot y)^2+2x\cdot y|y|^2+|y|^4}{|y|^2|x+y|^2}}=\frac{\sqrt{|x|^2|y|^2-(x\cdot y)^2}}{|y|\cdot|x+y|}$$
Similarly,
$$\sin Y=\frac{\sqrt{|x|^2|y|^2-(x\cdot y)^2}}{|x|\cdot|x+y|}$$
Thus,
$$\begin{align}\cos(2\pi-X-Y)&=\cos(X+Y)=\cos X\cos Y-\sin X\sin Y\\
&=\frac{|x|^2|y|^2+x\cdot y(|x|^2+|y|^2)+(x\cdot y)^2}{|x|\cdot|y|\cdot|x+y|^2}-\frac{|x|^2|y|^2-(x\cdot y)^2}{|x|\cdot|y|\cdot|x+y|^2}\\
&=\frac{x\cdot y|x+y|^2}{|x|\cdot|y|\cdot|x+y|^2}\\
&=\cos Z\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using induction, prove that $({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$. Question: Using the principle of mathematical induction, prove that for every integer $n\geq 1, ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.
My attempt:
Let $P(n): ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$ ($n\geq 1).$
BASE CASE:
$P(1): ({3^2}^1 -1)$ is divisible by $2^{1+2}$ but not by $2^{1+3}$ , that is, $8$ is divisible by $8$ but not by $16$, which is true.
$\therefore P(1)$ is true
INDUCTION HYPOTHESIS:
Let $P(n)$ be true for $n=k$.
$\therefore P(k)$ is true, that is, $({3^2}^k -1)$ is divisible by $2^{k+2}$ but not by $2^{k+3}$.
$\therefore$ Let ${3^2}^k -1=\lambda.2^{k+2}$
$\implies {3^2}^k=1+\lambda.2^{k+2}$
INDUCTIVE STEP:
$P(k+1):({3^2}^{k+1} -1)$ is divisible by $2^{k+1+2}$ but not by $2^{k+1+3}$
We have to prove that $P(k+1)$ is true
${3^2}^{k+1}-1={3^{{2^k}2}}-1=(1+\lambda.2^{k+2})^2-1=1+2\lambda.2^{k+2}+\lambda^2.2^{2(k+2)}-1=\lambda.2^{k+3}+\lambda^2.2^{k+3+k+1}=2^{k+3}(\lambda+\lambda^2.2^{k+1})$
$\therefore {3^2}^{k+1}-1$ is divisible by $2^{k+3}$
My problem: How do I prove that the number is not divisible by $2^{n+3}$? What should I add to the induction hypothesis part and the inductive step. I am not allowed to use congruence. I am allowed to use the properties of numbers and their division. I thought of doing it by proving that the number leaves a particular remainder on dividing by $2^{n+2}$ but then realized that it would be very cumbersome since the remainder can range from $1$ to $2^{n+3}-1$.
Please help.
| Supose that $3^{2^n}-1$ is not divisible by $2^{n+3}$. Then there are $q\in \mathbb{N}$ and $r\in \{1,2,3,\ldots, 2^{n+3}-1\}$ such that
$$
3^{2^n}-1 = q\cdot 2^{n+3}+r
$$
Then
\begin{align}
3^{2^{n+1}}-1 =& (3^{2^{n}})^2-1 \\
=& (3^{2^{n}}-1)(3^{2^{n}}+1) \\
=& (3^{2^{n}}-1)[(3^{2^{n}}-1)+2] \\
=& (q\cdot 2^{n+3}+r)[(q\cdot 2^{n+3}+r)+2]\\
=& q^2(2^{n+3})^2+2(2^{n+3})r+r^2+2q(2^{n+3})+2r\\
=& q^2(2^{2n+6})+(2^{n+4})r+r^2+q(2^{n+4})+2r\\
=& q^2\cdot(2^{n+2})\cdot(2^{n+4})+
(2^{n+4})\cdot r+r^2+q\cdot(2^{n+4})+2r\\
=&[q\cdot (2^{n+2})+r+q]\cdot(2^{n+4})+r^2+r
\end{align}
So what can you say about $r^2+r$? Hint.
A first case is simple. The second case has three subcases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\sum_{n=1}^{\infty}$ $\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$
MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
= Lim$_{k\rightarrow\infty}$$\sum_{k=1}^{n}\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$
$\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$=
$\frac{1}{6\left(n+1\right)}$-$\frac{2}{\left(n+2\right)}$+$\frac{9}{\left(n+3\right)2}$-$\frac{8}{\left(n+4\right)3}$
I don't think i can telescope it, And i don't know any other method
| Another approach , applicable to this type of sums in general, is to decompose the fraction as a sum
of fractions with Rising Factorials at the denominator.
$$
\eqalign{
& {{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}} = {{\left( {n + 1} \right)\left( {n + 2} \right) - 3n - 2} \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} = \cr
& = {{\left( {n + 1} \right)^{\,\overline {\,2\,} } - 3\left( {n + 1} \right) + 1} \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} = {1 \over {\left( {n + 3} \right)^{\,\overline {\,2\,} } }} - 3{1 \over {\left( {n + 2} \right)^{\,\overline {\,3\,} } }} + {1 \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }} \cr}
$$
That is of advantage because:
- a polynomial at the numerator (here the $n^2$) can always be expressed (through Stirling Numbers) into a "polynomial" in rising/falling "powers";
- the sum of Falling factorials parallels the integral of normal powers;
- you can easily express the sum also between finite bounds.
We have in fact this equivalence between Falling and Rising Factorials
$$
x^{\underline {\, - q\,} } = {1 \over {\left( {x + q} \right)^{\underline {\,q\,} } }} = {1 \over {\left( {x + 1} \right)^{\overline {\,q\,} } }}
$$
and the indefinite summation of the Rising Factorial is
$$
\sum\nolimits_x {{1 \over {\left( {x + 1} \right)^{\overline {\,q\,} } }}} = \sum\nolimits_x {x^{\underline {\, - q\,} } } = {1 \over {1 - q}}x^{\underline {\,1 - q\,} } + c = - {1 \over {\left( {q - 1} \right)\left( {x + 1} \right)^{\overline {\,q - 1\,} } }} + c
$$
as it easy to check by taking the Forward Finite Difference of both sides.
Therefore
$$
\sum\limits_{n = 1}^\infty {{1 \over {\left( {n + m} \right)^{\overline {\,q\,} } }}} \quad \left| {\;2 \le q} \right.\quad = {1 \over {\left( {q - 1} \right)\left( {m + 1} \right)^{\overline {\,q - 1\,} } }}
$$
and the result is
$$
\eqalign{
& \sum\limits_{n = 1}^\infty {{{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}}} = \cr
& = \sum\limits_{n = 1}^\infty {\left( {{1 \over {\left( {n + 3} \right)^{\,\overline {\,2\,} } }} - 3{1 \over {\left( {n + 2} \right)^{\,\overline {\,3\,} } }} + {1 \over {\left( {n + 1} \right)^{\,\overline {\,4\,} } }}} \right)} \cr
& = {1 \over {1\left( {3 + 1} \right)^{\overline {\,1\,} } }} - {3 \over {2\left( {2 + 1} \right)^{\overline {\,2\,} } }} + {1 \over {3\left( {1 + 1} \right)^{\overline {\,3\,} } }} = \cr
& = {1 \over 4} - {1 \over 8} + {1 \over {72}} = {{5} \over {36}} \cr}
$$
For more details, also refer to this and to this other related post.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
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Find $\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right).$ This is what my prof did:
Rewrite as $$x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\frac{x}{e}\left(1-e\cdot e^{-x\ln{\left(1+\frac{1}{x}\right)}}\right).$$
Here he uses that
$$\ln{(1+t)}=t-\frac{t^2}{2}+O(t^3),$$
so
$$\frac{x}{e}\left(1-e\cdot e^{-x\left(\frac{1}{x}-\frac{1}{2x^2}+O(\frac{1}{x^3})\right)}\right)=\frac{x}{e}\left(1-e^{\frac{1}{2x}+O\left(\frac{1}{x^3}\right)}\right).$$
So far so good. Now, he uses that $e^t=1+t+O(t^2)$ so that the expression aove becomes
$$\frac{x}{e}\left(1-\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\right)=-\frac{1}{2e}+O\left(\frac{1}{t}\right)$$
and now he lets $x\rightarrow\infty$ to get the answer. However, when he introduces the expansion of $e^t$, how does he get $O(1/x^2)$?
It is my understanding that I should plug in $t=\frac{1}{2x}+O\left(\frac{1}{x^3}\right)$ everywhere in $1+t+O(t^2),$ doing that I get
$$1+\frac{1}{2x}+O\left(\frac{1}{x^3}\right)+O\left(\left(\frac{1}{2x}+O\left(\frac{1}{x^3}\right)\right)^2\right),$$
Squaring that last ordo-term, the largest power of $x$ is $1/x^2$ and the smallest is $1/x^6.$ Why choose $1/x^2$?
| The whole argument must be understood for $t=1/x\to0$, so that the lowest exponents yield the largest terms ($t^2$ exceeds $t^4$ because $1$ exceeds $t^2$; if you prefer, $1/x^2$ exceeds $1/x^4$ because $x^4$ exceeds $x^2$).
Then
$$e^{t+O(t^3)}=1+t+O(t^3)+O\left(\left(t+O(t^3)\right)^2\right)=1+t+O(t^2)$$ by keeping the terms of lowest degree.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing volume using shell resulted in negative value? The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of $y=x^2$ and $y=2-x^2$, revolved about $x=2$
$$x^2=2-x^2→2x^2=2→\frac{2x^2}2=\frac{2}2→x^2=1$$
$$x=1, x=-1$$
Radius is $$r=2-x$$
Height is $$x^2-(2-x^2)$$
Finding the Integral,
$$\int_{\neg1}^12π(2-x)(x^2-(2-x^2 ))dx=2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)+C$$
Finding the limits,
$$\lim_{x→-1+}2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)=2π(\frac{4(-1)^3}3-4(-1)-\frac{2(-1)^4}3+2(-1)^2-(-1)^2+\frac{(-1)^4}6)=19.897$$
$$\lim_{x→1-}2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)=2π(\frac{4(1)^3}3-4(1)-\frac{2(1)^4}3+2(1)^2-(1)^2+\frac{(1)^4}6)=-13.614$$
And finally, computing the volume from the obtained limits
$$V=-13.614-19.897=-33.510$$
So, if someone could let me know if (or where) I went wrong, that would be great.
| In the region $-1 \leq x \leq1$, $2-x^2 > x^2$ (the $2-x^2$ curve is above the $x^2$ curve). So the height is $(2-x^2)-x^2$, not $x^2-(2-x^2)$.
This fix should simply change the sign of your answer -- which is exactly the issue you wanted to address!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$ QuestionProve that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$=
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$
My Approach R.H.S
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$=$\frac{x^{2^{k-1}\left(1-x^{2}\right)}}{\left(1-x^{2^{k}}\right)\left(1-x^{2^{k-1}}\right)}$
i just don't know what else i can do?
| $$\begin{align}
\frac{1}{1-x^{2^{k-1}}} - \frac{1}{1-x^{2^k}}
&= \frac{(1-x^{2^k}) - (1-x^{2^{k-1}})}{(1-x^{2^{k-1}})(1-x^{2^k})}\\
&= \frac{x^{2^{k-1}}(1-x^{2^{k-1}})}{(1-x^{2^{k-1}})(1-x^{2^k})}\\
&= \frac{x^{2^{k-1}}}{1-x^{2^k}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Derive the conditions $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ and $xy>-1$ for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \text{, }xy<1\\
\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \text{, }xy>-1
$$
But, How do I reach the conditions $xy<1$ for the first expression and $xy>-1$ for the second from the domain and range of the functions, provided we are only considering the principal value branch ?
My Attempt
$$
\tan^{-1}:\mathbb{R}\to \Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)
$$
$$
\text{Taking, }\alpha=\tan^{-1}x, \quad\beta=\tan^{-1}y\implies x=\tan\alpha,\quad y=\tan\beta\\
\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{x+y}{1-xy}\\
\text{We have, }-\pi<\tan^{-1}x+\tan^{-1}y=\alpha+\beta<\pi
$$
If $\tfrac{-\pi}{2}<\alpha+\beta<\tfrac{\pi}{2}$ we have,
$$
\alpha+\beta=\tan^{-1}\bigg(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\bigg)\implies \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}
$$
For the first expression, $xy\neq{1}$ as the denominator can not be equal to zero.
$$
\frac{-\pi}{2}<\tan^{-1}\frac{x+y}{1-xy}<\frac{\pi}{2}\text{ and }-\pi<\tan^{-1}x+\tan^{-1}y<\pi\\\implies\frac{-\pi}{2}<\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}<\frac{\pi}{2}
$$
I really dont see any clue which leads to the condition $xy<1$. I checked a similar question asked Inverse trigonometric function identity doubt, but it does not seem to clear how to get to the given conditions from the above proof.
Note: I am not looking for proving the statement is correct. I'd like to see how to reach the given conditions from the domain and range of the functions involved.
| Thanks @Rohan for the hint.
$$
\tan^{-1}:\mathbb{R}\to \Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)
$$
Taking,
$$
\alpha=\tan^{-1}x\implies x=\tan\alpha\text{ , where }\tfrac{-\pi}{2}<\alpha<\tfrac{\pi}{2}\\
\beta=\tan^{-1}y\implies{y}=\tan\beta\text{ , where }\tfrac{-\pi}{2}<\beta<\tfrac{\pi}{2}\\
$$
For,
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}
$$
$\implies-\pi<\alpha+\beta<\pi$.
$$
\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\frac{x+y}{1-xy}\\
$$
If $\tfrac{-\pi}{2}<\alpha+\beta<\tfrac{\pi}{2}$,
$$
\alpha+\beta=\tan^{-1}\bigg(\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\bigg)\implies \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}
$$
In the range $\tfrac{-\pi}{2}<\alpha+\beta<\tfrac{\pi}{2}$, we have $\cos(\alpha+\beta)>0$, $\cos\alpha>0$ and $\cos\beta>0.$
$$
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos\alpha\cos\beta-\cos\alpha\tan\alpha\cos\beta\tan\beta\\=\cos\alpha\cos\beta\Big(1-\tan\alpha\tan\beta\Big)>0\\\implies
1-\tan\alpha\tan\beta>0 \quad\bigg(\text{ as } \cos\alpha>0 \text{ and } \cos\beta>0\bigg)\\
\implies 1-xy>0\implies \color{red}{xy<1}
$$
For,
$$
\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}
$$
$$
\tfrac{-\pi}{2}<\alpha<\tfrac{\pi}{2}\\
\tfrac{-\pi}{2}<\beta<\tfrac{\pi}{2}\implies\tfrac{\pi}{2}>-\beta>\tfrac{-\pi}{2}\implies\tfrac{-\pi}{2}<-\beta<\tfrac{\pi}{2}
$$
$\implies -\pi<\alpha-\beta<\pi$
$$
\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\frac{x-y}{1+xy}
$$
If $\frac{-\pi}{2}<\alpha-\beta<\frac{\pi}{2}$,
$$
\alpha-\beta=\tan^{-1}\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\implies\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}
$$
In the range $\tfrac{-\pi}{2}<\alpha-\beta<\tfrac{\pi}{2}$, we have $\cos(\alpha-\beta)>0$, $\cos\alpha>0$ and $\cos\beta>0.$
$$
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos\alpha\cos\beta+\cos\alpha\tan\alpha\cos\beta\tan\beta\\
=\cos\alpha\cos\beta\Big(1+\tan\alpha\tan\beta\Big)>0\\
\implies 1+\tan\alpha\tan\beta>0\quad\Big(\text{ as }\cos\alpha>0\quad\&\quad\cos\beta>0\Big)\\
\implies1+xy>0\implies1>-xy\implies \color{red}{xy>-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
polar coordinate subtitution
$$\int\int_{(x^2+y^2) \le1} \left(\frac{x^4}{a^4}+ \frac{y^4}{b^4} \right)dxdy$$
I try to solve this multivariable integration, by substitute $x=ar \sin \theta$ and $y=br \sin \theta$
to $(\frac{x^4}{a^4}+ \frac{y^4}{b^4} )$ so it will produce $r^4 $ with Jacobin $abr$ but when I substitute to $x^2+y^2 \le1$
it won't produce $r^2$
$a^2r^2\sin^2 \theta+b^2r^2 \cos^2 \theta \le1$, so I cannot change it to polar coordinate.
Is my approach wrong? Can someone give me hint too!
| Just use plain old polar coordinates. Let $x = r\cos \theta$ and $y = r\sin \theta$, then everything boils down to
\begin{align}
\int^{2\pi}_0 \int^1_0\ \left( \frac{r^4 \cos^4\theta}{a^4}+\frac{r^4\sin^4\theta}{b^4}\right)\ rdrd\theta = \frac{1}{6} \int^{2\pi}_0 \frac{\cos^4 \theta}{a^4}+\frac{\sin^4\theta}{b^4}\ d\theta.
\end{align}
Additional:
\begin{align}
\int^{2\pi}_0 \cos^{2n}\theta\ d\theta =&\ \frac{1}{2^{2n}}\int^{2\pi}_0 \left( e^{i\theta}+e^{-i\theta}\right)^{2n} d\theta = \frac{1}{2^{2n}} \int^{2\pi}_0 \sum^{2n}_{k=0} \binom{2n}{k}e^{ik\theta} e^{-i(2n-k)\theta}\ d\theta\\
=&\ \frac{1}{2^{2n}}\sum^{2n}_{k=0}\binom{2n}{k} \int^{2\pi}_0 e^{i(2k-2n)\theta}\ d\theta = \binom{2n}{n}\frac{\pi}{2^{2n-1}} = \frac{(2n)! \pi}{(n!)^2 2^{2n-1}}
\end{align}
and
\begin{align}
\int^{2\pi}_0 \sin^{2n}\theta\ d\theta =&\ \frac{(-1)^n}{2^{2n}} \int^{2\pi}_0 (e^{i\theta}-e^{-i\theta})^{2n}\ d\theta = \frac{(-1)^n}{2^{2n}} \int^{2\pi}_0 \sum^{2n}_{k=0} \binom{2n}{k}(-1)^{2n-k}e^{i(2k-2n)\theta}\ d\theta\\
=&\ \frac{(-1)^n}{2^{2n}} \sum^{2n}_{k=0} \binom{2n}{k}(-1)^{2n-k} \int^{2\pi}_0 e^{i(2k-2n)\theta}\ d\theta = \frac{(2n)!\pi}{(n!)^22^{2n-1}}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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existence of solutions of $a^n+b^n+c^n=6^n$ $$2^2+4^2+4^2=6^2$$
$$3^3+4^3+5^3=6^3$$
There is no such thing for exponent 4. It's not hard to prove that the only integer solutions of $a^4+b^4+c^4=6^4$ are trivial ones.
For which $n$ there exist $a,b,c\in\mathbb{N}$ such that $a^n+b^n+c^n=6^n$?
(own question, a very probable answer is: $2$ and $3$ only).
| For $n \geq 7,$
$$ \left( \frac{5}{6} \right)^n < \frac{1}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$ Let $a, b, c$ be real numbers such that $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$. Prove that $a^2+b^2+c^2\gt \frac {1}{2018}$.
It looks like an easy question, but I thought for a while and could not figure it out. Can anyone give me some hints? Thank you.
| By using C-S we have :
$$|3a+28b+35c| \le \sqrt{3^2+28^2+35^2} \cdot \sqrt{a^2+b^2+c^2} \tag 1$$
and
$$|20a+23b+33c|\le \sqrt{20^2+23^2+33^2} \cdot \sqrt{a^2+b^2+c^2} \tag 2$$
Multiplying $(1)$ and $(2)$
$$|(3a+28b+35c)(20+23b+33c)| \le \sqrt{20^2+23^2+33^2} \cdot \sqrt{a^2+b^2+c^2} \cdot \sqrt{3^2+28^2+35^2} \cdot \sqrt{a^2+b^2+c^2} $$
$$\implies \underbrace{(3a+28b+35c)(20+23b+33c)}_{=1} \le 2018 \cdot (a^2+b^2+c^2) $$
Thus
$$a^2+b^2+c^2 > \frac{1}{2018}$$
And, the inequality is strict because $\frac ba =\frac {28}3$ and $\frac ba =\frac{23}{20}$ cannot be simultaneously true
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of Integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ Calculate integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ This was my exam question. I've tried many online math solvers and math programs but none were able to solve. If anybody has an answer would be helpful. Thanks
| I have found a solution in terms of a hypergeometric function for positive $x$:
$$
I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{3}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{3}{x^3}\right)\right) +C
$$
and verified the derivative equals the integrand (for positive $x$), and in fact for general $a$,
$$
\int \frac{x^2+1}{\sqrt{x^3+a}}dx
$$
is given by $$
I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{a}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{a}{x^3}\right)\right) + C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Solve $15\cdot x=27\pmod{18}$ EXERCISE:
Solve $15\cdot x\equiv 27\pmod{18}$
SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.
So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!
Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equiv 9\pmod{6}$
So, from now and then I can't understand how to proceed!
The book continue with this:
"We have a unique solution of $5\cdot x\equiv 9\pmod{6}$ which is":
$x=5\cdot 9 =45\equiv 3\pmod 6$ and $15\cdot 3\equiv 27\pmod{18}\equiv 9\pmod{18} $
The other solutions are:$(9,15)$. So all solutions are these: $(3,9,15)$
Can anyone explain me how the author of the book continue the exercise?I really can't understand how he ends up with these solutions!Is there another way to reach these solutions?
I would really appreciate a thorough explanation, since I've just started working on these type of exercises and I have to clear my mind on them.
Thanks for your time !
| The primary task is finding $5^{-1}$ modulo $6$, because once we find it, we multiply both sides by it to make $x$ have coefficient $1$ (therefore determining $x$). In general, here's how to find $z^{-1}$ modulo $a$.
*
*Want to find $y$ such that $zy \equiv 1 \pmod{a}$
*This is equivalent to: find $y$ and $b$ (modulo $6$) such that $zy - 1 = ab$, i.e. $xy + a(-b) = 1$. Equivalently, want to find $y$ and $b$ such that $zy + ab = 1$. These exists because of Bézout.
*To find these, use the Euclidean algorithm as in the following examples.
$1)$ $z = 24$, $a = 7$.
First divide $z$ by $a$: $24 = 3\cdot 7 + 3$. The quotient is $q_1 = 3$ and the remainder is $r_1 = 3$. Now divide $a$ by $r_1$: $7 = 2 \cdot 3 + 1$. Quotient is $q_2 = 2$ and remainder is $r_2 =1$. STOP (always stop when the remainder is $1$). Fill out a table.
$$\begin{array}{|c|c|} \hline y & b \\ \hline 1& 0\\ \hline 0 & 1 \\ \hline 1 & -3 \\ \hline -2 & 7 \\ \hline \end{array}$$
Here's how to fill out the table. Call $R_n$ the $n$th row. $R_1$ and $R_2$ are always as such. For $n\ge 3$, $R_n = R_{n-2} - q_{n-2} R_{n-1}$.
The last row has the numbers you want (so here, $y = -2$ and $b=7$).
$2)$ Your example ($a = 6$, $z = 5$).
$5 = 0\cdot 6 + 5$. So $q_1 = 0$ and $r_1 = 5$. Next $6 = 1\cdot 5 + 1$, so $r_2 = 1$ and $q_2 = 1$. STOP
$$\begin{array}{|c|c|} \hline y & b \\ \hline 1& 0\\ \hline 0 & 1 \\ \hline 1 & 0 \\ \hline -1 & 1 \\ \hline \end{array}$$
So $y = -1$ and $b = 1$.
$-1$ is $5$ modulo $6$. So $5^{-1}$ is $5$ modulo $6$. Now the author proceeds to find $x$ by multiplying both sides by $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Tricky convergence/divergence of a sum Let be $A \subset \mathbb{N}$ composed by all the positive integers that have NO $0$ in their decimal expression (so $A=\{1,2,3,\dots 9,11,12, \dots 19,21... \}$).
Show the convergence or divergence of $\sum_{n\in A} \frac{1}{n}$
| Define $S_n = [10^{n - 1}, 10^n) \cap A$ for $n \in \mathbb{N}_+$. Because for every $(a_n \cdots a_1 a_0)_{10} \in S_{n + 1}$, there is $(a_n \cdots a_1)_{10} \in S_n$, and for every $(a_{n - 1} \cdots a_0)_{10} \in S_n$, there is $(a_{n - 1} \cdots a_0 j)_{10} \in S_{n + 1}$ $(1 \leqslant j \leqslant 9)$, then$$
\sum_{k \in S_{n + 1}} \frac{1}{k} = \sum_{l \in S_n} \sum_{j = 1}^9 \frac{1}{10l + j} < \sum_{l \in S_n} \sum_{j = 1}^9 \frac{1}{10l} = \frac{9}{10} \sum_{l \in S_n} \frac{1}{l}.
$$
By induction,$$
\sum_{k \in S_n} \frac{1}{k} \leqslant \left( \frac{9}{10} \right)^{n - 1} \sum_{k \in S_1} \frac{1}{k} = \left( \frac{9}{10} \right)^{n - 1} \sum_{k = 1}^9 \frac{1}{k}. \quad n \in \mathbb{N}_+
$$
Therefore,$$
\sum_{m \in A} \frac{1}{m} = \sum_{n = 1}^\infty \sum_{k \in S_n} \frac{1}{k} \leqslant \sum_{n = 1}^\infty \left( \frac{9}{10} \right)^{n - 1} \sum_{k = 1}^9 \frac{1}{k} = 10 \sum_{k = 1}^9 \frac{1}{k} < +\infty.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Functional with 4th grade characteristic equation I get a functional example, which looks like this:
$\mathcal{F}(u) = \int_0^{\frac{\pi}{4}} (x^2 + 16u^2 - u''^2) dx $
with conditions: $u(0) = 1, u'(0) = 0, u(\frac{\pi}{4}) = 0, u(\frac{\pi}{4}) = -2 $
Using Euler-Lagrange equation :
$\frac{\partial F}{\partial u} - \frac{d^2}{dx^2} \cdot \frac{\partial F}{\partial u''} = 0 $
got equation:
$2u^{(4)} + 16u = 0$
$u^{(4)} +8u = 0$
I tried to solve it with characteristic equation
$\lambda^4 + 8 = 0$, but probably I failed.
These are my next steps. Could someone tell me whether they correct or not? And how to solve it?
$r = \lambda^2$
$ r^2 + 8 = 0 $
$\Delta = -4 \cdot 8 = -32 $
$\sqrt{\Delta} = \sqrt{-1 \cdot 32} = \pm t \cdot 4\sqrt{2}$
$r_1 = 2\sqrt{2} \cdot t$
$r_2 = - 2\sqrt{2} \cdot t$
$\lambda^2 = 2\sqrt{2} \cdot t $
$\lambda^2 = - 2\sqrt{2} \cdot t $
$\lambda_1 = \sqrt{2\sqrt{2} \cdot t}$
$\lambda_2 = - \sqrt{2\sqrt{2} \cdot t}$
$\lambda_3 = \sqrt{ - 2\sqrt{2} \cdot t}$
$\lambda_4 = - \sqrt{ - 2\sqrt{2} \cdot t}$
| Your method isn't very clear. You introduce a delta, then it goes away with no explanation. You seem to be using t where i is standard. Solving the equation $\lambda^4 + 8 = 0$ is pretty simple; just move the 8 to the other side, and then take the fourth root.
$\lambda^4 = 8^{(1/4)}\omega $
where $\omega^4 = -1$
Note that there are four solutions, corresponding to the four roots of -1.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big)$ is closed under multiplication. Here, we have that $d\equiv 1\bmod4$ and $$\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big):=\Big\{u+v\Big(\frac{1+\sqrt{d}}{2}\Big): u,v\in\Bbb Z\Big\}.$$
Here is my attempt so far:
Suppose $A:=u+v\Big(\frac{1+\sqrt{d}}{2}\Big)$ and $B:=w+x\Big(\frac{1+\sqrt{d}}
{2}\Big)$. Now, we consider the product $$\Big(u+v\Big(\frac{1+\sqrt{d}}{2}\Big)\Big)\cdot \Big(w+x\Big(\frac{1+\sqrt{d}}
{2}\Big)\Big)=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{4}+\frac{vxd}{4}+\frac{vx\sqrt{d}}{4}.$$
Now, write $d=4k+1$ for some $k\in\Bbb Z$, then
$$AB=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{4}+\frac{vx(4k+1)}{4}+\frac{vx\sqrt{d}}{4}\\
=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{4}+\frac{4vxk+vx}{4}+\frac{vx\sqrt{d}}{4}\\
=uw+ux\Big(\frac{1+\sqrt{d}}
{2}\Big)+\frac{vx}{2}+vxk+\frac{vx\sqrt{d}}{4}.$$
But this is where I'm stuck. Pointer on where to carry on please?
| \begin{eqnarray*}
\left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+(ux+vw) \left(\frac{1+ \sqrt{d}}{2}\right)+ vx \left(\frac{1+d +2\sqrt{d}}{4}\right)
\end{eqnarray*}
now use $d=4k+1$
\begin{eqnarray*}
\left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+kvx +(ux+vw+vx) \left(\frac{1+ \sqrt{d}}{2}\right).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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A faster way to evaluate $\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt$? I need to evaluate the integral
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt\tag1$$
After some workarounds I found the change of variable $t=2\sqrt{x^2-1}$, then
$$\int_1^\infty\frac{2\sqrt{1+(t/2)^2}}{t^3}\,\mathrm dt=\frac12\int_{\sqrt5/2}^\infty\frac{x^2}{(x^2-1)^2}\,\mathrm dx\\=\frac12\left[\frac{x}{2(1-x^2)}\bigg|_{\sqrt5/2}^\infty+\frac12\int_{\sqrt5/2}^\infty\frac{\mathrm dx}{x^2-1}\right]\\=\frac{\sqrt5}2+\frac18\int_{\sqrt5/2}^\infty\left(\frac1{x-1}-\frac1{x+1}\right)\,\mathrm dx\\=\frac{\sqrt5}2+\frac18\ln\left(\frac{\sqrt 5+2}{\sqrt 5-2}\right)\\=\frac{\sqrt5}2+\frac14\ln(\sqrt5+2)$$
But my intuition says that it must exists a more straightforward way to evaluate this integral. In fact, using Wolfram Mathematica, I get the equivalent[*] result
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt=\frac14(2\sqrt5+\operatorname{arsinh}(2))$$
[*] The equivalence can be seen from
$$\operatorname{arsinh}(x)=\ln(x+\sqrt{1+x^2})$$
My question: someone knows a faster way to evaluate manually this integral? Maybe a better change of variable?
| Enforcing a substitution of $x \mapsto 1/x$ we have
$$I = \int_0^1 \sqrt{4 x^2 + 1} \, dx.$$
This integral can be readily found using a hyperbolic substitution of $x \mapsto \frac{1}{2} \sinh x$. Doing so yields
\begin{align*}
I &= \frac{1}{2} \int_0^{\sinh^{-1} (2)} \cosh^2 x \, dx\\
&= \frac{1}{4} \int_0^{\sinh^{-1} (2)} [\cosh (2x) + 1] \, du \tag1\\
&= \frac{1}{4} \left [\frac{1}{2} \sinh (2x) + x \right ]_0^{\sinh^{-1} (2)}\\
&= \frac{1}{8} \sinh [2\sinh^{-1} (2)] + \frac{1}{4} \sinh^{-1} (2)\\
&= \frac{\sqrt{5}}{2} + \frac{1}{4} \sinh^{-1} (2) \tag2
\end{align*}
Explanation
(1) Using $\cosh^2 x = \frac{1}{2} (\cosh (2x) + 1)$
(2) Using $\sinh (2 \alpha) = 2 \sinh \alpha \cosh \alpha$ where $\alpha = \sinh^{-1} (2)$ such that $\sinh \alpha = 2$ and $\cosh \alpha = \sqrt{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$
find the :
$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$$
My Try :
$$\dfrac{1}{1+\dfrac{\sin 70^{\circ}}{\cos 70^{\circ}}}+\dfrac{1}{1+\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}}$$
$$\dfrac{\cos70^{\circ}}{\cos 70^{\circ}+\sin 70^{\circ}}+\dfrac{\cos20^{\circ}}{\cos 20^{\circ}+\sin 20^{\circ}}$$
now what do i do ?
| Let $x=20^{\circ}$
\begin{eqnarray}\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}&=&\dfrac{1}{1+\cot 20^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}\\&=& \dfrac{\sin x}{\sin x+\cos x}+\dfrac{\cos x}{\cos x+\sin x}\\ &=&1\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find $\lim_{n \to \infty} (a_n -\sqrt{n})$
Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find
$\displaystyle \lim_{n \to \infty} (a_n -\sqrt{n}).$
My Try:
$$\lim_n( \sqrt{a_{n}+n}-\sqrt{n}) \times \dfrac{( \sqrt{a_{n}+n}+\sqrt{n}) }{( \sqrt{a_{n}+n}+\sqrt{n})}=\lim_n \dfrac{1}{( \sqrt{\dfrac{1}{a_{n}}+\dfrac{n}{a_{n}^2}}+\sqrt{\dfrac{n}{a_{n}^2}})}.$$
Now what do I do?
| Let $b_n=a_n-\sqrt n$ therefore:$$b_{n+1}=\sqrt{b_n+\sqrt n+n}-\sqrt{n+1}=\dfrac{b_n+\sqrt n-1}{\sqrt{b_n+\sqrt n+n}+\sqrt{n+1}}=\dfrac{\dfrac{b_n-1}{\sqrt n}+1}{\sqrt{\dfrac{n+1}{n}}+\sqrt{\dfrac{b_n}{n}+\dfrac{1}{\sqrt n}+1}}\qquad\qquad\qquad(1)$$if we assume $\lim_{n\to\infty}b_n=l<\infty$ we have:$$l=\dfrac{1}{2}$$by substituting in $(1)$ and $n\to\infty$, but this is not a proof! We have now an intuition and a probable limit whether true or not. Instead here after we define $\epsilon_n=b_n-\frac{1}{2}$ and try to prove that $\lim_{n\to\infty}\epsilon_n=0$. By substituting $b_n=\epsilon_n+\frac{1}{2}$ in $(1)$ we obtain:$$\large\epsilon_{n+1}=\dfrac{\epsilon_n-\frac{1}{2}-\frac{1}{\sqrt n+\sqrt{n+1}}}{\sqrt{\epsilon_n+(\sqrt n+\frac{1}{2})^2+\frac{1}{4}}+\frac{1}{2}+\sqrt{n+1}}$$by taking absolute and applying some inequalities we have:
$$\large|\epsilon_{n+1}|=\left |\dfrac{\epsilon_n-\frac{1}{2}-\frac{1}{\sqrt n+\sqrt{n+1}}}{\sqrt{\epsilon_n+(\sqrt n+\frac{1}{2})^2+\frac{1}{4}}+\frac{1}{2}+\sqrt{n+1}}\right |\le
\dfrac{\left |\epsilon_n-\frac{1}{2}-\frac{1}{\sqrt n+\sqrt{n+1}}\right |}{\left |\frac{1}{2}+\sqrt{n+1}\right |}\le\dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\qquad\qquad\qquad (2)$$So far we have bounded $|\epsilon_{n+1}|$. Now we need a trick: we know that $|\epsilon_2|=|b_2-\frac{1}{2}|=|a_2-\sqrt 2-\frac{1}{2}|=\frac{1}{2}\le 1$ so for any $n\ge 2$:$$|\epsilon_{n+1}|\le \dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\le \dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{2}\le\dfrac{|\epsilon_n|+1}{2}\le 1$$this show that for $n\ge 2$ we have $|\epsilon|\le 1$. Now substituting this in $(2)$ gives us:$$|\epsilon_{n+1}|\le \dfrac{|\epsilon_n|+\frac{1}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\le \dfrac{\frac{3}{2}+\frac{1}{\sqrt n+\sqrt{n+1}}}{\frac{1}{2}+\sqrt{n+1}}\le\dfrac{4}{1+2\sqrt{n+1}}$$since $\lim_{n\to\infty}\dfrac{4}{1+2\sqrt{n+1}}=0$ so $\lim_{n\to\infty}|\epsilon_n|=0$ and this is what we wanted from beginning. Therefore:$$\Large\lim_{n\to\infty}a_n-\sqrt{n}=\frac{1}{2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| Take $u=x^{\frac{1}{5}}$ and $v=x^{\frac{1}{6}}$.
$\displaystyle u-1=\frac{u^5-1}{u^4+u^3+u^2+u+1}=\frac{x-1}{u^4+u^3+u^2+u+1}$
$\displaystyle v-1=\frac{v^6-1}{v^5+v^4+v^3+v^2+v+1}=\frac{x-1}{v^5+v^4+v^3+v^2+v+1}$
$\displaystyle \frac{u-1}{v-1}=\frac{v^5+v^4+v^3+v^2+v+1}{u^4+u^3+u^2+u+1}\to\frac{6}{5}$ as $x\to 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 6
} |
How do I find the quotient of the division of $x^{1993}+x^2+1$ by $x^2+x+1$? I have an exercise where I need to find the quotient of the division of polynomial $f=x^{1993}+x^2+1$ by $g=x^2+x+1$. I know to do it the long way but I think there is a trick to do it fast. By the way, I know the remainder is $0$.
| Just wrote $$X^{1993}+X^2+1=X^{1993}-X+X^2+X+1=X\left(X^{1992}-1\right)+X^2+X+1.$$
Since $$X^{1992}-1=(X^3-1)(X^{1989}+...+X^3+1)=(X-1)(X^2+X+1)(X^{1989}+...X^3+1),$$
we see that $X^{1993}+X^2+1$ is divisible by $X^2+X+1$.
The quotient it's $$X(X-1)(X^{1989}+...X^3+1)+1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factor $2x^3 + 21x^2 +27x$ I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks!
| $$2x^3+21x^2+27x\\=x(2x^2+21x+27)\\=\dfrac{x}{2}(4x^2+42x+54)\\=\dfrac{x}{2}(2x+18)(2x+3)\\=x(x+9)(2x+3)$$for factorizing $4x^2+42x+54$ we know that it must in form of $(2x+a)(2x+b)$. So:$$4x^2+42x+54=(2x+a)(2x+b)=4x^2+(a+b)2x+ab$$which implies$$ab=54\\a+b=21$$clearly the only numbers satisfying those equalities are $a=18$ and $b=3$ or vice versa.
Another way to attain such numbers when guessing them is not that easy is to solve the following equation$$4x^2+42x+54=0$$or$$2x^2+21x+27=0$$which leads to $$x=\dfrac{-21\pm\sqrt{{21^2-4\times 2\times 27}}}{4}=\dfrac{-21\pm 15}{4}$$therefore$$x_1=\dfrac{-21-15}{4}=-9\\x_2=-\dfrac{3}{2}$$so$$2x^2+21x+27=2(x+9)(x+\dfrac{3}{2})=(x+9)(2x+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find extremal values of $f(x,y)=x^2+4y^2$ on the set $D$. Find abs.min and max values of $f(x,y)=x^2+4y^2$ on the set $D=\{(x,y) \in \mathbb{R}^2; x^2 + (y+1)^2 \leq 4, y \geq -1, y \leq x+1\}$.
For the interior, it is as usually $f_x=0$ and $f_y=0$, and I got $(0,0)$ as a critical point, and $f(0,0) =0$.
For the boundary, we go along $y=x+1$ and $x \in [-2,0]$. Then $f(x,x+1)=5x^2+8x+4$ and we can see $f'(x)=10x+8$ where we get $10x+8=0 \iff x= \frac{-4}{5}$. Evaluating $f$ at the endpoints and critical points we get rel.min and max at points $$f(\frac{-4}{5},\frac{1}{5})=\frac{4}{5}, f(-2,-1)=8,f(0,1)=4.$$
Similarly we go along $y=-1$ and $x \in [-2,2]$, where $f(x,-1)=x^2+4$ and $f'(x)=2x$ and so $f'(x)=0 \iff x=0$. Evaluating $f$ we get $$f(-2,-1)=8,f(2,-1)=8,f(0,-1)=4.$$
Finally we go along $y=\sqrt{x^2-4}-1$ where $x \in[0,2]$. Then in a similar fashion $f(x,\sqrt{x^2-4}-1)=5x^2-12-8\sqrt{x^2-4}$ and its derivative equals zero iff $x=\pm \sqrt \frac{116}{25}$ and this is approximately 2.15 which will not be in the interval $[0,2]$. Thus maximum value is $8$ while the minimum is $0$. Can someone let me know if this approach is correct, or if I should use some other method?
Update: in the last step, we go along $y=\sqrt{4-x^2}-1$ where $x \in[0,2]$. Then in a similar fashion $f(x,\sqrt{4-x^2}-1)=5x^2+20-8\sqrt{x^2-4}$ and its derivative equals zero iff $x=\pm \sqrt \frac{84}{25}$. Now since $x \in[0,2]$ the value $x=- \frac{\sqrt{84}}{5}$ is not in the interval so we have only check $f(\frac{\sqrt{84}}{5},-\frac{1}{5})$.
| Your approach seems to be correct. However, note that you are able to in fact "talk" about extremum values since the function $f(x,y)$ is continuous on a compact set $D$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $n$th order determinant Solve $n$th order determinant:
$$
\begin{pmatrix}
1 & 2 & 3 & \cdots & n-2 & n-1 &n\\
2 & 3 & 4 & \cdots & n-1& n& 1\\
3 & 4 & 5 & \cdots & n &1 &2\\
\vdots& \vdots& \vdots& \ddots& \vdots& \vdots&\vdots\\
n-2 &n-1&n &\cdots &n-5 &n-4 &n-3\\
n-1 &n&1 &\cdots &n-4 &n-3 &n-2\\
n &1&2 &\cdots &n-3 &n-2 &n-1\\
\end{pmatrix}
$$
I tried adding $2$nd to $n$th row to the $1$st row, then the first row is:
$$
\begin{pmatrix}
\frac{n(n+1)}{2} &\frac{n(n+1)}{2} \cdots \frac{n(n+1)}{2}\\
\end{pmatrix}
$$
then I took out $\frac{n(n+1)}{2}$ from the first row to get
$$
\begin{pmatrix}
1 & 1 & \cdots& 1\\
\end{pmatrix}
$$
and then tried to work with that but I don't think this is the correct approach because anything I tried doesn't look nice.
| With the help of WA, we get this sequence
$$
1, -3, -18, 160, 1875, -27216, -470596, \dots
$$
oeis/A052182 tells us that the determinant is $(-1)^{n-1}n^{n-2}\dfrac{(n^2 + n)}{2}$, except for the sign.
| {
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"url": "https://math.stackexchange.com/questions/2629389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$
Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$
So basically one can evaluate all $a^2, 5b^3 \pmod {7}$ and look for all $(a,b) \in \mathbb Z_7\times \mathbb Z _7$ such that $a^2 \equiv 5b^3 \pmod 7$
Two questions:
*
*Is that the "best" way doing it?
*If I counted correctly there are exactly $7$ solutions. Is that a coincidence it equals to the modulus?
Thanks!
| If $7\mid x\iff7|y$
else $7\nmid xy$
In that case as $\phi(7)=6, x^3\equiv\pm1\pmod7$
If $x^3\equiv1,y^2\equiv5$
Now as $y\equiv\pm1,\pm2,\pm3\pmod7, y^2\equiv1,4,9\equiv2$
Hence, $y^2\not\equiv5$
If $x^3\equiv-1,y^2\equiv-5\equiv2\equiv3^2\implies y\equiv\pm3$
Now $x^3\equiv1\implies x\equiv3,5\equiv-2,6\equiv-1\pmod7$
| {
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"url": "https://math.stackexchange.com/questions/2629505",
"timestamp": "2023-03-29T00:00:00",
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$9^x=5$, solve for $5\cdot27^{-x-1}$ Question:
Given $9^x=5$
Compute the value of $5\cdot27^{-x-1}$
My attempt:
$9^x=(3^x)^2=5\Rightarrow3^x=\sqrt{5}$
$5\cdot27^{-x-1}=3^{2x}\cdot(3^{3-x}\div3^{3})=3^{2x}\cdot3^{-x}=3^{x}=\sqrt{5}$
My answer is incorrect and I wonder why.
| If you want to use your result
$$5.27^{-x-1}=\frac{9^x}{27.27^{x}}=\frac{1}{27}\frac{1}{3^x}=\frac{1}{27 \sqrt{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\int \frac{dx}{(a^2 + x^2)^2}$? I'm trying to use a trig substitution but I'm stuck. Here's what I did so far:
$$\int \frac{dx}{(a^2 + x^2)^2}$$
Let $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+sin^2\theta))^2} $$
$$\int \frac{a\cos \theta d\theta}{(a^2 \cos^2\theta)^2} =\int \frac{d\theta}{a^3cos^3\theta} $$
I don't know what to do anymore
| You can use that
$$
\int\frac{\text{d}x}{a^2+x^2}=\frac{1}{a}\text{arctan}\left(\frac{x}{a}\right)+K
$$
And then integrating by parts gives you
$$
\int\frac{\text{d}x}{a^2+x^2}=\left[\frac{x}{a^2+x^2}\right]+\int \frac{2x^2}{\left(a^2+x^2\right)^2}\text{d}x$$
$$=\left[\frac{x}{a^2+x^2}\right]+2\int \frac{\text{d}x}{a^2+x^2}-2a^2\int \frac{\text{d}x}{\left(a^2+x^2\right)^2}
$$
Then
$$
2a^3\int \frac{\text{d}x}{\left(a^2+x^2\right)^2}=\left[\frac{ax}{a^2+x^2}\right]+\text{arctan}\left(\frac{x}{a}\right)
$$
You finally obtain
$$
\int \frac{\text{d}x}{\left(a^2+x^2\right)^2}=\frac{1}{2a^3}\left(\frac{ax}{a^2+x^2}+\text{arctan}\left(\frac{x}{a}\right)\right)+C
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Expressing complex numbers in exponential form I do not know why I am having so many issues with complex numbers. I am basically trying to teach myself, but keep doubting myself and getting very frustrated. I've no way of checking my answers so I'm just lost. So I want to write the following in polar form $re^{i\theta}$ (although I thought that was called exponential form?) and $-\pi < \theta \leq \pi$.
My first question is expressing $(\cos(\frac{2\pi}{9}) + i\sin(\frac{2\pi}{9}))^3$. Using De Moivre, this is $$\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})$$ so it is $e^{i(\frac{2\pi}{3})}$ yes?
My next one is to express $\frac{2+2i}{-\sqrt{3} +i}$. So my attempt is to multiply by the complex conjugate and we get that this is $\frac{-\sqrt{3} +1}{2} + i\frac{-1 - \sqrt{3}}{2}$. Find $r$ which is the modulus which is $\sqrt2$ and $\theta$ is just something I cannot seem to find. I know if $z = a+ib$ then $\theta = \tan^{-1}(\frac{b}{a})$ but I'm stuck even with the $(\frac{b}{a})$ and am confusing myself of what quadrant to look in.
Lastly, I have no idea how to express the next one which is $\frac{4i}{3e^{(4+i)}}$.
Any help and direction at all is appreciated.
| $\frac{2+2i}{-\sqrt{3} +i} =$
$\frac {(2+2i)(-\sqrt{3} -i)}{(-\sqrt{3} +i)(-\sqrt{3} -i)}=$
$\frac {(-2\sqrt{3} +2)- (2\sqrt 3+2)i}{3+1}=$
$\frac {-\sqrt3 + 1}2 -\frac{\sqrt 3 + 1}2i=$
$r(\cos \theta + i\sin \theta)$
So we need to solve for $r\cos \theta = \frac {-\sqrt3 + 1}2$ and $r\sin \theta = -\frac{\sqrt 3 + 1}2$.
That's all.
We solve for $r$ by squaring and adding both terms:
$r^2\cos^2 \theta + r^2 \sin^2 \theta = (\frac {-\sqrt3 + 1}2)^2 + ( -\frac{\sqrt 3 + 1}2)^2$
$r^2(\cos^2 \theta + \sin^2 \theta) = \frac {3 - 2\sqrt 3 + 1}4 + \frac {3+2\sqrt 3 + 1}4$.
$r^2 = 1 - \frac {\sqrt 3}2 + 1+\frac {\sqrt 3}2$
$r^2 = 2$
$r = \sqrt 2$.
And we solve for $\theta$ by dividing both terms and taking the arctangent:
$\tan \theta = \frac {\sin\theta}{\cos\theta} = \frac {r*\sin\theta}{r*\cos\theta} = \frac {-\frac{\sqrt 3 + 1}2}{\frac {-\sqrt3 + 1}2}$
$=\frac {\sqrt 3 + 1}{-\sqrt3 + 1} =\frac {(\sqrt 3 + 1)(-\sqrt 3- 1}{(-\sqrt3 + 1)}=\frac {-4 - 2\sqrt 3}{2} = -2-\sqrt 3$
So $\theta = \arctan -2-\sqrt 3= -\frac 5{12}\pi$
S $\frac{2+2i}{-\sqrt{3} +i} = \sqrt 2*e^{-\frac 5{12}\pi i}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A fair six-sided die is rolled repeatedly I need help with this homework problem.
A fair six-sided die is rolled repeatedly. Each time the die is rolled, the number showing is written down. Let $X$ be the number of rolls until some number shows twice. The possible values of $X$ are $2,3,4,5,6,$ or $7$. For each of the following parts, explain your reasoning and express your answer as a fraction in reduced form.
Here is the solution I tried to derive.
*
*$\Pr(3) = \dfrac{30}{216}$
$$6 \times (6-1) \times 1 = 30$$ $$6^3 = 216$$
*
*$\Pr(4) = \dfrac{120}{1296}$
$$6 \times (6-1) \times (6-2) \times 1 = 120$$
$$6^4=1926$$
*
*$\Pr(5) = \dfrac{360}{7776}$
$$6 \times (6-1) \times (6-2) \times (6-3) \times 1 = 360$$
$$6^5=7776$$
Can someone help me out please? Thank
| You are on the right track, but the last roll of the die could be any of the numbers already rolled. Then
$$\text{Pr}(2)=\frac66\times\frac16=\frac{6}{36}=\frac{1}{6}$$
$$\text{Pr}(3)=\frac66\times\frac{(6-1)}{6}\times\frac26=\frac{60}{216}=\frac{5}{18}$$
$$\text{Pr}(4)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac36=\frac{360}{1296}=\frac{5}{18}$$
$$\text{Pr}(5)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac{(6-3)}{6}\times\frac46=\frac{1440}{7776}=\frac{5}{27}$$
$$\text{Pr}(6)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac{(6-3)}{6}\times\frac{(6-4)}{6}\times\frac56=\frac{3600}{46656}=\frac{25}{324}$$
$$\text{Pr}(7)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}\times\frac{(6-3)}{6}\times\frac{(6-4)}{6}\times\frac{(6-5)}{6}\times\frac66=\frac{4320}{279936}=\frac{5}{324}$$
Note that $\text{Pr}(2) + \text{Pr}(3) + \dots + \text{Pr}(7)=1$
Hope it help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638788",
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Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$ Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$
The only idea I have is that I could apply Cauchy-Schwartz but i don't see how, any hints?
| Using coordinate geometry
The perpendicular distance of $2x+4y=1$ from the origin is given by
$$d=\frac{1}{\sqrt{2^2+4^2}}=\frac{1}{\sqrt{20}}$$
By $\sqrt{x^2+y^2} \ge d$ gives
$$x^2+y^2 \ge \frac{1}{20}$$
| {
"language": "en",
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"source": "stackexchange",
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ODE Substitution Problem Have some practice exam questions and answers to work through, I'm almost there but get stuck.
what we have is:
$$
\begin{cases}
\dfrac{dx}{dt}=\dfrac{x^{3}-xt^{2}}{t^{3}} \\
x(1)=2 \\
\end{cases}
$$
and the (general solution) answer is:
$x(t) = \pm t\sqrt{\frac{2}{1-Dt^{4}}}$
now I figured that this is a substitution problem, using $y=\frac{x}{t}$ so that
$\frac{dx}{dt}=y^{3}-y$
and
$\frac{dy}{dt}=y+t\frac{dy}{dy}$
giving:
$y^{3}-2y = t\frac{dy}{dt}$
$\int{\frac{1}{y^3-2y}dy}=\int{\frac{1}{t}dt}$
$\ln{y^{3}-2y}=\ln{t}+C$
$y^{3}-2y=Dt$
and therefore:
$\frac{x^{3}}{t^{3}}-2\frac{x}{t}=Dt$
$x^3-2xt^{2}=Dt^{4}$
>
Here is where I get stuck... so I reverse engineered the solution to find that:
$x^{3}-2xt^{2}=x^{3}Dt^{4}$
and am now completely lost... where did that extra $x^{3}$ on the RHS of the equation come from??
| Assume that $\frac{dy}{y^3-2y}=\frac{dt}{t}$ is done alright. The anti derivative of $\frac{1}{t}dt$ is indeed $ln|t|$, however the anti derivative of $\frac{dy}{y^3-2y}$ cannot be $ln|y^3-2y|$ for the very simple reason, the derivative of $ln|y^3-2y|$ is not $\frac{1}{y^3-2y}$, in fact it is $\frac{3y^2-2}{y^3-2y}$ by Chain Rule!. So how to do $\int\frac{dy}{y^3-2y}$? The generic answer is using partial fraction decomposition (although this one could be done in a different way as well, see end of post).
So setting up the fractions (through factoring): $\frac{1}{y^3-2y}=\frac{A}{y}+\frac{B}{y-\sqrt{2}}+\frac{C}{y+\sqrt{2}}$. Multiply through by $y^3-2y$ and equating each equation with corresponding coefficients (which is 0,0,1 respectively from the numerator) gives $A+B+C=0,B\sqrt{2}-C\sqrt{2}=0,-2A=1$ results in $A=-0.5,B=1/4,C=1/4$ and you can now integrate $\int\frac{-0.5}{y}$+$\int\frac{0.25}{y-\sqrt{2}}$+$\int\frac{0.25}{y+\sqrt{2}}$$dy$. And done. Now in THIS (I repeat THIS!) particular case, you could have subbed in $\int\frac{dy}{y^3-2y}$the substitution $y=\frac{1}{v}$ and your integral would have become $\int\frac{-vdv}{1-2v^2}$ (verify!) which straight away would have become an $ln$ term. I will now leave it up to you to work out the rudiments. Good luck!
| {
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Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Please let me know how to approach this question, and if you attach full explanation, I will appreciate it. Thanks.
| To my surprise, this problem can be solved using geometry.
Identify the Euclidean plane with complex plane $\mathbb{C}$.
and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity.
Consider triangle $\triangle ABC$ with vertices at $A = x$, $B = y\omega$ and $C = z\omega^2$.
The sides of the triangle equal to
$$\begin{align}
a^2 = BC^2 &= | y\omega - z\omega^2|^2 = y^2 + yz+ z^2 = 1\\
b^2 = AC^2 &= | x - z\omega^2|^2 = x^2 + xz + z^2 = 4\\
c^2 = AB^2 &= | x - y\omega|^2 = x^2+xy+y^2 = 3
\end{align}
$$
Since $a^2 + c^2 = b^2$, this is a right angled triangle with area $\mathcal{A} = \frac12 ac
=\frac{\sqrt{3}}{2}$.
An alternate way to compute the area is cut the triangle into 3 pieces along the line $OA$, $OB$ and $OC$. This gives us
$$\mathcal{A}= \frac12 (xy + yz + xz)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4} (xy+yz+xz)$$
Combine these two results, we find:
$$xy+yz+xz = 2$$
Update
For a pure algebraic answer, one can substitute above expression of $a^2,b^2,c^2$ into Heron's formula for area of triangle, $$\mathcal{A} = \frac14\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$
simplify and obtain following algebraic identity
$$\begin{array}{rl}
3(xy+yz+zx)^2 =& \phantom{+0}((x^2+xz+y^2) + (y^2+yz+z^2) + (z^2+xz+x^2))^2\\
&-2((x^2+xz+y^2)^2 + (y^2+yz+z^2)^2 + (z^2+xz+x^2)^2)
\end{array}
$$
Using given values of $a, b, c$, we find
$$3(xy+yz+xz)^2 = (3+4+1)^2 - 2(3^2+4^2+1^2) = 64 -2(26) = 12 = 3(2)^2$$
This leads to the same conclusion $xy+yz+xz = 2$ as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
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Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
\end{align}
I'm a bit stuck on how to simplify from here? Was there a simpler way somewhere?
Also this one is giving me problems. I am thinking of changing the log forms first?:
$$\ln \frac{(2y+1)^5}{\sqrt{y^2 + 1}}= \ln(2y+1)^5 - \ln \sqrt{y^2 +1}=5 \ln(2y+1) - \frac{1}{2}\,\ln (y^2 +1)$$
so:
$$G'y = \frac{5}{2y+1} \cdot 2 - \frac{2y}{2(y^2 + 1)} = \frac{10}{2y+1} - \frac{y}{2 (y^2 +1)}$$
Is that right?
| $$\frac{1}{x + \sqrt{x^2 - 1}} \cdot (1 + \frac{2x}{2 \cdot \sqrt{x^2 - 1}})$$
$$=\frac{1}{x + \sqrt{x^2 - 1}}\frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}}$$
$$=\frac{1}{\sqrt{x^2 - 1}}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks like
$$
a - b = \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}},
$$
where $ a = \sqrt[3]{(n^{3}+n)} $ and $ b = \sqrt[3]{(n^3-n)} $.
It gave me
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} = \sum_{n=1}^{\infty} \frac{2n}{\sqrt[3]{(n^{3}+n)^{2}} + \sqrt[3]{n^{6} - n^{2}} + \sqrt[3]{(n^3-n)^{2}}}
$$
but for me it looks like a blind valley. I would appreciate every help, thank you.
| Using asymptotic calculus:
$$\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=n\left(\sqrt[3]{1+\frac{1}{n^2}}-\sqrt[3]{1-\frac{1}{n^2}}\right)=n\left(\left(1+\frac{1}{n^2}\right)^{1/3}-\left(1-\frac{1}{n^2}\right)^{1/3}\right)=n\left(\left(1+\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)\right)-\left(1-\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)\right)\right)=n\left(\frac{2}{3n^2}+o\left(\frac{1}{n^2}\right)\right)=\frac{2}{3n}+o\left(\frac{1}{n}\right)\sim\frac{2}{3n} (n\to\infty)$$
thus $\sum_{n=1}^{\infty}\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}$ diverges because $\sum_{n=1}^{\infty}\frac{2}{3n}$ diverges.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
Use Lagrange multiply to find the local extremes of $f$ Use Lagrange multiply to find the local extremes of $f$
Let $f(x,y)=y^2-4xy+4x^2$ and $x^2+y^2=1$
My work:
Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $f(x,y)=y^2-4xy+4x^2$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $g(x,y)=x^2+y^2-1$
We need define the function of Lagrange
$$L(x,y,\alpha)=f(x,y)+\alpha(g(x,y))$$
Then, calculating the critic point of $L$ we have
$\begin{equation}
-4y+8x+2x\alpha=0\\
2y-4x+2\alpha y=0\\
x^2+y^2-1=0
\end{equation}$
Then, solving the system, we have $x=0,\, \, y=0$
But i have a problem with this. this answer dont have sense because if $x=0$ and $y=0$ then $-1=0$
Can someone help me?
| If you solve the first two equations in your system of three equations for $\alpha$ you get
$$ \alpha=-4+2\frac{y}{x}=-1+2\frac{x}{y} $$
Substituting $u=\dfrac{y}{x}$ into $$ -4+2\frac{y}{x}=-1+2\frac{x}{y} $$
gives the equation
$$ 2u^2-3u-2=0 $$
with solutions $u=-\dfrac{1}{2}$ or $u=2$ leading to the result that $y=-\frac{1}{2}x$ or $y=2x$.
Substitute these into the third equation in your system of three equations and you will find eight solutions for $(x,y)$. Evaluate $f(x,y)=y^2-4xy+4x^2=(y-2x)^2$ at each of these eight solutions and you will find your extrema.
| {
"language": "en",
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If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad) I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:
If
$$x \cos\theta+y\sin\theta=a$$
$$x\sin\theta-y\cos\theta=b$$
then $$\tan\theta=\frac{bx+ay}{ax-by}$$
I've spent many hours trying.
Thanks in advance.
| We first try to find $\sin \theta $ in terms of $a, b, x, y$
Multiply equation first by $y$ and the equation $2^{nd}$ by $x$ and add the resultant equations to get $$\sin \theta= \frac {bx+ay}{x^2+y^2}$$
Now multiply the first equation by $x$ and second equation by $y$ to get following equations
$$x^2\cos \theta+xy\sin \theta=ax$$
And
$$xy\sin\theta-y^2\cos\theta=by$$
Hence we get $$\cos\theta=\frac {ax-by}{x^2+y^2}$$
By subtracting resultant $2^{nd}$ equation from the resultant $1^{st}$ equation.
Using these values of $\sin\theta$ and $\cos\theta$ we get $$\tan\theta = \frac {bx+ay}{ax-by}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let $a,b,c$ form a primitive Pythagorean triple, with $a$ odd. Prove that $(c-b)$ and $(c+b)$ share no common prime factors. Let $a,b,c$ form a primitive Pythagorean triple (meaning $(a,b,c) \in \mathbb{Z}^{3}$). Let $a$ be odd. Use a proof by contradiction to show that $(c-b)$ and $(c+b)$ share no common prime factors.
My Thoughts
So I began by assuming that $(c-b)$ and $(c+b)$ shares at least one common factor.
And so, for some prime $p$, $(c-b) = xp$ and $(c+b) = yp$, such that $(x,y) \in \mathbb{Z}^{2}$
So now, I thought maybe if we set this equal to either $a^2$ or $c^2 - b^2$ I could find a contradiction. But in either case, I couldn't come to any conclusion.
| Since $a$ is odd, $b$ is even . Because if $a=2a'+1$ is odd and if $b=2b'+1$ is also odd then $c^2=(2a'+1)^2+(2b'+1)^2$ has a remainder of $2$ modulo $4,$ which is impossible for a square.
So $c^2$ is odd, so $c$ is odd . So $c+b$ and $c-b$ are both odd.
Let $d$ be a common divisor of $c+b$ and $c-b$. Note that $d$ must be odd.
Then $d$ divides $(c+b)+(c-b)=2c$ and $d$ divides $(c+b)-(c-b)=2b.$ But $d$ is odd, so $d$ must divide $c$ and $b .$
Since $d$ divides $c+b$ and $c-b$ it divides the product $(c+b)(c-b)=a^2.$ But if $d$ is prime and $d$ divides $a^2$ then $d$ divides $a.$ This would make $d$ a common prime divisor of $a,b,c,$ which would make $(a,b,c)$ non-primitive.
| {
"language": "en",
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"source": "stackexchange",
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Compute $\iint_D(-2x-y)\cos((x-y)(-2x-y))dxdy$
Compute $$\iint_D(-2x-y)\cos((x-y)(-2x-y)) \ dxdy$$
Where $D$ has vertices at the points
$$P_1=\left(-1 + \frac{\pi}{6} , 2+\frac{\pi}{6} \right)$$
$$P_2=\left(-2 + \frac{\pi}{6} , 4+\frac{\pi}{6} \right)$$
$$P_3=\left(-2 + \frac{\pi}{4} , 4+\frac{\pi}{4} \right)$$
$$P_4=\left(-1 + \frac{\pi}{4} , 2+\frac{\pi}{4} \right)$$
I think I'm very close, but for the life of me I can't figure why I get the incorrect answer. Here is what I've done:
Plotting the points I get
The short lines, from left to right are $y=x+6$ and $y=x-3$. The long lines are $y=-2x+\pi/2$ and $y=-2x+3\pi/4.$ So I have that
\begin{array}{lcl}
-6 \leq &x-y& \leq 3 \\
-\frac{3\pi}{4} \leq &-2x-y& \leq -\frac{\pi}{2}
\end{array}
Setting $u=x-y$ and $v=-2x-y$ and solving the system I get
$$\left\{
\begin{array}{rcr}
x & = & \frac{1}{3}u-\frac{1}{3}v \\
y & = & -\frac{2}{3}u-\frac{1}{3}v \\
\end{array}
\right.$$
$\implies |J(u,v)|=1/3.$ Thus:
$$\iint_D(-2x-y)\cos((x-y)(-2x-y)) \ dxdy=\iint_Ev\cos{(uv)}|J(u,v)|\ du dv$$
$$= \frac{1}{3}\int_{-6}^{3}v\left(\int_{-3\pi/4}^{-\pi/2}\cos{(uv)} \ du\right) \ dv=\frac{2(3-\sqrt{2})}{9\pi}.$$
Where is the error?
| (Edited in response to OP's comment.)
Exact evaluation of the edited integral
You have correctly set up the relationship between $x,y$ and $u,v$.
\begin{align}
\begin{bmatrix}u \\ v\end{bmatrix} &= \begin{bmatrix}1 & -1 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \\
\begin{bmatrix} x \\ y \end{bmatrix} &= \begin{bmatrix}1 & -1 \\ -2 & -1 \end{bmatrix}^{-1} \begin{bmatrix} u \\ v \end{bmatrix} = \begin{bmatrix}1/3 & -1/3 \\ -2/3 & -1/3 \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix}
\end{align}
$|J(u,v)|=1/3$ is correct.
This major mistake is that you're integrating in much larger domain because there's a sign error in the boundary of $D$. In fact,
\begin{align}
D &= \left\{(x,y) \,\Bigg|\,
-6 \leq x-y \leq \color{red}{-3},
-\frac{3\pi}{4} \leq -2x-y \leq -\frac{\pi}{2}
\right\} \\
&= \left\{(u,v) \mid -6 \le u \le -3, -\frac{3\pi}{4} \leq v \leq -\frac{\pi}{2} \right\} \\
&\quad \iint_D(-2x-y)\cos((x-y)(-2x-y)) \ dxdy \\
&= \iint_D v\cos(uv) |J(u,v)| \ du dv \\
&= \int_{-3\pi/4}^{-\pi/2} \left( \int_{-6}^{-3} \frac13 v\cos(uv) \ du \right) dv \\
&= \int_{-3\pi/4}^{-\pi/2} \frac13 \Big[ \sin(uv) \Big]_{u=-6}^{u=-3} \ dv \\
&= \frac13 \int_{-3\pi/4}^{-\pi/2} (\sin(6v) - \sin(3v)) \ dv \\
&= \frac13 \Big[ \frac{\cos(3v)}{3} - \frac{\cos(6v)}{6} \Big]_{-3\pi/4}^{-\pi/2} \\
&= \frac{1}{18} \Big[ 2\cos(3v) - \cos(6v) \Big]_{-3\pi/4}^{-\pi/2} \\
&= \frac{1}{18} \Big[ -2 \cos \left( 3 \cdot \frac{3\pi}{4} \right) - \cos \left( 6 \cdot \frac\pi2 \right) \Big] \\
&= -\frac{\sqrt2-1}{18}
\end{align}
Exact evaluation of the original integral
\begin{align}
D &= \left\{(x,y) \,\Bigg|\,
-6 \leq x-y \leq \color{red}{-3},
-\frac{3\pi}{4} \leq -2x-y \leq -\frac{\pi}{2}
\right\} \\
&= \left\{(u,v) \mid -6 \le u \le -3, -\frac{3\pi}{4} \leq v \leq -\frac{\pi}{2} \right\} \\
&\quad \iint_D(-2x-y)\cos((-x-y)(-2x-y)) \ dxdy \\
&= \iint_D v\cos(\color{red}{(\frac{u}{3} + \frac{2v}{3})}v) |J(u,v)| \ du dv \\
&= \int_{-3\pi/4}^{-\pi/2} \left( \int_{-6}^{-3} \frac13 v\cos((\frac{u}{3} + \frac{2v}{3})v) \ du \right) dv \\
&= \int_{-3\pi/4}^{-\pi/2} \left[ \sin((\frac{u}{3} + \frac{2v}{3})v) \right]_{u=-6}^{u=-3} \ dv \\
&= \int_{-3\pi/4}^{-\pi/2} \left( \sin((\frac{2v}{3}-1)v) - \sin((\frac{2v}{3}-2)v) \right) \ dv
\end{align}
Note that you make a substitution error when rewriting $\cos(\dots)$.
Compare the results from Wolfram Alpha
| {
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Prove $15\mid 4^{2n}-1$ without induction
Prove without induction that $\forall n \in\mathbb Z$, $15\mid4^{2n}-1$.
$4^{2n} = (4^2)^n = 16^n$.
If $n=1,$ then $(4^2)^n-1=15$, and for $n=2$, it is $255$, which is divisible by $15$.
Using the congruence arithmetic, any $n$ in $4^{2n}-1$ can be expressed as a product of prime factors. Also, there is only one even prime, and any odd prime will be having for any positive integer $k$, an addition of $1+ k.$(even terms).
Say, for $n=5$, with $5=2^2+1$.
And it already proved that any even power of $n$ leads $4^{2n}-1$ to be a multiple of $15.$
edit Based on the selected answer, have observations, that are possibly implied in that answer, but not explicitly stated.
$4^{2n} -1 = (2^n -1)(2^n+1)(2^{2n}+1) = (2^n -1)(2^n+1)(4^n+1)$
Need consider both even and odd cases separately, for values w.r.t. two modulo $3,5$.
(A) $n$ is even :
(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv 1 \pmod 3 \implies (2^n-1) \equiv 0 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 2 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$
Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv 0 \pmod 3)(\equiv 2 \pmod 3)(\equiv 1 \pmod 3)$
As the first term of the product is a factor of $3$, so divisible by $3$.
(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: Take $n = 2n', 2^2 \equiv -1 \pmod 5\implies$$ 4^{n'} \equiv 1 \pmod 5 $ $\implies (2^n-1) \equiv 0 \pmod 5$
(b) $ 2^n+1$ :: Take $n = 2{n'}, 4^{n'} +1 \equiv 2 \pmod 5$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n \equiv 1 \pmod 5$
Finding the product of the three terms w.r.t. modulo $5$, is :
$(\equiv 0 \pmod 5)(\equiv 2 \pmod 5)(\equiv 1 \pmod 5)$
As the first term of the product is a factor of $5$, so divisible by $5$.
(B) $n$ is odd :
(i) check w.r.t. modulo $3$
(a) $ 2^n-1 :: 2\equiv -1 \pmod 3 \implies 2^n \equiv -1 \pmod 3 \implies (2^n-1) \equiv -2 \pmod 3$
(b) $ 2^n+1 :: 2^n +1 \equiv 0 \pmod 3$
(c) $ 4^n+1 :: 4 \equiv 1 \pmod 3 \implies 4^n \equiv 1 \pmod 3$
Finding the product of the three terms w.r.t. modulo $3$, is :
$(\equiv -2 \pmod 3)(\equiv 0 \pmod 3)(\equiv 1 \pmod 3)$
As middle term of the product is a factor of $3$, so divisible by $3$.
(ii) check w.r.t. modulo $5$
(a) $ 2^n-1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(b) $ 2^n+1$ :: $2 \equiv 2 \pmod 5\implies 2^n \equiv ........$
(c) $ 4^n+1 :: 4 \equiv -1 \pmod 5 \implies 4^n +1\equiv 0 \pmod 5$
Finding the product of the three terms w.r.t. modulo $5$, is :
$(....)(...)(\equiv 0 \pmod 5)$
As the last term of the product is a factor of $5$, so divisible by $5$.
Don't know how to handle the last case's , sub cases (a), (b) for checking w.r.t. modulo $5$.
| note that $$4^{2n}-1=(2^n-1)(2^n+1)(2^{2n}+1)$$
also possible: $$4^{2n}=16^n$$ and $$16\equiv 1 \mod 15$$ so $$16^n\equiv 1^n\equiv 1 \mod 15$$therefore $$16^n-1\equiv 0 \mod 15$$ so no factorization needed
| {
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"question_score": "2",
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} |
Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,120,240,300$ ? How do I find 240 and 300?
| hint: $\sin x = -\dfrac{\sqrt{3}}{2}$ would lead to the other two angles.
| {
"language": "en",
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How to win in this game You select 10 numbers from the set $\{2,3,\dots,12\}$.
You then continually roll 2 fair dice and sum them up, until your selection of 10 numbers come up.
For example if your selection was 7,7,7,7,8,8,8,6,6,6 (4 7's, 3 8's and 3 6's), and you roll the dice repeatedly and get 7,7,6,5,8,7,7,9,8,3,5,10,12,6,6,3,2,5,7,9,8, you may stop now because 4 7's have come up, 3 8's and 3 6's.
What is the best choice of 10 numbers so as to minimise the number of rolls ?
| Using a brute-force exhaustive search, and a recursion for each case, I get the following answer:
There are two equivalent optimal strategies, namely
$$4,5,6,6,7,7,7,8,8,9$$
$$5,6,6,7,7,7,8,8,9,10$$
yielding, for the expected number of rounds,
$$
e=\frac{a}{b} \approx 28.26676327
$$
where
$$
\begin{align*}
a&=71526610479792733682076713232552201067\\[4pt]
b&=2530413892848114144358747803518976000\\[4pt]
\end{align*}
$$
Remarks:
Intuitively, the multiplicities of the selected numbers should be in proportion to their probability of occurring in a single round.
Thus, for example, for the simpler game where you roll a single die, and get to guess $6$ numbers, the optimal selection is $1,2,3,4,5,6$.
If in each round you roll two dice, and get to guess $36$ numbers, I suspect the optimal selection is
$$2,
3,3,
4,4,4,
5,5,5,5,
6,6,6,6,6,
7,7,7,7,7,7,
8,8,8,8,8,
9,9,9,9,
10,10,10,
11,11,
12
$$
For the game in question, where in each round you roll two dice, and get to guess $10$ numbers, then, since the multiplicities for the selected values must be nonnegative integers, it's not possible for the multiplicities to be in proportion to the associated probabilities. Yet, it's intuitive that they should be approximately in those proportions.
| {
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"question_score": "8",
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} |
Number of solution set such that $a \geq b \geq c \geq d \geq 0$; $a, b, c, d \in \mathbb{Z}$ and $a + b + c + d = 10$ Given $(a, b, c, d)$ is a set of integers and $a \geq b \geq c \geq d \geq 0$. Find the number of solution sets for $a + b + c + d = 10$.
(This is problem from a competition, the answer key says its $23$, wherein my answer is $455$)
Solution
$x_1 + x_2 +1 + x_3 + 2 + x_4 + 3 = 10$
$x_1 + x_2 + x_3 + x_4 = 16$
$_{15}C_3$ = 455
| $a \geq b \geq c \geq d \geq 0$ give easily as follows
$$\begin{cases}a=10\text{ gives } 1\\a=9\text{ gives } 1\\a=8\text{ gives }2\\a=7\text{ gives }3\\a=6\text{ gives }4\\a=5\text{ gives }5\\a=4\text{ gives }5\\a=3\text{ gives }2
\end{cases}$$
examples: $8200,8110\text{ are the two solutions for a=8 }\\ 5500,5410,5320,5311,5221\text{ are the five solutions for a=5 }$
Thus there are $23$ solutions.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Integral $\int \sqrt{1+x^2}dx$ I was trying to do this integral
$$\int \sqrt{1+x^2}dx$$
I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form of $\int x^m(a+bx^n)^p\,dx$ and I spent a lot of time on it so I would like to see if it can be done this way and where did I go wrong.
$$\int(1+x^2)^\frac{1}{2}dx$$
$m=0$, $n=2$, $p=\frac{1}{2}$
Because $\frac{m+1}{n}+p \in \mathbb Z$ I used substitution $x^{-2}+1=t^2$.
From there I got:
$$-\frac{dx}{x^3}=t\,dt$$
$$x=\frac{1}{\sqrt {t^2-1}}$$
$$t=\frac{\sqrt{1+x^2}}{x}$$
I expanded the original with $x^4$:
$$\int \frac{x^4\sqrt{1+x^2}dx}{x^4}=\int \left(\frac{1}{\sqrt{t^2-1}}\right)^4t(-tdt)=\int\frac{-t^2dt}{(t^2-1)^2}$$
Now I used partial integration:
$u=t$,
$du=dt$,
$dv=\frac{-tdt}{(t^2-1)^2}$,
$v=\frac{1}{2(t^2-1)}$
Then
$$\begin{align}
\int\frac{-t^2dt}{(t^2-1)^2}
&=\frac{t}{2(t^2-1)}-\frac{1}{2}\int \frac{dt}{t^2-1}= \\
&=\frac{t}{2(t^2-1)}-\frac{1}{2}\frac{1}{2}\ln\frac{t-1}{t+1}= \\
&=\frac{\frac{\sqrt{1+x^2}}{x}}{2\left(\left(\frac{\sqrt{1+x^2}}{x}\right)^2-1\right)}-\frac{1}{4}\ln\frac{\frac{\sqrt{1+x^2}}{x}-1}{\frac{\sqrt{1+x^2}}{x}+1}= \\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}+x}=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}+x}\cdot\frac{\sqrt{1+x^2}-x}{\sqrt{1+x^2}-x}=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{4}\ln(\sqrt{1+x^2}-x)^2=\\
&=\frac{x\sqrt{1+x^2}}{2}-\frac{1}{2}\ln(\sqrt{1+x^2}-x)+C
\end{align}$$
The solution is
$$\frac{x\sqrt{1+x^2}}{2}+\frac{1}{2}\ln(x+\sqrt{1+x^2})+C$$
My solution looks very similar, so where did I go wrong?
| Solutions are same. Observe that $\sqrt{1+x^2}-x=\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}=\frac{1}{\sqrt{1+x^2}+x}$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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} |
$\sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)}$ How to calculate this sum
$$S_n = \sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)} ?$$
Note :
1)
$$S_n = n+\sum_{k=1}^n{\rm cotan}^2\left(\frac{(2k+1)\pi}{2n}\right)$$
2) $S_n$ seems to be equal to $\frac{n^2}{4}$.
3) I kwow how to calculate
$$\sum_{k=1}^n\frac{1}{\sin^2\left(\frac{k\pi}{2n}\right)}$$
with the roots of $(X+1)^{2n} - (X-1)^{2n}$.
| Hint 1: If you know the answer for $\sum\limits_{k=1}^{2n-1}\csc^2\left(\frac{k}{2n}\pi\right)$, then use
$$
\sum_{k=1}^n\csc^2\left(\frac{2k-1}{2n}\pi\right)+\sum_{k=1}^{n-1}\csc^2\left(\frac{k}{n}\pi\right)=\sum_{k=1}^{2n-1}\csc^2\left(\frac{k}{2n}\pi\right)
$$
Hint 2: $\frac{2n/z}{z^{2n}-1}$ has residue $1$ at $z=e^{\pi ik/n}$ and residue $-2n$ at $z=0$.
Apply Hint 2
$$
\begin{align}
\left(\frac{2i}{z-\frac1z}\right)^2\frac{2n/z}{z^{2n}-1}
&=\frac{-4z^2}{z^4-2z^2+1}\frac{2n/z}{z^{2n}-1}\\
&=\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
\end{align}
$$
has residue $\csc^2\left(\pi k/n\right)$ at $z=e^{\pi ik/n}$ except at $z=\pm1$. A bit of computation gives
$$
\begin{align}
\frac{2n}{z^{2n}-1}
&=\phantom{+}\frac1{z-1}-\frac{2n-1}2+\frac{(2n-1)(2n+1)}{12}(z-1)+O\!\left((z-1)^2\right)\\
&=-\frac1{z+1}-\frac{2n-1}2-\frac{(2n-1)(2n+1)}{12}(z+1)+O\!\left((z+1)^2\right)
\end{align}
$$
Therefore,
$$\newcommand{\Res}{\operatorname*{Res}}
\Res_{z=1}\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
=\frac14-\frac{4n^2-1}{12}=-\frac{n^2-1}3
$$
and
$$
\Res_{z=-1}\left(\frac1{(z+1)^2}-\frac1{(z-1)^2}\right)\frac{2n}{z^{2n}-1}
=\frac14-\frac{4n^2-1}{12}=-\frac{n^2-1}3
$$
Since the sum of the residues at all the singularities is $0$, we get that half the sum over the singularities except at $z=\pm1$ is
$$
\sum_{k=1}^{n-1}\csc^2\left(\frac{k}{n}\pi\right)=\frac{n^2-1}3
$$
Apply Hint 1
$$
\begin{align}
\sum_{k=1}^n\csc^2\left(\frac{2k-1}{2n}\pi\right)
&=\sum_{k=1}^{2n-1}\csc^2\left(\frac{k}{2n}\pi\right)
-\sum_{k=1}^{n-1}\csc^2\left(\frac{k}{n}\pi\right)\\
&=\frac{4n^2-1}3-\frac{n^2-1}3\\[9pt]
&=n^2
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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irrationality of $e$ and series I'm trying to understand a passage in the proof that e is an irrationale number.
We assume that $e$ is a rational number, $e= \frac{p}{q}$ in which $p$ and $q$ are two coprime numbers.
$$e= \sum_{k=0}^\infty \frac{1}{k!} \implies e-\sum_{k=0}^\infty \frac{1}{k!}=\frac{p}{q}-\sum_{k=0}^\infty \frac{1}{k!}=0 \\\implies \frac{p}{q}-\sum_{k=0}^q \frac{1}{k!}>0 \\\implies 0< \frac{p}{q}-\sum_{k=0}^q \frac{1}{k!}\\
\implies\sum_{k=0}^ \infty \frac{1}{k!} -\sum_{k=0}^q \frac{1}{k!} =\sum_{k=q+1}^\infty \frac{1}{k!} $$
Until this passage it is all clear. Then I don't understand why:
$$\sum_{k=q+1}^\infty \frac{1}{k!} <\frac{1}{q!}\cdot \frac{1}{q}$$
Then the proof goes on and it is quite simple to understand.
$$\sum_{k=q+1}^\infty \frac{1}{k!}= \frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots\\
=\frac{1}{q!}\cdot \left( \frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\cdots \right) \\
=\frac{1}{q!}\cdot\frac{1}{q}\cdot \left( \frac{q}{q+1}+\frac{q}{(q+1)(q+2)}+\cdots \right)$$
Why does $\left( \frac{q}{q+1}+\frac{q}{(q+1)(q+2)}+\cdots \right)<1$?
| \begin{align}\sum_{k=q+1}^\infty\frac1{k!}&=\frac1{q!}\left(\frac1{q+1}+\frac1{(q+1)(q+2)}+\cdots\right)\\&<\frac1{q!}\left(\frac1{q+1}+\frac1{(q+1)^2}+\cdots\right)\\&=\frac1{q!}\times\frac{\frac1{q+1}}{1-\frac1{q+1}}\\&=\frac1{q!}\times\frac1q.\end{align}
| {
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The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square.
The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square. List all possible values of $c$. (The letters are the digits of the base 6 number.)
I've rewritten $abcd_6$ into $216a + 36b + 6c + d$, and I know that $d$ can be $1,3,5$, but now I'm stuck. I guess I could start plugging in some numbers since $a,b,c,d \le 5$, but I feel like there would be a more efficient way.
| Work modulo $36$, test only the odd numbers $15$ through $35$:
$^\text{(we may omit $1$ through $13$, since these would not satisfy $a>0$)}$
$$\,$$
$$(15)^2 \equiv 9 \pmod{36} $$
$$(17)^2 \equiv 1 \pmod{36} $$
$$(19)^2 \equiv 1 \pmod{36}$$
$$(21)^2 \equiv 9 \pmod{36} $$
$$(23)^2 \equiv 25 \pmod{36}$$
$$(25)^2 \equiv 13 \pmod{36} $$
$$(27)^2 \equiv 9 \pmod{36} $$
$$(29)^2 \equiv 13 \pmod{36} $$
$$(31)^2 \equiv 25 \pmod{36}$$
$$(33)^2 \equiv 9 \pmod{36} $$
$$(35)^2 \equiv 1 \pmod{36} $$
$$\,$$
The possible remainders for odd perfect squares, modulo $36$, are $1$, $9$, $13$, and $25$.
In base $6$, these are equivalent to the last two digits being $$01_6,\,13_6,\,21_6,\,41_6$$
In particular, $$\boxed{\,c \, \in \, \left\{0,1,2,4\right\}\,\,}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the determinant of a special matrix I am having some difficulties to calculate a determinant. The matrix is an $n\times n$ one, but I will show the particular case $5\times 5$ in order to simplify the notation.
$\begin{pmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & x & x & x\\
1 & x & 0 & x & x\\
1 & x & x & 0 & x\\
1 & x & x & x & 0
\end{pmatrix}$
It seems easy but I am missing a step. If for every $i\geq 2$ I add $-x$ times the first row to the $i$th row then I obtain the following matrix:
$\begin{pmatrix}
0 & 1 & 1 & 1 & 1\\
1 & -x & 0 & 0 & 0\\
1 & 0 & -x & 0 & 0\\
1 & 0 & 0 & -x & 0\\
1 & 0 & 0 & 0 & -x
\end{pmatrix}$
and now I can get a recurrence: if $a_n$ is the $n$th determinant, then if I calculate the $n+1$th determinant in terms of the last row, I come up with the formula $a_{n+1}=\left (-1\right )^{n+2}\left (-x\right )^{n-1}-xa_n=-x^{n-1}-xa_n$.
Now, if I consider some small cases I get $a_1=0$, $a_2=-1$, $a_3=2x$, $a_4=-3x^2$, $a_5=4x^3$, $a_6=-5x^4$. It seems to be $a_n=\left (-1\right )^{n+1}\left (n-1\right )x^{n-2}$.
But my problem is that my recursive formula did not help me to prove it by induction. Since I don't think that the conjecture about the general term is wrong, I am for sure commiting a mistake on the recursive formula, but I can't see where exactly. Any help?
| An idea: subtract line two from lines 3,4,5 and cevelop by column 1:
$$\begin{vmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & x & x & x\\
1 & x & 0 & x & x\\
1 & x & x & 0 & x\\
1 & x & x & x & 0\end{vmatrix}=\begin{vmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & x & x & x\\
0 & x &-x & 0 & 0\\
0 & x & 0 &-x & 0\\
0 & x & 0 & 0 & -x\end{vmatrix}=-\begin{vmatrix}
1 & 1 & 1 & 1\\
x &-x & 0 & 0\\
x & 0 &-x & 0\\
x & 0 & 0 & -x\end{vmatrix}\stackrel{C_1\to C_1+(C_3+C_4)}=$$$${}$$
$$=-\begin{vmatrix}
3 & 1 & 1 & 1\\
x &-x & 0 & 0\\
0 & 0 &-x & 0\\
0 & 0 & 0 & -x\end{vmatrix}=x\begin{vmatrix}
3 & 1 & 1 \\
x &-x & 0 \\
0 & 0 &-x
\end{vmatrix}=-x^2\begin{vmatrix}3&1\\x&\!-x\end{vmatrix}=-x^2(-3x-x)=$$$${}$$
$$=x^3(x+3)$$
| {
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"source": "stackexchange",
"question_score": "1",
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Show that if $p(x)=a+bx+cx^2$ is a 2nd degree polynomial such that $p(1)=p(2)=p(3)=0$ then $p(x)=0$, using determinants.
Show that if $p(x)=a+bx+cx^2$ is a 2rd degree polynomial such that $p(1)=p(2)=p(3)=0$ then $p(x)=0$ (i.e. $a=b=c=0$), using the determinant of the matrix: $\left(\begin{array}{l}1&x&x^2\\1&y&y^2\\1&z&z^2\end{array}\right)$
Hey everyone. So I managed to prove that the determinant of the given matrix is $(y-x)(z-x)(z-y)$ and therefore the matrix is invertible whenever $x\neq z \lor y\neq x \lor z\neq y$
I've proved this claim using basic algebra but am confused on how to show it using the determinant. I tried showing it by plugging in the values of the polynomial by order:
$\left|\begin{array}{l}a&b&c\\a&2b&4c\\a&3b&9c\end{array}\right|$ (Column operations) $ =\left|\begin{array}{l}a&b&c+b+a\\a&2b&4c+2b+a\\a&3b&9c+3b+a\end{array}\right|$ $(p(1)=p(2)=p(3)=0)$$ = \left|\begin{array}{l}a&b&0\\a&2b&0\\a&3b&0\end{array}\right|=0$ but this does not help me. I would love to get some help on this question. Thanks in advance.
| Note that\begin{align}p(1)=p(2)=p(3)=0&\iff\left\{\begin{array}{l}a+b\times1+c\times1^2=0\\a+b\times2+c\times2^2=0\\a+b\times3+c\times3^2=0\end{array}\right.\\&\iff\begin{pmatrix}1&1&1\\1&2&2^2\\1&3&3^2\end{pmatrix}.\begin{pmatrix}a\\b\\c\end{pmatrix}=0.\end{align}So, if $(a,b,c)\neq(0,0,0)$, the determinant of that matrix will be $0$. But it isn't.
| {
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Show the polynomial $p(x,y) =y^3+xy^2+(x^2+x)y+(4x^3+2x)$ is irreducible in $\mathbb{Z}[x, y]$ I've thought about $p(x,y)$ being in $\mathbb{Z}[x][y]$ and tried to use Eisenstein's criterion, but I haven't had any luck. The best I've got is, we can suppose $p(x,y)$ is reducible and write it as
$$p(x,y)=(y+a)(y^2+by+c) = y^3 + (a+b)y^2 + (ab+c)y + ac$$
for some $a,b,c \in \mathbb{Z}[x]$.
Comparing coefficients we have
\begin{align}
&a+b=x \\
&ab+c = x^2+x \\
&ac = 4x^3+2x = 2x(2x^2+1)
\end{align}
From the last equation we have $a = 2x$ and $c=2x^2+1$ or alternatively $a = 2x^2+1$ and $c=2x$.
If $a=2x$, $c=2x^2+1$ then from the first equation, $2x+b=x$ so $b=-x$. From the second equation, $-2x^2+x-1=x^2+x$, but this doesn't hold for all $x$...
If $a=2x^2+1$, $c=2x$ then from the first equation, $2x^2+1+b=x$ so $b=-2x^2+x-1$. From the second equation, $(2x^2+1)(-2x^2+x-1) = x^2+x$, but again the l.h.s. and r.h.s. don't match since without even computing we see the l.h.s. is degree $4$.
Is there a better way to prove this?
| In $\mathbb{Z}[x][y]$, a generalization of Eisenstein's criterion will apply, since $\mathbb{Z}[x]$ is a UFD. Now, apply this generalization using the irreducible element $x \in \mathbb{Z}[x]$.
As an alternative: if $p$ were reducible, then its terms of highest total degree $4x^3 + x^2 y + x y^2 + y^3$ would have to be a product of the highest-degree terms of the factors. Now suppose the highest-degree terms of the factors are $ax^2 + bxy + cy^2$ and $dx + ey$, respectively. Then $(ax^2 + bxy + cy^2) (dx + ey) = 4x^3 + x^2 y + x y^2 + y^3$; now if we substitute $x := t$ and $y := 1$, we get $(at^2 + bt + c) (dt + e) = 4t^3 + t^2 + t + 1$, so $4 t^3 + t^2 + t + 1$ would have a rational root $-\frac{e}{d}$. But then, $d \mid 4$ and $e \mid 1$, and from here it is easy to get a contradiction.
The only other possibility would be if one of the factors were constant; however, then this constant would have to be $\pm 1$ since no other integer divides $p(x, y)$.
| {
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$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - \sqrt{x-4} = \sqrt{x-1} $
Squaring both Sides
$\mapsto x - 2 \sqrt{x^2 - 16}= -1$
$\mapsto \frac{x+1}{2} = \sqrt{x^2 - 16}$
Squaring both sides
$ \mapsto \frac{x^2 + 2x + 1}{4} = x^2 - 16$
$\mapsto 3x^2 -2x - 65 = 0$
Solving the quadratic equation
$$x = 5 OR \frac{-13}{3}$$
Checking in the initial equation we can see that $5$ is a valid root. But the second value that is given in the key to the book is $4$. How do I obtain $4$?
| At the stage where you divided by $\sqrt{x-4}$, you forgot that $\sqrt{x-4}$ could be $0$, which gives the solution $x=4$. You can only divide by $\sqrt{x-4}$ if $x-4\neq 0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Solve: $\sqrt{x-\frac1x}-\sqrt{1-\frac1x}=1-\frac1x$ for $ x\neq0$ $$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}{=1-\frac{1}{x}}, x\neq0$$
So my idea is that I moved the second sqrt to the RHS and left the first on the LHS. I then squared everything and ended up with $\frac{x^2-1}{x}=2-\frac{3}{x}+\frac{1}{x^2}$. I don't know what to do next? Can somebody give a hint?
| The domain gives $-1\leq x<0$ or $x\geq1$.
*
*$-1\leq x<0$.
We obtain:
$$\sqrt{\frac{(1-x)(x+1)}{-x}}=\frac{\left(\sqrt{1-x}\right)^2}{x}+\sqrt{\frac{1-x}{-x}}$$ or
$$\sqrt{\frac{x+1}{-x}}=\frac{\sqrt{1-x}}{x}+\frac{1}{\sqrt{-x}}$$ or
$$-\sqrt{-x(x+1)}=\sqrt{1-x}-\sqrt{-x}$$ or
$$\sqrt{-x}=\sqrt{1-x}+\sqrt{-x(1+x)}$$ or
$$-x=1-x-x-x^2+2\sqrt{x(x^2-1)}$$ or
$$x^2+x-1=2\sqrt{x(x^2-1)},$$ which is impossible because $x^2+x-1<0$ for $-1\leq x<0.$
*$x\geq1$.
We obtain:
$$\sqrt{\frac{(x-1)(x+1)}{x}}=\frac{x-1}{x}+\sqrt{\frac{x-1}{x}},$$ which gives
$$x=1$$ or
$$\sqrt{\frac{x+1}{x}}=\frac{\sqrt{x-1}}{x}+\sqrt{\frac{1}{x}},$$ which is
$$\sqrt{(x+1)x}=\sqrt{x-1}+\sqrt{x}$$ or
$$x^2+x=2x-1+2\sqrt{(x-1)x}$$ or
$$x^2-x-2\sqrt{(x-1)x}+1=0$$ or
$$\left(\sqrt{x^2-x}-1\right)^2=0$$ or
$$x^2-x-1=0,$$ which gives
$$x=\frac{1+\sqrt{5}}{2}$$ and we got the answer:
$$\left\{1,\frac{1+\sqrt5}{2}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving this ODE near x=0 I could not get the correct solution on this ODE:
$$(1+x^2)y''-xy'-8y=0$$ around $x=0$.
I tried to solve this using MacLaurin expansion on $y$ and forming recurrence relations between the coefficients.
Can anyone teach me how to solve this by steps? Thank you.
| Let $y(x) = \sum^\infty_{n=0} a_n x^n$. We see $$\sum^\infty_{n=2} n(n-1)a_nx^{n-2} + \sum^\infty_{n=2} n(n-1)a_nx^n - \sum^\infty_{n=1} n a_nx^n - \sum^\infty_{n=0} 8a_n x^n = 0.$$ By re-indexing and running off a few terms, we see $$\sum^\infty_{n=2} n(n-1)a_nx^{n-2} = \sum^\infty_{n=0} (n+2)(n+1)a_{n+2}x^n = 2 a_2 + 6a_3x + \sum^\infty_{n=2} (n+2)(n+1)a_{n+2}x^n.$$ We also need to run terms off a couple of the other sums so that all of them are indexed starting at $n = 2$: $$\sum^\infty_{n=0} 8a_n x^n = 8a_0 + 8a_1x + \sum^\infty_{n=2} 8a_n x^n$$ and $$\sum^\infty_{n=1} n a_nx^n = a_1x + \sum^\infty_{n=2} na_nx^n.$$ Putting all this together, we see $$(2a_2 - 8a_0) + (6a_3 - 9a_1)x + \sum^\infty_{n=2} [(n+2)(n+1)a_{n+2} + (n(n-1) - n - 8)a_n ]x^n = 0$$ or $$(2a_2 - 8a_0) + (6a_3 - 9a_1)x + \sum^\infty_{n=2} [(n+2)(n+1)a_{n+2} + (n+2)(n-4)a_n ]x^n = 0.$$ Thus the coefficients satisfy $$a_2 = 4a_0, \,\,\,\,\, a_3 = \tfrac 3 2 a_1, \,\,\,\,\, a_{n+2} = -\left(\tfrac{n-4}{n+1}\right)a_n, \,\,\,\, n \ge 2.$$
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Reduction to triple integral I have an integral
$$I = \int_{V}(x^2 + y^2)\, dV, $$
where $V = \{2ax \geqslant z^2 \} \cap \{ x^2 + y^2 \leqslant ax \}$.
Am I right, that
$$I = \int_{0}^{a} \int_{-\sqrt{2ax}}^{\sqrt{2ax}} \int_{-\sqrt{a^2/4 - (x-a/2)^2}}^{\sqrt{a^2/4 - (x-a/2)^2}} (x^2 + y^2)\, dy\,dz\,dx?$$
It seems to me, that this is not really right, because I've tried to integrate it and my answer is far from correct, however I can't find where is the mistake.
| Your region $V$ is the intersection of the parabolic plane and the cylinder.
$\hspace{13em}$
So it is not bad to begin with cylindrical coordinates:
$$ (x, y, z) = (r \cos\theta, r\sin\theta, z) $$
Then $V$ is written as
$$ V = \{ 2ar \cos\theta \geq z^2 \} \cap \{ r \leq a \cos\theta \} $$
The above description provides a natural limits for $r$ and $z$. For $\theta$, we read out that $\cos\theta \geq 0$ and hence $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} $. So
\begin{align*}
\int_V (x^2 + y^2) \, dxdydz
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{a\cos\theta} \int_{-\sqrt{2ar\cos\theta}}^{\sqrt{2ar\cos\theta}} r^3 \, dz dr d\theta \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{a\cos\theta} r^{7/2} \sqrt{8a\cos\theta} \, dr d\theta \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{4\sqrt{2} a^5}{9} \cos^5 \theta \, d\theta \\
&= \frac{64\sqrt{2}a^5}{135}
\end{align*}
On the other hand, if you choose to stick to the Cartesian coordinates, you can still have an option:
\begin{align*}
\int_V (x^2 + y^2) \, dxdydz
&= \int_{0}^{a} \int_{-\sqrt{ax-x^2}}^{\sqrt{ax-x^2}} \int_{-\sqrt{2ax}}^{\sqrt{2ax}} (x^2 + y^2) \, dz dy dx \\
&= \int_{0}^{a} \int_{-\sqrt{ax-x^2}}^{\sqrt{ax-x^2}} (x^2 + y^2) \sqrt{8ax} \, dy dx \\
&= \int_{0}^{a} \frac{4\sqrt{2}}{3} x^2 (2x + a) \sqrt{a\smash[b]{(a-x)}} \, dx \\
&= \frac{64\sqrt{2}a^5}{135}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $\int\frac{2+3x}{3x^2+4x-6}\,dx$? I am trying to solve
$$\int\frac{2+3x}{3x^2+4x-6}\,dx$$ using partial fraction decomposition but it fails, how to solve it?
| $\int\frac{2+3x}{3x^2+4x-6}=\frac{1}{2}\int\frac{6x+4}{3x^2+4x-6}=\frac{\ln(|3x^2+4x-6|)}{2}+K$
Since $6x+4$ is the derivative of $3x^2+4x-6$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the last 3 digits of $2015^{18}+2015^{19}+2015^{20}+...+2015^{2018}$. Obvious I would find the sum of the numbers mod 100, but how would I handle this when the powers are so large? Calculating one already seemds ridiculous.
| As $2015\equiv15\pmod{1000},$ we need $\displaystyle\sum_{r=18}^{2018}15^r\pmod{1000}$
As $(15^n,1000)=5^3$ for $n\ge3,$
let us find $\displaystyle S=\sum_{r=18-3}^{2018-3}15^r\pmod{1000/5^3}$
As $\displaystyle15\equiv-1\pmod8,S\equiv\sum_{r=18-3}^{2018-3}(-1)^r\equiv-1\pmod8$
$\displaystyle15^3S\equiv15^3\cdot-1\pmod{15^3\cdot8}\equiv-15^3\pmod{8\cdot5^3}$ as $5^3|15^3$
Now $-15^3=-225\cdot15=-3375\equiv-375\equiv1000-375\pmod{1000}$ as $S>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672805",
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"source": "stackexchange",
"question_score": "1",
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Prove that $\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix}=xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$ if $x+y+z=0$
If $x+y+z=0$, then prove that
$$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$
I can do it by Sarrus' law but how can I prove it by matrix operations without actually expanding the determinant ?
My Attempt
$$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&\frac{yb}{x}&\frac{zc}{x}\\
c&\frac{za}{y}&\frac{xb}{y}\\
b&\frac{xc}{z}&\frac{ya}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&-b-\frac{zb}{x}&-c-\frac{yc}{x}\\
c&-a-\frac{xa}{y}&-b-\frac{zb}{y}\\
b&-c-\frac{yc}{z}&-a-\frac{xa}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b+\frac{zb}{x}&c+\frac{yc}{x}\\
c&a+\frac{xa}{y}&b+\frac{zb}{y}\\
b&c+\frac{yc}{z}&a+\frac{xa}{z}\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c+\frac{yc}{x}\\
c&a&b+\frac{zb}{y}\\
b&c&a+\frac{xa}{z}\\
\end{vmatrix}+xyz\begin{vmatrix}
a&\frac{zb}{x}&c+\frac{yc}{x}\\
c&\frac{xa}{y}&b+\frac{zb}{y}\\
b&\frac{yc}{z}&a+\frac{xa}{z}\\
\end{vmatrix}\\
=xyz\bigg(\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}+\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&c\\
c&\frac{xa}{y}&b\\
b&\frac{yc}{z}&a\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{yc}{x}\\
c&\frac{xa}{y}&\frac{zb}{y}\\
b&\frac{yc}{z}&\frac{xa}{z}\\
\end{vmatrix}\bigg)\\
$$
I need to prove that the sum of last three terms is zero.
$$
\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&c\\
c&\frac{xa}{y}&b\\
b&\frac{yc}{z}&a\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{yc}{x}\\
c&\frac{xa}{y}&\frac{zb}{y}\\
b&\frac{yc}{z}&\frac{xa}{z}\\
\end{vmatrix}=\begin{vmatrix}
a&b&\frac{yc}{x}\\
c&a&\frac{zb}{y}\\
b&c&\frac{xa}{z}\\
\end{vmatrix}+\begin{vmatrix}
a&\frac{zb}{x}&\frac{-zc}{x}\\
c&\frac{xa}{y}&\frac{-xb}{y}\\
b&\frac{yc}{z}&\frac{-ya}{z}\\
\end{vmatrix}\\
$$
Solution by expansion
$$
\Delta=\begin{matrix}
xa&yb&zc&xa&yb\\
yc&za&xb&yc&za\\
zb&xc&ya&zb&xc\\
\end{matrix}=xyz(a^3+b^3+c^3)-abc(x^3+y^3+z^3)
$$
We have $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$ as $x+y+z=0$. Thus, $x^3+y^3+z^3=3xyz$
$$
\Delta=xyz(a^3+b^3+c^3)-abc(3xyz)=xyz(a^3+b^3+c^3-3abc)\\
=xyz\bigg[ a\big(a^2-bc\big)-b\big(ac-b^2\big)+c\big(c^2-ab\big) \bigg]\\
=xyz.\bigg[a\begin{vmatrix}
a&b\\
c&a\\
\end{vmatrix}-b\begin{vmatrix}
c&b\\
b&a\\
\end{vmatrix}+c\begin{vmatrix}
c&a\\
b&c\\
\end{vmatrix}\bigg]
=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$
| I don't think it is possible without expanding the determinants. The determinant on the right is $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$. The determinant on the left is $(a^3+b^3+c^3)xyz-abc(x^3+y^3+z^3)$. Subtracting the left side from the right side is equal to $abc(x^3+y^3+z^3-3xyz)=abc(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$. The result follows.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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Show that $\sum\limits_{r=1}^n\sum\limits_{k=1}^{n-r+1}(2r-1)k=\frac 13\sum\limits_{r=1}^nr\sum\limits_{s=1}^{n+1}s$
Question:
Show that
$$\sum_{r=1}^n\sum_{k=1}^{n-r+1}(2r-1)k=\frac 13 \sum_{r=1}^nr\sum_{s=1}^{n+1}s$$
purely by manipulating summation limits and summands, i.e. convert the original summation (on the left) to the summation on the right, without first evaluating the closed form, as shown below.
This question is related to Question $2435490$ on MSE.
From the solutions provided , it is clear that
$$\sum_{r=1}^n\sum_{k=1}^{n-r+1}(2r-1)k=\frac 1{12}n(n+1)^2(n+2)$$
This can also be written as
$$\frac 13\cdot \frac {n(n+1)}2\cdot\frac {(n+1)(n+2)}2$$
which is the closed form of
$$\frac 13 \sum_{r=1}^nr\sum_{s=1}^{n+1}s$$
| Not a solution.
Consider $n=4$
$LHS = 1(1+2+3+4) + 3(1+2+3) + 5(1+2) + 7 = 1(1+3+5+7) + 2(1+3+5) + 3(1+3)+ 4 = \sum\limits_{k=1}^n (n+1 - k)k^2 = \sum\limits_{k=1}^n (n+1)k^2 - k^3 $
Separately, $$\frac{1}{3}\sum\limits_{k=1}^n (n+1)k + k^3 = \frac{1}{3}(n+1)\sum\limits_{k=1}^n k + \frac{1}{3}(\sum\limits_{k=1}^n k)^2$$
$$= \frac{1}{3}(\sum\limits_{k=1}^n k )(n + 1 + \sum\limits_{k=1}^n k) $$
$$= \frac{1}{3}(\sum\limits_{k=1}^n k )(\sum\limits_{k=1}^{n+1} k)$$
So now all that remains is to prove:
$$\sum\limits_{k=1}^n ((n+1)k^2 - k^3) = \frac{1}{3}\sum\limits_{k=1}^n ((n+1)k + k^3)$$
Hope this helps a bit.
| {
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"source": "stackexchange",
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Verifying Newton's Binomial Series for $(1+x)^{−2}$
Verify the consistency of Newton’s Binomial series for the function
$(1+x)^{−2}$ in two ways:
$(a)$ by multiplying the usual geometric series for $(1 +x)^{−1}$with
itself
$(b)$ by differentiation of the series for $(1 +x)^{−1}$
$(a)$ I have that Newton's Binomial series says that
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
I have that the geometric series for $(1 +x)^{−1}$ is given by
$$\begin{align*}
\frac{1}{1+x}
&=\sum_{k=0}^{\infty}(-x)^k\\\\
&=-x^0+(-x)^1+(-x)^2+(-x)^3+...\\\\
&=1-x+x^2-x^3+...
\end{align*}$$
Thus
$$\begin{align*}
\left(\frac{1}{1+x}\right)^2
&=(1-x+x^2-x^3+...)(1-x+x^2-x^3+...)\\\\
&=(1-x+x^2-x^3+...)\\\\
& +(-x+x^2-x^3+...)\\\\
& +(x^2-x^3+...)\\\\
& +(-x^3+...)\\\\
&=1-2x+3x^2-4x^3+...
\end{align*}$$
Checking that this equals Newton's Binomial series
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
where $p=-2$
$$\begin{align*}
(1+x)^{-2}
&=1-2x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+···\\\\
&=1-2x+3x^2-4x^3+... \checkmark
\end{align*}$$
Is this a valid proof?
$(b)$
I have that $$\int_0^x\frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$$
but I'm not sure how to use this fact.
| The binomial expansion is
$$
\left( {1 + x} \right)^{\, - 2} = \sum\limits_{0\, \le \,k} {\left( \matrix{
- 2 \cr
k \cr} \right)x^{\,k} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
k + 1 \cr
k \cr} \right)x^{\,k} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)x^{\,k} }
$$
where "upper negation" of the binomial has been used
The multiplication of geometric series gives
$$
\eqalign{
& {1 \over {1 + x}}{1 \over {1 + x}} = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} x^{\,k} } \sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,j} x^{\,j} } = \cr
& = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,k + j} x^{\,k + j} } } \quad \left| \matrix{
\;0 \le k + j = s \hfill \cr
\;0 \le j = s - k\; \Rightarrow \;0 \le k \le s \hfill \cr} \right.\;\quad = \cr
& = \sum\limits_{0\, \le \,s} {\left( {\sum\limits_{0 \le k \le s} 1 } \right)\left( { - 1} \right)^{\,s} x^{\,s} } = \sum\limits_{0\, \le \,s} {\left( { - 1} \right)^{\,s} \left( {s + 1} \right)x^{\,s} } \cr}
$$
and differentiating the geometric series
$$
\eqalign{
& {d \over {dx}}\left( {1 + x} \right)^{\, - 1} = - \left( {1 + x} \right)^{\, - 2} = {d \over {dx}}\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} x^{\,k} } = \cr
& = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} k\,x^{\,k - 1} } = \sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k} k\,x^{\,k - 1} } = \cr
& = \sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k + 1} \left( {k + 1} \right)\,x^{\,k} } = - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)x^{\,k} } \cr}
$$
| {
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"source": "stackexchange",
"question_score": "1",
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Prove for any odd integer $n$, $n^2-1\equiv 1\pmod{8}$ Let, $\exists k \in \mathbb{Z}, n=2k+1\implies n^2-1= 4k(k+1) \pmod{8}$
Now, if $k$ is odd, $k+1$ is even, and vice-versa, so $4k(k+1) \pmod{8}=\forall k\, \exists k' \in \mathbb{Z},\,8k' \pmod{8}$
Unable to proceed further.
| The statement should have been$$ n^2-1\equiv 0\pmod{8}$$
Let $n=2k+1$, then $n^2 -1 =(2k+1)^2-1 = 4k^2 + 4k = 4k(k+1)$
Note that $ k(k+1)$ is always even.
Thus $n^2-1 = 4k(k+1)$ is a multiple of 8.
That implies$$ n^2-1\equiv 0\pmod{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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To find a remainder, when divided by $101$
There is a large number given in an exam. It is said to find a remainder when it is divided by $101$. The number is
$$1+11^{{(1+11+11^2+11^3+...+11^{100}~~)}^{(11^{101}~~+11^{102}~~+\cdots+11^{1000}~~)}}$$
We know $11$ and $101$ are itself prime then $(101,11)=1$ and $\phi(101)=100$
Now,
$$1+11+11^2+11^3+....+11^{100}\equiv r\mod 100 \\ \Rightarrow1+11+21+21*11+21^2+21^2*11+.....+(21^2)^{50}\equiv r \mod 100 \\ \Rightarrow (1+11+21+31+41+51+61+71+81-9)*10 \equiv r\mod 100 \\ \Rightarrow 0\equiv r\mod 100 \\ \therefore 0^{{(11^{101}+11^{102}+.....+11^{1000}~~)}}\equiv0 \mod 100 \\ \therefore 1+11^0\equiv1+1\equiv2 \mod 101$$
Is it correct? Any help is appreciated.
| You can use the geometric series formula $\displaystyle \frac{a(1-r^n)}{1-r}$, where $a$ is the first term, and $r$ is the common ratio.
For $1+11+\cdots+11^{100}$ this is $1 \cdot {{1-11^{100}} \over {1-11}}$. By Fermat's little theorem, $11^{101-1} \equiv 1 \pmod {101}$, so the whole expression is congruent to $1 \pmod {101}$.
For $11^{101}+11^{102}+\cdots+11^{1000}$ this is $11^{101} \cdot {{1-11^{1000}} \over {1-11}}$. Since $11^{101-1} \equiv 1 \pmod {101}$, then $11^{1000} \equiv (11^{100})^{10}$ $ \equiv 1 \pmod {101}$, so ${{1-11^{1000}} \over {1-11}}$ evaluates to $1 \pmod {101}$.
Therefore, the problem is simplified to $1+11^{1^1} \pmod {101}$, which is congruent to $12 \pmod {101}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$
Compute $$I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$$
The way I approach this problem was to:
*
*Set $u=\sqrt{x-2}$ and arrive at $$I=2\int\frac{u^2(u^2+1)}{\sqrt{u^2-4}}\mathrm du$$
*Set $u=2\sec t\implies\mathrm du=2\sec t\tan t\mathrm dt$ to get $I=8\int\sec^3(t)(4\sec^2(t)+1)\mathrm dt$
Now this integral involves $\sec^5t$ and $\sec^3t$ which does not make me very happy. I think there should be a simpler method. Can anyone show me simpler steps? (the first few steps/substitutions would suffice)
| Note: This is a long, long stretch.
Final Answer:$$\dfrac{16\ln\left(\left|\sqrt{\frac{x^2}4-1}\right|\right)+(x+10)\sqrt{x^2-4}}2+C$$
After applying long division on $\dfrac{(x+3)\sqrt{x^2-4}}{x-2}$ and rewriting the integral, you get:
$$5\int\frac{\sqrt{x^2-4}}{x-2}dx+\int\sqrt{x^2-4} dx$$
Let $x=2\sec(u)\rightarrow u=\text{arcsec}\left(\dfrac x2\right) \rightarrow dx=2\sec(u)\tan(u)du$.$$\int\frac{\sqrt{x^2-4}}{x-2}dx=\int\dfrac{2\sec(u)\sqrt{4\sec^2(u)-4}\tan(u)}{2\sec(u)-2}=2\left(\int\cos(u)\sec^2(u)du+\int\sec^2(u)du \right)$$
$$=2\left(\int\sec(u)+\int\sec^2(u)du \right)=2\Big{(}\ln\big{(}\tan(u)+\sec(u)\big{)}+\tan(u)\Big{)}$$
Substituting from $u$ to $x$:
$$2\Big{(}\ln\big{(}\tan(u)+\sec(u)\big{)}+\tan(u)\Big{)}=2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)+2\sqrt{\frac {x^2}4-1}$$
Then:
$$\int\sqrt{x^2-4}=x\sqrt{\frac {x^2}4-1}-2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)$$
So putting together our steps:
$$5\int\frac{\sqrt{x^2-4}}{x-2}dx+\int\sqrt{x^2-4} dx=$$$$5\cdot \left(2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)+2\sqrt{\frac {x^2}4-1}\right)+x\sqrt{\frac {x^2}4-1}-2\ln\left(\sqrt{\frac {x^2}4-1}+\frac x2\right)$$
Which finally helps you arrive at the answer. Whew!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Induction $(n^3+5n)$ is divisible by $6$
Here is what I have done
Assume $n=k$ is true
$\sum_{1}^{k} k =\dfrac{(k^3+5k)}{6}$
Now assume $n=k+1$ is true
$\sum_{1}^{k+1} k+1 =\dfrac{(k+1)^3+5(k+1)}{6}$
Then now
$\dfrac{(k+1)^3+5(k+1)}{6}=\sum_{1}^{k} k + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + (k+1)$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+5k)}{6} + \dfrac{(6k+6)}{6}$
$\dfrac{(k+1)^3+5(k+1)}{6}=\dfrac{(k^3+11k+6)}{6}$
But the other side doesnt equate (LHS)
$\dfrac{(k^3+3k^2+3k+1)+5k+5}{6}$
$\dfrac{(k^3+3k^2+8k+6)}{6}\ne\dfrac{(k^3+11k+6)}{6}$
I hope my method was clear enough so you can see where I went wrong. It would be much more use to me if you solved it as I learn better from looking at solutions and then applying them to other questions.
| I don't follow the induction step. First the base case when $n=1$ is clear since $6\mid 6$. Now suppose that the claim holds when $n=k$ for some $k\geq 1$. This means that $k^3+5k=6m$ for some $m\in \mathbb{N}$. We want to show that $(k+1)^3+5(k+1)$ is divisible by $6$. Expand
$$
(k+1)^3+5(k+1)=k^3+5k+6+3(k^2+k)
$$
and use the induction hypothesis, as well as the fact that the product of two consecutive integers is even.
Alternatively, we can express the polynomial in the binomial coefficient basis using the fact that
$$
n^k=\sum_{j=0}^k S(k,j)n^{\underline{j}}
=\sum_{j=0}^k S(k,j)\binom{n}{j}j!
$$
where $S(k,j)$ is the stirling number of the second kind to yield that
$$
n^3+5n=6\binom{n}{1}+6\binom{n}{2}+6\binom{n}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Spectral Decomposition of $A=PDP^{-1}$ We just crammed spectral decomposition into our last lecture of the quarter, and I'm quite confused by it.
The following question is on my homework:
Use the matrices P and D to construct a spectral decomposition of $A=PDP^{-1}$.
$A=\begin{bmatrix}-1 & -4 & -4\\-4 & -3 & 0\\-4 & 0 & 1\end{bmatrix}$, $P=\begin{bmatrix} u_1 & u_2 & u_3\end{bmatrix}=\begin{bmatrix}-\frac{2}{3} & \frac{1}{3} & -\frac{2}{3}\\\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3}\\\frac{2}{3} & \frac{2}{3} & -\frac{1}{3}\end{bmatrix}$, $D=\begin{bmatrix}5 & 0 & 0\\0 & -1 & 0\\0 & 0 & -7\end{bmatrix}$
I understand enough to get that $A=5u_1u_1^T-1u_2u_2^T-7u_3u_3^T$, but I don't get how to compute $u_1u_1^T$, $u_2u_2^T$, $u_3u_3^T$
| For example, compute
$$
u_1u_1^T = \pmatrix{- \frac 23 \\ \frac 13 \\ \frac 23 } \pmatrix{- \frac 23 & \frac 13 & \frac 23 } =\\
\pmatrix{\frac 49 & - \frac 29 & - \frac 49\\
- \frac 29 & \frac 19 & \frac 29\\
- \frac 49 & \frac 29 & \frac 49}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit at odd integer for $x$
$$\lim_{x\to a}\frac{1}{(a^2-x^2)^2}\cdot\left(\frac{a^2+x^2}{ax}-2\sin\frac{a\pi}{2}\sin\frac{\pi x}2\right)=?$$ if $a$ is an odd integer.
The way I set out is first assuming $a=1$ and seeing if I can spot some pattern. Now, if I rewrite this limit out as:
$$\lim_{x\to 1}\frac{1}{(1-x^2)^2}\cdot\left(\frac{1+x^2}{x}-2\sin\frac{\pi x}2\right)=?$$
Now, you can clearly see that the denominator is of the form $0^4$ (due to the $(1-x^2)^2$), whereas the numerator is, at max, $0^2$ (due to the $x^2$). Since denominator has a higher power of zero than the numerator, I believe that the limit won't exist.
However, my textbook says this limit exists and has a finite value. So, I wish to ask what is the fault in my reasoning.
| Note that for $a=1$ with $x=1+y$ and $y\to 0$
$$\frac{1}{ (1-x^2)^2 }\cdot\left(\frac{1+x^2}{x}-2\sin\frac{\pi x}2\right)
=\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{\pi y}2\right)=...$$
and by Taylor's series
$$\cos \frac{\pi y}2=1-\frac{\pi^2 y^2}8+o(y^2)$$
then
$$...=\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{\pi y}2\right)
=\\=\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2+\frac{\pi^2 y^2}4+o(y^2)\right)=\\=\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2-2-2y+\frac{\pi^2 y^2}4+o(y^2)}{y+1}\right)=\\=\frac{1}{ (y+2)^2 }\cdot\left(\frac{1+\frac{\pi^2 }4+o(1)}{y+1}\right)\to\frac14+\frac{\pi^2}{16}$$
For the general case just let $x=az$ with $z\to 1$ and note that
$$\lim_{x\to a}\frac{1}{(a^2-x^2)^2}\cdot\left(\frac{a^2+x^2}{ax}-2\sin\frac{a\pi}{2}\sin\frac{\pi x}2\right)=\\
=\lim_{z\to 1}\frac1{a^4}\frac{1}{(1-z^2)^2}\cdot\left(\frac{1+z^2}{z}-2\sin\frac{a\pi}{2}\sin\frac{\pi a z}2\right)$$
which can be solved with the same approach.
Notably note that for $z=1+y$ and $y\to 0$
$$\frac1{a^4}\frac{1}{ (1-z^2)^2 }\cdot\left(\frac{1+z^2}{z} -2\sin\frac{a\pi}{2}\sin\frac{\pi a z}2 \right)
=\frac1{a^4}\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{a \pi y}2\right)=...$$
and by Taylor's series
$$\cos \frac{a\pi y}2=1-\frac{a^2\pi^2 y^2}8+o(y^2)$$
then
$$...=\frac1{a^4}\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{a\pi y}2\right)
=\\=\frac1{a^4}\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2+\frac{a^2\pi^2 y^2}4+o(y^2)\right)=\\=\frac1{a^4}\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2-2-2y+\frac{a^2\pi^2 y^2}4+o(y^2)}{y+1}\right)=\\=\frac1{a^4}\frac{1}{ (y+2)^2 }\cdot\left(\frac{1+\frac{a^2\pi^2 }4+o(1)}{y+1}\right)\to\frac1{4a^4}+\frac{\pi^2}{16a^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Reduce $\frac{A(r)}{B^2(r)}(\frac{dr}{dt})^2+\frac{J^2}{r^2}-\frac{1}{B(r)} = -E$ for a slowly moving particle in a weak field I'm working in Weinberg's Gravitation and Cosmology and I'm having trouble seeing how he reduced a certain equation. He starts a derivation on p.185 going from a general metric in the standard form
$$d\tau^2=B(r)dt^2-A(r)dr^2-r^2d\theta^2-r^2\sin^2\theta d\varphi^2 \tag{8.4.1}.$$
From the geodesic (free-fall) equation $$\frac{d^2x^{\mu}}{dp^2}+\Gamma^\mu_{\nu\lambda}\frac{dx^\nu}{dp}\frac{dx^\lambda}{dp}=0$$ he works out 4 differential equations using a parameter p and the non-vanishing terms of the affine connection.
Then finds $$\frac{dt}{dp}=\frac{1}{B(r)} \tag{8.4.10},$$ angular momentum $$r^2\frac{d\varphi}{dp}=J\tag{8.4.11},$$ and energy $$A(r)(\frac{dr}{dp})^2+\frac{J^2}{r^2} - \frac{1}{B(r)}=-E \tag{8.4.13}.$$
Then eliminates the parameter p to get equations of motion back in terms of $t$.
He finds that:
$$r^2\frac{d\varphi}{dt}=JB(r)\tag{8.4.18}$$
$$\frac{A(r)}{B^2(r)}(\frac{dr}{dt})^2+\frac{J^2}{r^2 }-\frac{1}{B(r)}=-E\tag{8.4.19}$$
$$d\tau^2=EB^2(r)dt^2\tag{8.4.20}$$
Then he says:
For a slowly moving particle in a weak field $J^2/r^2$, $(dr/dt)^2$, $A-1$, and $B-1\approx2\phi$ will all be small, and to first order in these equations the above equations of motion become
$$r^2\frac{d\varphi}{dt}\approx J$$
$$\frac{1}{2}(\frac{dr}{dt})^2+\frac{J^2}{2r^2}+\phi\approx \frac{1-E}{2}$$
I'm sure this is just a question of algebra but for the life of me I can't see how eqn. 8.4.19 reduces to the last equation. I've wasted a ton of scrap paper trying to juggle the terms in 8.4.19 and letting $B^2=(1+2\phi)^2\approx(1+4\phi)$ but in never works out quite right. Maybe I'm missing something about how the approximations for slow motion and weak field interact with 8.4.19?
| If $B \approx 1 + 2\phi$ then $\frac{1}{B} \approx 1 - 2\phi$, hence the term before the equal sign becomes $(1 - 2\phi)$, hence
$$\frac{A}{B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{r^2} - (1 - 2\phi) = -E$$
$$\frac{A}{B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{r^2} + 2\phi = -E + 1$$
$$\frac{A}{2B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{2r^2} + \phi = \frac{-E + 1}{2}$$
It is also reasonable to assert that $\frac{A}{B^2} \approx 1$, which eventually brings you to
$$\frac{1}{2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{2r^2} + \phi = \frac{1 - E}{2}$$
More on the last assertion
$$\frac{A}{B^2} = \frac{A-1+1}{B^2} = \frac{1}{B^2}$$
Since $A-1$ is small. The rest is in the comments.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Summation Recurrence Relation How to solve this Summation Recurrence Relation: $$x_n=\sum_{i=1}^n a_ix_{n-i}\,,\,\,\,n\ge1$$where, $x_0=1$ and $a_n$ is some arbitrary sequence.
The right hand side of the recurrence looks partially like a discrete convolution and also like Cauchy's Product. I tried using Generating Functions, but I don't think they work because of the pesky $a_n$.
| As I understand the problem we have $A\cdot X = X$, displayed in matrix-arrangement
$$ \begin{array}{}
& * & \left [\begin{array}{} 1 \\ x_1 \\ x_2 \\ x_3
\end{array} \right] \\
\left [ \begin{array}{} 1 \\ a_1 &0 \\ a_2 & a_1 &0 \\ a_3 & a_2 & a_1 &0
\end{array} \right ]
& = &\left [\begin{array}{} 1 \\ x_1 \\ x_2 \\ x_3
\end{array} \right ]
\end{array} \tag 1 $$
This is, btw. an Eigenvector-problem: if we subtract on the left matrix A the identity we get $(A - I) \cdot X = 0 $ or in matrix-display
$$ \begin{array}{}
& * & \left [\begin{array}{} 1 \\ x_1 \\ x_2 \\ x_3
\end{array} \right] \\
\left [ \begin{array}{} 0 \\ a_1 &-1 \\ a_2 & a_1 &-1 \\ a_3 & a_2 & a_1 &-1
\end{array} \right ]
& = &\left [\begin{array}{} 0 \\ 0 \\ 0 \\ 0
\end{array} \right ]
\end{array} \tag 2$$
We can now separate the constant expression with the first column in $A-I$ from the rest and write
$$ \begin{array}{}
& & * & \left [\begin{array}{} x_1 \\ x_2 \\ x_3
\end{array} \right] \\
\left [\begin{array}{} a_1 \\ a_2 \\ a_3
\end{array} \right] + & \left [ \begin{array}{} -1 \\ a_1 &-1 \\ a_2 & a_1 &-1
\end{array} \right ]
& = &\left [\begin{array}{} 0 \\ 0 \\ 0
\end{array} \right ]
\end{array} \tag 3 $$
Let us denote the left column-vector as B and the left square-matrix as C then we have
$$ \begin{array}{rl} B + C\cdot X& = 0 & \text{and can rearrange for solving} \\
C \cdot X &= - B \\
X &= - C^{-1} \cdot B \end{array} \tag 4
$$
Solution: Using the symbolic feature in Pari/GP this gives (I used $x_1 ... x_4$ here)
$$ \begin{array} {}
\left[\begin{array} {rrrr}
a_1 \\
a_1^2+a_2 \\
a_1^3+2 \cdot a_2 \cdot a_1+a_3 \\
a_1^4+3 \cdot a_2 \cdot a_1^2+2 \cdot a_3 \cdot a_1+a_2^2+a_4 \\
\end{array} \right]& = &
\left[\begin{array} {rrrr}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{array} \right]
\end{array} \tag 5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$ C++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't look much alike. Any help will be appreciated. Thanks!
| $$\begin{align}(\color{red}{n^2}+\color{blue}{3n+1})^2&=(\color{red}{n^2})^2+2\color{red}{n^2}(\color{blue}{3n+1})+(\color{blue}{3n+1})^2\\&=n^4+2(3n^3+n^2)+(3n)^2+2(3n)(1)+1^2\\&=n^4+6n^3+2n^2+9n^2+6n+1\\&=n^4+6n^3+11n^2+6n+1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$ I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$
$\sum_{i=n}^{2n-1} 3+ \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ am I right?
$4 \sum_{i=0}^{2n-1-n} i+n $
$4 (\sum_{i=0}^{n-1} i+ \sum_{i=0}^{n-1} n )$ = $4(\frac{n(n-1)}{2}+n(n-1))$=$2(3n^2-3n)=6n^2-6n$
total= $3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $=$3n+ 6n^2-6n $, am I right???
| Alternatively:
$$\begin{align}\sum_{i=n}^{2n-1}(3+4i)&=\sum_{i=0}^{2n-1}(3+4i)-\sum_{i=0}^{n-1}(3+4i)=\\
&=3\cdot 2n+4\cdot \frac{2n(2n-1)}{2}-3n-4\cdot \frac{n(n-1)}{2}= \\
&=6n+8n^2-4n-3n-2n^2+2n=\\
&=6n^2+n=\\
&=n(6n+1).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Multivariable delta-epsilon limit proofs The Question asks: Find the delta epsilon limit if it exist
$$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$
What I've done so far:
$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$
Now, since $x ^2 \leq 2x^2 + y^ \implies \frac{x^2}{2x^2 + y^2} \leq 1$
Then:
$$ \frac{x^2 \mid y^3 \mid}{2x^2 + y^2} \leq \mid y^3 \mid = \sqrt{y^2}^3 = \quad ...$$
Here is where I am stuck, I can't make $\sqrt{y^2}^3$ into $\sqrt{ x^2 + y^2} $.
How should I proceed further?
Using Stewart's Calculus (8th ed) 14.2 Limit proof method
| For both $x,y\ne 0$, then
\begin{align*}
\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}&\leq\dfrac{|x|^{2}|y|^{3}}{2\sqrt{2}|x||y|}\\
&=\dfrac{1}{2\sqrt{2}}|x||y|^{2}\\
&\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),
\end{align*}
if either $x=0$ or $y=0$ (but not both), the above inequality is still true. So
\begin{align*}
\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),~~~~(x,y)\ne(0,0).
\end{align*}
Given $\epsilon>0$, for all $0<\sqrt{|x|^{2}+|y|^{2}}<2^{1/2}\epsilon^{1/3}$, then $\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2})=\dfrac{1}{2\sqrt{2}}(|x|^{2}+|y|^{2})^{3/2}<\epsilon$.
The limit is zero.
Another way:
\begin{align*}
\dfrac{x^{2}}{2x^{2}+y^{2}}\leq 1,~~~~(x,y)\ne(0,0),
\end{align*}
so
\begin{align*}
\dfrac{x^{2}|y|^{3}}{2x^{2}+y^{2}}\leq|y|^{3}=(y^{2})^{3/2}\leq(x^{2}+y^{2})^{3/2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why $\frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt0$? Why does the following inequality hold for $x > 0$?
$$
Q(x)<e^{-x^2/2}, \tag{1}
$$
where
$$
Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt. \tag{2}
$$
The following is my attempt: Define
$$
\begin{align}
f(x)
& \triangleq e^{-x^2/2} - Q(x) \\
& = e^{-x^2/2} - \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt \\
& = e^{-x^2/2} - \left( 1 - \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt \right) \\
& = e^{-x^2/2} - 1 + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-t^2/2} dt.
\tag{3}
\end{align}
$$
Then
$$
\begin{align}
f'(x)
& = -\frac{2x}{2} e^{-x^2/2} + \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \\
& = \left(-x+\frac{1}{\sqrt{2\pi}} \right) e^{-x^2/2}. \tag{4}
\end{align}
$$
When $0<x<\frac{1}{\sqrt{2\pi}}$, $f'(x)>0 \Rightarrow f(x)$ is a monotonically increasing function.
When $x=\frac{1}{\sqrt{2\pi}}$, $f'(x)=0$.
When $x>\frac{1}{\sqrt{2\pi}}$, $f'(x)<0 \Rightarrow f(x)$ is a monotonically decreasing function.
So $f(x)$ has a local maximum at $x = \frac{1}{\sqrt{2\pi}}$.
$$
\lim_{x \rightarrow \infty} f(x) = 0 - 0 =0.
$$
$$
f(0) = e^{-0^2/2} - Q(0) = 1 - \frac{1}{2} = \frac{1}{2}.
$$
Then I don't know how to continue. Any comments and answers are welcome. Thanks in advance.
| Your idea is quite good: if
$$
f(x)=e^{-x^2/2}-\frac{1}{\sqrt{2\pi}}\int_x^{\infty} e^{-t^2/2}\,dt
$$
then $f(0)=1/2$ and $\lim_{x\to\infty}f(x)=0$. On the other hand,
$$
f'(x)=-xe^{-x^2/2}+\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
$$
so $f$ is increasing in the interval $[0,1/\sqrt{2\pi}]$ and decreasing in the interval $[1/\sqrt{2\pi},\infty)$.
Hence $f(x)>0$ for every $x>0$.
When a function is decreasing over an interval $[a,b)$ (where $b$ can be $\infty$), then
$$
\lim_{x\to b}f(x)=\inf_{x\in[a,b)}f(x)
$$
Thus $\inf_{x\in[1/2\pi,\infty)}f(x)=0$. The infimum cannot be reached, or the function would be constant in some interval $[c,\infty)$, which it isn't, because the derivative is not zero over $(1/\sqrt{2\pi},\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let
\begin{eqnarray}
x= \sqrt[3]{a} \\
y= \sqrt[3]{b} \\
z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that
\begin{equation}
x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(x+z)(y+z)
\end{equation}
that turns out to be
\begin{equation}
(x+y+z-1)(x+y+z)(x+y+z-1)=3(x+y)(x+z)(y+z)
\end{equation}
plug in $x+y+z$,
we will get
\begin{equation}
\sqrt[3]{2}-1 - \sqrt[3]{\sqrt[3]{2}-1 } = 3(x+y)(x+z)(y+z)
\end{equation}
From now I stuck to solve for the rational numbers of $a,b,c$. Could anyone show me the way how to continue to solve it?
| Here is an alternative to denesting $(2^{1/3}-1)^{1/3}$. First, set $x^3=2$ so that$$x^3-1=1$$Factoring the left-hand side and isolating $x-1$ gives$$x-1=\frac 1{1+x+x^2}=\frac 3{3+x+3x^2}=\frac 3{(1+x)^3}$$Multiply both sides by $9$ to complete the cube$$9(x-1)=\left(\frac 3{1+x}\right)^3$$Cube root both sides and set $x=\sqrt[3]{2}$ gives$$\sqrt[3]{\sqrt[3]2-1}=\frac {3}{1+\sqrt[3]2}=1-\sqrt[3]2+\sqrt[3]4$$Hence$$\sqrt[3]{\sqrt[3]2-1}\color{blue}{=\sqrt[3]{\frac 19}-\sqrt[3]{\frac 29}+\sqrt[3]{\frac 49}}$$A similar technique can be done to show that$$\sqrt[3]{7\sqrt[3]{20}-1}=\sqrt[3]{\frac {16}9}-\sqrt[3]{\frac 59}+\sqrt[3]{\frac {100}9}$$
| {
"language": "en",
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"answer_count": 4,
"answer_id": 1
} |
$A289B$ is divisible by $90$, then what is $A+B$? It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?
My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
since it is divisible by $90$, the sum of the remainders of $A \times 1000$ and $2890$ being divided by 90, must be divisible by 90.
We have $mod(2890, 90) = 10$.
Now we may find $A$ such that $mod(A \times 10000, 90) = 80$.
$$mod(10000, 90) = 10 \implies 10000 = (90 \times 111 + 10) $$
so $A$ must be $8$, since
$$ 8 \times 10000 + 2890 = 8 \times (90 \times 111 + 10) + (90 \times 32 + 10) = 90 \times (888+32) + 90 $$
is divided by 90.
$$A+B = 8$$
Is this the best approach..? Thanks.
| If an integer is divisible $9$ then the sum of the digits is also divisible by $9$. So
$$A+2+8+9 +B \equiv A+B + 1 \equiv 0 \pmod{9}$$
shows $A+B \equiv_9 8$. Then since $B=0$ like you noted, $A+B=8$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
limit of a recursive sequence_1
I have this sequence
$$ \begin{cases}a_1= \alpha +2 & \alpha >0\\
a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} & n \ge1\end{cases}$$
and I want to know
$\lim_\limits{n\rightarrow \infty} a_n$
I can prove by induction that the sequence is a monotone increasing.
$$
a_1= \alpha +2
$$
$$
a_2= \alpha +\frac{3}{2}
$$
$$
a_3= \alpha^2+2\alpha +\frac{4}{3}
$$
$$
a_4= (\alpha^2+2\alpha +1)^3+\frac{1}{4}
$$
if we suppose that $$a_n>a_{n-1} \rightarrow a_n=\left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}+ \frac{1}{n}>a_{n-1} \rightarrow \left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}>a_{n-1}- \frac{1}{n}$$
Then, $$a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1}=a_{n+1}=\left(\left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}+ \frac{1}{n}-\frac{1}{n}\right)^n+\frac{1}{n+1}=\left(a_{n-1}- \frac{1}{n-1}\right)^{n(n-1)}+\frac{1}{n+1}>\left(a_{n-1}- \frac{1}{n}\right)^{n}+\frac{1}{n+1}$$
$$a_{n+1}>\left(a_{n-1}- \frac{1}{n}\right)^{n}+\frac{1}{n+1}=a_n \rightarrow a_{n+1}>a_n$$
We can say that $a_{n+1}>a_n>0$
$$\lim_{n\rightarrow \infty} a_n=\sigma$$ can be a $\alpha \in R$ or $\infty$.
In particular we can say that:
$$a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} \rightarrow a_{n+1}- \frac{1}{n+1}=\left(a_n- \frac{1}{n}\right)^n \rightarrow \sigma= \sigma^n \rightarrow \sigma=+ \infty$$
| You can try unfolding the recursion:
\begin{align}
a_{n+1} &= \left(a_n - \frac1n\right)^n + \frac{1}{n+1}\\
&= \left(a_{n-1} - \frac1{n-1}\right)^{(n-1)n} + \frac{1}{n+1}\\
&= \left(a_{n-2} - \frac1{n-2}\right)^{(n-2)(n-1)n} + \frac{1}{n+1}\\
&\cdots\\
&= \left(a_{2} - \frac1{2}\right)^{2\cdot3 \cdots(n-1)n} + \frac{1}{n+1}\\
&= \left(a_1 - 1\right)^{n!} + \frac{1}{n+1}\\
&= \left(\alpha +1\right)^{n!} + \frac{1}{n+1}\\
\end{align}
Letting $n \to\infty$ gives $\lim_{n\to\infty} a_{n+1} = +\infty$ because $\alpha + 1 > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve DE: $x^{3}y''' + 6x^{2}y'' + 4xy' - 4y = 1 - x - 2x^{2}$ Find particular solution for:
$x^{3}y''' + 6x^{2}y'' + 4xy' - 4y = 1 - x - 2x^{2}$
$\forall x > 0$
I believe that this problem can be solved by means of the method of undetermined coefficients. Something like this:
$x^3y_1''' + 6x^2y_1'' + 4y_1' -4y_1 = 1 - x \rightarrow y_1 = C_1 - C_2x$
$x^3y_2''' + 6x^2y_2'' + 4y_2' -4y_2 = -2x^2 \rightarrow y_2 = - \frac{2}{3}C_3x^3$
Unfortunately, I don't really know all of the steps. Can someone provide further explanation on this method of solving differential equations?
| Its been a while. I believe this is called an Euler equation.
Try assuming a homogeneous solution of the form $y=x^n$ and work from there. You should end up having to solve a cubic equation for $n$ that will provide you with three homogeneous solutions. Some linear combination will be in your final answer.
$x^3 y''' + 6x^2 y'' + 4xy' - 4y = 0$
$x^3 (x^n)''' + 6x^2 (x^n)'' + 4x(x^n)' - 4(x^n) = 0$
$x^3 n(n-1)(n-2)x^{n-3} + 6x^2 n(n-1)x^{n-2} + 4xnx^{n-1} - 4x^n = 0$
$n(n-1)(n-2)x^{n} + 6n(n-1)x^{n} + 4nx^{n} - 4x^n = 0$
$n(n-1)(n-2) + 6n(n-1) + 4n - 4 = 0$
$n^3 + 3n^2 - 4 = 0$
$(n-1)(n+2)^2 = 0$
This implies that your homogeneous solutions are $y = x^1$, $y=x^{-2}$ and, because a double root, the third solution is of the form $y=\ln(x)x^{-2}$.
Your homogeneous solution is therefore $y(x)=Ax + Bx^{-2}+ C\ln(x)x^{-2}$.
Now you can do your magic for the particular solutions.
Then evaluate the constants with the IVP, if you have one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)d\left(\frac{\sin x}{1+\cos x}\right)$$
$$=\int_{0}^{\frac{\pi}{2}}x^2d\log\left(\frac{\sin x}{1+\cos x}\right)=x^2\log\left(\frac{\sin x}{1+\cos x}\right)|_{0}^{\frac{\pi}{2}}-2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{\sin x}{1+\cos x}\right)dx$$
$$=0+2\int_{0}^{\frac{\pi}{2}}x\log\left(\frac{1+\cos x}{\sin x}\right)dx=2\int_{0}^{\frac{\pi}{2}}x\log\left(1+\cos x\right)dx-2\int_{0}^{\frac{\pi}{2}}x\log\left(\sin x\right)dx$$
$$=2\int_{0}^{\frac{\pi}{2}}x\log\cot \left(\frac{x}{2}\right)dx=8\int_{0}^{\frac{\pi}{4}}x\log\cot xdx$$
but I can't proceed next step,help me,thanks.
| I will present an evaluation that makes use of the following two Euler sums:
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3) \qquad \text{and} \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} = \frac{23}{16} \zeta (3) - \pi \mathbf{G}.$$
Here $\mathbf{G}$ is Catalan's constant. For a proof of the first, see either here or Eq. (646) in this link. For a proof of the second, see Eq. (659) of this link.
Begin by enforcing a substitution of $x \mapsto 2 \arctan x$. Then
$$\int_0^{\frac{\pi}{2}} \frac{x^2}{\sin x} \, dx = 4 \int_0^1 \frac{\arctan^2 x}{x} \, dx.\tag1$$
Since $\displaystyle{\arctan x = \sum_{n = 0}^\infty \frac{(-1)^n x^{2n + 1}}{2n + 1}}$ for $|x| < 1$, applying the Cauchy product to the product between the two inverse tangent functions one obtains the following Maclaurin series expansion for $\arctan^2 x$ of:
$$\arctan^2 x = \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) x^{2n},$$
where $H_n$ is the $n$th Harmonic number.
Substituting the Maclaurin series expansion for $\arctan^2 x$ into (1), after changing the order of the summation with the integration one obtains
\begin{align}
\int_0^{\frac{\pi}{2}} \frac{x^2}{\sin x} \, dx &= 4 \sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n} \left (H_{2n} - \frac{1}{2} H_n \right ) \int_0^1 x^{2n - 1} \, dx\\
&= -2 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} + \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2}\\
&= -2 \left (\frac{23}{16} \zeta (3) - \pi \mathbf{G} \right ) - \frac{5}{8} \zeta (3)\\
&= 2 \pi \mathbf{G} - \frac{7}{2} \zeta (3),
\end{align}
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 2
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Simplistic Odd Collatz formulas I've been experimenting with the Collatz for a couple of months.
I just want to give you some formulas and show tables that give the same results for odd inputs. I have also several new functions, but ill only provide a few of them here. I tweaked the formulas. It seems very strange that all of them give the same outputs. My notes are a bit messy, so bear with me if I didnt provide all the details.
\begin{array}{|c|c|c|c|}\hline
n & 3n & 3n+1 & \frac{3n+1}{2} \\ \hline
1 & 3 & 4 & 2 \\ \hline
3 & 9 & 10 & 5 \\ \hline
5 & 15 & 16 & 8 \\ \hline
7 & 21 & 22 & 11 \\ \hline
9 & 27 & 28 & 14 \\ \hline
11 & 33 & 34 & 17 \\ \hline
13 & 39 & 40 & 20 \\ \hline
\end{array} orginal collatz formula, dividing by two.
\begin{array}{|c|c|c|c|}\hline
n & 2n & \lfloor\frac{n}{2}\rfloor & 2n-\lfloor\frac{n}{2}\rfloor \\ \hline
1 & 2 & 0 & 2 \\ \hline
3 & 6 & 1 & 5 \\ \hline
5 & 10 & 2 & 8 \\ \hline
7 & 14 & 3 & 11 \\ \hline
9 & 18 & 4 & 14 \\ \hline
11 & 22 & 5 & 17 \\ \hline
13 & 26 & 6 & 20 \\ \hline
\end{array} a new(?) formula.
\begin{array}{|c|c|c|c|}\hline
n & \lfloor\frac{n}{2}\rfloor & n+1 & \lfloor\frac{n}{2}\rfloor + n+1 \\ \hline
1 & 0 & 2 & 2 \\ \hline
3 & 1 & 4 & 5 \\ \hline
5 & 2 & 6 & 8 \\ \hline
7 & 3 & 8 & 11 \\ \hline
9 & 4 & 10 & 14 \\ \hline
11 & 5 & 12 & 17 \\ \hline
13 & 6 & 14 & 20 \\ \hline
\end{array} yet another.
Here $n$ is a positive integer. The floor function omits the fractional part if the result is real.
The functions from the tables above:
$f(n) = \frac{3n+1}{2}$
$f(n) = 2n-\lfloor\frac{n}{2}\rfloor$
$f(n) = \lfloor\frac{n}{2}\rfloor + n+1$
The latter formula is the one i've been working on lately.
We can change it again,
$\lfloor\frac{n}{2}\rfloor+n+1 = \frac{n}{2}+\frac{2n}{2}+\frac{2}{2} = \frac{3n+2}{2}=\lfloor\frac{3n}{2}\rfloor+1$.
That gets us back to the formula in the first table. They look quite similar.
The function
$f(n) = 2n-\lfloor\frac{n}{2}\rfloor$
is also quite interesting. In a "geometric" standpoint we can look at it like doubling the $n$ and then removing the half of the total, here's a simple figure showing this:
I have never seen this formula before, and wonder if I can use this notation:
$f(n)=\lfloor\frac{n}{2}\rfloor+\begin{cases} n+1 & n\equiv 1\pmod2 \\ 0 & n\equiv 0\pmod2 \end{cases}$
See, for odd and even we can halve the input in one go. And then add the result of the cases.
| Sure, there's nothing strange going on here: these formulas are all equivalent. (Can you prove it?)
Hint: if $n$ is odd then $\lfloor \frac{n}{2} \rfloor = \frac{n}{2} - \frac{1}{2}$.
| {
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"url": "https://math.stackexchange.com/questions/2714923",
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"source": "stackexchange",
"question_score": "2",
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Why is $y_{n+1}=\frac{1}{2}(y_n+\sqrt{\frac{1}{2^{2n}}+y_n^2})$ giving the inverse of $\pi$? A simple and interesting recursion:
$$y_{n+1}=\frac{1}{2}(y_n+\sqrt{\frac{1}{2^{2n}}+y_n^2})$$
has these curious solutions
$$y_1=-\infty,y_{\infty}=\frac{1}{2\pi}$$
$$y_1=-\frac{1}{2},y_{\infty}=\frac{2}{3\pi}$$
$$y_1=0,y_{\infty}=\frac{1}{\pi} $$
$$y_1=\frac{1}{2},y_{\infty}=\frac{2}{\pi}$$
Cannot find it in the literature as such and it does not look like coming from AGM, but I suspect elliptic integrals. Still cannot start from anywhere for some time. Any ideas?
| The closed form of the limit doesn't come from elliptic integral, but from half-angle formula for cotangent function.
To illustrate this result, we will only condiser the case where $y_1 > 0$.
Consider the auxillary sequence $a_n = 2^n y_n$, it satisfies
$$a_{n+1} = a_n + \sqrt{1 + a_n^2}$$
If $a_n = \cot(\theta)$ for some $\theta \in (0,\frac{\pi}{2})$, then
$$a_{n+1} = \frac{\cos\theta}{\sin\theta} + \sqrt{1 + \frac{\cos^2\theta}{\sin^2\theta}} = \frac{1+\cos\theta}{\sin\theta} = \cot\frac{\theta}{2}$$
Using this and the assumption $y_1 > 0$, we find following closed form expression of $y_{n+1}$.
$$2^{n+1}y_{n+1} = a_{n+1} = \cot\left(\frac{\cot^{-1}(2y_1)}{2^n}\right)$$
This leads to
$$y_\infty = \lim_{n\to\infty} y_{n+1} = \lim_{n\to\infty} \frac{1}{2^{n+1}}
\cot\left(\frac{\cot^{-1}(2y_1)}{2^n}\right) = \frac{1}{2\cot^{-1}(2y_1)}\quad\text{ whenever}\quad y_1 > 0$$
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:
$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$
holds
I don't see room to use derivatives, since it seems a little messy to calculate the $\lim_{x\to 0}$ of $\frac{2\ln(\cos{x})}{x^2}$
(which, I think, is necessary in order to make usage of derivatives).
Any hints? I've already tried cross multiplying but it doesn't lead anywhere, unless I missed something.
| Note that since $e^x$ is strictly increasing
$$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1\iff\ln \cos x< \frac{x^4}{24}- \frac{x^2}{2}\iff \cos x<e^{\frac{x^4}{24}- \frac{x^2}{2}}$$
and since
*
*$e^x>1+x$
*$\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$
(refer to Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$)
we have
$$e^{\frac{x^4}{24}- \frac{x^2}{2}}>1-\frac{x^2}{2}+\frac{x^4}{24}>\cos x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
| $$
\begin{align*}
2x^2 - 9x + 9 &= 2\left(x^2 - \frac{9}{2}x + \frac{9}{2}\right)
\\&= 2\left(x^2 - 2\frac{9}{4}x + \frac{9}{2}\right)
\\&= 2\left(x^2 - 2\frac{9}{4}x + \frac{9}{2} + \left(\frac{9}{4}\right)^2 - \left(\frac{9}{4}\right)^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 + \frac{9}{2} - \left(\frac{9}{4}\right)^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 - \left(-\frac{9}{2} + \left(\frac{9}{4}\right)^2\right)\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 - \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right) - \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)\left(\left(x-\frac{9}{4}\right) + \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)
\\&= 2\left(x-\frac{9}{4} - \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)\left(x-\frac{9}{4} + \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)
\end{align*}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show the equivalence of norms on polynomial space Let $\|ax+b\|^2:=2a^2+3b^2+2ab$ be a norm on the space of all polynomials of degree one, where $a,b\in\mathbb{R}$.
Hence, $\|ax+b\|_{L^2} ^2:=\int_0^1(ax+b)^2\,\mathrm{d}x$ be another norm ($L^2$-norm).
I want to show equivalence, in the sense that, for all polynomials of degree one
$\exists\, c>0\colon \|\cdot\|\le c\|\cdot \|_{L^2}$ holds and vice versa.
The first part was not too hard,
\begin{align*}
\|ax+b\|_{L^2}^2 &= \int_0^1 (ax+b)^2\,\mathrm{d}x = a^2/3+ab+b^2\le (a+b)^2-ab\le (a+b)^2+\frac{a^2+b^2}{2}\\
&\le \frac{3}{2}a^2+2ab+\frac{3}{2}b^2\le 2a^2+2ab+3b^2=\|ax+b\|^2
\end{align*}
Is there any way to do it better? I noticed, that I got $c=1$, is that in some way significant?
The other way was resistant against all my attempts.
$\|ax+b\|^2=2a^2+2ab+3b^2=\ ...$
Here, I thought that finding a real constant $\alpha$ such that $\alpha(a^3/3+ab+b^2)\ge 2a^2+2ab+3b^2$ would yield success, since then I can find a suitable antiderivative to connect the norms. I was not successful, nevertheless.
| Both ways will always be possible for any two norms you care to choose. This is since the space of polynomials of degree less than or equal to one is finite dimensional, hence all norms are equivalent.
In this particular case, you've shown $\lVert ax + b \rVert_{L^2} \leq \lVert ax + b \rVert$. The other direction is not very pretty, but I think the following works. Pick $\frac{1}{2} < \lambda < \frac{1}{\sqrt{3}}$. Then we have that $\mu = \frac{1}{3} - \lambda^2 > 0$ and $\nu = 1 - \frac{1}{4\lambda^2} > 0$. We can then check that
\begin{align*}
\lVert ax + b \rVert_{L^2}
&= \left(\lambda a + \frac{1}{2 \lambda}b\right)^2 + \mu a^2 + \nu b^2
\end{align*}
Now pick $c$ such that $c > 2$, $c \mu > 2$ and $c \nu > 3$. Then we have
\begin{align*}
\lVert ax + b \rVert_{L^2}
&= \frac{1}{c} \left[ c\left(\lambda a + \frac{1}{2 \lambda}b\right)^2 + c\mu a^2 + c\nu b^2 \right] \\
&\geq \frac{1}{c} \left[ 2\left(\lambda a + \frac{1}{2 \lambda}b\right)^2 + 2 a^2 + 3 b^2 \right] \\
&= \frac{1}{c} \left[ 2\lambda^2 a^2 + 2ab + \frac{1}{2\lambda^2} b^2 + 2a^2 + 3b^2 \right] \\
&\geq \frac{1}{c} \left[2a^2 + 3b^2 + 2ab\right] \\
&= \frac{1}{c} \lVert ax + b \rVert^2
\end{align*}
Like I said, not very pretty. Hence why having a general theorem saying all norms on a finite dimensional space are equivalent is a nice thing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rewriting $\lfloor x \rfloor+\lfloor-x\rfloor$ How can I prove that
$$\lfloor x\rfloor + \lfloor -x\rfloor = \begin{cases} \phantom{-}0 &, x \in \mathbb{Z} \\ -1 &, x \notin \mathbb{Z} \end{cases}$$
where $x$ is a real number?
| Consider the following equation: $\lfloor x \rfloor - \lfloor x \rfloor = 0.$ Doh, of course, because $\lfloor x \rfloor = \lfloor x \rfloor$. The question is then: when does $\lfloor x \rfloor = |\lfloor -x \rfloor|$? (The straight bars here represent the absolute value function).
If $x$ is positive but not an integer, then $\lfloor -x \rfloor = -(\lfloor x \rfloor + 1)$, so $|\lfloor x \rfloor| < |\lfloor -x \rfloor|$. For example, if $x = \pi \approx 3.14159$, then $-x \approx -3.14159$ and $\lfloor -x \rfloor = -4$, and obviously $3 < |-4|$.
Similarly, if $x$ is negative but not an integer, then $\lfloor -x \rfloor = -(\lfloor x \rfloor + 1)$ just the same, but $|\lfloor x \rfloor| > |\lfloor -x \rfloor|$. Example: $x = -\frac{3}{2}$. Then $$\lfloor -x \rfloor = \left\lfloor \frac{3}{2} \right\rfloor = 1,$$ and obviously $|-2| > 1$.
But if $x$ is an integer, be it positive or negative or 0, then $\lfloor x \rfloor = |\lfloor -x \rfloor|$. For example, if $x = -7$, then $\lfloor x \rfloor = -7$ and the absolute value of that is 7, the same as the absolute value of $x$ itself, and $\lfloor x \rfloor$ as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How has this result been obtained?? We have the quadratic equation
$$ 2y^2 -(1+x)y + x = 0 $$
The author derives the following result from the above equation -
$$ y' = \frac{y-1}{4y-1-x} = \frac{(x-3)y-x+1}{x^2-6x+1} $$
Now, I understand the first part. It has been obtained by simply differentiating w.r.t to $x$ and then solving for $y'$. But, I am unable to obtain the second part.
I have tried lot of different substitutions, but haven't been able to arrive at the result. Please help!!
| $$(x^2-6x+1)(y-1)-(4y-x-1)((x-3)y-x+1)\\
=-2x^2-4xy^2-4xy+6x+12y^2+2yx^2-6y\\
=(6-2x)(2y^2-(1+x)y+x)\\=0.$$
The identity is valid.
To establish it, solve the initial equation for $y$
$$y=\frac{x+1\pm\sqrt{x^2-6x+1}}4$$
and plug in the fraction, to get
$$y'=\frac{x-3\pm\sqrt{x^2-6x+1}}{\pm4\sqrt{x^2-6x+1}}=\frac{\pm(x-3)\sqrt{x^2-6x+1}+x^2-6x+1}{4(x^2-6x+1)}=\frac{4(x-3)y+x^2-6x+1-(x-3)(x+1)}{4(x^2-6x+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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