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Integration of $\sqrt{1-x^2}$ by parts. Interpretation of 2nd term. I was wondering about the anti-derivative of $\sqrt{1-x^2}$
Method for solving it is given here. This solution is fine, but I was bit confused about last term in first line i.e.
$$\int \left(\frac{d}{dx}\sqrt{1-x^2}\int dx\right)dx$$
$$=\int x\cdot\frac{d}{dx}\sqrt{1-x^2}\cdot\ dx$$
Can't it be written as
$$\int x\cdot\ d\sqrt{1-x^2}$$
i.e. cancelling out both $dx$
Now if we substitute $1-x^2 = t^2$ then the integral would become
$$\int \sqrt{1-t^2}\cdot\ dt$$
Which is integral $I$ itself
This implies
$$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{d}{dx}\sqrt{1-x^2}\int dx\right)dx$$
$$=x \sqrt{1-x^2}-I$$
Or
$$I=\frac{x}{2}\sqrt{1-x^2}+C$$
I know there is something wrong. Might be cancelling $dx$ was the wrong step. But same result is obtained by this alternate method:
$$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{d}{dx}\sqrt{1-x^2}\int dx\right)dx$$
$$=x\sqrt{1-x^2}+\int\frac{x^2}{\sqrt{1-x^2}}dx$$
Now in second term again if we put $1-x^2=t^2$ which gives
$$dx=\frac{t}{-\sqrt{1-t^2}}dt$$
And so
$$I=x\sqrt{1-x^2}+\int\frac{1-t^2}{t}\cdot\frac{t}{-\sqrt{1-t^2}}dt$$
$$=x\sqrt{1-x^2}-\int \sqrt{1-t^2}\cdot\ dt$$
$$=x\sqrt{1-x^2}-I$$
Which again give same result
$$I=\frac{x}{2}\sqrt{1-x^2}+C$$
What step am I doing incorrectly? Is there an algebraic mistake? Please help. Thanks in advance. :)
| \begin{align}
I & = \displaystyle\int \sqrt{1-x^2} \\
& = x\sqrt{1-x^2}+\displaystyle \int \dfrac{x^2}{\sqrt{1-x^2}} \\
& = x\sqrt{1-x^2}+\displaystyle \int \dfrac{x^2+1-1}{\sqrt{1-x^2}} \\
& = x\sqrt{1-x^2}+\arcsin\left(x\right)-I \\
& =\dfrac{x\sqrt{1-x^2}}{2}+\dfrac{\arcsin\left(x\right)}{2}+C
\end{align}
EDIT:
in your first method you were cancelling dx which is wrong and in your so called alternative method you assumed $\sqrt{1-t^2}=I$ which is wrong because you are working with indefinite integral here not definite
so,
you can't say $\displaystyle\int {f(x)} dx=\displaystyle\int {f(t)} dt$ where $t$ is some transformation of $x$
But yes in case it was definite integral then above will stand true because changing variable i.e, from $x$ to $t$ will also change the bounds (aka limits ) in a way that value of integral(area under curve) remain the same and indefinite integrals lacks those bounds.so both integrals (integrals before change of variable and that after) wouldn't be equal
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $F_n$ is the Fibonacci sequence, show that $F_n < \left(\frac 74\right)^n$ for $n\geq 1.$ Recall that the Fibonacci sequence is defined by $F_0 = 0$, $F_1 = 1$, and $F_n = F_{n−1} + F_{n−2}$ for $n ≥ 2$. Prove that:
$$\forall \,\, n ≥ 1 ,\,\, F_n < \left(\frac 74\right)^n$$
In this question I understand how to do the basis step.
In the induction step I know that you have to assume that n=k but I am having trouble figuring out on how to do that. Could someone please explain how to do this question.
| The proposition that you're trying to prove is that $F_n<(\frac{7}{4})^n$
For $n = 0$, this is trivial; $0 < (\frac{7}{4})^0$
For $n = 1$, we have $1 < (\frac{7}{4})^1$
For your induction step, you assume that for all k < n, $F_k<(\frac{7}{4})^k$
So $F_{n-2}<(\frac{7}{4})^{n-2}$ and $F_{n-1}<(\frac{7}{4})^{n-1}$
$F_{n} = $
$F_{n-2}+ F_{n-1}$ <
$(\frac{7}{4})^{n-2} + (\frac{7}{4})^{n-1}$ =
$(\frac{7}{4})^{n-2} + \frac 7 4 (\frac{7}{4})^{n-2} = $
$\frac 4 4(\frac{7}{4})^{n-2} + \frac 7 4 (\frac{7}{4})^{n-2} $=
$\frac {11} 4 (\frac{7}{4})^{n-2}$=
$\frac {44} {16} (\frac{7}{4})^{n-2}$<
$\frac {49} {16} (\frac{7}{4})^{n-2}$=
$ (\frac{7}{4})^{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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I need the steps to solving this limit without using l´Hopital rule I've tried many ways of solving this limit without using l'Hopital and I just can't figure it out. I know the answer is $3/2 \sin (2a).$
$$\lim_{x\,\to\,0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x$$
Thank you!
| Perhaps not the most elegant, but:
$\lim_\limits{x\to0} \frac{\sin(a+x)\sin(a+2x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a + 3x) + cos x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \sin^2(a)} x\\
\lim_\limits{x\to 0} \frac{\frac 12 (-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x) - \frac 12 (1-\cos 2a)} x\\
\lim_\limits{x\to 0} \frac{-\cos(2a)\cos 3x + \sin(2a)\sin (3x) + \cos x - 1 +\cos 2a} {2x}\\
\lim_\limits{x\to 0} \frac{\cos(2a)(1-\cos 3x)}{2x} - \frac {1-\cos x}{2x} + \frac { \sin(2a)\sin (3x)}{2x}$
The first term evaluates to $0,$ the second term evaluates to $0,$ the third term evaluates to $\frac 32 \sin 2a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix}
1 && 7 && -2 \\
0 && 3 && -1 \\
0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity.
What I have so far:
$\det(A - \lambda I) = \det\begin{bmatrix}
1-\lambda && 7 && -2 \\
0 && 3-\lambda && -1 \\
0 && 0 && 2-\lambda \end{bmatrix}$
I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me
$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).
Following that I move on to the geometric multiplicity:
$ A - 3I = \begin{bmatrix}
-2 && 7 && -2 \\
0 && 0 && -1 \\
0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && -\frac{7}{2} && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
\frac{7}{2}\\
1 \\
0\end{bmatrix} s$ which has a geometric multiplicity of 1.
$ A - 2I = \begin{bmatrix}
-1 && 7 && -2 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
1 && 0 && -5 \\
0 && 1 && -1 \\
0 && 0 && 0 \end{bmatrix}$ which yields
$\begin{bmatrix}
x \\
y \\
z \end{bmatrix} = \begin{bmatrix}
5\\
1 \\
1\end{bmatrix} s$ which has a geometric multiplicity of 1.
Finally,
$ A - I = \begin{bmatrix}
0 && 7 && -2 \\
0 && 2 && -1 \\
0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix}
0 && 1 && 0 \\
0 && 0 && 1 \\
0 && 0 && 0 \end{bmatrix}$
This is where I am stuck, I'm not sure what is the resulting $\begin{bmatrix}
x \\
y \\
z \end{bmatrix}$ that column of 0's is confusing me.
Any help would be appreciated.
Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix}
1 \\
0 \\
0 \end{bmatrix}$ ? or is that wrong?
| That RRE $\begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}$ means $y=0$ and $z=0$, so $$\begin{bmatrix} x\\y\\z \end{bmatrix} =\begin{bmatrix} x\\0\\0 \end{bmatrix} =x\begin{bmatrix} 1\\0\\0 \end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Number of $6$-digit numbers made up of the digits $1$, $2$, and $3$ with no digit occurring $3$ or more times consecutively? Find the number of 6-digit numbers made up of the digits $1$, $2$, and $3$ that have no digit occur three or more times consecutively. (For example, $123123$ would count, but $123111$ would not.)
We know the total number of possibilities without the occurrence restriction is $3^6 = 729$.
In order to eliminate the possibilities that can't be used, I set a block of 3 numbers as my 3 consecutive digits block. This block would have 3 different combinations. I then split the number of eliminations into 2 cases: one where the block of 3 is in the front, and one where the block of 3 is in the middle.
For case 1: I computed $3\cdot 24$, with $3$ being the ways to change the block of 3, and $24$ being the number of ways to change the other 3 numbers while not having another block of 3 consecutive.
For case 2: I computed $3^4$, ignoring the restriction of case 1.
Summing the 2 cases, I got $72+ 3\cdot 81 = 315$, I then subtracted from $729$ and got $414$, which turns out to be incorrect. What am I doing wrong?
| There are $3^6 = 729$ possible sequences. From these, we must subtract those in which three consecutive digits are the same.
Observe that a prohibited sequence must begin in one of the first four positions. Let $A_1, A_2, A_3, A_4$ be the set of outcomes in which three consecutive digits beginning in the first, second, third, and fourth positions, respectively, are the same.
The set of outcomes that violate the restriction is $A_1 \cup A_2 \cup A_3 \cup A_4$. By the Inclusion-Exclusion Principle,
\begin{align*}
|A_1 \cup A_2 \cup A_3 \cup A_4| & = |A_1| + |A_2| + |A_3| + |A_4|\\
& \quad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_4| - |A_3 \cap A_4|\\
& \quad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\
& \quad - |A_1 \cap A_2 \cap A_3 \cap A_4|
\end{align*}
$|A_1|$: There are three ways of selecting the number that occupies the first three positions and three ways of selecting the number that occupy each of the last three positions. Hence, $|A_1| = 3^4$. By symmetry,
$$|A_1| = |A_2| = |A_3| = |A_4| = 3^4$$
$|A_1 \cap A_2|$: Since the first three and the second three positions overlap, the first four positions must be filled with the same number. There are three ways to select this number and three ways to fill each of the remaining two positions. Hence, $|A_1 \cap A_2| = 3^3$. By symmetry,
$$|A_1 \cap A_2| = |A_2 \cap A_3| = |A_3 \cap A_4| = 3^3$$
$|A_1 \cap A_3|$: Since the first three and the third three positions overlap, the first five positions must be filled with the same number. There are three ways to select this number and three ways to fill the remaining position. Hence, $|A_1 \cap A_3| = 3^2$. By symmetry,
$$|A_1 \cap A_3| = |A_2 \cap A_4| = 3^2$$
$|A_1 \cap A_4|$: Since the first three and last three positions do not overlap, the first three positions can be filled in three ways, as can the last three. Hence,
$$|A_1 \cap A_4| = 3^2$$
$|A_1 \cap A_2 \cap A_3|$: Since these sets overlap, the first five positions must be filled with the same number. There are three ways to choose this number and three ways to fill the remaining position. Thus, $|A_1 \cap A_3 \cap A_3| = 3^2$. By symmetry,
$$|A_1 \cap A_2 \cap A_3| = |A_2 \cap A_3 \cap A_4| = 3^2$$
$|A_1 \cap A_2 \cap A_4|$: Since the first three positions overlap with the second three and the second three positions overlap with the last three positions, all six positions must be the same. There are three ways to select the number that fills all six positions. Hence, $|A_1 \cap A_2 \cap A_4| = 3$. By symmetry,
$$|A_1 \cap A_2 \cap A_4| = |A_1 \cap A_3 \cap A_4| = 3$$
$|A_1 \cap A_2 \cap A_3 \cap A_4|$: All six positions must be filled with the same number. Since there are three ways to select the number,
$$|A_1 \cap A_2 \cap A_3 \cap A_4| = 3$$
By the Inclusion-Exclusion Principle,
\begin{align*}
|A_1 \cup A_2 \cup A_3 \cup A_4| & = 3^4 + 3^4 + 3^4 + 3^4\\
& \quad - 3^3 - 3^2 - 3^2 - 3^3 - 3^2 - 3^3\\
& \quad + 3^2 + 3 + 3 + 3^2\\
& \quad - 3\\
& = 237
\end{align*}
Hence, there are $729 - 237 = 492$ admissible sequences.
| {
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"url": "https://math.stackexchange.com/questions/2728002",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$
Evaluate$$\prod^{100}_{k=1}\left[1+2\cos \frac{2\pi \cdot 3^k}{3^{100}+1}\right]$$
My attempt:
$$1+2\cos 2\theta= 1+2(1-2\sin^2\theta)=3-4\sin^2\theta$$
$$=\frac{3\sin \theta-4\sin^3\theta}{\sin \theta}=\frac{\sin 3\theta}{\sin \theta}$$
I did not understand how to solve after that. Help required.
| Let $$z=\cos\bigg(\frac{2\pi}{3^n+1}\bigg)+i\sin\bigg(\frac{2\pi}{3^n+1}\bigg)$$
Then $z^{3^n+1}=1$ and also $\displaystyle 2\cos \bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)=z^{3^k}+\frac{1}{z^{3^k}}$
Write $$\prod^{n}_{k=1}\bigg[1+2\cos\bigg(\frac{2\pi\cdot 3^k}{3^n+1}\bigg)\bigg]$$
$$=\bigg(1+z^3+\frac{1}{z^3}\bigg)\bigg(1+z^9+\frac{1}{z^9}\bigg)\cdots \cdots \bigg(1+z^{3n}+\frac{1}{z^{3n}}\bigg)$$
$$=\frac{(1+z^3+z^6)(1+z^9+z^{18})\cdots \cdots (1+z^{3^n}+(z^{3^n})^2)}{z^{3+9+\cdots \cdots +3^n}}$$
Multiply both Nr and Dr by $(1-z^3)$
$$=\frac{1-z^{3^{n+1}}}{(1-z^3)\cdot z^{3\frac{(3^n-1)}{2}}}= \frac{1-z^{-3}}{-(1-z^3)\cdot z^{-3}}=1.$$
$\text{Simplification}:\;\; $ From $z^{3n+1}=1\Rightarrow z^{3n}=z^{-1}$
And $$z^{3\frac{(3^n-1)}{2}}=z^{-3}\cdot z^{3\frac{(3^+1)}{2}}=z^{-3}\cdot \bigg(z^{\frac{(3^+1)}{2}}\bigg)^3=-z^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show there are infinitely many positive integers $k$ such that $\phi(n)=k$ has exactly two solutions where $n$ is a positive integer
Show that there are infinitely many positive integers $k$ such that the equation $\phi(n)=k$ has exactly two solutions, where $n$ is a positive integer.
Not entirely sure where to start or finish with this problem. I know that $\phi$ notation states that $\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_S})$
but other than that I am lost.
| Let $k = 2 \times 3^{6m + 1}$. We claim that if $m > 0$, then $\phi(n) = k$ has exactly $2$ solutions. (For $m = 0$, it has four solutions: $7$, $9$, $14$, and $18$.)
Let $n = p_1^{a_1} p_2^{a_2} \cdots p_i^{a_i}$, where $p_j$ is a prime number, and $a_j$ is a positive natural number. Then $n$ is a solution if and only if
$$
p_1^{a_1 - 1} p_2^{a_2 - 1} \cdots p_i^{a_i - 1} (p_1 - 1)(p_2 - 1) \cdots (p_i - 1) = k.
$$
We note that if $n$ has at least two odd prime factors $p$ and $q$, then $(p - 1)(q - 1)$ is a factor of $\phi(n)$. Hence $4$ is a factor of $\phi(n)$, and sol $n$ is not a solution.
Thus any solution $n$ is of the form $n = 2^\alpha \cdot p^\beta$ where $p$ is some odd prime number, and $\alpha$ and $\beta$ are natural numbers, possibly equal to $0$. If $\alpha > 2$, then $\phi(8) = 4$ is a factor of $\phi(n)$, and so $n$ is not a solution. Thus $\alpha \in \{0, 1, 2\}$.
Suppose that $\alpha = 2$. Then if $\beta > 0$, we have that $2(p - 1)$ is a factor of $\phi(n)$, and so $4$ is a factor of $\phi(n)$, and so $n$ is not a solution. Thus if $\alpha = 2$, then $\beta = 0$, and so $n = 4$. But $\phi(4) = 2$ is not divisible by $3$, and so it is not equal to $2 \times 3^{6m + 1} = k$.
We thus must have that either $n = p^\beta$, or $n = 2 p^\beta$. In each case, we have that
$$
\phi(n) = p^\beta - p^{\beta - 1}.
$$
We thus must solve the equation
$$
p^{\beta - 1} (p - 1) = 2 \times 3^{6m + 1}.
$$
If $\beta = 1$, then this is equivalent to $p - 1 = 2 \times 3^{6m + 1}$. Not we note that $2 \times 3^{6m + 1} + 1 \equiv 2 \times 3 + 1 \equiv 0 \pmod 7$, and so this implies that $p = 7$. This corresponds to $m = 0$. We're considering the case where $m > 0$, so we can thus assume from now on that $\beta > 1$.
Since $\beta > 1$, we note that $p \mid 2 \times 3^{6m + 1}$, and so we have that $p = 3$. We thus must solve the equation
$$
2 \times 3^{\beta - 1} = 2 \times 3^{6m + 1}.
$$
This clearly has the unique solution $\beta = 6m + 2$, and so the two solutions to the equation $\phi(n) = k$ are given by $n = 3^{6m + 2}$, and $n = 2 \times 3^{6m + 2}$.
| {
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"url": "https://math.stackexchange.com/questions/2734926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why can't I cancel $2x-3$ from $(2x-3)(x+5)=9(2x-3)$? Why are these simplifications wrong?
$$\begin{align}
(2x-3)(x+5)=9(2x-3) &\quad\to\quad \frac{(2x-3)(x+5)}{2x-3} = \frac{9(2x-3)}{2x-3} \quad\to\quad x+5 = 9\\[4pt]
x(x+2)=x(-x+3) &\quad\to\quad \frac{x(x+2)}{x} = \frac{x(-x+3)}{x} \quad\to\quad x+2=-x+3
\end{align}$$
| From here
$$(2x-3)(x+5)=9(2x-3)$$
we can observe that $2x-3=0$ is a solution and for $2x-3\neq 0$ we can cancel out and obtain
$$(2x-3)(x+5)=9(2x-3)\iff x+5=9\iff x=4$$
thus the solutions for the original equation are $x=\frac32$ and $x=4$.
As an alternative note that
$$(2x-3)(x+5)=9(2x-3)\iff 2x^2+7x-15=18x-27 \iff2x^2-11x+12=0$$
and then
$$x_{1,2}=\frac{11\pm\sqrt{121-96}}{4}=\frac{11\pm5}{4}=\frac32,4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How solve the following PDE? I want to solve the following PDE:$$
\begin{cases}
u_{tt}-c^2u_{xx}=0, \quad x\in\mathbb{R},\ t\geq x\\
u(x,x)=φ(x), \quad x\in\mathbb{R}\\
u_t(x,x)=0, \quad x\in\mathbb{R}
\end{cases}
$$
where $φ:\mathbb{R} \to \mathbb{R}$, $φ\in C^1(\mathbb{R})$.
Thanks for your help.
| $\def\d{\mathrm{d}}$Case 1: $c^2 \neq 1$. Make substitution $(y, s) = (x - c^2 t, t - x)$, then$$
u_t = u_s - c^2 u_y, \quad u_{tt} = u_{ss} - 2c^2 u_{ys} + c^4 u_{yy}, \quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss},
$$
and the equations become$$
\begin{cases}
(1 - c^2) u_{ss} - (c^2 - c^4) u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right), \quad y \in \mathbb{R}\\
(u_s - c^2 u_y)\bigr|_{s = 0} = 0. \quad y \in \mathbb{R}
\end{cases} \tag{1}
$$
Because $u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right)$, then$$
u_y\bigr|_{s = 0} = \frac{\d}{\d y} \left( φ\left( \frac{y}{1 - c^2} \right) \right) = \frac{1}{1 - c^2} φ'\left( \frac{y}{1 - c^2} \right),
$$
and (1) becomes$$
\begin{cases}
u_{ss} - c^2 u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ\left( \dfrac{y}{1 - c^2} \right), \quad y \in \mathbb{R}\\
u_s\bigr|_{s = 0} = \dfrac{c^2}{1 - c^2} φ'\left( \dfrac{y}{1 - c^2} \right). \quad y \in \mathbb{R}
\end{cases} \tag{2}
$$
Thus\begin{align*}
u(y, s) &= \frac{1}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) + φ\left( \frac{y - cs}{1 - c^2} \right) \right) + \frac{1}{2c} \int_{y - cs}^{y + cs} \dfrac{c^2}{1 - c^2} φ'\left( \dfrac{ξ}{1 - c^2} \right) \,\d ξ\\
&= \frac{1}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) + φ\left( \frac{y - cs}{1 - c^2} \right) \right) + \frac{c}{2} \left( φ\left( \frac{y + cs}{1 - c^2} \right) - φ\left( \frac{y - cs}{1 - c^2} \right) \right)\\
&= \frac{1 + c}{2} φ\left( \frac{y + cs}{1 - c^2} \right) + \frac{1 - c}{2} φ\left( \frac{y - cs}{1 - c^2} \right),
\end{align*}
and$$
u(x, t) = \frac{1 + c}{2} φ\left( \frac{x + ct}{1 + c} \right) + \frac{1 - c}{2} φ\left( \frac{x - ct}{1 - c} \right).
$$
Case 2: $c^2 = 1$. Make substitution $(y, s) = (x, t - x)$, then$$
u_t = u_s, \quad u_{tt} = u_{ss}, \quad u_{xx} = u_{yy} - 2u_{ys} + u_{ss},
$$
and the equations become$$
\begin{cases}
2u_{ys} - u_{yy} = 0, \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ(y), \quad y \in \mathbb{R}\\
u_s\bigr|_{s = 0} = 0, \quad y \in \mathbb{R}
\end{cases} \tag{3}
$$
which implies$$
\begin{cases}
2u_s - u_y = η(s), \quad y \in \mathbb{R},\ s > 0\\
u\bigr|_{s = 0} = φ(y), \quad y \in \mathbb{R}
\end{cases} \tag{4}
$$
where $η(s)$ is a function to be determined. By the method of characteristics, the general solution to (4) is$$
u(y, s) = \frac{1}{2} \int_0^s η(ξ) \,\d ξ + φ\left( y + \frac{s}{2} \right),
$$
and plugging it back to (3) yields$$
0 = u_s\bigr|_{s = 0} = \frac{1}{2} η(s) + \frac{1}{2} φ(y) = 0. \quad \forall y \in \mathbb{R}
$$
For the existence of solutions to the original equations, $φ$ must be a constant (namely $-η(0)$) and$$
u(x, t) = \frac{1}{2} \int_0^{t - x} η(ξ) \,\d ξ - \frac{1}{2} η(0).
$$
It can be verified that $u(x, t)$ with the form above are indeed solutions.
| {
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"url": "https://math.stackexchange.com/questions/2736043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-2)\gt0}$
${3-x^2-2x\gt0}$
${-x^2-2x+3\gt0}$
${-1(x^2+2x-3)\gt0}$
$-1\frac{(x-1)(x+3)}{-1}\gt0/-1$
${(x-1)(x+3)\lt0}$
At this point I really do not know what to do after this point or if I really even went about it the right way. Thank you for the help.
| $x > 3
\implies x-2 > 1
\implies x(x-2) > 3
\implies 1 > \dfrac{3}{x(x-2)}
$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Inconsistency for solving $x' = x^{1/2}$ The proposed system $x' = x^{1/2}$ can be solved easily to obtain $x(t) = \frac{1}{4} (t^2 + t c + c^2)$, where $c$ is the integration constant.
However, differentiate the newly-found $x(t)$, one gets: $x(t)' = \frac{1}{2}t+\frac{1}{4}c$. This implies that $x^{1/2} = \frac{1}{2}t+\frac{1}{4}c$. However,
$$(x^{1/2})^2 = \left(\frac{1}{2}t+\frac{1}{4}c\right)^2 = \frac{1}{4}\left(t^2 + \frac{1}{2}tc + c^2\right) \neq \frac{1}{4} \left(t^2 + t c + c^2\right) = x(t)$$
Does anyone know why the inconsistency occurs? I understand there is another solution, but it is also inconsistent.
| By separation of variables:
$$
\frac{dx}{\sqrt{x}}=dt
$$
so
$$
\sqrt{x}=\frac{1}{2}(t+c)
$$
and finally
$$
x=\frac{1}{4}(t+c)^2=\frac{1}{4}(t^2+2ct+c^2)
$$
Can you spot your error?
$\displaystyle x=\frac{1}{4}(t^2+{\Large\color{red}{2}}ct+c^2)$
| {
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How to find the equation of the tangent to the parabola $y = x^2$ at the point (-2, 4)? This question is from George Simmons' Calc with Analytic Geometry. This is how I solved it, but I can't find the two points that satisfy this equation:
$$
\begin{align}
\text{At Point P(-2,4):} \hspace{30pt} y &= x^2 \\
\frac{dy}{dx} &= 2x^{2-1} \\
&= 2x = \text{Slope at P.}
\end{align}
$$
Now, the equation for any straight line is also satisfied for the tangent:
$$
\begin{align}
y - y_0 &= m(x - x_0) \\
\implies y - y_0 &= 2x (x - x_0) \\
\text{For point P, } x_0 &= -2 \text{ and } y_0 = 4 \\
\implies y - 4 &= 2x(x+2)\\
\implies y - 4 &= 2x^2 + 4x\\
\implies y &= 2x^2 + 4x +4\\
\end{align}
$$
This is where the problem occurs. If I were to try to solve for $y$ using:
$$
y = ax^2+bx+c \implies y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
I'd get:
$$
\begin{align}
y &= 2x^2+4x+4 \text{ and, at x-intercept: }\\
x &= \frac{-4 \pm \sqrt{4^2 - (4\times2\times4)}}{2\times2} \\
x &= \frac{-4 \pm \sqrt{16 - 32}}{4} \\
x &= \frac{-4 \pm 4i}{4} \\
x &= -1 \pm i
\end{align}
$$
Is this the correct direction, or did I do something wrong?
| The others did point out your error, so I will just add the way I'd do it:
The tangent line we are looking for is in the form of $$g(x)=ax+b$$ for the function $$f(x)=x^2$$ at $x=-2$. We know that their derivate and their value most be equal at the given point, so we have that $$a=2*(-2)=-4$$ and $$(-2)^2=-4(-2)+b$$ $$4=8+b$$ $$b=-4$$ So the eqution for the tangent line is $$y=-4x-4$$
I like this method because I do not need to remember to the equation of the line through a given point.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sides of $(1)$ by $(x-1)$ we get that $1-x^d\mid 1-x^n$ which is the final result
Is this an ok proof?
| You can always do:
$f(y)=1+y+ \dots +y^{r-1}$ so that $yf(y)=y+y^2+\dots +y^r$ and $yf(y)-f(y)=(y-1)f(y)=y^r-1$
Then put $y=x^d$ with $dr=n$ and obtain $(x^d-1)f(x^d)=x^n-1$ and by construction $f(x^d)$ is a polynomial.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still too complicated for one to work out during the exam. So I think there should exist some better approaches to handle this identity without too much direct computation.
In addition, this identity is supposed to be true:
$$ \frac{x^7+y^7+z^7}{7}=\frac{x^2+y^2+z^2}{2}\times\frac{x^5+y^5+z^5}{5} .$$
| Write $S_n=x^n+y^n+z^n$. Then $S_0=3$, you are given $S_1=0$ and are charged to prove that $S_5=(5/6)S_2S_3$. Define
$$F(t)=\sum_{n=0}^\infty S_nt^n.$$
Then
$$ F(t)=\frac1{1-xt} + \frac1{1-yt} + \frac1{1-zt} = \frac{3+2e_1t+e_2t^2}
{1-e_1t+e_2t^2-e_3t^3} $$
where $e_1=x+y+z$, $e_2=xy+xz+yz$ and $e_3=xyz$. Then $e_1=0$ so
\begin{align}
F(t)&=\frac{3+e_2t^2}
{1+e_2t^2-e_3t^3}=(3+e_2t^2)\sum_{k=0}^\infty(-1)^k(e_2t^2-e_3t^3)^k\\
&=(3+e_2t^2)(1-e_2t^2+e_3t^3+e_2^2t^4-2e_2e_3t^5+\cdots)\\
&=3-2e_2t^2+3e_3t^3+2e_2^2t^4-5e_2e_3t^5+\cdots.
\end{align}
Therefore $S_2=-2e_2$, $S_3=3e_3$ and $S_5=-5e_2e_3$ etc.
| {
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Largest integer $n$ such that $3^n$ divides every $abc$ with $a$, $b$, $c$ positive integers, $a^2+b^2=c^2$, and $3|c$
Let $P$ denote the set { $abc$ : $a$, $b$, $c$ positive integers, $a^2+b^2=c^2$, and $3|c$}. What is the largest integer $n$ such that $3^n$ divides every element of $P$?
I first saw that as $3|c\implies c=3k$ where $k\in \mathbb{N}$
$\implies a^2+b^2=9k^2$
Now this means $a^2\equiv r1(mod \space 9)$ & $b^2\equiv r2(mod \space 9)$
where $r1+r2=9k_1$ where $k_1\in \mathbb{N}$
In order to get the largest integer $n$ s.t $3^n|\space \text{every element of P}$
We have to assume $a=3k$ and $b=3k$
Thus $abc=3^3\times(some \space number \ne a \space multiple \space \space of \space 3)$
Thus $n \space should \space be =3$
But the answer here is 4
I can't figure out the solution.Please help.
P.S. Is there any geometrical angle in this problem meaning can this $a^2+b^2=c^2$ be considered a circle somehow and proceed? I can't think of that approach.
OK here we have a solution, lets look at it:
As @ChristianF suggested as an answer to another of my question,
Lemma: If $a^2+b^2=c^2$ then at least one of integers $a,b,c$ is divisible by $3$.
Proof: If $3\mid c$ we are done. Say $3$ doesn't divide $c$. Then $$a^2+b^2\equiv 1 \pmod 3$$
So if $3$ doesn't divide none of $a$ and $b$ we have $$2\equiv 1 \pmod 3$$ a contradiction.
Also suggested by @Christian Blatter
Modulo $3$ only $0$ and $1$ are squares, hence $x^2+y^2=0$ mod $3$ implies $x=y=0$ mod $3$. It follows that all three of $a$, $b$, $c$ are divisible by $3$. Canceling this common factor we obtain $a'^2+b'^2=c'^2$ which is only possible if at least one of $a'$, $b'$, $c'$ is $=0$ mod $3$.
We will follow along the lines:
If $3|c \implies c=3k$ for some $k \in \mathbb{N}$
Now this means according to @Christian Blatter,Modulo $3$ only $0$ and $1$ are squares, hence $x^2+y^2=0$ mod $3$ implies $x=y=0$ mod $3$. It follows that all three of $a$, $b$, $c$ are divisible by $3$.
Hence $a=3k_1,b=3k_2$ for some $k_1,k_2 \in \mathbb{N}$
Now we get $9k_1^2+9k_2^2=9k^2$
Cancelling the 9 from the above from LHS and RHS we get
$k_1^2+k_2^2=k^2$
which is analogous to the fact that
we obtain $a'^2+b'^2=c'^2$ which is only possible if at least one of $a'$, $b'$, $c'$ is $=0$ mod $3$
Thus one of $k_1,k_2$ is still a multiple of 3
$\implies$ $a\times b \times c= 3^4 \times K$
Hence for $n=4$, which is the largest integer, $3^n|a\times b \times c$
Thanks to @ChristianF and @Christian Blatter for the insight
| First, convince yourself that every solution of $a^2+b^2=c^2$ has $abc$ a multiple of 3.
Then convince yourself that every solution with $c$ a multiple of 3 must be such that $(a/3)^2+(b/3)^2=(c/3)^2$ with $a,b,c$ all integers (I think you've already done this, so the first sentence above is all you really need).
| {
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Minimum length of the hypotenuse
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Prove that the minimum length of the hypotenuse is $(a^{2/3}+b^{2/3})^{3/2}$.
My Attempt
$\frac{x}{y}=\frac{a}{CM}=\frac{AN}{b}$
$$
\frac{x}{y}=\frac{AN}{b}\implies y=\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h(x)=x+y=x+\frac{xb}{\sqrt{x^2-a^2}}
$$
$$
h'(x)=1+\frac{\sqrt{x^2-a^2}.b-xb.\frac{x}{\sqrt{x^2-a^2}}}{x^2-a^2}=1+\frac{x^2b-a^2b-x^2b}{(x^2-a^2)^{3/2}}\\
=1+\frac{-a^2b}{(x^2-a^2)^{3/2}}=\frac{(x^2-a^2)^{3/2}-a^2b}{(x^2-a^2)^{3/2}}
$$
$$
h'(x)=0\implies (x^2-a^2)^{3/2}=a^2b\implies (x^2-a^2)^{3}=a^4b^2\\
\implies x^6-3x^4a^2+3x^2a^4-a^6=a^4b^2\implies x^6-3x^4a^2+3x^2a^4-a^6-a^4b^2=0\\
$$
How do I proceed further and find $h_{min}$ without using trigonometry ? Or is there anything wrong with my calculation ?
| we have $$\cos(\alpha)=\frac{a}{x}$$ and $$\sin(\alpha)=\frac{b}{y}$$ so we get
$$1=\frac{a^2}{x^2}+\frac{b^2}{y^2}$$ with this equation you ca eliminate $x$ or $y$ in $$c=x+y$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$ Show that $x^3-a$ is irreducible in $\mathbb{Z}_7$ unless $a=0$ or $\pm1$
My Idea:
Suppose $x^3-a=(Ax+b)(Bx^2+cx+d).$
Then $A=B=1$ or $A=B=-1$ WLOG $A=B=1.$
Then $x^3-a=x^3+x^2(b+c)+x(bc+d)+bd\\
\Rightarrow c+b=0,d+bc=0,bd=-a.$
I can't go further from here. Help...Thank You!
| Let $F$ be any field, and let
$p(x) = x^3 + ax^2 + bx + c \in F[x] \tag 1$
be any cubic polynomial over $F$. Then we have the following
Fact: $p(x)$ is reducible in $F[x]$ if and only if it has a zero in $F$.
Proof of Fact: Clearly if $p(x)$ has a zero $z \in F$, then $p(x) = (x - z)q(x)$ where $q(x) \in F[x]$ is of degree $2$; thus $p(x)$ is reducible. Now if $p(x)$ is assumed reducible, we may write $p(x) = r(x)s(x)$ where precisely one of $r(x), s(x) \in F[x]$ is of degree one, since $\deg r + \deg s = \deg p = 3$. But a factor $\alpha x + \beta$ of degree $1$ yields a root $-\beta/\alpha$, an thus we are done. End of Proof of Fact.
So the question becomes, for which $a \in \Bbb Z_7$ does $x^3 - a$ have no zeroes. Now it is just a matter of simple arithmetic to discover which $a \in \Bbb Z_7$ are not perfect cubes. We have
$0^3 = 0, \; 1^3 = 1, \; 2^3 = 1, \; 3^3 = 6 = -1, \; 4^3 = 1, \; 5^3 = 6 = - 1, \; 6^3 = (-1)^3 = -1; \tag 2$
we see that $0, 1, -1$ are perfect cubes, but that
$a = 2, 3, 4, 5, 6 \tag 3$
are not cubes in $\Bbb Z_7$; thus $x^3 - a$ is irreducible for $a$ given by (3).
By the way, $x^3 = xx^2$ is trivially reducible, and
$x^3 - 1 = (x - 1)(x^2 + x + 1), \tag 4$
$x^3 + 1 = (x + 1)(x^2 - x + 1). \tag 5$
| {
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"timestamp": "2023-03-29T00:00:00",
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British Maths Olympiad (BMO) 2006 Round 1 Question 5, alternate solution possible? The question states
For positive real numbers $a,b,c$ prove that
$(a^2 + b^2)^2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b)$
After some algebraic wrangling we can get to the point where:
$(a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4 ≥ 2c^2(a^2 + b^2)$
At this point if we take the $LHS - RHS$ we can write the expression as the sum of squares proving the inequality.
I was wondering, is it possible to divide both sides by $c^2(a^2 + b^2)$ and show somehow that
$((a^2 + b^2)^2 + (a + b)^2(a − b)^2 + c^4)/(c^2(a^2 + b^2)) ≥ 2$
I tried but was not able to.
| Suppose that $a,b,c$ can form the sides of a triangle. Let $s=\frac{a+b+c}{2}$ be the semiperimeter. The inequality becomes $$
(a^2+b^2)^2\ge2s\cdot 2(s-a)\cdot 2(s-b)\cdot 2(s-c)
$$
or by Heron's formula, $$a^2+b^2\ge 4A$$
where $A$ is the area of the triangle. If $\theta$ is the angle between sides $a$ and $b$, this reduces to $$
a^2-2ab\sin\theta+b^2\ge0,$$ and we have$$
a^2-2ab\sin\theta+b^2\ge a^2-2ab+b^2=(a-b)^2\ge0$$
In the case where $a,b,c$ do not form a triangle, exactly one of the factors on the right-hand is negative or one of the factors is $0$, so the inequality is trivial.
| {
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Geometric solution? Given coordinates of $A$, $B$, $C$, find $M$ on $y=x$ minimizing $AM+BM+CM$ I have the problem:
Let be given three points $A(1,2)$, $B(3,4)$, $C(5,6)$. Find point $M$ on the line $y=x$ so that sum of distances $P=AM+BM+CM$ is smallest.
I tried. We have
$$P=\sqrt{(x-1)^2 + (x-2)^2} + \sqrt{(x-3)^2 + (x-4)^2} +\sqrt{(x-5)^2 + (x-6)^2}.$$
We know that
$$\sqrt{a^2 + b^2}+\sqrt{c^2 + d^2} \geqslant \sqrt{(a+c)^2 + (b+d)^2}.$$
The sign of equal occur when and only when $\dfrac{a}{c}=\dfrac{b}{d}$.
We have
\begin{align*}
\sqrt{(x-1)^2 + (x-2)^2} + \sqrt{(x-5)^2 + (x-6)^2} & = \sqrt{(x-1)^2 + (x-2)^2} + \sqrt{(5-x)^2 + (6-x)^2} \\
& \geqslant \sqrt{(x-1 + 6-x)^2 + (x-2 + 5-x)^2}\\
& \geqslant \sqrt{34}.
\end{align*}
The sign of equal occur $$\dfrac{x-1}{6-x}=\dfrac{x-2}{5-x} \Leftrightarrow x=\dfrac{7}{2}.$$
Another way
$$\sqrt{(x-3)^2 + (x-4)^2} =\sqrt{2x^2 - 14 x + 25} = \sqrt{2}\sqrt{\left (x-\dfrac{7}{2}\right)^2 + \dfrac{1}{4} } \geqslant \dfrac{1}{\sqrt{2}}.$$
The sign of equal occur $ x=\dfrac{7}{2}.$
Therefore, the least of the expression $P $ is $\dfrac{1}{\sqrt{2}}+\sqrt{34}$ at $x=\dfrac{7}{2}.$
How can I solve this problem geometrically?
| Let's $a$ be a line $y=x$.
So, we claim that for the basis $M$ of the perpendicular dropped from $B$ to $a$ sum $AM + BM + CM$ is the smallest. It is easy to demonstrate this using additional point $A'$ which is symmetrical to the point $A$ with respect to the line $a$. One may see that $A'MC$ is the line segment (because of the symmetry of points $A$, $C$ with respect to point $B$). Thus, if $M' \neq M$ is arbitrary chosen point on $a$, then
\begin{align}
AM' + BM' + CM' = & A'M' + M'C + BM' > \\ &A'M + MC + BM = AM + BM + CM,
\end{align}
so $M$ is our desired point. It's easy to calculate its coordinates, which are $(\frac{7}{2},\frac{7}{2})$ exactly as in your answer.
| {
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Show that if $2x + 4y = 1$ where x and y are real numbers. Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$
I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!
If $x=2$ and $y=4$, then $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$ iff $$\frac{x}{2}=\frac{y}{4}$$, then $$20\ge(2x+4y)^2$$
$$-4.47\le2x+4y\le4.47$$
| Your work is correct up to $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$
From which you get $$ 20(x^2+y^2)\ge (1)^2$$
Therefore, $$x^2+y^2\ge \frac{1}{20}$$
At this point your are done.
| {
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How to solve this complex limits at infinity with trig? Please consider this limit question
$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$
How should I solve this? I have no idea where to start please help.
| HINT
Note that
$$\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}=\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{\frac{a}2\cdot \frac{\sin\left(\frac{a}{2x}\right)}{\frac{a}{2x}}}$$
| {
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} |
Prove that following recursively defined sequence converges and its limit is 1/2
\begin{cases}a_1 = 1\\ \\ \displaystyle a_{1+n} = \sqrt{\sum_{i=1}^n a_i}\end{cases}
Prove that $\{\frac{a_n}{n}\}$ is convergent and its limit is $\frac12$
My proof: We can recursively define relation between $a_{1+n}$ & $a_n$ as
$a_{1+n} = \sqrt{a_n^2+a_n}$
Also, it is evident that $a_{1+n} > a_n > 1$
Thus, $a_{n}^2 - a_{n-1}^2 + a_{n} - a_{n-1} = a_{n} > 1 > 1/4$
Thus, $a_{n+1} = \sqrt{a_{n}^2 + a_{n}} > \sqrt{a_{n-1}^2+a_{n-1}+1/4} = a_{n-1} + 1/2 $
Similarly, we can show $a_{n+1} = \sqrt{a_{n}^2 + a_n} < a_n + 1/2$
So, we have $a_1 + \frac{n-1}2 < a_{n+1} < a_1 + \frac{n+1}2$
Using Sandwich Theorem we got limit of $\frac{a_n}n$ as $\frac12$
Is my proof correct.
I will be grateful if you can provide alternative proofs
| As pointed out in question, the sequence satisfy the recurrence relation $a_{n+1} = \sqrt{a_n^2 + a_n}$.
From this, it is easy to see $a_n$ is strictly increasing and positive.
One consequence of this is $a_n$ cannot be bounded.
Assume the contrary, then $a_n$ converges to some finite value $M > 0$. Since the map $x \mapsto \sqrt{x(x+1)}$ is continuous, the "limit" $M$ will then satisfy $M = \sqrt{M^2 + M} > M$ which is absurd.
Another consequence is
$$a_n < a_{n+1} < \sqrt{a_n^2 + a_n + \frac14} = a_n + \frac12
\quad\implies\quad 1 < \frac{a_{n+1}}{a_n} < 1 + \frac1{2a_n}$$
Since $a_n$ is unbounded, this leads to
$\displaystyle\;\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1$ and hence
$$\lim_{n\to\infty} \frac{a_{n+1}-a_{n}}{(n+1)-n} =
\lim_{n\to\infty}\frac{a_{n+1}^2-a_n^2}{a_{n+1}+a_n}
= \lim_{n\to\infty}\frac{a_n}{a_n + a_{n+1}}
= \lim_{n\to\infty}\frac{1}{1 + \frac{a_{n+1}}{a_n}}
= \frac{1}{1+1} = \frac12$$
By Stolz-Cesàro, we can conclude
$$\lim_{n\to\infty} \frac{a_n}{n} \quad\text{exists and equals to}\quad \lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)-n} = \frac12$$
| {
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Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\
\end{vmatrix}
I tried getting the eigenvalues for A =
\begin{vmatrix}
0 & 0 & 0 & \cdots &0&0& n-1 \\
0 & 0 & 0 & \cdots &0&0& n-2\\
0 & 0 & 0 & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&0&1 \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\
\end{vmatrix}
For $a=0$ , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $
$m(0)>=n-2$
However, I was not able to determine the other eigenvalues.
Testing for different values of n :
for $n=2$ :
$D_2 = a^2-1$
for $n=3$ :
$D_3 = a^3 -5a$
$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .
However I'm aware that testing for different values of $n$ is not enough to generalize the formula.
Thanks in advance.
| You can just go for calculating the characteristic polynomial $\chi_{-A}$ of minus the matrix $A$ (the one at $a=0$), then your determinant will be $\chi_{-A}[a]$. As you already found that the rank of $A$ is$~2$ (if $n\geq2$; otherwise it is $0$) the coefficients of $\chi_{-A}$ in all degrees less than $n-2$ are zero (as its coefficient of degree $n-r$ is the sum of all principal $r$-minors of$~A$. Since $A$ has zero trace, one has $$\chi_{-A}=X^n+0x^{n-1}+c_nX^{n-1}$$ where $c_n$ is the sum of all principal $2$-minors of$~A$, which is easily seen to be
$$ c_n=-\sum_{k=0}^{n-1}k^2=-\frac{2n^3-3n^2+n}6.$$
Therefore $\det(D_n)=a^n+c_na^{n-2}=a^{n-2}(a^2+c_n)$, as you guessed (with $c_n\leq0$, so the roots of $\chi_{-A}$ are real: $\pm\sqrt{-c_n}$ and $0$ with multiplicity $n-2$).
| {
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Probability of drawing a king immediately after an ace among $5$ cards drawn Five cards are drawn one by one from a standard deck of $52$ cards. What is the probability of drawing a king immediately after an ace?
The number of ways for taking $5$ cards, one by one, from a deck of $52$ is $52.51.50.49.48$
The number of ways for taking the Ace of spades and, immediately after, the king of spades, is $50.49.48.4$?? If this is correct, how to generalize for any ace and king??
| The number of sequences of five cards is
$$P(52, 5) = 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48$$
If an ace immediately precedes a king, there are four positions in the ace-king subsequence can begin, four possible suits for the ace, four possible suits for the king, and $50 \cdot 49 \cdot 48$ ways of selecting the remaining cards in the hand, which gives
$$4 \cdot 4 \cdot 4 \cdot 50 \cdot 49 \cdot 48$$
However, we have counted hands with two places in which an ace immediately precedes a king twice, once for each way of designating each such subsequence as the one in which an ace immediately precedes a king. We only want to count such hands once, so we must subtract them from the total.
If there are two subsequences in which a king appears immediately after an ace, there are three objects to arrange, the two ace-king subsequences and the other card. There are $\binom{3}{2}$ ways to choose the positions of the ace-king subsequences, four ways to choose the suit of the ace and four ways to choose the suit of the king in the first such subsequence, three ways to choose one of the remaining aces and three ways to choose one of the remaining kings for the second such subsequence, and $48$ ways to choose the other card. Hence, there are
$$\binom{3}{2} \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 48$$
such hands.
Therefore, the probability of drawing a king immediately after an ace among five cards drawn from a standard deck is
$$\frac{4 \cdot 4 \cdot 4 \cdot 50 \cdot 49 \cdot 48 - \binom{3}{2} \cdot 4 \cdot 4 \cdot 3 \cdot 3 \cdot 48}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}$$
| {
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Continuity of functions of two variables Find the points where the function $f(x,y)$ is continuous where
$$f(x,y)=\left\{ \begin{array}{ll}
\frac{x^2\sin^2y}{x^2+2y^2} & \mbox{if $(x,y)\not=(0,0)$};\\
0 & \mbox{if $(x,y)=(0,0)$}.\end{array} \right.$$
What I attempted: Here $f(x,y)$ is continuous at all the points $(x,y)\not=(0,0)$
We will check continuity at the point $(x,y)=(0,0)$
$f(x,y)=0$ is well defined. Now,
$$\lim_{(x,y)\to (0,0)} \frac{x^2\sin^2y}{x^2+2y^2}$$
Here $0\le \frac{x^2\sin^2y}{x^2+2y^2}\le \frac{x^2y^2}{x^2+2y^2}$ (As $\sin^2y \le y$) [Am I correct here?]
If we let $x=r\cos\theta$, $y=r\sin\theta$, then $$\frac{x^2y^2}{2x^2+y^2}=\frac{r^2 \cos^2\theta r^2 \sin^2\theta}{r^2+r^2 \sin^2\theta}=\frac{r^2 \sin^2\theta \cos^2\theta }{1+\sin^2\theta}\le r^2$$ (As $\frac{\sin^2\theta}{1+\sin^2\theta}\le1$, $\cos^2\theta\le 1$ )
As $(x,y)\to (0,0)$, $r\to 0$. Therefore using Sandwich Theorem we should have $$\lim_{(x,y)\to (0,0)} \frac{x^2\sin^2y}{x^2+2y^2}=0$$
So, it is continuous at $(0,0)$.
I am not sure whether it is correct or not.The question was asked in an interview and I used the above technique. However, they did not pointed my mistake but just gave an another problem of an similar kind where $f(x,y)=\frac{x^2+\sin^2y}{2x^2+y^2}$ for all $(x,y) \not= (0,0)$ and $0$ for the origin.I tried the same method but failed. Then they asked me to leave.
| Let $f(x)=\sin^2 x$ and $g(x)=x$. Observe $f(0)=g(0)=0$. Furthermore, $f'(x)=2\sin x \cos x=\sin(2x)$ and $g'(x)=1$. But $f'(x)=\sin(2x) \leq 1=g'(x)$. Therefore, $f(x) \leq g(x)$ for $x \geq 0$. So your inequality is correct.
But this is making things too difficult. This is a Squeeze Theorem problem 'in reverse', meaning while you tend to eliminate the bounded trig function, here you bound the rest. Note that $x^2 \leq x^2 + 2y^2$, we have $\dfrac{x^2}{x^2+2y^2} \leq 1$. Therefore,
$$
0 \leq \dfrac{x^2\sin^2 y}{x^2+2y^2} = \dfrac{x^2}{x^2+2y^2} \cdot \sin^2 y \leq \sin^2y
$$
and $\sin^2 y \to 0$ as $(x,y) \to (0,0)$. Therefore by Squeeze Theorem, $\dfrac{x^2\sin^2 y}{x^2+2y^2} \to 0$ as $(x,y) \to (0,0)$. Note that you also want to say that the function is continuous elsewhere. Namely, $x^2 \sin^2 y$ and $x^2+2y^2$ are continuous functions for $(x,y) \neq (0,0)$ and hence $\dfrac{x^2\sin^2 y}{x^2+2y^2}$ is a continuous function for $(x,y) \neq (0,0)$. Therefore, the function $f(x,y)$ is continuous on all of $\mathbb{R}^2$.
| {
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Solving the problem : $z z_x + z_y = 0, \quad z(x,0) = x^2$ Exercise :
For the problem :
$$\begin{cases} zz_x + z_y = 0 \\ z(x,0) = x^2\end{cases}$$
derive the solution :
$$z(x,y) = \begin{cases} x^2, \quad y = 0\\ \frac{1+2xy - \sqrt{1+4xy}}{2y^2}, \quad y \neq 0 \; \text{and} \; 1+4xy >0 \end{cases}$$
When do shocks develop ? Use the Taylor series for $\sqrt{1+\epsilon}$ about $\epsilon= 0$ to verify that $\lim_{y\to 0} z(x,y) = x^2$.
Attempt :
$$\frac{\mathrm{d}x}{z} = \frac{\mathrm{d}y}{1} = \frac{\mathrm{d}z}{0}$$
We yield the integral curves :
$$\frac{\mathrm{d}y}{1} = \frac{\mathrm{d}z}{0} \implies z_1 = z $$
$$\frac{\mathrm{d}x}{z} = \frac{\mathrm{d}y}{1} \implies z_2 = x -zy$$
Thus, the general solution will involve a $F \in C^1$ function, such that :
$$z(x,y) = F(x-zy)$$
For $y=0$ :
$$z(x,0) = F(x) \Rightarrow F(x) = x^2$$
How would one proceed now to find the second branch of the solution ?
(The shock-taylor part is easy)
| The general solution is :
$$z(x,y) = F(x-zy)\qquad\text{OK}.$$
$F$ is an arbitrary function, to be determines according to the boundary condition.
Condition : $\quad z(x,0)=x^2=F(x-0y)$
So, the function $F$ is determined :
$$F(X)=X^2\qquad\text{any }X$$
We put this function into the general solution where $X=x-zy$.
$$z(x,y)=F(x-zy)=(x-zy)^2$$
$z=x^2-2xyz+y^2z^2$
$y^2z^2-(2xy+1)z+x^2=0$
Solving for $z$ leads to :
$$z=\frac{2xy+1\pm\sqrt{(2xy+1)^2-4x^2y^2}}{2y^2}$$
$$z=\frac{2xy+1\pm\sqrt{1+4xy}}{2y^2}$$
For $y\to 0$ :
Let $\quad 4xy=\epsilon>0 \quad$ because $1+4xy>0$.
$$z=\frac{\frac{\epsilon}{2}+1\pm\sqrt{1+\epsilon}}{2y^2}=\frac{\frac{\epsilon}{2}+1\pm\sqrt{1+\epsilon}}{2\left(\frac{\epsilon}{4x}\right)^2}=8x^2\left(\frac{\frac{\epsilon}{2}+1\pm\sqrt{1+\epsilon}}{\epsilon^2}\right)$$
$\sqrt{1+\epsilon}\simeq 1+\frac12\epsilon-\frac18\epsilon^2+...$
$$z\simeq 8x^2\left(\frac{\frac{\epsilon}{2}+1\pm\left(1+\frac12\epsilon-\frac18\epsilon^2+...\right)}{\epsilon^2}\right)$$
Case of sign $+$ :
$z\simeq 8x^2\left(\frac{2+\epsilon-\frac18\epsilon^2+...}{\epsilon^2}\right)\to\infty\quad$ when $\epsilon\to 0$. This case is rejected because $z\to x^2$.
Case of sign $-$ :
$z\simeq 8x^2\left(\frac{\frac18\epsilon^2+...}{\epsilon^2}\right)\to x^2\quad$when $\epsilon\to 0$. This agrees. Thus the final result is :
$$z=\frac{2xy+1-\sqrt{1+4xy}}{2y^2}$$
| {
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Finding integers and the limit of a sequence - from a 100 sequence and more (Russian book) Consider a sequence $a_{1}=1$ and for every $k>1$ integer $a_k=a_{k-1}+\dfrac{1}{a_{k-1}}$.
a) How many positive integers $n$ are there, satisfying that $a_n$ be an integer?
b) Find the limit (if there is) of the $a_k$ sequence!
Please help! It looks like that for a) the answer is 2. Thanks in advance!
| Let's prove that
$$
1 \le a_n \le n
$$
Since $a_1 = 1$ and $a_{n+1} > a_n$ we clearly have $a_n \ge 1$. We can prove that $a_n \le n$ by induction. We have $a_1 = 1 \le 1$ and induction step is
$$
a_{n+1} = a_n + \frac{1}{a_n} \le n + \frac{1}{a_n} \le n + 1,
$$
since $a_n \ge 1$.
About question a). If $a_n$ is an integer then $1/a_n$ is an integer only if $a_n = 1$. That gives us case $a_2 = 2$. Also by definition $a_1 = 1$. There is no other integers in this sequence. Let $a_n = \frac{p}{q}$ where $p$ and $q$ are coprime and $q>1$. Then
$$
a_{n+1} = a_n + \frac{1}{a_n} = \frac{p^2 + q^2}{pq}.
$$
So we must have $p^2 + q^2 = m\cdot pq$ for some $m \in \mathbb{N}$. But that's not possible if $p$ and $q$ are coprime, because $q$ must divide $p^2$ and $q > 1$. So there are only two indices ($n=1,2$) for which $a_n$ is integer.
Now regarding question b). Since
$$
a_{n+1} = a_n + \frac{1}{a_n} \ge \frac{1}{a_n} + \frac{1}{a_{n-1}} + \ldots + \frac{1}{a_1} \ge \frac{1}{n} + \frac{1}{n-1} + \ldots + 1 = H_n
$$
where $H_n$ is a harmonic series, which clearly diverges.
| {
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Series 1 - 1/2^2 + 1/3 - 1/4^2 + 1/5 - 1/6^2 + 1/7... This is the exercise 2.7.2 e) of the book "Understanding Analysis 2nd edition" from Stephen Abbott, and asks to decide wether this series converges or diverges:
$$ 1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots $$
I have noticed that
$$\dfrac{1}{3}< 1-\dfrac{1}{2^2}+\dfrac{1}{3};\\
\dfrac{1}{3}+\dfrac{1}{5}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5};\\
\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}\\
$$
which is true in general because
$$1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{36}-\dfrac{1}{64}-\dots=1-\sum_{n=1}^\infty \dfrac{1}{(2n)^2}=1-\dfrac{1}{4}\sum_{n=1}^\infty\dfrac{1}{n^2}=1-\dfrac{\pi^2}{24}>0.
$$
Thus
$$
\sum_{n=1}^\infty \dfrac{1}{2n+1}<1-\dfrac{1}{2^2}+\dfrac{1}{3}-\dfrac{1}{4^2}+\dfrac{1}{5}-\dfrac{1}{6^2}+\dfrac{1}{7}-\dfrac{1}{8^2}\dots
$$
Finally, as the series $\sum_{n=1}^\infty \dfrac{1}{2n+1}$ diverges, so does the series requested.
My two questions are:
I) Is this reasoning correct?
II) Can this exercise be done without using that $\sum_{n=1}^\infty\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$? I would like to find a solution with more elementary tools.
Thank you.
| Don't be swayed by the particular case. You have the following
Lemma Let $\sum a_n$ be a series (resp. $\sum b_n$ be a convergent series), then the intertwining $x_{2n}=a_n$ and $x_{2n+1}=b_n$ is convergent iff $\sum a_n$ is so.
Proof Let us, for short, suppose the indexing starting from zero. Then, we get
$$
\sum_{n=0}^{2N+1}x_n=\sum_{n=0}^N a_n + \sum_{n=0}^N b_n
$$
this proves the equivalence
$$
\sum x_n\mbox{ converges }\Longleftrightarrow \sum a_n\mbox{ converges }
$$
end of proof
Here
$$
1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\ldots
$$
diverges while
$$\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}\ldots
$$
converges. Hence your series diverges.
| {
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show whether $\frac {xy}{x^2+y^2}$ is differentiable in $0$ or not? (multivariable) Q: $f(x,y)=\frac {xy}{x^2+y^2}$ if $(x,y)\not=(0.0)$, and $0$ if $(x,y)=(0,0)$.
Is $f$ differentiable at $(0,0)$?
Attempt: $$\lim_{(x,y) \to (0,0)} \frac {f(x,y)-f(0,0)}{||(x,y)-(0,0)||} = \lim_{(x,y) \to (0,0)} \frac {\frac{xy}{x^2+y^2}}{\frac {\sqrt {x^2+y^2}}{1}}$$
$$=\lim_{(x,y) \to (0,0)} \frac {xy}{(x^2+y^2)^{3/2}}.$$
Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Then,
$$ \frac {xy}{(x^2+y^2)^{3/2}}= \frac {r^2\cos(\theta)\sin(\theta)}{r^{3/2}}=\sqrt r\cos(\theta)\sin(\theta).$$
What should be the next step? If I use polar coordinates, how can I transform $$\lim_{(x,y) \to (0,0)}\rightarrow\lim_{(r,\theta) \to (?,?)}$$??
| Hint:
AM-GM gives
$$\left|\frac{xy}{x^2+y^2}\right| = \frac{|x||y|}{x^2+y^2} \le \frac{\frac{x^2+y^2}2}{x^2+y^2} = \frac12$$
with equality when $x=y$.
Therefore $\lim_{(x,y)\to (0,0)} f(x,y) \ne 0$ so $f$ is not continuous at $(0,0)$.
| {
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If $0Given that $f(x)=e^x(x^2-6x+12)-(x^2+6x+12),\;\;x>0$ is an increasing function. I want to prove that:
If $0<x<\infty$, then $0<\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}<\frac{x}{12}$.
Here is what I have done:
If $0<x<\infty$, then by Mean Value Theorem, $\exists\; c\in(0,x)$ such that
$$f'(c)=\frac{e^x(x^2-6x+12)-(x^2+6x+12)- 0}{x- 0}>0$$
but how do I get the desired inequality? Can anyone help out?
| We'll prove that $$0<\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}$$ or
$$\frac{1}{e^x-1}>\frac{2-x}{2x},$$ which is obvious for $x\geq2$.
But, for $0<x<2$ we need to prove that
$$e^x-1<\frac{2x}{2-x}$$ or $f(x)>0,$ where
$$f(x)=\ln(x+2)-\ln(2-x)-x.$$
Indeed, $$f'(x)=\frac{x^2}{4-x^2}>0,$$ which says
$$f(x)>\lim_{x\rightarrow0^+}f(x)=0$$ and the left inequality is proven.
By the same way we can prove a right inequality.
Indeed, we need to prove that
$$\frac{1}{e^x-1}<\frac{x}{12}+\frac{1}{x}-\frac{1}{2}$$ or
$$e^x-1>\frac{12x}{x^2-6x+12}$$ or $g(x)>0,$ where
$$g(x)=x-\ln(x^2+6x+12)+\ln(x^2-6x+12)$$ and since
$$g'(x)=\frac{x^4}{(x^2+6x+12)(x^2-6x+12)}>0,$$ we obtain:
$$g(x)>\lim_{x\rightarrow0^+}g(x)=0$$ and we are done.
| {
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USAMO 2018: Show that $2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$ Here is question 1 from USAMO 2018 Q1 (held in April):
Let $a,b,c$ be positive real numbers such that $a+b+c = 4 \sqrt[3]{abc}$.
Prove that: $$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2$$
This question is on symmetric polynomials. I recall as many facts as I can think of regarding inequalities. (This test was closed book).
*
*AM-GM inequality: $a + b + c \geq 3 \sqrt[3]{abc}$ so the equality there is strange.
*quadratic mean inequality suggests $3(a^2 + b^2 + c^2) > a + b + c $.
*The $\min(a^2,b^2,c^2)$ on the left side makes things difficult since we can't make it smaller.
*I am still looking for other inequalities that might work.
It's tempting to race through this problem with the first solution that comes to mind. I'm especially interested in some kind of organizing principle.
| Without loss of generality, let $a\le b=ax\le c=axy, x\ge 1, y\ge 1$.
Then:
$$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a+ax+axy=4\sqrt[3]{a(ax)(axy)} \Rightarrow 1+x+xy=4x^{\frac23}y^{\frac13} \qquad (1)$$
Also:
$$a+b+c = 4 \sqrt[3]{abc} \Rightarrow a^2+b^2+c^2=16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \qquad (2)$$
Plugging $(2)$ and then $(1)$ into the given inequality:
$$2(ab+bc+ca) + 4 \min(a^2,b^2,c^2) \geq a^2 + b^2 + c^2 \overbrace{\Rightarrow}^{(2)}\\
2(ab+bc+ca)+4a^2\ge 16\sqrt[3]{(abc)^2}-2(ab+bc+ca) \Rightarrow \\
(a(ax)+(ax)(axy)+(axy)a)+a^2\ge 4\sqrt[3]{(a(ax)(axy))^2} \Rightarrow\\
x+x^2y+xy+1\ge 4x\sqrt[3]{xy^2} \overbrace{\Rightarrow}^{(1)}\\
4x^{\frac23}y^{\frac13}+x^2y\ge 4x^{\frac43}y^{\frac23} \Rightarrow\\
\left(x^{\frac23}y^{\frac13}\right)^2-4\left(x^{\frac23}y^{\frac13}\right)+4\ge 0 \Rightarrow \\
\left(x^{\frac23}y^{\frac13}-2\right)^2\ge 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Determinant of the matrix associated with the quadratic form If A is the matrix associated with the quadratic form $4x^2+9y^2+2z^2+8yz+6zx+6xy$ then what is the determinant of A? I don't know how to solve quadratic form of a matrix pls help me
| Consider the matrix $$A= \begin{pmatrix}
4 & 3 & 3 \\
3 & 9 & 4 \\
3 & 4 & 2 \\
\end{pmatrix}.$$
Note that the coefficients relating to $x^2,y^2$ and $z^2$ lie on its diagonal. The other entries correspond to half of the coefficients of the interaction terms. Multiplying this matrix with $\begin{pmatrix}
x&y&z \\
\end{pmatrix}^T$ and $\begin{pmatrix}
x\\y\\z \\
\end{pmatrix}$ yields
\begin{align} \begin{pmatrix}
x&y&z \\
\end{pmatrix}^T A \begin{pmatrix}
x\\y\\z \\
\end{pmatrix} &= \begin{pmatrix}
x&y&z \\
\end{pmatrix}^T \begin{pmatrix}
4 & 3 & 3 \\
3 & 9 & 4 \\
3 & 4 & 2 \\
\end{pmatrix}\begin{pmatrix}
x\\y\\z \\
\end{pmatrix}\\
&=\begin{pmatrix}
x&y&z \\
\end{pmatrix}^T \begin{pmatrix}
4x + 3y + 3z \\
3x + 9y + 4z \\
3x + 4y + 2z \\
\end{pmatrix}\\
&= 4x^2 +3xy + 3xz +3xy +9y^2+4yz+3xz+4yz +2z^2\\
&= 4x^2 +9y^2 +2z^2 +6xy + 6xz + 8yz.
\end{align}
Hence, $A$ is the matrix we are looking for which produces the desired function.
However, we are interested in the determinant of $A$. Luckily, $A$ is a $3 \times 3$ matrix. Hence, we can use the rule of Sarrus to calculate the determinant:
\begin{align}
\det(A)= A&= \begin{vmatrix}
4 & 3 & 3 \\
3 & 9 & 4 \\
3 & 4 & 2 \\
\end{vmatrix}\\
&= 4\cdot 9 \cdot 2 +3\cdot4\cdot3 +3\cdot4\cdot3 - 3\cdot9\cdot3 -3\cdot3\cdot2 -4\cdot4\cdot4\\
&= 72 +36+36 - 81-18 -64\\
&=-19.
\end{align}
Hence, det$(A)=-19$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ How to get the exact value of $\sin(x)$ if $\sin(2x) = \frac{24}{25}$ ?
I checked various trigonometric identities, but I am unable to derive $\sin(x)$ based on the given information.
For instance:
$\sin(2x) = 2 \sin(x) \cos(x)$
| Refer to the diagram below.
$AD$ is the angle bisector of the right-angled triangle $\Delta ABC$.
Given $BC=24$ and $AC=25$.
Let $\angle DAB = x$.
From $\Delta ABC$ we see that $\sin(2x) = \dfrac{24}{25}$.
Now, by angle bisector theorem, $BD:DC = 7:25$.
Therefore, $BD = \dfrac{7}{7+25} \times 24 = \dfrac{21}4$.
Observing that $AB:BD = 4:3$, we see that $AD = \dfrac{35}4$
Therefore, $\sin x = \dfrac {BD} {DA} = \dfrac 3 5$.
The other possible $x$ would be $x+180^\circ$, since the period of sine is $360^\circ$, making $\sin x = -\dfrac35$.
In conclusion, $\sin x = \pm \dfrac 35$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Maximizing $3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. What went wrong?
Let $f(x) = 3\sin^2 x + 8\sin x\cos x + 9\cos^2 x$. For some $x \in \left[0,\frac{\pi}{2}\right]$, $f$ attains its maximum value, $m$. Compute $m + 100 \cos^2 x$.
What I did was rewrite the equation as $f(x)=6\cos^2x+8\sin x\cos x+3$. Then I let $\mathbf{a}=\left<6\cos x,8\cos x\right>$ and $\mathbf{b}=\left<\cos x,\sin x\right>$.
Using Cauchy-Schwarz, I got that the maximum occurs when $\tan x=\frac{4}{3}$, and that the maximum value is $10\cos x$. However, that produces a maximum of $9$ for $f(x)$, instead of the actual answer of $11$.
What did I do wrong, and how do I go about finding the second part? Thanks!
| Using the identity $$\cos^2 x=\frac {1+\cos 2x}{2}$$ and $$2\sin x\cos x=\sin 2x$$ the question changes to finding minimum value of the function
$$6+3\cos 2x+4\sin 2x$$
And now using a standard result that the range of a function $a\sin \alpha\pm b\cos \alpha$ is $[-\sqrt {a^2+b^2},\sqrt {a^2+b^2}]$
Hence the range of the given expression becomes $[1,11]$
Hope you can continue further
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Summation of $\sum_{r=1}^{n} \frac{\cos (2rx)}{\sin((2r+1)x \sin((2r-1)x}$ Summation of $$S=\sum_{r=1}^{n} \frac{\cos (2rx)}{\sin((2r+1)x) \sin((2r-1)x)}$$
My Try:
$$S=\sum_{r=1}^{n} \frac{\cos (2rx) \sin((2r+1)x-(2r-1)x)}{\sin 2x \:\sin((2r+1)x \sin((2r-1)x}$$
$$S=\sum_{r=1}^{n} \frac{\cos (2rx) \left(\sin((2r+1)x \cos (2r-1)x-\cos(2r-1)x)\sin(2r+1)x\right)}{\sin 2x \:\sin((2r+1)x \sin((2r-1)x)}$$
$$S=\sum_{r=1}^n \frac{\cos(2rx)}{\sin 2x}\left(\cot(2r-1)x-\cot(2r+1)x\right)$$
Any clue here?
| Good question! Here is one possible approach. First we rewrite the numerator as
$$
\cos(2rx) = \cos[(r+\frac{1}{2})x+(r-\frac{1}{2})x] = \cos\frac{2r+1}{2}x\cos\frac{2r-1}{2}x - \sin\frac{2r+1}{2}x\sin\frac{2r-1}{2}x
$$
and the denominator as
$$
4\sin\frac{2r+1}{2}x\cos\frac{2r+1}{2}x\cdot \sin\frac{2r-1}{2}x\cos\frac{2r-1}{2}x.
$$
Therefore the original summation can be written as
$$
S_n = \frac{1}{4}\sum_{r=1}^n \frac{1}{\sin\frac{2r+1}{2}x\sin\frac{2r-1}{2}x} -\frac{1}{4}\sum_{r=1}^n \frac{1}{\cos\frac{2r+1}{2}x\cos\frac{2r-1}{2}x}.
$$
Now we can use the trick that multiply a factor $\frac{\sin x}{\sin x}$ and use $$\sin x = \sin\frac{2r+1}{2}x\cos\frac{2r-1}{2}x - \sin\frac{2r-1}{2}x\cos\frac{2r+1}{2}x.$$ For the first sum,
$$
\frac{\sin x}{\sin\frac{2r+1}{2}x\sin\frac{2r-1}{2}x} = \cot \frac{2r-1}{2}x-\cot \frac{2r+1}{2}x.
$$
For the second sum,
$$
\frac{\sin x}{\cos\frac{2r+1}{2}x\cos\frac{2r-1}{2}x} = \tan \frac{2r+1}{2}x-\tan \frac{2r-1}{2}x.
$$
Thus the original sum is
$$
S_n = \frac{1}{4\sin x}(\cot \frac{1}{2}x-\cot \frac{2n+1}{2}x-\tan \frac{2n+1}{2}x+\tan \frac{1}{2}x).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Probability approach in the expected payoff of a dice game I am trying to understand the problem of expected payoff of a dice game explained here. I can roll the dice up to three times, but after each one I decide if I want to try once again or not. The idea is to find an optimal strategy that maximizes the expected payoff (expected number of spots; they do not sum up).
Let's say I know the optimal strategy: If in the first roll I get 1,2 or 3 I roll the dice once again. If in the second I get 1,2,3 or 4 I roll the dice once again. The end.
I wanted to calculate the expected number using probability $P(X=k)$ that I end up with $k$ spots. I am probably doing something wrong because I don't get the correct answer. My reasoning is the following:
$$P(X=\{1,2,3\}) = \underbrace{\frac{3}{6} \cdot \frac{4}{6} \cdot\frac{1}{6}}_{\text{3rd roll}}$$
$$P(X=4) = \underbrace{\frac{1}{6}}_{\text{1st roll}}+\underbrace{\frac{3}{6} \cdot \frac{4}{6} \cdot\frac{1}{6}}_{\text{3rd roll}}$$
$$P(X=5) = \underbrace{\frac{1}{6}}_{\text{1st roll}}+\underbrace{\frac{3}{6}\cdot\frac{1}{6}}_{\text{2nd roll}}+\underbrace{\frac{3}{6} \cdot\frac{4}{6}\cdot\frac{1}{6}}_{\text{3rd roll}}$$
$$P(X=6) = \frac{1}{6} + \frac{3}{6}\cdot\frac{1}{6}+\frac{3}{6} \cdot \frac{4}{6} \cdot\frac{1}{6}$$
Expected number is $E[X] = \sum_{k=1}^{6}kP(X=k) \approx 4.58$, which is not correct.
I appreciate any help.
| Using your method:
$1$-roll game:
$$E(X)=\sum_{k=1}^6 kP(X=k)=1\cdot \frac{1}{6}+2\cdot \frac16+3\cdot \frac16+4\cdot \frac16+5\cdot \frac16+6\cdot \frac16=3.5.$$
$2$-roll game:
$$E(X)=\sum_{k=1}^6 kP(X=k)=1\cdot \frac{3}{36}+2\cdot \frac{3}{36}+3\cdot \frac{3}{36}+\\
4\cdot \left(\frac16+\frac{3}{36}\right)+5\cdot \left(\frac16+\frac{3}{36}\right)+6\cdot \left(\frac16+\frac{3}{36}\right)=4.25.$$
Note: The player rerolls if the first roll was $1,2,3$.
$3$-roll game:
$$E(X)=\sum_{k=1}^6 kP(X=k)=1\cdot \frac{12}{216}+2\cdot \frac{12}{216}+3\cdot \frac{12}{216}+\\
4\cdot \left(\frac{4}{36}+\frac{12}{216}\right)+5\cdot \left(\frac16+\frac{4}{36}+\frac{12}{216}\right)+6\cdot \left(\frac16+\frac{4}{36}+\frac{12}{216}\right)=4\frac23.$$
Note: The player rerolls if the first roll is $1,2,3,4$ and the second roll is $1,2,3$. You did the other way around!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Without-loss-of-generality question Going through solutions of IMO'09. Bumped into a without-loss-of-generality assumption that I can't comprehend.
Here's the statement of the problem:
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c$. Prove that
$$
\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}\leq\frac{3}{16}.
$$
Here's how they start in the solution:
We prove the homogenized inequality
$$
\frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
for all positive real numbers $a,b,c$. Without loss of generality we choose $a+b+c=1$. Thus, the problem is equivalent to prove for all $a,b,c>0$, fulfilling this condition, the inequality
$$
\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}\leq\frac{3}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).
$$
And so on... If someone's interested in the rest, it's problem A2, solution 2 in this pdf: https://www.imo-official.org/problems/IMO2009SL.pdf
Why is there no loss of generality in such a choice of $a,b,c$?
| There is no loss of generality because the inequality
$$
\frac{(a+b+c)^2}{(2a+b+c)^2}+\frac{(a+b+c)^2}{(2b+c+a)^2}+\frac{(a+b+c)^2}{(2c+a+b)^2}\leq\frac{3}{16}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
is homogenized. This means: if you multiply $a,b$ and $c$ with the same constant $r$, the inequality does not change. Now give yourself some $a,b$ and $c$ where you have free choice of selecting these variables. You will have the sum $s = a + b+ c$. Now multiply $a,b$ and $c$ with the same constant $r=1/s$, then in the new (multiplied) variables you have $1 = a + b+ c$ which means you can always take that choice.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer. QUESTION: Find all positive integers $x,y$ such that $\frac{x^2-1}{xy+1}$ is a non-negative integer.
ATTEMPT (AND SOME SOLUTIONS): So, for a positive integer $x$, $x^2-1\ge0$. In the case $x^2-1=0$ i.e $x=1$ (since it's a positive integer), we get
$\frac{x^2-1}{xy+1}=0$ which is a non-negative integer, regardless of $y$. So, one solution is $(x,y)=(1,k)$ where $k$ is a positive integer.
Now, let's say $y=1$. Then $\frac{x^2-1}{xy+1}=\frac{x^2-1}{x+1}=x-1$ which is always a non-negative integer so $(x,y)=(k,1)$ is also a solution. However, I don't know how to find the other or prove that those are the only ones.
| $x^2-1=(x-1)(x+1)$
$\frac{x^2-1}{xy+1}$ can be an integer if: | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 1
} |
Finding the sum of $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, $\cos\frac{5π}{7}$ by first finding a polynomial with those roots
Without using tables, find the value of $$\cos\frac{\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{5\pi}{7}$$
This is a very common high school trigonometric problem, and the usual way to solve this is by repeated application of trigonometric identities. But I thought of a bit different approach.
Somehow, if we can find a polynomial whose roots are the three terms of the above expression, then we can apply Vieta's formula to find the value.
So please help me with it. (Any hint will be appreciated.)
| $\text{Using Complex Number}$
Let $z=e^{\frac{\pi i}{7}},$ so $z^7=-1.$ Let $Q$ be the desired quantity. Then
$$2Q=z+\frac{1}{z}+z^3+\frac{1}{z^3}+z^5+\frac{1}{z^5} = \frac{z^{10}+z^8+z^6+z^4+z^2+1}{z^5} = \frac{z^{12}-1}{z^5(z^2-1)}$$
$$=\frac{-z^5-1}{z^7-z^5} = \frac{-z^5-1}{-1-z^5} = 1$$ $$\therefore Q=\frac{1}{2}\ \ \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_0^1x(\tan^{-1}x)^2~\textrm{d}x$
Evaluate $\int\limits_0^1x(\tan^{-1}x)^2~\textrm{d}x$
My Attempt
Let, $\tan^{-1}x=y\implies x=\tan y\implies dx=\sec^2y.dy=(1+\tan^2y)dy$
$$
\begin{align}
&\int\limits_0^1x(\tan^{-1}x)^2dx=\int\limits_0^{\pi/4}\tan y.y^2.(1+\tan^2y)dy\\
&=\int\limits_0^{\pi/4}\tan y.y^2dy+\int\limits_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-\int2y.\log|\sec y|dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
&=\bigg[y^2.\log|\sec y|-y^2.\log|\sec y|+\int\frac{\tan y\sec y}{\sec y}y^2dy\bigg]+\int_0^{\pi/4}\tan^3y.y^2dy\\
\end{align}
$$
How do I proceed further and solve the integration?
| $$\int_ {0}^{1}x(arctan(x))^2dx$$
First solve the integral without boundaries
$$\int x(arctan(x))^2dx$$
Apply Integration By Parts, where $u=\arctan^2(x),v^{\prime}=x$
$$=\arctan^2(x)\frac{x^2}{2}-\int \frac{2\arctan(x)}{1+x^2} (\frac{x^2}{2})dx$$
$$=\frac12 x^2\arctan^2(x)-\int \frac{x^2\arctan(x)}{x^2+1}dx$$
Note that $\int \frac{x^2\arctan(x)}{x^2+1}dx= x\arctan(x)-\arctan^2(x)+\ln|\frac{1}{\sqrt{1+x^2}}|+\frac12\arctan^2(x)$
So,
$$=\frac12 x^2\arctan^2(x)-(x\arctan(x)-\arctan^2(x)+\ln|\frac{1}{\sqrt{1+x^2}}|+\frac12\arctan^2(x))$$
After simplifying we get,
$$=\frac12 x^2\arctan^2(x)-x\arctan(x)+\frac12\arctan^2(x)-\ln|\frac{1}{\sqrt{1+x^2}}|$$
Now compute the boundaries and we get
$$=\frac{\ln(2)}{2}+\frac{\pi^2}{16}-\frac{\pi}{4}$$
$$\int_ {0}^{1}x(arctan(x))^2dx=\frac{\ln(2)}{2}+\frac{\pi^2}{16}-\frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2794560",
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"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$
If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$
My approach:
$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$
$$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$
$$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$
I can not proceed further, please help.
| Hint for your last equation:
$$\sin(2\theta)=2sin(\theta)\cos(\theta)$$
so $$\sin(2\theta)^2=4\sin^2(\theta)\cos^2(\theta)$$
and you can eliminate $\theta$
So you get $$4\left(1-\cos(\theta)^2\right)\cos(\theta)^2=\frac{4}{3}\frac{m^2-1}{m}$$
Solve this for $\cos(\theta)$
I know this, when you get $\theta$ then you can compute $\alpha$ with the equations above!
Expanding the term
$$\cos(\alpha-3\theta)\sin(\theta)^3-\sin(\alpha-3\theta)\cos(\theta)^3$$
we get
$$4\, \left( \sin \left( \theta \right) \right) ^{3}\cos \left( \alpha
\right) \left( \cos \left( \theta \right) \right) ^{3}-3\, \left(
\sin \left( \theta \right) \right) ^{3}\cos \left( \alpha \right)
\cos \left( \theta \right) +4\, \left( \sin \left( \theta \right)
\right) ^{4}\sin \left( \alpha \right) \left( \cos \left( \theta
\right) \right) ^{2}- \left( \sin \left( \theta \right) \right) ^{4
}\sin \left( \alpha \right) -4\, \left( \cos \left( \theta \right)
\right) ^{6}\sin \left( \alpha \right) +3\, \left( \cos \left( \theta
\right) \right) ^{4}\sin \left( \alpha \right) +4\, \left( \cos
\left( \theta \right) \right) ^{5}\cos \left( \alpha \right) \sin
\left( \theta \right) - \left( \cos \left( \theta \right) \right) ^{
3}\cos \left( \alpha \right) \sin \left( \theta \right)=0
$$
This term can we solve for $$\alpha$$
$$\alpha=-\arctan \left( 3\,{\frac {\cos \left( \theta \right) \sin \left(
\theta \right) \left( 2\, \left( \sin \left( \theta \right) \right)
^{2}-1 \right) }{6\, \left( \sin \left( \theta \right) \right) ^{4}-6
\, \left( \sin \left( \theta \right) \right) ^{2}+1}} \right)
$$ and now you can use your $$\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Divisibility of Sum of Equally Spaced Binomial Coefficients According to a numerical calculation I did for small values of $k$, it appears that the following is true.
$$4|\left[\sum_{j=1}^{n-1}\binom{3n}{3j}\right]$$ or $$\sum_{j=1}^{n-1}\binom{3n}{3j}=4p, p\in\mathbb{Z}$$
Ex.
If $n=2, \binom{6}{3}=20=4\cdot 5$
If $n=3, \binom{9}{3}+\binom{9}{6}=2\cdot 84=168=4\cdot 42$
If $n=4, \binom{12}{3}+\binom{12}{6}+\binom{12}{9}=2\cdot 220+924=1364=4\cdot341$
If $n=5, \binom{15}{3}+\binom{15}{6}+\binom{15}{9}+\binom{15}{12}=2\cdot 455+2\cdot 5005=10920=4\cdot 2730$
Is there a way to prove this? Using induction, as above I've shown the base case is true. Then if we assume that
$$S_m=\sum_{j=1}^{m-1}\binom{3m}{3j}=4q, q\in\mathbb{Z}$$
Then
$$S_{m+1}=\sum_{j=1}^{m}\binom{3m+3}{3j}=?$$
And I have no idea how to go forward. Perhaps its not true? Is there a counterexample?
| Let $$S=\sum_{j=1}^{n-1}\binom {3n}{3j}$$
Adding terms for $j=0$ and $j=n$ gives
$$\begin{align}
S+2
&=\sum_{j=0}^n \binom {3n}{3j}\\
&=\sum_{r=0}^{3n}\binom {3n}r-\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}+\binom {3n}{3j+2}\right]\\
&=2^{3n}+2\Re\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}e^{i2\pi/3}+\binom {3n}{3j+2}e^{i4\pi/3}\right]\\
&=8^n+2\Re\left[\sum_{j=0}^{n-1}\binom {3n}{3j+1}e^{i2\pi(3j+1)/3}+\binom {3n}{3j+2}e^{i2\pi(3j+2)/3}\right]\\
&=8^n+2\Re\left[\underbrace{\binom {3n}{3n}+\sum_{j=0}^{n-1}\sum_{k=1}^3\binom {3n}{3j+k}e^{i2\pi(3j+k)/3}}-\sum_{j=0}^{n}\binom {3n}{3j}\right]\\
&=8^n+2\Re\left[\qquad\ \ \quad\overbrace{\sum_{r=0}^{3n}\binom {3n}r e^{i2\pi r/3}}\qquad \qquad -\left(S+2\right)\right]\\
&=8^n+2\Re\big[\big(1+e^{i2\pi /3}\big)^{3n}-\left(S+2\right)\big]\\
&=8^n+2\Re\big[\big(e^{i\pi /3}\big)^{3n}\big]-2\big(S+2\big)\\
&=8^n+2\Re \big[e^{i\pi n}\big]-2\big(S+2\big)\\
&=8^n+2(-1)^n-2\big(S+2\big)\\
3\big(S+2\big)
&=2(-1)^n+8^n\\
3S&=2(-1)^n+8^n-6\\
&\equiv 2(-1)^n+0+2\mod 4\\
&\equiv 2\pm 2\mod 4\\
&\equiv 0\mod 4\\
S=\sum_{j=1}^{n-1}\binom {3n}{3j}&\equiv 0\mod 4\\
\Longrightarrow 4\bigg|&\sum_{j=1}^{n-1}\binom {3n}{3j}\; \blacksquare\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Suppose we are dealt five cards from an ordinary 52-card deck. What is the probability that Suppose we are dealt five cards from an ordinary 52-card deck. What is the
probability that
(c) we get no pairs (i.e., all five cards are different values)?
(d) we get a full house (i.e., three cards of a kind, plus a different pair)?
Solution:
(c) The number of ways this can happen is equal to $\frac{52 \cdot 48 \cdot44 \cdot 40 \cdot 36}{5!}$. Therefore $1317888/{52\choose5}$.
(d) The number of ways this can happen is equal to $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$. Therefore the probability is $3744/{52\choose 5}$
I don't understand the solutions visually.
For (c) I don't understand why they do $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36$. I suppose they divide by $5!$ because there are five categories of cards?
For (d) I don't understand how they got $(13)(12) {4 \choose 3}{4 \choose 2} = 3744$.
|
Suppose we are dealt five cards from an ordinary $52$-card deck. What is the probability that we get no pairs?
We must select cards from five of the thirteen ranks. For each selected rank, we must select one of the four suits. Hence, the number of favorable cases is
$$\binom{13}{5}4^5$$
Since there are $\binom{52}{5}$ ways to select five of the $52$ cards in the deck,
$$\Pr(\text{five cards of different ranks}) = \frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
As for the given solution: The first card that is selected can be any of the $52$ cards in the deck. Since the second card that is selected must be of a different rank, it can be selected in $48$ ways. Since the third card that is selected must be of a different rank than each of the first two cards, it can be selected in $44$ ways. Continuing in this way, we get $52 \cdot 48 \cdot 44 \cdot 40 \cdot 36$ ordered selections of five cards of different ranks. However, the order of selection does not matter, so we must divide by the $5!$ orders in which the same five cards could be selected, so the number of favorable cases is
$$\frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$
Dividing by $\binom{52}{5}$ gives the probability that each card is of a different rank.
You should check that
$$\binom{13}{5}4^5 = \frac{52 \cdot 48 \cdot 44 \cdot 44 \cdot 36}{5!}$$
Suppose we are dealt five cards from an ordinary $52$-card deck. What is the probability that we get a full house?
There are $13$ ways to select the rank from which three cards are selected and $\binom{4}{3}$ ways to select three of the four cards of that rank. There are $12$ ways to select the rank from which two cards are selected and $\binom{4}{2}$ ways to select two cards of that rank. Hence, the number of favorable cases is
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$
Since there are $\binom{52}{5}$ ways to select five cards from the deck,
$$\Pr(\text{full house}) = \frac{\dbinom{13}{1}\dbinom{4}{3}\dbinom{12}{1}\dbinom{4}{2}}{\dbinom{52}{5}}$$
| {
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} |
Exponential equation with double radical I'm trying ti solve this exponential equation:
$(\sqrt{2+\sqrt{3}})^x+(\sqrt{2-\sqrt{3}})^x=2^x$.
Here my try:
$\sqrt{2+\sqrt{3}}=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ and
$\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$
So i get this relation:
$\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}}$
Using the substitution $t=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}$ the equation can be written as
$(t)^x+(\frac{1}{t})^x=2^x$.
And from this i wrote:
$t^{2x}-(2t)^x-1=0$.
Now i don't how to proceed. Any suggestions?
Thanks!
| Hint:
Write $s= \sqrt{2+\sqrt{3}}$, then $\sqrt{2-\sqrt{3}} = {1\over s}$
Then $$s^x+({1\over s})^x\geq 2 \implies x\geq 1$$
| {
"language": "en",
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"answer_count": 3,
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} |
Evaluating $\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$ I want to evaluate $$\int_0^1\frac{\ln^2(1+x^2)}{x^4}dx$$
My attempt: Letting
$$I(\alpha,\beta)=\int_0^1\frac{\ln(1+\alpha^2x^2)\ln(1+\beta^2x^2)}{x^4}dx$$
$$
\begin{aligned}
I_{12}''(\alpha,\beta)&=\int_0^1\frac{4\alpha\beta}{(1+\alpha^2x^2)(1+\beta^2x^2)}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}\int_0^1\frac{\alpha^2}{1+\alpha^2x^2}-\frac{\beta^2}{1+\beta^2x^2}dx\\
&=\frac{4\alpha\beta}{\alpha^2-\beta^2}(\alpha\arctan\alpha-\beta\arctan\beta)
\end{aligned}
$$
$$
I=\int_0^1\int_0^1I_{12}''(\alpha,\beta)d\beta d\alpha
$$
But I can't go further.
| Taking integration by parts,
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= -\frac{1}{3}\log^2 2 + \frac{4}{3}\int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx. $$
Now
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2(1+x^2)} \, dx
= \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx - \int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx, $$
and the first integral is easily computed by integration by parts:
$$ \int_{0}^{1} \frac{\log(1+x^2)}{x^2} \, dx
= -\log 2 + \frac{\pi}{2}. $$
The second integral is trickier, and plugging $x=\tan\theta$ and utilizing the expansion
\begin{align*}
\log \sec\theta
= -\log \left| \frac{e^{i\theta} + e^{-i\theta}}{2} \right|
&= \log 2 - \operatorname{Re}\log(1+e^{2i\theta}) \\
&= \log 2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\cos(2n\theta),
\end{align*}
we have
\begin{align*}
\int_{0}^{1} \frac{\log(1+x^2)}{1+x^2} \, dx
&= 2 \int_{0}^{\frac{\pi}{4}} \log \sec\theta \, d\theta \\
&= \frac{\pi}{2}\log 2 - \underbrace{ \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} }_{=G,}
\end{align*}
where $G$ is Catalan's constant. Combining altogether, we obtain
$$ \int_{0}^{1} \frac{\log^2(1+x^2)}{x^4} \, dx
= \frac{1}{3}\left( 4G - \log^2 2 - 4\log2 - 2\pi\log2 + 2\pi \right). $$
Remark. Of course, some CAS can deal with this integral. For instance, Mathematica 11 yields
| {
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"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Which integers can be written as $x^2+2y^2-3z^2\ $?
For which integers $n$ has the diophantine equation $$x^2+2y^2-3z^2=n$$ solutions ?
These theorems
https://en.wikipedia.org/wiki/15_and_290_theorems
do not apply because the given quadratic form is not positive (or negative) definite. It seems that the quadratic form is universal (for every integer $n$ a solution exists) , but I have no idea how this can be proven.
| We claim that any $n\in\mathbb{Z}$ can be written as $x^2+2y^2-3z^2$.
Note that any such integer $n$ is $0$ or it is equal to a $4^a\cdot 2^b\cdot d$ where $a$ is a non negative integer, $b\in\{0,1\}$ and $d$ is a signed odd number. Then we consider the following cases.
0) If $n=0$ then let $x=0$, $y=0$ and $z=0$.
1) If $n=d=2k+1$ then let $x=k+1$, $y=k$ and $z=k$:
$$x^2+2y^2-3z^2=(k+1)^2-k^2=2k+1=n.$$
2) If $n=2d=2(2k+1)$ then let $x=k$, $y=k+1$ and $z=k$:
$$x^2+2y^2-3z^2=2(k+1)^2-2k^2=4k+2=n.$$
3) If $n=4^ak$ and $k$ can be written as $x_k^2+2y_k^2-3z_k^2$ then
let $x=2^ax_k$, $y=2^ay_k$ and $z=2^az_k$:
$$x^2+2y^2-3z^2=4^a(x_k^2+2y_k^2-3z_k^2)=4^ak=n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How to show the double sum identity $\sum ^{n}_{i=1}\sum ^{i}_{j=1}i-j$ = $\dfrac {1}{6}n\left( n-1\right) \left( n+1\right) $ I am not sure how to go about showing this: $\sum ^{n}_{i=1}\sum ^{i}_{j=1}i-j$ = $\dfrac {1}{6}n\left( n-1\right) \left( n+1\right) $
It is a bit like the formula for $\sum ^{n}_{i=1}i^{2}$
and this $\sum_{i=1}^n \sum_{j=1}^i \frac{i-j}{nm} + \sum_{i=1}^{n} \sum_{j=i}^m \frac{j-i}{nm} = \frac{2 n^2 - 3 n m + 3 m^2 - 2}{6m}$
As I really do not know how to proceed.
| I would suggest an inductive proof to that identity. Assume the induction hipotesis
$$
\sum^{n}_{i=1}\sum^{i}_{j=1}i-j= \dfrac{1}{6}n\left( n-1\right) \left( n+1\right).
$$
It's easy to verify identity for $n=1$, $n=2$ and $n=3$. Consider the scheme.
$$
\begin{array}{rl}
\sum^{n}_{i=1}\sum^{i}_{j=1}i-j=&0
\\
+&0+1
\\
+&0+1+2
\\
+&0+1+2+3
\\
+&0+1+2+3+4
\\
&\vdots \;\;\;\,\vdots \;\;\;\,\vdots \;\;\;\, \vdots \;\;\;\, \vdots \;\;\;\, \ddots
\\
+&0+1+2+3+4+\cdots +i
\\
&\vdots \;\;\;\,\vdots \;\;\;\,\vdots \;\;\;\, \vdots \;\;\;\,\vdots \;\;\;\, \quad \;\;\;\, \vdots \;\;\;\,\ddots
\\
+&0+1+2+3+4+\cdots +i+\cdots+(n-1)
\end{array}
$$
Note by scheme that
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \left[\sum^{n}_{i=1}\sum^{i}_{j=1}i-j\right]+\big[ 1+2+3+\ldots +n\big]
\end{align}
By induction hipotesis we have
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \left[\dfrac{1}{6}\left( n-1\right)n\left( n+1\right)\right]+\big[ 1+2+3+\ldots +n\big]
\end{align}
And since we know that $\dfrac{1}{2}n(n+1)$ is the result of the sum $1+2+3+\ldots +n$ we have
\begin{align}
\sum^{n+1}_{i=1}\sum^{i}_{j=1}i-j &= \dfrac{1}{6}( n-1)n(n+1)+ \dfrac{1}{2}n(n+1)
\\
&= \dfrac{1}{6}(n-1)n(n+1)+ \dfrac{3}{6}n(n+1)
\\
&= \Big[(n-1)+ 3\Big]\dfrac{1}{6}n( n+1)
\\
&= \dfrac{1}{6}n\cdot (n+1)(n+2)
\\
&= \dfrac{1}{6}[\color{red}{(n+1)}-1]\cdot [\color{red}{(n+1)}][\color{red}{(n+1)}+1]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Convex sum order If I have a strictly convex function $f(x)$ with $f''(x)>0$
and if I know that for some $a\le b \le c$ and $x \le y \le z$ I have
$$a+b+c = x+y+z$$
$$f(a)+f(b)+f(c)=f(x)+f(y)+f(z)$$
can I conclude that at least one of the following must be true about the order?
$$ a \le x \le y \le b \le c \le z$$
$$ x \le a \le b \le y \le z \le c$$
Perhaps an equivalent question is to consider coincident centroids of two triangles with vertices on the convex curve, looking something like this diagram with pairs of vertices on the inside of the order
though I realise that I cannot push this analogy too far as it would not be necessarily true if say the curve was a circle and the triangles equilateral
| In geometry a triangle $(ABC)$ on the plane $xOy$ with coordinates $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ has its gravity center (baricenter) the point
$$
G=\left(\frac{a_1+b_1+c_1}{3},\frac{a_2+b_2+c_2}{3}\right).
$$
Hence what you ask is:
Given a curve (function) $c_f:y=f(x)$ or $M=(x,f(x))$, with positive curvature, find all triangles $(ABC)$ inscribed in $c_f$ such that they have the same gravity center.
In geometry, the gravity center of (ABC) is defined as the point of intersection of all the dimedians i.e.
In $A$ edge the dimedian is the sigment $AM_1$ with $M_1$ being the median of $BC$ side. In $B$ edge the coresponding dimedian is $BM_2$, where $M_2$ is the median of $AC$ and in $C$ edge the sigment $CM_3$.
For a given baricenter there are finite number of triangles inscribed in $c_f$ with the same baricenter $G=\left(\frac{a+b+c}{3},\frac{f(a)+f(b)+f(c)}{3}\right)$.
A theorem of Leibniz says
Theorem.
If $M$ is an arbitrary point of plane of the triangle $(ABC)$ and $G$ is the baricenter, then
$$
MA^2+MB^2+MC^2=3MG^2+\frac{1}{3}(AB^2+BC^2+CD^2)
$$
Hence your broblem reduces to find all points $A=(a,f(a))$, $B=(b,f(b))$, $C=(c,f(c))$ of curve $y=f(x)$ such
$$
GA^2+GB^2+GC^2=\frac{1}{3}(AB^2+BC^2+CA^2)
$$
or equivalently for given $f$ find all $a,b,c$ such
$$
(g_1-a)^2+(g_2-f(a))^2+(g_1-b)^2+(g_2-f(b))^2+(g_1-c)^2+(g_2-f(c))^2=\frac{1}{3}\left((a-b)^2+(f(a)-f(b))^2+(b-c)^2+(f(b)-f(c))^2+(c-a)^2+(f(c)-f(a))^2\right),\tag 1
$$
where
$$
g_1=\frac{a+b+c}{3}\textrm{ and }g_2=\frac{f(a)+f(b)+f(c)}{3}.\tag 2
$$
Hence we have three equations with three unknowns $a,b,c$. Hence the number of points will be finite.
For example assume the parabola $y=x^2$. Then equations $(2)$ become
$$
a=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
$$
b=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
or
$$
a=\frac{1}{2}\left(-c+3g_1+\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
$$
b=\frac{1}{2}\left(-c+3g_1-\sqrt{3}\sqrt{-c^2+2cg_1-3g_1^2+2g_2}\right)
$$
The value of $c$ is given from
$$
\frac{4 a^3}{3}-\frac{2 a^2 b^2}{3}-\frac{2 a^2 c^2}{3}-\frac{2 a b}{3}-\frac{2 a c}{3}+\frac{4 b^3}{3}-\frac{2 b^2 c^2}{3}-\frac{2 b c}{3}+\frac{4 c^3}{3}=0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Someone's Inequality ? $(a + b)^2 \le 2(a^2 + b^2) $ For real $a, b$ then $(a + b)^2 \le 2(a^2 + b^2) $
This fairly trivial inequality crops up a lot in my reading on (Lebesgue) integration, is it named after someone ? It extends rather obviously for positive reals to $a^2 + b^2 \le (a + b)^2 \le 2(a^2 + b^2) $.
Proof (if you need it):
$0 \le (a - b)^2 = a^2 + b^2 -2ab \implies 2ab \le a^2 + b^2$
$(a + b)^2 = a^2 + b^2 + 2ab$ which by previous $ \le 2(a^2 + b^2) $.
Application:
If $f, g$ are positive functions then $(f + g)^2$ is integrable $\iff$ $f^2, g^2$ are integrable since $f^2 + g^2 \le (f + g)^2 \le 2(f^2 + g^2) $ pointwise.
| This inequality can as weel be seen as a particular case of the equivalence between $p$-norms in $\Bbb R^n$. Indeed for $1<p\leq q<\infty$, it holds
$$\|x\|_q \leq \|x\|_p \leq n^{1/p-1/q}\|x\|_q$$
In the particular case $n=2$, $p=1$ and $q=2$ we get
$$(|a|+|b|)=\|(a,b)\|_1\leq 2^{1-1/2}\|(a,b)\|_2 =\sqrt{2(a^2+b^2)},$$
which is even slightly tighter as $|a+b| \leq |a|+|b|$.
I should nevertheless point out that the equivalence between $p$-norms is proved using the Cauchy-Schwarz (or more generally the Hölder) inequality mentioned by @Ihf
| {
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"question_score": "3",
"answer_count": 3,
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} |
Simplify $\arccos{\sqrt{ 2\over 3}}−\arccos{\frac{\sqrt6+1}{2\sqrt3}}$
Prove $$\arccos{\sqrt{ 2\over 3}}−\arccos{\dfrac{\sqrt6+1}{2\sqrt3}} = \dfrac\pi6$$
How to proceed with this question? I have tried changing them to $\arctan$ and applying $\arctan a- \arctan b$ but ended up getting some numbers which cant be simplied further.
| Let $$\alpha = \arccos{\sqrt2\over \sqrt3}\;\;\;\;{\rm and}\;\;\;\;\beta =\arccos{\sqrt6+1\over 2\sqrt3}$$
so $$ \cos \alpha = \sqrt{2\over 3}\;\;\;\;{\rm and}\;\;\;\;\cos \beta = {\sqrt{6}+1\over 2\sqrt{3}}$$
and
$$ \sin \alpha = \sqrt{1-{2\over 3}} ={1\over \sqrt{3}}\;\;\;\;{\rm and}\;\;\;\;\sin \beta = \sqrt{1-{7+2\sqrt{6}\over 12}} = \sqrt{{5-2\sqrt{6}\over 12}}$$
so $$\cos (\alpha -\beta) = \sqrt{2\over 3} {\sqrt{6}+1\over 2\sqrt{3}}+{1\over \sqrt{3}}\sqrt{{5-2\sqrt{6}\over 12}}=$$
$$ = {\sqrt{12}+\sqrt{2}+\sqrt{5-2\sqrt{6}}\over 6} $$
$$ = {2\sqrt{3}+\sqrt{2}+\sqrt{(\sqrt{3}-\sqrt{2})^2}\over 6} $$
$$ = {\sqrt{3}\over 2} $$
so $\alpha -\beta = \pi/6$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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} |
equation of a plane in space though 3 points We have the points $M(1,2,3),N(2,1,5), P(4,3,2)$ is it a correct way to find the equation of the plane using the determinant $\Delta$
$$\Delta=\begin{array}{|cccc|}
x & y & z & 1\\
1& 2 & 3 & 1\\
2 & 1 & 5 & 1\\
4 & 3 & 2 & 1
\end{array}=0$$
If so, what is the intuition behind this method of finding equation of a plane?
Thanks in advance!
| $$\Delta=\begin{array}{|cccc|}
x & y & z & 1\\
1& 2 & 3 & 1\\
2 & 1 & 5 & 1\\
4 & 3 & 2 & 1
\end{array}=x\begin{array}{|ccc|}2&3&1\\1&5&1\\3&2&1\end{array}-y\begin{array}{|ccc|}1&3&1\\2&5&1\\4&2&1\end{array}+z\begin{array}{|ccc|}1&2&1\\2&1&1\\4&3&1\end{array}-\begin{array}{|ccc|}1&2&3\\2&1&5\\4&3&2\end{array}=0$$
Compare this with $P:Ax+By+Cz=D$. So each of the above 3x3 determinant is a coefficient of the equation of the plane.
Since you're given $M(1,2,3)$, $N(2,1,5)$ and $P(4,3,2)$, you can easily find out that $\vec{MN}=(1,-1,2)$ and $\vec{MP}=(3,1,-1)$. The normal vector to the plane is then $\vec{n}=\vec{MN}\times \vec{MP}=(-1,7,4)$. Define an arbitrary point $K(x,y,z)$ that is on the plane.
Expand $\Delta$: $$\Delta=x[2(5-2)-3(1-3)+1(2-3\cdot 5)]-y[(5-2)-3(2-4)+(4-20)]+z[(1-3)-2(2-4)+(6-4)]-[(2-15)-2(4-20)+3(6-4)]=0\\\implies-x+7y+4z-25=0\\\iff -1(x-1)+7(y-2)+4(z-3)=0\\\iff \vec{n}\,\bullet\vec{MK}=0.$$
This means the normal vector $\vec{n}$ defined by the cross product of two vectors from 3 points is orthogonal to arbitrary vectors on the plane. Calculation of $\Delta$ addresses these reasoning.
| {
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What is $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac1k\right) / \left(\sum_{k=0}^n \frac1{2k+1}\right)$? I have the following problem:
Evaluate
$$ \lim_{n\to\infty}{{1+\frac12+\frac13 +\frac14+\ldots+\frac1n}\over{1+\frac13 +\frac15+\frac17+\ldots+\frac1{2n+1}}} $$
I tried making it into two sums, and tried to make it somehow into an integral, but couldn't find an integral.
The sums I came up with,
$$ \lim_{n\to\infty} { \sum_{k=1}^n {\frac1k} \over {\sum_{k=0}^n {\frac{1}{2k+1}}}} $$
| Hint Denote the $n$th harmonic number by $$H_n := 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}.$$
Then, the numerator of the given ratio is $H_n$, and the denominator can be written as
\begin{align*}
1 + \tfrac{1}{3} + \tfrac{1}{5} + \cdots + \tfrac{1}{2 n + 1}
&= \left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{2 n}\right) - \left(\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \cdots + \tfrac{1}{2 n}\right) + \tfrac{1}{2 n + 1} \\
&= \left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{2 n}\right) - \tfrac{1}{2}\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{n}\right) + \tfrac{1}{2 n + 1} \\
&= H_{2 n} - \tfrac{1}{2} H_{n} + \frac{1}{2 n + 1} .
\end{align*}
Now, using appropriate Riemann sum estimates gives that $$H_n = \log n + O(1).$$
Additional hint So, the denominator is $$\log (2 n) - \tfrac{1}{2} \log n + O(1) = \tfrac{1}{2} \log n + O(1),$$ and so the ratio is $$\frac{\log n}{\tfrac{1}{2} \log n} + O((\log n)^{-1}) = 2 + O((\log n)^{-1}) .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
$f \in \mathrm{End} (\mathbb{C^2})$ $f(e_1)=e_1+e_2$ $f(e_2)=e_2-e_1$. Eigenvalues of f and the bases of the associated eigenspaces Let $f \in \mathrm{End} (\mathbb{C^2})$ be defined by its image on the standard basis $(e_1,e_2)$:
$f(e_1)=e_1+e_2$
$f(e_2)=e_2-e_1$
I want to determine all eigenvalues of f and the bases of the associated eigenspaces.
First of all how does the transformation matrix of $f$ look like?
Is it
$\begin{pmatrix}1 &-1 \\1 &1 \end{pmatrix}$?
| If one represents the standard basis $e_1$, $e_2$ in the usual form
$e_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \tag 1$
$e_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \tag 2$
and writes the matrix of $f$ as
$[f] = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}, \tag 3$
then we have, since
$f(e_1) = e_1 + e_2, \; f(e_2) = e_2 - e_1, \tag 4$
$\begin{pmatrix} 1 \\ 1 \end{pmatrix} = e_1 + e_2 = [f]e_1 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \alpha \\ \gamma \end{pmatrix}, \tag 5$
and
$\begin{pmatrix} -1 \\ 1 \end{pmatrix} = e_2 - e_1 = [f]e_2 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \beta \\ \delta \end{pmatrix}, \tag 6$
from which it immediately follows that
$\alpha = \gamma = \delta = 1, \tag 7$
$\beta = -1; \tag 8$
thus
$[f] = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}, \tag 9$
as anticipated by our OP user567319. Once we have (9), it is an easy matter to find the eigenvalues of $f$, sincd they must satisfy
$0 = \det([f] - \lambda I) = \det \left ( \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} - \lambda I \right ) = \det \left (\begin{bmatrix} 1 - \lambda & -1 \\ 1 & 1 - \lambda \end{bmatrix} \right )$
$= (1 - \lambda)^2 + 1 = \lambda^2 - 2\lambda + 2; \tag{10}$
it follows from (10), using the quadratic formula, that
$\lambda = \dfrac{1}{2}(2 \pm \sqrt{-4}) = \dfrac{1}{2}(2 \pm 2 i) = 1 \pm i; \tag{11}$
it is now an easy matter to find the eigenvectors, satisfying as they do
$\begin{pmatrix} \lambda \mu \\ \lambda \nu \end{pmatrix} = \lambda \begin{pmatrix} \mu \\ \nu \end{pmatrix} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{pmatrix} \mu \\ \nu \end{pmatrix} = \begin{pmatrix} \mu - \nu \\ \mu + \nu \end{pmatrix}, \tag{12}$
whence
$\lambda \mu = \mu - \nu, \tag{13}$
$\lambda \nu = \mu + \nu; \tag{14}$
from (13),
$(1 - \lambda) \mu = \nu; \tag{15}$
it follows that, taking $\mu = 1$, the eigenvectors are
$\begin{pmatrix} 1 \\ \mp i \end{pmatrix} = \begin{pmatrix} 1 \\ 1 - \lambda \end{pmatrix}, \; \lambda = 1 \pm i; \tag{16}$
we check the case $\lambda = 1 + i$:
$\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{pmatrix} 1 \\ -i \end{pmatrix} = \begin{pmatrix} 1 + i \\ 1 - i \end{pmatrix} = (1 + i)\begin{pmatrix} 1 \\ -i \end{pmatrix}; \tag{17}$
a check of the case $\lambda = 1 - i$ follows from this by complex conjugation, since $[f]$ is a real matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$ $a, b, c ∈ \mathbb{R}+.$
WLOG assume $a \leq b \leq c.$ I tried substitution: $x=\frac{1} {1+a^2}, y=\frac{1} {1+b^2}, z=\frac{1} {1+c^2},$ so $x \geq y \geq z$ and $(1-x)+(1-y)+(1-z)=2 \to x+y+z=1.$
We want to prove $ax+by+cz \leq \sqrt{2}.$ This somewhat looks like Cauchy-Schwarz so I tried that: $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq (ax+by+cz)^2.$ The problem becomes $(a^2+b^2+c^2)(x^2+y^2+z^2) \geq 2,$ since $a,b,c,x,y,z>0.$
Expressing $a,b,c$ in terms of $x,y,z$: $(\frac {1}{x} + \frac {1}{y} + \frac {1}{z} - 3)(x^2+y^2+z^2)$
$= x+y+z+\frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 2.$
$\to \frac{y^2}{x}+\frac{z^2}{x}+\frac{x^2}{y}+\frac{z^2}{y}+\frac{x^2}{z}+\frac{y^2}{z}-3(x^2+y^2+z^2) \geq 1.$ Stuck here. Thinking about using AM-GM but not sure how. Help would be greatly appreciated.
| Let $$B:=\frac{1} {1+a^2} + \frac{1} {1+b^2} + \frac{1} {1+c^2}$$
From: $$A:=\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2$$
we get $A+B =3$ so $B =1$.
Now by Cauchy inequality we have $$A\cdot B \geq \big(\underbrace{\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2}}_{C}\big)^2$$
So we have $C^2\leq 2$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2822937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
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} |
How do I apply integrating factor to solve this differential equation? $$x \frac{dy}{dx} + y = -2x^6y^4$$
I tried to find the general solution by dividing both sides by $x$ or $x^6$ but no solution I could get.. Do I even solve it with integrating factor?
| Contrary to other answers, you CAN find an integrating factor and manipulate your ODE via substitutions.
We have the ODE :
$$xy' + y = -2x^6y^4$$
Dividing both sides by $-\frac{1}{3}xy^4$, we yield :
$$-\frac{3y'}{y^4} - \frac{3}{xy^3} = 6x^5$$
Let $v(x) = \frac{1}{y^3(x)}$ and then this gives $v'(x) = -\frac{3y'(x)}{y^4(x)} $ and then the differential equation becomes :
$$v'(x) - \frac{3v(x)}{x} = 6x^5$$
Let $μ(x) = e^{\int -\frac{3}{x}\rm d x} = \frac{1}{x^3}$ and then multiply both sides by $μ(x)$ :
$$\frac{v'(x)}{x^3}-\frac{3v(x)}{x^4}=6x^2$$
By substituting $-\frac{3}{x^4}=\big(\frac{1}{x^3}\big)'$ we have :
$$\frac{v'(x)}{x^3}-\bigg(\frac{1}{x^3}\bigg)'v(x)=6x^2$$
Now, we shall apply the reverse product rule : $f\frac{\rm d g}{\rm d x}+g\frac{\rm d f}{\rm dx} = \frac{\rm{d}}{\rm d x}(f\;g)$ :
$$\int \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{v(x)}{x^3}\bigg)\mathrm{d}x=\int6x^2\mathrm{d}x \implies v(x) = x^3(2x^3+c_1)$$
Now, you can substitute $v(x) = \frac{1}{y^3(x)}$ and solve for $y(x)$ to yield the final result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2823820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx = \frac 14 + \frac {\pi}8$
Find for how many values of $n$. $I_n=\int_0^1 \frac {1}{(1+x^2)^n} \, dx=\frac 14 + \frac {\pi}8$
My attempt (integration by parts):
\begin{align}
I_n & = \int_0^1 \frac 1{(1+x^2)^n}\,dx = \left. \frac {x}{(1+x^2)^n} \right|_0^1+2n\int_0^1 \frac {x^2+1-1}{(1+x^2)^{n+1}}\,dx \\[10pt]
& =\frac 1{2^n}+2n \times I_n-2n\times I_{n+1}\implies I_{n+1}= \frac 1{2^{n+1}n}+\frac {2n-1}{2n}I_n.
\end{align}
where $I_1=\frac {\pi}{4}.$
Finding out that $I_2=\frac {\pi}{8}+\frac 14.$ But how do I find out that this is the only solution?
| Note that the sequence $(I_n)$ is decreasing, hence there is at most one value of $n$ such that $I_n=\frac{1}{4}+\frac{\pi}{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ Solve $\cos x +\cos y - \cos(x+y)=\frac 3 2$ where $x,y\in [0,\pi]$.
I am trying to solve this but I am stuck. I know that $x=y=\pi/3$ is a solution but how do I show this is the only one? I think there are no others! Hints would be appreciated
| \begin{align}
\cos x +\cos y - \cos(x+y) &= \frac 32 \\
\cos x + \cos y - \cos x \cos y + \sin x \sin y
&= \frac 32 \\
(1 - \cos y)\cos x + \sin y \sin x &= \frac 32 - \cos y \\
\end{align}
Note that
$$\sqrt{(1-\cos y)^2 + (\sin y)^2} = \sqrt{1 - 2\cos y + \cos^2y + \sin^2 y}
= \sqrt{2(1 - \cos y)}
= 2\sin \dfrac y2$$
$$\dfrac{1-\cos y}{2 \sin \dfrac y2} = \sin \dfrac y2$$
$$ \dfrac{\sin y}{2 \sin \dfrac y2}
= \cos \dfrac y2$$
So
\begin{align}
\sin \dfrac y2\cos x + \cos \dfrac y2 \sin x
&= \frac{\frac 32 - \cos y}{2 \sin \dfrac y2} \\
\sin \left(x + \dfrac y2 \right)
&= \frac{3 - 2\cos y}{4 \sin \dfrac y2} \\
\sin \left(x + \dfrac y2 \right)
&= \frac{1 + 4\sin^2 \dfrac y2}{4 \sin \dfrac y2} \\
\sin \left(x + \dfrac y2 \right)
&= \sin \dfrac y2 + \frac 14 \csc \dfrac y2\\
\end{align}
We can show that, for $y \in [0, \pi]$, the minimum acceptable value of
$\sin \dfrac y2 + \frac 14 \csc \dfrac y2$ is $1$. The rest is pretty straight forward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2827457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that $n^4-4n^2$ is divisible by 3 for all integers $n\geq1$. For the induction case, we should show that $(n+1)^4-4(n+1)^2$ is also divisible by 3 assuming that 3 divides $n^4-4n^2$. So,
$$ \begin{align} (n+1)^4-4(n+1)&=(n^4+4n^3+6n^2+4n+1)-4(n^2+2n+1)
\\ &=n^4+4n^3+2n^2-4n-3
\\ &=n^4+2n^2+(-6n^2+6n^2)+4n^3-4n-3
\\ &=(n^4-4n^2) + (4n^3+6n^2-4n)-3
\end{align}$$
Now $(n^4-4n^2)$ is divisible by 3, and $-3$ is divisible by 3. Now I am stuck on what to do to the remaining expression.
So, how to show that $4n^3+6n^2-4n$ should be divisible by 3? Or is there a better way to prove the statement in the title? Thank you!
| Because $$n^4-4n^2=n^4-n^2-3n^2=n(n-1)n(n+1)-3n^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2828422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
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An interesting proof that $\sin^2(x) + \cos^2(x) = 1$ (using only series, no trigonometry). This question concerns an interesting proof of the fact that $\sin^2(x) + \cos^2(x) = 1$, but only using the series that defines them, not any trigonometry. So define
$$
s(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots
$$
and
$$
c(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots
$$
Step 1: we prove that $s' = c$ and $c' = -s$. This can be done by differentiating the series componentwise:
$$
s'(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots = c(x),
$$
and
$$
c'(x) = - x + \frac{x^3}{3!} - \frac{x^5}{5!} + \ldots = -s(x).
$$
Step 2: we prove that $(s^2+c^2)' = 0$. Using the chain-rule on both terms and then using our result of step 1 we compute:
$$
(s^2+c^2)' = 2s \cdot s' + 2c \cdot c' = 2sc+2c(-s) = 0.
$$
Step 3: we prove that $s^2 + c^2 = 1$. The idea here is to use step 2, to obtain something like
$$
s^2 + c^2 = \int (s^2 + c^2)'\,dx = \int 0 \, dx = 1.
$$
However, I cannot figure out the details of this last step. In particular, as far as I know, $\int (s^2 + c^2)'\,dx = s^2+c^2 + C$, and $\int 0\,dx = C'$. What happens with these constants?
| Your last step is unnecessarily complicated. In Step 2, you show that
$$ (s^2 + c^2)' = 0 \implies (s^2 + c^2)(x) = C, $$
where $C$ is some constant. In particular,
$$ (s^2 + c^2)(0) = C. $$
But, directly from the power series definitions of $s$ and $c$, we have
$$ s(0) = \frac{0}{1!} - \frac{0^3}{3!} + \frac{0^5}{5!} + \dotsb = 0
\quad\text{and}\quad
c(0) = 1 - \frac{0^2}{2!} + \frac{0^4}{4!} + \dotsb = 1.$$
Therefore
$$ (s^2 + c^2)(x)
= (s^2 + c^2)(0)
= s(0)^2 + c(0)^2
= 0^2 + 1^2
= 1. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2830475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Obtain variance from density of random variable X is a random variable with the density
$$F_X(x) = \begin{cases}2 x^{-2}, & x \in (1,2)\\
0 & \operatorname{otherwise}\end{cases}$$
How can I find the
$VAR(3X^2-5)$
Normal variation is computed from equation
$E(X^2)-(EX)^2 = VAR^2(X)$
So instead of multiplication my function by X i need to multiplicate by $3X^2-5$?
$E(3X^2-5)^2 =\int_1^2(3x^2-5)^2*2x^{-2} =15.5
$
$(EX)^2 =\int_1^2(3x^2-5)*2x^{-2} =1$
$VAR(3X^2-5)=\sqrt{14.5}$
Should it be like that?
| First, using the fact that $Var(aX)=a^2Var(X)$ and $Var(X+c)=Var(X)$, note that $Var(3X^2-5)=9Var(X^2)$ and $Var(X^2)=E(X^4)-(E(X^2))^2$.
$E(X^4)=\int_1^2 x^4 .2x^{-2}=14/3$, similarly $E(X^2)=2$ and so $Var(X^2)=2/3$
and $Var(3X^2-5)=6$
Your approach is also correct but you are making a calculation error, that is $Var(3X^2-5)=E((3X^2-5)^2) - (E(3X^2-5))^2$ where $E((3X^2-5)^2)=\int_1^2(3x^2-5)2x^{-2}=7$ and $E(3X^2-5)=1$ and so $Var(3X^2-5)=6$. .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does $(x+y)^m=x^m+y^m+z^m$ imply $(x+y+z)^m=(x+z)^m+(y+z)^m$? Let $x,y,z,m\in\mathbb{N}$, and $x,y,z,m>0$, and also $x>y$.
My problem is to understand if, under these sole hypotheses, we can prove that
$(x+y)^m=x^m+y^m+z^m \Longrightarrow (x+y+z)^m=(x+z)^m+(y+z)^m.$
If yes, how can we prove it?
If not, which other hypotheses are needed, in order to make the implication true?
EDIT: I am also interested in the softer versions of the statement, i.e. whether we can prove or not that
$(x+y)^m=x^m+y^m+z^m \Longrightarrow (x+y+z)^m\lessgtr (x+z)^m+(y+z)^m,$
and, if not, which additional conditions we need to make the statement(s) true.
| Expanding the given condition using binomial we get,
$$ z^m = {m\choose{1}} xy^{m-1} + {m\choose{2}} x^2y^{m-2} +...........+{m\choose{m-1}} x^{m-1}y $$
Now if we expand the claim, we get,
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = z^m + \Bigg( {m\choose{1}} xz^{m-1} + {m\choose{2}} x^2z^{m-2} +...........+{m\choose{m-1}} x^{m-1}z \Bigg) + \Bigg( {m\choose{1}} yz^{m-1} + {m\choose{2}} y^2z^{m-2} +...........+{m\choose{m-1}} y^{m-1}z \Bigg)
$$
where $a, b, c$ are natural numbers such that $a+b+c = m$ and $a,b,c \neq m$.
We already know the value of $z^m$ so we can substitute that,
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = \Bigg( {m\choose{1}} xy^{m-1} + {m\choose{2}} x^2y^{m-2} +...........+{m\choose{m-1}} x^{m-1}y \Bigg) + \Bigg( {m\choose{1}} xz^{m-1} + {m\choose{2}} x^2z^{m-2} +...........+{m\choose{m-1}} x^{m-1}z \Bigg) + \Bigg( {m\choose{1}} yz^{m-1} + {m\choose{2}} y^2z^{m-2} +...........+{m\choose{m-1}} y^{m-1}z \Bigg)
$$
We can simplify this to,
$$
\sum \frac{m!}{a!b!c!} x^ay^bz^c = \Bigg( {m\choose{1}} ( xy^{m-1} + yz^{m-1} + zx^{m-1} ) +
{m\choose{2}} (x^2y^{m-2} + y^2z^{m-2} + z^2x^{m-2})...........+{m\choose{m-1}}( x^{m-1}y^ + y^{m-1}z + z^{m-1}x ) \Bigg)
$$
I can't see how to prove that $LHS=RHS$ from here. In fact, since you are not sure whether your claim was true, this seems to indicate that the claim is false.
Can anyone try something from here?
Edit As the other answer says the claim is clearly contradicting FMT.
| {
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"url": "https://math.stackexchange.com/questions/2832185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.
I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.
| Given $m$ is the product of four consecutive integers.
$$m=p(p+1)(p+2)(p+3)$$where $p$ is an integer
we need to show that $p(p+1)(p+2)(p+3)+1$ is a perfect square
Now,$$p(p+1)(p+2)(p+3)+1=p(p+3)(p+1)(p+2)+1$$
$$=(p^2+3p)(p^2+3p+2)+1$$
$$=(p^2+3p+1)(p^2+3p+2)-(p^2+3p+2)+1$$
$$=(p^2+3p+1)(p^2+3p+1+1)-(p^2+3p+2)+1$$
$$=(p^2+3p+1)(p^2+3p+1)+(p^2+3p+1)-p^2-3p-2+1$$
$$=(p^2+3p+1)(p^2+3p+1)=(p^2+3p+1)^2$$
So, $m+1$ is a perfect square where $m$ is the product of four consecutive integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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} |
Suppose $b$ a random number in the interval $(-3,3)$ Which the probability of the equation $x^2+bx+1=0$ have a least a real root?
Suppose $b$ a random number in the interval $(-3,3)$ Which the probability of the equation $x^2+bx+1=0$ have a least a real root?
I don't know how to start this exercise. Can someone help me?
| Hint: The polynomial $f(x) = x^2+bx+1$ has a real root if there is an $x$ s.t. $f(x) \le 0$ is satisfied [make sure you can see why].
The minimum of $f(x)$ is achieved when $df(x)/dx = 2x+b$ is 0, at $x=\frac{-b}{2}$.
The values of $b \in (-3,3)$ s.t. the inequality $f(-\frac{b}{2}) = \frac{b^2}{4} - \frac{b^2}{2}+1 =-\frac{b^2}{4}+1 \le 0$ is satisfied
are $|b| \ge 2$.
Note that $|b| \ge 2$ holds for precisely $\frac{1}{3}$ of $b \in (-3,3)$, so the probability is $\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2834160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving $\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$
Prove for every $a, b, c \in\mathbb{R^+}$, given that $abc=1$ :
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3(a+b+c+1)$$
I tried using AM-GM or AM-HM but I can't figure it out.
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\root3\of{\Big(a+\frac{1}{b}\Big)^2\Big(b+\frac{1}{c}\Big)^2\Big(c+\frac{1}{a}\Big)^2}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(\Big(a+\frac{1}{b}\Big)\Big(b+\frac{1}{c}\Big)\Big(c+\frac{1}{a}\Big)\bigg)^{2/3}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(abc+a+b+c+\frac1a+\frac1b+\frac1c+\frac{1}{abc}\bigg)^{2/3}$$
$$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq 3\bigg(2+a+b+c+\frac1a+\frac1b+\frac1c\bigg)^{2/3}$$
| By Cauchy we have $$\Big(a+\frac{1}{b}\Big)^2 + \Big(b+\frac{1}{c}\Big)^2 + \Big(c+\frac{1}{a}\Big)^2 \geq \underbrace{{1\over 3}\bigg(a+b+c+\frac1a+\frac1b+\frac1c\bigg)^{2}}_B$$
By AM-GM and assumption $abc=1$ we have $$\frac1a+\frac1b+\frac1c \geq 3$$ so $$B \geq {1\over 3}\underbrace{\bigg(a+b+c+3\bigg)^{2}}_C$$
Let $x=a+b+c\geq 3$ then we have to check if $$C\geq 9(x+1)$$ or $$(x+3)^2\geq 9(x+1)\iff x(x-3)\geq 0$$
which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Equation involving floor function and fractional part function How to solve $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} = \{x\} + \frac{1}{3}$ , where $\lfloor \rfloor$ denotes floor function and {} denotes fractional part. I did couple of questions like this by solving for {x} and bounding it from 0 to 1. So here we will have, $$0\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} - \frac{1}{3}\lt1$$ adding throughout by $\frac{1}{3}$ $$\frac{1}{3}\le\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2x \rfloor} \lt\frac{4}{3}$$ Now I am stuck.
| Notice that (1) the values $\lfloor x \rfloor, \lfloor 2 x \rfloor$ depend only on the value of the integer $\lfloor 2 x \rfloor$ and (2) $x \geq 1$. On the other hand, for, e.g., $x \geq 5$, we have $\frac{1}{\lfloor x \rfloor} + \frac{1}{\lfloor 2 x \rfloor} \leq \frac{1}{5} + \frac{1}{10} = \frac{3}{10} < \frac{1}{3}$, leaving only finitely many integer values $\lfloor 2 x \rfloor$ to check.
| {
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"timestamp": "2023-03-29T00:00:00",
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I need to find the real and imaginary part of this $Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}$ I have a test tomorrow and i have some troubles understanding this kind of problems, would really appreciate some help with this
$$ Z_{n}=\left (\frac{ \sqrt{3} + i }{2}\right )^{n} + \left (\frac{ \sqrt{3} - i }{2}\right )^{n}
$$
$$
Z_{n}\epsilon \mathbb{C}
$$
| The above two can be written using binomial theorem
$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{i}{2}\right)^k$$
And
$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{-i}{2}\right)^k$$
We rewrite the second as
$$\sum_{k=0}^n \binom{n}{k}\left(\frac{3}{4}\right)^\frac{n-k}{2}\left(\frac{i}{2}\right)^k(-1)^k$$
Such that even k terms are the same and odd k terms are opposite signs
Combining the sums we have
$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^\frac{n-2k}{2}\left(\frac{i}{2}\right)^{2k}$$
$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^{\frac{n}{2}-k}\left(\frac{-1}{4}\right)^{k}$$
$$\sum_{k=0}^{\frac{n}{2}} 2\binom{n}{2k}\left(\frac{3}{4}\right)^{\frac{n}{2}}\left(\frac{-1}{3}\right)^{k}$$
Which calculating the sum from wolfram alpha gives, $$Re(Z_n)=2cos(\frac{n\pi}{6})$$
And
$$Im(Z_n)=0$$
The phase would flip between $\pi$ and $0$ and the function is always real. (Which I should have realized by seeing that this clearly resembles the cos function)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate the integral $\int \frac{\sec x}{\sqrt{3+\tan x}}dx$ Evaluate the following integral.$$\int \frac{\sec x}{\sqrt{3+\tan x}}dx$$
On putting $t=\tan x$ I am getting $$\int\frac{1}{\sqrt{(t^2+1)(t+3)}}dt$$.
How should i proceed from here.
| We are given:
$$ y = \int\frac{\sec x\,dx}{\sqrt{3+\tan x}}. $$
First, change the variable of integration from $x$ to $u=\sqrt{3+\tan x}$; we have that
\begin{align*}
du &= \frac{\sec^{2}x\,dx}{2\sqrt{3+\tan x}} \\
\therefore dy &= \frac{2\,du}{\sec x} \\
&= \frac{2\,du}{\sqrt{1+(u^{2}-3)^{2}}}.
\end{align*}
Now, if there exists $a$ with $u=a t$ such that
$$ 1 + (u^{2}-3)^{2} = r^{2}(1-t^{2})(1-k^{2}t^{2}) $$
for some $r,k$, then changing the variable of integration from $u$ to $t$ turns our integral into an incomplete elliptic one of the first kind.
To find such $a,r,k$, expand the powers and rewrite $u$ as $at$ in the previous equation to yield
$$a^{4}t^{4}-6a^{2}t^{2}+10=r^{2}k^{2}t^{4}-r^{2}(1+k^{2})t^{2}+r^{2}.$$
Equating coefficients of like powers immediately yields
\begin{align*}
r^{2}=10 && 6a^{2}=r^{2}(1+k^{2}) && a^{4}=r^{2}k^{2}
\end{align*}
which gives us (but not before some sacrifice)
\begin{align*}
k=\frac{3\pm i}{\sqrt{10}}&&
k^{2}=\frac{4\pm 3i}{5}&&
a=\sqrt{3\pm i}.
\end{align*}
So now return to our integral
$$ y = 2\int\frac{du}{\sqrt{1+(u^{2}-3)^{2}}} $$
and substitute $u$ for $t$, whereby it becomes
\begin{align*}
y &= 2a\int\frac{dt}{\sqrt{r^{2}(1-t^{2})(1-k^{2}t^{2})}}\\
&=\frac{2a}{r}\int\frac{dt}{\sqrt{(1-t^{2})(1-k^{2}t^{2})}}\\
&=\frac{2a}{r}\mathrm{F}(t;k)+C.
\end{align*}
Finally, we undo our subsitutions,
\begin{align*}
t &= \frac{\sqrt{3+\tan x}}{a} \\
&= \sqrt{\frac{3\mp i}{10}}\sqrt{3+\tan x}
\end{align*}
expand
$$ \frac{2a}{r} = \sqrt{\frac{6\pm2i}{5}}, $$
and pass to Legendre's notation, to yield
$$ y=\int\frac{\sec x\,dx}{\sqrt{3+\tan x}}=\sqrt{\frac{6\pm2i}{5}}\mathrm{F}\left(\arcsin\left(\sqrt{\frac{3\mp i}{10}}\sqrt{3+\tan x}\right)\Bigg|\frac{4\pm3i}{5}\right)+C. \tag{$\star$}\label{star}$$
I have gone over the above a handful of times now, and can't find any mistakes.
One thing concerned me though: this is similar enough to David G. Stork's provided answer for me to believe it is not entirely incorrect, but at the same time, it is different enough for me to suspect it is not entirely correct either.
If one examines the differences however, one finds that
$$ \cos x \sqrt{(1+3i)+(3-i)\tan x}\sqrt{(-3-i)(\tan x +i)} = \pm(1+3i)$$
is locally constant everywhere it is defined. Thus \eqref{star} is at least everywhere a constant multiple of the other answer.
This behaviour is probably due to the participation of square-root and tangent in the integrand and whatnot, though I don't know enough complex analysis. Just mind any possible singularities in your path of integration...
| {
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"timestamp": "2023-03-29T00:00:00",
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Find symmetric matrix containing no 0's, given eigenvalues I'm preparing for a final by going through the sample exam, and have been stuck on this:
$$Produce\ symmetric\ matrix\ A ∈ R^{3×3},\ containing\ no\ zeros.\ \\
A\ has\ eigenvalues\ λ_1 = 1,\ λ_2 = 2,\ λ_3 = 3$$
I know $A = S^{-1}DS$, where A is similar to the diagonal matrix D, and S is orthogonal.
The diagonal entries of D are the eigenvalues of A.
I also know that A & D will have the same determinant, eigenvalues, characteristic polynomial, trace, rank, nullity, etc. I am not sure where to go from here though. How cna A be found with only the two pieces of information? It seems like too little information is given...
| You are correct in observing that "too little" information is given in the sense that there are infinitely many such matrices. But you need to produce just one. So start with the diagonal matrix $D = \operatorname{diag}(1,2,3)$ and conjugate it by a simple (but not too simple) orthogonal matrix $S$. You don't want $S$ to be a block matrix because then $S^{-1} D S$ will also be a block matrix and so will have zeroes. For example, we can take
$$ S = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix}, S^{-1} = S^T = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} $$
and define
$$ A = S^{-1} D S = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} & \frac{2}{\sqrt{6}} & -\frac{4}{\sqrt{6}} \\ -\frac{3}{\sqrt{2}} & \frac{3}{\sqrt{2}} & 0 \end{pmatrix} \\
= \begin{pmatrix} \frac{1}{3} + \frac{2}{6} + \frac{3}{2} & \frac{1}{3} + \frac{2}{6} - \frac{3}{2} & \frac{1}{3} - \frac{4}{6} \\ \frac{1}{3} + \frac{2}{6} - \frac{3}{2} & \frac{1}{3} + \frac{2}{6} + \frac{3}{2} & \frac{1}{3} - \frac{4}{6} \\ \frac{1}{3} - \frac{4}{6} & \frac{1}{3} - \frac{4}{6} & \frac{1}{3} + \frac{8}{6} \end{pmatrix} =
\begin{pmatrix} \frac{13}{6} & -\frac{5}{6} & -\frac{1}{3} \\
-\frac{5}{6} & \frac{13}{6} & -\frac{1}{3} \\
-\frac{1}{3} & -\frac{1}{3} & \frac{5}{3} \end{pmatrix}. $$
Then $A$ is symmetric and has eigenvalues $1,2,3$ (because it is similar to $D$).
| {
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"timestamp": "2023-03-29T00:00:00",
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In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$. In $\triangle ABC$, we have $AB = 14$, $BC = 16$, and $\angle A = 60^\circ$. Find the sum of all possible values of $AC$.
When I use the Law of Cosines, I get a quadratic like expected. However, when I use Vieta's Formulas to get the sum of the roots, I get the wrong answer. Please help me figure out what I did wrong.
| What you did wrong was forgetting that there is a negative solution!
When you use the Law of Cosines, you get
$$\begin{align}16^2 &= 14^2 + x^2 - 2(14)(x)cos(60^\circ) \\
16^2 - 14^2 &= x^2 - 28x\cdot(\frac{1}{2}) \\
16^2 - 14^2 &= x^2 - 14x \\
60 &= x^2 - 14x \\
0 &= x^2 - 14x - 60. \end{align}$$
Then we realize that one of the roots is negative leaving us with the answer
$$\boxed{7 + \sqrt{109}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846205",
"timestamp": "2023-03-29T00:00:00",
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Find the quotient and remainder Find the quotient and remainder when $x^6+x^3+1$ is divided by $x+1$
Let $f(x)=x^6+x^3+1$
Now $f(x)=(x+1).q(x) +R $ where r is remainder
Now putting $x=-1$ we get $R=f(-1)$
i.e $R=1-1+1=1$
Now $q(x)=(x^6+x^3)/(x+1)$
But what I want to know if there is another way to get the quotient except simple division.
| You're on the right track:
$$
x^6+x^3=x^3(x^3+1)=x^3(x+1)(x^2-x+1)
$$
Therefore
$$
q(x)=(x^6+x^3)/(x+1)=x^3(x^2-x+1)=x^5-x^4+x^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\quad (a+b)^2 \le 2(a^2+b^2) = 2\quad \Rightarrow\quad a+b\le \sqrt{2}$
But using Titu's Lemma I get $\quad \frac1a+\frac1b \ge \frac{4}{a+b}\quad \Rightarrow\quad \frac1a+\frac1b \ge 2\sqrt{2}$
I'm stuck here, any hint?
| Calling $a = \cos u, b = \sin u\;\;$ we have
$$
\left(\cos u + \frac{1}{\cos u}\right)+\left(\sin u + \frac{1}{\sin u}\right)\ge 2\sqrt{\left(\cos u + \frac{1}{\cos u}\right)\left(\sin u + \frac{1}{\sin u}\right)} = 2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}
$$
Now examining
$$
f(u) = \frac{(\cos^2 u+1)}{\cos u}\frac{(\sin^2 u + 1)}{\sin u}
$$
by symmetry considerations, the feasible minimum is at $u = u_0 = \frac{\pi}{4}\;\;$ (because $\sin u_0 = \cos u_0\;$)
giving the value
$$
f(\frac{\pi}{4}) = \frac 92
$$
then following we have
$$
2\sqrt{\frac{(\cos^2 u+1)(\sin^2 u + 1)}{\sin u\cos u}}\ge 2\sqrt{\frac 92} = 3\sqrt2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Suppose $x$ is an integer such that $3x \equiv 15 \pmod{64}$. Find remainder when $q$ is divided by $64$. Suppose $x$ is an integer such that $3x \equiv 15 \pmod{64}$. If $x$ has remainder $2$ and quotient $q$ when divided by $23$, determine the remainder when $q$ is divided by $64$.
I tried a couple things. By division algorithm, know $x = 23q + 2$. So $3(23q + 2) \equiv 15 \pmod{64}$. Not sure how to go from there.
Another thing I tried is $3x = 64q + 15$. If we let $q$ be zero, then $x$ is obviously $5$. This also doesn't wind up being that helpful, and I think $x$ can have other values asides from $5$.
| Continuing from where you stopped:
$$\begin{align} 3(23q + 2) &\equiv 15 \pmod{64} \Rightarrow \\
69q+6 &\equiv 15 \pmod{64} \Rightarrow \\
69q &\equiv 9 \pmod{64} \Rightarrow \\
64q+5q &\equiv 9 \pmod{64} \Rightarrow \\
5q &\equiv 9 \pmod{64} \Rightarrow \\
5q\cdot 13 &\equiv 9\cdot 13 \pmod{64} \Rightarrow \\
65q &\equiv 117 \pmod{64} \Rightarrow \\
64q+q &\equiv 64+53 \pmod{64} \Rightarrow \\
q &\equiv 53 \pmod{64}.
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimum of $\left(a + b + c + d\right)\left(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\right)$ If $a$, $b$, $c$, $d$ are positive integers, find the minimum value of
$$P = \left(a + b + c + d\right)\left(\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}\right)$$
and the values of $a$, $b$, $c$, $d$ when it is reached.
My try:
$$\left.
\begin{array}{l}
a + b + c + d \ge 4\sqrt[4]{abcd}\\
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4\sqrt[4]{\frac{64}{abcd}}
\end{array}
\right\}
\Rightarrow P \ge 32\sqrt{2}$$
I have used mean inequalities, but that doesn't mean that I have found the minimum value. Also, I have found a similar exercise here (exercise #5), but the author shows that $P \ge 64$, which is greater than what have I found.
Can you help me solve the problem, please? Thanks!
| If you want to use the AM-GM Inequality, it can be done as follows. Observe that
$$a+b+c+d=a+b+2\left(\frac{c}{2}\right)+4\left(\frac{d}{4}\right)\geq 8\sqrt[8]{ab\left(\frac{c}{2}\right)^2\left(\frac{d}{4}\right)^4}$$
and that
$$\frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}=\frac{1}{a}+\frac{1}{b}+2\left(\frac{2}{c}\right)+4\left(\frac{4}{d}\right)\geq 8\sqrt[8]{\left(\frac1a\right)\left(\frac1b\right)\left(\frac{2}{c}\right)^2\left(\frac{4}{d}\right)^4}\,.$$
However, using the Cauchy-Schwarz Inequality is probably the easiest way. (The equality holds iff there exists $\lambda >0$ such that $(a,b,c,d)=(\lambda,\lambda,2\lambda,4\lambda)$.)
| {
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find all the points on the line $y = 1 - x$ which are $2$ units from $(1, -1)$ I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?
1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$
2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$
3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$
4.) combine, $2x^2 -6x +5 = 4$
5.) general form, $2x^2 -6x +1$
6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$
7.) Vertex form, $2(x-1.5)^2 -3.5$
8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$
9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$
10.) $x = 0.18$ and $2.82$
When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?
| You have
$2x^2 - 6x + 1 = 0$
This looks good.
then I would say
$x = \frac {3\pm\sqrt 7}{2}$ is simpler than $x = 1.5 \pm \sqrt {1.75}$
Plug each value of $x$ into $y = 1-x$ to find $y.$
$y = 1 - \frac {3+\sqrt 7}{2} = \frac {-1-\sqrt 7}{2}\\y = 1 - \frac {3-\sqrt 7}{2} = \frac {-1+\sqrt 7}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\log _{24}48$ if $\log_{12}36=k$ Find $\log _{24}48$ if $\log_{12}36=k$
My method:
We have $$\frac{\log 36}{\log 12}=k$$ $\implies$
$$\frac{\log 12+\log 3}{\log 12}=k$$ $\implies$
$$\frac{\log3}{2\log 2+\log 3}=k-1$$ So
$$\log 3=(k-1)t \tag{1}$$
$$2\log 2+\log 3=t$$ $\implies$
$$\log 2=\frac{(2-k)t}{2} \tag{2}$$
Now $$\log _{24}48=\frac{\log 48}{\log 24}=\frac{4\log 2+\log 3}{3\log 2+\log 3}=\frac{2(2-k)+k-1}{3\left(\frac{2-k}{2}\right)+k-1}=\frac{6-2k}{4-k}$$
is there any other approach?
| You can convert all to the smallest common base $2$:
$$\log_{12}36=\frac{\log_{2}36}{\log_2 12}=\frac{2+2\log_2 3}{2+\log_2 3}=k \Rightarrow \log_2 3=\frac{2k-2}{2-k}.$$
Hence:
$$\log _{24}48=\frac{\log_2 48}{\log_2 24}=\frac{4+\log_23}{3+\log_23}=\frac{4+\frac{2k-2}{2-k}}{3+\frac{2k-2}{2-k}}=\frac{6-2k}{4-k}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Question about substitutions in the double integral $\int_0^1\int_0^1 \frac{-\ln xy}{1-xy} dx\, dy = 2\zeta(3)$ \begin{align}
\int_0^1\int_0^1 \frac{-\ln xy}{1-xy} dx\, dy = 2\zeta(3) \tag{1}
\end{align}
Since,
\begin{align}
\int_0^1 \frac{1}{1-az} dz = \frac{-\ln (1-a)}{a}
\end{align}
and putting $a = 1-xy$, $(1)$ becomes:
\begin{align}
\int_0^1\int_0^1 \int_0^1 \frac{1}{1-z(1-xy)} dx\, dy\, dz = 2\zeta(3) \tag{2}
\end{align}
with $t = 1-z(1-xy) \implies dt = -(1-xy) \, dz$. When $z=0$, $t=1$ and when $z=1$, $t=xy.$ Making these substitutions in $(2)$:
\begin{align}
-\int_1^{xy}\int_0^1 \int_0^1 \frac{1}{t(1-xy)} dx\, dy\, dt \tag{3}
\end{align}
Is $(3)$ still equal to $2\zeta(3)$? I think $(3)$ will be a function of $x$ and $y$, but after making these substitutions the integral should stay the same.
| By geometric series, your integral equals
$$-\int^1_0\int^1_0(\ln x+\ln y)\sum^\infty_{k=0}(xy)^kdydx$$
Due to the symmetry in $x$ and $y$, this equals
$$-2\int^1_0\int^1_0(\ln x)\sum^\infty_{k=0}(xy)^kdydx$$
Since the series converges uniformly, we can integrate it termwise to obtain
$$-2\sum^\infty_{k=0}\int^1_0 x^k\ln x\cdot\frac{1}{1+k} dx$$
Since $\int^1_0 x^k\ln xdx=-\frac1{(1+k)^2}$, this simplifies to
$$2\sum^\infty_{k=0}\frac{1}{(1+k)^3} =2\zeta(3)$$ by definition.
Prove $\int^1_0 x^k\ln xdx=-\frac1{(1+k)^2}$:
By the substitution $t=-(k+1)\ln x$,
$$-\int_\infty^0 e^{\frac{-kt}{k+1}}\frac{t}{k+1}\cdot\frac{-1}{1+k}e^{\frac{-t}{k+1}}dt=-\frac1{(k+1)^2}\int^\infty_0te^{-t}dt=-\frac1{(k+1)^2}(1!)=\color{red}{-\frac1{(k+1)^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality ???
Let $a,b,c$ are positive real numbers such that $a+b+c=3$.
Prove that
$$\sum\frac{(7a^{3}+3)(b+c)}{7a+3} \geq 6 $$
I try to prove $LHS \geq \sum\frac{9}{5}a+\frac{1}{5}$ but don't succeed
| Yes, you are right, the TL method does not help here.
But $uvw$ helps.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$\sum_{cyc}(7a^3+3u^3)(3u-a)(7b+3u)(7c+3u)\geq6u^3\prod_{cyc}(7a+3u)$$ and we see that our inequality it's $f(w^3)\geq0,$ where
$$f(w^3)=-343\cdot3w^6+A(u,v^2)w^3+B(u,v^2).$$
We see that $f$ is a concave function, which says that it's enough to prove our inequality for an extreme value of $w^3$, which happens in the following cases.
*
*$w^3\rightarrow0$.
Let $c\rightarrow0$ and $b=3-a$, where $0<a<3$.
We obtain:
$$(3-a)^2a^2\geq0;$$
2. Two variables are equal.
Let $b=a$ and $c=3-2a$, where $0<a<1.5.$
We obtain:
$$a^2(a-1)^2(39+70a-49a^2)\geq0.$$
Done!
A proof by LCF.
Let $f(x)=\frac{(7x^3+3)(x-3)}{7x+3}.$$
Hence, $$f''(x)=\frac{42(x+1)(49x^3-42x^2-3x-24)}{(7x+3)^3}<0$$ for all $0<x<1$ and we need to prove that
$$\frac{f(a)+f(b)+f(c)}{3}\leq f\left(\frac{a+b+c}{3}\right).$$
Thus, by the Vasc's LCF Theorem it's enough to prove the last inequality for $b=a\leq1$ and $c=3-2a\geq1.$
After these substitutions we obtain $$a^2(3-2a)(39+70a-49a^2)\geq0,$$ which is true even for all $0<a<1.5.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi}{20}$$
and that,
$$\cos\frac{\pi}{20}-\cos\frac{3\pi}{20}=-2\sin\frac{2\pi}{20}\sin\frac{\pi}{20}$$
So,
$$2\sin\frac{2\pi}{20}(\cos\frac{\pi}{20}-\sin\frac{\pi}{20})=\frac{\sqrt2}{2}=\sin\frac{5\pi}{20}$$
Unfortunately, I can't find a way to continue this, any ideas or different ways of proof?
*Taken out of the TAU entry exams (no solutions are offered)
| Let $A,B,C,\ldots,T$ be points on a circle with diameter $1$ dividing it into $20$ equal arcs. Let $RG$ intersect $BO, BK, KF$ at $U,V,W$, respectively.
It is easy to get the following equalities
$$\angle URO = \angle OUR = \angle BUV = \angle UVB = \angle WVK = \angle KWV = \angle FWG = \angle WGF = \frac 25 \pi,$$
so in particular $OR=OU$, $UB=VB$, $VK=WK$, and $WF=FG$.
Moreover
\begin{align*}
OR & = \sin \dfrac{3\pi}{20},\\
OB & = \sin \dfrac{7\pi}{20} = \cos \dfrac{3\pi}{20},\\
KB & = \sin \dfrac{9\pi}{20} = \cos \frac{\pi}{20}, \\
KF & = \sin \frac \pi 4 = \dfrac{\sqrt 2}{2}, \text{ and }\\
FG & = \sin \dfrac{\pi}{20}.
\end{align*}
Therefore
\begin{align*}
\frac{\sqrt 2}{2} & = KF \\ & = KW + FG \\ & = KV + FG \\ & = BK - BV + FG \\
& = BK - BU + FG \\ & = BK - (OB - OU) + FG \\ & = BK - OB + OU + FG \\ & = BK - OB + OR + FG \\ & = FG + BK + OR - OB \\ & = \sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 2
} |
Partial sums of the series $\sum\limits_{k\geq1}\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}$ The series $\sum\limits_{k=1}^{\infty}\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}$ is divergent. I am interested in its partial sums to do some computations based on them.
I tried to multiply $\sqrt{2k+\sqrt{4k^2-1}}$ by $\sqrt{2k-\sqrt{4k^2-1}}$ and to divide by the same term, but I have got no telescoping sum for evaluation. Maybe I should use others technics.
Now my question is what is the closed form of the sum $$\sum_{k=1}^{n}\frac{1}{\sqrt{2k+\sqrt{4k^2-1}}}\ ?$$
| Using the identity
$$\sqrt{a+\sqrt{b}}=\sqrt{\frac{1}{2} \left(a+\sqrt{a^2-b}\right)}+\sqrt{\frac{1}{2} \left(a-\sqrt{a^2-b}\right)}$$
where: $a = 2 k$ and $b=4 k^2-1$, along with evaluating a telescoping series, we find that
$$\begin{align}
\color{red}{\sum _{k=1}^n \frac{1}{\sqrt{2 k+\sqrt{4 k^2-1}}}}&=\sum _{k=1}^n \frac{\sqrt{2}}{\sqrt{-1+2 k}+\sqrt{1+2 k}}\\\\
&=\sum _{k=1}^n \frac{\sqrt{2}
\left(\sqrt{2 k+1}-\sqrt{2 k-1}\right)}{\left(\sqrt{-1+2 k}+\sqrt{1+2 k}\right) \left(\sqrt{2 k+1}-\sqrt{2 k-1}\right)}\\\\
&=\frac{\sum _{k=1}^n
\left(\sqrt{2 k+1}-\sqrt{2 k-1}\right)}{\sqrt{2}}\\\\
&=\frac{-1+\sqrt{1+2 n}}{\sqrt{2}}\\\\
&=\color{red}{-\frac{1}{\sqrt{2}}+\frac{\sqrt{1+2 n}}{\sqrt{2}}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to get the shaded region of the rectangle? I have this problem:
So my development was:
Denote side of rectangle with: $2a, 2b$.
So, $4ab= 64, ab = 16$
Denote shaded region with $S$
Denote area of triangle $DGH = A_1$ and triangle $FBE = A_2$.
So, $A_1 + A_2 + S = 64$
$S = 64 - A_1 - A_2$
The triangles $A_1, A_2$ are congruent because $LAL$ congruence criterion.
The area of $A_1$ and $A_2$, is the same and i got it with this way:
Since, the $\angle{GDH} = 90$ and the median from this angle to the base $HG$, that is the altitude of the triangle $DGH$, will measure the half of the $HG$ side.
And the $HG$ side by Pythagorean theorem, will be $\sqrt{a^2 + b^2}$, that will be the base of the triangle.
And the altitude will be: $\frac{\sqrt{a^2 + b^2}}{2} $,
So the Area of $A_1 = \frac{a^2 + b^2}{4}$
So, $A_1 + A_2 = \frac{a^2 + b^2}{2}$
Then, $64 - (\frac{a^2 + b^2}{2}) = S$
And, $-(a^2 - 8ab + b^2) = 2S$
And I have not been able to continue from here, what should I do? Thanks in advance.
|
If you notice that if you combine two right triangles then they occupy the area of $\dfrac14$ of the total area.
So, the area of the shaded region is $=64-\dfrac14(64)=64-16=48$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Show that $\sum_{n=1}^\infty 2^{2n}\sin^4\frac a{2^n}=a^2-\sin^2a$ Show that $$\sum_{n=1}^\infty 2^{2n}\sin^4\frac a{2^n}=a^2-\sin^2a$$
I am studying for an exam and I bumped into this question. It's really bothering me because I really don't have any clue what to do. Does it have anything to do with the Cauchy condensation? Can somebody help me I would really appreciate it
| Solution
Notice that
\begin{align*}
2^{2n}\sin^4\frac a{2^n}&=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\sin^2\frac a{2^n}\\&=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\left(1-\cos^2\frac a{2^n}\right)\\&=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n}\cdot\sin^2\frac a{2^n}\cos^2\frac a{2^n}\\&=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n-2}\sin^2\frac a{2^{n-1}}.
\end{align*}
Hence, the partial sum
\begin{align*}
&\sum_{n=1}^{m}\left(2^{2n}\sin^4\frac a{2^n}\right)\\=&2^2\sin^2\frac a{2}-\sin^2a+2^4\sin^2\frac a{2^2}-2^2\sin^2\frac a{2}+ \cdots+ 2^{2m}\sin^2\frac a{2^m}-2^{2m-2}\sin^2\frac a{2^{m-1}}\\=&2^{2m}\sin^2\frac a{2^m}-\sin^2a.
\end{align*}
Let $m \to \infty$. We obtain $$\sum_{n=1}^{\infty}\left(2^{2n}\sin^4\frac a{2^n}\right)=\lim_{m \to \infty}\left(\frac{\sin\dfrac a{2^m}}{\dfrac{a}{2^m}}\cdot a\right)^2-\sin^2 a=(1\cdot a)^2-\sin^2 a=a^2-\sin^2 a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve: $2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1\Bigl) = 0$ The question says to find the value of $x$ if, $$2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl)= 0$$
My approach:
I rewrote the expression as,
$$2^x\Bigl(2^x-1\Bigl) + \frac{2^x}{2}\Bigl(\frac{2^x}{2} -1 \Bigl) + .... + \frac{2^x}{2^{99}} \Bigl(\frac{2^x}{2^{99}} - 1 \Bigl)= 0$$
I then took $\bigl(2^x\bigl)$ common and wrote it as,
$$2^x \Biggl[ \Bigl(2^x - 1\Bigl) + \frac{1}{2^1}\Bigl(2^x -2^1\Bigl) + \frac{1}{2^2}\Bigl(2^x - 2^2\Bigl) + \;\ldots + \frac{1}{2^{99}} \Bigl(2^x - 2^{99}\Bigl)\Biggl] = 0$$
After further simplification I got,
$$\frac{2^x}{2^{99}} \Biggl[ \Bigl(2^x\cdot2^{99} - 2^{99}\Bigl) + \Bigl(2^x \cdot 2^{98} - 2^{99}\Bigl) + \ldots + \bigl(2^x -2^{99}\bigl)\Biggl] = 0$$
Taking $-2^{99}$ common I got,
$$-2^x \Biggl[ \Bigl( 2^{x+99} + 2^{x+98} + \ldots + 2^{x+2} + 2^{x+1} + 2^x \Bigl)\Biggl]= 0$$
Now the inside can be expressed as $$\sum ^ {n= 99} _{n=1} a_n$$ Where $a_n$ are the terms of the GP.
Thus we can see that either $$-2^x= 0$$ Or, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
Since the first condition is not poossible, thus, $$\sum ^ {n= 99} _{n=1} a_n = 0$$
So,
$$2^{x + 99} \Biggl(\frac{1-\frac{1}{2^{100}}}{1-\frac{1}{2}} \Biggl) = 0$$
Either way once I solve this, I am not getting an answer that is even in the options. The answers are all in the form of logarithmic expressions.
Any help would be appreciated. We have to find the value of $x$.
| \begin{align}
2^x\Bigl(2^x-1\Bigl) + 2^{x-1}\Bigl(2^{x-1} -1 \Bigl) + .... + 2^{x-99}\Bigl(2^{x-99} - 1 \Bigl) &= 0 \\
\sum_{i=0}^{99}2^{x-i}\left(2^{x-i}-1\right) &= 0 \\
\sum_{i=0}^{99}\left[\left(2^{x-i}\right)^2-2^{x-i}\right] &= 0 \\
\sum_{i=0}^{99}\left(2^{x-i}\right)^2-\sum_{i=0}^{99}2^{x-i} &= 0 \\
\sum_{i=0}^{99}2^{2x-2i}&=\sum_{i=0}^{99}2^{x-i} \\
\dfrac{2^{2x}}{\sum_{i=0}^{99}2^{2i}}&=\dfrac{2^{x}}{\sum_{i=0}^{99}2^{i}} \\
2^x&=\dfrac{\sum_{i=0}^{99}2^{2i}}{\sum_{i=0}^{99}2^{i}} \\
&= \dfrac{4^{100}-1}{4-1}\cdot\dfrac{2-1}{2^{100}-1} \\
&= \dfrac{4^{100}-1}{3\left(2^{100}-1\right)}\\
x &= \log_2\left(\dfrac{4^{100}-1}{3\left(2^{100}-1\right)}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
how can I calculate the probability to get triples or better when throwing n 6-sided dice? I've been banging my head on a wall with this question. I'm designing a game and would like to implement a loot system inspired by a game called "Vermintide" where players roll a certain number of dice and gain loot according to the result.
I want my players to get rewards depending on the result of the dice. but I need to evaluate the quality of items, and the number of dice they get to throw based on the probability.
Essentially if they get doubles they get a standard item, if they get triples they get a magic item etc...
I heard of the "birthday problem" and "multinomials" but most of what I could find about it seemed to be very particular cases (I found birthday problem for 2 or more but not for 3 or more, which seems much less trivial).
Is there a smart way to go about this problem which I first thought was going to be trivial, and the more I search the more it seems complicated.
| Hint
You can compute the probability of not getting triple. The number $f_n$ of possibilities of not getting triples in $n$ throws is:
$$f_1 = \binom{6}{1}\\
f_2 = \binom{6}{2}2!+\binom{6}{1}\\
f_3=\binom{6}{3}3!+\binom{6}{1}\binom{5}{1}\frac{3!}{2!}\\
f_4=\binom{6}{4}4!+\binom{6}{1}\binom{5}{2}\frac{4!}{2!}+\binom{6}{2}\frac{4!}{2!2!}\\
f_5=\binom{6}{5}5!+\binom{6}{1}\binom{5}{3}\frac{5!}{2!}+\binom{6}{2}\binom{4}{1}\frac{5!}{2!2!}\\
f_6=\binom{6}{6}6!+\binom{6}{1}\binom{5}{4}\frac{6!}{2!}+\binom{6}{2}\binom{4}{2}\frac{6!}{2!2!}+\binom{6}{3}\frac{6!}{2!2!2!}\\
f_7=\binom{6}{1}\frac{7!}{2!}+\binom{6}{2}\binom{4}{3}\frac{7!}{2!2!}+\binom{6}{3}\binom{3}{1}\frac{7!}{2!2!2!}\\
f_8=\binom{6}{2}\frac{8!}{2!2!}+\binom{6}{3}\binom{3}{2}\frac{8!}{2!2!2!}+\binom{6}{4}\frac{8!}{2!2!2!2!}\\
f_9=\binom{6}{3}\frac{9!}{2!2!2!}+\binom{6}{4}\binom{2}{1}\frac{9!}{2!2!2!2!}\\
f_{10} = \binom{6}{4}\frac{10!}{2!2!2!2!}+\binom{6}{5}\frac{10!}{2!2!2!2!2!}\\
f_{11}=\binom{6}{5}\frac{11!}{2!2!2!2!2!}\\
f_{12}=\binom{6}{6}\frac{12!}{2!2!2!2!2!2!}$$
For $k>13$ we have $f_k=0$
The number of all possible throws $\omega_n$ is:
$$\omega_n=6^n$$
Now you can compute the probability of not getting triple:
$$p_n=\frac{f_n}{\omega_n}$$
and probability of getting triple:
$$q_n=1-p_n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Gauss elimination. Where did I go wrong?
Gaussian elimination with back sub:
So my starting matrix:
\begin{bmatrix}
1 & -1 & 1 & -1
\\2 & 1 & -3 & 4
\\2 & 0 & 2 & 2
\end{bmatrix}
multiply the 2nd and 3rd row by -1 * (first row):
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 3 & -5 & 6
\\0 & 2 & 0 & 4
\end{bmatrix}
then add -1(third row) to the 2nd row->
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 2 & 0 & 4
\end{bmatrix}
add -2(2nd row) to the third row ->
\begin{bmatrix}
1 & -1 & 1 & -1
\\0 & 1 & -5 & 2
\\0 & 0 & 10 & 0
\end{bmatrix}
But then this seems to have no solution because $10z = 0$.... ugh
EDIT
As I was writing this, it occurred to me that $z = 0$, $y = 2$, $x = 1$. Is that right?
| I'd use a more systematic method:
\begin{align}
\begin{bmatrix}
1 & -1 & 1 & -1\\
2 & 1 & -3 & 4\\
2 & 0 & 2 & 2
\end{bmatrix}
&\to
\begin{bmatrix}
1 & -1 & 1 & -1\\
0 & 3 & -5 & 6\\
0 & 2 & 0 & 4
\end{bmatrix}
&&\begin{aligned} R_2&\gets R_2-2R_1 \\ R_3&\gets R_3-2R_1 \end{aligned}
\\ &\to
\begin{bmatrix}
1 & -1 & 1 & -1\\
0 & 1 & -5/3 & 2\\
0 & 2 & 0 & 4
\end{bmatrix}
&& R_2\gets\tfrac{1}{3}R_2
\\ &\to
\begin{bmatrix}
1 & -1 & 1 & -1\\
0 & 1 & -5/3 & 2\\
0 & 0 & 10/3 & 0
\end{bmatrix}
&& R_3\gets R_3-2R_2
\\ &\to
\begin{bmatrix}
1 & -1 & 1 & -1\\
0 & 1 & -5/3 & 2\\
0 & 0 & 1 & 0
\end{bmatrix}
&& R_3\gets\tfrac{3}{10}R_3
\\ &\to
\begin{bmatrix}
1 & -1 & 0 & -1\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 0
\end{bmatrix}
&& \begin{aligned} R_2 &\gets R_2+\tfrac{5}{3}R_3 \\ R_1&\gets R_1-R_3\end{aligned}
\\ &\to
\begin{bmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & 2\\
0 & 0 & 1 & 0
\end{bmatrix}
&& R_1\gets R_1+R_2
\end{align}
The solution, which is explicit when the RREF is reached, is
\begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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if $a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$
If $a+b$ is an positive integer and $a\ge b$ and
$a^3+b^3 +3(a^2+b^2) - 700(a+b)^2 = 0$ then find the sum of all possible values of $a+b$.
I tried a lot to solve it, i came to a step after which i was not able to proceed forward, the step was $(a+b)^3 - 41\cdot17 (a+b)^2 - 3ab(a+b) - 6ab = 0$
Please help me to solve this.
| Here is a start, not a full answer.
Let $s=a+b$ and $p=ab$. Then $3 p (s + 2) = (s - 697) s^2$ as WA tells us.
Then $s+2$ divides $(s - 697) s^2$ and so $s+2$ divides $2796 = 2^2 \cdot 3 \cdot 233$.
| {
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"url": "https://math.stackexchange.com/questions/2871998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
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compute the summation $\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$ compute the summation
$$\sum_ {n=1}^\infty \frac{2n-1 }{2\cdot4\cdots(2n)}= \text{?}$$
My attempts : i take $a_n =\frac{2n-1}{2\cdot4\cdots(2n)}$
Now
\begin{align}
& = \frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n} \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{2n}{2\cdot4\cdot6\cdots2n} -\frac{1}{2\cdot4\cdot6\cdots2n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{1\cdot2\cdot3\cdots(n-1)} -\frac{1}{2(1\cdot2\cdot3\cdots n}\right) \\[10pt]
& =\sum_ {n=1}^\infty \left(\frac{1}{(n-1)!} - \frac{1}{2}\sum_ {n=2}^\infty \frac{1}{n!}\right) \\[10pt]
& = e - \frac{1}{2} (e- 1)= \frac{1}{2}(e+1)
\end{align}
Is it correct ???
if not correct then any hints/solution will be appreciated..
thanks in advance
| Using telescopic approach with$$a_n=\dfrac{1}{2\cdot4\cdot6\cdot\cdots\cdot2n}$$we have $$S=\sum_{n=1}^{\infty}{a_n-a_{n-1}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomial Division: dividing by a double root I have this fairly interesting problem that is based on Polynomial Division and/or factor/remainder theorem.
Determine the values of $a$ and $b$ such that $ax^4 + bx^3 -3$ is divisible by $(x-1)^2$.
This is interesting because the root we are dividing by is a double root, so its difficult to get $2$ equations and $2$ unknowns.
(note that one cannot use ideas from Calculus)
The only approach I have been able to come up with is to expand the perfect square divisor to a full quadratic and perform a brute force division, but it did not really lead me to a solution.
| Method one (compare the coefficients):
$$\begin{eqnarray}ax^4+bx^3-3 &=& (x-1)^2(cx^2+dx+e)\\
&=& (x^2-2x+1)(cx^2+dx+e)\\
&=& cx^4+(d-2c)x^3+(e-2d+c)x^2+(-2e+d)x+e\end{eqnarray}$$
From here we see that:
$$
\begin{eqnarray*}
c&=&a\\
e&=& -3\\
-2e+d&=& 0\implies d=-6\\
e-2d+c&=&0 \implies c=-9\implies a=-9\\
d-2c&=&b\implies b=-12
\end{eqnarray*}$$
Method two (Vieta formulas), $x_1=x_2=1$:
$$ 2+x_3+x_4 = -{b\over a}$$
$$ 1+2(x_3+x_4)+x_3x_4 = 0$$
$$ 2x_3x_4+ x_3+ x_4 =0$$
$$ x_3x_4 = -{3\over a}$$
from 2. and 3. equation we get
$$x_3+x_4 = -{2\over 3}\;\;\;{\rm and}\;\;\;x_3x_4 ={1\over 3}$$
From 4. equation we get $a=-9$ and from 1. equation we get $b=-12$.
Method three Since $1$ is root of a polynomial $p(x)=ax^4+bx^3-3$ we get $b=3-a$ so we have
$$p(x)=ax^4+3x^3 -ax^3-3 $$
$$= ax^3(x-1)+3(x-1)(x^2+x+1) $$
$$= (x-1)\underbrace{\Big(ax^3+3(x^2+x+1)\Big)}_{q(x)} $$
Now since $1$ double root we have also $q(1)=0$ so $a+9=0$.
Method four: (Horner schema)
$$\begin{array}{cccccc}
& a & b & 0 & 0 & -3 \\ \hline
1 & & a & a+b & a+b & a+b \\ \hline
& a & a+b & a+b & a+b & \color{red}{a+b-3} \\ \hline
1 & & a & 2a+b & 3a+2b & \\ \hline
& a & 2a+b & 3a+2b & \color{red}{4a+3b} & \\
\end{array}$$
Both red expressions must be zero...
Method five: Direct (long) division. What you are left ($1$. degree polynomial ) must be identical to $0$, so you get $2$ equations again...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Are there any 2 primitive pythagorean triples who share a common leg? So is it possible for:
$\gcd(a,b,c)=1$
$a^2+b^2=c^2$
and
$\gcd(a,d,e)=1$
$a^2+d^2=e^2$
?
| This system of equations:
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&z^2+b^2=x^2\end{aligned}\right.$$
Solutions have the form:
$$b=4tkp^2s^2$$
$$a=2(t^2-k^2)p^2s^2$$
$$z=4k^2s^4-t^2p^4$$
$$c=2(t^2+k^2)p^2s^2$$
$$x=4k^2s^4+t^2p^4$$
$t,k,p,s$ - integers asked us.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$ Solve: $(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$
Is my solution correct:
Answer given in the book : $y(x+1)=c_1+c_2\log(x+1)+\log3(x+1)$
For my later reference: link wolfram alpha
| Robert pointed out your mistake..
Here is another approach
$$(x+1)^3y''+3(x+1)^2y'+(x+1)y=6\log(x+1)$$
divide by $x+1$
$$(x+1)^2y''+3(x+1)y'+y=6\frac {\log(x+1)}{x+1}$$
on the left there is a derivative
$$((x+1)^2y')'+((x+1)y)'=6\frac {\log(x+1)}{x+1}$$
Integrate
$$(x+1)^2y'+(x+1)y=6\int \frac {\log(x+1)}{x+1} dx$$
$$(x+1)^2y'+(x+1)y=3\log^2(x+1)+K_1$$
Divide again by $x+1$
$$(x+1)y'+y=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$
$$((x+1)y)'=3\frac {\log^2(x+1)}{x+1}+K_1 \frac 1 {x+1}$$
Integrate again
$$(x+1)y=3\int\frac {\log^2(x+1)}{x+1}+K_1 \int \frac {dx} {x+1}$$
$$(x+1)y=\log^3(x+1)+K_1 \ln |{x+1}|+K_2$$
$$ \boxed {y=\frac {\log^3(x+1)}{x+1}+K_1 \frac {\ln |{x+1}|}{x+1}+\frac {K_2}{x+1}}$$
| {
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"url": "https://math.stackexchange.com/questions/2881378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of $\sum \frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$ Study the convergence of the following series as $\alpha \in \mathbb{R}$
$$\sum_{n=1}^{\infty}\frac{\left|\sin\left(\left(1-\frac{1}{n}\right)^n-\frac{1}{e}\right)\right|^{\alpha}}{e-\left(1+\frac{1}{n}\right)^n}$$
Maybe this series is quite simple but gives me hard times how to interpret it asymptotically. Could comparison test work?
| Hint for the denominator, you can write:
\begin{align*}
\left(1+\frac{1}{n}\right)^n
&= \exp\left( n \log\left(1+\frac{1}{n}\right)\right) \\
&= \exp\left( n\left(\frac{1}{n} - \frac{1}{2 n^2} + o\left(\frac{1}{n^2} \right)\right)\right) \quad (n\rightarrow \infty) \\
&= e^1\left( \exp\left( - \frac{1}{2 n} + o\left(\frac{1}{n} \right)\right) \right)\quad (n\rightarrow \infty) \\
&= e^1\left(1 - \frac{1}{2n} + o\left(\frac{1}{n} \right) \right) \quad (n\rightarrow \infty) \\
\end{align*}
Therefore:
$$ e - \left(1+\frac{1}{n}\right)^n \underset{n\rightarrow\infty}{ \sim \frac{e}{2n} } $$
Using the same method for the numerator, you might be able to discuss the convergence of the series.
| {
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"timestamp": "2023-03-29T00:00:00",
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Derive the following identity $1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}$.
Count the elements of the following set
$$A=\{(x,y,z): 1\leq x,y,z \leq n+1, z>\max\{x,y\}\}.
$$
From this derive the following identity:
$$1^2+2^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6}.$$
In the same manner find the formula for $1^k + 2^k + \ldots + n^k$ for $k=3,4$.
It is easy to see that $|A| = 1^2 + 2^2 + \ldots + n^2$, since from the sum rule we have $$|A| = \sum_{i=0}^{n+1} |\{(x,y,i): 1\leq x,y,z \leq n+1, i> \max\{x,y\}\}| = \sum_{i=0}^{n+1} i^2$$ (as we can choose $x$ and $y$ in $i \times i$ ways for each $i$).
However I can't see why is $|A|$ equals $\dfrac{n(n+1)(2n+1)}{6}$.
| We can prove it by induction:
Step 1: check it for base case (n=1)
If we replace $n$ by one, we get: $$ 1^2=\frac{1\times 2 \times 3}{6}$$
which holds $\checkmark$
Step 2: assume it is true for $n=k$ and check if holds for $n=k+1$
If it holds for $n=k$, then we have:
$$1^2+2^2+…+k^2=\frac{k(k+1)(2k+1)}{6}$$
Now we show it also holds for $n=k+1$. By replacing $n$ by $k+1$ we should show:
$$1^2+2^2+…+(k+1)^2=\frac{(k+1)(k+2)(2(k+1)+1)}{6}$$
or
$$1^2+2^2+…+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$$
(I want to skip some parts here, but its straight forward). Now the right hand side can be reshaped as:
$$\frac{(k+1)(k+2)(2k+3)}{6}=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
which proves it holds for $n=k+1$ if it holds for $n=k$ $\checkmark$
This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\sqrt {4x^2+7x}-2x}$$
Dividing numerator and denominator by x:
$$=\lim_{x\to -\infty} \frac{7}{\sqrt{4+\frac{7}{x}}-2}$$
$$= \frac{7}{\sqrt{4+\frac{7}{-\infty}}-2}$$
$$= \frac{7}{\sqrt{4+0}-2}$$
$$=\frac{7}{2-2}$$
$$=\infty$$
Conclusion: Limit does not exist.
Why is my solution wrong?
Correct answer: $\frac{-7}{4}$
| hint When $x$ goes to $-\infty,$ it becomes negative .
on the other hand,
we have $$\boxed{\sqrt{x^2}=|x|}$$
the mistake you made can be corrected by $$\sqrt{(-x)^2}=-x$$.
In the denominator, factor out by $(-x)^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ x,y ∈\Bbb{Z} $ find $x$ and $y$ given: $2x^2-3xy-2y^2=7$ We are given an equation:
$$2x^2-3xy-2y^2=7$$
And we have to find $x,y$ where $x,y ∈\Bbb{Z}$.
After we subtract 7 from both sides, it's clear that this is quadratic equation in its standard form, where $a$ coefficient equals to 2, $b=-3y$ and $c=-2y^2-7$. Thus discriminant equals to $9y^2+16y^2+56=25y^2+56$.
Now $x = \frac{3y±\sqrt{25y^2+56}}{4}$ , $x$ to be an integer $\sqrt{25y^2+56}$ has to be an integer too. I substituted $y$ as nonnegative integer first, because the answer wouldn't differ anyway as it is squared and found that when $y=1$, $\sqrt{25y^2+56}=9$, so we get $x = \frac{3±9}{4}$ and when we have plus sign we get $x = \frac{3+9}{4}=3$. So there we have it, $x=3, y=1$ is on of the solution. But $y=-1$ will also work because $y$ is squared, again, we get $x = \frac{-3±9}{4}$, if we have minus sign we have $x=-3$. Thus another solution, leaving us with:
$$ x=3, y=1$$
$$ x=-3, y=-1$$
I checked other integers for $y$ but none of them lead to solution where $x$ is also an integer. But here's problem, how do I know for sure that these two solutions are the only solutions, I can't obviously keep substituting $y$ as integers, as there are infinitely many integers. So that's why I came here for help.
| Notice $$2x^2-3xy-2y^2=(2x+y)(x-2y)=7.$$
Therefore, we have the four cases:
1) $2x+y=1, x-2y=7.$ Thus $x=\dfrac{9}{5},y=-\dfrac{13}{5}$, which are not integers.
2) $2x+y=7, x-2y=1.$ Thus $x=3,y=1$, which is a group of proper solution.
3) $2x+y=-1, x-2y=-7.$ $x=-\dfrac{9}{5},y=\dfrac{13}{5}$, which are not integers.
4)$2x+y=-7, x-2y=-1.$ Thus $x=-3,y=-1$, which is a second group of proper solution.
As a result, we have find two group of integer solution that $$x=3, y=1,$$ or $$x=-3,y=-1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How does $f(x)= x \sin(\frac{\pi}{x})$ behave? I think this function is increasing for $x>1$ but wanted to find the reason. So I thought about taking the derivative:
$f(x)= x \sin(\frac{\pi}{x})$
Aplying the chain an the product rule, we get:
$f'(x)= \sin(\frac{\pi}{x})-\frac{\pi}{x} \cos (\frac{\pi}{x})$
The function is increasing if the derative is more than or equal to $0$, so:
$\sin(\frac{\pi}{x})-\frac{\pi}{x} \cos (\frac{\pi}{x}) \ge 0$
$\sin(\frac{\pi}{x}) \ge \frac{\pi}{x} \cos (\frac{\pi}{x}) $
Since $ \cos ( x) > 0$, if $ 0< x < \pi$, $ \cos (\frac{\pi}{x}) > 0 $, because $ 0<\frac { \pi}{x}< \pi$.
$ \tan (\frac{\pi}{x}) \ge \frac{\pi}{x}$
I get to this point and don't know how to continue. I'd like you to help me or give me a hint, or maybe see a different way of showing it. Anyway, thanks.
| Try the Maclaurin series for tangent:
$$\tan x = x + \frac{x^3}{3} + \frac{2 x^5}{15} + \ldots$$
All terms are positive when $x > 0$, so $\tan x > x$. This proves that $\tan \frac{\pi}{x} > \frac{\pi}{x}$ for $0 < x < \frac{\pi}2$.
Alternatively, define $f(x) = \tan x - x$. Then $f(0) = 0$ and $f'(x) = \sec^2 x - 1 > 0$ for $0 < x < \frac{\pi}2$, so $f(x)$ is increasing, and $f(x) > 0$ on $0 < x < \frac{\pi}2$.
Now, here's the catch: remember that tangent is not defined at $x = \frac{\pi}2$. However, the argument of your tangent function is $\frac{\pi}{x}$. So, for $x > 2$, this fraction decreases from $x = 2$ to $x = \infty$, and in fact, approaches zero. Therefore, you are safe with your argument showing that the original function is increasing for $x > 2$. For $1 < x \leq 2$, you need to be a bit more careful.
| {
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"timestamp": "2023-03-29T00:00:00",
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Last digit of sequence of numbers
We define the sequence of natural numbers
$$
a_1 = 3 \quad \text{and} \quad a_{n+1}=a_n^{a_n},
\quad
\text{ for $n \geq 1$}.
$$
I want to show that the last digit of the numbers of the sequence $a_n$ alternates between the numbers $3$ and $7$. Specifically, if we symbolize with $b_n$ the last digit of $a_n$, I want to show that
$$
b_n =
\begin{cases}
3, & \text{if $n$ is odd}, \\
7, & \text{if $n$ is even}.
\end{cases}
$$
There is a hint to prove that for each $n \in \mathbb{N}$, if $a_n \equiv 3 \pmod{5}$ then $a_{n+1} \equiv 2 \pmod{5}$ and if $a_n \equiv 2 \pmod{5}$ then $a_{n+1} \equiv 3 \pmod{5}$.
First of all, if we take $a_n \equiv 3 \pmod{5}$, then $a_{n+1}=3^3\pmod{5} \equiv 2 \pmod{5}$.
If $a_n \equiv 2 \pmod{5}$, then $a_{n+1}=2^2 \pmod{5}=4$. Or am I doing something wrong?
And also how does it follow, if we have shown the hint, that $b_n$ is $3$ when $n$ is odd, and $7$ if $n$ is even?
| The mistake you are making is that if $a_n \equiv 2 \pmod 5$ it's not true that $a_n^{a_n} \equiv 2^2 \pmod 5$. The reason behind this is that the exponents aren't repeating in blocks of $5$, but instead in blocks of $\phi(5) = 4$, in your case. Indeed by Fermat's Little Theorem we have that $a_n^4 \equiv 1 \pmod 5$. Thus you need to find $a_n \pmod 4$ first.
This isn't hard to do, as $a_1 = 3$. Thus whenever it's raised to an odd power we get that $a_n \equiv -1 \pmod 4$. Hence we have:
$$a_n \equiv a_{n-1}^{a_{n-1}} \equiv a_{n-1}^{-1} \pmod 5$$
Now use the fact that $2$ is the modular inverse of $3$ modulo $5$ to conclude that:
$$a_n \equiv \begin{cases}
3 \pmod 5, & \text{if $n$ is odd} \\
2 \pmod 5, & \text{if $n$ is even}
\end{cases}$$
Finally note that $a_n \equiv 1 \pmod 2$ and use Chinese remainder Theorem to conclude that:
$$a_n \equiv \begin{cases}
3 \pmod{10}, & \text{if $n$ is odd} \\
7 \pmod{10}, & \text{if $n$ is even}
\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$ , find $f(x)$
Find $f(x)$ if $f(x)-2f(\frac{x}{2})+f(\frac{x}{4})=x^2$, where $x, f(x)\in (-\infty , \infty)$ and $f(x)$ is continuous.
| $y=mx^2+c$ is one solution where $c\in R$
$$mx^2-2\cdot m \frac{x^2}{4}+m \frac{x^2}{16}=x^2$$
$$m=\frac{16}{9}$$
$$y=\frac{16}{9}x^2+c$$
As mentioned by @lulu if $f(x)$ is a solution than $f(x)+c$ is also a solution.
| {
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Qualitative inspection of solutions to $x^{4}-2x+1=0$ Consider the following polynomial
$$
x^{4}-2x+1=0
$$
Is it possible to check if there is or there is not a solution in $x\in\left]0,1\right[$ without explicitly evaluating the expression? What other tests are there to qualitatively classify the solutions for this polynomial?
| Since $x=1$ works,
$$
x^4 - 2x + 1 = (x-1)p(x)\quad [p \in \mathbb R[x]_3].
$$
Now
$$
x^4 - 2x +1 = x^2(x^2 -1) + (x-1)^2 = (x-1)(x-1+x^2(x+1)) = (x-1)(x^3 + x^2 + x - 1).
$$
Then $p(x) = x^3 + x^2 + x-1$. Since
$$
p(1)= 2 >0, p(0) = -1 < 0,
$$
by intermediate value theorem, $p$ has a root in $(0,1)$, hence so does $x^4 - 2x + 1$.
| {
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Tough Divisibility Problem
When the five digit number $2A13B$ is divided by $19$, the remainder is $12$. Determine the remainder of $3A21B$ when divided by $19$.
$$2A13B \equiv 12 \pmod{19}$$
$$20000 + 1000A + 100 + 30 + B \equiv 12 \pmod{19}$$
$$ 5 + 12A + 5 + 11 + B \equiv 12 \pmod{19}$$
$$ 21+ 12A+ B \equiv 12 \pmod{19}$$
$$ 12A+ B + 9 \equiv 0 \pmod{19}$$
This is where I'm stuck.
| \begin{align}30\,000+1\,000A+200+10+B\equiv x\pmod{19}&\iff-1+12A+10+10+B\equiv x\pmod{19}\\&\iff12A+B\equiv x\pmod{19}.\end{align}Therefore, since $12A+B+9\equiv0\pmod{19}$, take $x=10$.
| {
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Evaluate: $ \int \frac{\sin x}{\sin x - \cos x} dx $ Consider
$$ \int \frac{\sin x}{\sin x - \cos x} dx $$
Well I tried taking integrand as $ \frac{\sin x - \cos x + \cos x}{\sin x - \cos x} $ so that it becomes,
$$ 1 + \frac{\cos x}{\sin x - \cos x} $$
But does not helps.
I want different techniques usable here.
| Let $$I_{1} = \int \frac{\sin x}{\sin x - \cos x}dx, \quad I_{2} = \int \frac{\cos x}{\sin x - \cos x}dx$$
Then $$I_{1}-I_{2} = \int 1\, dx = x + c_{1}$$
$$I_{1}+I_{2} = \int \frac{\sin x + \cos x}{\sin x - \cos x} dx = \ln(\sin x - \cos x) + c_{2} $$
Then solve simultaneously
$$I_{1} = \frac{1}{2}\left(x+ \ln(\sin x - \cos x) \right) + c$$
| {
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"answer_id": 3
} |
Finding Jordan canonical form of a matrix given the characteristic polynomial I am trying to find the Jordan canonical form of a matrix $A$ given its characteristic polynomial. Suppose $A$ is a complex $5\times 5$ matrix with minimal polynomial $X^5-X^3$. The end goal of the problem is to find the characteristic polynomial of $A^2$ and the minimal polynomial of $A^2$.
I know that since the minimal polynomial of a matrix divides the characteristic polynomial of a matrix, then $A$ has the same minimal and characteristic polynomial, namely $X^5-X^3$. Now I am trying to find the JCF (Jordan Canonical Form) of $A$ to make it easier to compute $A^2$, since $A$ is conjugate to its JCF. So, since the characteristic polynomial of $A$ splits into $X^3(X+1)(X-1)$, then I know that the Jordan canonical form will have three Jordan blocks, 2 of size 1 corresponding to $1$ and $-1$ and one of size 3. Now, my problem is that I can't figure out the form of this third Jordan block. How do I know that it has the from $$\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$$
or the form
$$ \begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix}$$
or the form
$$\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}$$
Thanks for all your help!
| We have the minimal polynomial is $X^3(X^2-1)$. Over $\Bbb C$, the exponent of the irreducible factor $(x-a)$ in the minimal polynomial gives the size of the largest Jodan block. Thus we have a $3\times 3$ Jordan block corresponding to the eigenvalue $0$. The only possibility is the middle one:
$$A=\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right).$$
To see this, note that the other two cases you gave have smaller minimal polynomials. For example, if
$$A=\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right),$$
then $A$ is annihilated by $X^2(X^2-1)$. If its not clear why, notice that an $n\times n$ matrix with all zeros and ones on the super diagonal is nilpotent with minimal polynomial $X^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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