Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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When is $m^2k^2(c^2+1)^2-4mc(c^2-c+1)$ a perfect square? Suppose, $m,k,c$ are positive integers
Conjecture : The expression $$m^2k^2(c^2+1)^2-4mc(c^2-c+1)$$ is a perfect square if and only if $m=k=1$
In the case $m=k=1$ , we get $(c-1)^4$ which is a perfect square ($0$ and $1$ are allowed)
The hard part is to show that otherwise the expression cannot be a perfect square. I tried to compare $(mk(c^2+1)\pm 1)^2$ with the given expression but this led to nowhere. The conjecture is true for $m,k,c\le 1\ 600$
I arrived at this problem by trying to prove that for positive integers $a,b,c$ with $c^2+1\mid a+b$ and $ab\mid c(c^2-c+1)$ we have $a=c$ , $b=c^2-c+1$ or vice versa.
| If we write $d=\gcd(m,4c(c^2-c+1))$, then there necessarily exist positive integers $a_1$ and $a_2$ such that $m=da_1^2$ and
$$
mk^2(c^2+1)^2-4c(c^2-c+1) = d a_2^2.
$$
Writing $t=a_1k$, we thus have
$$
t^2 (c^2+1)^2 -a_2^2 = \frac{4c(c^2-c+1)}{d}
$$
and so, taking $a_3=tda_2$ and multiplying by $t^2d^2$,
$$
\left( t^2 d (c^2+1) \right)^2 - a_3^2 = 4 t^2dc(c^2-c+1).
$$
If we define $a_4=t^2dc^2-2c+t^2d$, then
$$
\left( t^2 d (c^2+1) \right)^2 - a_4^2 = 4 t^2dc(c^2+1)-4c^2.
$$
If $t=d=1$, we are led to $a_3=a_4$ and so to the case $m=k=1$. Otherwise, $t^2d>1$ and hence necessarily have that $a_3 > a_4$. Checking that $a_3$ and $a_4$ have the same parity, it follows that $a_3 \geq a_4+2$. But
$$
\left( t^2 d (c^2+1) \right)^2 - (a_4+2)^2 = 4 t^2dc(c^2-c+1) - (4t^2d + 4(c-1)^2) < 4 t^2dc(c^2-c+1).
$$
This completes the proof. This argument is essentially Runge's method in disguise.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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The limit of $S_n=\sum_{k=1}^{n} \frac{kn}{k+n^3}$ Given $n>1$ and $S_n=\sum\limits_{k=1}^{n} \frac{kn}{k+n^3}$
Calculate $\lim\limits_{n \to \infty} S_n $
Well its obvious that $\frac{n^2}{n^3+n} \leq S_n\leq \frac{n^3}{n^3+1}$
$S_n$ converge to a limit $l$ such that $0\leq l \leq 1$
How can we determine the value of $l$ ?
| You can also get the result if you only estimate the denominator in order to get better bounds:
$$\frac{1}{2} \stackrel{n\to\infty}{\longleftarrow}\frac{n(n+1)}{2(1+n^2)} = \frac{n}{n+n^3} \sum \limits_{k=1}^n k \leq S_n \leq \frac{n}{1+n^3} \sum \limits_{k=1}^n k = \frac{n^2(n+1)}{2(1+n^3)} \stackrel{n\to\infty}{\longrightarrow} \frac{1}{2}\, . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2863962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
| The triangle has area
$$
\frac12 \times \sqrt2 \times (1- \sqrt2/2) = \frac{\sqrt2 - 1}{2},
$$
so the total polygon has area
$$
2\sqrt{2} - \frac{\sqrt2 - 1}{2} = \frac{3}{2}\sqrt{2} + \frac12 = \frac{3\sqrt{2} + 1}{2},
$$
at least in lowest terms. The result is $a + b + c + d = 8$, contrary to the 10 you claimed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Reduction of Order Leads to Non-Elementary Integral
If $u_1=x+1$ is a solution of $$xu''-(x+1)u'+u=0$$ find another linearly independent solution using reduction of order.
I let $u_2=(x+1)v(x)$ be the second solution. Hence
$$v''(x^2+x)-v'(x^2+1)=0$$ Let $w=v'$, so
\begin{align}
\frac{dw}{dx}(x^2+x)-w(x^2+1)&=0 \\
\frac{dw}{dx}&=\frac{(x^2+x)^{-1}(x^2+1)}{w^{-1}} \\
\text{ln}(w)&=\int \frac{x^2+1}{x^2+x} \ dx \\
\text{ln}(w)&=\int 1-\frac{1}{x}+\frac{2}{x+1} \ dx \\
\text{ln}(w)&=x+\text{ln}\left(\frac{(x+1)^2}{x}\right)+C \\
w&=C_1\frac{e^x(x+1)^2}{x} \\
v&=C_1\int \frac{e^x(x+1)^2}{x} \ dx
\end{align}
Where $$C_1\int \frac{e^x}{x} \ dx$$ cannot be solved. How do I find $v$?
| Your work is correct up to the partial fractions
$$ \frac{v''}{v'} = \frac{x^2+1}{x^2+x} = 1 + \frac{1-x}{x(x+1)} = 1 + \frac{1}{x} - \frac{2}{x+1} $$
Integrating this gives
$$ \ln(v') = x + \ln x - 2\ln(x+1) $$
$$ \implies v' = \frac{xe^x}{(x+1)^2} = \frac{e^x}{x+1}-\frac{e^x}{(x+1)^2} $$
This has the form $e^xf(x) + e^xf'(x) = (e^xf(x))'$, therefore
$$ v(x) = \frac{e^x}{x+1} $$
I've ignored the integration constants, since they're already included in the general solution, which is
$$ u(x) = (x+1)(c_1 + c_2v(x)) = c_1(x+1) + c_2e^x $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Transformation restriction to commutative matrices Let $T: \Bbb M_{3x3}(\Bbb R) \rightarrow \Bbb M_{3x3}(\Bbb R)$ be a linear transformation such that $T(B) = AB$ where:
$$ A=
\begin{pmatrix}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \\
\end{pmatrix}
$$
now let $W = \{B|AB=BA\}$ be the group of all matrices that commute with $A$.
I want to
*
*find the minimal polyonimal for $T|_w$.
*prove that $W$ is the direct sum of two spaces $W = W_1 \oplus W_2$
my idea so far was:
*
*Let P be any matrix:$$ P =\begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{pmatrix} $$
then we could look at the transformation as $\Bbb R^9 \rightarrow \Bbb R^9$ like that:
$$ T\begin{pmatrix}
a \\
d \\
g \\
b \\
e \\
h \\
c \\
f \\
i \\
\end{pmatrix} = \begin{pmatrix}
\begin{pmatrix}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \\
\end{pmatrix} & & \\
& \begin{pmatrix}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \\
\end{pmatrix} & \\
& & \begin{pmatrix}
2 & -1 & -1 \\
-1 & 2 & -1 \\
-1 & -1 & 2 \\
\end{pmatrix} \\
\end{pmatrix} \begin{pmatrix}
a \\
d \\
g \\
b \\
e \\
h \\
c \\
f \\
i \\
\end{pmatrix} $$
*Also, since A is real and symmetric - then it is similar to the diagonal matrix $$ [A]_o=
\begin{pmatrix}
0 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3 \\
\end{pmatrix}
$$
which implies that the minimal polynomial for the transformation $T$ is $m_A(x) = x(x-3)$
my idea was that the restriction to $W$ is in fact any matrix that commute with $A$, which means that $T|_w = A$, therefore this is a direct sum of 2 different eigenspace, but i'm not sure about it...
Can you help me to better understand the restriction?
Thanks..
| $W$ is a vector space and not a group. Note that $T=A\otimes I_3$ -if we stack the matrices row by row-. It is easy to see that $T(W)\subset W$. Moreover $spectrum(T)=\{0,0,0,3,3,3,3,3,3\}$; cf.
https://en.wikipedia.org/wiki/Kronecker_product
We may assume that $A=diag(0,3,3)$ -because $(P\otimes I)(A\otimes I)(P\otimes I)^{-1}=(PAP^{-1})\otimes I$-.
Thus $W=\{diag(a,R);a\in\mathbb{R},R\in M_2\}$ has dimension $5$. Moreover $W=W_1\oplus W_2$ where $W_1=span(E_{1,1}),W_2=span(E_{2,2},E_{2,3},E_{3,2},E_{3,3})$ and $T_{|W_1}=0,T_{|W_2}=3I$.
Finally, $T_{|W}$ has $2$ eigenvalues $0,3$ and $spectrum(T_{|W})=\{0,3,3,3,3\}$; its minimal polynomial divides $x(x-3)$; conclusion: its minimal polynomial is $x(x-3)$ again.
| {
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Finding the basis for the nullspace, the row space, and the column space of the given matrix.
Find the basis for the nullspace, the row space, and the column space of the given matrix.$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}$$
My Try
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 3 & 2 & 5 & 2 \\ -1 & 2 & 1 & 3 \\ 1 & 1 & 2 & 2 \end{bmatrix}_{R_2\rightarrow R_2-3R_1\\R_3\rightarrow R_3+R_1\\R_4\rightarrow R_4-R_1}$$
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & -4 & -4 & -1 \\ 0 & 4 & 4 & 4 \\ 0 & -1 & -1 & 1 \end{bmatrix}_{R_2\rightarrow \frac{R_2}{-1}\\R_3\rightarrow \frac{R_3}{4}}$$
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & -1 & -1 & 1 \end{bmatrix}_{R_3\rightarrow 4R_3-R_2\\R_4\rightarrow4R_4+R_2}$$
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$
I got solution as inconsistent. How to proceed further?
I also referred to $1^{st}$ question of This
| We don't say that the RREF is not consistent in that case.
We need another step to obtain
$$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix}\to \begin{bmatrix} 1 & 2 & 3 & 1 \\ 0 & 4 & 4 & 1 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
therefore a basis for the nullspace is given by solving $Ax=0$ that is $$(1,1,-1,0)$$
For the column space, a basis is formed by the columns of the original matrix containing the pivots in the RREF.
For the row space, as a basis we can select the first three rows in the RREF.
can you see why?
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of solution of $x^4-5x^3+(\lambda+2)x^2-5x+1=0$
Consider the bi-quadratic equation $E:x^4-5x^3+(\lambda+2)x^2-5x+1=0$ then, the real values of $\lambda$ so that $E$ has four different solutions is?
My attempts:
As $x=0$ is not a solution for any $\lambda$ hence we divide by $x^2$,\begin{align}x^2+\dfrac{1}{x^2}-5x-5\dfrac{1}{x}+\lambda+2&=0\\\bigg(x+\dfrac{1}{x}\bigg)^2-5\bigg(x+\dfrac{1}{x}\bigg)&=-\lambda\\t(t-5)&=-\lambda\end{align}
Here, $x+\dfrac{1}{x}=t$ cannot lie between $[-2,2]$.
Now we draw the graph of $-t(t-5)$ and $\lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+\lambda=0$, each of the roots of $Q$ gives two $x's$.
From GRAPH HERE, my solution is $\lambda\in(-\infty,-14]\cup[6,\frac{25}{4})$
But this wrong according to answer provided, please help.
| Let $f(x)=-\frac{x^4-5x^3+2x^2-5x+1}{x^2}.$
We need to find all values of $\lambda,$ for which the equation
$$\lambda=f(x)$$ has four different roots.
Now, $$f'(x)=\frac{(2-x)(x-1)(2x-1)(x+1)}{x^3},$$ which says that
$$\lambda<f(-1)$$ or $$f(1)<\lambda<\min\left\{f\left(\frac{1}{2}\right),f(2)\right\},$$
which gives the answer:
$$(-\infty,-14)\cup(6,6.25).$$
| {
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Matrix vector form. Is this in the correct form? I have this question:
Write the linear system
$$\begin{array}{rcr}-2x_1+x_2-4x_3 & = & 1 \\ x_1-2x_2 & = & -3 \\ x_1+x_2-4x_3 & = & 0 \end{array}$$
in the matrix-vector form $A\mathbf{x}=\mathbf{b}$.
Is this what they want?
$$
x_1*
\begin{bmatrix}
-2 \\ 1 \\ 1
\end{bmatrix}
+
x_2*
\begin{bmatrix}
1 \\ -2 \\ 1
\end{bmatrix}
+
x_3*
\begin{bmatrix}
-4 \\ 0 \\ -4
\end{bmatrix}
=
\begin{bmatrix}
1 \\ -3 \\ 0
\end{bmatrix}
$$
| basically!
$
A = \begin{pmatrix} -2 & 1 & -4 \\ 1 & -2 & 0 \\ 1 & 1 & -4 \end{pmatrix}
$
and $b = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$ yielding $Ax = b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Shorter Way to Solve $\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$ The limit to find is
$$\lim_{x\to 0} \frac{x^3-\ln^3(1+x)}{\sin^2x-x^2}$$
What I've tried was using the factorisation for $a^3-b^3$ and $a^2-b^2$ like so:
$$\lim_{x\to 0}\frac{(x-\ln(1+x))(x^2+x\ln(x+1)+\ln^2(x+1))}{(\sin x-x)(\sin x+x)}$$
but I don't know how to make a product of this so I get something meaningful (like a standard limit or something).
Lastly I've tried l'Hospital's rule and it works in the end, but after the second time the numerator and denominator are pretty long. Is there maybe a shorter solution to this?
| Using $\sin(x)=x-\frac{x^3}{6}+\mathcal{O}(x^5)$ and $\ln(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$:
$$\begin{aligned}\frac{x^3-\ln^3(1+x)}{\sin^2(x)-x^2}
&=\frac{x^3-\left(x-\frac{x^2}{2}+\mathcal{O}(x^3)\right)^3}{\left(x-\frac{x^3}{6}+\mathcal{O}(x^5)\right)^2-x^2}\\
&=\frac{x^3-\left(x^3- \frac{3 x^4}{2}+\mathcal{O}(x^5)\right)}{\left(x^2-\frac{x^4}{3}+\mathcal{O}(x^6)\right)-x^2}\\
&=\frac{\frac{3 x^4}{2}+\mathcal{O}(x^5)}{-\frac{x^4}{3}+\mathcal{O}(x^6)}
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Ratio of areas determined by a square inscribed in the corner of a right triangle I’m having trouble working out how to algebraically get to the answer of this question. (See original image below.)
A square is drawn in the corner of a right-angled triangle with side lengths $a$, $b$, and [hypotenuse] $c$, as shown.
Which expression gives the ratio of the unshaded area [inside the triangle, but outside the square] to the shaded area [of the square] in all cases?
*
*(A) $1:1$
*(B) $c:(a+b)$
*(C) $a b: c^2$
*(D) $( a + b )^2 : 2 c^2$
*(E) $c^2 : 2 a b$
Apparently the answer is $c^2 : 2 a b$ (choice E), but how?
Your help is greatly appreciated! Thank you in advance.
(Please ignore the pen marks! They are incorrect assumptions a friend made on the diagram.)
| Let the side of the square is $x$. Then, using similarity of triangles, $\frac{b-x}{x}=\frac{x}{a-x}$ or $(b-x)(a-x)=x^2$; $ab-ax-bx+x^2=x^2$; $ab=x(a+b)$ or $x=\frac{ab}{a+b}$. Thus, the shaded area is $\frac{(ab)^2}{(a+b)^2}$. The unshaded area is $$0.5x(b-x)+0.5x(a-x)=0.5\left(\frac{ab}{a+b}(b-\frac{ab}{a+b})+\frac{ab}{a+b}(a-\frac{ab}{a+b}\right)=\frac{ab(b^2+a^2)}{2(a+b)^2}=\frac{abc^2}{2(a+b)^2}$$ so the final ratio is $$\frac{2ab}{c^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that inequality $\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$ Let $a;b\ge 0$. Prove that inequality $$\frac{2ab}{a+b}+\sqrt{\frac{a^2+b^2}{2}}\ge \sqrt{ab}+\frac{a+b}{2}$$
My try: $LHS-RHS=\frac{2ab}{a+b}-\frac{a+b}{2}+\sqrt{\frac{a^2+b^2}{2}}-\sqrt{ab}\ge 0$
Or $-\frac{\left(a-b\right)^2}{2\left(a+b\right)}+\frac{\left(a-b\right)^2}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge 0$
Or $\left(a-b\right)^2\left(\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}-\frac{1}{2\left(a+b\right)}\right)\ge 0$
Then i can't prove $\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$
I squared the two sides of the inequality but the exponential is hard to solve. And I can see $HM+QM\ge AM+GM$ with $n=2$ so is that true for $n=i$?
| I think this is the simplest solution:
So you are stuck here:
$$\frac{1}{2\left(\sqrt{\frac{a^2+b^2}{2}}+\sqrt{ab}\right)}\ge \frac{1}{2\left(a+b\right)}$$
...which is equivalent to:
$$\sqrt\frac{a^2+b^2}{2}+\sqrt{ab}\le a+b\tag{1}$$
Apply Jensen inequality:
$$\frac{f(x_1)+...+f(x_n)}{n}\le f(\frac{x_1+...+x_n}{n})$$
...for concave function $f(x)=\sqrt{x}$ and $x_1=\frac{a^2+b^2}{2}$, $x_2=ab$:
$$\frac{\sqrt\frac{a^2+b^2}{2}+\sqrt{ab}}2\le \sqrt{\frac{\frac{a^2+b^2}{2}+ab}{2}}\tag{2}$$
Simplify (2) and you get (1) directly.
| {
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"timestamp": "2023-03-29T00:00:00",
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$x^4+x^3+x^2+x+1$ irreducible over $\mathbb F_7$ This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\\ac+b+d=1\\bc+ad=1\\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
| HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-\dfrac 14(-2x^2+(\sqrt5-1)x-2)(2x^2+(\sqrt5+1)x+2)$$
Since $\mathbb F_7^2=\{1,2,4,0\}$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.
Besides that a linear factor is not possible is checked easily.
| {
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Number of non negative integral values of $n,n\le 10$ so that a root of the equation $n^2\sin^2 x-2\sin x-(2n+1)=0$ lies in the interval $[0,\pi/2]$ Find the number of non negative integral values of $n,n\le 10$ so that a root of the equation $n^2\sin^2 x-2\sin x-(2n+1)=0$ lies in the interval $[0,\pi/2]$
As $x\in[0,\pi/2]$ so $\sin x\in[0,1]$ and letting $\sin x=t$,we get $n^2t^2-2t-2n-1=0$ I am stuck here.
| $n^2 t^2 -2t - 2n -1=0$ is a quadratic in $t$. Solving it for $t$, we get:
$$
\begin{align}
t &= \frac{2 \pm \sqrt{4-4n^2(-2n-1)}}{2n^2}
\\ &= \frac{1 \pm \sqrt{1+n^2(2n+1)}}{n^2}
\end{align}
$$
It can easily be checked that $n=0$ does not lead to $t \in [0,1]$ so we may assume that $n \neq 0$.
Now for $t \in [0,1]$ we must have:
$$
0 \leq \frac{1 \pm \sqrt{1+n^2(2n+1)}}{n^2} \leq 1
\\ \implies 0 \leq 1 \pm \sqrt{1+n^2(2n+1)} \leq n^2
\\ \implies -1 \leq \pm \sqrt{1+n^2(2n+1)} \leq n^2 -1
\\ \implies 1 \leq 1+n^2(2n+1) \leq n^4 -2n^2 +1
\\ \implies 0 \leq n^2(2n+1) \leq n^4 -2n^2
\\ \implies 0 \leq 2n+1 \leq n^2 -2
$$
For $0 \leq n$ we automatically have $0 \leq 2n+1$. Solving the quadratic inequality, we get $3 \leq n$.
| {
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} |
Given the equation $\sum_{k=11}^{99} \left[x + \frac{k}{100} \right] = 765$ find $[10 \, x]$.
Given the equation $\sum_{k=11}^{99} \left[x + \frac{k}{100} \right] = 765$ find $[10 \, x]$.
I have tried this problem in many ways and I think it uses the identity
$$[x]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\right]+\left[x+\frac{3}{n}\right]+.....+\left[x+\frac{n-1}{n}\right]= [nx]$$
Could someone please help me
and keep in mind x is a real number and $[\cdot]$ denotes the greatest integer function
| This is a well known identity known as Hermite's identity. A simple proof is here (Although, Hermite's original proof relied on identities involving exponents)
https://en.wikipedia.org/wiki/Hermite%27s_identity
For general setup, we can also do the following.
\begin{eqnarray*}
S &=& \sum_{k=m+1}^{n-1}{\left\lfloor x+\frac{k}{n} \right\rfloor} \\
&=& l \left\lfloor x \right\rfloor + (n-1-m) \left( 1+ \left\lfloor x \right\rfloor\right) \\
\end{eqnarray*}
\begin{eqnarray*}
\left\lfloor x \right\rfloor &\ge& \frac{S}{n-m-1} \\
&=& \left\lceil \frac{S}{n-m-1}\right\rceil
\end{eqnarray*}
Using this, we can solve $l$,
\begin{eqnarray*}
l &=& (n-m-1) \left( 1+ \left\lfloor x \right\rfloor\right) -S
\end{eqnarray*}
Now observe that,
\begin{eqnarray*}
\left\lfloor x+ \frac{k}{n} \right\rfloor &=& \begin{cases} \lfloor x \rfloor, &0\le k \le m+l \\ 1+ \lfloor x \rfloor, &k > m+l \end{cases}.
\end{eqnarray*}
For any $\epsilon \in [0,1]$, s.t., $q=n\epsilon, q \in \mathbb{Z}_{+}$,
\begin{eqnarray*}
\left\lfloor x+ \frac{k}{\epsilon n} \right\rfloor &=& \begin{cases} \lfloor x \rfloor, &0\le k \le \lfloor \epsilon(m+l) \rfloor \\ 1+ \lfloor x \rfloor, &k > \lfloor \epsilon (m+l) \rfloor \end{cases}.
\end{eqnarray*}
Now then the desired partial Hermite sum, $S_{\epsilon} (x)= \lfloor \epsilon n x \rfloor = \sum_{k=0}^{n\epsilon-1}{\left\lfloor x+\frac{k}{n \epsilon} \right\rfloor}$, will be,
\begin{eqnarray*}
S_{\epsilon} (x) &=& \sum_{k=0}^{n\epsilon-1}{\left\lfloor x+\frac{k}{n \epsilon} \right\rfloor} \\
&=&\left(1+ \left\lfloor \epsilon (m+l) \right\rfloor \right) \lfloor r \rfloor + \left(1+ \lfloor r \rfloor \right) \left(n-1- \left\lfloor \epsilon (m+l) \right\rfloor \right) \\
&=& \left(1+ \left\lfloor \epsilon (m+l) \right\rfloor \right) \left\lceil \frac{S}{n-m-1}\right\rceil + \left(1+ \left\lceil \frac{S}{n-m-1}\right\rceil \right) \left(n-1- \left\lfloor \epsilon (m+l) \right\rfloor \right)
\end{eqnarray*}
The running example has $S=765,\epsilon=0.1, n=100, m=10$. Substitution will give $l=36, \lfloor x \rfloor =8$ , and $S_{\epsilon} (x) = \lfloor 10x \rfloor =85$, matching what Ronald arrived at.
Note: (1+floor(0.1*(10+36)))*8+(1+8)floor(0.1(100-1-(10+36)))=85.
Note: I was merely trying to generalize this problem. Hermite's identity is quite useful and several partial sums like these are possible, by simple manipulations.
On the fun side, it is easy to play with different $\epsilon$ and $m$ values. As a special case, when $m=0$ and $\epsilon=1$, we can get $S_{0}$,the standard Hermite identity. Here is a short matlab code snippet
function [Se,xfloor,l]= partial_hermite_puzzle(S,m,n,epsilon)
% Se -- Partial Hermite sum (0 to epsilon*n-1)
% S -- Hermite Sum
xfloor=ceil((S-n+m+1)/(n-m-1));
l=-S+(n-m-1)*(xfloor+1);
Se=(1+floor(epsilon*(m+l)))*xfloor+(1+xfloor)*floor(epsilon*(n-1-(m+l)));
| {
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} |
Integer solutions to $x^3=y^3+2y+1$?
Find all integral pairs $(x,y)$ satisfying $$ x^3=y^3+2y+1.$$
My approach:
I tried to factorize $x^3-y^3$ as $$(x-y)(x^2 + xy + y^2)=2y+1,$$ but I know this is completely helpless. Please help me in solving this problem.
| we first consider only $x> y\ge0$
suppose $x=y+a$
then the equation is $(y+a)^3=y^3+2y+1$
$y^3+3ay^2+3a^2y+a=y^3+2y+1$
$3ay^2+(3a^2-2)y+(a-1)=0$
however, as $a\ge1$, $3a^2-2\ge0$,$a-1\ge0$
If $y>0$, then $3ay^2+(3a^2-2)y+(a-1)>0$, so $y=0$
then, $x^3=1$
$(x,y)=(1,0)$ are only integral solution
| {
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} |
Let $k$ be a natural number $\ge$ 3. Find the value of the constants $x$ and $y$ such that $A^k = xA +yI$ (Both x and y depend on k). So the question states that the matrix
$ A = \begin{bmatrix} 4 & -3 \\ 2 & -1 \end{bmatrix}$
characteristic polynomial is $ p(x) = x^2 - 3x + 2.\\$
I am supposed to find x and y when k is natural $\ge 3$ that $A^k = xA + yI$
So I tried to solve for both x and y by using the Cayley-Hamilton Theorem for A which you get:
$p(A) = A^2 - 3A + 2I => A^2 = 3A - 2I$
which then I tried to use this to find a pattern among the first few values of k, which lead me to:
$k= 1 => A = A-0I\\ k=2 => A^2 = 3A - 2I \\ k=3 => A^3 = A^2A = (3A-2I)A = 3A^2-2A = 3(3A-2I)-2A = 7A - 6I \\ k=4 => A^4 = 15A - 14I \\ k=5 => A^5 = 31A -30I \\ k=6 => a^6 = (31A -30I)A = 31(3A-2I)-30A = 63A - 62I \\ k=7 => A^7 = 63(3A-2I)-62A = 127A - 126I$
here is where I got lost because I stopped being able to find a pattern.
Is there an easier way to be able to find the values of x and y in terms of k?
| Write $A^k=x_kA+y_kI$. Then $A^{k+1}=x_kA^2+y_kA=x_k(3A-2I)+y_kA=(3x_k+y_k)A-2x_kI$. In other words:
$$\begin{bmatrix}x_{k+1} \\ y_{k+1} \end{bmatrix}=\begin{bmatrix} 3 & 1 \\ -2 & 0 \end{bmatrix} \begin{bmatrix}x_{k} \\ y_k \end{bmatrix}$$
Diagonalizing the matrix above yields
$$\begin{bmatrix} 3 & 1 \\ -2 & 0 \end{bmatrix}=\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}$$
which implies that
$$\begin{bmatrix}x_{k} \\ y_k \end{bmatrix}=\begin{bmatrix} -1 & -1 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 2^k \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}\begin{bmatrix}x_0 \\ y_0 \end{bmatrix}$$
| {
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"answer_id": 0
} |
Find all matrices that commute with $A$
Given $$A = \begin{bmatrix}
3 & 1 &0 \\
0 &3 & 1\\
0 &0 & 3
\end{bmatrix}$$ find matrices $B$ such that $AB=BA$.
Trivially $B=A^{-1}$ and $B=kI$ are the solutions
Also we have Characteristic Polynomial as
$$A^3-9A^2+27A-27I=0$$ $\implies$
$$(A-3I)^3=0$$
Is it possible to find other $B's$ using above Nilpotency of $A-3I$?
| Since$$A=3\operatorname{Id}_3+\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$and every matrix commutes with $3\operatorname{Id}_3$, you're after the matrices that commute with$$\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}.\tag1$$A simple computation shows that\begin{multline}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}-\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}=\\=\begin{pmatrix}d & e-a & f-b \\ g & h-d & i-e \\ 0 & -g & -h\end{pmatrix}\end{multline}and therefore the matrix$$\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$$commutes with $(1)$ if and only if$$\left\{\begin{array}{l}d=g=h=0\\a=e=i\\f=b.\end{array}\right.$$Therefore, the answer to your question is:$$\left\{\begin{pmatrix}a&b&c\\0&a&b\\0&0&a\end{pmatrix}\,\middle|\,a,b,c\in\mathbb{R}\right\}.$$
| {
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} |
Let $G=\mathbb{Z}/24\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. Consider the quotient group $H=G/\langle (10,3,2)\rangle$. Determine a direct product of cyclic groups that is isomorphic to $H$.
The Smith Normal form can be used to find the invariant factors in the structure theorem for finitely generated abelian groups. I set up the following matrix $$\begin{pmatrix} 24 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 3 \\ 10 & 3 & 2\end{pmatrix}$$ but I was unable to find the SMF of this matrix.
I know the following about the structure of $H$:
$|\langle (10,3,2)\rangle|=12$
Also $|G|=24*6*3$ so by Lagrange's Theorem, $|H| = 36$.
By the fundamental theorem of finitely generated abelain groups, $H$ may have the following structure:
$\mathbb{Z}/36\mathbb{Z}$
$\mathbb{Z}/18\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$
$\mathbb{Z}/12\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$
$\mathbb{Z}/6\mathbb{Z}\times\mathbb{Z}/6\mathbb{Z}$
I am not sure how to identify the correct group structure without the invariant factors from the SMNF.
Edit: Here is some of my work by hand. Am I proceeding correctly?
\begin{align*}
&\,
\begin{pmatrix}
24 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 3 \\
10 & 3 & 2
\end{pmatrix}
\sim
\begin{pmatrix}
10 & 3 & 2 \\
24 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 3
\end{pmatrix}
\sim
\begin{pmatrix}
2 & 10 & 3 \\
0 & 24 & 0 \\
0 & 0 & 6 \\
3 & 0 & 0
\end{pmatrix}
\\
\sim&\,
\begin{pmatrix}
2 & 10 & 1 \\
0 & 24 & 0 \\
0 & 0 & 0 \\
3 & 0 & -3
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 2 & 10 \\
0 & 0 & 24 \\
0 & 0 & 0 \\
-3 & 3 & 0
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 2 & 10 \\
0 & 0 & 24 \\
0 & 0 & 0 \\
0 & 9 & 30
\end{pmatrix}
\\
\sim&\,
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 24 \\
0 & 0 & 0 \\
0 & 9 & 30
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 0 & 0 \\
0 & 9 & 30 \\
0 & 0 & 24 \\
0 & 0 & 0
\end{pmatrix}
\end{align*}
(Original image here.)
| The Smith Form has diagonal entries 1, 3, 12 so that the quotient is isomorphic to $Z_3\times Z_{12}$.
| {
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Possibilty of a number $n$ to be a multiple of $x$ that has a remainder of $y$ Is there a formula or solution for this :
"$n$ is a number that is a multiple of $x$ and when divided by $b$ has remainder of $y$".
Is there a fast way to finding this number?
| The short answer is "No". Here is the long answer.
$n \equiv 0 \pmod x \implies n = \alpha x \ \text{for some integer $\alpha$}$
$n \equiv y \pmod b \implies \alpha x \equiv y \pmod b$
Let $g = \gcd(x, b)$. Then there is no solution unless $g \mid y$.
If $g \mid y$...
Let $X = \dfrac xg, \quad Y = \dfrac yg, \quad B = \dfrac bg.$
Then $\alpha X \equiv Y \pmod B$ and $\gcd(X,B) = 1$.
So there exists integers $\xi, \beta$ such that $\xi X + \beta B = 1$.
Then $\xi Y \equiv \alpha \xi X \equiv \alpha - \alpha \beta B \equiv \alpha \pmod B$.
That is $\alpha \equiv \xi Y \pmod B$
Example.
Find a number $n$ that is a multiple of $20$ and leaves a remainder of $24$ when divided by $34$.
We have $x = 20, \quad y=24, \quad b=34$.
$n \equiv 0 \pmod{20} \implies n = 20 \alpha \ \text{for some integer $\alpha$}$.
$n \equiv 24 \pmod{34} \implies 20 \alpha \equiv 24 \pmod{34}$
Let $g = \gcd(20, 34) = 2$. Then there is no solution unless $2 \mid 24$. Which it does.
Let $X = \dfrac{20}{2} = 10, \quad Y = \dfrac{24}{2}=12, \quad B = \dfrac{34}{2}=17.$
Then $10 \alpha \equiv 12 \pmod{17}$ and $\gcd(10,17) = 1$.
So there exists integers $\xi, \beta$ such that $10 \xi + 17 \beta = 1$.
\begin{array}{r|rrr|l}
& 17 & 1 & 0 & 17 = (1)17 + (0)10\\
-1 & 10 & 0 & 1 & 10 = (0)17 + (1)10\\
\hline
-1 & 7 & 1 & -1 & 7 = (1)17 + (-1)10\\
-2 & 3 & -1 & 2 & 3 = (-1)17 + (2)10 \\
\hline
& 1 & 3 & -5 & 1 = (3)17 + (-5)10
\end{array}
We find $\xi = 3, \quad \beta = -5$
So $\alpha \equiv -5 \cdot 12 = -60 \pmod{17} = 8 \pmod{17}$
and $n = 20\alpha = 160$
CHECK:
$160$ is a multiple of $20$
$160 = 4 \cdot 34 + 24$
| {
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Solve the equation $\frac{\sqrt {3}}{2}\sin(x) -\cos x=\cos^2x$ Solve the equation $\frac{\sqrt {3}}{2}\sin x-\cos x=\cos^2x$
My approach $\cos^2x=1-\sin^2x $
$\frac{\sqrt {3}}{2}\sin x-\cos^2x =\cos x$
$\frac{\sqrt {3}}{2}\sin x+\sin^2x -1=\cos x$
$\sqrt {3}\sin x+2\sin^2x-2=2\cos x$
Though the equation comes in form of $\sin x$ from here onward after squaring still not getting the answer.
| Hint; Use the so-called Weierstrass Substitution:
$$\sin(x)=\frac{2t}{1+t^2}$$
$$\cos(x)=\frac{1-t^2}{1+t^2}$$
$$\tan(\frac{x}{2})=t$$
and you don't need square the equation.
You will get the equation
$$-\sqrt {3} \left( \sqrt {3}t+{t}^{2}+2 \right) \left( -3\,t+\sqrt {3}
\right)
=0$$
| {
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At what angle from the ground should it be fired so that it travels the maximum distance in the air? A projectile is going to be fired from a cannon on level ground.
At what angle from the ground should it be fired so that it travels the maximum distance in the air?
Details and Assumptions:
*
*There is downward gravitational acceleration $g.$
*Neglect air resistance.
If the ball is released with initial speed $V$ at an angle of $\theta$ to the horizontal, the trajectory of the ball is
$y \; = \; x\tan\theta - \tfrac{g}{2V^2}x^2\sec^2\theta \; = \; x\tan\theta - \tfrac{1}{2a}x^2\sec^2\theta $
where $a = \tfrac{V^2}{g}$. Then the ball hits the ground again when $y=0$, which happens when $x = 2a\sin\theta\cos\theta$. Thus how should I continue to solve this?
|
If the ball is released with initial speed $V$ at an angle of $\theta$
to the horizontal, the trajectory of the ball is $ y \; = \;
> x\tan\theta - \tfrac{g}{2V^2}x^2\sec^2\theta \; = \; x\tan\theta -
> \tfrac{1}{2a}x^2\sec^2\theta $ where $a = \tfrac{V^2}{g}$. Then the
ball hits the ground again when $y=0$, which happens when $x =
> 2a\sin\theta\cos\theta$.
Continuing what you said
The length of the flightpath of the ball is
$ \begin{align} D(\theta) & = \; \int_0^{2a\sin\theta\cos\theta} \sqrt{1 + (y')^2}\,dx \; = \; \int_0^{2a\sin\theta\cos\theta}\sqrt{1 + \big(\tan\theta - \tfrac{1}{a}x\sec^2\theta\big)^2}\,dx \\ & = \; \frac{1}{a\cos^2\theta} \int_0^{2a\sin\theta\cos\theta}\sqrt{a^2\cos^4\theta + (x - a\sin\theta\cos\theta)^2}\,dx \; = \; \frac{1}{a\cos^2\theta}\int_{-a\sin\theta\cos\theta}^{a\sin\theta\cos\theta}\sqrt{a^2 \cos^4\theta + x^2}\,dx \\ & = \; a\cos^2\theta \int_{-\tan\theta}^{\tan\theta} \sqrt{1+y^2}\,dy \; = \; 2a\cos^2\theta\int_0^\theta \sec^3u\,du \\ & = \; 2a\cos^2\theta \times \tfrac12\left[\tan\theta\sec\theta + \ln(\sec\theta + \tan\theta)\right] \; = \; a\left[\sin\theta + \cos^2\theta\ln(\sec\theta + \tan\theta)\right] \end{align} $
and
$ D'(\theta) \; = \; a\left[\cos\theta + \cos\theta - 2\sin\theta\cos\theta\ln(\sec\theta+\tan\theta)\right] \; = \; 2a\cos\theta\big[1 - \sin\theta\ln(\sec\theta + \tan\theta)\big] $
Thus $D(\theta)$ will be maximized when $\sin\theta\ln(\sec\theta+\tan\theta) = 1$ (there is also a local minimum at $\theta = \tfrac12\pi$, and a global minimum at $\theta = 0$). Solving this last equation numerically, we obtain $\theta = 0.98551473786$, or $\boxed{56.4658^\circ}$.
| {
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Prime Number Congruence Modulo 8 Proof I want to rigorously prove that:
$$p_n=2\Biggl(\Bigl\lfloor \frac{p_n+1}{8}\Bigr\rfloor+\Bigl\lfloor \frac{p_n+3}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+5}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+7}{8} \Bigr\rfloor\Biggr)-1+\delta(n,1) \quad\quad\quad\quad\quad\quad(0)$$
So far what has convinced me is the observations as follows:
$$\frac{p_n-5}{2}-2\Biggl(\Bigl\lfloor \frac{p_n-1}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n-5}{8} \Bigr\rfloor\Biggr)+\frac{1}{2}\delta(n,1) \in {\{0,1}\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(1)$$
$$\frac{p_n+1}{2}-2\Biggl(\Bigl\lfloor \frac{p_n+1}{8} \Bigr\rfloor+\Bigl\lfloor \frac{p_n+5}{8} \Bigr\rfloor\Biggr)-\frac{3}{2}\delta(n,1) \in {\{0,1}\}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad(2)$$
$$\Bigl\lfloor \frac{n+1}{8} \Bigr\rfloor+\Bigl\lfloor \frac{n+3}{8} \Bigr\rfloor+\Bigl\lfloor \frac{n+5}{8} \Bigr\rfloor+\Bigl\lfloor \frac{n+7}{8} \Bigr\rfloor=\Bigl\lfloor \frac{n+1}{2} \Bigr\rfloor \,\,\forall n \in \mathbb N\quad\quad\quad\quad\quad\quad\quad\quad\quad\,\,\,\,(3)$$
$(3)$ explains why the RHS of $(0)$ must be odd, and $(0)$ being the sum of the expressions in lemmas $(1)$ & $(2)$ * show why the RHS of $(0)$ is equal to $p_n$, but this is as far as I can get without a text reference to something specifically relevant.
*I have manipulated these based on considerations of what a congruence relation implies, having the property of translation
| Here is an alternative way. It is well known that
$$\lfloor nx\rfloor =\sum_{i=1}^n\,\left\lfloor x+\frac{i-1}{n}\right\rfloor\text{ for all }x\in\mathbb{R}\text{ and }n\in\mathbb{Z}_{>0}\,.$$
Take $n:=8$ and $x:=\dfrac{p}{8}$ if $p$ is an odd prime. Thus, we get
$$p=\Biggl\lfloor 8\left(\frac{p}{8}\right)\Biggr \rfloor =\sum_{i=1}^8\,\left\lfloor\frac{p+i-1}{8}\right\rfloor=\sum_{i=1}^8\,\left\lfloor\frac{p+i}{8}\right\rfloor-1\,.$$
Now, since $p\equiv 1\pmod{2}$, we obtain
$$\left\lfloor\frac{p+2i-1}{8}\right\rfloor=\left\lfloor\frac{p+2i}{8}\right\rfloor\text{ for }i=1,2,3,4\,.$$
This shows that
$$p=2\,\sum_{i=1}^4\,\left\lfloor\frac{p+2i-1}{8}\right\rfloor-1\text{ for an odd prime }p\,.$$
You just have to check the case $p=2$ separately.
| {
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proof with induction $\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$ prove with induction:
$$\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$$
I'm stuck on
$$n[\sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} (n-k+1)^n]= \sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$$
$$n[\sum_{k=1}^{n-1} (-1)^{k-1} \binom{n-1}{k-1} (n-k+1)^{n-1}]=\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n$$
$$\sum_{k=0}^{n} (-1)^{k-1} k \binom{n}{k} (n-k+1)^{n-1} = \sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n $$
| Let's assume
$$\sum_{k=0}^n (-1)^k \binom{n}{k} (n-k+1)^n = n!$$
Do the following change of variable so that it looks easier:
$$k \leftarrow n-i$$
So that the equation that needs to proved is now
$$\sum_{i=0}^n (-1)^{n-i} \binom{n}{n-i} (i+1)^n = n!$$
Notice that
$$\binom{n}{n-i}=\binom{n}{i}$$
So why not write it as
$$\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} (i+1)^n = n!$$
The induction step that needs to be proved is
\begin{equation}
(n+1)!=
\sum_{i=0}^{n+1} (-1)^{n+1-i} \binom{n+1}{i} (i+1)^{n+1}
\end{equation}
\begin{align}
(n+1)! &= (n+1)n! \\
&= (n+1)\sum_{i=0}^n (-1)^{n-i} \binom{n}{i} (i+1)^n \\
&= (n+1)\sum_{i=0}^n (-1)^{n-i} \frac{n!}{i!(n-i)!} (i+1)^n \\
&= \sum_{i=0}^n (-1)^{n-i} \frac{(n+1)n!}{i!(n-i)!} (i+1)^n \\
&= \sum_{i=0}^n (-1)^{n-i} \frac{(n+1)!}{i!(n-i)!(n+1-i)} (n+1-i)(i+1)^n \\
&= \sum_{i=0}^n (-1)^{n-i} \frac{(n+1)!}{i!(n+1-i)!} (n+1-i)(i+1)^n \\
&= \sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (n+1-i)(i+1)^n \\
&= \sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (n+2-(i+1))(i+1)^n \\
&= \sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (n+2)(i+1)^n
-\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)(i+1)^n \\
&= (n+2)\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n
+\sum_{i=0}^n (-1)^{n+1-i} \binom{n+1}{i} (i+1)^{n+1} \\
\end{align}
But ( see here why )
\begin{equation}
\sum_{i=0}^n (-1)^{n-i} \binom{n+1}{i} (i+1)^n
=
(n+2)^n
\end{equation}
So
\begin{align}
(n+1)! &= (n+2)(n+2)^n +\sum_{i=0}^n (-1)^{n+1-i} \binom{n+1}{i} (i+1)^{n+1}\\
&= (n+2)^{n+1} +\sum_{i=0}^n (-1)^{n+1-i} \binom{n+1}{i}(i+1)^{n+1}\\
&= \sum_{i=0}^{n+1} (-1)^{n+1-i} \binom{n+1}{i}(i+1)^{n+1}
\end{align}
| {
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Factoring $x^n + 1$. By the Fundamental Theorem of Algebra, every polynomial of degree $n$ can be factored into a product of $n$ linear polynomials.
As an example, since the polynomial $ x^5 +1$ has the five complex roots $$\tag{1} -1,\quad e^{\frac{\pi i}{5}}, \quad e^{\frac{-\pi i}{5}}, \quad e^{\frac{3\pi i}{5}}, \quad e^{\frac{-3\pi i}{5}},$$ we can write $$\tag{2} x^5+1=(x+1)(x- e^{\frac{\pi i}{5}})(x- e^{\frac{-\pi i}{5}})(x- e^{\frac{3\pi i}{5}})(x- e^{\frac{-3\pi i}{5}}).$$
Multiplying several terms in $(2)$ gives $$\tag{3} x^5+1=(x+1)(x^2 -2\cos{\frac{\pi}{5}}x + 1)(x^2 -2\cos{ \frac{3\pi}{5}}x+1).$$
Now, $(3)$ is of course a specific example of the general theorem (which also relies on the structure of $\mathbb{C}$):
Any polynomial with real coefficients can be factored into a product of real linear and real quadratic polynomials.
Generalising $(3)$ to any odd value $n\in \mathbb{N}$, gives the following formula:
$$\tag{4} x^n +1 = (x+1) \left(x^2 -2\big(\cos{\frac{\pi}{n}}\big)x + 1\right) \left(x^2 -2\big(\cos{\frac{3\pi}{n}}\big)x + 1\right)\cdots \left(x^2 -2\big(\cos{\frac{(n-2)\pi}{n}}\big)x + 1\right) .$$
This result is interesting to me, as it suggests that even in $\mathbb{R}$ the cosine function is, in some vague sense, built in to exponentiation.
Question: I am wondering if either the general formula $(4)$ or a specific case, such as $(3)$, could be derived using only real analytical tools (and maybe some ring theory?): i.e., without complex numbers and Euler’s formula?
| The case $\,n=5\,$ can be solved algebraically by first factoring $\,x^5+1$ $ = (x+1)(x^4-x^3+x^2-x+1)\,$, then the second factor is a palindromic polynomial which reduces to a quadratic in $\,x+1/x\,$:
$$
\begin{align}
x^4-x^3+x^2-x+1 &= x^2\left(\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)+1\right) \\
&= x^2\left(\left(x+\frac{1}{x}\right)^2 - \left(x+\frac{1}{x}\right) - 1\right) \\
&= x^2\left(\left(x+\frac{1}{x}\right)-z_1\right)\left(\left(x+\frac{1}{x}\right)-z_2\right) \\
&= \left(x^2-z_1 x + 1\right)\left(x^2-z_2 x + 1\right)
\end{align}
$$
In the latter $\,z_{1,2}=\dfrac{1\pm\sqrt{5}}{2}\,$ are the roots of $\,z^2-z-1=0\,$, and the expression matches the trigonometric form. In fact, this is one way to prove that $\,\cos \dfrac{\pi}{5}=\dfrac{1+\sqrt{5}}{4}\,$.
Similarly, the case $\,n=7\,$ factors as:
$$
x^7+1=(x+1)\left(x^2-z_1 x + 1\right)\left(x^2-z_2 x + 1\right)\left(x^2-z_3 x + 1\right)
$$
Here $\,z_{1,2,3}\,$ are the roots of the cubic $\,z^3-z^2-2z+1=0\,$.
For odd $\,n\,$ the problem reduces to factoring a polynomial of degree $\,\dfrac{n-1}{2}\,$, however that is not solvable by radicals in general for $\,n \gt 9\,$.
| {
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"answer_count": 1,
"answer_id": 0
} |
Evaluate: $\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}$
Evaluate
$$
\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} \cdotp
$$
My attempt:
$$
I
= \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}
= \int \frac{\mathrm{d}x}{x\sqrt{\left(x + \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}
$$
I thought completing the square would bring the integrand into some form, but it did not. Please help.
| For integrals of the form $$\dfrac1{(x+a)\sqrt{(x+b)^2+c^2}},$$
Choose $x+b= c
\tan y,$ to reach at an integral of the form $$\dfrac1{A\cos y+B\sin y}$$
Now $A\cos y+B\sin y=\sqrt{A^2+B^2}\sin (y+\arctan \dfrac AB)=\sqrt{A^2+B^2}\cos(y-\arctan\dfrac AB)$
| {
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How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.
How can I graph $f(x)=\lfloor{x^2}\rfloor$ when the domain is $ℝ^{-}$.
I know that by definition $(\lfloor{x}\rfloor=m) ≡ (m≤x<m+1)$, so it follows that $(\lfloor{x^2}\rfloor=m ≡ (m≤x^2<m+1) ≡ (\sqrt{m}≤x^2<\sqrt{m+1})$; from this I can do
$\lfloor{x^2}\rfloor=
\begin{cases}
...\\
-3, & -3≤x^2<-2 \\
-2, & -2≤x^2<-1 \\
-1, & -1≤x^2<0 \\
0, & 0≤x^2<1 \\
1, & 1≤x^2<2 \\
2, & 2≤x^2<3 \\
3, & 3≤x^2<4 \\
...
\end{cases}
$
=
$\begin{cases}
...\\
-3, & -3≤x^2<-2 \\
-2, & -2≤x^2<-1 \\
-1, & -1≤x^2<0 \\
0, & 0≤x<1 \\
1, & 1≤x<\sqrt{2} \\
2, & \sqrt{2}≤x<\sqrt{3} \\
3, & \sqrt{3}≤x<2 \\
...
\end{cases}
$
Everything is normal until I "solve" for x in the negative intervals of the domain, which results in this
$\lfloor{x^2}\rfloor=
\begin{cases}
...\\
-3, & 3i≤x<2i \\
-2, & 2i≤x<i \\
-1, & i≤x^2<0 \\
0, & 0≤x<1 \\
1, & 1≤x<\sqrt{2} \\
2, & \sqrt{2}≤x<\sqrt{3} \\
3, & \sqrt{3}≤x<2 \\
...
\end{cases}
$
I don't know how to graph between those intervals.
Nonetheless when i see a graph of this function it seems that the function is symmetrical to the y axis.
Could you explain me, please, how can I graph the function in those negative intervals?
Thanks in advance.
| Well, by the trivial inequality, we have that $x^2$ is nonnegative. Note that you are solving for the floor of the $y$ outputs, not the $x$ inputs. Also, $\text{floor}(-x)^2=\text{floor}(x^2),$ so both sides are symmetrical.
| {
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Show that $0$ has multiplicity $3$ in $M-4I$? Question from an exam:
Consider the matrix $M=$
\begin{bmatrix} 5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3\end{bmatrix}
Show that $4$ is an eigen value of the above matrix with multiplicity $3$.
I considered $M-4I$ from which I got
\begin{bmatrix} 1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&1&1&1\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
By elementary row operations:
$$M-4I=$$
\begin{bmatrix} 1&1&1&1&1&1\\0&0&0&0&0&0\\0&0&0&0&0&0\\1&1&1&0&1&0\\1&1&1&1&0&0\\1&1&1&0&0&-1\end{bmatrix}
So $0$ is an eigen value with multiplicity at least $2$ since $M-2I$ has two zero rows.
How to show that $0$ has multiplicity $3$ in $M-4I$?
If one can show how to proceed after this,I will be really grateful.
Please help
| If it only had multiplicity $2$, $M-4I$ would have rank $4$, i.e. four of its rows would be linearly independent. Since the top three rows are the same, that would have to be the last four rows. But $(row\ 3) - (row\ 4) - (row\ 5) + (row\ 6) = 0$, so they are linearly dependent.
| {
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"source": "stackexchange",
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An expression involving trig functions and binomial coefficients Prove that
\begin{eqnarray}\sum_{i=0}^k5^{k-i-1}\binom{k}{i}\left(\frac{(37+62\cos\frac{2\pi}{5}+56\cos\frac{4\pi}{5})^i}{\left(2\sin\frac{\pi}{5}\right)^{2(k-i-1)}}+\frac{(37+62\cos\frac{4\pi}{5}+56\cos\frac{8\pi}{5})^i}{\left(2\sin\frac{2\pi}{5}\right)^{2(k-i-1)}}\right)\end{eqnarray}
is an integer. Can we further simplify the above expression? Thanks.
| Using the same approach as @drhab in his answer, I arrived to a slightly different result
$$\color{blue}{t_k=\frac{b}{5}\left(a+\frac 5b\right)^{k}+\frac{d}{5}\left(c+\frac5d\right)^{k}}$$ Now, using the values of the trigonometric terms, we have
$$a=\frac{3}{2} \left(5+\sqrt{5}\right)\qquad b=\frac{1}{2} \left(5-\sqrt{5}\right)\qquad c=\frac{3}{2} \left(5-\sqrt{5}\right)\qquad d=\frac{1}{2} \left(5+\sqrt{5}\right)$$
$$a+\frac 5b=2 \left(5+\sqrt{5}\right)\qquad c+\frac5d=2 \left(5-\sqrt{5}\right)$$ leading the the sequence
$$\{1,8,80,960,12800,179200,2560000,36864000,532480000,7700480000,111411200000\}$$
We could make $t_k$ more fancy in terms of $\phi$, the golden ratio.
| {
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"source": "stackexchange",
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} |
How do I solve 6b) and 6c) if my solution for 6a) is a consistent system of linear equations? $$\begin{cases}
2x_1-x_2= dx_1 \\
2x_1-x_2+x_3=dx_2 \\
-2x_1+2x_2+x_3=dx_3
\end{cases}
$$
a) Is it possible for the system to be inconsistent? Explain?
b) For what values of d will the system have infinitely many solutions?
c) Solve the system when it has infinitely many solutions?
For my solution in part a),
$$ \left[
\begin{array}{ccc|c}
1&0&0&0\\
0&1&0&0\\
0&0&1&0
\end{array}
\right] $$
Hence, it is a unique set of solutions i.e. the system cannot be inconsistent. So how is it possible to get infinitely many solutions in 6b) and 6c)?
| Using the Gaussian elimination method to solve a set of linear equations,
From the equations, you have given,
\begin{cases}
2x_1-x_2= dx_1 \\
2x_1-x_2+x_3=dx_2 \\
-2x_1+2x_2+x_3=dx_3
\end{cases}
We can arrive at this augmented matrix,
\begin{bmatrix}
\begin{array}{ccc|c}
-d+2&-1&0&0\\
2&-1-d&1&0\\
-2&2&1-d&0
\end{array}
\end{bmatrix}
Using row transformations,
\begin{bmatrix}
\begin{array}{ccc|c}
-2 & 2 & 1-d &0 \\
2 & -1-d & 1 &0\\
-d+2 & -1 & 0 &0\\
\end{array}
\end{bmatrix}
\begin{equation}
\downarrow
\end{equation}
\begin{bmatrix}
\begin{array}{ccc|c}
-2 & 2 & 1-d &0 \\
0 & 1-d & 2-d &0\\
0 & 0 & (1-d)(2-d) -2(2-d) &0\\
\end{array}
\end{bmatrix}
For it to have infinite solutions,
\begin{equation}
(1-d)(2-d) -2(2-d) = 0
\end{equation}
\begin{equation}
d = 2, -1
\end{equation}
If $d$ takes the above value, then you will end up with a free variable ($x_3$)
\begin{equation}
x_2 = \frac{(d-2)x_3}{1-d}
\end{equation}
\begin{equation}
x_1 = \frac{1}{2}(2x_2 + (1-d)x_3)
\end{equation}
| {
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Closed form solution for this integral? I'm hitting a road block in finding an expression (closed form preferably) for the following integral:
\begin{equation}
\int^{+\infty}_0 x^b \left ( 1-\frac{x}{u} \right )^c \exp(-a x^3) dx
\end{equation}
where $a,b$ are positive constants; $b>1$ is an odd multiple of $0.5$, while $c$ is a positive or negative odd multiple of $0.5$; $u$ is a (positive) parameter.
Things I have considered or tried:
*
*look up in tables (Gradshsteyn and Ryzhik): there are very few explicit results for integrals involving $\exp(-a x^3)$ (or for the other factors after transforming via $y=x^3$). Also, tabulated results involving $\exp(-a x^p)$ for more general $p$ do not include the other factors $x^b (1-x/u)^c$. One exception is (3.478.3):
\begin{equation}
\int^{u}_0 x^b (u-x)^c \exp(-a x^3) dx,
\end{equation}
but the limits of integration do not match with my case;
*there is a closed form solution (3.478.1) for the simpler integral
\begin{equation}
\int^{+\infty}_0 x^{d-1} \exp(-a x^3) dx = \frac{a^{-d/3}}{3} \Gamma(d/3).
\end{equation}
(NB: there is also an expression for the indefinite integral.)
A binomial expansion of $[1-(x/u)]^n$ for integer $n$ would produce a solution in series form. However, in my case, the exponents $b$ and $c$ are strictly half-integer. For the same reason, integration by parts does not lead to a simpler integral without the factor $[1-(x/u)]^c$;
*Wolfram Math online did not produce a result;
*the integral is an intermediate step in a longer analysis, so numerical solution (with given values for the parameter) is not practical.
Grateful for any pointers or solution.
| Long Comment:
Since it was explicitly asked for, Mathematica 11.3 when provided with
$$\text{Integrate}\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>1\land a>0\land u>0\right]$$
gives up the monster
$$\text{ConditionalExpression}\left[\frac{1}{6} \left(2 a^{\frac{1}{3} (-b-c)} \left(\frac{\left(-\frac{1}{u}\right)^c \Gamma \left(\frac{1}{3} (b+c+1)\right) \, _3F_3\left(\frac{1}{3}-\frac{c}{3},\frac{2}{3}-\frac{c}{3},-\frac{c}{3};\frac{1}{3},\frac{2}{3},-\frac{b}{3}-\frac{c}{3}+\frac{2}{3};-a u^3\right)}{\sqrt[3]{a}}-c u \left(-\frac{1}{u}\right)^c \Gamma \left(\frac{b+c}{3}\right) \, _3F_3\left(\frac{1}{3}-\frac{c}{3},\frac{2}{3}-\frac{c}{3},1-\frac{c}{3};\frac{2}{3},\frac{4}{3},-\frac{b}{3}-\frac{c}{3}+1;-a u^3\right)+\frac{\pi \sqrt[3]{a} (-1)^c 3^{\frac{3}{2}-c} u^{2-c} \Gamma (c+1) \Gamma \left(\frac{1}{3} (b+c-1)\right) \, _3F_3\left(\frac{2}{3}-\frac{c}{3},1-\frac{c}{3},\frac{4}{3}-\frac{c}{3};\frac{4}{3},\frac{5}{3},-\frac{b}{3}-\frac{c}{3}+\frac{4}{3};-a u^3\right)}{\Gamma \left(\frac{c-1}{3}\right) \Gamma \left(\frac{c}{3}\right) \Gamma \left(\frac{c+1}{3}\right)}\right)+\frac{3^{-b-c-\frac{1}{2}} u^{b+1} \Gamma (c+1) \left(\frac{4 \pi ^2 \Gamma (b+1)}{\Gamma \left(\frac{1}{3} (b+c+2)\right) \Gamma \left(\frac{1}{3} (b+c+3)\right) \Gamma \left(\frac{1}{3} (b+c+4)\right)}+\frac{(-1)^c \Gamma \left(-\frac{b}{3}-\frac{c}{3}\right) \Gamma \left(\frac{1}{3} (-b-c-1)\right) \Gamma \left(\frac{1}{3} (-b-c+1)\right)}{\Gamma (-b)}\right) \, _3F_3\left(\frac{b}{3}+\frac{1}{3},\frac{b}{3}+\frac{2}{3},\frac{b}{3}+1;\frac{b}{3}+\frac{c}{3}+\frac{2}{3},\frac{b}{3}+\frac{c}{3}+1,\frac{b}{3}+\frac{c}{3}+\frac{4}{3};-a u^3\right)}{\pi }\right),\Re(c)>-1\right] $$,
including the condition that the only half integer negative value $c$ can take is $-\frac{1}{2}$ given the above assumptions.
Further Thoughts...
By differentiating $e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c$ w.r.t $x$ and then integrating between $0$ and $\infty$ we obtain the integral identity
$$\frac{c}{u} \int_0^\infty e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^{c-1} \, dx =b \int_0^\infty e^{-a x^3} x^{b-1} \left(1-\frac{x}{u}\right)^c \, dx-3 a \int_0^\infty e^{-a x^3} x^{b+2} \left(1-\frac{x}{u}\right)^c \, dx$$
since $\left[e^{-a x^3} x^b \left(1-\frac{x}{u}\right)^c \right]_0^\infty=0$
Mathematica again gives the condition that $c>-1$ with the following assumptions
$$\text{Integrate}\left[e^{-a x^3} x^{b} \left(1-\frac{x}{u}\right)^c,\{x,0,\infty \},\text{Assumptions}\to b>0\land a>0\land u>0\right]$$.
which implies $c$ must be positive for your integral to exist.
| {
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"question_score": "8",
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Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.
$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$
The solution uses the ratio test and says:
$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $
I understand every step until here
$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$
How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
|
How can all n's on the left site become 1/n?
This is a standard method when finding the limit of such sequences.
You factor out the highest power of $n$ to cancel it out.
So if we would like to involve one more step befor the one you do not follow it would be:
$\frac{(n+1)(n+1)}{(2n+2)(2n+1)}\cdot\left(\frac{n+1}{n}\right)^n=\frac{n^2(1+1/n)(1+1/n)}{n^2(2+2/n)(2+1/n)}\cdot\left(\frac{n+1}{n}\right)^n$
Also note, that $\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^n=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e$.
| {
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Prove this inequality for given conditions For all $x,y>0$, $$\frac{1}{(x+1)^2} + \frac{1}{(y+1)^2} \ge \frac{1}{xy+1}$$
I can only think of substituting $x+1$ with $a$ and $y+1$ with $b$. Then the inequality turns into
$$(a^2 + b^2) (ab-a-b +2) \ge a^2b^2$$
I can proceed no further. Please help.
| After eliminating denominators and simplifying, the inequality reduces to:
$$
x^3y+xy^3-x^2y^2-2xy+1 \ge 0 \;\;\iff\;\; xy(x^2+y^2) - x^2y^2-2xy+1 \ge 0
$$
Using that $\,xy \ge 0\,$ and $\,\color{blue}{x^2+y^2 \ge 2xy}\,$, the above follows from:
$$
xy(x^2+y^2) - x^2y^2-2xy+1 \color{blue}{\ge 2x^2y^2}-x^2y^2-2xy+1 = (xy-1)^2 \ge 0
$$
| {
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"url": "https://math.stackexchange.com/questions/2902350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Prove equality involving binomial coefficients I was solving a probability problem and I got a different answer than the one given in the book. Seems the authors were using a different way of counting/arguing.
For the two answers to be equal, the following equality should hold true.
$$\sum_{k=0}^{b-1} \binom{a+k-1}{a-1} p^a (1-p)^k = \sum_{k=a}^{a+b-1} \binom {a+b-1}{k} p^{k} (1-p)^{a+b-k-1}$$
How can this be proved?
And the problem itself was: in a series of Bernoulli trials with probability for success in a single trial equal to $p$, what is the probability to get $a$ successes before getting $b$ failures?
I think you guys will reverse engineer how I counted and how the authors counted.
| Starting from
$$\sum_{k=0}^{b-1} {a+k-1\choose a-1} p^a (1-p)^k
= \sum_{k=a}^{a+b-1} {a+b-1\choose k} p^k (1-p)^{a+b-k-1}$$
we simplify to
$$\sum_{k=0}^{b-1} {a+k-1\choose a-1} p^a (1-p)^k
= \sum_{k=0}^{b-1} {a+b-1\choose a+k} p^{a+k} (1-p)^{b-k-1}$$
or
$$\sum_{k=0}^{b-1} {a+k-1\choose a-1} (1-p)^k
= \sum_{k=0}^{b-1} {a+b-1\choose a+k} p^k (1-p)^{b-k-1}.$$
We get for the LHS
$$\sum_{k\ge 0} {a+k-1\choose a-1} (1-p)^k
[[0\le k\le b-1]]
\\ = \sum_{k\ge 0} {a+k-1\choose a-1} (1-p)^k
[z^{b-1}] \frac{z^k}{1-z}
\\ = [z^{b-1}] \frac{1}{1-z}
\sum_{k\ge 0} {a+k-1\choose a-1} (1-p)^k z^k
\\ = [z^{b-1}] \frac{1}{1-z} \frac{1}{(1-(1-p)z)^a}.$$
The RHS is
$$\sum_{k=0}^{b-1} p^k (1-p)^{b-k-1}
[z^{b-1-k}] \frac{1}{(1-z)^{a+k+1}}
\\ = [z^{b-1}] \frac{1}{(1-z)^{a+1}}
\sum_{k=0}^{b-1} p^k (1-p)^{b-k-1}
\frac{z^k}{(1-z)^{k}}.$$
There is no contribution to the coefficient extractor in front when
$k\gt b-1$ and we may extend $k$ to infinity, getting
$$(1-p)^{b-1} [z^{b-1}] \frac{1}{(1-z)^{a+1}}
\sum_{k\ge 0} p^k (1-p)^{-k}
\frac{z^k}{(1-z)^{k}}
\\ = (1-p)^{b-1} [z^{b-1}] \frac{1}{(1-z)^{a+1}}
\frac{1}{1-pz/(1-p)/(1-z)}
\\ = (1-p)^{b-1} [z^{b-1}] \frac{1}{(1-z)^a}
\frac{1}{1-z-pz/(1-p)}
\\ = [z^{b-1}] \frac{1}{(1-(1-p)z)^a}
\frac{1}{1-(1-p)z-pz}
\\ = [z^{b-1}] \frac{1}{1-z} \frac{1}{(1-(1-p)z)^a}.$$
The LHS and the RHS are seen to be the same and we may conclude.
Remark. The first one is the easy one and follows by
inspection. The Iverson bracket may be of interest here as an example
of the method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
| Hint : You wrote $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0$ but you forgot to study what is on the bottom. $(x+4)$ changes sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Finding the sum $\sum_{k=1}^rk^2\binom {n-k}{r-k}$ I was stuck while finding the given summation.
$$\sum_{k=1}^rk^2\binom {n-k}{r-k}$$
Since $n$ and $r$ are both constants, so I have first converted the above summation into this:
$$\sum_{k=1}^rk^2\binom {n-k}{n-r}$$
I have no idea how to proceed next. Any help will be appreciated.
| We have
$$\sum_{k=1}^r k^2 {n-k\choose r-k}
= \sum_{k=0}^r k^2 [z^{r-k}] (1+z)^{n-k}
\\ = [z^r] \sum_{k=0}^r k^2 z^k (1+z)^{n-k}
= [z^r] (1+z)^n \sum_{k=0}^r k^2 z^k (1+z)^{-k}.$$
There is no contribution to the coefficient extractor
when $k\gt r$ and we may continue with
$$[z^r] (1+z)^n \sum_{k\ge 0} k^2 z^k (1+z)^{-k}
= [z^r] (1+z)^n
\frac{(z/(1+z))(1+z/(1+z))}{(1-z/(1+z))^3}
\\ = [z^r] (1+z)^n
z (1+z)^2 (1+z/(1+z))
= [z^r] (1+z)^n z(1+z)(1+2z)
\\ = [z^{r-1}] (1+z)^{n+1} (1+2z).$$
This is for $r\ge 2$
$$\bbox[5px,border:2px solid #00A000]{
{n+1\choose r-1} + 2 {n+1\choose r-2}.}$$
Alternatively,
$$\frac{n-r+3}{r-1} {n+1\choose r-2} + 2{n+1\choose r-2}
= \frac{n+r+1}{r-1} {n+1\choose r-2}.$$
Here we have used the fact that with formal power series
$$\sum_{k\ge 0} k^2 w^k = \frac{w(1+w)}{(1-w)^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Solving Homogenous First Order ODE Using Substitution Problem
Find the general solution of the following homogeneous equation.
$$ ty' = y + \sqrt { t^2 - y^2} $$
Attempt
I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, which suggests rearranging the equation into the form:
$$ y' = ... $$
where $ y = tv $ and $ y' = v + tv' $ (v is the substitution variable).
I haven't been able to get too far in the procedure to make the substitution:
$$ y' = \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{\frac{1}{t}}{\frac{1}{t}} \frac{y + \sqrt { t^2 - y^2}}{t} $$
$$ y' = \frac{y}{t} + \frac{\sqrt { t^2 - y^2}}{t} $$
Notes
Can someone please help me reduce the RHS of the above equation to the form $ y' = f(\frac{y}{t}) $ (an equation in terms of $ (\frac{y}{t}) $), using the substitution $ y = tv $? For bonus points, please provide a solution to the ODE so myself and other viewers can check their answers!
Thanks in advance!
Solution
With the help of Isham, I was able to get the solution.
$$ ty' = y + \sqrt { t^2 - y^2 } $$
It is implied that $ t^2 - y^2 \geq 0 $, and so $ t^2 \geq y^2 \gt 0 $ or $ \lt 0 $.
$$ ty' - y = \sqrt { t^2 - y^2} $$
$$ \frac {1}{\sqrt {t^2} } (ty' - y) = \frac {1}{\sqrt {t^2} } \sqrt { t^2 - y^2} $$
$$ \frac {1}{t} (ty' - y) = \sqrt { \frac{t^2}{t^2} - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - \frac{y^2}{t^2}} $$
$$ y' - \frac{y}{t} = \sqrt { 1 - (\frac{y}{t})^2} $$
$$ y' = \sqrt { 1 - (\frac{y}{t})^2} + \frac{y}{t} $$
Let $ v = \frac{y}{t} $. Then, $ y = tv $ and so $ y' = \frac{dy}{dt} = v + tv' $. Substituting for $ v $ in the above equation we get:
$$ v + tv' = \sqrt {1 - v^2} + v $$
$$ tv' = \sqrt {1 - v^2} $$
Note, since $ t^2 \geq y^2 $, $ \frac{y^2}{t^2} = v^2 \leq 1 $, so there are no equilibrium cases to consider.
$$ t\frac{dv}{dt} = \sqrt {1 - v^2} $$
$$ \frac{1}{\sqrt { 1 - v^2}}dv = \frac{1}{t}dt $$
$$ \int\frac{1}{\sqrt { 1 - v^2}}dv = \int\frac{1}{t}dt $$
$$ arcsin(v) = ln|t| + C_1, C_1 \in \mathbb R $$
$$ \frac {y}{t} = \sin{(ln|t| + C_1)} $$
$$ y = t \sin{(ln|t| + C_1)} $$
| According with the suggested
$$
y'=v+\sqrt{1-v^2}
$$
with
$$
v = \frac yt
$$
so
$$
y' = v + t v'
$$
then
$$
v + t v' = v +\sqrt{1-v^2}
$$
or
$$
\frac{dv}{\sqrt{1-v^2}} = \frac{dt}{t}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2912091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
let $a_{1} = 0, a_{2} =1$ and $a_{n+2} = \frac{(n+2)a_{n+1} - a_{n}}{n+1}$. Prove that $\lim\limits_{n \to +\infty} a_{n} = e$ let $a_{1} = 0$, $a_{2} =1$ and $a_{n+2} = \frac {(n+2)a_{n+1} - a_{n}} {(n+1)}$.
Prove that $\lim a_{n} = e$
Knowing that I have proved that $a_{n+1} -a_{n} = 1/n!$, then I shall do what next?
| Since you've proved that $a_{n+1}-a_n = \frac{1}{n!}$ you have that $$a_n = \frac{1}{1!}+ \ldots + \frac{1}{(n-1)!}$$
So $$\lim\limits_{n \to +\infty} a_n = \lim\limits_{n \to \infty} \left( \frac{1}{1!} + \ldots + \frac{1}{(n-1)!} \right) = \sum\limits_{n=1}^{+\infty} \frac{1}{n!} = e-1$$
Are you sure the correct answer is $e$? Maybe it was just $a_0 = 0$ and $a_1 = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ r^2 = a^2\cos^2(\theta) + b^2\sin^2(\theta) $ [Using pythagoras theorem]
$$ Area = \int_{0}^{\pi/2} \int_{0}^{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}} rdrd\theta $$
$$ = 1/2 \int_{0}^{\pi/2} r^2 \Big|_{0}^{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}} d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} (a^2\cos^2(\theta)+b^2\sin^2(\theta)) d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} ((a^2 - b^2)\cos^2(\theta)+b^2) d\theta $$
$$ = 1/4 \int_{0}^{\pi/2} (a^2 - b^2)(1+ \cos(2\theta)) d\theta +2b^2 d\theta $$
I am getting $$ \pi/8 (a^2 + b^2).$$ But the correct answer is $ \pi ab/4 $
| Set $x=ar\cos \theta, y=br\sin \theta.$ The Jacobian is $abr$ and we compute the area
$$\mathcal {A}= \int_0^{\pi/2} \int_0^1 abr \;d r \;d\theta,$$ which is $\frac{ab\pi}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Showing that the graph of $\frac{\cos(\pi x/2)}{1-x^2}$ decreases over the interval $(0,2)$. How would one justify the behaviour of the graph of
$$\frac{\cos(\frac{\pi}{2}x)}{1-x^2}$$
over the interval $(0,2)$?
More precisely, how can it be shown that the graph is decreasing over that interval?
| The derivative of $f(x)$ takes the following form
\begin{equation}
f'(x) = \frac{g(x)}{(1-x^2)^2}
\end{equation}
\begin{equation}
g(x) = -\frac{\pi}{2} \sin (\frac{\pi}{2}x)(1-x^2)+2x( \cos (\frac{\pi}{2}x))
\end{equation}
PS: The function $f(x)$ is not defined at $1$
Since the denominator of $f'(x) > 0$ on $[0,1[ \cup ]1,2]$, we will study $g(x)$. Get it's derivative
\begin{equation}
g'(x) = -\frac{\pi^2}{4} \cos (\frac{\pi}{2}x)(1-x^2)
+
\pi x \sin (\frac{\pi}{2}x)
+
2\cos (\frac{\pi}{2}x)
-\pi x \sin (\frac{\pi}{2}x)
\end{equation}
which is
\begin{equation}
g'(x) = -\frac{\pi^2}{4} \cos (\frac{\pi}{2}x)(1-x^2)
+
2\cos (\frac{\pi}{2}x)
\end{equation}
Factorize to get
\begin{equation}
g'(x) = [-\frac{\pi^2}{4}(1-x^2)
+
2]\cos (\frac{\pi}{2}x)
\end{equation}
You could further factorize as
\begin{equation}
g'(x) = \big(x - \sqrt{1 - \frac{8}{\pi^2}} \big)\big(x + \sqrt{1 - \frac{8}{\pi^2}} \big)\cos (\frac{\pi}{2}x)
\end{equation}
On the interval $[0,1]$
In the interval $[0,1]$, we know that both $\cos(\frac{\pi}{2})$ and $x + \sqrt{1 - \frac{8}{\pi^2}}$ are positive. So in $[0,1]$ the sign of $g'(x)$ depends on $x - \sqrt{1 - \frac{8}{\pi^2}}$. Obviously we get that $g'(x) < 0$ in $[0,x_0]$ and $g'(x) > 0$ in $[x_0,1]$ where $x_0 = \sqrt{1 - \frac{8}{\pi^2}}$. So there is a minimum at $x_0$. But $g(0) = g(1) = 0$. So $g(x) < 0$ in $[0,1]$.
On the interval $[1,2]$
In the interval $[1,2]$, we have that both $x + \sqrt{1 - \frac{8}{\pi^2}}$ and $x + \sqrt{1 + \frac{8}{\pi^2}}$ are positive so the sign of $g'(x)$ depends on $\cos \frac{\pi}{2}x$ which is negative in $[1,2]$. Hence, $g(x)$ is decreasing decreasing in $[1,2]$, but $g(1) = 0$, hence $g(x)$ is also negative in $[1,2]$.
Concluding on $[0,2]$
We conclude that in $[0,2]$,
\begin{equation}
g(x) \leq 0
\end{equation}
Hence $f(x)$ is decreasing in $[0,2]$, excluding $x = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$ Prove that $f(x)=\frac{ax^2+x-2}{a+x-2x^2}$ has the range $ℝ$ when $x\inℝ$ if $a \in [1,3]$
My working:
Let $f(x)=\frac{ax^2+x-2}{a+x-2x^2}=y$
therefore,
$$(a+2y)x^2+(1-y)x-(2+ay)=0$$
Now, for $x$ to have real values, the discriminant of the above equation must not be less than zero.
So,
$$(1+8a)y^2+(14+4a^2)y+(1+8a)\ge 0$$
Now, somehow from this, we have to land on the condition for $a$ which I am unable to do.
| The minimum of the polynomial $ax^2+bx+c$ , where $a$ is positive, is $\frac{4ac - b^2}{4a}$. To see this, note that $ax^2 + bx + c = \frac{(4a^2x^2 + 4abx + b^2)}{4a} -\frac{b^2}{4a} + c = \frac 1{4a} (2ax + b)^2 + \frac{4ac - b^2}{4a}$, so the smallest value is when the square is zero, then the value is $\frac{4ac - b^2}{4a}$.
In other words, the polynomial is positive everywhere if and only if $4ac > b^2$.
Note that if $1 \leq a \leq 3$, then $(1+8a) > 0$, so you can apply the given criterion to conclude that it is enough to check that $4 \times (1+8a) \times (1+8a) > (14 + 4a^2)^2$ whenever $1 \leq a \leq 3$.
Can you check this?
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving periodic equation $\cos 7\theta=\cos 3\theta+\sin 5\theta$.
Here I have this trigonometric equation $\cos 7\theta=\cos 3\theta+\sin 5\theta$.
I approached the problem as follows :
$\cos 7\theta=\cos 3\theta+\sin 5\theta$
$\implies \cos 7\theta-\cos 3\theta=\sin 5\theta$
$\implies 2\sin 5\theta\sin(-2\theta)=\sin 5\theta$
From there on we get,
$\sin 2\theta=-\dfrac{1}{2}$ and $\sin 5\theta=0$
So, we deduce that $\theta=-\dfrac{\pi}{12}+\pi k$ and $\theta=\dfrac{2}{5}\pi n$ where $k$ and $n$ are integers.
However, Wolfram|Alpha doesn't seem to agree with me. They present $\dfrac{\pi}{5}$ as a solution which is not included in my solution.
Can someone help me out?
| $$\sin 5\theta=0$$
$\sin x =0$ has solutions $x=n\pi$
Thus we get $$5\theta=n \pi$$
$$\theta= \frac{n\pi}{5}\hspace{10pt} n\in Z$$
Also $$\sin 2\theta=-\dfrac{1}{2}$$
$$2\theta=2n\pi-\frac{\pi}{6} \implies \theta=n\pi-\frac{\pi}{12}\hspace{10pt} n\in Z$$
$$2\theta=(2n+1)\pi+\frac{\pi}{6} \implies \theta=n\pi+\frac{7\pi}{12}\hspace{10pt} n\in Z$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Decomposition of polynomials like $1+x^4$ What is the "trick" to get from $1+x^4$ to $$(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)?$$ Of course I can calculate it's true, but I don't understand what steps to take to get from one to the other.
Next to this specific question, I am also looking for the general rules for such decompositions?
Cheers!
| By the factor theorem, the linear of $x^n + 1$ are of the form $x - a$, where $a$ is one of the $n$th roots of $-1$ in $\mathbb{C}$. These form complex conjugate pairs, and when two such paired factors multiply together, we get an irreducible quadratic over $\mathbb{R}$.
We know that $-1 = e^{i\pi}$, which makes the four roots of unity
\begin{align*}
e^{i\frac{\pi}{4}} &= \cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \\
e^{i\frac{\pi}{4}} &= \cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \\
e^{i\frac{\pi}{4}} &= \cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \\
e^{i\frac{3\pi}{4}} &= \cos\left(\frac{3\pi}{4}\right) + i\sin\left(\frac{3\pi}{4}\right) = -\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}.
\end{align*}
Thus, we have
\begin{align*}
x^4 + 1 &= \left(x + \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)\left(x + \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right)\left(x - \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right)\left(x - \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) \\
&= \left(\left(x + \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}\right)\left(\left(x - \frac{1}{\sqrt{2}}\right)^2 + \frac{1}{2}\right) \\
&= (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1).
\end{align*}
Using similar methods, we can also factorise $x^{2n} + 1$. For example,
$$x^6 + 1 = (x^2 + \sqrt{3}x + 1)(x^2 + 1)(x^2 - \sqrt{3}x + 1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16 \\
9^2 &= \phantom{1}8\cdot10+1 = 81 \\
15^2 &= 14\cdot16+1 = 225
\end{align}$$
and so on.
Then, to know any number raised to the power of $2$, you can multiply the previous number $(x-1)$ by the next one $(x+1)$, and add $1$.
So, to know the squared root of a number like $64$, you have to substract $1$ ($63$) and look for two numbers multiplied are $63$ and subtracted are $2$:
$$x \cdot y = 63 \qquad x - y = 2$$
Solving the equation you get $9$ and $7$ (or $-7$ and $-9$). The number between these is the square root ($8$).
I don't know if this apply for power of $3$.
Does this fact/theorem/relation has a name or something?
| If you were to subtract 1 from both sides, you would get
$$ x^2 - 1 = (x + 1)(x - 1)$$
This is a simpler case of the more general factoring of the difference of squares:
$$ x^2 - b^2 = (x + b)(x - b)$$
For more information see Wikipedia's Difference of two squares article.
| {
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How many integers from 1 through 1000 are divisible by 3 and by at least one of 2,5,7, and 11? So I got 294 for this question, though I'm not a hundred percent sure but let me explain my process:
I did
to find all the multiples:
(floor of each)
1000/6 = 166
1000/15 = 66
1000/21 = 47
1000/33 = 30
166+66+47+30 = 309
to eliminate the repeats:
(floor of each)
66/6 = 11
47/6 = 7
30/6 = 5
47/15 = 3
30/15 = 2
30/21 = 1
11+7+5+3+2+1 = 29
to add in the repeats that were eliminated twice over:
(floor of each)
5/2 = 2
7/2 = 3
11/2 = 5
7/5 = 1
11/5 = 2
11/7 = 1
2+3+5+1+2+1 = 14
309 - 29 + 14 = 294
| If you consider the range $3$ to $999$, divide by $3$, then we find that your question is equivalent to the following simpler question:
how many integers between $1$ and $333$ are divisible by at least one of $2$, $5$, $7$, or $11$?
For this question we may use inclusion-exclusion.
Multiples of $2,5,7,11$ with much overcounting:
$$\left\lfloor \frac{333}{2} \right\rfloor + \left\lfloor \frac{333}{5} \right\rfloor + \left\lfloor \frac{333}{7} \right\rfloor + \left\lfloor \frac{333}{11} \right\rfloor $$
Subtract these terms:
$$ \left\lfloor \frac{333}{10} \right\rfloor + \left\lfloor \frac{333}{14} \right\rfloor + \left\lfloor \frac{333}{22} \right\rfloor + \left\lfloor \frac{333}{35} \right\rfloor + \left\lfloor \frac{333}{55} \right\rfloor + \left\lfloor \frac{333}{77} \right\rfloor $$
Add back these terms:
$$ \left\lfloor \frac{333}{70} \right\rfloor + \left\lfloor \frac{333}{110} \right\rfloor + \left\lfloor \frac{333}{154} \right\rfloor + \left\lfloor \frac{333}{385} \right\rfloor $$
Since $770>333$, there are no multiples of $770$ that we have to worry about having overcounted.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3
I am trying to find the Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ with centre $1$ that converges at $3$.
I was thinking that if we want the series to converge at $3$, we are interested in the region $\sqrt{2}<|z-1|<\sqrt{5}$. Hence,
\begin{align}
f(z)&=-\frac{1}{2i-1-(z-1)}+\frac{1}{-i-1-(z-1)} \\
&=-\frac{1}{2i-1}\left(\frac{1}{1-\frac{z-1}{2i-1}}\right)-\frac{1}{z-1}\left(\frac{1}{1-\frac{-1-i}{z-1}}\right) \\
&=-\frac{1}{2i-1}\sum_{n=0}^{\infty}\left(\frac{z-1}{2i-1}\right)^n-\frac{1}{z-1}\sum_{n=0}^{\infty}\left(\frac{-1-i}{z-1}\right)^n \\
&=-\sum_{n=0}^{\infty}\left(\frac{(z-1)^n}{(2i-1)^{n+1}}+\frac{(-1-i)^n}{(z-1)^{n+1}}\right)
\end{align}
Is this solution correct? This question is different to others I have done in the past.
| $$
\begin{align}
f(z)
&=\frac1{z-2i}-\frac1{z+i}\\
&=\frac1{w+1-2i}-\frac1{w+1+i}\\
&=\frac1{1-2i}\sum_{k=0}^\infty(-1)^k\left(\frac w{1-2i}\right)^k
-\frac1w\sum_{k=0}^\infty(-1)^k\left(\frac{1+i}w\right)^k\\
&=-\sum_{k=0}^\infty\frac{w^k}{(-1+2i)^{k+1}}-\sum_{k=0}^\infty\frac{(-1-i)^k}{w^{k+1}}\\
&=-\sum_{k=0}^\infty\frac{(z-1)^k}{(-1+2i)^{k+1}}-\sum_{k=1}^\infty\frac{(-1-i)^{k-1}}{(z-1)^k}\\
\end{align}
$$
Your series looks correct and converges for $\sqrt2\lt|z-1|\lt\sqrt5$.
| {
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"url": "https://math.stackexchange.com/questions/2929847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $A = x - y + z$ if $3x + 5y + 7z = 29$ and $x , y , z \in \mathbb{Z}^+$ I've found the answer by trial and error $(x = 4 , y=2,z=1) \Rightarrow$ $A = 3$. I tried to solve it using modular arithmetic but it didn't work .
$$3x + 5y + 7z = 29 $$
$$ 3x + 5y+7z \equiv 29 \mod 3$$
$$ 2y +z \equiv 2 \mod 3 $$
$$ y= -k , z=2k + 2$$
putting $ y= -k , z=2k + 2$ to original equation leads to $x = -3k + 5$ . Then $A = -3k+5 +k +2k+2 = 7$ . What's wrong about my answer?
| Starting from your 3.rd line. For some integer $t$ we have $$2y+z-2 = 3t\implies z = 2+3t-2y$$
So $$3x+5y+7(2+3t-2y) = 29\implies 3x = 15-21t+9y$$
so $$x = 5-7t+3y$$
thus $$A = 5-7t+3y -y+2+3t-2y=-4t+7$$
Since $2y+z\geq 3\implies 3t+2\geq 3\implies t\geq 1$
Since $z\leq 4$ and $y\leq 5$ we get $3t+2\leq 14$ so $t\leq 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}$ Greetings I saw here (among the last integrals) that: $$\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}=\pi C$$
Where $C$ is Catalan's constant. Did this integral appear here before? (my quick search did not found anything).
I gave it a try and got stuck. Denoting the integral as $I$ and using that $\ln \left(\frac ab\right)=\ln a- \ln b\ $ we have: $$I=K(1,1)-K(1,-1)-K(-1,1)+K(-1,-1)$$ Where $$K(a,b)=\int_0^1\frac{\ln(1+ax)\ln(1+bx^2)}{x}dx$$ Differentiating under the integral sign: $$\frac{\partial^2}{\partial a \partial b}K(a,b)=\int_0^1 \frac{x^2}{(1+ax)(1+bx^2)}\,dx$$ By partial fractions we get: $$\frac{1}{a^2+b}\left(\int_0^1 \frac{ax}{bx^2+1}\,dx -\int_0^1 \frac{1}{bx^2+1}\,dx +\int_0^1 \frac{1}{ax+1} \,dx\right)$$
$$=\frac{1}{a^2+b}\left(\frac{a\ln(1+b)}{2b}-\frac{\arctan \left(\sqrt b\right)}{\sqrt{b}} +\frac{\ln(1+a)}{a}\right) $$ And now since $K(0,b)=K(a,0)=0$ $$K(a,b)=\frac12\int_0^a \int_0^b \frac{x\ln(1+y)}{y(x^2+y)}\,dy\,dx-\int_0^a \int_0^b \frac{\arctan \left(\sqrt y\right)}{\sqrt{y}(x^2+y)}\,dy\,dx +\int_0^a \int_0^b \frac{\ln(1+x)}{x(x^2+y)}\,dy\,dx$$ Is there a clever way to solve this?
Another way is to start by using: $$-\frac12\ln\left(\frac{1-x}{1+x}\right)=\sum_{n=1}^\infty \frac{x^{2n+1}}{2n+1}$$ $$I=4\sum_{n,k=1}^\infty \frac{1}{(2n+1)(2k+1)}\int_0^1 x^{4n+2k+2}\,dx=4\sum_{n,k=1}^\infty \frac{1}{(2n+1)(2k+1)(4n+2k+3)}$$ But I dont know how to deal with this series. I would appreciate some help with this integral!
| It's always the same story,
$$J=\int_0^1 \ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)\frac{dx}{x}$$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$$
J =2\int_0^1 \frac{\ln\left(\frac{x^2+1}{2x}\right)\ln x}{x^2-1}\,dx
$$
For $x\in [0;1]$ define the function $R$,
\begin{align}R(x)&=\int_0^x \frac{\ln t}{t^2-1}\,dt\\
&=\int_0^1 \frac{x\ln( tx)}{t^2x^2-1}\,dt\\
\end{align}
Observe that $R(0)=0$.
\begin{align}J = {} & 2\left[R(x)\ln\left(\frac{x^2+1}{2x}\right)\right]_0^1-2\int_0^1\int_0^1 \frac{(x^2-1)\ln(tx)}{(x^2+1)(t^2x^2-1)}\,dt\,dx\\
= {} & -2\int_0^1\int_0^1 \frac{(x^2-1)\ln(tx)}{(x^2+1)(t^2x^2-1)}\,dt\,dx\\
= {} & -2\int_0^1\int_0^1 \frac{(x^2-1)\ln t}{(x^2+1)(t^2x^2-1)}\,dt\,dx-2\int_0^1\int_0^1 \frac{(x^2-1)\ln x}{(x^2+1)(t^2x^2-1)}\,dt\,dx\\
= {} & \int_0^1\left[\frac{1-t^2}{t(1+t^2)}\ln\left(\frac{1+tx}{1-tx}\right)-\frac{4\arctan x}{t^2+1}\right]_{x=0}^{x=1}\ln t\,dt-{}\\
&\int_0^1 \left[\frac{1-x^2}{x(1+x^2)}\ln\left(\frac{1+tx}{1-tx}\right)\right]_{t=0}^{t=1}\ln x\,dx\\
= {} & 4\times \frac{\pi}{4}\times -\int_0^1\frac{\ln t}{1+t^2}\,dt\\
= {} & \boxed{\pi\text{G}}
\end{align}
$\text{G}$ is the Catalan constant.
PS:
The idea is always the same, rewrite the integral as $\displaystyle \int_0^1 A(x)\ln x\ln(B(x))\,dx$, $A,B$ rational fraction functions. Then consider $\displaystyle R(x)=\int_0^x A(t)\ln t\,dt$ and finally perform integration by parts. If you know that the result is not too complicated you're pretty sure the process will work ;)
PS2:
Actually, $\displaystyle \int_0^1 A(x)\left(\sum_{n=1}^N\beta_n\ln(B_n(x))+\sum_{n=1}^M \delta_n \arctan(C_n(x))\right)\,dx$ with $\beta_n,\delta_n$ real numbers, $A,B_n,C_n$ rational fractional function will work too if the result is supposed to be not too complicated.
(see Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ for another miraculous evaluation of integral. )
| {
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"url": "https://math.stackexchange.com/questions/2934903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How to show that $\dfrac{n^3 + 2n}{3}$ is an integer Show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer
My try:
Let P(n) be the statement that $n^3 + 2n$ is divisible by $3$.
Base step:
When $n = 0$ we have $0^3 + 0 = 0 = 3 \times 0$
So, the base case true.
Inductive hypothesis:
Assume that $P(k)$ is true.
which means $\dfrac{k^3 + 2k}{3}$ is divisible by $3$ and $\dfrac{k^3 + 2k}{3}=p$ for some integer $p$.
Now we need to show that $P(k+1)$ is true.
$(k+1)^3+2(k+1)$ and we will show that this divisible by $3$.
Proof:
$(k+1)^3+2(k1)=k^3+3k^2+3k+1+2k+2$
$.$
$.$
$.$
$.$
$=3(p+k^2+k+1)$
As $p+k^2+k+1$ is an integer we have that $(k+1)^3+2(k+!)$ is divisible by $3$.
Is my above attempt correct?
Did I show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer?
| $n^3 +2n=n(n^2+2) $, now, check that all square numbers will leave a remainder $0$(means divisible by $3$) or, $1$ when we divide $n$ by $3$. Hence, when $3|n$ we have $3|n(n^2+2)$ and when $n=3k+1$ or $3k+2$, then $3|n^2+2\implies 3|n(n^2+2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $(q^k n^2 \text{ is perfect }) \iff (D(q^k)D(n^2) = 2s(q^k)s(n^2))$ only true for odd perfect numbers $q^k n^2$? (Preamble: This question is an offshoot of this earlier MSE post.)
The title says it all.
Is $\bigg(q^k n^2 \text{ is perfect }\bigg) \iff \bigg(D(q^k)D(n^2) = 2s(q^k)s(n^2)\bigg)$ only true for odd perfect numbers $q^k n^2$?
Here,
$$D(x) = 2x - \sigma(x)$$
is the deficiency of $x$,
$$s(x) = \sigma(x) - x$$
is the sum of the aliquot divisors of $x$, and $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$, the set of positive integers.
IN RESPONSE TO A CLARIFICATION FROM mathlove
That is:
If $\gcd(y,z)=1$, is the biconditional "$yz$ is perfect $\iff D(y)D(z)=2s(y)s(z)$" always true?
| MY ATTEMPT
For even perfect numbers $2^{p-1}(2^p - 1)$, I get
$$D(2^p - 1)D(2^{p-1}) = (2(2^p - 1) - (2^p))(1) = 2^{p+1} - 2 - 2^p = 2^p - 2$$
$$2s(2^p - 1)s(2^{p-1}) = 2(2^p - (2^p - 1))(2^p - 1 - 2^{p-1}) = 2(1)(2^{p-1} - 1) = 2^p - 2.$$
Thus, the equation
$$D(2^p - 1)D(2^{p-1}) = 2s(2^p - 1)s(2^{p-1}) = 2^p - 2$$
is true.
Therefore, the required relationship
$$D(q^k)D(n^2) = 2s(q^k)s(n^2)$$
holds for both even and odd perfect numbers.
Here is my question:
Does this proof suffice?
Added October 02 2018
Note that it would appear as though we have the corresponding equation
$$D(q^k)D(n^2) = 2s(q^k)s(n^2) = q^k - 1$$
for odd perfect numbers. We show here that this assumption is false.
Assuming that $D(q^k)D(n^2) = 2s(q^k)s(n^2) = q^k - 1$, we get
$$\frac{2(q^k - 1)}{(q - 1)}s(n^2) = q^k - 1,$$
since $s(q^k) = \sigma(q^k) - q^k = \sigma(q^{k-1})$.
This simplifies to
$$s(n^2) = \frac{q-1}{2}$$
or
$$\frac{\sigma(n^2)}{n^2} - 1 = \frac{s(n^2)}{n^2} = \frac{q-1}{2n^2}.$$
Using the following results from this paper:
$$\frac{8}{5} < \frac{\sigma(n^2)}{n^2}$$
and
$$\frac{q}{n^2} \leq \frac{q^k}{n^2} < \frac{2}{3},$$
we get a contradiction, as follows:
$$\frac{3}{5} = \frac{8}{5} - 1 < \frac{\sigma(n^2)}{n^2} - 1 = \frac{s(n^2)}{n^2} = \frac{q-1}{2n^2} < \frac{q}{2n^2} \leq \frac{q^k}{2n^2} < \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$ are the singular values of $A$ along with the negative signs.
The eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$ are the singular values of $A$ along with the negative signs.
Here $A$ is an $n \times n$ matrix, has $n$ singular values. Here we are consider both $\lambda $ and $- \lambda$ if $\lambda$ is an singular value of $A$. Thus we have a set of $2n$ elements and we have to show that these are the eigenvalues of $\begin{pmatrix}0&A\\A^*&0\\ \end{pmatrix}$.
Need some hint to proceed with the problem.
| Let $A$ be a $(n \times n)$-matrix and let $B$ be the matrix $$B = \begin{pmatrix} 0 & A \\ A^* & 0
\end{pmatrix}.$$
Let $A = U_1 \Sigma U_2^*$ be a singular value decomposition of $A$. Then we can decompose $B$ into a product $B = USU^*$ where $U$ is unitary and $S$ has a simpler structure than $B$. Let us show this explicitly:
We have
$$
B = \begin{pmatrix} 0 & U_1 \\ U_2 & 0 \end{pmatrix} \begin{pmatrix} 0 & \Sigma^* \\ \Sigma & 0 \end{pmatrix} \begin{pmatrix} 0 & U_2^* \\ U_1^* & 0 \end{pmatrix},
$$
so we can choose
$$ S = \begin{pmatrix} 0 & \Sigma^* \\ \Sigma & 0 \end{pmatrix}, \quad U = \begin{pmatrix} 0 & U_1 \\ U_2 & 0 \end{pmatrix}.$$
This means that the eigenvalues of $B$ are exactly the eigenvalues of $S$.
Let $\epsilon_i$ be the $i$-th unit vector of dimension $2n$ and $\sigma_1,\dots,\sigma_{n}$ be the singular values of $A$ resp. the entries of the diagonal matrix $\Sigma$. Now try to consider the vectors $\epsilon_k+\epsilon_{n+k}$ and $\epsilon_k - \epsilon_{n+k}$ for $k = 1,\dots,n$. What happens if you multiply them with our matrix $S$?
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\binom{n}{2}+\binom{n}{3}\cdot7+\binom{n}{4}\cdot7^2+....+7^{n-2}\in \mathbb{Z}$ for all $n\geq 2$
If $X=\{8^n-7n-1:n\in \mathbb{N}\}$ and $Y=\{49(n-1):n\in \mathbb{N}\}$ then show that $X⊂Y$
While solving the problem
$$8^n-7n-1=(1+7)^n-7n-1\\
=49\bigg[\binom{n}{2}+\binom{n}{3}\cdot7+\binom{n}{4}\cdot7^2+...+\binom{n}{n-1}.7^{n-3}+7^{n-2}\bigg]\\
$$
How do I know that the expression inside the bracket is an integer for all $n\geq 2$ ?
| Another way to expand using
$$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1}\right) \tag{1}$$
is
$$8^n-7n-1=8^n-1-7n=\color{red}{(8-1)}\left(8^{n-1}+8^{n-2}+...+8+1\right)-\color{red}{7}n=\\
7\left(8^{n-1}+8^{n-2}+...+8+1-\color{blue}{n}\right)=...$$
there are $\color{blue}{n}$ terms $8^{n-1},8^{n-2},...,8,1$, thus
$$...=7\left(8^{n-1}-\color{blue}{1}+8^{n-2}-\color{blue}{1}+...+8-\color{blue}{1}+1-\color{blue}{1}\right)=\\
7\left[\color{red}{(8-1)}\left(8^{n-2}+8^{n-3}+...+8+1\right)+\color{red}{(8-1)}\left(8^{n-3}+8^{n-4}+...+8+1\right)+...+\color{red}{7}\right]=\\
49\left[\left(8^{n-2}+8^{n-3}+...+8+1\right)+\left(8^{n-3}+8^{n-4}+...+8+1\right)+...+1\right]$$
Your version is a lot shorter and, as per the comments, correct one (binomial coefficients are integers). Sometimes it's good to have more versions to validate the answer.
| {
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Least value of $L$ for $\sqrt{x^2+ax}-\sqrt{x^2+bx}0$
If $a$ and $b$ are positive real numbers such that $a-b=2$ , then the smallest value of the constant $L$ for which $\sqrt{x^2+ax}-\sqrt{x^2+bx}<L$ for all $x>0$.
This question is similar to this one, but I don't want to apply the concept of limit. I want to use application of derivative in solving this problem.
I thought of putting $\sqrt{x^2+ax}-\sqrt{x^2+bx}=L$ and then using $\frac{dL}{dx}=0$, but the equation is coming in the form of $a+b$ which I am not able to solve.
| We have $$\sqrt{x^2+ax}-\sqrt{x^2+bx}=\frac{(x^2+ax)-(x^2+bx)}{\sqrt{x^2+ax}+\sqrt{x^2+bx}}=\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}$$
for all $x>0$. Because $\sqrt{1+\frac{a}{x}}>1$ and $\sqrt{1+\frac{b}{x}}>1$ for all $x>0$, we get
$$\sqrt{x^2+ax}-\sqrt{x^2+bx}< \frac{a-b}{2}.$$
This shows that $L\leq \frac{a-b}{2}$. As
$$\lim_{x\to\infty}\left(\sqrt{x^2+ax}-\sqrt{x^2+bx}\right)=\lim_{x\to\infty}\frac{a-b}{\sqrt{1+\frac{a}{x}}+\sqrt{1+\frac{b}{x}}}=\frac{a-b}{2},$$
we must have $L=\frac{a-b}{2}$. (By the way, you cannot avoid using limits in this problem because $L$ is attained as $x$ goes to infinity. Taking derivative of $\sqrt{x^2+ax}-\sqrt{x^2+bx}$ will not give you a maximizing point on $(0,\infty)$ because the maximum does not exist.)
| {
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what is the characteristic polynomial and minimal polynomial of A and B? What is the characteristic polynomial and minimal polynomial of $A$ and $B$ ?
$A=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},B=\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{pmatrix}$
My attempt : for $A$) , $ch_A(x) = (x-a)^4$ , $m_A = (x-a)^2(x-a)$
for $B)$ $ch_B(x) = (x-a)^4$ , $m_B = (x-a)^2(x-a)^2$
where $ch$ and $m$ denote the characteristic and minimal polynomial.
Is my answer is correct or not ???
Any hints/solution will be apprciated
thanks u
| Set
$N_A = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}; \tag 1$
then
$N_A^2 = 0; \tag 2$
further note that
$A - aI = N_A; \tag 3$
thus
$(A - aI)^2 = 0, \tag 4$
which means that
$m_A = (x-a)^2, \tag 5$
since $A$ can satisfy no polynomial of degree 1; indeed,
$cA + dI = \begin{bmatrix} ca + d & c & 0 & 0 \\ 0 & ca + d & 0 & 0 \\ 0 & 0 & ca + d & 0 \\ 0 & 0 & 0 & ca + d \end{bmatrix} = 0 \Longleftrightarrow c = d = 0; \tag 6$
thus $m_A(x) = (x - a)^2$ is the polynomial of least degree satisfied by $A$; hence, minimal.
It it easy to see by direct and simple calculation that
$e_1 = (1, 0, 0, 0)^T, \; e_3 = (0, 0, 1, 0)^T, \; e_4 = (0, 0, 0, 1) \tag 7$
are eigenvectors of $A$, each with eigenvalue $a$, and that
$(A - aI) e_2 = e_1, \tag 8$
that is, $e_2 = (0, 1, 0, 0)^T$ is a generalized eigenvector of $A$ corresponding to eigenvalue $a$; therefore $a$ is an eigenvalue of algebraic multiplicity $4$ (tho' of geometric multiplicity $3$); thus the characteristic polynomial of $A$ is
$c_A(x) = (x - a)^4, \tag 9$
which of course may also be had as
$c_A(x) = \det(A - xI). \tag{10}$
As for $B$, we set
$N_B = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}; \tag{11}$
then
$N_B^2 = 0; \tag{12}$
therefore
$(B - aI)^2 = N_B^2 = 0, \tag{13}$
so
$m_B(x) = (x -a)^2; \tag{14}$
we can see that
$\deg m_B(x) \ne 1 \tag{15}$
by more or less the same logic that was used above to show $\deg m_A(x) \ne 1$.
Finally, we see that $e_1$ and $e_3$ are eigenvectors of $A$ corresponding to $a$, and that here $e_2$ and $e_4$ are generalized eigenvectors for $a$; it follows now that $a$ is of algebraic multiplicity $4$, and we conclude that
$c_B(x) = (x - a)^4. \tag{16}$
| {
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"url": "https://math.stackexchange.com/questions/2946605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability that exactly 2 balls are white
What I did.
I let $X$ be number of withdraws before $x$ white balls. We can call our succes in this case to be gettin a white ball and probability is of course $p = \frac{3}{6} = \frac{1}{2}$. So we see $X$ is negative binomial r.v with $n=4$ trials. So,
$$ P(X=2) = { 4 - 1 \choose 2 - 1} \left( \frac{1}{2} \right)^2 \left( \frac{1}{2} \right)^2 $$
Which gives
$$ P(X=2) = \boxed{\dfrac{3}{16} }$$
Am I interpreting the problem correctly?
| $P(2W|4) = \binom{4}{2}\cdot (\frac{1}{2})^4$
$ = 6\cdot \frac{1}{16} = \frac{3}{8}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding values to make an equation with a $x^4$ into a perfect square For what values of p and q is the expression $$x^4+ 6x^3+ 13x^2+ px + q$$ a perfect
square?
The answer I got didn't seem correct so if someone break it down to me how about going this problem it would be appreciated
| Hint: write like this
$$x^4+ 6x^3+ 13x^2+ px + q= (x^2+ ax +b)^2$$
$$=x^4+ 2ax^3+ (2b+a^2)x^2+ 2abx + b^2$$
So $a =3$ and $2b+9 = 13$ so $b= 2$. Then $p = 2ab =12$ and $q=b^2 =4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
continued fraction development of the numbers of the following form, $\sqrt{n^2-1}$
Determine the continued fraction development of the numbers of the
following form, $\sqrt{n^2-1}$, with $n>1$ an integer.
I wasn't sure how to tackle this, so I just tried to write it out and tried to find $a_1...a_n$.
$$\alpha_0=\sqrt{n^2-1}=n-1 +\sqrt{n^2-1}+n-1$$where $n-1$ is $a_0$ and $\sqrt{n^2-1}+n-1$ is {$\alpha_0$}.
This brings us to $\alpha_1=\frac{1}{\{\alpha_0\}}=\frac{1}{\sqrt{n^2-1}+n-1}=\frac{\sqrt{n^2-1}-n+1}{n^2-1-(n-1)^2}=\frac{\sqrt{n^2-1}-n+1}{2n-2}$
But $a_1$ would be $0$ here, so I don't think I'm on the right track.
| $a_0=\lfloor \sqrt {n^2-1} \rfloor = n-1$
The repeating part is $1,2(n-1)$. We know when we get to $2a_0$ in a square root we have hit the repeat. You can write
$$\sqrt{n^2-1}=n-1+(\sqrt {n^2-1}-(n-1))\\=n-1+\frac 1{\frac 1{\sqrt {n^2-1}-(n-1)}}\\=n-1+\frac 1{\frac {\sqrt{n^2-1}+n-1}{n^2-1-(n-1)^2}}\\=n-1+\frac 1{\frac {\sqrt{n^2-1}+n-1}{2n-2}}$$
and the first $1$ comes out in the continued fraction. Keep going and you will get $2n-2$ at the next step.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to get the Taylor polynomial arctan with error? We have to find Taylor polynomial $P_2(x)$ of $\arctan(x)$ of grade 2 from point $x = 0$
My answer was
If $f(x) = \arctan(x)$ then $f(0)=0$
And $f'(x)= \frac{1}{1+x^2}$ where $f'(0) = 1$
And $f''(x) = \frac{-2x}{(1+x^2)^2}$ where $f''(0) = 0$
And $P_2(x) = x - \frac{1}{3}x^3 + \frac{1}{5}x^5$
Then we have to use the Taylor theorem to show that $P_2(x) - \frac{1}{3}x^3 < \arctan(x) < P_2(x) + \frac{1}{12}x^3$
when $0 < x ≤ 1$
What should I do to get it? I know to calculate an error we use
$E_n(x)= \frac{f^{n+1}(s)}{(n+1)}(x-a)^{n+1}$
Where am I wrong exactly?
| A simpler approach is as follows. We know that if $f(x) = \arctan(x)$, then $f' = (1 + x^2)^{-1}$. Thus
$$ \arctan(x) = \int_0^x \frac{1}{1 + t^2} dt + C$$
for some constant $C$. As $\arctan(0) = 0$, we even have that $C = 0$.
A Taylor expansion for $(1 + t^2)^{-1}$ is straightforward, as this is a geometric series
$$ \frac{1}{1 + t^2} = 1 - t^2 + t^4 - t^6 + \cdots$$
Thus the Taylor expansion for $\arctan(x)$ centered at $0$ is given by
$$ \arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + \cdots$$
For $x \geq 0$, this is an alternating series. For $0 \leq x < 1$, this is an alternating series with strictly decreasing (in magnitude) terms. Thus consecutive partial sums bound the actual value (as in the alternating series test for convergence).
Thus
$$ x - x^3/3 \leq \arctan(x) \leq x - x^3/3 + x^5/5.\tag{1}$$
This is stronger than the bound you were asked to show.
Nonetheless, you could go from this bound to the prescribed bound (if desired) by noting that for $0 \leq x < 1$, we have that $x^5/5 < x^4/4$, and further that $x^4/4 < x^3/4$, so that
$$\arctan(x) \leq x - x^3/3 + x^5/5 \leq x - x^3/3 + x^4/4 \leq x - x^3/3 + x^3/4$$,
which simplifies to
$$ \arctan(x) \leq x - x^3/12.$$
This is sort of silly, but this shows that the simpler bound in $(1)$ is stronger than the requested bound.
| {
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"url": "https://math.stackexchange.com/questions/2959258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Series of Nested Radicals
I can't seem to find a way, squaring the expression would make more terms and would make it harder, I guess there must be something to do with the first and last terms as they sum to 100? or maybe difference of to squares, but i can't solve it.
|
Useful fact:
$$
\sqrt{a\pm \sqrt{b}}=
\sqrt{\frac{a+\sqrt{a^2-b}}{2}} \pm
\sqrt{\frac{a-\sqrt{a^2-b}}{2}} \tag{1}$$
Proof:
\begin{align}
RHS^2 &=
\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2} \pm 2
\sqrt{\left( \frac{a+\sqrt{a^2-b}}{2} \right)
\left( \frac{a-\sqrt{a^2-b}}{2} \right)} \\
&= a \pm 2\sqrt{\frac{a^2-(a^2-b)}{4}} \\
&= a \pm 2\sqrt{\frac{b}{4}} \\
&= a \pm \sqrt{b} \\
&= LHS^2 \\
LHS &= RHS \qquad (a^2 \ge b \ge 0)
\end{align}
Let
$$S_{\pm}=\sum_{j=1}^{n^2-1} \sqrt{n \pm \sqrt{j}} \tag{2}$$
and reverse the order of summation by taking $\, k=n^2-j$, then
$$S_{\pm}=\sum_{k=1}^{n^2-1} \sqrt{n \pm \sqrt{n^2-k}} \tag{3}$$
Now by $(1)$, $(2)$ and $(3)$,
$$S_{\pm}=\frac{S_+ \pm S_-}{\sqrt{2}} \implies \frac{S_+}{S_-}=1+\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2960855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to prove $\sqrt{2} + \sqrt[3]{2}$ is an algebraic number I need to show that this is an algebraic number by showing that it is a solution to:
$x^{6} - 9x^{4} - 4x^{3} + 27x^{2} + 36x -23 = 0.$
I'm really struggling with this one, I tried squaring and cubing $\sqrt{2} + \sqrt[3]{2}$, but I only end up with more squares and cubes.
| Set
$\alpha = \sqrt 2 + \sqrt[3] 2; \tag 1$
then,
$\alpha - \sqrt 2 = \sqrt [3] 2; \tag 2$
$(\alpha - \sqrt 2)^3 = 2; \tag 3$
$\alpha^3 - 3 \sqrt 2 \alpha^2 + 6 \alpha - 2\sqrt 2 = 2; \tag 4$
$-\sqrt 2(2 + 3\alpha^2) = 2 - 6\alpha - \alpha^3; \tag 5$
$2(2 + 3\alpha^2)^2 = (2 - 6\alpha - \alpha^3)^2. \tag 6$
If all we want to do is show $\alpha$ algebraic over $\Bbb Q$, we can stop here; (6) is evidently a $6$ degree polynomial over $\Bbb Q$;
also, a finer inspection of (6) shows $\alpha$ to be an algebraic integer, since the leading coefficient, of $\alpha^6$, is $1$; to see exactly what polynomial we get, however, we must expand out each side of (6):
$2(4 + 12\alpha^2 + 9\alpha^4) = 4 + 36\alpha^2 + \alpha^6 - 24\alpha - 4\alpha^3 + 12\alpha^4; \tag 7$
$8 + 24\alpha^2 + 18\alpha^4 = 4 + 36\alpha^2 + \alpha^6 - 24\alpha - 4\alpha^3 + 12\alpha^4; \tag 8$
finally,
$\alpha^6 - 6\alpha^4 - 4\alpha^3 + 12 \alpha^2 -24\alpha - 4 = 0, \tag 9$
a sixth-degree polynomial satisfied by $\alpha$, in agreement with that obtained by Jack D'urizio.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
$4ac-b^2\leq 3a(a+b+c)$ in quadratic Suppose that the polynomial $(b+c)x^2+(a+c)x+(a+b)$ doesn't have real roots, where $a,b,c\in\mathbb{R}$. Prove that $4ac-b^2\leq 3a(a+b+c)$.
The quadratic not having real roots means that $$(a+c)^2-4(b+c)(a+b)<0$$ which translates to $(a^2+2ac+c^2)-4(b^2+ab+ac+bc)<0$, or
$$a^2+c^2-2ac-4b^2-4ab-4bc<0$$
which is still quite far from the inequality in question. We need to eliminate the $c^2$ term, which might be possible using a square form like $(c-b)^2\geq 0$, but it doesn't really get us closer.
| in standard order $a^2, b^2, c^2, bc, ca, ab,$ you are given
$$ -a^2 + 4b^2 - c^2 + 4bc +2ca+ 4 ab > 0 $$
The following is positive semidefinite (rank one Hessian matrix)
$$ 25a^2 + 4b^2 + c^2 - 4bc -10ca+ 20 ab \geq 0 $$
It is, in fact, simply $(5a+2b-c)^2$
Add to get
$$ 24a^2 + 8b^2 + 0 c^2 + 0bc -8ca+ 24 ab > 0 $$
or
$$ 24a^2 + 8b^2 -8ca+ 24 ab > 0 $$
Divide by $8$ to get
$$ 3a^2 + b^2 -ca+ 3 ab > 0 $$
================================================
As to how to find this: the (a,b,c) Hessian matrix of $$ t(3a^2 + b^2 -ca+ 3 ab) - (-a^2 + 4b^2 - c^2 + 4bc +2ca+ 4 ab), $$
restricted to $t>0,$ turned out to have determinant $$ -2(t+1)(t-8)^2 $$
so the only positive $t$ giving a non-negative determinant was $t=8$
================================================
PARI/GP is free software,
? a = [ 6,3,-1; 3,2,0; -1,0,0]
%1 =
[ 6 3 -1]
[ 3 2 0]
[-1 0 0]
? b = [ -2,4,2;4,8,4;2,4,-2]
%2 =
[-2 4 2]
[ 4 8 4]
[ 2 4 -2]
? h = t * a - b
%8 =
[6*t + 2 3*t - 4 -t - 2]
[3*t - 4 2*t - 8 -4]
[ -t - 2 -4 2]
? d = matdet(h)
%11 = -2*t^3 + 30*t^2 - 96*t - 128
? factor(d)
%12 =
[t - 8 2]
[t + 1 1]
?
=============================================
| {
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"url": "https://math.stackexchange.com/questions/2961241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$
The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\sqrt{1+\tan^2\theta}}=\frac{1}{\sqrt{1+\frac{81}{4}}}=\frac{2}{\sqrt{85}}
$$
$$
\cos\theta=\frac{\sin\theta}{\tan\theta}=\frac{\frac{3}{4}}{\frac{9}{2}}=\frac{3}{4}.\frac{2}{9}=\frac{1}{6}
$$
Why is it getting confused here ?
| Because you have two different problems. Obviously (from your calculations which are correct) it can't be $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ at the same time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Limit $\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$
What is the limit of:
$$\lim_\limits{x\to0}{\tan{x}-\sin{x}\over x^3}$$
So what I tried is:
$$\lim_\limits{x\to0}{{\sin{x}\over\cos{x}}-\sin{x}\over x^3}=\lim_\limits{x\to0}{{\sin{x}}({1\over\cos x}-1)\over x\cdot x^2}$$
From here, using the rule $\lim_\limits{x\to0}{\sin{x}\over{x}}=1$ it remains to evaluate
$$\lim_\limits{x\to0}{{1\over\cos{x}}-1\over x^2}.$$
I tried changing it a lot of ways but it always gets messier so I'm not sure what to apply here to complete the question.
| $$\frac{\tan x - \sin x}{x^3} = \frac{\sin x - \sin x\cos x}{x^3\cos x} = \frac{\sin x}x \cdot\frac{1-\cos x}{x^2}\cdot\frac 1{\cos x}\to 1\cdot \frac 12\cdot \frac 1{\cos 0} = \frac 12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
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Prove $\sum_{1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2 $
Prove that $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2$$
I have found that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} < \pi / 2 $ with integrating from $1$ to infinity but integral isn’t allowed.
| $$
\frac{1}{n\sqrt n} = \frac{1}{\sqrt n(n+\sqrt n)}+\frac{1}{n(n+\sqrt n)}
$$
hence
$$
\frac{1}{\sqrt n(n+\sqrt n)}<\frac{1}{n^{\frac 32}}
$$
NOTE
$$
\sum_{n=3}^{\infty} \frac{1}{n^{\frac 32}} < 1.24
$$
now adding $\frac{1}{\sqrt 1(1+\sqrt 1)}+\frac{1}{\sqrt 2(2+\sqrt 2)}+\sum_{n=3}^{\infty} \frac{1}{n^{\frac 32}} < 1.95$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help find the mistake in this problem of finding limit (using L'Hopital)
Evaluate $$\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right).$$
Attempt
\begin{align*}
&\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)\\
= &\lim_{x \to 0} \left(\frac{1}{x}-\cot{x}\right)\left(\frac{1}{x}+\cot{x}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right)\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right)\\
= &\lim_{x \to 0} \left(\frac{\sin{x}+x\cos{x}}{x\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{\sin{x}-x\cos{x}}{x\sin{x}}\right).
\end{align*}
Both the terms are in $\frac00$ form. So applying L'Hopital on both the limits we have,
$$= \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\sin{x}}{x\cos{x}+\sin{x}}\right).$$
The second term is in $\frac00$ form. So applying L'Hopital on the second limit we have,
\begin{align*}
= &\lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \lim_{x \to 0}\left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& \lim_{x \to 0} \left(\frac{2\cos{x}-x\sin{x}}{x\cos{x}+\sin{x}}\right) \times \left(\frac{x\cos{x}+\sin{x}}{2\cos{x}-x\sin{x}}\right)\\
=& 1
\end{align*}
The correct answer is $\dfrac23$ which can be found using series expansion. But I think I'm making a conceptual mistake in one of the above steps. Could you please point out to the specific step where I've committed a mistake in above solution?
| Similar work:
$$\begin{align}\lim_{x \to 0} \left(\frac{1}{x^2}-\cot^2x\right)&=\lim_{x \to 0} \left(\frac{1}{x^2}-\frac{1-\sin^2x}{\sin^2x}\right)=\\
&=\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{1}{\sin^2x}+1\right)=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\right)+1=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^2\sin^2x}\cdot \frac{\sin^2x}{x^2}\right)+1=\\
&=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)+1.
\end{align}$$
According to the algebraic limit theorem, you can express the limit as a product of two existing limits:
$$\begin{align}\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)+1=&\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^3}\right)}_{-\frac16}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x}\right)}_{=2} +1\stackrel{LR}{=}&\\
\lim_{x \to 0}\left(\frac{\cos x-1}{3x^2}\right)\cdot \lim_{x \to 0}\left(\frac{\cos x+1}{1}\right) +1\stackrel{LR}{=}&\\
\lim_{x \to 0}\left(\frac{-\sin x}{6x}\right)\cdot 2 +1=&\\
-\frac13+1=&\frac23.\end{align}$$
However you can not express:
$$-\frac13=\lim_{x \to 0}\left(\frac{(\sin x-x)(\sin x+x)}{x^4}\right)=\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}0}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}4}\right)}_{=\infty} \ \ \text{OR}\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}1}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}3}\right)}_{=\infty} \ \ \text{OR}\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}2}\right)}_{=0}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}2}\right)}_{=\infty} \ \ \text{OR}\\
\underbrace{\lim_{x \to 0}\left(\frac{\sin x-x}{x^\color{red}4}\right)}_{=\infty}\cdot \underbrace{\lim_{x \to 0}\left(\frac{\sin x+x}{x^\color{blue}0}\right)}_{=0}.$$
because all are the indeterminate form of $0\cdot \infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
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What is the coefficient of the term $x^3y^5$, as a result of the binomial expansion of the following term? We have the term
$(1+xy+y^2)^n$
If we expand it using the binomial theorem, why is the factor of the term $x^3y^5$ the following: $4{n\choose 4}$? (The binomial coefficient n choose 4 multiplied by 4)
| Yes, for n = 5, use the binomial theorem by substituting $xy + y^2$ for "$a$" and then expand $(1 + a)^5 = 1 + 5a + 10a^2 + 10a^3 + 5a^4 + a^5$.
Then substitute back $xy + y^2$ for $a$.........
Here, the only term which will give $x^3y^5$ is $5a^4$ whereby again using the binomial theorem on $5(xy + y^2)^4$ we get......
$5x^4y^4 + 5(4)x^3y^5 + 5(6)x^2y^6 + 5(4)xy^7 + 5y^8$
We get $20x^3y^5$ which is in keeping with $4\binom{5}{4}$
So to answer you question in a general sense, expanding a trinomial involves more than one application of the binomial theorem for some of the terms.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality proof as a part of induction As a part of an induction proof, I need to prove that if $\frac{3x+2}{x+2} < x$, then $\frac{3x+2}{x+2} > \frac{11x+10}{5x+6}$. I need to prove this for $x>0$ if that matters so basically $x$ has to be greater than 2. I have tried manipulating this inequality in a number of ways but it always leads to a dead end. How can I prove this?
Edit:
Managed to prove this like following:
From the assumption we get that $x>2$. So considering this
$$\frac{11x+10}{5x+6} < \frac{11x+10}{4x+8} < \frac{12x+8}{4x+8} = \frac{3x+2}{x+2}$$
| Hint: For $$x>0$$ we get that $$\frac{3x+2}{x+2}<x$$ is equivalent to $$0<x^2-x-2$$ and this is $$0<(x+1)(x-2)$$ so it must be $$x>2$$
and now we will Show that
$$\frac{3x+2}{x+2}>\frac{11x+10}{5x+6}$$ is true for $$x>2$$
this is $$(3x+2)(5x+6)>(11x+10)(x+2)$$
Expanding and collecting like terms we get
$$4x^2-4x-8>0$$
dividing by $4$ we get
$$x^2-x-2>0$$
and this is
$$(x+1)(x-2)>0$$ true for $$x>2$$!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
CAN and BAN estimators of $\sigma^2$ Let $X_1,X_2,..X_n \sim^{\text{i.i.d}} N(0,\sigma^2)$
Show that $T_n=\frac{1}{n}\sum_{i=1}^{n} X_i^2$ is a Consistent and Asymptotically Normal estimator(CAN) as well as the Best Asymptotic Normal Estimator(BAN) of $\sigma^2$.
Indeed, $E(T_n)=\sigma^2$ and $V(T_n)=\frac{2 \sigma^4}{n}$.
So, $T_n$ is a CAN estimator.
But to show that it is a BAN estimator , $V(T_n)$ should be equal to $n$ times the Rao Cramer lower bound.
But I am failing to show that.
| CRB
Let's write down the PDF
\begin{equation}
f_{X_i}(x_i \vert \sigma^2) \sim \frac{1}{\sqrt{2\pi \sigma^2}}\exp(-\frac{x_i^2}{2\sigma^2} )
\end{equation}
Assuming independence, we have that
\begin{equation}
L(\sigma^2) = f(x_1 \ldots x_n \vert \sigma^2)
=
f(x_1 \vert \sigma^2) \ldots f(x_n \vert \sigma^2)
=
\frac{1}{(2\pi \sigma^2)^{n/2}}
\exp(- \frac{1}{2\sigma^2}\sum_{i=1}^n x_i^2)
\end{equation}
Take the log likelihood
\begin{equation}
l(\sigma^2)
=
\log L(\sigma^2)
=
-\frac{n}{2}
\log(2 \pi \sigma^2)
- \frac{1}{2\sigma^2}\sum_{i=1}^n x_i^2
\end{equation}
Deriving w.r.t $\sigma^2$, we have
\begin{equation}
l'(\sigma^2)
=
-\frac{n}{2}
\frac{1}{\sigma^2}
+ \frac{1}{2\sigma^4}\sum_{i=1}^n x_i^2
\end{equation}
Derive again
\begin{equation}
l''(\sigma^2)
=
\frac{n}{2}
\frac{1}{\sigma^4}
- \frac{1}{\sigma^6}\sum_{i=1}^n x_i^2
\end{equation}
Take the expectation now
\begin{equation}
E(l''(\sigma^2))
=
E \big(
\frac{n}{2}
\frac{1}{\sigma^4}
- \frac{1}{\sigma^6}\sum_{i=1}^n x_i^2
\big)
\end{equation}
The only random part here is $x_i^2$, hence
\begin{equation}
E(l''(\sigma^2))
=
\frac{n}{2}
\frac{1}{\sigma^4}
- \frac{1}{\sigma^6}\sum_{i=1}^n E x_i^2
\end{equation}
But $E x_i^2 = \sigma^2$ so
\begin{equation}
E(l''(\sigma^2))
=
\frac{n}{2}
\frac{1}{\sigma^4}
- \frac{1}{\sigma^6}\sum_{i=1}^n \sigma^2
=
\frac{n}{2}
\frac{1}{\sigma^4}
- \frac{n}{\sigma^4}
=
-
\frac{n}{2\sigma^4}
\end{equation}
So, the Fisher information is
\begin{equation}
F = - E(l''(\sigma^2)) = \frac{n}{2\sigma^4}
\end{equation}
Then the CRB is
\begin{equation}
CRB = \frac{1}{F} = \frac{2\sigma^4}{n} = V(T_n)
\end{equation}
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
} |
Given that $f(x) = 1 - 3/(x+2) + 3/(x+2)$ while $x ≠- 2$, show that $f(x) = (x^2 + x + 1)/(x+2)^2$ What type of problem is this and what are the first step(s) needed to tackle it?
Given that:
$$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}, \quad \text{with }x ≠ -2$$
Show that:
$$f(x) = \frac{x^2 + x +1}{(x + 2)^2}$$
It's fairly basic, clearly, but from my attempts so far it appears that finding common denominators doesn't take you from the first form to the second.
Should something be factorised?
| $$f(x) = 1-\frac{3}{x + 2}+\frac{3}{(x +2)^2}; x ≠ 2$$
$$\implies f(x) = \frac{(x+2)^2}{(x+2)^2}-\frac{3(x+2)}{(x + 2)^2}+\frac{3}{(x +2)^2}$$
$$\implies f(x) = \frac{(x+2)^2-3(x+2)+3}{(x+2)^2}$$
Just expand and then add up the values in the numerator...
| {
"language": "en",
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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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How to show that $x_{n+1} = \frac{x_n^4 + 1}{5x_n}$ bounded below and above by $1\over 5$ and $2$
Given a sequence
$$
\begin{cases}
x_{n+1} = \frac{x_n^4 + 1}{5x_n} \\
x_1 = 2 \\
n \in \mathbb N
\end{cases}
$$
Prove it has lower bound at $1\over 5$ and upper bound at $2$
I've tried to find a closed form for the recurrence relation, but couldn't arrive at anything. Also:
$$
x_{n+1} = \frac{x_n^4+1}{5x_n}=\frac{2(x_n^4 +1)}{2\cdot5x_n}=\frac{2}{5x_n}\cdot\frac{x_n^4+1}{2} \ge\frac{2}{5x_n}\sqrt{x_n^4\cdot1} =\frac{2x_n}{5}
$$
So I got:
$$
x_{n+1} \ge \frac{2x_n}{5} \tag1
$$
I have no ideas how to proceed. I'm not even sure it's valid to use AM-GM here.
So my main questions are:
*
*Does this recurrence have a closed form?
*What else should I try to solve the problem?
Please note this is precalculus. I'm not allowed to use calculus.
Update
Using $(x_n^2 - 1)^2 > 0$ i get the same result as in $(1)$. Expanding the terms only shows that the sequence is greater than $0$:
$$
x_{n+1} \ge 2\cdot \left(2\over 5\right)^n
$$
which is tending to $0$ with growing $n$.
Update 2
Consider the following expressions:
$$
x_1 = 2 \\
x_2 = \frac{x_1^3}{5} + \frac{1}{5x_1} \\
\dots \\
x_{n+1} = \frac{x_n^3}{5} + \frac{1}{5x_n} \\
$$
Multiply both sides of each expression by some $z$ in the power of $n$:
$$
z\cdot x_1 = 2\cdot z \\
z^2\cdot x_2 = \left(\frac{x_1^3}{5} + \frac{1}{5x_1}\right)z^2 \\
\dots \\
z^{n+1}\cdot x_{n+1} = \left(\frac{x_n^3}{5} + \frac{1}{5x_n}\right) \cdot z^{n+1} \\
$$
Now sum them up:
$$
\sum_{k=1}^{n+1}x_k\cdot z^k = 2z + {1 \over 5}\left( \sum_{k=2}^{n+1}x_{k-1}^3z^k + \sum_{k=2}^{n+1}{z^k\over 5x_{k-1}} \right) = \\
= 2z + {1\over 5z} \left(\sum_{k=1}^{n}x_k^3z^k + \sum_{k=1}^{n}{z^k\over x_k}\right)
$$
Now define:
$$
G(z) = \sum_{k=1}^{n+1}x_k\cdot z^k
$$
From this point there may be a way to express RHS in terms of $G(z)$ but i couldn't handle that.
Update 3
This goes beyond precalculus level but anyway here is another observation inspired by @amam_Abdallah.
Define $x_{n+1} = f(x_n)$ if this function have fixed points then:
$$
\overline{x} = \frac{\overline{x}^3}{5} + \frac{1}{\overline{x}} \iff \\
\iff \overline{x} = \sqrt[^3]{5\overline{x} - {1\over \overline{x}}}
$$
This equation has two solutions:
$$
\overline{x} = \sqrt{{5\over 2} \pm {\sqrt{21} \over 2}}
$$
Perhaps this will lead someone to ideas on how to use that fact.
Update 4
Some more thoughts on the sequence:
$$
x_{n+1} = \frac{1}{5}x_n^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}x_{n-1}^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \\
= \frac{1}{5}\left(\frac{1}{5}\left(\frac{1}{5}x_{n-2}^3 + \frac{1}{5x_{n-2}} \right)^3 + \frac{1}{5x_{n-1}}\right)^3 + \frac{1}{5x_n} = \dots
$$
Can this somehow be wrapped into something in the form of $\prod \dots$ or $\sum \dots$?
| hint
write $x_{n+1}$ as
$$x_{n+1}=\frac{x_n^3}{5}+\frac{1}{5x_n}$$
and prove it by induction.
there are two fixed points satisfying
$$L^4-5L^2+1=0$$
$$L_1^2=\frac{5-\sqrt{21}}{2}\approx 0.7$$
$$L_2^2=\frac{5+\sqrt{21}}{2}\approx 4.7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Systems of Equations degree 2 Given the system
$$\begin{aligned}
x^2+y^2+\frac{\sqrt 3}{2}xy&=32\\
x^2+z^2+\frac{1}{2}xz&=16\\
y^2+z^2&=16,\end{aligned}$$
find the value of $(xy+\sqrt{3}xz+2yz).$
The answer is $32,$ so I think there will be a really nice solution to this.
I tried completing the square for each pair $x$ and $y$ but I could not find anything, I tried adding all three equations but I got nothing.
| I guess the two factors (1/2) shouldn't be there, which mean we have instead
\begin{align}
x^2+y^2+\sqrt{3}xy&=32\tag{A}\\
x^2+z^2+xz&=16\tag{B}\\
y^2+z^2&=16\tag{C}.
\end{align}
The rationale being that it gives a much nicer set of coefficients when completing the square
\begin{align}
(x+\frac{\sqrt3}2y)^2+\frac14y^2&=32\\
(x+\frac12z)^2+\frac34z^2&=16\\
\end{align}
and we have accidental cancellation when eliminating $y,z$, giving the biquartic $37x^4-896x^2+4096=0$, and hence a nice form of the four real solutions
$$
(x,y,z)=\left(8 \sqrt{\frac{7-2\sqrt{3}}{37}},
4(3 + \sqrt3)\sqrt{\frac{7-2\sqrt3}{111}},
4\sqrt{\frac{21 - 6\sqrt3}{37}}\right),\text{ and similar}
$$
and it is easy to check $(xy+\sqrt{3}xz+2yz)^2=2^{10}$. Indeed, it follows immediately from noting the LHS of
$$
\mathrm{(A)^2+(B)^2+(C)^2-2\times(A)\times(B)-2\times(B)\times(C)-2\times(C)\times(A)}
$$
is $-(xy+\sqrt{3}xz+2yz)^2$, which I suppose is the only way to solve this within the 120 seconds limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2978918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $x,y,z>0$ and $x+y+z=1$ then $\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$ If $x,y,z>0$ and $x+y+z=1$,
then: $$\frac{xyz}{(1-x)(1-y)(1-z)}\le\frac{1}{8}$$
$$\frac{xyz}{(1-x)(1-y)(1-z)}=\frac{x}{(1-x)}\frac{y}{(1-y)}\frac{z}{(1-z)}$$
Let $\frac{x}{(1-x)}=a$, $\frac{y}{(1-y)}=b$, $\frac{z}{(1-z)}=c$
I am stuck here.
| You have: $$x+y+z=1$$
So, write: $$1-x=y+z$$
Using inequality $AP\geq GP$, $$y+z\geq 2\sqrt{yz}$$
So, $$(1-x)\geq2 \sqrt{yz}$$ Similarly, $$(1-y)\geq2 \sqrt{xz} \hspace{0.5cm} {and} \hspace{0.5cm} (1-z)\geq2 \sqrt{xy}$$
Multiplying all: $$(1-x)\cdot (1-y) \cdot (1-z) \geq 8 \cdot xyz$$
Divide:$$\frac{(1-x)\cdot (1-y)\cdot (1-z)}{xyz}\geq 8$$ Or,
$$\frac{xyz}{(1-x)(1-y)(1-z)}\leq \frac 1 8$$ In-equality will be reversed on dividing (why?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2980320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Probability mass function of max and min function
You repeatedly draw a number at random from the numbers $1,2,...,10$.
Let $X$ be the number of drawings until the number $1$ appears and $Y$
be the number of drawings until the number $10$ appears. What is the
joint probability mass function of $X$ and $Y$? What are the
probability mass functions of $\max (X,Y)$ and $\min (X,Y)$?
Attempt
$X$ and $Y$ are geometric both with parameter $\dfrac{1}{10}$ since each number is equally likely to occur when picking a number at random. We need to find $P(X=x, Y=y)$. Here I get confused as to whether make the assumption that both $X$ are $Y$ are independent. Can we assume this? IF so , then
$$ P(X=x,Y=y) = (0.9)^{x-1} 0.1 (0.9)^{y-1} 0.1 $$
Now, for the next part notice that $max(X,Y)=n$ iff $X \leq n \; \text{or} Y \leq n$.hence,
$$ P(\max(Y,X) = n) = P(X \leq n) + P(Y \leq n) = \sum_{x=1}^n (0.9)^{x-1} 0.1 + \sum_{y=1}^n (0.9)^{y-1} 0.1 $$
Similarly,
$$ P( \min (Y,X) = m) = O(Y \geq n) + (X \geq n) = \sum_{y=n}^{\infty} (0.9)^{y-1} 0.1 + \sum_{ x =n}^{\infty} (0.9)^{x-1} 0.1 $$
Am I correct?
| $X$ and $Y$ are not independent since if $P(X=1,Y=1) =0$ but $P(X=1)>0$ and $P(Y>1)>0$.
if $1\le x < y$,
$$Pr(X=x, Y=y) = (0.8)^{x-1}(0.1)(0.9)^{y-x-1}(0.1)=0.8^{x-1}(0.9)^{y-x-1}(0.1)^2$$
That is the first $x-1$ times can be anything besides the two numbers, followed by $1$, then the following $y-x-1$ times we can get any number besides $10$.
Try to work out what happens when $x=y$ and for the case where $x>y$.
Notice that we can't conclude $\max(X,Y)=n \iff X \le n\lor Y \le n$ as if $X=n-2$ and $Y=n-1$, then we ahve $X \le n \lor Y \le n$ being true but $\max(X,Y)=n-1$.
\begin{align}
Pr(\max(X,Y)=n) &= Pr(Y > X \land Y=n) + Pr(X > Y \land X=n) \\
&= \sum_{x=1}^{n-1}Pr(X=x,Y=n) + \sum_{y=1}^{n-1}Pr(X=n,Y=y)
\end{align}
and
\begin{align}
Pr(\min(X,Y)=n) &= Pr(Y > X \land X=n) + Pr(X > Y \land Y=n) \\
&= \sum_{y=n+1}^{\infty}Pr(X=n,Y=y) + \sum_{y=n+1}^{\infty}Pr(X=x,Y=n)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2983701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all $2\times 2$ matrices that commute with $AX = XA$? Find all $2\times 2$ matrices that commute with $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$ where $AX = XA$.
Solution and ask:
$$ X = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$
$$AX = \left( \begin{array}{cc} 1\cdot a + 1\cdot c & 1\cdot b + 1\cdot d \\ 0\cdot a + 0\cdot c & 0\cdot b + 0\cdot d \end{array} \right) = \left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right)$$
$$XA = \left( \begin{array}{cc} a\cdot 1 + b\cdot 0 & a\cdot 1 + b\cdot 0 \\ c\cdot 1+d\cdot 0 & c\cdot 1+d\cdot 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
Since $AX = XA$, we obtain
$$\left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
so that $a+c = a$, $b+d = a$, $0 = c$ and $0 = c$.
Is the calculation correct?
What is the value of $a,b,c,d$ ?
$c = 0, a = ?, b = ?, d = ?$
| Here is a synthetic solution. Let $V$ denote the vector space of $2\times2$ matrices. Let $T:V\to V$ be the linear operator $TX=AX-XA$. The kernel of $T$ consists of eigenvectors of $T$ with eigenvalue $0$. Observe that $A$ is diagonalizable with eigenvalues $0$ and $1$ (well, this is obvious since $A$ is triangular, so the eigenvalues are the diagonal entries, and since they are distinct, $A$ must be diagonalizable).
From this thread, we know that the eigenspace of $T$ with eigenvalue $0$ is spanned by $u_0^t v_0$ and $u_1^t v_1$, where $u_\lambda$ is an eigenvector of $A^t$ with eigenvalue $\lambda\in\{0,1\}$, and $v_\lambda$ is an eigenvector of $A$ with eigenvalue $\lambda\in\{0,1\}$. We can take $$u_0=\begin{pmatrix}0\\1\end{pmatrix},\ u_1=\begin{pmatrix}1\\1\end{pmatrix},\ v_0=\begin{pmatrix}-1\\1\end{pmatrix},\ v_1=\begin{pmatrix}1\\0\end{pmatrix}.$$
So,
$$u_0^tv_0=\begin{pmatrix}0&-1\\0&1\end{pmatrix},\ u_1^tv_1=\begin{pmatrix}1&1\\0&0\end{pmatrix}.$$
Thus, every $X\in \ker T$ must be of the form
$$a\ u_0^tv_0+b\ u_1^tv_1=\begin{pmatrix}b&-a+b\\0&a\end{pmatrix}.$$
This gives the same answer as Servaes' solution.
| {
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"url": "https://math.stackexchange.com/questions/2984447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplifying fraction containgin square root Say we have an expression like
$$\frac{\sqrt{\frac{k}{g}}}{k}=\frac{1}{\sqrt{gk}}$$
How do we get from the left hand side, to the right hand side? If we simplify the square root
$$\sqrt{\frac{k}{g}}=\frac{\sqrt{k}}{\sqrt{g}}$$
Then
$$\frac{\sqrt{k}}{\sqrt{g}}\times \frac{1}{k}=\frac{\sqrt{k}}{k\sqrt{g}}$$
Then multiply through by $\sqrt{k}$ and cancel out $k$?
$$\frac{\sqrt{k}}{k\sqrt{g}}=\frac{k}{k\sqrt{g}\sqrt{k}}=\frac{1}{\sqrt{gk}}$$
Is that correct?
| $$\frac{\sqrt{\frac{k}{g}}}{k}=\frac{\sqrt \frac kg}{\sqrt\frac{k^2}{1}}=\sqrt\frac{k \over g}{k^2 \over 1}=\sqrt\frac{k}{gk^2}=\sqrt\frac{1}{gk}={\sqrt{1} \over
\sqrt{gk}}={1 \over \sqrt{gk}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Struggle with finding equality condition in an AM-GM inquality problem Given $0<x,y,z<1$. Prove that
$$\dfrac{1-x}{1+y+z}+\dfrac{1-y}{1+z+x}+\dfrac{1-z}{1+x+y}\geq 3(1-x)(1-y)(1-z).$$
I've thought that the equality holds when $x=y=z$ and I'm thinking ways to use AM-GM inequality. However, it turns out that
$$3\dfrac{1-x}{1+2x}=3(1-x)^3 \Longleftrightarrow (1-2x+x^2)(1+2x)=1\Longleftrightarrow 2x^3-3x^2=0\Longleftrightarrow x=\dfrac{3}{2},$$
which is totally wrong due to $0<x,y,z<1$! Thank you a lot!
| AM-GM seems like an unlikely choice in this problem. I am presenting a different approach.
Note that
\begin{align}\frac{1-x}{1+y+z}-(1-x)(1-y)(1-z)&=\frac{1-x}{1+y+z}\big(1-(1-y-z+yz)(1+y+z)\big)
\\&=\frac{1-x}{1+y+z}\big(y^2(1-z)+z^2(1-y)+yz\big)\geq 0\end{align}
for all $x,y,z\in[0,1]$. Similarly,
\begin{align}\frac{1-y}{1+z+x}-(1-x)(1-y)(1-z)\geq 0\end{align}
and
\begin{align}\frac{1-z}{1+x+y}-(1-x)(1-y)(1-z)\geq 0.\end{align}
Summing all three inequalities above yields
$$\frac{1-x}{1+y+z}+\frac{1-y}{1+z+x}+\frac{1-z}{1+x+y}\geq 3(1-x)(1-y)(1-z)$$
for all $x,y,z\in[0,1]$. There are only two equality cases, i.e., $x=y=z=0$ and $x=y=z=1$.
| {
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"url": "https://math.stackexchange.com/questions/2987384",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then...
if 1, $\alpha_1$, $\alpha_2$, $\alpha_3$, $\ldots$, $\alpha_{n-1}$ are nth roots of unity then
$$\frac{1}{1-\alpha_1} + \frac{1}{1-\alpha_2} + \frac{1}{1-\alpha_3}+\ldots+\frac{1}{1-\alpha_n} = ?$$
Now this has the solution too, but I do not understand the last step of the solution so here is the solution from the book.
$1,\, \alpha_1, \, \alpha_2,\, \ldots, \, \alpha_n$ are the $n^\text{th}$ of unity. These are the roots of $x^n-1=0$
Let $y=\frac{1}{1-\alpha}$ where $\alpha = \alpha_1, \, \alpha_2, \, \alpha_3, \, \ldots, \, \alpha_n$
$$1 - \alpha = \frac{1}{y} \Rightarrow \alpha = \frac{y-1}{y}.$$
But $\alpha$ is a root of $x^n-1=0 \therefore \alpha^n=1 \Rightarrow (y-1)^n = y^n$
$$\Rightarrow y^n - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = y^n \\
\Rightarrow - _nC_1y^{n-1} + _nC_2y^{n-2}-\ldots+(-1)^n = 0.$$
Sum of roots
$$\frac{1}{1-\alpha_1}+\frac{1}{1-\alpha_2}+\ldots+\frac{1}{1-\alpha_{n-1}} = \frac{_nC_2}{_nC_1} = \frac{n-1}{2}$$
So this last part from "Sum of roots" I do not understand. I cannot see how this last shape relates to this binomial theorem notation. Can anyone help?
| The sum of the roots of $a_nx^{n}+a_{n-1}x^{n-1}+\cdots +a_0$ is $-\frac {a_{n-1}} {a_n}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a quadratic equation using roots
If $x_1$ and $x_2$ are the roots of
$$ax^2+bx+c=0$$
then $x_1^3$ and $x_2^3$ are the roots of which equation?
So I tried by solving this for $x_{1/2}$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_{1/2}=\large{-b\pm{\sqrt{4ac}}\over2a}$
and from here:
$$\begin{align}x_1^3&=\bigg({-b+{\sqrt{4ac}}\over2a}\bigg)^3\\&={(\sqrt{4ac}-b)^2(\sqrt{4ac}-b)\over8a^3}\\&={(4ac-2b\sqrt{4ac}+b^2)(\sqrt{4ac}-b)\over8a^3}\\&={4ac\sqrt{4ac}-4abc-8abc-2b^2\sqrt{4ac}+b^2\sqrt{4ac}-b^3\over8a^3}\\&={4ac\sqrt{4ac}-12abc-b^2\sqrt{4ac}-b^3\over8a^3}\end{align}$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.
| HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Dividing numbers with exponents and powers Can someone please help me simplify the below problem. I am trying to help my niece
$${24C^4 \over 4C^{12}}$$
I believe that the simplified answer is $6/C^8$ because $4$ goes into $24$ $6$ times and $6$ remains as the numerator. Also, you need to subtract $4$ from $C^4$ which becomes $1$ and subtract $4$ from $C^{12}$ and it equals $C^8$. however my niece thinks it is: $1/(6C^8)$. what is the correct answer?
| One approach I've used for someone beginning to work with this type of situation is the following, although I'd probably replace it with something similar and simpler, such as $\frac{24C^2}{4C^5},$ to actually illustrate the idea.
$$ \frac{24C^4}{4C^{12}} \; = \; \frac{24 \cdot C^4}{4 \cdot C^{12}} \; = \; \frac{24}{4} \cdot \frac{C^4}{C^{12}} $$
$$ = \; \; \frac{24}{4} \cdot \frac{C \cdot C \cdot C \cdot C \cdot 1 \cdot 1\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1\cdot 1}{C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C} $$
$$ = \;\; \frac{24}{4} \cdot \frac{C}{C} \cdot \frac{C}{C} \cdot \frac{C}{C} \cdot \frac{C}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} $$
$$ = \;\; \frac{6}{1} \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} \cdot \frac{1}{C} $$
$$ = \;\; \frac{6}{1} \cdot \frac{1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 }{C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C \cdot C}$$
$$ = \;\; \frac{6}{1} \cdot \frac{1}{C^8} \;\; = \;\; \frac{6}{C^8} $$
What you want to do is go over several elaborate expansions like this and let the student discover the shortcuts that can be taken, and then when the student is comfortable with simple things like this (integer times a letter to a positive integer power, divided by an integer times a letter to a positive integer power), begin branching out to things like $\frac{24x^3y^5}{6x^2y^8}.$ All this should not take more than 5 to 15 minutes, after which you can begin talking about the relevance of the laws of exponents to the shortcuts the student has observed, as well as mixing in things like $C^3 \cdot C^4 = (CCC)\cdot (CCCC) = CCCCCCC = C^{3+4}$ $(3\;C$'s followed by $4\;C$'s is $7\;C$'s), if this had not already been done at some previous time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a\mid b^2, b\mid c^2, c\mid a^2\Longrightarrow abc\mid (a+b+c)^k$ Is it true that if $a,b,c$ are positive integers such that $a\mid b^2, b\mid c^2, c\mid a^2$, then $abc$ always divides $(a+b+c)^6$?
If not (I couldn’t find a counterexample), then $abc\mid (a+b+c)^7$ is always true? Or in general, find the minimum value of $k$, such that $abc\mid (a+b+c)^k$.
I proved that $a,b,c\mid (a+b+c)^6$, but unfortunetly that is not enough, since $a,b,c$ are not necessarily relative primes.
| If you expand out $(a + b + c)^6$, then all terms are multiples of some $a^ib^jc^k$, with $i+j+k=n$. Now, at least one of $i$, $j$, and $k$ is at least $\frac{n}{3}$.
Now, we have $abc|a^3b|a^3c^2|a^7$, and symmetrically $abc|b^7$, $abc|c^7$. Thus, we have an upper bound on the solution: $abc|(a+b+c)^{21}$.
To bring that down, we're going to have to do a bit more fiddling with our indices.
Now, for any $a^ib^jc^k$, if $i, j, k > 0$, then $abc|a^ib^jc^k$, so we just need to worry about the terms where one of them is zero. Since everything's symmetrical, we can just deal with the $k = 0$ case.
So, we need to know when $abc|a^ib^j$, with $i, j > 0$ (we've already dealt with the case where only one is non-zero above), But notice that we can divide out by an $ab$, so we just need to know when $c|a^{i-1}b^{j-1}$. Since $c|a^2$, $i \geq 3$ suffices. If $i = 2$, then we have $c|a^2|ab^2$, so $i = 2, j \geq 3$ also suffices. If $i = 1$, then we have $c | a^2 | b^4$, so $i = 1, j \geq 5$ suffices. In particular, $i + j \geq 5$ suffices.
Putting all of those together, whenever $i + j + k \geq 7$, one of the following (or the cyclic permutations of them) holds:
*
*$i \geq 7$, in which case $abc|a^7|a^ib^jc^k$
*$i, j, k \geq 1$, in which case $abc|a^ib^jc^k$ since $a|a^i, b|b^j, c|c^k$.
*$k = 0$, $i \geq 3$, in which case $abc|a^3b|a^ib^jc^k$.
*$k = 0$, $i = 2$, $j \geq 2$, in which case $abc|a^3b|a^2b^3|a^ib^jc^k$.
*$k = 0$, $i = 1$, $j \geq 5$, in which case $abc|a^3b|ab^5|a^ib^jc^k$.
Thus, we have $abc|(a+b+c)^7$.
Now, we just need to check if $6$ suffices (given your question, I assume you already have a counterexample to show that $5$ does not).
But notice that, of our options above, 2, 3, 4, and 5 apply to the $i+j+k=6$ case equally, so we know that $abc$ divides all terms of $(a+b+c)^6$, except possibly $a^6$, $b^6$, and $c^6$. We're going to do some symmetry stuff, so we'll start by just considering $a^6$. So we need to know if $bc|a^5$. We know that $bc|a^6$. For any integer $d$ dividing any one of (hence all of) $a, b, c$, say with $x,y,z$ the largest integers such that $p^x|a$,$p^y|b$,$p^z|c$, we have:
$d^x|a|b^2$, so $x \leq 2y$ (and symmetrically, $y \leq 2z$, $z \leq 2x$). If $g = gcd(b,c)$, then $g^2|bc$ and $g^2|c^2|a^4|a^5$, so $bc|a^5$ if $\frac{bc}{g^2}|\frac{a^5}{g^2}$, so with $b^* = \frac{b}{g}$, $c^* = \frac{c}{g}$, $b^*$ and $c^*$ are coprime, and we wish to know if $b^*c^*|\frac{a^5}{g^2}$, so it suffices to show that $b^*|\frac{a^5}{g^2}$ and $c^*|\frac{a^5}{g^2}$. Multiplying everything by $g^2$, we wish to show that $bg|a^5$ and $cg|a^5$. But $g | c$, so $bg|bc|c^3|a^5$, and $cg|c^2|a^4|a^5$, so indeed $bg|a^5$ and $cg|a^5$, hence $bc|a^5$, hence $abc|a^6$, so indeed, $abc|(a+b+c)^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$2^{2n+1} +1$ divisible by $3$ $2^{2n+1} +1$ divisible by $3$
Inducción over n
Let $n=1$, then: $$2^{2(1)+1}+1=2^{3}+1=9$$
It works!
Hip. $$2^{2n+1} +1=3k$$
Then we have to show: $$2^{2n+3} +1=3k$$
Any guess?
| HINT
$$2^{2n+3} +1=2^2 \cdot 2^{2n+1} +1\stackrel{Hyp.}=2^2(3k-1)+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$
If a parabola touches the line $y=x$ and $y=-x$ at $A(3,3)$ and $B(1,-1)$, then find the focus, axis of the parabola and its directrix.
What I thought: Since the 2 tangents are perpendicular,the origin must lie on the directrix and the line joining A and B is a focal chord. Don't know how to proceed from here..
| The most general form of a parabola is $$Ax^2+Bxy + Cy^2 + Dx +Ey +F=0$$
where $$B^2 = 4AC \tag{1}$$.
Letting $(x,y)=(3,3)$, we get $$9A + 9B + 9C + 3D + 3E + F = 0 \tag{2}$$
Letting $(x,y)=(1, -1)$, we get $$A - B + C + D - E + F = 0 \tag{3}$$
\begin{align}
Ax^2+Bxy + Cy^2 + Dx +Ey +F=0
&\implies 2Ax + By + Bxy' + 2Cyy' + D + Ey' = 0 \\
&\implies (Bx + 2Cy + E)y' + (2Ax + By + D) = 0
\end{align}
Letting $(x,y,y') = (3,3,1)$, we get $(3B + 6C + E) + (6A + 3B + D) = 0$,
which implies $$6A + 6B + 6C + D + E = 0 \tag{4}$$
Letting $(x,y,y') = (1,-1,-1)$, we get $-(B - 2C + E) + (2A - B + D) = 0$,
which implies $$2A - 2B + 2C + D - E = 0\tag{5}$$
Solving equations $(1)$ through $(5)$ and letting $A=1$, we get
$$(A,B,C,D,E,F) = (1, -4, 4, -12, 6, 9)$$
So, the equation of the parabola becomes
$$x^2 - 4xy + 4y^2 - 12x + 6y + 9 = 0$$
Added because of something that I found out later.
If $A=0$, then $B^2=4AC$ implies $B=0$ and the equation becomes $Cy^2 + Dx +Ey +F=0$.
If $A \ne 0$, then $A < 0$ implies
$$(-A)x^2 +(-B)xy + (-C)y^2 + (-D)x + (-E)y + (-F) = 0$$
and $B^2=4AC \iff (-B)^2 = 4(-A)(-C)$
So we may as well assume that $A = a^2 > 0$. Then
\begin{align}
Ax^2 + Bxy + Cy^2
&= \dfrac{1}{4A}(4Ax^2 + 4ABxy + B^2y^2) \\
&= \dfrac{1}{4A}(2Ax+By)^2 \\
&= (ax+by)^2 \\
\end{align}
where $b = \dfrac{B}{2a}$.
So we can write the most general form of a parabola as $$(ax + by)^2 + Dx +Ey +F=0$$
We can now argue much as I did above and get the same answer without being annoyed by the nonlinear equation $B^2 = 4AC$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2999165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Number of unique ways to select $3$ balls from $4$ red balls, $2$ green balls and $1$ blue ball Suppose there is a jar with $7$ balls: $4$ red, $2$ green and $1$ blue.
In how many unique ways can we choose $3$ balls?
Just by considering all the possibilities, I found out the answer is $6$. Namely:
*
*red, red, red
*red, red, green
*red, red, blue
*green, green, red
*green, green, blue
*blue, green, red
I was wondering if it is possible to express the solution in terms of the amount of different colors (in this case $3$) and the quantity per color. (in this case $4$, $2$ and $1$)
| Consider the polynomial
$$\eqalign{p(x)&:=(1+x+x^2+x^3+x^4)(1+x+x^2)(1+x)\cr &=(1-x^5)(1-x^3)(1-x^2)(1-x)^{-3}\cr &=(1-x^5)(1-x^3)(1-x^2)\sum_{k\geq0}{2+k\choose k}x^k\ .\cr}\tag{1}$$
For each admissible choice of $r\in[0 .. 4]$ red balls, $g\in[0 .. 2]$ green balls, and $b\in[0 .. 1]$ blue balls the expansion of this polynomial creates a term $x^r\,x^g\,x^b=x^{r+g+b}$. Now we want $r+g+b=3$. It follows that we have to extract $N:={\rm coeff}[x^3]$ on the last line of $(1)$. Since
$$(1-x^5)(1-x^3)(1-x^2)=1-x^2-x^3+{\rm higher\ terms}$$
it follows that
$$N={2+3\choose3}-{2+1\choose1}-{2+0\choose0}=6\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3001464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with a system of inequalities with absolute values I'm trying to solve this system on inequalities
$$
\left\{
\begin{array}{c}
|x-3|<2x \\
|2x+5|>3
\end{array}
\right.
$$
The steps I'm taking are:
Finding the absolute values sings, so for
$x-3 \geq 0$ we have $x \geq 3$ therefore
$$|x-3| =
\left\{
\begin{array}{c}
x-3 & \text{for $x \geq 3$} \\
-x+3 & \text{for $x < 3$}
\end{array}
\right. $$
and
$2x+5 \geq 0$ we have $x \geq \frac{-2}{5}$ therefore
$$|2x+5| =
\left\{
\begin{array}{c}
2x+5 & \text{for $x \geq \frac{-2}{5}$} \\
-2x-5 & \text{for $x<\frac{-2}{5}$} \\
\end{array}
\right. $$
So I build a few systems with the complete inequalities, for the first one we have:
$$
\left\{
\begin{array}{c}
x \geq 3 \\
x-3<2x = x>-3
\end{array}
\right. $$
So the solution here would be $x>3$, then:
$$
\left\{
\begin{array}{c}
x<3 \\
-x+3<2x = x>1
\end{array}
\right. $$
The solution would be $1<x<3$. Then
$$
\left\{
\begin{array}{c}
x \geq \frac{-2}{5} \\
2x+5>3 = x>-1
\end{array}
\right. $$
So the solution of the system is $x>-1$, then
$$
\left\{
\begin{array}{c}
x< \frac{-2}{5} \\
-2x-5>3 = x<-4
\end{array}
\right. $$
And the solution is $x<-4$
Now the solution my book gives is x>1 for the initial system. But I can't find that one. I can't get a solution at all. I tried finding a common point between the 4 solutions I found (as if it was a 4-inequalities system), but there isn't one really. What am I doing wrong?
| Since $0\le\left|x-3\right|$ and $\left|x-3\right|\le2x$ we conclude $2x\ge0$ and then $x\ge0$.
From $\left|x-3\right|<2x$ and since $x\ge0$, we get $-2x< x-3<2x$ and therefore we have $\left\{\begin{array}{c}x-3<2x \\x-3>-2x\end{array}\right.$ and therefore $\left\{\begin{array}{c}x>-3 \\3x>3\end{array}\right.$ and then $\left\{\begin{array}{c}x>-3 \\x>1\end{array}\right.$, so $x>1$ which satisfies the initial statement $x\ge0$.
So far we see $x$ has to be greater than $1$ to satisfy $\left|x-3\right|<2x$.
Let us consider the next inequality $|2x+5|>3$ which leads to $2x+5>3$ or $2x+5<-3$. Then we get $x>-1$ or $x<-4$. And finally if we combine this with the result of first inequality, $x>1$ we get $x>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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What is the limit of $\lim_{n\to \infty} (1 - \frac{1}{4})(1 - \frac{1}{9})(1 - \frac{1}{16}) \cdots (1 - \frac{1}{(n+1)^2})$? What is the evaluation of the following infinite series?
$$\lim_{n\to \infty} \left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right)\left(1 - \frac{1}{16}\right) \cdots \left(1- \frac{1}{(n+1)^2}\right)$$
I've tried to simplify each expression which left me with:
$$\lim_{n\to \infty} \frac{3\times8\times15\times24\times\cdots\times((n+1)^2-1)}{4\times9\times16\times25\times\cdots\times(n+1)^2}$$
Is this a good way to approach this problem?
| $\frac{1\times 3}{2\times 2}\cdot\frac{2\times 4}{3\times 3}\cdot\frac{3\times 5}{4\times 4}\cdot\frac{4\times 6}{5\times 5}\cdots$
Do you see it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
solve $\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$ I need to find
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35}$$
Looking at the graph, I know the answer should be $\frac{20}{17}$, but when I tried solving it, I reached $0$.
Here are the two ways I approached this:
WAY I:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{x^2}(2- \frac{50}{x^2})} {\require{cancel} \cancel{x^2}(2+ \frac{3}{x}-\frac{35}{x^2})}
=\frac{2-2}{\frac {42}{5}}=0
$$
WAY II:
$$\lim_{x\rightarrow -5} \frac{2x^2-50}{2x^2+3x-35} =
\lim_{x\rightarrow -5} \frac{\require{cancel} \cancel{2}(x^2- 25)} {\require{cancel} \cancel{2}(x^2+ \frac{3}{2}x-\frac{35}{2})}
=\lim_{x\rightarrow -5} \frac{{\require{cancel} \cancel{(x-5)}}(x+5)}{{\require{cancel} \cancel{(x-5)}}(x+3.5)}= \frac{-5+5}{-5+3.5}=0
$$
What am I doing wrong here?
Thanks!
| Hint: Try factorization!
$$
\frac{2x^2-50}{2x^2+3x-35}=\frac{2(x^2-25)}{(1/2)(4x^2+6x-70)}=\frac{4(x-5)(x+5)}{(2x+10)(2x-7)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$
Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?
| This is 2nd tough-degree exercise in induction. Observe that both the first and last elements change, so if $\;k=n\;$ we get
$$(n+1)^2+(n+2)^2+\ldots+(2n)^2=\frac{n(2n+1)(7n+1)}6\;\;(**)$$
then for $\;k=n+1\;$ we get
$$(n+1+1)^2+(n+3)^2+\ldots+(2n)^2+(2n+1)^2+\overbrace{(2n+2)^2}^{=(2(n+1))^2}=\frac{(n+1)(2n+3)(7n+8)}6$$
Now, we prove the last one assuming the first one. Developing left side:
$$(n+2)^2+\ldots+(2n)^2+(2n+1)^2+(2n+2)^2=(n+1)^2+\ldots+(2n+2)^2-(n+1)^2=$$
$$=(n+1)^2+\ldots+(2n)^2+(2n+1)^2+(2n+2)^2-(n+1)^2\stackrel{\text{Ind. Hyp.}}=$$
$$=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+4(n+1)^2-(n+1)^2=$$
$$=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2=\frac{(2n+1)\left[n(7n+1)+12n+6\right]+18(n+1)^2}6=$$
$$=\frac{(2n+1)(7n^2+13n+6)+18n^2+36n+18}6=\frac{(n+1)(2n+3)(7n+8)}6$$
The last equality you can prove it opening parentheses and equating coefficients, or showing the left side has as roots $\;-1,\,-3/2,\,-8/7\;$
Alternative way to prove:
$$(n+1)^2+(n+2)^2+\ldots+(n+n)^2=$$
$$=n^2+\ldots+n^2+2n+4n+\ldots+2nn+1+2^2+\ldots+n^2=$$
$$n\cdot n^2+2(1+2+\ldots+n)n+\frac{n(n+1)(2n+1)}6=$$
$$=n^3+n^2(n+1)+\frac{n(n+1)(2n+1)}6=\ldots$$
| {
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"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*\sqrt{1-\frac{3}{x}}}$
5) $\frac{3x}{x-x(\sqrt{1-\frac{3}{x}})}$
6) $\frac{3}{1-(\sqrt{1-\frac{3}{x}})}$
Now I would just take the limit, it would result in $\frac{3}{1-1}$ which would be undefined. For some reason, the $x$ in the denominator of step 5 should turn into $-(-x)$ which in turn would be positive and therefore be $\frac{3}{1+\sqrt{1=\frac{3}{x}}}$ which would equal $\frac{3}{2}$.
I really don't get it. Apparently the $-\infty$ would mean that $\sqrt{x^2}$ = $-x$. We didn't even evaluate the limit yet.. how does that turn into $-x$, just because we know the limit is negative does not mean we evaluated it yet..., why not simplify until there is no more simplification to be done, which is what I did in my steps, which would evaluate to undefined?
Would love some help, thanks!
| $a^2-b^2=(a-b)(a+b).$
$y:=-x$ , and consider $\lim y \rightarrow + \infty.$
$\sqrt{y^2+3y}-y= \dfrac{(y^2+3y)-y^2}{\sqrt{y^2+3y}+y}=$
$\dfrac{3y}{\sqrt{y^2+3y}+y}= \dfrac{3y}{y(\sqrt{1+3/y}+1)}$
$=\dfrac{3}{\sqrt{1+3/y}+1}.$
Take the limit.
| {
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General solution of $\sin 2x+\cos x=0$
Solve the trigonometric equation $\sin 2x+\cos x=0$
My Attempt
$$
2\sin x\cos x+\cos x=0\implies\cos x=0 \text{ or }\sin x=\frac{-1}{2}=\sin\frac{-\pi}{6}\\
x=(2n+1)\frac{\pi}{2} \text{ or }x=n\pi+(-1)^n\frac{-\pi}{6}\\
x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}
$$
Reference
$$
\cos x=-\sin 2x=\cos\Big(\frac{\pi}{2}+2x\Big)\implies x=2n\pi\pm\Big(\frac{\pi}{2}+2x\Big)\\
-x=2n\pi+\frac{\pi}{2}\text{ or }3x=2n\pi-\frac{\pi}{2}\implies x=2m\pi-\frac{\pi}{2}\text{ or }x=\frac{2m\pi}{3}-\frac{\pi}{6}
$$
But my reference gives the solution $x=2n\pi-\dfrac{\pi}{2}$ or $x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}$. I understand how it is reached and both represent the same solutions. But, how do I derive the solution in my reference from what I found in my attempt ? i.e.,
How to derive
$$
\bigg[x=n\pi+\frac{\pi}{2} \text{ or }x=2n\pi-\frac{\pi}{6}\text{ or }x=2n\pi+\pi+\frac{\pi}{6}\bigg]\\
\implies \bigg[x=2n\pi-\dfrac{\pi}{2}\text{ or }x=\dfrac{2n\pi}{3}-\dfrac{\pi}{6}\bigg]
$$
| $$\sin(2x)+\cos(x)=0$$
$$2\sin(x)\cos(x)+\cos(x)=0$$
$$\cos(x)\left(2\sin(x)+1\right)=0$$
so you have two sets of solutions:
$$\cos(x)=0,\,\sin(x)=-\frac{1}{2}$$
EDIT:
firstly they have:
$$x=2n\pi-\pi/2=\pi(2n-1/2)$$
and you have:
$$x=(2n+1)\pi/2$$
let:
$$\pi/2(4n-1)=\pi/2(2m+1)$$
so:
$$4n-1=2m+1$$
$$m=(4n-2)/2=2n-1$$
so for all integer values of $n$, $m$ is also an integer and so they are equivalent?
| {
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Prove for all $a, b, c \in \mathbb {R^+}$ ${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$ is true
Prove for all $a, b, c \in \mathbb {R^+}$ $${a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}} \gt \sqrt{a+b+c}$$ is true
I have been studying about inequalities and how to prove different inequalities, and I stumbled over this question which I failed to answer. I failed to see a connection between $\sqrt{a+b+c} \text{ and} {a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}}$. Is there any way I can prove this question and if there is can you please help. Thank you.
| You may proceed as follows using convexity of $\frac{1}{\sqrt{x}}$:
*
*$\sum_{k=1}^3 \lambda_k f(x_k) \geq f\left( \sum_{k=1}^3 \lambda_k x_k\right)$ with $\sum_{k=1}^3 \lambda_k = 1$, $\lambda_1,\lambda_2,\lambda_3 \geq 0$
$$
\begin{eqnarray*} {a\over\sqrt{a+b}} + {b\over\sqrt{b+c}} + {c\over\sqrt{c+a}}
& = & (a+b+c)\sum_{cyc}\frac{a}{(a+b+c)(a+b)} \\
& \stackrel{Jensen}{\geq} & (a+b+c)\frac{1}{\sqrt{\frac{1}{a+b+c}\sum_{cyc}a(a+b)}} \\
& \color{blue}{>} & (a+b+c)\frac{1}{\sqrt{\frac{1}{a+b+c}\sum_{cyc}a(a+b\color{blue}{+c})}} \\
& = & (a+b+c)\frac{1}{\sqrt{\frac{1}{a+b+c}(a+b+c)^2}} \\
& = & \color{blue}{\sqrt{a+b+c}}
\end{eqnarray*}
$$
| {
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Values of $a$ for which $p(x)$ has a complex roots? Let $p(x) = x^4 + ax^3 + bx^2 + ax + 1$. Suppose $x = 1$ is a root of $p(x)$, then find the range of values of $a$ such that $p(x)$ has complex (non-real) roots?
My approach: Using the fact that $1$ is a root, I am able to deduce $b = -2(a+1)$. Then I wrote $$p(x) = (x-1)(x-z)(x-\bar{z})(x-k)$$ where $z = c + id, d\neq 0$ and $k$ is some other real roots. Now I plan to expand and compare the coefficients but have been unable to do so in a fruitful manner (there just too many variables to get to a proper estimate of range of $a$ )
| When a polynomial is "palindromic", reading the same way forwards and backwards, it must also have palindromic factors. Thus
$x^4+ax^3+bx^2+ax+1=(x^2+cx+1)(x^2+dx+1)$
If $x=1$ is to be a root then it must give $c=-2$ or $d=-2$ in one of the factors, as no other choice has the root $x=1$. Taking $c$ as $-2$ we then get the complex roots if $-2<d<2$, corresponding to a negative discriminate for the $x^2+dx+1$ factor.
Expanding the product $(x^2+cx+1)(x^2+dx+1)$ and equating this to $x^4+ax^3+bx^2+ax+1$ then implies $a=c+d=d-2$ so the solution to the problem is $-4<a<0$.
| {
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Little inequality After deconditioning and calculating some stuff I obtained this inequality, which looks simple but I can't get to group the termens conveniently.
$x^3 + y^3 + z^3 + 6(x^2z + y^2x + z^2y) ≥ 12xyz + 3(x^2y + y^2z+z^2x)$ for $x,y,z≥0$.
I tried breaking it in more inequalities , but the last one is not always true like this:
$4(x^2z + y^2x + z^2y)≥12xyz$ from MA - MG
$x^3 + y^3 + z^3≥x^2y + y^2z+z^2x$
and this remains:
$x^2z + y^2x + z^2y≥x^2y + y^2z+z^2x$ which doesn't always hold? or does it?
| Buffalo Way helps here very well.
Indeed, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus,
$$\sum_{cyc}(x^3+6x^2z-3x^2y-4xyz)=6(u^2-uv+v^2)x+u^3-3u^2v+6uv^2+v^3\geq0.$$
It's interesting that your first step gives a right inequality because after using AM-GM
$$\sum_{cyc}4x^2z\geq12xyz$$ it's enough to prove that
$$\sum_{cyc}(x^3+2x^2z-3x^2y)\geq0,$$ which we can prove by BW again:
Let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus,
$$\sum_{cyc}(x^3+2x^2z-3x^2y)=2(u^2-uv+v^2)x+u^3-3u^2v+2uv^2+v^3\geq0$$
and we are done!
| {
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Inequality for $a,c,b$ positive real numbers with $a + b + c = 1$ Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{1}{2a^2+bc}+\frac{1}{2b^2+ac}+\frac{1}{2c^2+ab}\geq \frac{1}{\sqrt{a^2-ab+b^2}}+\frac{1}{\sqrt{b^2-bc+c^2}}+\frac{1}{\sqrt{c^2-ca+a^2}}$$
I have a solution
$$
\left(2 a^{2}+b c\right)\left(2 b^{2}+c a\right)-(a+b+c)^{2}\left(a^{2}-a b+b^{2}\right)=-(a-b)^{2}\left(a^{2}+b^{2}+c^{2}+3 b c\right) \leq 0
$$
From here on ($a+b+c=1$) it should be,
$$
\left(2 a^{2}+b c\right)\left(2 b^{2}+c a\right) \leq a^{2}-a b+b^{2}
$$
Applying AM-GM, we have
$$
\begin{aligned}
\frac{1}{2 a^{2}+b c}+\frac{1}{2 b^{2}+c a} & \geq \frac{2}{\sqrt{a^{2}-a b+b^{2}}} \\
\frac{1}{2 c^{2}+a b}+\frac{1}{2 b^{2}+c a} & \geq \frac{2}{\sqrt{b^{2}-b c+c^{2}}} \\
\frac{1}{2 a^{2}+b c}+\frac{1}{2 c^{2}+a b} & \geq \frac{2}{\sqrt{c^{2}-c a+a^{2}}}
\end{aligned}
$$
I need another solution. Thanks all
| Another way, but in one line :)
$$\sum_{cyc}\frac{1}{2a^2+bc}-\sum_{cyc}\frac{1}{\sqrt{a^2-ab+b^2}}=$$
$$=\sum_{cyc}\left(\frac{a+b+c}{2a^2+bc}-\frac{2}{b+c}\right)+\sum_{cyc}\left(\frac{2}{a+b}-\frac{1}{\sqrt{a^2-ab+b^2}}\right)=$$
$$=\sum_{cyc}\frac{b^2+c^2+ab+ac-4a^2}{(2a^2+bc)(b+c)}+\sum_{cyc}\frac{2\sqrt{a^2-ab+b^2}-a-b}{(a+b)\sqrt{a^2-ab+b^2}}=$$
$$=\sum_{cyc}\tfrac{(c-a)(2a+c)-(a-b)(2a+b)}{(2a^2+bc)(b+c)}+\sum_{cyc}\tfrac{3(a-b)^2}{(a+b)\sqrt{a^2-ab+b^2}(2\sqrt{a^2-ab+b^2}+a+b)}=$$
$$=\sum_{cyc}(a-b)\left(\tfrac{2b+a}{(2b^2+ac)(a+c)}-\tfrac{2a+b}{(2a^2+bc)(b+c)}\right)+\sum_{cyc}\tfrac{3(a-b)^2}{2(a+b)(a^2-ab+b^2)+(a+b)^2\sqrt{a^2-ab+b^2}}=$$
$$=\sum_{cyc}\tfrac{(a-b)^2(2a^2b+2ab^2+3abc-2c^2a-2c^2b)}{(a+c)(b+c)(2b^2+ac)(2c^2+ab)}+\sum_{cyc}\tfrac{3(a-b)^2}{2(a+b)(a^2-ab+b^2)+(a+b)^2\sqrt{a^2-ab+b^2}}\geq$$
$$\geq\sum_{cyc}\tfrac{(a-b)^2(abc-2c^2a-2c^2b)}{(a+c)(b+c)(2b^2+ac)(2c^2+ab)}+\sum_{cyc}\tfrac{3(a-b)^2}{2(a+b)(a^2-ab+b^2)+(a+b)^3}=$$
$$=\sum_{cyc}\tfrac{(a-b)^2(abc-2c^2a-2c^2b)}{(a+c)(b+c)(2b^2+ac)(2c^2+ab)}+\sum_{cyc}\tfrac{(a-b)^2}{(a+b)(a^2+b^2)}=$$
$$=\sum_{cyc}\tfrac{(a-b)^2((abc-2c^2a-2c^2b)+(a+c)(b+c)(2a^2+bc)(2b^2+ac))}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}\geq$$
$$\geq\sum_{cyc}\tfrac{(a-b)^2((abc-2c^2a-2c^2b)+c(a+b+c)(2a^2+bc)(2b^2+ac))}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}=$$
$$=\sum_{cyc}\tfrac{(a-b)^2c(ab(a+b)(a^2+4ab+b^2)-2abc(a^2+b^2)+c^2(a+b)(2a^2-ab+2b^2)+abc^3)}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}\geq$$
$$\geq\sum_{cyc}\tfrac{(a-b)^2c(ab(a+b)(a^2+b^2)-2abc(a^2+b^2)+(a+b)(a^2+b^2)c^2)}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}=$$
$$=\sum_{cyc}\tfrac{(a-b)^2c(a^2+b^2)(ab(a+b)-2abc+(a+b)c^2)}{(a+c)(b+c)(a+b)(a^2+b^2)(2b^2+ac)(2c^2+ab)}\geq0.$$
| {
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Series convergence: $\sin (n \frac{\pi}{2})$
Determine whether the following series :
$$\sum_{n=1}^\infty \sin \left(\frac{n\pi}{2}\right) \frac{n^2+2}{n^3 +n}$$
converges absolutely, conditionally or diverges.
I know that for even natural numbers the expression will equal zero and that for odd values of $n$ the value of $\sin$ will go from $1$ to $-1$.
Could I theoretically reduce this series into a subseries:
$$\sum_{n=0}^\infty \sin \left(\frac{(2n+1)\pi}{2}\right) \frac{(2n+1)^2+2}{(2n+1)^3 +2n + 1}$$
And then treat it as if it were a standard alternating series?
| HINT
*
*Note that
$$
\frac{n^2+2}{n^3+n}
= \frac{1}{n} \times \frac{n^2+2}{n^2+1}
= \frac{1}{n} \left[ 1 + \frac{1}{n^2+1} \right]
= \Theta(1/n)
$$
Does this series converge?
*Adding $\sin(n\pi/2)$ in the front effectively kills all the even-$n$ terms and makes an alternating series out of the odd ones -- does the alternating series converge?
| {
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$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\frac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$?
$\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ true for all $x$ ?
This expression is found without mentioning the domain of $x,y,z$, but I don't think its true for all $x,y,z$ as the case with the expression for $\tan^{-1}x+\tan^{-1}y$, but I have trouble proving it.
So what is the complete expression for $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z$ ?
\begin{align}
\tan^{-1}x+\tan^{-1}y+\tan^{-1}z&=
\begin{cases}\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy < 1 \\[1.5ex]
\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y>0 \\[1.5ex]
-\pi + \tan^{-1}\left(\dfrac{x+y}{1-xy}\right)+\tan^{-1}z, &xy>1,\:\:x,y<0 \end{cases}\\
&=
\end{align}
| We have that by addition formula
$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$$
then
$$(\tan^{-1}x+\tan^{-1}y)+\tan^{-1}z=\tan^{-1}\frac{\frac{x+y}{1-xy}+z}{1-\frac{(x+y)z}{1-xy}}$$
the simplify to the given identity, which could be not well defined for (to check)
*
*$1-xy=0$
and is certainly not well defined for
*
*$1-xy-yz-zx=0$
| {
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For all $n>1$ there are positive $a+b=n$ such that $a+ab+b\in\mathbb P$ For all integers $n>1$ there are positive integers $a,b$ such that $a+b=n$ and such that $a+ab+b\in\mathbb P$.
Tested for all $n\leq 1,000,000$. Hopefully, someone can explore and explain the heuristics about this conjecture.
This question is related to:
Any odd number is of form $a+b$ where $a^2+b^2$ is prime
Does every power of two arise as the difference of two primes?
Most even numbers is a sum $a+b+c+d$ where $a^2+b^2+c^2=d^2$
Natural numbers large enough can be written as $ab+ac+bc$ for some $a,b,c>0$
$\{a+b|a,b\in\mathbb N^+\wedge ma^2+nb^2\in\mathbb P\}=\{k>1|\gcd(k,m+n)=1\}$
Even numbers has the form $a+b$ where $\frac{a^2+b^2}{2}$ is prime
Is every positive integer greater than $2$ the sum of a prime and two squares?
It's about a relation $R\subseteq \mathbb N^m$, a function
$f:\mathbb N^m\to \mathbb N$, and an image of a restriction
$\operatorname{Im}(f|R)$.
In Goldbachs conjecture the relation is $p,q\in\mathbb P$, the function is $(p,q)\mapsto p+q$ and the image of the restriction is
$2\mathbb N\setminus\{2\}$.
Maybe some of the conjectures can be generalized?
| This is only a partial answer. Reformulating according to the comment of Crostul, we want to find $AB-1 \in \mathbb P$. Other than the even prime $2$, all primes are odd, so at least one of $A,B$ must be even. For $A+B=N,\ N\ge 6$, any prime generated will have the form $6m\pm 1$. So it is necessary (but not sufficient) that for every $N$, there are some $A,B$ such that $AB-1=6m\pm 1$ for the conjecture to be true.
$AB-1=6m\pm 1\Rightarrow AB\equiv (0,2) \mod{6}$. Any $N\ge 6$ can be split into two addends with that property. The following table lists values of residues $\mod{6}$ for $N,A,B$ that satisfy $N=A+B\mod{6}$ and $AB\equiv (0,2) \mod{6}$ (up to the order of $A,B$).
$$\begin{array}{ccc}
\ N&A&B \\
0&0&0 \\
&2&4 \\ \\
1&0&1 \\
&3&4 \\ \\
2&0&2 \\
3&0&3 \\
&1&2 \\
&4&5 \\ \\
4&0&4 \\
5&0&5 \\
&2&3 \\
\end{array}$$
This shows that any $N$ can be split into addends $A,B$ such that $AB\equiv (0,2) \mod{6}$. In every case, it is possible to obtain addends such that $AB\equiv 0 \mod{6}$. Interestingly, only for $N\equiv (0,3)\mod6 \Rightarrow N\equiv 0 \mod3$ is it possible to obtain addends such that $AB\equiv 2 \mod{6}$. This means that $AB-1$ can generate numbers of the form $6m-1$ from any $N$, but can generate numbers of the form $6m+1$ only if $N\equiv 0\mod3$. It remains open at this point whether the numbers of the form $6m\pm 1$ obtained from a particular $N$ will necessarily feature a prime.
| {
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sum of $\sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$
Finding sum of $\displaystyle \sum^{n}_{k=0}(-1)^k\cdot \frac{\binom{n}{k}}{\binom{k+3}{k}}$
Try: Using
$$\int^{1}_{0}x^m\cdot (1-x)^ndx = \frac{m!\cdot n!}{(m+n+1)!}=\frac{1}{(m+n+1)\binom{m+n}{n}}.$$
So $$ \frac{1}{(k+4)\binom{k+3}{k}}=\int^{1}x^3\cdot (1-x)^kdx$$
So our sum is $$\sum^{n}_{k=0}(-1)^k(k+4)\binom{n}{k}\int^{1}_{0}x^3(1-x)^kdx$$
$$ = \int^{1}_{0}x^3\sum^{n}_{k=0}(-1)^kk\binom{n}{k}(1-x)^kdx+4\int^{1}_{0}x^3\sum^{n}_{k=0}(-1)^k\binom{n}{k}(1-x)^k$$
$$ = -n\int^{1}_{0}x^{n+2}(1-x)dx+4\int^{1}_{0}x^{n+3}dx$$
$$ = -n\bigg[\frac{1}{n+3}-\frac{1}{n+4}\bigg]+\frac{4}{n+4} = -\frac{n}{(n+3)(n+4)}+\frac{4}{n+4} = $$
but answer given in book as $\displaystyle \frac{3}{n+3}$
Could some help me how to solve it, Thanks
| I think it is simpler to utilize binomial identities:
$$\dfrac{\binom nk}{\binom{k+3}k}=\dfrac{n!3!}{(n-k)!(k+3)!}=\dfrac6{(n+1)(n+2)(n+3)}\binom{n+3}{k+3}$$
$$\sum_{k=0}^n\dfrac{\binom nk}{\binom{k+3}k}=\dfrac6{(-1)^3(n+1)(n+2)(n+3)}\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$
$$\sum_{k=0}^n\binom{n+3}{k+3}(-1)^{k+3}$$
$$=\sum_{r=0}^{n+3}\binom{n+3}r(-1)^r-\sum_{r=0}^2\binom{n+3}r(-1)^r$$
$$=(1-1)^n-\sum_{k=0}^2\binom{n+3}k(-1)^k=?$$
| {
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$\text{SL}_2(\mathbb Z)$ acts on upper plane $\mathbb H$. What kind of points have non-trivial stabilizer? And how many orbits are there? $\text{SL}_2(\mathbb Z)$ acts on upper plane $\mathbb H= \{z \in \mathbb{C} | \Im(z) > 0 \}$ via Mobius transformation. $$
\text{ For } \gamma =\begin{bmatrix}
a &b \\c&d \end{bmatrix} \in\text{SL}_2(\mathbb Z), \ \gamma z =\begin{bmatrix}
a &b \\c&d \end{bmatrix}\cdot z = \frac{az +b}{cz+d} $$
Stabilizer of $z$ means set $\{\gamma \in \text{SL}_2(\Bbb Z), \gamma z=z\}$.
I want to know what kind of points have non-trivial stabilizer and the number of orbits.
My effort:
For $z \in \Bbb H$, suppose $z = x + i y$.
$\text{For } \gamma =\begin{bmatrix} a &b \\c&d \end{bmatrix}, \gamma z =\frac{az+b}{cz+d}=z\iff az+b=cz^2+dz$
$$\iff ax+ayi+b = cx^2-cy^2+2cxyi + dx +dyi$$
$$\iff ay=2cxy+dy\ \&\ ax+b=cx^2-cy^2+dx$$
$$\iff a=2cx+d \ \&\ ax+b=cx^2-cy^2+dx$$
$$\implies b=-c(x^2+y^2),\ \gamma =\begin{bmatrix} 2cx+d &-c(x^2+y^2) \\c&d \end{bmatrix}.$$
$$\gamma \in \text{SL}_2(\Bbb Z),\ (2cx+d)\times d-(-c(x^2+y^2))\times c=1 $$
$$\implies (cx+d)^2+(cy)^2=1.$$
Then how to proceed? Thanks in advance.
I haven't learnt modular form yet, and I don't know if these help:
Good description of orbits of upper half plane under $SL_2 (Z)$
Orbit of complex unit $i$ under moebius tranformation in $SL_2(\mathbb{Z})$
Edit:
GTM$105$, Serge Lang, SL$_2(\mathbb R)$ might help.
Comment:
It's an exercise of section about group action on set, and before this section the book just introduces definition and basic property of group, so this problem is a bit more difficult than I thought.
| At some point you have $b=-c(x^2+y^2)$ and $a=2cx+d$. If $c=0$, then $b=0$, in which case the element $\begin{pmatrix} a&b\\c&d\end{pmatrix}$ of the projective special linear group $\operatorname{PSL}_2(\Bbb Z)$ of this form is the identity element (recalling that $\operatorname{PSL}_2(\Bbb Z)=\operatorname{SL}_2(\Bbb{Z})/\{\pm I\}$), so we are not interested in this case. So, $c$ is assumed to be non-zero, and can without loss of generality assumed to be positive.
That is, $x=\frac{a-d}{2c}$, so that $$-\frac{b}{c}=x^2+y^2=\frac{(a-d)^2}{4c^2}+y^2.$$
Since $y>0$, we have
$$y=\frac{\sqrt{4-(a+d)^2}}{2c},$$ which means $d=-a$ or $d=-a\pm1$.
In the case $d=-a$, we must have $x=\frac{a}{c}$ and $y=\frac{1}{c}$.
For a given point $z=x+yi$ of this form that is fixed by a non-trivial $\gamma\in\operatorname{PSL}_2(\Bbb{Z})$, we can assume that $$\gamma=\begin{pmatrix}a&b\\c&-a\end{pmatrix}\wedge bc=-(a^2+1).\tag{1}$$ That is, $\gamma^2$ is the identity of $\operatorname{PSL}_2(\Bbb{Z})$. The points in $\Bbb H$ with a stabilizer of order $2$ are of the form $z=x+yi$, where
$$x=\frac{a}{c}\wedge y=\frac{1}{2c}$$
for some integer $a$ and for some integer $c>0$, such that $a^2+1$ is divisible by $c$ (so that there exists $b\in \Bbb Z$ such that $bc=-(a^2+1)$). The number of orbits for a given $a$ is $\sigma_0(a^2+1)$, where $\sigma_0$ is the divisor counting function.
For the case $d=-a+1$, we see that $x=\frac{2a-1}{2c}$ and $y=\frac{\sqrt{3}}{2c}$. Then, $z=x+yi$ is fixed by
$$\gamma=\begin{pmatrix}a&b\\c&-a+1\end{pmatrix}\wedge bc=-(a^2-a+1).\tag{2}$$
Note that $\gamma^3$ is the identity of $\operatorname{PSL}_2(\Bbb{Z})$. Thus, for a given $a$, there are corresponding $\sigma_0(a^2-a+1)$ points $z\in \Bbb H$.
For the case $d=-a-1$, we see that $x=\frac{2a+1}{2c}$ and $y=\frac{\sqrt{3}}{2c}$. Then, $z=x+yi$ is fixed by
$$\gamma=\begin{pmatrix}a&b\\c&-a-1\end{pmatrix}\wedge bc=-(a^2+a+1).\tag{3}$$
Note that $\gamma^3$ is the identity of $\operatorname{PSL}_2(\Bbb{Z})$. Thus, for a given $a$, there are corresponding $\sigma_0(a^2+a+1)$ points $z\in \Bbb H$. (It can be seen that this case is identical to the previous case via the transformation $a\mapsto a-1$.)
For example, $x=\frac{3}{5}$ and $y=\frac{1}{5}$ fit the bill (with $a=3$, $b=-2$, $c=5$, and $d=-a=-3$) for a point with stabilizer of order $2$, with $$\gamma=\begin{pmatrix}3&-2\\5&-3\end{pmatrix}.$$
For the same $a=3$, there are three more points with stablizers of order $2$, i.e., with $c=1$, $c=2$, and $c=10$. That is, for $a=3$, we have in total four points with non-trivial stabilizers $\gamma$ of order $2$: $z=3+i$ with $\gamma=\begin{pmatrix}3&-10\\1&-3\end{pmatrix}$, $z=\frac{3}{2}+\frac{i}{2}$ with $\gamma=\begin{pmatrix}3&-5\\2&-3\end{pmatrix}$, $z=\frac{3}{5}+\frac{i}{5}$ with $\gamma=\begin{pmatrix}3&-2\\5&-3\end{pmatrix}$, and $z=\frac{3}{10}+\frac{i}{10}$ with $\gamma=\begin{pmatrix}3&-1\\10&-3\end{pmatrix}$.
For the same $a=3$, there are also four points with stabilizers of order $3$. That is, for $a=3$, we have four points with non-trivial stabilizers $\gamma$ of order $3$: $z=\frac52+\frac{\sqrt{3}i}{2}$ with $\gamma=\begin{pmatrix}3&-7\\1&-2\end{pmatrix}$, $z=\frac{5}{14}+\frac{\sqrt{13}i}{14}$ with $\gamma=\begin{pmatrix}3&-1\\7&-2\end{pmatrix}$, $z=\frac{7}{2}+\frac{\sqrt{3}i}{2}$ with $\gamma=\begin{pmatrix}3&-13\\1&-4\end{pmatrix}$, and $z=\frac{7}{26}+\frac{\sqrt{3}i}{26}$ with $\gamma=\begin{pmatrix}3&-1\\13&-4\end{pmatrix}$.
In conclusion, there are three kinds of points $z\in\Bbb H$---those with trivial stabilizers, those with stabilizers of order $2$, and those with stabilizers of order $3$. The stablizers in non-trivial cases are generated by $\gamma$ given in (1) and (2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
obtain matrix $ A$ if $X$ and $b$ given
for the right and detailed answer refer to user9077 answer
| Since $\begin{pmatrix}1\\0\end{pmatrix}+c\begin{pmatrix}0\\1\end{pmatrix}$ are all solutions to $Ax=\begin{pmatrix}1\\3\end{pmatrix}$, then we have
$$
\begin{pmatrix}1\\3\end{pmatrix}=Ax=A\begin{pmatrix}1\\0\end{pmatrix}+cA\begin{pmatrix}0\\1\end{pmatrix}.
$$
The above is true for any $c$. This can only happen when $A\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$. From this we know that the second column of $A$ is $\begin{pmatrix}0\\0\end{pmatrix}$.
We also have $A\begin{pmatrix}1\\0\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}$. So the first column of $A$ is $\begin{pmatrix}1\\3\end{pmatrix}$. Therefore
$$A=\begin{pmatrix}1&0\\3&0\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3035528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The distributive law $4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
|
$4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$
This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.
but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
The distributive law works even if $x$ and/or $y$ are not integers.
Keep one of the factors together in a first step, and apply distributivity twice:
$$\begin{align}
(\color{blue}{x}+\color{red}{y})(x+y) & =\color{blue}{x}(x+y)+\color{red}{y}(x+y) \\ & =\color{blue}{x}x+\color{blue}{x}y+\color{red}{y}x+\color{red}{y}y
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving $\lim_{n\to\infty}\sqrt{n^2+n}-n=\frac{1}{2}$
Prove using the definition of a limit that
$$\displaystyle{\lim_{n\to\infty}\underbrace{\sqrt{n^2+n}-n}_{s_n}}=\frac{1}{2}.$$
Proof:
Let $\epsilon>0$ and $n\in\mathbb{N}$, then
$$\left|\sqrt{n^2+n}-n-\frac12\right|\stackrel{(*)}{=}\left|\frac{n-\sqrt{n^2+n}}{2(n+\sqrt{n^2+n})}\right|=\left|\frac{n}{2(n+\sqrt{n^2+n})^2}\right|=\left|\frac{n}{4n^2+4n\sqrt{n^2+n}+2n}\right|<\left|\frac{n}{4n\sqrt{n^2+n}+2n}\right|=\left|\frac{1}{4\sqrt{n^2+n}+2}\right|$$
Since $\forall n\in\mathbb{N}\,\,\sqrt{n^2+n}>\sqrt{n}\,$ then
$$\left|\frac{1}{4\sqrt{n^2+n}+2}\right|<\left|\frac{1}{4\sqrt{n}}\right|<\epsilon\quad\text{ if }\quad n>\frac{1}{16\epsilon^2}.$$
Would this be correct?
(*) maybe a simpler way would be to recognize that the numerator is $|s_n|<1?$
| We know that $$\sqrt{n^2+n}-n{=(\sqrt{n^2+n}-n){\sqrt{n^2+n}+n\over \sqrt{n^2+n}+n}\\={n\over \sqrt{n^2+n}+n}\\={1\over 1+\sqrt{1+{1\over n}}}}$$therefore$$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|{={1\over 2}-{1\over 1+\sqrt{1+{1\over n}}}<\epsilon}$$which means that $${ 1+\sqrt{1+{1\over n}}}<{1\over {1\over 2}-\epsilon}\\\sqrt{1+{1\over n}}<{{1\over 2}+\epsilon\over {1\over 2}-\epsilon}\\{1+{1\over n}}<{{1\over 4}+\epsilon^2+\epsilon\over {1\over 4}+\epsilon^2-\epsilon}\\{1\over n}<{2\epsilon\over {1\over 4}+\epsilon^2-\epsilon}$$therefore by choosing $$n>{{1\over 4}+\epsilon^2-\epsilon\over 2\epsilon}$$or even $n>{1\over 8\epsilon}$ for small enough $\epsilon$ we obtain $$\left|{1\over 1+\sqrt{1+{1\over n}}}-{1\over 2}\right|<\epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3040207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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