Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) =
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.
| We have $\sqrt{3}+i=2e^{i\pi/6}$ and $\sqrt{3}-i=2e^{-i\pi/6}$. So
\begin{eqnarray*}
z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m} = 2^{n-m} e^{ i \pi (n+m) /6}.
\end{eqnarray*}
So we require $ n+m \equiv 0 \pmod{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\int_{/6}^{/2}\sin^3(x+\sin3x+\cos3x)\,dx 4$ Evaluate
$\int_{/6}^{/2}\sin^3(x+\sin3x+\cos3x)\,dx </4$
I tried to use the fact that the $\sin^3x<\sin x$. It does not help. This qesion is proposed by Jalil Hajimir to RMM.
| There is more than likely a much simpler solution.
Use the fact that, over the considered range the cube of the sine is smaller than the cube of its argument. Developing the later and using a couple of integration by parts plus triple angle formulae, you end for
$$I=\int \left(x+\sin3x+\cos3x\right)^3\,dx$$
$$I=\frac{x^4}{4}+\frac{3 x^2}{2}+x^2 \sin (3 x)-x^2 \cos (3 x)+\frac{2}{3} x \sin (3
x)+\frac{5}{18} \sin (3 x)+\frac{1}{12} \sin (6 x)-\frac{1}{18} \sin (9
x)+\frac{2}{3} x \cos (3 x)-\frac{1}{2} x \cos (6 x)-\frac{5}{18} \cos (3
x)-\frac{1}{18} \cos (9 x)$$ and, using the bounds
$$J=\int_{\frac \pi 6}^{\frac \pi 2} \left(x+\sin3x+\cos3x\right)^3\,dx=\frac{ 5 \pi ^4+18 \pi ^2 -90 \pi-216} {324}\approx 0.512207$$ which is even smaller than $\frac \pi 6$.
| {
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"url": "https://math.stackexchange.com/questions/3499831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Derivative of sum of two functions I have to find $\frac{dy}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$ but would like to find where I made a mistake in my solution.
Here is my work:
\begin{align}
(f+g)'=& \ f'+g', f = x \sqrt{x}, g= \frac{1}{x^2 \sqrt x}\\ f=& \ x \cdot x^{1/2} = x^{3/2}
\\ f'=& \ \frac{3}{2x\sqrt{x}} \\ \\
g =& \ x^{-2}\cdot x^{-1/2} \\ g =& \ x^{-5/2} \\ g' =& \ -\frac{5}{2x^3 \sqrt{x}} \\ \\ f'+g' =& \ \frac{3}{2x\sqrt{x}} - \frac{5}{2x^3 \sqrt{x}} \\ =& \ \frac{3x^2-5}{2x^3 \sqrt x} \\ \frac{dy}{dx}=& \ \frac{\sqrt x (3x^2-5)}{2x^4}
\end{align}
| $$\frac{d}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$$$$=\frac{d}{dx}\left(x\sqrt{x}\right)+\frac{d}{dx}\ \left(\frac{1}{x^{2}\sqrt{x}}\right)$$$$=\frac{d}{dx}\left(x^{\frac{3}{2}}\right)+\frac{d}{dx}\left(x^{\large-\frac{5}{2}}\right)$$$$=\frac{3}{2}\large x^{\frac{1}{2}}-\frac{5}{2}\large x^{-\frac{7}{2}}$$$$=\frac{3}{2}\sqrt{x}-\frac{5}{2\sqrt{x^{7}}}$$
In most high school calculus books the power rule for derivatives is proven used binomial theorem, but binomial theorem is not helpful for all arbitrary real powers.
For more information about the derivation of the function $x^{n}$ for all real $n,x$ refer to the link
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the result of diagonalization of a matrix, determine the two invertible matrices. Determine $3$ by $3$ invertible matrices P and Q, such that
$$P\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{array} \right)Q = \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array} \right).
$$
I tried to diagonalize the first matrix but the result diagonalized matrix I got is
$$\left( \begin{array}{ccc}
0 & 0 & 0 \\
0 & -\sqrt{2} & 0 \\
0 & 0 & \sqrt{2}
\end{array} \right).
$$
The signal doesn't seem to be right...
| As it stands, the question does not require $P$ and $Q$ to be inverses of each other, or real orthogonal, etc.. So, there is no need to diagonalise the matrix on the left by similarity or by congruence. In fact, since the matrix on the LHS has a zero diagonal and the one on the RHS is positive semidefinite, they are neither similar nor congruent to each other. Thus there does not exist any pair of invertible real matrices $P$ and $Q$ that solves the equation with $P=Q^{-1}$ or $P=Q^T$.
The equation is satisfied if and only if
$$
\pmatrix{1&0\\ 0&1\\ 1&0}\pmatrix{0&1&0\\ 1&0&1}
=\pmatrix{0&1&0\\ 1&0&1\\ 0&1&0}
=P^{-1}\pmatrix{1\\ &1\\ &&0}Q^{-1}
=P^{-1}\pmatrix{1&0\\ 0&1\\ 0&0}\pmatrix{1&0&0\\ 0&1&0}Q^{-1}.
$$
So, it suffices to pick $P$ and $Q$ such that
$$
P^{-1}=\pmatrix{1&0&\ast\\ 0&1&\ast\\ 1&0&\ast},
\ Q^{-1}=\pmatrix{0&1&0\\ 1&0&1\\ \ast&\ast&\ast}.
$$
For instance, we may set
$$
P^{-1}=\pmatrix{1&0&1\\ 0&1&0\\ 1&0&0},
\ Q^{-1}=\pmatrix{0&1&0\\ 1&0&1\\ 1&0&0},
$$
so that
$$
P=\pmatrix{0&0&1\\ 0&1&0\\ 1&0&-1},
\ Q=\pmatrix{0&0&1\\ 1&0&0\\ 0&1&-1}.
$$
Alternatively, you may also continue your work. Suppose you have found two invertible matrices $P_1$ and $Q_1$ such that
$$
P_1\pmatrix{0&1&0\\ 1&0&1\\ 0&1&0}Q_1=\pmatrix{0\\ &-\sqrt{2}\\ &&\sqrt{2}}.
$$
Then
$$
\pmatrix{0\\ &-\frac{1}{\sqrt{2}}\\ &&\frac{1}{\sqrt{2}}}P_1\pmatrix{0&1&0\\ 1&0&1\\ 0&1&0}Q_1=\pmatrix{0\\ &1\\ &&1},
$$
So, if we apply a further permutation to flip the first and the last diagonal entries on the RHS, we obtain
$$
\pmatrix{0&0&1\\ 0&1&0\\ 1&0&0}\pmatrix{0\\ &-\frac{1}{\sqrt{2}}\\ &&\frac{1}{\sqrt{2}}}\pmatrix{0&1&0\\ 1&0&1\\ 0&1&0}Q_1\pmatrix{0&0&1\\ 0&1&0\\ 1&0&0}=\pmatrix{1\\ &1\\ &&0}.
$$
Hence we may set
$$
P=\pmatrix{0&0&1\\ 0&1&0\\ 1&0&0}\pmatrix{0\\ &-\frac{1}{\sqrt{2}}\\ &&\frac{1}{\sqrt{2}}}P_1,
\quad Q=Q_1\pmatrix{0&0&1\\ 0&1&0\\ 1&0&0}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $a_n:a_{n+1}$ if $a_n=\int_0^{\pi/2}\cos^nx.\cos nx.dx$
Find $a_n:a_{n+1}$ if $a_n=\int_0^{\pi/2}\cos^nx.\cos nx.dx$
Attempt 1
$$
a_{n+1}=\int_0^{\pi/2}\cos^{n+1}x.\cos(n+1)x.dx=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x-(n+1)\int\cos^{n}x.\frac{\sin (n+1)x}{(n+1)}.dx\\
=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x-\int\cos^{n}x.{\sin (n+1)x}.dx\\
=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x-\bigg[\frac{-\cos(n+1)x}{n+1}.\cos^nx-n\int\cos^{n-1}x.\frac{-\cos(n+1)x}{n+1}.dx\bigg]\\
=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x+\frac{\cos(n+1)x}{n+1}.\cos^nx+\frac{n}{n+1}\int\cos^{n-1}x.\big[\cos nx.\cos x-\sin nx\sin x\big]dx\\
=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x+\frac{\cos(n+1)x}{n+1}.\cos^nx+\frac{n}{n+1}\bigg[\color{red}{\int\cos^nx\cos nxdx}+\int\cos^{n-1}x.\sin x.\sin nx.dx\bigg]\\
=\frac{\sin (n+1)x}{n+1}.\cos^{n+1}x+\frac{\cos(n+1)x}{n+1}.\cos^nx+\frac{n}{n+1}\bigg[\color{red}{a_n}+\int\cos^{n-1}x.\sin x.\sin nx.dx\bigg]
$$
I don't think it is leading anywhere, is there any trickier way to see the solution ?
Note: The solution given in my reference is $2:1$ and I understand that $$
\int_0^{\pi/2}\sin^nx.dx=\int_0^{\pi/2}\cos^nx.dx=\begin{cases}
\dfrac{(n-1)(n-3)....2}{n(n-2)....1}\quad\text{if $n$ is odd}\\
\dfrac{(n-1)(n-3)....1}{n(n-2)....2}\quad\text{if $n$ is even}
\end{cases}
$$
Thanx @Math1000,
Attempt 2
$$
a_n = \int_0^{\frac\pi2}\cos^n x\cos nx\ \mathsf dx = \frac{1}{2.2^n} \int_0^{\frac{\pi }{2}} \left(e^{i x}+e^{-i x}\right)^n \left(e^{i n x}+e^{-i n x}\right) \, dx\\
I_1=\int_0^{\frac{\pi }{2}} \left(e^{i x}+e^{-i x}\right)^n e^{i n x}.dx=\bigg[(e^{i x}+e^{-i x})^n\frac{e^{inx}}{in}\bigg]_0^{\pi/2}-n\int(e^{i x}+e^{-i x})^{n-1}\frac{e^{i n x}}{in}.dx\\
=\frac{1}{in}-n\int(e^{i x}+e^{-i x})^{n-1}\frac{e^{i n x}}{in}.dx\\
$$
$$
I_2=\int_0^{\frac{\pi }{2}} \left(e^{i x}+e^{-i x}\right)^n e^{-i n x}.dx=\bigg[(e^{i x}+e^{-i x})^n\frac{e^{-inx}}{-in}\bigg]_0^{\pi/2}-n\int(e^{i x}+e^{-i x})^{n-1}\frac{e^{-i n x}}{-inx}.dx\\
=\frac{-1}{in}+n\int(e^{i x}+e^{-i x})^{n-1}\frac{e^{-i n x}}{in}.dx\\
a_n=\frac{i}{4}\bigg(\int_0^{\pi/2}(e^{i x}+e^{-i x})^{n-1}e^{i n x}.dx-\int_0^{\pi/2}(e^{i x}+e^{-i x})^{n-1}e^{-i n x}.dx\bigg)\\
=\frac{i}{2^{n+1}}\int_0^{\pi/2}(e^{i x}+e^{-i x})^{n-1}\Big(e^{i n x}-e^{-i n x}\Big).dx
$$
| $\newcommand{\d}[1]{\, \mathrm{d}#1}$
It's easier to start by expanding $\cos{(n+1)x}$.
\begin{align*}
a_{n+1} &= \int_0^\frac{\pi}{2} \cos^{n+1}{x}\cos{(n+1)x} \d{x} \\
&= \int_0^\frac{\pi}{2} \cos^{n+1}{x}(\cos{nx}\cos{x} - \sin{nx}\sin{x}) \d{x} \\
&= \int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} - \int_0^\frac{\pi}{2} \sin{x}\cos^{n+1}{x}\sin{nx} \d{x}
\end{align*}
Let's first look at the second term. Integrating by parts yields:
\begin{align*}
\int_0^\frac{\pi}{2} \sin{x}\cos^{n+1}{x}\sin{nx} \d{x} &= \left[-\frac{1}{n+2}\cos^{n+2}{x}\sin{nx}\right]_{x = 0}^{x = \frac{\pi}{2}} + \frac{n}{n+2}\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} \\
&= \frac{n}{n+2}\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x}
\end{align*}
Back to $a_{n+1}$:
\begin{align*}
a_{n+1} &= \int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} - \frac{n}{n+2}\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} \\
&= \frac{2}{n+2}\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x}
\end{align*}
We now examine the term in RHS. Integrating by parts again:
\begin{align*}
\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} &= \left[\frac{1}{n}\cos^{n+2}{x}\sin{nx}\right]_{x = 0}^{x = \frac{\pi}{2}} + \frac{n+2}{n}\int_0^\frac{\pi}{2}\sin{x}\cos^{n+1}{x}\sin{nx} \d{x} \\
&= \frac{n+2}{n}\int_0^\frac{\pi}{2}\sin{x}\cos^{n+1}{x}\sin{nx} \d{x} \\
&= \frac{n+2}{n}\left[-\frac{1}{n}\sin{x}\cos^{n+1}{x}\cos{nx}\right]_{x = 0}^{x = \frac{\pi}{2}} + \frac{n+2}{n^2}\int_0^\frac{\pi}{2} (\cos^{n+2}{x} - (n+1)\sin^2{x}\cos^n{x})\cos{nx} \d{x} \\
&= \frac{n+2}{n^2}\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} - \frac{(n+1)(n+2)}{n^2}\int_0^\frac{\pi}{2} \sin^2{x}\cos^n{x}\cos{nx} \d{x} \\
&= \frac{n+2}{n^2}\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} - \frac{(n+1)(n+2)}{n^2}\int_0^\frac{\pi}{2} (1 - \cos^2{x})\cos^n{x}\cos{nx} \d{x} \\
&= \left(\frac{n+2}{n}\right)^2\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} - \frac{(n+1)(n+2)}{n^2}\color{red}{\int_0^\frac{\pi}{2} \cos^n{x}\cos{nx} \d{x}}
\end{align*}
Therefore:
\begin{align*}
&\left(1 - \left(\frac{n+2}{n}\right)^2\right)\int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} = -\frac{(n+1)(n+2)}{n^2}a_n \\
&\implies \int_0^\frac{\pi}{2} \cos^{n+2}{x}\cos{nx} \d{x} = \frac{n+2}{4}a_n
\end{align*}
Substituting back finally yields the desired result:
$$
a_{n+1} = \frac{1}{2}a_n
$$
| {
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"url": "https://math.stackexchange.com/questions/3504581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $\frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$.
Given positives $x$ and $y$ such that $x^2 + y^2 = xy + 1$, prove that $$\large \frac{x}{x^2 + y} + \frac{y}{y^2 + x} \le 1$$
Well, I have provided my solution below, which is clearly not the most straightforward and convenient one. I would be greatly appreciated if you could come up with any other solutions.
| We need to prove that $$(x+y)(x-y)^2\geq1-x^2y^2$$ or $$(x+y)(x-y)^2\geq(x-y)^2(x^2+y^2)$$ or $$(x+y)^2(x^2-xy+y^2)\geq(x^2+y^2)^2$$ or $$(x-y)^2xy\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to start solving this Logarithmic problem? If $abc= 2^6$, $a, b, c \ge 0$, $\log_2 (a)\log_2 (bc)+\log_2 (b)\log_2 (c)= 10$, find $\sqrt{((\log_2 (a))^2 + (\log_2 (b))^2 + (\log_2 (c))^2}$
| Using the substitution suggested in the comments gives the equations
\begin{cases}
A+B+C=6\\
AB+AC+BC=10
\end{cases}
We can then square the first equation to give
\begin{align}
(A+B+C)^2
&=A^2+B^2+C^2+2(AB+AC+BC)\\
&=A^2+B^2+C^2+2\cdot10\\
&=A^2+B^2+C^2+20\\
&=6^2\\
&=36\\
\end{align}
Thus we get
$$A^2+B^2+C^2=36-20=16$$
Note that you want the quantity
$$\sqrt{A^2+B^2+C^2}=\sqrt{16}=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| Apply the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ to rewrite the equation
$$\frac\pi6=\tan^{-1}\frac{11}{x} -\tan^{-1} \frac{1}{x}
=\tan^{-1}\frac{\frac{10}x}{1+\frac{11}{x^2}}$$
Then, take $\tan(\cdot)$ on both sides along with $\tan\frac\pi6=\frac1{\sqrt3}$,
$$\frac{11}{x^2}+\frac{10\sqrt3}x+1=0$$
which is a quadratic equation in $1/x$. Solve to obtain
$$x=5\sqrt3\pm8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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probability that two of these boxes contain exactly $2$ and $3$ balls is
If $10$ different balls are to be placed in $4$ distinct boxes at random, then the probability that two of these boxes contain exactly $2$ and $3$ balls is
What I tried:
Total number of ways $\displaystyle 4^{10}$
probability that two of these boxes contain exactly $2$ and $3$ balls is
$\displaystyle \binom{4}{2}\cdot 2^5$
So required probability $$\frac{\binom{4}{2}\cdot 2^5}{4^{10}}$$
But answer given as $\displaystyle \frac{945}{2^{10}}$
| We have the following variants to distribute balls into the boxes:
$$
\underbrace{2+3+0+5}_{A_1}, \quad \underbrace{2+3+1+4}_{A_2}, \quad \underbrace{2+3+2+3}_{A_3}.
$$
And the probability that two of these box contain exactly $2$ and $3$ balls is the sum of probabilities for these variants to distribute balls:
$$
\mathbb P(A)=\mathbb P(A_1)+\mathbb P(A_2)+\mathbb P(A_3).
$$
Find $\mathbb P(A_1)$. There are $\dfrac{10!}{2!\cdot 3!\cdot 0!\cdot 5!}$ ways to chose balls for each box and $4!$ ways to arrange boxes: box for $2$ balls can be chosen by $4$ ways, box for $3$ balls - by $3$ ways, for $0$ balls - by $2$ ways and the rest box is for $5$ balls. So
$$
\mathbb P(A_1) = \frac{10! \cdot 4!}{2!\cdot 3! \cdot 0!\cdot 5! \cdot 4^{10}} = \frac{60480}{4^{10}},
$$
The same way
$$
\mathbb P(A_2) = \frac{10! \cdot 4!}{2!\cdot 3! \cdot 1!\cdot 4! \cdot 4^{10}} = \frac{302400}{4^{10}}.
$$
For $A_3$, we need to chose pair of boxes for two balls by $\binom{4}{2}$ ways instead of $4!$. So
$$
\mathbb P(A_3) = \frac{10! \cdot \binom{4}{2}}{2!\cdot 3! \cdot 2!\cdot 3! \cdot 4^{10}} = \frac{151200}{4^{10}}.
$$
Finally
$$
\mathbb P(A)=\frac{514080}{4^{10}}=\frac{16065}{2\cdot 4^7}\approx 0,490264893.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding an angle in equilateral triangle
Given $\triangle ABC$ ($AC=AB$). $X$ - the point on side $AC$ such as $AX=BC$. $\angle A = 20^0$. Find $\angle XBC$.
Here is my attempt:
Let side $AB = a$, then side $BC = 2a \sin10^0$.
Construct $B_1X \parallel BC$.
Similar triangles $\triangle BAC \sim \triangle B_1AX$ gives $B_1X=4a\sin^2 10^0$.
$XK \perp BC$. From $\triangle CXK$: $XK=XC\cdot \cos 10^0=(a-2a\sin 10^0)\cos10^0$.
$BK= \frac{B_1X+BC}{2}=2a\sin^2 10^0+a\sin 10^0$.
$\tan XBK = \frac{KX}{BK}= \frac{\cos 10^0(1-2\sin 10^0)}{\sin 10^0(1+2\sin 10^0)}= \cot 10^0 \cdot \frac{(1-2\sin 10^0)}{(1+2\sin 10^0)}$
Then I find perfect solution of this problems by @Seyed in this post Find $x$ angle in triangle.
That's why I have a question: is $\tan 70^0$ equal $\cot 10^0 \cdot \frac{(1-2\sin 10^0)}{(1+2\sin 10^0)}$ or I have a mistake in my attempt?
| Let $Y\in CX$,$Z\in AB$ and $X'\in AY$ such that $BY=ZY=ZX'.$
Thus, $$\measuredangle ZX'Y=\measuredangle X'YZ=180^{\circ}-\measuredangle ZYB-\measuredangle BYC=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ},$$
which gives $$\measuredangle AZX'=\measuredangle ZX'Y-\measuredangle A=40^{\circ}-20^{\circ}=20^{\circ},$$ which says $$AX'=ZX'=BC,$$ which gives $$X'\equiv X.$$
Id est, $$\measuredangle XBC=\measuredangle ABC-\measuredangle XBZ=80^{\circ}-\frac{1}{2}\cdot20^{\circ}=70^{\circ}.$$
By the way, you are right:
$$\tan70^{\circ}=\cot10^{\circ}\cdot\frac{1-2\sin10^{\circ}}{1+2\sin10^{\circ}}.$$
Indeed, $$\cot10^{\circ}\cdot\frac{1-2\sin10^{\circ}}{1+2\sin10^{\circ}}=\cot10^{\circ}\cdot\frac{\sin30^{\circ}-\sin10^{\circ}}{\sin30^{\circ}+\sin10^{\circ}}=$$
$$=\cot10^{\circ}\cdot\frac{2\sin10^{\circ}\cos20^{\circ}}{2\sin20^{\circ}\cos10^{\circ}}=\cot20^{\circ}=\tan70^{\circ}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $m$ such that $x^4 - (2m - 1)x^2 + 4m -5 = 0$ has real roots Consider the equation:
$$ x ^ 4 - (2m - 1) x^ 2 + 4m -5 = 0 $$
with $m \in \mathbb{R}$. I have to find the values of $m$ such that the given equation has all of its roots real.
This is what I did:
Let $ u = x^2, \hspace{.25cm} u\ge 0$
We get:
$$ u ^ 2 - (2m - 1)u + 4m -5 = 0 $$
Now since we have
$$ u = x ^ 2$$
That means
$$x = \pm \sqrt{u}$$
That means that the roots $x$ are real only if $u \ge 0$.
So we need to find the values of $m$ such that all $u$'s are $\ge 0$. If all $u$'s are $\ge 0$, that means that the sum of $u$'s is $\ge 0$ and the product of $u$'s is $ \ge 0 $. Using Vieta's formulas
$$S = u_1 + u_2 = - \dfrac{b}{a} \hspace{2cm} P = u_1 \cdot u_2 = \dfrac{c}{a}$$
where $a, b$ and $c$ are the coefficients of the quadratic, we can solve for $m$. We get:
$$S = - \dfrac{-(2m - 1)}{1} = 2m - 1$$
We need $S \ge 0$, so that means $m \ge \dfrac{1}{2}$ $(1)$
$$P = \dfrac{4m - 5 }{1} = 4m - 5$$
We need $P \ge 0$, so that means $m \ge \dfrac{5}{4}$ $(2)$
Intersecting $(1)$ and $(2)$ we get the final answer:
$$ m \in \bigg [ \dfrac{5}{4}, \infty \bigg )$$
My question is: Is this correct? Is my reasoning sound? Is there another way (maybe even a better way!) to solve this?
| You need also to consider $\Delta$ to be positive in order for the solutions to be real.
$\Delta = (2m-1)^2-4(4m-5)=4m^2-4m+1-16m+20=4m^2-20m+21$
$m_{1,2}=\frac{10 \pm \sqrt{100-84}}{4}=\frac{10 \pm 4}{4}=\{\frac{3}{2},\frac{7}{2} \}$
Thus $\Delta \geq 0 \Leftrightarrow m \in (-\infty, \frac{3}{2}] \cup [\frac{7}{2},\infty)$
Take for example $m=2$: now $m \geq \frac{5}{4}$ but the equation
$$
u^2-(2 \cdot 2 -1)u+4\cdot 2-5=0 \\
u^2-3u+3=0
$$
has not real solutions $\frac{3 \pm \sqrt{-3}}{2}$.
Hence we need $m \in [\frac{7}{2},\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
I thought about MVT , what i did is :
let $g(x) = \arctan\left(x\right) $
so by MVT there is $c\in (0,x)$ such that :
$g'(c) = \frac{g\left(x\right)-g\left(0\right)}{x-0} \implies \frac{1}{c^2+1} = \frac{\arctan\left(x\right)}{x}$
now i need to prove that $\frac{x}{c^2+1}>\frac{x}{\sqrt{x^2+1}}$
any hint how to prove this ?
thanks
| Let $f(x)= \arctan x - \frac{x}{\sqrt{1+x^2}}$. Evaluate
$$f'(x)=\frac{\sqrt{1+x^2}-1}{(1+x^2)^{3/2}}>0$$
So, $f(x)$ is strictly increasing for $x>0$. Thus, $f(x) > f(0) = 0$, or
$$\arctan x > \frac{x}{\sqrt{1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$
show that
$$\csc x \cot x=\frac{a^2-b^2}{4ab}$$
(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the answer.
| Rearrange
$$\frac{a^2-b^2}{4ab}=\frac{(a+b)(a-b)}{(a+b)^2-(a-b)^2}
=\frac{1}{\frac{a+b}{a-b}-\frac{a-b}{a+b}}$$
Then, substitute $\frac{a+b}{a-b}=\frac{\csc x}{\cot x}$,
$$\frac{a^2-b^2}{4ab}=\frac1{\frac{\csc x}{\cot x}-\frac{\cot x}{\csc x}}
=\frac{\csc x\cot x}{\csc^2 x-\cot^2x}=\csc x\cot x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3520401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate matrix by using Cayley-Hamilton theorem Calculate matrix $B = A^{10}-3A^9-A^2+4A$ using Cayley-Hamilton theorem on $A$.
$$A = \begin{pmatrix}
2 & 2 & 2 & 5 \\
-1 & -1 & -1 & -5 \\
-2 & -2 & -1 & 0 \\
1 & 1 & 3 & 3
\end{pmatrix}$$
Now, I've calculated the characteristic polynomial of $A$:
$P_A(\lambda) = \lambda^4-3\lambda^3+\lambda^2-3\lambda$
So I know that $P(A) = 0 \rightarrow A^4-3A^3+A^2-3A = 0$, hereby $0$ is a $4 \times 4$ matrix.
$B = A^{10}-3A^9-A^2+4A = A^4 \cdot A^6 - 3A^3 \cdot A^6 + A^2 \cdot (-1) -3A + 7A $
Can I go further from here?
I tried doing polynomial division $B/P(A)$, but i stopped halfway since the numbers were getting too big and it didn't seem to get me to the right solution.
How do I transform $B$, so that I can use $P(A)=0$ and calculate B.
| By polynomial division, we see that
\begin{align}
x^{10}-3x^9-x^2+4x = (x^4-3x^3+x^2-3x)(x^6-x^4+x^2+1)+x
\end{align}
which means
\begin{align}
A^{10}-3A^9-A^2+4A = (A^4-3A^3+A^2-3A)(A^6-A^4+A^2+1)+A = A.
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$ where $x, y, z \ge 0$?
For $x, y, z \in \mathbb R^+\cup\{0\} $, prove that
$$(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$$ and
$$(x+y+z)^{(x+y+z)^2}x^{x^2}y^{y^2}z^{z^2} \ge (x+y)^{(x+y)^2}(y+z)^{(y+z)^2}(z+x)^{(z+x)^2}$$
[Edited] I currently solved the first one. Please check the procedures I made:
$$x(x+y+z)=x^2+xy+xz \le x^2+xy+xz+yz=(x+y)(z+x) $$
$$x^x(x+y+z)^{x} \le (x+y)^x(z+x)^x \cdots (1)$$ In the same way,
$$y^y(x+y+z)^y \le (x+y)^y(y+z)^y \cdots (2)$$
$$z^z(x+y+z)^z \le (y+z)^z(z+x)^z \cdots (3)$$
When we multiply all the three inequalities (1)~(3), since $x, y, z\ge0$:
$$(x+y+z)^{x+y+z}x^xy^yz^z \le (x+y)^{x+y}(y+z)^{y+z}(z+x)^{z+x}$$
However, I do not have any ideas to prove the second inequality although I spent about a whole week to solve this. I'm very confused since the direction of the sign of the second inequality is opposite to the previous one.
Also, this question belongs to the calculus. How can I apply the calculus in the proof? Thanks for your help.
| It should be $x>0$, $y>0$ and $z>0$, otherwise $x^x$ is not defined for $x=0$.
Let $f(z)=\sum\limits_{cyc}(x+y)\ln(x+y)-\sum\limits_{cyc}x\ln{x}-(x+y+z)\ln(x+y+z).$
Thus, $$f'(z)=\ln(x+z)+1+\ln(y+z)+1-\ln{z}-1-\ln(x+y+z)-1=$$
$$=\ln\frac{(x+z)(y+z)}{z(x+y+z)}=\ln\left(1+\frac{xy}{z(x+y+z)}\right)>0.$$
Thus, $$f(z)\geq \lim_{z\rightarrow0^+}f(z)=0$$ and we are done!
The second problem we can solve by the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
| There is one trick which is very useful when you face products, quotients, powr,.. : logarithmic differentiation.
$$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}\implies \log(f(x))=\frac 12 \log(3-x^2)-\log(3+x)$$
$$\frac{f'(x)}{f(x)}=\frac 12\frac{-2x}{3-x^2}-\frac 1{3+x}=-\frac{3 (x+1)}{(x+3) \left(3-x^2\right)}$$
Now
$$f'(x)=f(x) \times \frac{f'(x)}{f(x)}$$ Just simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b^2\sin^2 t}}$$
Now let $$p=\cos(2t)$$
then we get a single variable function as:
$$h(p)=\frac{1}{\sqrt{2}}\frac{(a+b)+p(a-b)}{\sqrt{a^2+b^2+p(a^2-b^2)}}$$
where $p \in [-1, 1]$
Now we can find critical point and find minimum.
Is there a better approach, i tried lagrange multipliers but very tedious
| $$(a+b)(ax^2+by^2)=p^2+ab$$ where $p=\sqrt{a^2x^2+b^2y^2}$
Now assuming $ab>0,$
$$\dfrac{p^2+ab}p\ge2\sqrt{p\cdot\dfrac{ab}p}$$ using AM-GM inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Computation of the $2n$-th order determinant Compute the determinant of the $2n-\text{th}$ order.
$$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\\2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\end{vmatrix}$$
My attempt:
I noticed the two following blocks:
$\begin{vmatrix}0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\3&0&\ldots&0&0&0&0&\ldots&0&2\end{vmatrix}\;\&\;\begin{vmatrix}2&0&\ldots&0&0&0&0&\ldots&0&3\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\end{vmatrix}$
I switched the blocks because I was dealing with the determinant of the even-order:
$$\begin{vmatrix}2&0&\ldots&0&0&0&0&\ldots&0&3\\0&2&\ldots&0&0&0&0&\ldots&3&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&\ldots&2&0&0&3&\ldots&0&0\\0&0&\ldots&0&2&3&0&\ldots&0&0\\0&0&\ldots&0&3&2&0&\ldots&0&0\\0&0&\ldots&3&0&0&2&\ldots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&3&\ldots&0&0&0&0&\ldots&2&0\\3&0&\ldots&0&0&0&0&\ldots&0&2\end{vmatrix}$$
Then I saw we can subtract $j-\text{th column}$ multiplied by $-\frac{3}{2}$ from the $(n-j+1)-\text{column}\;\forall j\in\{1,\ldots,2n\}$
Then I got a $\text{lower-triangular}$ matrix with entries $-\frac{5}{2}$ on the main diagonal.
My final result is: $$D_{2n}=\left(-\frac{5}{2}\right)^{2n}=\left(\frac{5}{2}\right)^{2n}$$
Is this correct?
| A recursion can also be helpful here:
*
*$n=1$: $D_2 = \begin{vmatrix} 2 & 3 \\ 3 & 2 \end{vmatrix} = -5$
*For $n >1$, expanding $D_{2n}$ along the first column gives:
$$D_{2n} = 2\cdot D_{2(n-1)}\cdot 2 - 3 \cdot D_{2(n-1)}\cdot 3 = -5D_{2(n-1)}$$
It follows
$$D_{2n} = (-5)^{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Diagonalizability of a matrix
Show that $$ A :=\begin{pmatrix} 1 & 0 & 0 \\ -2 & 1 & 2 \\ -2 & 0 & 3 \end{pmatrix}$$ is diagonalizable.
What I did:
First, I determined the characteristic polynomial $$\chi_A(X) = \det(X \cdot E_3-A)=(X-3)(X-1)(X-1)=X^3-5X^2+7X-3,$$
so the eigenvalues are $3$ and $1$.
I then determined the eigenspaces of each eigenvalue:
$$X=3: \left(\begin{array}{@{}ccc|c@{}}
2 & 0 & 0 & 0 \\
-2 & 2 & 2 & 0 \\
-2 & 0 & 0 & 0 \\
\end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}}
2 & 0 & 0 & 0 \\
0 & 2 & 2 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right),$$ so $x_1=0$, $x_2=-x_3$, $x_3=x_3$ and thus $V_3(C) = \left< \begin{pmatrix} 0\\-1\\1 \end{pmatrix} \right>$.
Analogous:
$$X=1: \left(\begin{array}{@{}ccc|c@{}}
0 & 0 & 0 & 0 \\
-2 & 0 & 2 & 0 \\
-2 & 0 & -2 & 0 \\
\end{array}\right) \leadsto \left(\begin{array}{@{}ccc|c@{}}
-2 & 0 & 2 & 0 \\
0 & 0 & -4 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}\right),$$ so $x_1=x_3=0$, $x_2=x_2$ and thus $V_1(A) = \left< \begin{pmatrix} 0\\1\\0 \end{pmatrix} \right>$.
It now follows that $\dim(V_3(C)) + \dim(V_1(A)) = 1+1=2 \lt 3 = \dim(A)$ and because of the $\lt$, A shouldn't be diagonalizable, but it is.
So where's the mistake? Thanks in advance!
| When $X=1$ the matrix becomes $$XI-A=I-A=\begin{bmatrix}0 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & 0 & -2\end{bmatrix}$$ which has rank 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3527328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim\limits_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$
Consider two sequences $(a_n)_{n\ge 1}$ and $(b_n)_{n\ge 1}$ of
positive real numbers such that
$$\lim_{n\to\infty} \frac {a_{n+1}}{n^2\cdot a_n}=x>0$$
and
$$\lim_{n\to\infty} \frac {b_{n+1}}{n\cdot b_n}=y>0$$
Evaluate:
$$\lim_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$$
My thoughts: Intuitively $\dfrac{a_{n+1}}{a_n} \sim xn^2$ and $\dfrac{b_{n+1}}{b_n} \sim yn$, so $a_n \sim x \sqrt[n]{(n!)^2}$ and $b_n \sim y\sqrt[n]{n!}$.
Since $\lim\limits_{n\to \infty} \dfrac{n}{\sqrt[n]{n!}} = e$, we should get:
$$\sqrt[n+1]{\frac{a_{n+1}}{b_{n+1}}} \sim \frac{x}{y}\cdot \sqrt[n+1]{(n+1)!} \sim\frac{x}{y}\cdot \frac{n+1}{e}$$
and the final answer should be $\dfrac{x}{ye}$. But how do we write this with sound arguments?
| I think I completed a proof thanks to Gary's idea with the $c_n$ substitution and the ratio limits. Bear with me. First, let's make a little different substitution $c_n = \sqrt[n]{\dfrac{a_n}{b_n}}$ and use the following criterion.
Lemma (Cauchy-d'Alembert): If $x_n > 0,\ (\forall) n \in \mathbb{N}$ and the limit $l = \lim\limits_{n\to \infty} \dfrac{x_{n+1}}{x_n}$ exists, then $\sqrt[n]{x_n}$ converges to $l$.
Since:
$$\lim_{n \to \infty}\frac{\dfrac{a_{n+1}}{(n+1)^{n+1}\cdot b_{n+1}}}{\dfrac{a_n}{n^n\cdot b_n}} = \lim_{n \to \infty} \left(\frac{a_{n+1}}{n^2\cdot a_n} \cdot \frac{n\cdot b_n}{b_{n+1}} \cdot \frac{n^{n+1}}{(n+1)^{n+1}}\right) = \frac{x}{y \cdot e}$$
we can apply Cauchy D'Alembert for
$$\dfrac{c_n}{n} = \sqrt[n]{\frac{a_n}{n^n\cdot b_n}} \to \frac{x}{y \cdot e}$$
From this we also have
$$\lim\limits_{n\to\infty}\, \frac {c_{n+1}}{c_n}=\lim\limits_{n\to\infty}\, \left[\frac {c_{n+1}}{n+1}\cdot\frac n{c_n}\cdot\frac {n+1}n\right]=1$$
and
$$\lim\limits_{n\to\infty}\, \left(\frac {c_{n+1}}{c_n}\right)^n=\lim\limits_{n\to\infty}\, \left(\frac {a_{n+1}}{n^2\cdot a_n}\cdot\frac {n\cdot b_n}{b_{n+1}}\cdot n \cdot \sqrt[n]{\frac {b_n}{a_n}}\right)^{\frac n{n+1}}=\frac xy\cdot\frac {y\cdot e}x=e$$
Therefore, we can compute the required limit as follows:
$$
\begin{aligned}
\lim_{n\to\infty}\, \left(\sqrt[n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right) &= \lim_{n\to\infty}\, \left(c_{n+1}-c_n\right) \\
&=\lim_{n\to\infty}\, \left[c_n\cdot\left(\frac {c_{n+1}}{c_n}-1\right)\right]\\
&= \lim_{n\to\infty}\ \left[\frac {c_n}n\cdot\frac {e^{\ln\left(\dfrac {c_{n+1}}{c_n}\right)}-1}{\ln\left(\dfrac {c_{n+1}}{c_n}\right)}\cdot\ln\left(\frac {c_{n+1}}{c_n}\right)^n\right] \\
&= \frac x{y\cdot e}\cdot 1\cdot\ln e \\
&=\frac x{y\cdot e}
\end{aligned}
$$
This completes the solution. And interesting remark is that choosing $a_n= (n!)^2$ and $b_n = n!$, we get the following limit (Lalescu):
$$\lim_{n\to\infty}\, \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\frac 1{e}$$
| {
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"url": "https://math.stackexchange.com/questions/3528462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$ Find the maximum value of $\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}$ as $x$, $y$, and $z$ range over all positive real numbers.
My first instinct was to apply AM-GM on each factor in the denominator since every variable is a positive real number. I also thought that it is just right to do so because minimizing the denominator would maximize the expression. The answer I got was $\frac{1}{2880}$, but the answer key states that it is $\frac{1}{5120}$. How come?
| By AM-GM
$$\frac{xyz}{(1+5x)(4x+3y)(5y+6z)(z+18)}=$$
$$=\frac{xyz}{\left(1+3\cdot\frac{5x}{3}\right)(4x+3\cdot y)(5y+3\cdot2z)(z+3\cdot6)}\leq$$
$$\leq \frac{xyz}{4\sqrt[4]{1\cdot\left(\frac{5x}{3}\right)^3}\cdot4\sqrt[4]{4x\cdot y^3}\cdot4\sqrt[4]{5y\cdot(2z)^3}\cdot4\sqrt[4]{z\cdot6^3}}=\frac{1}{5120}.$$
The equality occurs for $$1=\frac{5x}{3},$$ $$4x=y,$$ $$5y=2z$$ and $$z=6$$ or $$(x,y,z)=\left(\frac{3}{5},\frac{12}{5},6\right),$$ which says that we got a maximal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3529987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \sin(2x+y) + \cos(y)\sin(y) = 1$$ But this idea didn't give me anything. Also if I sum I have problem... but this is high school problem so it must have some easy solution.
| Hint:
$$1=(1-\cos y)^2+(1-\sin y)^2=1-2\cos y-2\sin y+1$$ gives you $y$. And at the same time, $2x$.
Alternatively:
$$\frac{\sin 2x+\sin x}{\cos 2x+\cos x}=\frac{2\sin\dfrac{2x+y}2\cos\dfrac{2x-y}2}{2\cos\dfrac{2x+y}2\cos\dfrac{2x-y}2}=\tan\frac{2x+y}2=1$$ so that
$$2x+y=\frac\pi4+k\pi.$$
Then
$$\cos\left(\frac\pi2+2k\pi-y\right)+\cos y=1$$
or
$$2\cos\left(\frac\pi4+k\pi\right)\cos\left(\frac\pi4+k\pi-y\right)=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Compute$\int_0^{2\pi}\frac{\cos^2\theta }{ |re^{i\theta} -z|^2}d\theta$
$r\gt0$, Compute
$$\int_0^{2\pi}\frac{\cos^2\theta }{ |re^{i\theta} -z|^2}d\theta$$
when $|z|\ne r$
The problem is related to Poisson kernel and harmonic function, but I don't know how to start, $\cos^2\theta =1/2 (1+\cos 2\theta )$.
| Express $z$ also in polar form, $z = s^{i\alpha}$. Then,
$$|re^{i\theta} -z|^2=r^2+s^2-2rs\cos(\theta-\alpha)$$
and, with the variable change $t= \theta - \alpha$, the integral reads,
$$I=\int_0^{2\pi}\frac{\cos^2\theta }{ |re^{i\theta} -z|^2}d\theta
=\int_0^{2\pi}\frac{\cos^2(t+\alpha)}{ r^2+s^2-2rs\cos t}dt$$
Expand the numerator,
$$\cos^2(t+\alpha)= \frac12+\frac12 \cos2\alpha \cos2t+\frac12\sin2\alpha\sin 2t$$
to decompose the integral into three manageable pieces,
$$I=I_1+I_2+I_3\tag 1$$
where
$$I_1=\frac12\int_0^{2\pi}\frac{ dt}{ r^2+s^2-2rs\cos t}=\frac{\pi}{|r^2-s^2|}
$$
$$I_2=\frac12\int_0^{2\pi}\frac{\cos2\alpha\cos 2t\> dt}{ r^2+s^2-2rs\cos t}
=\frac {\pi\cos2\alpha}{2r^2s^2|r^2-s^2|}\left(r^4+s^4-|r^4-s^4|\right)$$
$$I_3= \sin2\alpha\int_0^{2\pi}\frac{\cos t\sin tdt}{ r^2+s^2-2rs\cos t}
=0$$
(see derivations at end.) Plug the results into (1) to obtain
$$I=\int_0^{2\pi}\frac{\cos^2\theta d\theta }{ |re^{i\theta} -z|^2}
=\frac{\pi}{|r^2-s^2|}\left( 1+\cos2\alpha\frac{r^4+s^4-|r^4-s^4|}{2r^2s^2}\right)$$
Note that the results for $r>s$ and $r<s$ are respectively,
$$I_{r>s} =\frac{\pi}{r^2-s^2}\left( 1+\frac{s^2}{r^2}\cos2\alpha \right),\>\>\>\>\>
I_{r<s} =\frac{\pi}{s^2-r^2}\left( 1+\frac{r^2}{s^2}\cos2\alpha \right)$$
PS: Use $u = \tan\frac t2$ to integrate $I_1$ and use the result in $I_2$,
$$I_1=\frac12\int_0^{2\pi}\frac{ dt}{ r^2+s^2-2rs\cos t}
=\int_0^{\infty}\frac{du}{ (r-s)r^2+(r+s)^2u^2}$$
$$=\frac{2}{r^2-s^2}\tan^{-1}\left(\frac{r+s}{r-s}\right)\bigg|_0^\infty
=\frac{\pi}{|r^2-s^2|}
$$
$$I_2=\frac12\int_0^{2\pi}\frac{\cos2\alpha\cos2t\> dt}{ r^2+s^2-2rs\cos t}
=\cos2\alpha\left(\int_0^{2\pi}\frac{\cos^2t\> dt}{ r^2+s^2-2rs\cos t}-I_1\right)$$
$$=\cos2\alpha\left[\left(\frac{(r^2+s^2)^2}{2r^2s^2}-1\right) I_1 -\frac{1}{4r^2s^2} \int_0^{2\pi} (r^2+s^2-2rs\cos t)dt\right]$$
$$=\cos2\alpha\left[\frac{(r^4+s^4)^2}{2r^2s^2}\frac{\pi}{|r^2-s^2|}
-\frac{(r^2+s^2)\pi}{2r^2s^2}\right]$$
$$=\frac {\pi\cos2\alpha}{2r^2s^2|r^2-s^2|}\left(r^4+s^4-|r^4-s^4|\right)\tag 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Generalized version of calculating expected minimum value rolled until $5$ is obtained from a fair die The following is a modified interview question.
Given an $n$-sided fair die where $n\geq 1.$
You roll a die until you get a $m$ where $1\leq m\leq n$.
Calculate the expected value of the minimum rolled.
The original interview question takes $n=6$ (standard fair die) and $m=5$.
I mange to solve the problem and I reproduce my attempt below.
The expected minimum value rolled is $\frac{137}{60}$ because if $X$ is the minimum value rolled up to and including $5,$ then
$$P(X=x) = \frac{1}{x(x+1)} \quad \text{for }x=1,2,3,4 \quad \text{and} \quad P(X=5) = \frac{1}{5}.$$
So,
$$E(X) = \sum_{x=1}^5 xP(X=x) = \frac{137}{60}.$$
I am trying to solve the generalized version of the problem.
By the same spirit, let $Y$ be the minimum value rolled up to and including $m.$ Then
$$P(Y=y) = \frac{(y-1)!}{(y+1)!} = \frac{1}{y(y+1)} \quad \text{for }y =1,2,...,m-1 \quad \text{and}\quad P(Y=m) = \frac{1}{m}.$$
Therefore,
$$E(Y) = \sum_{y=1}^m y P(Y=y) = 1 + \sum_{y=1}^{m-1} \frac{1}{y+1} = 1 + \sum_{y=2}^{m} \frac{1}{y}.$$
| A slight variant of angryavian's answer above:
Let $T$ be the number of rolls before the first appearance of $m.$ We have $$P(T = t) = \left(\frac{n-1}{n}\right)^{t}\cdot \frac{1}{n},\ \ t=0,1,2,\ldots.$$
We have $$\mathbb{E}[Y \ | \ T=t] = \sum_{y=0}^{m-1} P(Y> y \ | \ T=t) = \sum_{y=0}^{m-1} \left( \frac{n-y-1}{n-1} \right)^t$$
The last equation comes from the fact that, under the condition $T=t,$ we have $Y>y$ if and only if $X_1, X_2, \ldots, X_t$ are all greater than $y$ (where these $X_i$ may be in $\{1,\ldots, n\} \setminus \{ m\}.$
By the Law of Total Expectation, we have
$$\mathbb{E}[Y] = \mathbb{E}\left( \mathbb{E}[Y \ | \ T=t]\right) = \sum_{t=0}^{\infty} \left( \left(\frac{n-1}{n}\right)^{t}\cdot \frac{1}{n} \ \sum_{y=0}^{m-1} \left( \frac{n-y-1}{n-1} \right)^t \right)$$
$$ = \frac{1}{n} \sum_{y=0}^{m-1} \sum_{t=0}^{\infty} \left(\frac{n-y-1}{n} \right)^t = \frac{1}{n} \sum_{y=0}^{m-1} \frac{1}{1 - \frac{n-y-1}{n}} = \sum_{y=0}^{m-1} \frac{1}{y+1} = H_m$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Greatest common divisor for $x+1$ and $2x+1$? I was just wondering, what would the gcd of $x+1$ and $2x+1$ be?
I know that $x+1$ is not even or odd, whereas, $2x+1$ is odd. Is the gcd $1$?
| If $d(x)|x+1$ and $d(x)|2x+1$ then $d(x)|(2x+1)-(x+1)= x$. And so $d(x)|x$ and $d(x)|x+1$ so $d(x)|(x+1) -x = 1$ so $d(x)|1$. And assuming we are talking $\mathbb Z[x]$ this is as far as we go and $d(x) =1$ is the only (positive) common divisor of $2x+1$ and $x+1$. So the greatest common divisor of $2x + 1$ and $x+1$ is $1$.
In general you have the rule $\gcd(a,b) = \gcd(a \pm k b, b)$ so $\gcd(2x+1, x+1) = \gcd(x, x+1) = \gcd(x, 1) = 1$.
Euclid's algorithm confirm:.
$(2x+1) = (x+1) + (x)$
$(x+ 1) = (x) + (1)$
$(x) = (1)\cdot x + 0$ so $\gcd(2x+1, x) = 1$.
As does Bezout's Lemma:
$2\cdot(x+1) - 1\cdot(2x+1) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=2-2=0$$
But on the right hand I have inrcrased $\sqrt{x^2-2x}$ rather then deceased
Another attempt:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
| The trick is to add and cancel an $x$ term:
$$\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\right) =\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-x +x-\sqrt{x^2-2x}\right)$$
and compute each limit separately:
$$\lim_{x\to \infty} \left(\sqrt[3]{x^3+2x}-x\right) = \lim_{x\to \infty} \frac{2x}{\sqrt[3]{(x^3+2x)^2}+x\sqrt[3]{x^3+2x}+x^2} = 0$$
and $$\lim_{x\to \infty} \left(x-\sqrt{x^2-2x}\right) = \lim_{x \to \infty} \frac{2x}{x+\sqrt{x^2-2x}} = \lim_{x\to \infty} \frac{2}{1+\sqrt{1-\frac{2}{x}}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
remainder when $a_{1000}$ is divided by $1000$
If $a_{1}=7,a_{2}=7^7,a_{3}=7^{7^{7}}.$ Then the remainder when
$a_{1000}$ is divided by $1000$
what i try
$a_{1}=7=1\mod(1000)$ and
$a_{2}=7^7=1^7\mod(1000)=1\mod(1000)$
from using modulo theorem
$a_{3}=7^{7^{7}}=1^7\mod(1000)=1\mod(7)$
can we say that $a_{1000}= 1\mod(1000)$
Help me to solve it please
| $a_1=7\equiv\color{red}7\bmod 1000$.
$a_2=7^7=823543\equiv543\bmod 1000$.
$7^4=2401\equiv401\bmod1000,$ so $7^{20}=(7^4)^5\equiv401^5=(400+1)^5\equiv1\bmod1000$.
Since $a_2\equiv543\bmod1000,$ $a_2\equiv3\bmod20$. Therefore, $a_3=7^{a_2}\equiv7^3=343\bmod1000$.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How many pairs of $(x,y,z)$ for $x+y+2z=n$
How many pairs of $(x,y,z)$ for $x+y+2z=n$ when $n$ is an odd integer $\geq 5$
For $n=5$ there are 2 pairs. $n=7$ there are 6 pairs. $n=9$ there are 12 pairs. $n=11$ there are 20 pairs.
$an^2 + bn + c = f(n)$
$25a + 5b + c = 2$
$49a + 7b + c = 6$
$81a + 8b + c = 12$
$f(n) = (n^2)/4 - n + 3/4 = (n-3)(n-1)/4$
Is there a better way to solve it?
| put z=2 then x+y=n-2 has n-3 ways (x can be anything from 1 to n-3).
then put z=4 then x+y=n-4 has n-5 ways (x can be anything from 1 to n-5).
.
.
.
.
put z =n-3 then x+y =3 have 2 ways.
total number of cases = 2+4+ 6.......n-3 = 2(1+2+3....$\frac{n-3}{2}$) = 2 $\frac{(\frac{n-3}{2})(\frac{n-1}{2})}{2}$ = $\frac{(n-3)(n-1)}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Product fractions limit $\lim_{n\to \infty} (\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}})^{1/n^2}$ I need some help with this limit
$$\lim_{n\to \infty} \left(\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}}\right)^{1/n^2}$$
This problem appears in a chapter about Riemann sums, so I think I must split the fractions in the parantheses:
$$\frac{1^1 \times 2^2\times ... \times n^n}{n^{1+2+...+n}} = \frac{1^1}{n^1}\times \frac{2^2}{n^2}\times ... \times \frac{n^n}{n^n}$$
Now I am stuck.
| $$\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)^{\frac{1}{n^{2}}}=\exp\left(\frac{1}{n^{2}}\ln\left(\frac{1^{1}\cdot2^{2}\cdot\cdot\cdot n^{n}}{n^{\left(1+2+...+n\right)}}\right)\right)$$$$=\exp\left(\frac{1}{n^{2}}\ln\left(\prod_{k=1}^{n}\left(\frac{k}{n}\right)^{k}\right)\right)=\exp\left(\frac{1}{n}\sum_{k=1}^{n}\ln\left(\left(\frac{k}{n}\right)^{\frac{k}{n}}\right)\right)$$$$=\exp\left(\int_{0}^{1}\ln\left(\left(x\right)^{x}\right)dx\right)=\exp\left(\int_{0}^{1}x\ln\left(\left(x\right)\right)dx\right)$$$$=\exp\left(\frac{x^{2}}{2}\ln\left(\left(x\right)\right)\Big|_0^1-\frac{1}{2}\int_{0}^{1}xdx\right)$$$$=\color{red}{\exp\left(\frac{-1}{4}\right)\simeq0.778800783071}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.
| I found
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2 = xy(x-y)^2+x^2(z+x-y)^2+yz^2(y+z) \geqslant 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solutions of p-Laplace equation I found that for the following problem
\begin{cases}
-\Delta_p u = 1,&x\in B_1(0)\\
u = 0,\quad &x\in\partial B_1(0)
\end{cases}
where $B_1(0)$ is the unitary ball of $\mathbb{R}^N$ and $\Delta_p u = \operatorname{div}(\|\nabla u\|^{p-2}\nabla u) $,
there exists a weak solution of the form:
$ u(x) = C_{N,p} (1-\|x\|^{\frac{p}{p-1}})$ with $C_{N,p}\in\mathbb{R}$ a constant depending just on the dimension of the ball $N$ and the value of $p>2$.
The problem is that I cannot find the constant $C_{N,p},$ and I need it for the case $N=3$.
| Assume the given solution $u$ and note that it is smooth almost everywhere.
Compute the partial derivative (define $C:=C_{N,p}$)
\begin{align}
\partial_i u
&= -C \partial_i |x|^\frac{p}{p-1}
\\
&= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \partial_i |x|
\\
&= -C \frac{p}{p-1} |x|^{\frac{p}{p-1}-1} \frac{x_i}{|x|}
\\
&= -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x_i
\,.
\end{align}
Gather for the gradient
\begin{align}
\nabla u
&= -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x \,.
\end{align}
The norm of the gradient (assume $C>0$)
\begin{align}
| \nabla u |
&= C \frac{p}{p-1} |x|^\frac{1}{p-1} \,.
\end{align}
The factor needed for the $p$-Laplacian
\begin{align}
| \nabla u |^{p-2} \nabla u
&= \left( C \frac{p}{p-1} |x|^\frac{1}{p-1} \right)^{p-2}
\left( -C \frac{p}{p-1} |x|^{\frac{1}{p-1}-1} x \right)
\\
&= - \left( C \frac{p}{p-1} \right)^{p-1} |x|^{\frac{p-2}{p-1}+\frac{1}{p-1}-1} x
\\
&= - \left( C \frac{p}{p-1} \right)^{p-1} x \,.
\end{align}
For the $p$-Laplacian we need to compute the divergence of the above computed factor
\begin{align}
\Delta_p u
&= \nabla \cdot \left( | \nabla u |^{p-2} \nabla u \right)
\\
&= \nabla \cdot \left( - \left( C \frac{p}{p-1} \right)^{p-1} x \right)
\\
&= -N \left( C \frac{p}{p-1} \right)^{p-1}
\\
&\overset{!}{=} -1
\\
\implies C &= \frac{p-1}{p} N^\frac{1}{1-p} = C_{N,p} \,.
\end{align}
Note, that there are no special cases for $p>2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find values of $x$, such as $\log_3 \sqrt{x+3}−\log_3(9−x^2) < 0$ The Function is $$f(x) = \log_3\sqrt{(x+3)}−\log_3(9−x^2)$$
and I need to figure out arguments for which $$ f(x) < 0 $$
So I calculated the domain of function which is $ D: (-3;3)$
However I am still unable to solve $f(x) < 0$
I simplified $ \log_3\sqrt{(x+3)}−\log_3(9−x^2) < 0$ to $\log_3\frac{\sqrt{(x+3)}}{(9-x^2)} < 0$
which gets me to $$ \frac{\sqrt{(x+3)}}{9-x^2} < 1$$
But the solutions of the above equation are complex numbers and I definitely should get them as my result. So, what I am doing wrong here?
Would apprecite every answer
| Continue with $ \frac{\sqrt{x+3}}{9-x^2} < 1$, yet with the domain restriction $-3<x<3$. Then, examine the roots of the equation
$$\sqrt{x+3} = 9-x^2$$
Square and factorize to get
$$(x+3)(x^3-3x^2-9x + 26)=0$$
Since $x>-3$, the valid roots come from the second factor. Let $x= t+1$ to depress the second factor as
$$t^3-12t+15=0$$
which has three real roots. Use the analytical root expressions below for the cubic equation of the form $t^2+pt+q=0$,
$$t_k = 2\sqrt{\frac{-p}3}\cos\left( \frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{-3}p} \right) -\frac{2\pi k}3 \right), \>\>\>k=0,1,2$$
The valid solutions for $x$ are
$$x_1=1+4\cos\left(\frac13\cos^{-1}\frac{15}{16}+\frac{\pi}3\right)=2.577$$
$$ x_2=1-4\cos\left(\frac13\cos^{-1}\frac{15}{16}\right)= -2.972$$
Then, it is straightforward to verify that $ \frac{\sqrt{x+3}}{9-x^2} < 1$ if the values of $x$ are in the range,
$$x_2 < x < x_1$$
| {
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"url": "https://math.stackexchange.com/questions/3547692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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About Harmonic Numbers and a sequence in OEIS I'd been looking for a series in OEIS and the one that fits better (resembles, at least) is one in this link in Examples: http://oeis.org/A082687. There, as we can see, it is equal to
$$H'(2n) = H(2n)-H(n) = \cdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
Also, I know that $H$ represents harmonic numbers. However here is the thing: I'm confused, since the harmonic numbers should be only numbers not a series. Then I searched at Wiki about these "harmonic series", but they seem not to exist. Bellow (1), there is something about Hilbert Matrix, maybe that is it, but I'm not sure as I do not even know what this is.
Could someone give me a clue about what these $H$s are and/or tell what (1) means?
Thanks
| Here is how I would write the expression you give in (1):
$$\overline H_{2n} = H_{2n} - H_n. \tag2$$
Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$ while $\overline H_n$ denotes the $n$th skew-harmonic number $\sum_{k = 1}^n (-1)^{k + 1}/k$.
In proving (2), we have
\begin{align}
\overline H_{2n} &= \sum_{k = 1}^{2n} \frac{(-1)^{k + 1}}{k}\\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}\\
&= 1 + \left (\frac{1}{2} - 1 \right ) + \frac{1}{3} + \left (\frac{1}{4} - \frac{1}{2} \right ) + \frac{1}{5} + \left (\frac{1}{6} - \frac{1}{3} \right ) + \cdots\\
&\qquad \cdots + \frac{1}{2n - 1} + \left (\frac{1}{2n} - \frac{1}{n} \right )\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} \right ) - \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right )\\
&= \sum_{k = 1}^{2n} \frac{1}{k} - \sum_{k = 1}^n \frac{1}{k}\\
&= H_{2n} - H_n,
\end{align}
as required to prove.
Checking that we indeed get the first few terms in OEIS A082687, since
\begin{align}
\overline H_2 &= \sum_{k = 1}^2 \frac{(-1)^{k + 1}}{k} = \frac{1}{2}\\
\overline H_4 &= \sum_{k = 1}^4 \frac{(-1)^{k + 1}}{k} = \frac{7}{12}\\
\overline H_6 &= \sum_{k = 1}^6 \frac{(-1)^{k + 1}}{k} = \frac{37}{60}\\
\overline H_8 &= \sum_{k = 1}^8 \frac{(-1)^{k + 1}}{k} = \frac{533}{840}
\end{align}
and selecting the numerators for each of these quantities we have the sequence: $\{1, 7, 37, 533, \ldots\}$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate this determinant of a $2\times 2$matrix? This matrix arises from a homework problem which our professor gave.
We need to find the determinant of this matrix.
Does there exist any simple way to find the determinant of this matrix?
$\begin{pmatrix}
x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) && (1-p)(2+\frac{l}{x-n+2-l})\\
(1-q)(2+\frac{l}{x-n+2-l}) && x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})
\end{pmatrix}$
Here $n=pq$ and $l=\phi(n)+1$.
Is there any software which can calculate this large determinant?
One of my friends got $x-n+2-l$ as a factor of this determinant.
Is there any simple way to calculate this determinant?
I am stuck.
My try:
$R_1\to R_1-R_2$ gives
$\begin{pmatrix}
x-pq-p+q+2 && -x+pq+q-p-2\\
(1-q)(2+\frac{l}{x-n+2-l}) && x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})
\end{pmatrix}$
| Call your matrix $M$ and let $s=x+2-n$. Then
\begin{aligned}
M&=\pmatrix{
x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) &(1-p)(2+\frac{l}{x-n+2-l})\\
(1-q)(2+\frac{l}{x-n+2-l}) &x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})}\\
&=\pmatrix{x-pq-p+3 &1-p\\ 1-q &x-pq-q+3}
+\frac{x+2-n}{x+2-n-l}\pmatrix{1-q&1-p\\ 1-q&1-p}\\
&=sI+\pmatrix{1-p&1-p\\ 1-q&1-q}
+\frac{s}{s-l}\pmatrix{1-q&1-p\\ 1-q&1-p}.
\end{aligned}
(The second equality above is borrowed from J.G.'s answer.)
If $l=2s$, then $\dfrac{s}{s-l}=-1$. Hence
$$
M=\pmatrix{s+q-p&0\\ 0&s+p-q}
$$
and
$$
\det(M)=s^2-(p-q)^2.
$$
If $l\ne2s$, let $t=\dfrac{s-l}{2s-l},\ p'=\dfrac{1-p}{t}$ and $q'=\dfrac{1-q}{t}$. Then
$$
M=\pmatrix{s+tp'+(1-t)q'&p'\\ q'&s+tq'+(1-t)p'}
$$
and hence
$$
\det(M)=\left(s+tp'+(1-t)q'\right)\left(s+tq'+(1-t)p'\right)-p'q'.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to prove that $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ I am asked to simplify the following expression:
$$\tan \left( \frac{\pi}{2} - \theta \right)$$
The book gives me the answer $\cot \theta$ but, when I try to derive that formula using
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1 \mp \tan A \cdot \tan B}$$
I get $\tan \frac{\pi}{2}$ as one of the terms (which is undefined).
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1 \mp \tan A \cdot \tan B} = \frac{\tan \frac{\pi}{2} - \tan \theta}{1 + 0 \cdot \tan \theta}$$
How do I proceed?
Thank you.
| The formula for $\tan(A\pm B)$ is obtained by dividing the formula for $\sin(A\pm B)$ by the one for $\cos(A\pm B)$, cancelling a factor of $\cos A\cos B$ from each to obtain the usual expression. As this is an illegal division by $0$ if $A=\frac{\pi}{2}$, it's easiest just to go back to the tangent's definition, viz.$$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{\sin\left(\frac{\pi}{2}-\theta\right)}{\cos\left(\frac{\pi}{2}-\theta\right)}=\frac{\cos\theta}{\sin\theta}=\cot\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A sequence $(a_n)_{n\ge 1}$ such that $a_1>0$ and $a_{n+1}=a_n-\ln(1+a_n)$ Let $(a_n)_{n\ge 1}$ be a sequence such that $a_1>0$ and $$a_{n+1}=a_n-\ln(1+a_n)$$
a) Prove that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$ and $\lim\limits_{n\to \infty} (n^{2019}a_n)=0$.
b) If $a_1<2$, prove that $a_{12} \in (0,10^{-300})$.
It is easy to see that $a_n >0$, $\forall n\in \mathbb{N}$. I considered the function $f:(0,\infty)\to \mathbb{R}$, $f(x)=x-\ln(1+x)-\frac{x^2}{2}$ and I could easily prove that this function is strictly decreasing, so it follows that $$x-\ln(1+x)<\frac{x^2}{2}, \forall x>0 \tag{*}$$
Now from $(*)$ we get that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$.
I couldn't make much further progress. I could show that $\lim\limits_{n\to \infty}a_n=0$, but then I got stuck.
| From
$ a_{n+1}
<\frac{a_n^2}{2}
$
we get
$ a_{n+2}
<\frac{a_{n+1}^2}{2}
<\frac{(\frac{a_{n}^2}{2})^2}{2}
=\frac{a_{n}^4}{8}
=\frac{a_{n}^4}{2^3}
$
and
$ a_{n+3}
<\frac{a_{n+2}^2}{2}
<\frac{(\frac{a_{n}^4}{2^3})^2}{2}
=\frac{a_{n}^8}{2^7}
$.
By induction
we can show that
$ a_{n+m}
<\frac{a_{n}^{2^m}}{2^{2^{m}-1}}
$.
If this holds that
$ a_{n+m+1}
<\frac{a_{n+m}^2}{2}
<\frac{(\frac{a_{n}^{2^m}}{2^{2^{m}-1}})^2}{2}
=\frac{a_{n}^{2^{m+1}}}{2^{2^{m+1}-2+1}}
=\frac{a_{n}^{2^{m+1}}}{2^{2^{m+1}-1}}
$.
If $a_1 < 2$
then,
since
$x-\ln(1+x)$
is increasing,
$a_2
< 2-\ln(3)
< .902
$
If $m=10$ and $n=2$ then
$ a_{12}
<\frac{a_2^{2^{10}}}{2^{2^{10}-1}}
<\frac{.902^{1024}}{2^{1023}}
<1.501 × 10^{-354}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove if parallelogram law worked in a quadrilateral that shape is parallelogram Image with my solution
We have a quadrilateral ABCD. And we know: $$a^2+b^2+c^2+d^2=p^2+q^2$$ Prove this shape is parallelogram.
I write law of cosines between sides and diameters. Then I write sum of them and with the formula that the question give to us, We have this:
$$ab.cos\hat B+bc.cos\hat C+cd.cos\hat D+da.cos\hat A=0$$
And I don't know what should I do.
| Let $K$, $L$, $M$, $N$, $P$ and $Q$ be a midpoints of $AB,$ $BC,$ $CD,$ $DA$, $AC$ and $BD$ respectively.
Thus, since $PLQN,$ $PKQM$ and $KLMN$ are parallelograms, by your work we obtain:
$$PL^2+LQ^2+QN^2+NP^2=PQ^2+LN^2$$ or
$$AB^2+CD^2=2PQ^2+2LN^2.$$
Also, $$KP^2+PM^2+MQ^2+QK^2=PQ^2+KM^2$$ or
$$BC^2+AD^2=2PQ^2+2KM^2.$$
Id est, $$AB^2+CD^2+BC^2+AD^2=$$
$$=4PQ^2+2(LN^2+KM^2)=4PQ^2+2(KL^2+LM^2+MN^2+NK^2)=$$
$$=4PQ^2+AC^2+BD^2$$ and since
$$AB^2+CD^2+BC^2+AD^2=AC^2+BD^2,$$ we obtain $$PQ=0,$$ $$P\equiv Q$$ and $ABCD$ is a parallelogram.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find all triples $(x,y,z)\in \Bbb{R}$ that satisfy the following conditions: The question is to find all real numbers solutions to the system of equations:
*
*$y=\Large\frac{4x^2}{4x^2+1}$,
*$z=\Large\frac{4y^2}{4y^2+1}$,
*$x=\Large\frac{4z^2}{4z^2+1}$,
This seems simple enough, so I tried substituting the values of x, y and z into the different equations but I only ended up with a huge degree 8 equation which clearly doesn't seem like the right approach. I really have no idea on how to go about solving this if substitution is not the answer.
Any help would be greatly appreciated :)
| $\bullet\; $ Clearly $x=y=z=0$ are the solution of system of eqn
$\bullet\; $ If $x,y,z\neq 0\; $ Then
*
*$\displaystyle y=\frac{4x^2}{4x^2+1}\Longrightarrow \frac{1}{y}=1+\frac{1}{4x^2}$,
*$\displaystyle z=\frac{4y^2}{4y^2+1}\Longrightarrow \frac{1}{z}=1+\frac{1}{4y^2},$
*$\displaystyle x=\frac{4z^2}{4z^2+1}\Longrightarrow \frac{1}{x}=1+\frac{1}{4z^2}$,
Adding all three
$\displaystyle \bigg(1-\frac{1}{2x}\bigg)^2+\bigg(1-\frac{1}{2y}\bigg)^2+\bigg(1-\frac{1}{2z}\bigg)^2=0$
which is possible when
$\displaystyle 1-\frac{1}{2x}=0,1-\frac{1}{2y}=0,1-\frac{1}{2z}=0$
System of equation have $\displaystyle (x,y,z)=\bigg(\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigg)$ as solution also
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How many ways to distribute $m$ distinguishable passengers in $n$ cars when $m>n$? I know there are versions of this question on here, but I'm looking to relate this concept to permutation. Let me explain: If we have $10$ seats and $7$ distinguishable people, and we want to find all ways $7$ people can sit on $10$ seats (they can't sit on top of each other), the formula for that is $$P(10,7)=\frac{10!}{(10-7)!}$$.
Now take the converse, so there are $7$ seats and $10$ distinguishable people, I want to count the ways these $10$ people can sit on $7$ chairs such that at least one person sits on each chair. So I first thought the following:
The first person has $7$ choices to sit on, the second person has $6$... and the $7^{th}$ person has 1 choice. This leaves us $3$ people who can sit on any of the 7 chairs.
Therefore we resort to the formula above (since $3<7$), so we the total ways $10$ people can sit on $7$ chairs with no chair empty is this:
$$7!\cdot \frac{7!}{(7-3)!}$$
This only worked because in the second round because we had fewer people than chairs. So can we generalize this? In other words, how can find a formula for sitting $m$ people in $n$ cars when $m>n$ such that there is at least $1$ person per car and no car can have more than $2$ people than any other cars (sort of like quasi evenly layering a cake).
|
In how many ways can ten distinct objects be placed in seven distinct boxes if no box is left empty?
Method 1: The number $10$ can be partitioned into seven parts in three ways.
\begin{align*}
10 & = 4 + 1 + 1 + 1 + 1 + 1 + 1\\
& = 3 + 2 + 1 + 1 + 1 + 1 + 1\\
& = 2 + 2 + 2 + 1 + 1 + 1 + 1
\end{align*}
Four objects placed in one box and one object apiece placed in each of the other boxes: Choose which box receives four objects, choose which four objects it receives, then distribute the remaining six objects to the remaining six boxes so that one object is placed in each of those boxes. This can be done in
$$\binom{7}{1}\binom{10}{4}6!$$
ways.
Three objects placed in one box, two objects placed in another box, and one object apiece placed in each of the other boxes: Choose which box receives three objects, choose which three objects it receives, choose which of the other boxes receives two objects, choose which two of the remaining objects it receives, then distribute the remaining five objects to the remaining five boxes so that one object is placed in each of those boxes. This can be done in
$$\binom{7}{1}\binom{10}{3}\binom{6}{1}\binom{7}{2}5!$$
ways.
Two objects apiece place in three boxes and one object apiece placed in each of the remaining boxes: Choose which three boxes receive two objects each. Suppose the boxes are lined up from left to right. Place two objects in the leftmost box that has been selected to receive two objects, two of the remaining objects in the middle box that has been selected to receive two objects, and two of the remaining objects to be placed in the rightmost box that has been selected to receive two objects. Distribute the remaining four objects to the remaining four boxes so that one object is placed is placed in each of those boxes. This can be done in
$$\binom{7}{3}\binom{10}{2}\binom{8}{2}\binom{6}{2}4!$$
ways.
Total: Since these three cases are mutually exclusive and exhaustive, the number of ways of distributing ten distinct objects to seven distinct boxes so that no box is left empty is
$$\binom{7}{1}\binom{10}{4}6! + \binom{7}{1}\binom{10}{3}\binom{6}{1}\binom{7}{2}5! + \binom{7}{3}\binom{10}{2}\binom{8}{2}\binom{6}{2}4!$$
Method 2: We use the Inclusion-Exclusion Principle.
If there were no restrictions, we would have seven choices for each of the ten objects. Therefore, there are $7^{10}$ ways to distribute ten distinct objects to seven distinct boxes without restriction.
From these, we must subtract those distributions in which at least one box is left empty. There are $\binom{7}{k}$ ways to select $k$ boxes to be left empty and $(7 - k)^{10}$ ways to distribute the objects to the remaining $7 - k$ boxes. Thus, by the Inclusion-Exclusion Principle, the number of ways ten distinct objects can be distributed to seven distinct boxes if no box is left empty is
$$\sum_{k = 0}^{7} (-1)^k\binom{7}{k}(7 - k)^{10} = 7^{10} - \binom{7}{1}6^{10} + \binom{7}{2}5^{10} - \binom{7}{3}4^{10} + \binom{7}{4}3^{10} - \binom{7}{5}2^{10} + \binom{7}{6}1^{10} - \binom{7}{7}0^{10}$$
In how many ways can $m$ distinct objects be placed in $n$ distinct boxes if no box is left empty, where $m \geq n$?
Apply the Inclusion-Exclusion Principle.
$$\sum_{k = 0}^{n} (-1)^k\binom{n}{k}(n - k)^m$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve a differential equation $(x^2y+y^5)dx+(x^3-xy^4)dy=0$?
Solve the following differential equation:
$$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
I noticed that it is not exact, since:
$$
(x^2y+y^5)'_y=x^2+5y^4\ne3x^2-y^4=(x^3-xy^4)'_x
$$
Then I tried to express $y'$:
$$
y'=\frac{x^2y+y^5}{xy^4-x^3}
$$
And now I don't understand what type of differential equation it is and how to solve it.
Could someone help me?
| $$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
Rearrange terms:
$$y^4(ydx-xdy)+x^2(ydx+xdy)=0$$
Divide by $x^2$:
$$-y^4d\left (\frac y x \right ) +d(xy)=0$$
Divide by $(xy)^2$ and integrate:
$$\frac {y^2}{x^2}d\left (\frac y x \right )=\dfrac {d(xy)}{x^2y^2}$$
$$\frac {y^3}{x^3}=-\dfrac {3}{xy}+C$$
Finally, multiply by $x^3y$:
$$ \boxed {{y^4}+ {3}{x^2}+Cx^3y=0}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I get the inverse of this equation? The equation is the following:
$$y = \frac{x}{x-2}$$
What I did is:
$$y = \frac{x}{x-2}$$
$$y \times (x-2) = x$$
$$\frac{x-2}{x} = \frac{1}{y}$$
$$\frac{x}{x}-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{2}{x} = \frac{1}{y}$$
$$1-\frac{x}{2} = y$$
I've got a lot of doubts about that last step. Besides that, the result according to the textbook is the following:
$$ f^{−1}(x)=-\frac{2x}{x−1} $$
| \begin{align}
y &= \frac{x}{x-2}
.
\end{align}
\begin{align}
y&=\frac{x+2-2}{x+2}
,\\
y&=1-\frac 2{x+2}
,\\
y-1&=-\frac 2{x+2}
,\\
-\frac{y-1}2&=\frac 1{x+2}
,\\
x+2 &=\frac 2{1-y}
,\\
x &=\frac 2{1-y}-2;
,\\
x &=\frac{2\,y}{1-y}
,
\end{align}
so the answer is indeed
\begin{align}
f^{-1}(x)=\frac{2x}{1-x}
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Let a,b and c be the roots of the equation $x^3-9x^2+11x-1=0$ and $s=\sqrt a +\sqrt b +\sqrt c$
Then Find the value of $s^4-18s^2-8s$
$$s=\sqrt a + \sqrt b +\sqrt c$$
$$s^2=a+b+c+2(\sqrt {ab} +\sqrt {bc} +\sqrt {ac})$$
I can’t seem to find a way around this obstacle. Squaring it again gives another $\sqrt {ab}+\sqrt {bc} +\sqrt {ac}$, and is seemingly never ending.
This drove me to the conclusion that the value of s cannot actually be found, and the question must be manipulated to get the required from. I don’t know how to do that though.
| You should use the following properties of the roots of a cubic equation ax^3 + bx^2 + cx + d = 0:
a+b+c = -b/a = -(-9)/1 = 9
abc = -d/a = 1
and
ab + ac + bc = c/a = 11/1 =11
From these you can find:
(a+b+c)^2 = 81 = a^2+b^2+c^2 + 2 (ab + ac + bc) = a^2+b^2+c^2 + 2 * 11
=> a^2+b^2+c^2 = 81 - 22 = 59
Now expand s^2 and s^4 using these expressions and you should be one step closer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Greatest value of $xz$ in expression
Let $x,y,z,t\in\mathbb{R}$ and $x^2+y^2=9,z^2+t^2=4$ and $xt-yz=6$. Then greatest value of $xz$ is
What i try
I am trying to solve without trigonometric substution
$$(xt-yz)^2+(xz+yt)^2=(x^2+y^2)(z^2+t^2)$$
$6^2+(xz+yt)^2=4\cdot 9\Longrightarrow (xz+yt)^2=0$
$$xz=-yt$$
How do i solve it Help me please
| Notice that:
$$4(x^2+y^2) + 9(t^2 + z^2) = 12(xt - yz)$$
and this is equivalent with
$$(2x - 3t)^2+ (2y + 3z)^2 = 0$$
Thus $z=-\frac{2}{3}y$, and we want to maximize $-\frac{2}{3}xy$. This is simple, given that $x^2+y^2=9$ because:
$$0\leq (x+y)^2\Rightarrow -2xy \leq x^2+y^2=9$$
Therefore:
$$zx=-\frac{2}{3}xy\leq 3$$
with equality when $(x,z)=\left(\pm \frac{3}{\sqrt{2}},\pm \sqrt{2}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\binom{n}{0} + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n} = \frac{2^{n + 1} - 1}{n + 1}$. I have to prove the identity:
$$\binom{n}{0} + \dfrac{1}{2} \binom{n}{1} + \dfrac{1}{3} \binom{n}{2}
+ ... + \dfrac{1}{n+1} \binom{n}{n} = \dfrac{2^{n + 1} - 1}{n + 1}$$
First thing I tried was to think about a combinatorial proof, but I couldn't interpret the identity in any way that makes sense. So, how should I prove this?
| Hint: Integrate the binomial theorem
$$(1+x)^n = \sum_{k=0}^n\binom nk x^k$$
from $x=0$ to $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}$ I need to calculate the following sum:
$$
\sum \limits_{n = 1}^{\infty} \frac{\cos 2n}{n^2}
$$
I've tried adding an imaginary part and differentiating:
$$
f(x) = \sum \limits_{k = 1}^{\infty} \frac{\cos 2xk + i \sin 2xk}{k^2} \\
f(x) = \sum \limits_{k = 1}^{\infty} \frac{e^{2ixk}}{k^2} \\
f'(x) = \sum \limits_{k = 1}^{\infty} \frac{2i e^{2ixk}}{k} \\
f''(x) = - 4\sum \limits_{k = 1}^{\infty} e^{2ixk} \\
f''(x) = -4\frac{e^{2ix}}{1 - e^{2ix}}
$$
Where $f(x)$ if a function of which I need to find the value at $x = 1$.
After differentiating once I get
$$
f'(x) = \frac{\log \left( 1 - e^{kx} \right)}{k} + C
$$
(k is just some constant), and I can't integrate once more as I'll get an integral logarithm which I don't want to work with.
Is there any more pleasant way to calculate the aforementioned sum?
| Subtract $\sum_n\frac1{n^2}=\frac{\pi^2}6$ to obtain
\begin{eqnarray}
\sum_{n=1}^\infty\frac{\cos2n}{n^2}-\frac{\pi^2}6
&=&
\sum_{n=1}^\infty\frac{1-\cos2n}{n^2}
\\
&=&
-2\sum_{n=1}^\infty\left(\frac{\sin n}n\right)^2\;.
\end{eqnarray}
Now note that $\frac1\pi\frac{\sin n}n$ is the $n$-th coefficient in the Fourier series of a rectangular pulse with period $2\pi$ and length $2$:
$$
\frac1{2\pi}\int_{-1}^1\mathrm e^{-\mathrm inx}\mathrm dx=\frac1\pi\frac{\sin nx}n\;.
$$
By Parseval’s theorem we have
$$
\sum_{n=-\infty}^\infty\left(\frac1\pi\frac{\sin nx}n\right)^2=\frac1{2\pi}\int_{-1}^1\mathrm dx=\frac1\pi\;,
$$
and thus
\begin{eqnarray}
\sum_{n=1}^\infty\left(\frac{\sin n}n\right)^2
&=&
\frac12\left(\sum_{n=-\infty}^\infty\left(\frac{\sin nx}n\right)^2-1\right)
\\
&=&
\frac{\pi-1}2\;.
\end{eqnarray}
Substituting above yields
\begin{eqnarray}
\sum_{n=1}^\infty\frac{\cos2n}{n^2}
&=&
\frac{\pi^2}6-2\cdot\frac{\pi-1}2
\\
&=&\frac{\pi^2}6-\pi+1\;.
\end{eqnarray}
| {
"language": "en",
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Find $a, b \in \mathbb{R}$ such that the polynomial $f(x)=x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$ has a triple root. I am given the following polynomial:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b = 0$$
with $a, b \in \mathbb{R}$.
I have to find $a$ and $b$ such that the given polynomial has a triple root.
I know that if a polynomial has a triple root $\alpha$ then we have:
$$f(\alpha) = 0$$
$$f'(\alpha) = 0$$
$$f''(\alpha) = 0$$
And in previous exercises (where the degree of the polynomial was $4$, not $5$) I could use $f''(\alpha) = 0$ to find $2$ $\alpha$'s and find $a$ and $b$ for each $\alpha$. I could do that because the second derivative of a $5$th degree polynomial is a quadratic so I could find the $\alpha$'s. Here, it's different. We have:
$$f(x) = x^5 + 7x^4 + 19x^3 + 26x^2 + ax + b$$
$$f'(x) = 5x^4+28x^3+57x^2+52x+a$$
$$f''(x) = 20x^3+84x^2+114x+52$$
And when I try to solve $f''(\alpha) = 0$ I get:
$$20 \alpha ^ 3 + 84 \alpha^2 + 114 \alpha + 52 = 0$$
And I don't know how to find the $\alpha$'s. So, what approach should I use to solve this exercise?
| Simplify the equation as
$$10 a^ 3 + 42 a^2 + 57a + 26= 0$$
which may be factorized as follows,
$$(10 a ^ 3 + 20a^2)+ (22 a^2 + 57a + 26 ) $$
$$= 10(a + 2) a^2+ (a + 2)(22a + 13 ) = (a + 2)(10^2+22a + 13 ) =0$$
which yields $a = -2$.
Alternatively, the cubic equation can always be solved with the Cardano's formula. First, make the variable change $x = t - \frac{42}{3\cdot 10} = t - \frac{7}{5}$ to write the equation in the depressed form $t^3+pt+q=0$, i.e.
$$t^3-\frac9{50}t+\frac{27}{250}=0$$
Then, apply the Cardano's formula to get the analytical solution $$t = \sqrt[3]{-\frac q2+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}=-\frac35$$
which gives the same result $a= -2$. The 'standard' approach may appear brute force, but it is guaranteed to yield the solution.
| {
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"answer_id": 4
} |
Numerical methods: forward difference operator $${ \text{If }y = \frac{1}{{(4 x + 1)} {(4 x + 5)} {(4 x + 9)}}, }\\ { \text{find } z = \Delta ^ {2} y .}$$
Can someone help with this? I tried it using factorial notation but apparently my answer is partially correct.
Any help would be much appreciated.
| By induction, for $a,b\in \mathbb{R}$ with $a\ne 0$ and denoting forward difference by $\Delta$, we have
$$
{\Delta}^n (ax+b)^{-1} = (-1)^n\cdot n! \cdot a^n \cdot \prod_{k=0}^{n} \left(ax+b +ak\right)^{-1}
$$So, your question can be rewritten as
$$
z = \Delta^2 y = \frac{1}{2!\cdot 4^2}\Delta^4 (4x+1)^{-1} = \frac{4!\cdot 4^4}{2!\cdot 4^2}\cdot \prod_{k=0}^{4} \left(4x+1 +4k\right)^{-1}
$$
$$
= \frac{192}{(4x+1)(4x+5)(4x+9)(4x+13)(4x+17)}
$$
| {
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Stuck on Some Derivative in Derivation of Faà di Bruno's Formula I have no problem with the underlying idea of Faà di Bruno's Formula as generalization of composite function derivatives of single-variable functions. Where I am not sure that there is some recursive operation related to chain rule. Here I will directly delve into the part I am stuck on:
Let $f(u)=y$ and $g(x)=u$ be single-variable functions of class $C^n$ defined on $\mathbb{R}$. We can write the first derivative of the composite function $f(g(x))$ with respect to $x$, which is obvious, as such: $$\frac{dy}{dx}= \frac{du}{dx} \frac{dy}{du}$$
The problematic part shown in the box below comes up when trying to take the second derivative as follows: $$\frac{d^2y}{dx^2}= \frac{d^2u}{dx^2} \frac{dy}{du}+ \frac{du}{dx} \fbox{$\frac{d}{dx} \left( \frac{dy}{du} \right)$}$$
I did not know how to take derivative of $\frac{dy}{du}$ with respect to $x$ at first. Then noting to use $\frac{dy}{dx}= \frac{du}{dx} \frac{dy}{du}$ as a chain rule operator where we can plug other functions between parentheses $($ and $)$: $$\frac{d \left(\ \right)}{dx}= \frac{du}{dx} \frac{d \left(\ \right)}{du}$$
However, even though it works out to give the correct result, which is as shown: $$\frac{d^2y}{dx^2}= \frac{d^2u}{dx^2} \frac{dy}{du}+ \left(\frac{du}{dx} \right)^2 \frac{d^2y}{du^2}$$
And something feels wrong about such a symbolic manipulation to me when taking $\frac{dy}{du}$ as some other function, say $h(x)=z$, of which we know that it contains $g(x)$ function itself so that we can plug it in the chain rule, otherwise no need to apply chain rule in the first place. It is also obvious that we cannot represent $h(x)$ contains $g(x)$ as $h(g(x))$. Thus I end up here in some kind of circular argument. If all make sense, where do I fail to comprehend? Is it mathematically valid of my treatment of chain rule in such an operator fashion?
| The derivation of $\frac{d^2 y}{dx^2}$ is sound.
We obtain
\begin{align*}
\color{blue}{\frac{d^2 y}{dx^2}}
&=\frac{d}{dx}\left(\frac{dy}{dx}\right)\\
&=\frac{d}{dx}\left(\frac{dy}{du}\,\frac{du}{dx}\right)\\
&=\left(\frac{d}{dx}\left(\frac{dy}{du}\right)\right)\frac{du}{dx}+\frac{dy}{du}\,\left(\frac{d}{dx}\left(\frac{du}{dx}\right)\right)\\
&=\left(\frac{d^2y}{du^2}\,\frac{du}{dx}\right)\frac{du}{dx}+\frac{dy}{du}\,\frac{d^2u}{dx^2}\\
&\,\,\color{blue}{=\frac{d^2y}{du^2}\,\left(\frac{du}{dx}\right)^2+\frac{dy}{du}\,\frac{d^2u}{dx^2}}\tag{1}
\end{align*}
in accordance with OPs derivation.
It is convenient to make a plausibility check by calculating a specific example in two ways. We consider
\begin{align*}
y&=f(u)=3\sin (u)+u^2+3\\
u&=g(x)=4x^2+1\\
\end{align*}
We have according to (1)
\begin{align*}
\frac{d^2 y}{dx^2}=&\left(\frac{d^2}{du^2}\left(3\sin(u)+u^2+3\right)\right)\,\left(\frac{d}{dx}\left(4x^2+1\right)\right)^2\\
&\quad+\left(\frac{d}{du}\left(3\sin(u)+u^2+3\right)\right)\,\left(\frac{d^2}{dx^2}\left(4x^2+1\right)\right)\\
&=\left(\frac{d}{du}\left(3\cos(u)+2u\right)\right)\left(8x\right)^2+(3\cos(u)+2u)\frac{d}{dx}\left(8x\right)\\
&=\left(-3\sin(u)+2\right)64x^2+(3\cos(u)+2u)8\\
&=\left(-3\sin\left(4x^2+1\right)+2\right)64x^2+\left(3\cos\left(4x^2+1\right)+2\left(4x^2+1\right)\right)8\\
&\,\,\color{blue}{=-192x^2\sin\left(4x^2+1\right)+24\cos\left(4x^2+1\right)+192x^2+16}\tag{2}
\end{align*}
On the other hand we obtain
\begin{align*}
\frac{d^2 y}{dx^2}=&\frac{d^2}{dx^2}\left(3\sin\left(4x^2+1\right)+\left(4x^2+1\right)^2+3\right)\\
&=\frac{d}{dx}\left(8x\cdot 3\cos\left(4x^2+1\right)+2\left(4x^2+1\right)8x\right)\\
&=\frac{d}{dx}\left(24x\cos\left(4x^2+1\right)+64x^2+16x\right)\\
&=\left(24\cos\left(4x^2+1\right)+24x\left(-\sin\left(4x^2+1\right)\right)8x\right)+192x^2+16\\
&\,\,\color{blue}{=-192x^2\sin\left(4x^2+1\right)+24\cos\left(4x^2+1\right)+192x^2+16}
\end{align*}
in accordance with (2).
| {
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"answer_count": 2,
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How can I find the volume of this 4-dimensional ball?
*
*Let $B$ be the $4$-dimensional ball
$$
\left\{(x, y, z, w): x^2 + y^2 + z^2 + w^2 \leq R^2 \right\}
$$
in $\mathbb{R}^4$. I want to make the change of variables
$$
x=r \cos \phi, \quad y=r \sin \phi, \quad z=\rho \cos \theta, \quad w=\rho \sin \theta
$$
to compute the volume of $B$, can you help? Thanks...
$\bf{Added:}$ The substitution above means that $x^2 + y^2 = r^2$, $z^2 + w^2 = \rho^2$, so $r^2 +\rho^2 = R^2$. Therefore $r = R \cos u$, $\rho = R \sin u$, where $u \in [0, \pi/2]$, since $r$, $\rho\ge 0$. Also $\phi$, $\theta \in [0, 2 \pi]$.
| I've taught this: the main thing is to do induction by two dimensions, using polar coordinates. So: we know the volume of the $n$ ball of radius $R$ is $c_n R^n.$ Next, the unit ball in dimension $n+2.$
So, on the unit disc in dimension 2, at a point $(r, \theta)$ we integrate the $n$ volume of a ball with radius $\sqrt {1-r^2},$ which becomes $c_n (1-r^2)^{n/2}$
Our new number, called $c_{n+2},$ becomes
$$ c_{n+2} = \int_0^{2 \pi} \int_0^1 c_n (1-r^2)^{n/2} r dr d\theta = 2 \pi c_n \int_0^1 (1-r^2)^{n/2} r dr$$
An antiderivative is
$$ \frac{-1}{n+2} \left( 1 - r^2 \right)^{\frac{n+2}{2}} \; , \; \; $$
and this is to be evaluated between $0$ and $1.$ This vanishes at endpoint $1,$ what remains is the negation of $\frac{-1}{n+2}.$ Put back the coefficients, we reach the simple
$$ c_{n+2} = \frac{2 \pi c_n}{n+2} $$
For example, $c_1 = 2$ for the line segment of "radius" 1, or length 2. Then
$$ c_3 = \frac{2 \pi 2}{3} = \frac{4 \pi }{3}, $$
and the volume of the ball with radius $R$ is $ \frac{4 \pi R^3}{3}$
Well, $c_2 = \pi$ for the area of the unit disc, $\pi R^2 \; . \;$ So
$$ c_4 = \frac{2 \pi \pi}{4} = \frac{ \pi^2 }{2}, $$
and the 4-ball of radius $R$ has 4-volume $\frac{\pi^2 R^4}{2}$
| {
"language": "en",
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The $x$-coordinate of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$.
The $x$-coordinates of the two points $P$ and $Q$ on the parabola $y^2=8x$ are roots of $x^2-17x+11$. If the tangents at $P$ and $Q$ meet at $T$, then find the distance of $T$ from the focus.
The point of intersection of tangents is (GM of abscissa, AM of ordinate)
Hence x coordinate of the point is
$$x=-\sqrt{\alpha \beta}$$
$$x=-\sqrt {11}$$
And y coordinate will be
$$y=\frac{y_1+y_2}{2}$$
But $y^2=8 \alpha$ and $y^2=8\beta$
$$y=\frac{2\sqrt {2\alpha} +2\sqrt {2\beta}}{2}$$$$y=\sqrt 2 (\sqrt{\alpha+\beta+2\sqrt {\alpha\beta}})$$
$$y=\sqrt 2 (17+2\sqrt 11)$$
As you may have already realised, I am going wrong. But I can’t seem to pinpoint it.
| Note that
$$x_1+x_2=17,\>\>\>\>\>x_1x_2 =11$$
Then, the intersection coordinates of the two tangents are (GM of abscissa, AM of ordinate),
$$x=\sqrt{x_1x_2} = \sqrt{11}$$
$$y = \frac{y_1+y_2}2=\sqrt2(\sqrt{x_1}+\sqrt{x_2})=\sqrt2\sqrt{x_1+x_2+2\sqrt{x_1x_2} }
=\sqrt{34+4\sqrt{11}}$$
Since the focus of $y^2=8x$ is $(2,0)$, the distance is thus,
$$\sqrt{(x-2)^2+y^2} =\sqrt{(\sqrt{11}-2)^2+34+4\sqrt{11}}=\sqrt{49}=7$$
| {
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An interesting question of algebraic manipulation If $a+b+c+d+e=0$ then prove that $a^3+b^3+c^3+d^3+e^3=3(abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde)$ Extend this argument to n integers such that if $a_1+a_2+a_3+\cdots a_n=0$ then $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
My try: At first I tried it for the base case. $a+b+c=0$ then it is a well-known fact that $a^3+b^3+c^3=3abc$, which proves the base case.Let us assume that the given statement is true for some $n$. $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
$$a_1^3+a_2^3+a_3^3\cdots a_n^3+a_{n+1}^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
Subtract the two equations we get,
$$a_{n+1}^3=3\left[(a_1a_2a_{n+1}+a_1a_3a_{n+1}+\cdots a_1a_na_{n+1})+(a_2a_3a_{n+1}+a_2a_4a_{n+1}+\cdots a_2a_na_{n+1})+\cdots +(a_{n-1}a_na_{n+1})\right]$$
I am stuck here. I cannot proceed further. Can please anybody help me?
| Instead of claiming it is "well known", actually prove it.
$0^3 = (a+b+c)^3 =$
$(a^3 + b^3 + c^3)+ 3(a^2b + a^2c + b^2a + b^2 c + c^2a + c^2b) + 6abc=$
$(a^3 + b^2 + c^3) +3((a^2b + a^2c + abc) + (b^2a+b^2c + bac) + (c^2a+c^2b + cab))- 3abc=$
$(a^3 + b^3 + c^2) + 3(a(ab+ac+bc) + b(ba + bc + ac) + c(ac+bc + ab))-3abc=$
$(a^3 + b^3 + c^2) + 3(a+b+c)(ab+ac+bc)-3abc=$
$(a^3 + b^3 + c^2) + 3*0*(ab+ac+bc)-3abc=(a^3 + b^3 + c^2) -3abc$ so $a^3 + b^3 + c^2 = -3abc$
It's much the same with multiple values but it's a variable tracking night mare:
$(a + b + c+d + e)^3 = $
$(a^3 + b^3 + c^3 + d^3 + e^3) + 3(a^2b + a^2c + a^2d + a^2e + b^2a+b^2c+b^2d +b^2e + c^2a+c^2b + c^2d + c^2e + d^2a+d^2b + d^2c+d^2e+e^2a+e^2b +e^2c +e^2d) + 6(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)=$
$(a^3 + b^3 + c^3 + d^3 + e^3) + 3((a^2b + a^2c + a^2d + a^2e + abc + abd+abe+acd+ace + ade) + (b^2a+b^2c+b^2d +b^2e+bac + bad+bae+bcd+bce+bde) + (c^2a+c^2b + c^2d + c^2e+cab+cad+cae+cbd+cbe+cde) + (d^2a+d^2b + d^2c+d^2e+dab + dac+dae+dbc+dbe+dce)+(e^2a+e^2b +e^2c +e^2d+eab+eac+ead + ebc+ebd+ecd)) - 3(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)=$
$(a^3 + b^3 + c^3 + d^3 + e^3) + 3(a+b+c+d+e)(ab+ac+ad+ae+bc+bd+be+cd+ce) - 3(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)=0$
$(a^3 + b^3 + c^3 + d^3 + e^3) - 3(abc + abd + abe + acd+ace+ade + bcd+ bce+bde + cde)$
Doing it with multiple variables is much the same (but extremely easy to get lost in the notation):
$(\sum a_i)^3 = (\sum a_i^3) + 3(\sum_{j\ne k} a_j^2a_k) + 6(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=$
$(\sum a_i^3) + 3(\sum_{\lnot(i=j=k)}a_ia_ja_k)- 3(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=$
$(\sum a_i^3) + 3[\sum a_i](\sum_{j\ne k} a_ja_k) - 3(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=$
$(\sum a_i^3) - 3(\sum_{i,j,k\text{ distict}} a_ia_ia_k)=0$
| {
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Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$
We have that $$a^2 + 4b^2 + c^2 = a^2 + (a + b + c + 1)b^2 + c^2 = (b^2 + a)a + (b + 1)b^2 + (b^2 + c)c$$
$$\implies \sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} = \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\frac{(a + b + c)^2}{(b^2 + a)a + (b + 1)b^2 + (b^2 + c)c}$$
$$ \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left(\frac{a}{b^2 + a} + \frac{1}{b + 1} + \frac{c}{b^2 + c}\right)$$
Furthermore, $\dfrac{a}{c^2 + a} + \dfrac{c}{a^2 + c} = 1 - \dfrac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}$, $$\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left[\frac{1}{b + 1} - \frac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}\right] + \frac{1}{3}$$
Then I don't know what to do next.
| The idea is indeed that of using Cauchy-Schwarz. The problem is you went from a homogeneous expression ($a^2+4b^2+c^2$) to non-homogeneous terms and that may be harder to prove than the original inequality.
Instead, I would apply Cauchy-Schwarz like this:
$$\frac{(a+b+c)^2}{a^2+4b^2+c^2} \leq \frac{a^2}{a^2+b^2}+\frac{b^2}{2b^2}+\frac{c^2}{b^2+c^2}=\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}+\frac{1}{2}$$
and summing cyclically:
$$
\begin{aligned}
\sum \frac{(a+b+c)^2}{a^2+4b^2+c^2} &\leq \sum\left(\frac{a^2}{a^2+b^2}+\frac{c^2}{b^2+c^2}\right)+\frac{3}{2}\\
&=\sum\left(\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}\right)+\frac{3}{2}\\
&=3+\frac{3}{2}=\frac{9}{2}
\end{aligned}
$$
| {
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Given $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ prove $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$ Given a sequence $(a_n)_{n\geq 1}$ such that $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ where $ n\geq1$
prove that $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$
I tried to prove that $a_n$ is increasing but it I just can't get around to it. Thank you in advance!
| So for given $a_n$, the next term $a_{n+1}$ is one of the roots of
$$X^2-(a_n+1)X-a_n, $$
i.e.,
$$\tag1\begin{align}a_{n+1}&=\frac{a_n+1\pm\sqrt{(a_n+1)^2+4a_n}}{2}\\&=\frac{a_n+1\pm\sqrt{(a_n+3)^2-8}}{2},\end{align}$$
which unfortunately gives us two choices. But there are some restrictions: As we (seem to) want a sequence of real numbers, we must have $(a_n+3)^2\ge 8$ for all $n$, i.e., $$\tag2a_n\ge 2\sqrt 2-3\approx -0.17\quad \text{or}\quad a_n\le -2\sqrt 2-3\approx -5.8.$$
If $a_n>0$ and we take the plus-branch in $(1)$, then we have $a_{n+1}\ge a_n+1$ from the first line in $(1)$.
As long as we take the plus-branch throughout, we thus find that $a_n\ge n$.
Assume we do not always take the plus-branch in $(1)$ and let $n$ be the first time we take the minus-branch. Then $$ \begin{align}a_{n+1}&=\frac{a_n+1-\sqrt{(a_n+3)^2-8}}2\\&=\frac{(a_n+1)^2-(a_n+3)^3+8}{2(a_n+1+\sqrt{(a_n+3)^2-8})}\\
&=\frac{-4a_n}{2(a_n+1+\sqrt{(a_n+3)^2-8})}\\
&<\frac{-4a_n}{2(a_n+1+\sqrt{(a_n+3)^2})}\\
&=-1+\frac{2}{a_n+2}\\&\le -1+\frac2{n+2}\\&\le -\frac13\\&<2\sqrt 2-3,\end{align}$$
contradicting $(2)$.
We conclude that we always take the plus-branch, i.e.,
$$a_{n+1}=\frac{a_n+1+\sqrt{(a_n+3)^2-8}}{2}$$
and (as seen above)
$$ a_n\ge n.$$
Now
$$\begin{align}a_{n+1}-a_n&=\frac{1-a_n+\sqrt{(a_n+3)^2-8}}2\\
&= 2+\frac{\sqrt{(a_n+3)^2-8}-(a_n+3)}2\\
&=2-\frac{4}{\sqrt{(a_n+3)^2-8}+(a_n+3)}\end{align}$$
and as $a_n\to\infty$, the desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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${2 x^{1-y}(1 - x^y) \over y (1-x)(1+x^{1-y})} \leq 1$ for $0The title says it all. I'd like to prove the inequality
$${2 x^{1-y}(1 - x^y) \over y (1-x)(1+x^{1-y})} \leq 1$$
for $0<x<1, 0<y<1$. I'd be happy just showing that the left hand side is bounded. Mathematica seems to believe that the inequality holds, and some formal asymptotics also suggest it. For example, fix $y$, and send $x$ to $1$. Writing $x^y = e^{y \log x} = 1 + y \log x + O((\log x)^2)$, we obtain that
$${1-x^y \over 1-x} \leq {y \log x + O((\log x)^2) \over \log x} = y + O(\log x) \to y$$
as $x \to 1$. Thus, for fixed $y$, the left hand side converges to $1$ as $x \to 1$. Similarly, for fixed $x$, as $y \to 0$, $1 - x^y = -y \log(x) + O(y^2)$, and so as $y \to 0$ the expression on the left converges to
$${2 x \log x \over (1-x)(1+x)}.$$
Since $\log x \leq 1-x$ for all $x \in (0,1)$ and $2x \leq 1 + x$ for all $x \in (0,1)$, this is bounded by $1$.
| Rewrite our inequality in the following form.
$$\frac{2x+y-xy}{2-y+xy}\geq x^{1-y}$$ or $f(x)\geq0,$ where
$$f(x)=\ln(2x+y-xy)-\ln(2-y+xy)-(1-y)\ln{x}.$$
But $$f'(x)=\frac{2-y}{2x+y-xy}-\frac{y}{2-y+xy}-\frac{1-y}{x}=$$
$$=-\frac{y(1-y)(2-y)(x-1)^2}{x(2-y+xy)(2x+y-xy)}<0.$$
Thus, $$f(x)>f(1)=0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem. I have been losing lots of points for minor details that I've been unable to see. Does the following proof hold? Am I making any major (or minor) errors?
\begin{align*}
\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|&\leq\left|\frac{1}{2+\sqrt{x}}\right|+\left|\frac{1}{3}\right|<\epsilon~~~\mbox{(by triangle inequality)}\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right|+\frac{1}{3}<\epsilon\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right| < \epsilon-\frac{1}{3}\\
&\implies \frac{1}{2} < \epsilon-\frac{1}{3}~~~~\mbox{(Because, }\sqrt{x}~\mbox{only a real number when } x\geq 0.)\\
&\implies 1<2(\epsilon-\frac{1}{3})\\
&\implies \left|x-1\right|<2\epsilon-\frac{2}{3}=\delta~~~~\mbox{(Because, choosing }x~s.t.~0<x<2\implies~-1<x-1<1)\\
\end{align*}
$\therefore \left|x-1\right|<\delta\implies\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|<\epsilon$ and $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
| What you did cannot possibly work. Since, for any $x\in\mathbb R$,$$\left\lvert\frac1{2+\sqrt x}\right\rvert+\frac13\geqslant\frac13,$$ if $\varepsilon\in\left(0,\frac13\right)$, then there is no $\delta>0$ such that$$\lvert x-1\rvert<\delta\implies\left\lvert\frac1{2+\sqrt x}\right\rvert+\frac13<\varepsilon.$$Note that\begin{align}\left\lvert\frac1{2+\sqrt x}-\frac13\right\rvert&=\left\lvert\frac{1-\sqrt x}{3\left(2+\sqrt x\right)}\right\rvert\\&\leqslant\frac{\left\lvert\sqrt x-1\right\rvert}6\\&=\frac{\left\lvert\left(\sqrt x-1\right)\left(\sqrt x+1\right)\right\rvert}{6\left(\sqrt x+1\right)}\\&\leqslant\frac{\left\lvert x-1\right\rvert}6.\end{align}So, for each $\varepsilon>0$, take $\delta=6\varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there a formula for expanding these kind of things (what kind of expression is it, by the way?)?
This is my solution:
$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$
$= \displaystyle\frac{[(\sqrt5+2)^3+(\sqrt5-2)^3][(\sqrt5+2)^3-(\sqrt5-2)^3]}{8\sqrt5}$
$=\displaystyle\frac{(\sqrt5+2+\sqrt5-2)[(\sqrt5+2)^2\color{red}{+}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2](\sqrt5+2-\sqrt5+2)[(\sqrt5+2)^2\color{red}{-}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2]}{8\sqrt5}$
$=\displaystyle\frac{[2\sqrt5(5+4\sqrt5+4+\color{red}{5-4}+5-4\sqrt5+4][4(5+4\sqrt5+4\color{red}{-(5-4)}+(5-4\sqrt5+4)]}{8\sqrt5}$
$=\displaystyle\frac{2584\sqrt5}{8\sqrt5}$
$=323$
Because of the multiplication, I still got the same answer as given in the book. However, is the book or I correct in terms of the positive/negative signs(in red)?
| The book solution used the formulas for the sum and difference of two cubes,
$x^3+y^3=(x+y)(x^2-xy+y^2)$ and $x^3-y^3=(x-y)(x^2+xy+y^2),$
with $x=\sqrt5+2$ and $y=\sqrt5-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can I solve this other than using Newton's sums or Newton's identities?
$\begin{cases}
&x + y + z = 12\\& x^2 + y^2 + z^2 = 12
\\ & x^3 +y^3 + z^3 = 12
\end{cases}$
If $x,$ $y,$ and $z$ satisfy the system of equations above, what is the value of $x^4+y^4+z^4?$
People told me that this can be solved by using Newton's sums or Newton's identities which I don't know how. Does anybody know how to do that or any other method to solve this?
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $$u=4,$$
$$3v^2=\frac{12^2-12}{2}=66,$$ which gives $$v^2=22.$$
Also, $$12=x^3+y^3+z^3=27u^3-27uv^2+3w^3,$$ which gives $$w^3=220$$ and use
$$x^4+y^4+z^4=81u^4-108u^2v^2+18v^4+12uw^3.$$
I got $1992.$
I used the known $uvw$'s substitutions: https://artofproblemsolving.com/community/c6h278791
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$. If $P_k=(\alpha)^k+(\beta)^k$, $k\ge 1$, then prove that-
a) $P_5=11$$
b) $P_1+P_2+P_3+P_4+P_5 =26$
For the first part
$$\alpha^5+\beta ^5$$
$$=(\alpha^3+\beta ^3)^2-2(\alpha \beta )^3$$
I found the value of $\alpha^3+\beta^3=4$
So $$16-2(-1)=18$$ which doesn’t match.
In the second part depends on the value obtained from part 1, so I need to get that cleared up.
I checked the computation many times, but it might end up being just that. Also, is there a more efficient way to do this?
| if $$x^2-x-1=0~~~(1)$$ has roots as $a,b$ then $P_k=a^k+b^k,P_0=2,P_1=a+b=1$ and
$$a^2-a-1=0~~~(2),~~ b^2-b-1=0~~~(3)$$
Multiply Eq.(2) once by $a^k$ and (3) by $b^k$. Adding these two Eqs. you get
$$(a^{k+2}+b^{k+2})-(a^{k+1}+b^{k+1})-(a^k+b^k)=0 \implies P_{k+2}= P_{k+1}+P_k~~~(4)$$
So $P_5=P_4+P_3, P_4=P_3+P_2, P_3=P_2+P_1, P_2=P_1+P_0=a+b+2=3$$ $$ \implies P_3=3+1=4, \implies P_4=4+3=7, P_5=7+4=11$
So $P_1+P_2+P_3+P_4+P_5=1+3+4+7+11=26$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Euclid's Lemma in the proof that $\mathbb{Z}[\sqrt{-5}]$ is not euclidean In proving, I induced that $3\gamma + (2+\sqrt{-5})\delta=1$. Textbook says that multiplying both sides by $2-\sqrt{-5}$ implies $2-\sqrt{-5}$ is multiple of 3. I don't know why. Can you explain the calculation.
| If you multiply both sides by $2 - \sqrt{-5}$ you get
$$ \begin{align} (2 - \sqrt{-5}) \cdot 3 \cdot \gamma + (2 - \sqrt{-5}) \cdot (2 + \sqrt{-5}) \delta &= 2 - \sqrt{-5} \\
(2 - \sqrt{-5}) \cdot 3 \cdot \gamma + 3 \cdot 3 \delta &= 2 - \sqrt{-5} \\
3\cdot ((2 - \sqrt{-5}) \cdot \gamma + 3 \delta) &= 2 - \sqrt{-5} \end{align}$$
From the first to the second line we use $(2 - \sqrt{-5})(2 + \sqrt{-5}) = 4+5 = 9 = 3 \cdot 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
how to prove this inequality with $a,b,c\in [1,3]$ let $a,b,c\in [1,3]$,show that
$$3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{45}{a+b+c}\ge 16\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)$$
I had found this simaler problem,https://artofproblemsolving.com/community/c6h615194
I can't prove this it
| $$3\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)+\frac{45}{a+b+c} ≥ 16\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)$$
Let:
$\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)=A$
$\frac{9}{a+b+c}=B$
$2\big(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\big)=C$
We use inequality:
$$C=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}≥\frac{9}{2(a+b+c)}=\frac{B}{2}$$
So we must show:
$3A+5B ≥ 16 C≥ 8B$, or $3A +5B≥ 8B$, or $A≥B$
Or:
$\big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\big)≥\frac{9}{a+b+c}$
Which is easy to prove.
Note:If you want the proof of inequality used, ask it as a question, I will answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx$
Evaluate
$$ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx,$$
My attempt : $u=\tan\frac{x}{2} \rightarrow x=2\arctan(u) \rightarrow \frac{2}{1+u^2}du=dx$
$$\sin x= \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{2u}{1+u^2}$$
$$\cos x= \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}= \frac{1-u^2}{1+u^2}$$
I was looking to see if this is the right way to go and also i am not sure how to evaluate the boundaries . Could you help me out ?
| Hint:
$$2\sin x+\cos x=\frac{4}{5}(\sin x+2\cos x)+\frac{3}{5}(2\sin x-\cos x)$$
This makes the integral $\displaystyle \int\left(\frac{4}{5}\mathrm dx-\frac{3}{5}\mathrm d(\ln|\sin x+2\cos x|)\right)$ with the same limits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Evaluate the following integral :$\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$ Evaluate the following integral :
$$I=\int\limits_0^{\infty}\frac{\log (1+x^{4})}{\sqrt{x}(1+x)}dx$$
I was tried use change variable ,
If I use $x=y^2$ integral becomes :
$$I=2\int\limits_0^{\infty}\frac{\log (1+x^{8})}{1+x^{2}}dx$$
From here I have one idea the derivative under sing integral but I got I difficult integration :
$$I=2\int\limits_0^{\infty}\frac{x^{8}}{(1+ax^{8})(1+x)}dx$$
I already to see you hints or solution!
| Let
$$I(a)=\int_0^{\infty}\frac{\ln(a^8+x^4)}{\sqrt x(1+x)}dx=2\int_0^{\infty}\frac{\ln(a^8+x^8)}{1+x^2}$$
Then
$$I(0)=2\int_0^{\infty}\frac{\ln(x^8)}{1+x^2}dx=2\int_{\infty}^0\frac{\ln(y^{-8})}{1+\frac1{y^2}}\left(-\frac{dy}{y^2}\right)=-2\int_0^{\infty}\frac{\ln(y^8)}{1+y^2}dy=-I(0)=0$$
And
$$\begin{align}I^{\prime}(a)&=2\cdot8a^7\int_0^{\infty}\frac{dx}{(a^8+x^8)(1+x^2)}dx=8a^7\int_{-\infty}^{\infty}\frac{dx}{(a^8+x^8)(1+x^2)}\\
&=2\pi i\cdot8a^7\left(\frac1{(a^8+1)(2i)}+\sum_{n=0}^3\frac1{8a^7e^{\pi i(2n+1)\cdot7/8}(1+a^2e^{\pi i(2n+1)/4})}\right)\\
&=\frac{8\pi a^7}{a^8+1}+2\pi i\sum_{n=0}^3\frac{-e^{-\pi i(2n+1)/8}}{a^2+e^{-\pi i(2n+1)/4}}\\
&=\frac{8\pi a^7}{a^8+1}-\frac{2\pi i}{(-2i)}\sum_{n=0}^3\left(\frac1{a+ie^{-\pi i(2n+1)/8}}-\frac1{a-ie^{-\pi i(2n+1)/8}}\right)\end{align}$$
So
$$\begin{align}I(1)&=I(0)+\int_0^1I^{\prime}(a)da\\
&=\pi\int_0^1\frac{8a^7}{a^8+1}da+\pi\sum_{n=0}^3\int_0^1\left(\frac1{a+ie^{-\pi i(2n+1)/8}}-\frac1{a-ie^{-\pi i(2n+1)/8}}\right)da\\
&=\pi\ln2+\left.\pi\sum_{n=0}^3\left(\ln\left(a+ie^{-\pi i(2n+1)/8}\right)-\ln\left(a-ie^{-\pi i(2n+1)/8}\right)\right)\right|_0^1\\
&=\pi\ln2+\pi\ln\left(\frac{\cos^2\frac{\pi}{16}\cos^2\frac{3\pi}{16}}{\sin^2\frac{\pi}{16}\sin^2\frac{3\pi}{16}}\right)\\
&=\pi\ln2+\pi\ln\left(\frac{\left(1+\cos\frac{\pi}8\right)\left(1+\cos\frac{3\pi}8\right)}{\left(1-\cos\frac{\pi}8\right)\left(1-\cos\frac{3\pi}8\right)}\right)\\
&=\pi\ln2+\pi\ln\left(\frac{\left(2+\sqrt{2+\sqrt2}\right)\left(2+\sqrt{2-\sqrt2}\right)}{\left(2-\sqrt{2+\sqrt2}\right)\left(2-\sqrt{2-\sqrt2}\right)}\right)\\
&=2\pi\ln\left(\left(2+\sqrt{2+\sqrt2}\right)\left(2+\sqrt{2-\sqrt2}\right)\right)\\
&=4\pi\ln\left(\sqrt2+\sqrt{2+\sqrt2}\right)\end{align}$$
WolframAlpha seems to agree with this result at least numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $z=\cos \theta + i \sin \theta$, express $\displaystyle \frac {1}{1-z \cos \theta}$ in the form $a+i\cdot b$. In order to get rid of the $z$, should I substitute $z=\cos\theta + i\sin \theta$ into the complex number, or what conjugate should I multiply the complex number by?
I have tried substituting $z=\cos\theta + i\sin \theta$ into the complex number, but only got this far:
$\displaystyle \frac {1}{1-z \cos \theta}$
$= \displaystyle\frac{1}{1-(\cos\theta + i\sin\theta)(\cos\theta)}$
$=\displaystyle\frac{1}{1-\cos^2\theta-i\cos\theta\sin\theta}$
As for multiplying the complex number by a conjugate, I have used $\big(\displaystyle\frac {1}{z}-\frac{1}{\cos\theta}\big)$, $\big(\displaystyle z-\frac{1}{z}\big)$ and $(1+z\cos\theta)$ but to no avail.
I have only learnt de Moivre's theorem, and I haven't learnt $\cosθ+i\sinθ=e^{iθ}$, so I would appreciate if this question can be solved in the simplest way possible. But other methods are welcome.
| Note that $z=e^{i\theta}$, $z=e^{-i\theta}$ and $\cos\theta = \frac{z+\bar z}2$
$$\frac {1}{1-z \cos \theta}=\frac {1}{1-z \cos \theta}\cdot\frac {1-\bar z \cos \theta}{1-\bar z \cos \theta}$$
$$=\frac {1-e^{-i\theta}\cos \theta}{1-(\bar z +z)\cos \theta + \cos^2\theta}$$
$$=\frac {1-(\cos\theta - i\sin\theta)\cos \theta}{1-2\cos\theta\cos \theta + \cos^2\theta}$$
$$=\frac {1-\cos^2\theta + i\sin\theta\cos \theta}{1- \cos^2\theta}$$
$$=1+ i\cot\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\sum_{n=0}^{\infty} \frac{n! a^n}{(n+1)^{n+1}}<\pi \sqrt{\frac{2}{e(e-a)}}$ Show $$\sum_{n=0}^{\infty} \frac{n! a^n}{(n+1)^{n+1}}<\pi \sqrt{\frac{2}{e(e-a)}}$$ where $a<e$.
This problem comes from the original version:
$$\int_0^{\infty} \frac{{\rm d}x}{e^x-ax}<\pi \sqrt{\frac{2}{e(e-a)}}(a<e),$$ which can deduce the present one, since
\begin{align*} \int_0^{+\infty} \frac{{\rm d}x}{e^x-a x}&=\int_0^{+\infty} \frac{e^{-x}}{1-a xe^{-x}}{\rm d}x\\ &=\int_0^{+\infty} \sum_{n=0}^{\infty} a^nx^ne^{-(n+1)x}{\rm d}x\\ &= \sum_{n=0}^{\infty} \int_0^{+\infty}a^nx^ne^{-(n+1)x}{\rm d}x\\ &=\sum_{n=0}^{\infty}\frac{n!a^n}{(n+1)^{n+1}} \end{align*}
| A much tighter bound will be derived:
$$(1) \quad S(a):=\sum_{n=0}^\infty \frac{n! \, a^n}{(n+1)^{n+1}} \le \frac{\sqrt{2}\pi}{a}
\Big(\frac{1}{\sqrt{1-a/e}} - g(a)\Big) \quad ,$$
where
$$ g(a)=1 + 2 c \log \big(\frac{1+\sqrt{1-a/e}}{2}\big) , \quad c= \frac{\sqrt{2}\, e}{\pi} - 1$$
The bound of the OP follows from setting $g(a)=0$, which is allowable because for all $0\le a < e,\quad g(a)<0.$ However, this action makes the expression blow up at the origin, which is a still a bound, but only good as $a \to e.$ However, if the region in which in which one is most interested is $a \to e,$ then replace the offending denominator of $a$ with $e.$ The expression (1) is very accurate. It is an exact match at $a=0$ and preserves the asymptote as $a \to e.$ For $a=e-1/10,$ the error is less than 1.2% .
With an index shift $S(a)$ can be rewritten as
$$ S(a) = \frac{1}{a} \sum_{n=1}^\infty \frac{n!}{n^{n+1}}\,a^n \, .$$
Via Stirling's formula one can derive
$$\frac{n!}{n^{n+1}}= \sqrt{2} \pi e^{-n} \Big( \frac{1}{2^{2n}} \binom{2n}{n} \Big) \Big( 1 + \frac{c}{n} + \frac{25}{1152n^2} -...\Big) \, , \quad c=\frac{5}{24}$$
I've chosen to write it this way because every knows the quantity in the first big parentheses ~ $1/\sqrt{n}.$ We can neglect the terms after the first if we make $c$ large enough, converting the asymptotics into a bound. We leave it undertermined, for now. Thus
$$ S(a) \le \Big(\sum_{n=0}^\infty x^n \frac{1}{2^{2n}} \binom{2n}{n} + c \sum_{n=0}^\infty \frac{x^n}{n} \frac{1}{2^{2n}} \binom{2n}{n} \Big)\, , \text{ where } x=a/e.$$ The sums are solvable in closed form, and we get the expression of (1), albeit with $c$ undetermined. However, we know that $S(0) = 1$ (the usual 0^0=1 convention is followed, and it also follows from the integral form.) Using L'Hopital's rule for expressions looking like 0/0, we get the equation
$$ 1 = \frac{1+c}{2} \sqrt{2} \,\frac{\pi}{e}, $$ which gives us that value for $c.$ This particular value of $c$ also satisfies
$$ \frac{n!}{n^{n+1}} \le \sqrt{2} \pi e^{-n} \Big( \frac{1}{2^{2n}} \binom{2n}{n} \Big) \Big( 1 + \frac{c}{n} \Big) $$ with equality for $n=1.$ Furthermore this $c$ is about 0.223657, which is only slightly larger than $5/24.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that this function is continuous at $(0,0)$ I need to prove that the function is continuous at $(0,0)$.
$$f(x,y)=\frac{2x^2y^2}{x^2+y^2}, \text{ if } (x,y)\neq(0,0)$$
and $0 \text{ if } (x,y)=(0,0)$
I'm trying to prove that $$\lim_{(x,y)\to (0,0)}\frac{2x^2y^2}{x^2+y^2}=0$$
but I don't know how to relate it to $$0\leq \sqrt{x^2+y^2}\leq \delta.$$ Finally I have $$2(x^2+y^2)\leq \epsilon.$$ Any ideas that can help me?
| Since $x^2,y^2\leqslant x^2+y^2$,$$0\leqslant\frac{2x^2y^2}{x^2+y^2}\leqslant2\frac{(x^2+y^2)^2}{x^2+y^2}=2(x^2+y^2).\tag1$$So, by the squeeze theorem,$$\lim_{(x,y)\to(0,0)}\frac{2x^2y^2}{x^2+y^2}=0.$$
Or, given $\varepsilon>0$, you can take $\delta=\sqrt{\frac\varepsilon2}$. If $\sqrt{x^2+y^2}<\delta$, then, by $(1)$,$$\frac{2x^2y^2}{x^2+y^2}<2\sqrt{\frac\varepsilon2}^2=\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is it possible to find where two curves have a common value? Suppose I have;
$$f(x) = x^2 - 42x + 364$$
$$g(y) = y^2 - 35y + 364$$
Computing out values I find they have a common value
$$f(6) = g(8) = g(27) = f(36) = 148$$
Is there a way of finding these values?
Working through based on received help...
(1) $$f(x) = g(y)$$
(2) $$f(x) - g(y) = x^2 - 42x + 364 - y^2 + 35y - 364 = 0$$
(3) $$x^2 - 42x - y^2 + 35y = 0$$
Complete the squares...
(4) $$(x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4}$$
(5) $$4 ((x^2 - 42x + 21^2) - (y^2 - 35y + \frac{35^2}{4}) = 21^2 - \frac{35^2}{4} )$$
(6) $$(4x^2 - 168x + 42^2) - (4y^2 - 140y + 35^2) = 42^2 - 35^2$$
(7) $$(2x - 42)^2 - (2y - 35)^2 = 539$$
Considering (7) as Fermat's factorization method;
(8) $$a^2 - b^2 = c$$
(9) $$(a + b)(a - b) = 539$$
(10) $$(2x - 42 + 2y - 35)(2x - 42 - 2y + 35) = 539$$
(11) $$(2x + 2y - 77)(2x - 2y - 7) = 539$$
Then, considering (11) as a simple product of two terms gives;
(12) $$d * e = 539$$
(13) $$(2x - 2y - 7) = d$$
(14) $$(2x + 2y - 77) = 539/d$$
| As the abovementioned 2 functions are continuous, you can find infinitely many common values in the intersection of their ranges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\int_{x^2+y^2+z^2
Calculate the integral: $$\int_{x^2+y^2+z^2<x+y+z} (x^2+y^2+z^2)dxdydz$$
Of course calculating the integral $\int_{a}^{b}\int_{c}^{d}\int_{e}^{f} (x^2+y^2+z^2)dxdydz$ would be very easy. However I have a problem because I must find $a,b,c,d,e,f$. I tried to use substitution: $$\begin{cases} x=r \cos \alpha \cos \beta \\ y=r \cos \beta \sin \alpha \\ z=r \sin \beta \end{cases} \text{ or } \begin{cases} x=r \cos \alpha \\ y=r \sin \alpha \\ z=z \end{cases}$$
But every of them proved to be ineffective.
Is there anyone who have some tips what I can do to find limits of integrity?
| Note that \begin{align*} x^2 + y^2 + z^2 < x+y+z \,\,\,\,\, &\Longleftrightarrow \,\,\,\,\, (x^2-x) + (y^2-y) + (z^2-z) < 0\\
&\Longleftrightarrow \,\,\,\,\, (x-1/2)^2 + (y-1/2)^2 + (z-1/2)^2 < 3/4.
\end{align*} Thus you are integrating over a sphere centered not at the origin, but at $(1/2,1/2,1/2)$ which should inform your substitution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Lagrange Multipliers - Solving system of equation $$\begin{array}{ll} \text{extremize} & \overbrace{xy + xz + yz}^{=: f(x,y,z)}\\ \text{subject to} & x^2 + y^2 + z^2 = 1\end{array}$$
I get the following Lagrange system of equations
$y + z - 2x\lambda = 0\qquad(1)$
$x + z - 2y\lambda = 0\qquad(2)$
$x + y - 2z\lambda = 0\qquad(3)$
$x^2 + y^2 + z^2 = 1\qquad(4)$
if i do (1) + (2) + (3) and divide by 2 i get $x +y+z = \lambda(x + y + z)$ which means $\lambda = 1$ or $x+y+z=0$
I can properly conclude the $\lambda = 1$ case but i'm not being able to develop the case $x+y+z=0$. Any tips on how to procede?
| $$1-(xy+xz+yz)=\sum_{cyc}(x^2-xy)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0,$$
which says that $$xy+xz+yz\leq1.$$
The equality occurs for $x=y=z=\frac{1}{\sqrt3},$ which says that we got a maximal value.
Also, (it's the case, for which you look) $$\frac{1}{2}+xy+xz+yz=\frac{1}{2}\sum_{cyc}(x^2+2xy)=\frac{1}{2}(x+y+z)^2\geq0,$$
which says $$xy+xz+yz\geq-\frac{1}{2}.$$
The equality occurs for $z=0$, $x=\frac{1}{\sqrt2}$ and $y=-\frac{1}{\sqrt2},$ which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T
\begin{pmatrix}
10 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25 \\
\end{pmatrix}$$
In the above formula, $A^T$ is the transpose of matrix A.
At present, I don't have a good way. I only know that the reference answer of matrix A is
$$\left(\begin{array}{ccc}
\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
0 & 0 & 1 \end{array}\right)$$
| Call the matrix $$C'=\begin{bmatrix}\frac{35}{2}&\frac{5\sqrt 3}{2}\\\frac{5\sqrt 3}{2}&\frac{25}{2}\end{bmatrix}=\begin{bmatrix}a&b\\b&d\end{bmatrix}$$ Apply the 'diagonalization of a symmetric matrix by congruence' formula (which comes from 'completing the square' in the corresponding quadratic form). If $$E=\begin{bmatrix}1&-\frac{b}{a}\\0&1\end{bmatrix}$$ then $$E^TC'E=\begin{bmatrix}a&0\\0&\frac{\det C'}{a}\end{bmatrix}.$$ Thus $$E=\begin{bmatrix}1&\frac{-\sqrt 3}{7}\\0&1\end{bmatrix}$$
and $$E^TC'E=\begin{bmatrix}\frac{35}{2}&0\\0&\frac{80}{7}\end{bmatrix}.$$ Let
$$F=\begin{bmatrix}\frac{2}{\sqrt 7}&0\\0&\frac{\sqrt 7}{2}\end{bmatrix}$$ Then
$$F^TE^TC'EF=\begin{bmatrix}10&0\\0&20\end{bmatrix}.$$ Let G be the 3 x 3 matrix whose upper left 2 x 2 block is $EF$ and whose element in row 3 and column 3 is 1 and whose other elements are 0. Then $$G^TCG=\begin{bmatrix}10&0&0\\0&20&0\\0&0&25\end{bmatrix}.$$ Thus $A=G^{-1}$
| {
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"url": "https://math.stackexchange.com/questions/3601237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On bounds for the deficiency of $m^2$, where $p^k m^2$ is an odd perfect number with special prime $p$ Hereinafter, call a number $N$ perfect if $N$ satisfies $\sigma(N)=2N$, where
$$\sigma(x)=\sum_{d \mid x}{d}$$
is the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of aliquot divisors of $x$ by $s(x)=\sigma(x)-x$.
Let $n = p^k m^2$ be an odd perfect number given in Eulerian form, that is, $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
We shall use the following results in deriving bounds for $D(m^2)$:
It turns out that it is possible to express $\gcd(m^2,\sigma(m^2))$ as an integral linear combination of $m^2$ and $\sigma(m^2)$, in terms of $p$ alone.
To begin with, write
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{s(p^k)}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$
Now, using the identity
$$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$
where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2) - (2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)}$$
so that we get
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{s(p^k)}=2m^2 - pD(m^2)=2(1-p)m^2 + p\sigma(m^2).$$
At once, since $s(p^k) \geq 1$, we have the upper bound:
$$D(m^2) \leq 2m^2 - pD(m^2) \implies D(m^2) \leq \frac{m^2}{(p+1)/2}.$$
Equality holds if and only if the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
We now attempt to derive a lower bound for $D(m^2)$ (in terms of $p$, $m^2$ and $\sigma(m^2)$), using the result discussed in this recent MSE question:
From the result
$$s(a)s(b) + (a + b) \leq s(ab)$$
which holds when $\gcd(a,b)=1$, $a>1$, and $b>1$, then setting $a=p^k$ and $b=m^2$, we obtain
$$s(p^k)s(m^2) + (p^k + m^2) \leq s(p^k m^2) = p^k m^2$$
$$\implies 1 + s(p^k)s(m^2) \leq (p^k m^2 - (p^k + m^2) + 1) = (p^k - 1)(m^2 - 1) = (p - 1)(m^2 - 1)s(p^k)$$
$$\implies 1 \leq \bigg((p-1)(m^2 - 1) - s(m^2)\bigg)s(p^k)$$
Multiplying both sides by $D(m^2)$ and dividing through by $s(p^k)$, we get
$$\frac{D(m^2)}{s(p^k)} \leq D(m^2)\cdot{\bigg((p-1)(m^2 - 1) - s(m^2)\bigg)}.$$
But we know from a previous calculation that
$$\frac{D(m^2)}{s(p^k)}=2m^2 - pD(m^2)=2(1-p)m^2 + p\sigma(m^2).$$
Hence, we have the lower bound
$$\frac{2(1-p)m^2 + p\sigma(m^2)}{(p-1)(m^2 - 1) - (\sigma(m^2) - m^2)} \leq D(m^2).$$
Summarizing, we have the bounds:
$$\frac{2(1-p)m^2 + p\sigma(m^2)}{(p-1)(m^2 - 1) - (\sigma(m^2) - m^2)} \leq D(m^2) \leq \frac{m^2}{(p+1)/2}.$$
Here are my questions:
(1) Does anybody here have any bright ideas on how to simplify the lower bound for $D(m^2)$?
(2) Are these bounds best-possible?
| We can get a better bound.
To get a better bound, we need a better inequality than $\sigma(x)-x\ge 1$.
So, let us find a better inequality on $\sigma(m^2)$.
To find a better lower bound, let us consider $m$ of the form $PQ$ where $P\lt Q$ are distinct primes.
Then, we have
$$\begin{align}\sigma(m^2)&\ge (1+P+P^2)(1+Q+Q^2)
\\\\&=1+P+P^2+Q+Q^2+PQ(P+Q+1)+P^2Q^2
\\\\&\ge 1+2+2^2+3+3^2+m(2+3+1)+m^2
\\\\&=m^2+6m+19\end{align}$$
from which we have
$$m^2-\sigma(m^2)\le -6m-19$$
Using this, we get, similarly as you did,
$$\begin{align}&s(p^k)s(m^2) -s(p^km^2)\le -m^2+p^k(-6m-19)
\\\\&\implies s(p^k)s(m^2)\le p^km^2-m^2+p^k(-6m-19)
\\\\&\implies s(p^k)s(m^2)\le (p^k-1)(m^2-1)+p^k(-6m-18)-1
\\\\&\implies s(p^k)s(m^2)\le (p-1)(m^2-1)s(p^k)+p^k(-6m-18)-1
\\\\&\implies p(6m+18)+1\le ((p-1)(m^2-1)-s(m^2))s(p^k)\end{align}$$
Multiplying the both sides by $\frac{D(m^2)}{s(p^k)}$ gives
$$\frac{D(m^2)}{s(p^k)}(p(6m+18)+1)\le ((p-1)(m^2-1)-s(m^2))D(m^2)$$
from which we get
$$\frac{(2(1-p)m^2 + p\sigma(m^2))(p(6m+18)+1)}{(p-1)(m^2-1)-(\sigma(m^2)-m^2)}\le D(m^2)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with polynomial divisilibity proof.
Show that for any natural $k$, $m$, $n$ polynomial $x^{3k}+x^{3m+1}+x^{3m+2}$ is divisible by $x^2+x+1$.
My observation:
$$x^a=(x^{a-2}-x^{a-3})(x^2+x+1)+x^{a-3}$$
so,
$$x^{3k}+x^{3m+1}+x^{3m+2}=(x^{3k-2}-x^{3k-3})(x^2+x+1)+x^{3k-3}+(x^{3m-1}-x^{3m-2})(x^2+x+1)+x^{3m-2}+(x^{3n}-x^{3n-1})(x^2+x+1)+x^{3n-1}$$
We can factor $x^2+x+1$, but i don't see any conclusion after that.
| Observe that $q(x)=x^2+x+1$ divides $p(x)=x^{3k}+x^{3m+1}+x^{3n+2}$ if and only if the roots of $q(x)$ are roots of $p(x)$.
Since $x^3-1 = (x-1)(x^2+x+1)$, the roots of $q(x)$ are $\alpha$ and $\beta$ and they are the primitive cubic roots of unity and obviously the following holds:
\begin{gather}
\alpha^3 = \beta^3 =1\\
\alpha^2+\alpha+1 = \beta^2+\beta+1=0
\end{gather}
Now we simply have:
$$
p(\alpha)=\alpha^{3k}+\alpha^{3m+1}+\alpha^{3n+2} = (\alpha^{3})^k+\alpha(\alpha^{3})^m+\alpha^2(\alpha^{3})^n =1+ \alpha+ \alpha^2 =0
$$
and the same thing happens to $\beta$. So $p(\alpha)=p(\beta)=0$ and $x^2+x+1\ | \ p(x)$ for all $n,m,k$
| {
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"url": "https://math.stackexchange.com/questions/3603365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find delta for the limit I'm having difficulty to solve this problem:
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$.
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displaystyle \lim_{x \to -1} \frac{1}{\sqrt{x^2+1}} = \frac{1}{\sqrt 2}.}$
I've started as this:
$$ \left
|\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right|
= \left|\left(\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right)\frac{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)} \right| = \left| \frac{\frac{1}{x^2+1} - \frac 12}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}
\right| < \epsilon.$$
But I do not know how can I finish solving this to find x.
| You should proceed with your approach a little further. Note that in the last step the denominator is not less than $1/\sqrt{2}$ which itself exceeds $1/2$ and therefore $$\left|\dfrac{\dfrac{1}{1+x^2}-\dfrac{1}{2}}{\dfrac{1}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{2}}}\right|<2 \left|\frac{1}{1+x^2}-\frac{1}{2}\right|=\frac{|1-x^2|}{\sqrt{1+x^2}}$$ And the right most expression does not exceed $|1-x^2|$ and hence our job is done if we can ensure that $$|1-x^2|<\epsilon$$ or $$|x+1||x-1|<\epsilon $$ To control the factor $|x-1|$ you can note that $$|x+1|<1\implies |x-1|\leq |x+1|+|-2|<3$$ Thus if $0<|x+1|<1$ then we have $|x^2-1|<3|x+1|$ and the last expression can be made less than $\epsilon$ if $|x+1|<\epsilon/3$. It follows that we can choose $\delta=\min(1,\epsilon/3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
I used the fact that
$$(A+B+C)^2=A^2+AB+AC+BA+B^2+BC+CA+CB+C^2$$
and multiply this with $(A+B+C)$ and I got
$$A^3+A^2B+A^2C
+ABA+AB^2+ABC
+ACA+ACB+AC^2
+BA^2+BAB+BAC
+B^2A+B^3+B^2C
+BCA+BCB+BC^2
+CA^2+CAB+CAC
+CBA+CB^2+CBC
+C^2A+C^2B+C^3$$
I want to simplify this and my question is the terms $AB$ and $BA$ equal so I can add them to be $2AB$ or not necessary since the $A$ and $B$ are matrices. Thanks
| You cannot simplify the full expansion, which has $27$ terms, because matrix multiplication is non-commutative ($AB\ne BA$ in general).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
| Just another way is to note for $t \in (0, 1)$,
$$\frac{1+t^2}{1-t^2} \geqslant \frac{27t+11}{16} \iff (3t-1)^2(3t+5)\geqslant 0$$
Hence
$$\sum_{cyc} \frac{1+a^2}{1-a^2}\geqslant \frac1{16}\sum_{cyc}(27a+11) = \frac{15}4 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that $DE\perp EF$.
Question: Point $D$ lies inside $\Delta ABC$ such that $\angle DAC=\angle DCA=30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE\perp EF$.
My approach: Let $\angle CDF=\alpha$ and $\angle EDC=\beta$.
Now in $\Delta FDA$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FA}{\sin(120^\circ-\alpha)}\\\implies FD=\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}.$$
Again in $\Delta FDC$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FC}{\sin \alpha}\\\implies FD=\frac{1}{2}.\frac{FC}{\sin \alpha}.$$
Thus, $$\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}=\frac{1}{2}.\frac{FC}{\sin \alpha}\\\implies \frac{\sin \alpha}{\sin(120^\circ-\alpha)}=\frac{FC}{FA}=\frac{1}{2}\\\implies \tan \alpha=\frac{1}{\sqrt 3}\implies \alpha=30^\circ.$$
Thus $\angle ADF=90^\circ$. Now let $CD$ extended meet $AB$ at $J$. Thus $\angle ADJ=60^\circ.$ Now observe that if we can prove that points $A,D$ and $E$ are collinear, then we can conclude that $\angle EDC=\beta=60^\circ$. Hence we will be done.
I tried to use Menalaus Theorem to prove the same, but it was of no use.
Also I tried to use coordinate bash. Consider $\Delta CDA$. Observe that $\Delta CDA$ is isosceles with $CD=BA$. Let $DO$ be the angular bisector of $\angle ADC$. Thus $DO$ is also the perpendicular bisector of $AC$. Now let $O$ be the origin and let $AC$ be the x-axis. Thus clearly $DO$ represents the y-axis. Now let $DA=s$. Thus clearly $A=\left(-\frac{\sqrt 3}{2}s,0\right),C=\left(\frac{\sqrt 3}{2}s,0\right)$ and $D=\left(0,\frac{s}{2}\right)$. Now let $B=(a,b)$, thus $$E=\left(\frac{a}{2}+\frac{\sqrt{3}}{4}s, \frac{b}{2}\right).$$
Now slope of $AB=m_1=\frac{2b}{2a+\sqrt 3s}$ and slope of $DB=m_2=\frac{2b-s}{2a}$. Now since the angle between $AB$ and $DB=60^\circ,$ thus we have $$\sqrt 3=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$ After this I haven't found anything significant.
So, how to proceed after this?
| Let $AD=DC=p$ and $\measuredangle BAD=\alpha$.
Thus, $$AC=p\sqrt3,$$
$\frac{BD}{\sin\alpha}=\frac{p}{\sin60^{\circ}},$ which gives $$BD=\frac{2p\sin\alpha}{\sqrt3},$$ $\frac{AB}{\sin(60^{\circ}+\alpha)}=\frac{p}{\sin60^{\circ}},$ which gives $$AB=\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}$$ and
$$\vec{DE}\cdot\vec{FE}=\frac{1}{2}\left(\vec{DB}+\vec{DC}\right)\left(\frac{1}{3}\vec{AC}+\frac{1}{2}\left(-\vec{AC}+\vec{AB}\right)\right)=$$
$$=\frac{1}{12}(\vec{DB}+\vec{DC})(3\vec{AB}-\vec{AC})=\frac{1}{12}\left(3\vec{DB}\cdot\vec{AB}-\vec{DB}\cdot\vec{AC}+3\vec{DC}\cdot\vec{AB}-\vec{DC}\cdot\vec{AC}\right)=$$
$$=\frac{1}{12}\left(3\cdot\frac{2p\sin\alpha}{\sqrt3}\cdot\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}\cdot\cos60^{\circ}-\frac{2p\sin\alpha}{\sqrt3}\cdot p\sqrt3\cdot\cos(90^{\circ}+\alpha)\right)+$$
$$+\frac{1}{12}\left(3\cdot p\cdot\frac{2p\sin(60^{\circ}+\alpha)}{\sqrt3}\cdot\cos(60^{\circ}+\alpha)-p\cdot p\sqrt3\cdot\cos30^{\circ}\right)=$$
$$=\frac{p^2}{12}\left(2\sin\alpha\sin(60^{\circ}+\alpha)+2\sin^2\alpha+2\sqrt3\sin(60^{\circ}+\alpha)(\cos(60^{\circ}+\alpha)-\frac{3}{2}\right)=$$
$$=\frac{p^2}{12}\left(\cos60^{\circ}-\cos(60^{\circ}+2\alpha)+1-\cos2\alpha+\sqrt3\sin(120^{\circ}+2\alpha)-\frac{3}{2}\right)=$$
$$=\frac{p^2}{12}\left(-2\cos30^{\circ}\cos(30^{\circ}+2\alpha)+\sqrt3\sin(120^{\circ}+2\alpha)\right)=0$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find $x,y \in \mathbb C $ such that $x^5+y^5=275, x+y=5$. My attempt:
Let $x, y$ be roots of
$$t^2-5t+p=0$$
I got
$$p=\frac{5 \pm \sqrt{-431}}{2}$$
using Vieta's relations.
Now I just calculated x and y using the quadratic formula in $t^2-5t+p=0$.
But I got two very hideous complex values of x and y. So I think my answer may be wrong.
The values I got are
$$x= \frac{5+\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253+5\sqrt{-431}}{2}}}{2}$$ or
$$x= \frac{5+\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}, y=\frac{5-\sqrt{\frac{253-5\sqrt{-431}}{2}}}{2}$$
Please post your own solutions too if you have any different solutions.
| $$x^3+y^3=(x+y)^3-3xy(x+y)=5^3-3xy(5)$$
$$x^2+y^2=(x+y)^2-2xy=5^2-2xy$$
$$(x^3+y^3)(x^2+y^2)=x^5+y^5+x^2y^2(x+y)$$
$$\iff(125-15xy)(25-2xy)=275+5x^2y^2$$
$$\iff(xy)^2-25xy+114=0$$
For the rest, follow How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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What is the solution of the following definite integral? I encountered this integral in my calculations (picture here):
$$
\int_{f=0}^{f=\pi/2} \frac
{\left(n_0 \sqrt{1 - \left(\frac{n_0}{n_1} \sin f \right)^2} - n_1 \cos f\right)^2}
{\left(n_0 \sqrt{1 - \left(\frac{n_0}{n_1} \sin f \right)^2} + n_1 \cos f\right)^2}
2 ~ \sin f ~ \cos f ~\mathrm{d}f
$$
Here, $n_0$ and $n_1$ are constants, such that $n_0$<$n_1$. Both constants are real and positive.
I tried to find solution in Table of integrals,series and products and by using Wolfram mathematica, but I have not managed to find it.
| This seems to be the Lambertian reflectance for parallelly polarised light passing from medium 0 to medium 1, so let us introduce $n = \frac{n_1}{n_0} > 1$ and write your integral as
\begin{align}
r_\text{p} (n) &= \int \limits_0^{\pi/2} \left(\frac{n^2 \cos(f) - \sqrt{n^2 - \sin^2(f)}}{n^2 \cos(f) + \sqrt{n^2-\sin^2(f)}}\right)^2 2 \sin(f) \cos(f) \, \mathrm{d} f \\
&\!\!\!\!\!\!\stackrel{\sin^2(f) = t}{=} \int \limits_0^1 \left(\frac{1 - \frac{n^2 \sqrt{1-t}}{\sqrt{n^2-t}}}{1 + \frac{n^2 \sqrt{1-t}}{\sqrt{n^2-t}}}\right)^2 \, \mathrm{d} t \stackrel{\frac{n \sqrt{1-t}}{\sqrt{n^2-t}} = u}{=} 2 n^2 (n^2 - 1) \int \limits_0^1 \frac{u (1-nu)^2}{(1+nu)^2(n+u)^2(n-u)^2} \, \mathrm{d} u \\
&= 2 n^2 (n^2 - 1) \int \limits_0^1 \left[\frac{8 n^3(n^4+1)}{(n^4-1)^3(1+nu)} - \frac{4n^3}{(n^4-1)^2 (1+nu)^2} \right. \\
&\phantom{= 2 n^2 (n^2 - 1) \int \limits_0^1 \left[\vphantom{\frac{8 n^3(n^4+1)}{(n^4-1)^3(1+nu)}}\right.} \left. - \, \frac{n^2-1}{(n^2+1)^3 (n-u)} + \frac{(n^2-1)^2}{4(n^2+1)^2n(n-u)^2} \right. \\
&\phantom{= 2 n^2 (n^2 - 1) \int \limits_0^1 \left[\vphantom{\frac{8 n^3(n^4+1)}{(n^4-1)^3(1+nu)}}\right.} \!\left. - \, \frac{n^2+1}{(n^2-1)^3(n+u)} - \frac{(n^2+1)^2}{4(n^2-1)^2 n (n+u)^2}\right] \mathrm{d} u \, .
\end{align}
The remaining integrals are elementary and after some simplification we end up with
$$ r_\text{p} (n) = 1 - \frac{4 n^3 (n^2+2n-1)}{(n^2-1)(n^2+1)^2} + \frac{16 n^4 (n^4+1)}{(n^2-1)^2(n^2+1)^3} \ln(n) - \frac{4 n^2 (n^2-1)^2}{(n^2+1)^3} \operatorname{arcoth} (n) \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expansion of $(a+b+c+d+e+....)^n$, but with all coefficients equal to 1. I'm looking for a formula to calculate the sum of $(a+b+c+d+...)^n$ but with coefficients equal to 1.
For example in $(a+b+c)^2$. I want the sum of $a^2 + b^2 + c^2 + ab + bc + ca$. And for $(a+b+c+d)^3$,
I want the sum of $a^3 + b^3 + c^3 + a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc$.
Similarly, I want the sum of expansion of $(a+b+c+d+e+....)^n$ with coefficient equal to 1.
I tried to find the pattern by expanding the $(a+b+c)^2$. And I found that its formula is $a^2 + b^2 + c^2$ + $\frac{((a+b+c)^2 - a^2-b^2-c^2)}{2}$. This works for the power of 2. But fails on other powers.
Any help will be really appreciated.
| For any $k$ variables $x_1, \ldots, x_k$, let $P_n(x_1,\ldots,x_k)$ be the sum of all monomials over them for degree $n$.
$$P_n(x_1,\ldots,x_k) \stackrel{def}{=} \sum_{\sum_{j=1}^{k} e_j = n} \prod_{j=1}^k x_j^{e_j}, \quad\text{where}\quad e_1,\ldots,e_k \in \mathbb{N}$$
Multiply by $s^n$ and sum over $n$, we find
$$\begin{align}
P(s) \stackrel{def}{=} \sum_{n=0}^\infty s^n P_n(\cdots)
&= \sum_{n=0}^\infty \sum_{\sum_{j=1}^{k} e_j = n} \prod_{j=1}^k (sx_j)^{e_j}\\
&= \sum_{e_1=0}^\infty\cdots\sum_{e_k=0}^\infty \prod_{j=1}^k (sx_j)^{e_j}
= \prod_{j=1}^k \sum_{e_j=0}^\infty (sx_j)^{e_j}\\
&= \prod_{j=1}^k \frac{1}{1 - s x_j}\end{align}$$
Apply partial fraction decomposition to RHS, we obtain
$$P(s) = \sum_{j=1}^k \frac{1}{1-sx_j} \prod_{\ell=1,\ne j}^k \frac{1}{1 - \frac{x_\ell}{x_j}}
= \sum_{j=1}^k \frac{x_j^{k-1}}{1-s x_j}\prod_{\ell=1,\ne j}^k \frac1{x_j - x_\ell}$$
Expand both sides and compare coefficients of $s^n$, we can represent $P_n$ as a sum of $k$ rational functions.
$$P_n(x_1,\ldots,x_k) = \sum_{j=1}^k x_j^{n+k-1}\prod_{\ell=1,\ne j}^k \frac{1}{x_j - x_\ell}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
sum of power floor $ S(n,m) = \sum_{i = 1}^{n} \Bigl \lfloor \frac{n}{i^m} \Bigr \rfloor i^m $ How do I calculate
$$ S(n,m) = \sum_{i = 1}^{n} \Bigl \lfloor \frac{n}{i^m} \Bigr \rfloor i^m $$
This can be simplifies to
$$ S(n, m) = \sum_{i = 1}^{\sqrt[m] n} \Bigl \lfloor \frac{n}{i^m} \Bigr \rfloor i^m = \sum_{i = 1}^{\sqrt[m] n} n - \sum_{i = 1}^{\sqrt[m] n}n \bmod i^m$$
Example
$$S(100,2) = \sum_{i=1}^{10} \Bigl \lfloor \frac{n}{i^2} \Bigr \rfloor i^2 = 100 + 100 + 99 + 96 + 100 + 72 + 98 + 64 + 81 + 100 = 910$$
More work.
Let's divide the $\Bigl \lfloor \frac{n}{i^{2}} \Bigr \rfloor $ into intervals of $1, 2, 3, \cdots$ Like this. and calculate the sum.
i.e from interval $\sqrt{\frac{n}{1}} \ge i \gt \sqrt{\frac{n}{2}} \quad $ like this and the summation becoms.
$$S(n,2) = \sum_{1}^{\bigl \lfloor \frac{n}{2} \bigr \rfloor} \Bigl \lfloor \frac{n}{i^2} \Bigr \rfloor i^2 + \sum_{i = \bigl \lfloor \frac{n}{2} \bigr \rfloor}^{\lfloor n \rfloor} 1 . i^2 $$
like so we can divide it even more but nothing converges to a smaller value and the multiplier of $i^2$ keeps on increasing.
$$S(n,2) =
\sum_{i = \bigl \lfloor \frac{n}{2} \bigr \rfloor}^{\lfloor n \rfloor} 1 . i^2 +
\sum_{i = \bigl \lfloor \frac{n}{3} \bigr \rfloor}^{\bigl \lfloor \frac{n}{2} \bigr \rfloor} 2 . i^2 +
\sum_{i = \bigl \lfloor \frac{n}{4} \bigr \rfloor}^{\bigl \lfloor \frac{n}{3} \bigr \rfloor} 3 . i^2 + \cdots$$
Is there a way to generalize this?
| Hint
Using the division algorithm, we can write $a=kq+r$, where $0 \leq r < k$. Then,
$$\frac{a}{k}=q+\frac{r}{k} \implies q=\lfloor\frac{a}{k}\rfloor.$$
So
$$a=k\lfloor\frac{a}{k}\rfloor+r \implies \lfloor\frac{a}{k}\rfloor k=a-r.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$
But I can't progress after this.
| $$0 = x^2 + 4 y^2 + 16 z^2 - 2 (4yz + 2zx + xy) = (x-y-2z)^2 + 3 (y-2z)^2 $$
So, as noted $y = 2z$ and $x = y+2z = 4z,$ so
$$ 48 = x^2 + 4 y^2 + 16 z^2 = 16 z^2 + 16 z^2 + 16 z^2 = 48 z^2 $$
and $z = \pm 1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3622114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.
My proof is the following:
\begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}
Could anyone find a simpler non-calculator proof without using big numbers?
| Claim 1: $2.3<\ln 10.$
Claim 2: $\ln 1.7<8/15$
Both these claims can be proven easily via Taylor series, etc.
Now, using the above inequalities, we have $1.7^{69}<e^{69\cdot \frac{8}{15}}<10^{16},$ or, multiplying $10^{69}$ on both sides, $17^{69}<10^{85}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
System of congruences where $\gcd(m, n)\ne1$ I have to solve this system of congruences:
$$
\begin{cases}
x^2+2x+2\equiv 0\pmod{10}\\
7x\equiv 20\pmod{22}
\end{cases}
$$
after some calculations
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cases}
$$
since $x\equiv 6\pmod{2}$ and $x\equiv 0\pmod{2}\\$ are equal, we get:
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 2\pmod{5}\\
x\equiv 0\pmod{2}\\
x\equiv 6\pmod{11}\\
\end{cases}
$$
$$
\begin{cases}
x\equiv 1\pmod{5}\\
x\equiv 6\pmod{11}\\
\end{cases}\implies x\equiv 46\pmod{55}
$$
$$
\begin{cases}
x\equiv 0\pmod{2}\\
x\equiv 2\pmod{5}\\
\end{cases}\implies x\equiv 2\pmod{10}
$$
but, $\gcd(55,10)\ne1$, so I cannot apply the Chinese theorem. What have I done wrong?
| \begin{cases}
x\equiv 0\;(mod\;2)\\
\hline
x\equiv 1\;(mod\;5)\\
x\equiv 2\;(mod\;5)\\
\hline
x\equiv 6\;(mod\;11)\\
\end{cases}
So you want a solution modulo $2 \cdot 5 \cdot 11 = 110$.
This is how I would solve it.
\begin{array}{r|rrr}
& 2 & 5 & 11 \\
\hline
55 & 1 & 0 & 0\\
22 & 0 & 2 & 0\\
10 & 0 & 0 & -1 \\
\hline
\end{array}
Note that the top row is the three prime moduli that we are using.
The left column is
$\dfrac{2 \cdot 5 \cdot 11}{2} = 55 \quad $,
$\dfrac{2 \cdot 5 \cdot 11}{5} = 22 \quad$, and
$\quad \dfrac{2 \cdot 5 \cdot 11}{11} = 10$.
The remaining entries show $55, 22, 10$ modulo $2, 5, 11$.
The goal is to multiply $55, 22$ and $10$ by the appropriate integers so that the three diagonal elements are all $1$.
*
*$55$ already gives us a diagonal element of $1$.
*Since $2 \cdot 3 \equiv 1 \pmod 5$, we change $22$ to $22 \cdot 3 = 66$.
*Since $-1 \cdot -1 \equiv 1 \pmod{11}$, we change $10$ to $10 \cdot (-1) = -10$.
\begin{array}{r|rrr}
& 2 & 5 & 11 \\
\hline
55 & 1 & 0 & 0\\
22 & 0 & 2 & 0\\
10 & 0 & 0 & -1 \\
\hline
55 & 1 & 0 & 0\\
66 & 0 & 1 & 0\\
-10 & 0 & 0 & 1 \\
\hline
\end{array}
We use those numbers, $55, 66, -10$ as follows
$\left. \begin{align}
x &\equiv 0 \pmod 2 \\
x &\equiv 1 \pmod 5 \\
x &\equiv 6 \pmod{11}
\end{align} \right\}
\iff x \equiv 0(55) + 1(66) + 6(-10) \equiv 6 \pmod{110}$
$\left. \begin{align}
x &\equiv 0 \pmod 2 \\
x &\equiv 2 \pmod 5 \\
x &\equiv 6 \pmod{11}
\end{align} \right\}
\iff x \equiv 0(55) + 2(66) + 6(-10) \equiv 72 \pmod{110}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Mod arithmetic and divisibility The question comes from a textbook
Prove that $3x^5+5x^3+7x$ is divisible by 15 for any integer $x$. I want to do this through mod arithmetic.
The first thing that I did was to try and solve it through mod 3 and 5. If they are congruent to 0 for both, then it is divisible by 15.
So I tried to solve it mod 5 first.
$3x^5+5x^3+7x$ is congruent to $3x^5+x^3+2x$
If x (1 to 5) when subbed in gives 0(mod 5), then all $x$ values will give 0(mod 5).
However when I come to sub in $x=2$, it gives $96+8+4$ which doesn't give 0(mod 5). Have I done or assumed anything wrong up to this point?
| Note since $5 \equiv 0 \pmod{5}$, then $5x^3 \equiv 0 \pmod{5}$, but in your statement of
$3x^5+5x^3+7x$ is congruent to $3x^5+x^3+2x$
you have that $5x^3 \equiv x^3 \pmod{5}$ instead (but you did correctly simplify $7x \equiv 2x \pmod{5}$). This is only true for $x \equiv 0 \pmod{5}$. Also, it's why your substitution of $x = 2$ didn't work properly. Note with the correction simplification of $3x^5 + 2x$, you for $x = 2$, you get a result of $96 + 4 = 100$, which is divisible by $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$ I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities.
For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^4 \leq \Big (\frac{x + a + b + x + a - b+ x - a + b+ x - a - b}{4}\Big)^4 = $$
$$= x^4.$$
So $maxf(x) = x^4$.
But i don't know how to solve is for $minf(x)$, which i need to find.
Sure do i can find structure of square difference in $f(x)$ and we can rewrite our equality:
$$f(x) = (x^2 - (a + b)^2) (x^2 - (b - a)^2).$$
But i don't know what to do next.
UPD: We need to find extremum on $x$ via fixing $a, b$. I understand that $max$ is found wrong way. How can i do it correctly?
| $$f(x)=(x^2-(a^2+b^2))^2+(b^2-a^2)^2-(a^2+b^2)^2\ge(b^2-a^2)^2-(a^2+b^2)^2=-4a^2b^2$$
The equality occurs for $x^2=a^2+b^2$, which is always possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Quartic polynomials of a complex variable I want to answer the following:
The equation
$$
a \bigg(z + \frac{1}{z}\bigg)^2
+ b \bigg(z + \frac{1}{z}\bigg)
+ c = 0
$$
has four solutions. Find which quartics can be put in this form after apply a linear change of vars.
| $a(z+\frac{1}{z})^2+b(z+\frac{1}{z})+c = 0$
This is a quartic polynomial, if we multipy by $z^2$ we obtain
$az^4+bz^3+(2a+c)z^2+bz+a = 0$
This quartic can be related to the general quartic polynomial
$f(x) = Ax^4+Bx^3+Cx^2+Dx+E$
Now compare the coefficient of the two polynomials
$az^4+bz^3+(2a+c)z^2+bz+a$ and $Ax^4+Bx^3+Cx^2+Dx+E$
Are you looking for a way to transform the general quartic to this form $az^4+bz^3+(2a+c)z^2+bz+a$ ?
Now we are looking for a quartic that can be put in that form, since it has $3$ variables $a,b,c$, then reduce the polynomials and compare with a depressed quartic $y^4+py^2+qy+r$, equate their coefficient then put $a,b,c$ in terms of the order coefficient
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 0
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Least $x$ Such That $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$ A friend recently texted me the following:
Compute the least $x$ such that $\lfloor x^2\rfloor -\lfloor x\rfloor ^2=10$.
Is there a way to do what my friend is asking analytically? I graphed it on Desmos, and got $x\approx 5.91608$. I realized that this should be (and indeed appears to be) $\sqrt{35}$, as then we have $35-25=10$. To show this, I tried my typical way of solving floor problems, which is to break up $x=I+F$, where $I=\lfloor x\rfloor$ and $F=x-\lfloor x\rfloor$. After simplifications, I got $\lfloor F(2I+F)\rfloor =10$. But then upon any further manipulation, I get back the original equation. Is there some further manipulation I am not aware of that lets me solve this analytically? Or am I on the wrong track entirely? Should I perhaps try to prove that $\sqrt{35}$ is the least value of $x$ satisfying the above condition? Sorry if I’m missing something obvious.
| $$x = I+ F \qquad (I \in \mathbb Z, \quad 0 \le F < 1).$$
$$\lfloor 2IF+F^2\rfloor =10 \tag{A.}$$
A quick check shows that $I$ needs to be positive.
Let's first consider what happens when
$$F^2 + 2IF = 10 \tag{B.}$$
By the quadratic equation,
$$F = \dfrac{-2I+\sqrt{4I^2+40}}{2}
= -I+\sqrt{I^2+10}$$
This clearly implies $F \ge 0$. We also need
\begin{align}
F &< 1 \\
-I+\sqrt{I^2+10} &< 1 \\
\sqrt{I^2 + 10} &< I + 1 \\
I^2 + 10 &< I^2 + 2I + 1 \\
10 &< 2I + 1 \\
I &\ge 5
\end{align}
Which leads, unsurprisingly, t0 $x = \sqrt{I^2+10}$ with the restriction $I \ge 5$.
For $I = 5$, we get $F = -5 + \sqrt{35}$. So $x = \sqrt{35}$.
$\lfloor x^2 \rfloor -\lfloor x\rfloor ^2 = 35 - 25 = 10$
In general, there will be a solution when
$$10 \le \lfloor 2IF+F^2\rfloor < 11$$
So our more general equation must have the form
$$ F^2 + 2IF = 10 + \epsilon$$
where $0 \le \epsilon < 1$. Then we would get
$$F = -I+\sqrt{I^2+10 + \epsilon}$$
and $$x = \sqrt{I^2+10 + \epsilon}$$
where $0 \le \epsilon < 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is the number of nonzero digits always at least $m$? if $k,m$ is give postive integers,for any postive integer $p$, if we define
$$p\cdot\dfrac{k^m-1}{k-1}=a_{i}k^{i}+a_{i-1}k^{i-1}+\cdots+a_{1}\cdot k+a_{0}$$
where $a_{i}\in\{0,1,2,\cdots,k-1\}$,and let set $A=\{i|a_{i}\neq 0\}$,show that
$$|A|\ge m$$
this problem it see interesting,such if $k=2,m=3$ I found $p=1,2,3,4,\cdots 10$ it is clear true,But How to prove this General
for example: for $k=2,m=3$,it is clear $p=2^a,a\ge 1$ is right,and we only consider $p$ is prime,
and
(1)$p=3$,then $3(2^2+2^1+2^0)=21=2^4+2^2+2^0$ so $A|=3=m$
(2)$p=5$,then $5(2^2+2^1+2^0)=35=2^5+2^1+2^0$,so $A|=3$
(3):$p=7$,then $7(2^2+2^1+2^0)=49=2^5+2^4+2^0$ so $|A|=3$
(4):$p=11$,then $11(2^2+2^1+2^0)=77=2^6+2^3+2^2+2^0$ so $|A|=4\ge 3=m$
(5):$p=13$,then $13(2^2+2^2+2^0)=2^6+2^4+2^3+2^1+2^0$,so $|A|=5\ge m$
(5):$p=17$,then $13(2^2+2^2+2^0)=119=2^6+2^5+2^4+2^2+2^1+2^0$,so $|A|=6\ge m$
in general: let $p=2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{k}},0\le a_{1}<a_{2}<\cdots<a_{k},a_{i}\in N$
| Notice that you actually claim that the number:
$$
N=N(p,k,m)=p\cdot\dfrac{k^m-1}{k-1}=p(k^0+k^1+k^2+\dots+k^{m-1})
$$
Has at least $m$ non-zero digits in number base $k$, where $a_{i}\in\{0,1,2,\cdots,k-1\}$ are its digits.
To prove the claim for $p\lt k$ is easy. Notice that for this case, we have:
$$
N=\sum_{i=0}^{m-1}p \cdot k^i
$$
Where $a_i=p\le k-1$ for $i=0,\dots,m-1$, implying $|A|=m$.
This proves the $p\lt k$ case.
It remains to prove the $p\ge k$ case.
$(\star):$ Notice that (where $a\in\{1,2,\dots,k-1\}$ and $r\in\mathbb N$)
$$
a\cdot k^r\equiv\{a\cdot k^0,a\cdot k^1,\dots,a\cdot k^{m-1}\}\pmod{ N}
$$
Now suppose that a counter-example exists: (Proof by contradiction)
$$\begin{align}
N&=a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\\
N&\equiv a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\pmod{N}
\end{align}$$
Where $i_1,i_2,\dots,i_t$ are distinct and $t\le m-1$.
But because of $(\star)$, looking at RHS modulo $N$ we have at most (for all $p\ge k$):
$$
a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\le \sum_{i=1}^{m-1} (k-1)k^i = k(k^{m-1}-1)\lt N
$$
Which implies $a_{i1}k^{i1}+a_{i2}k^{i2}+\dots+a_{it}k^{it}\not \equiv N$, which is a contradiction.
This proves the $p\ge k$ case.
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
How to analyze the equation $(x-y)^2=2\big( (x+y)-2\sqrt{xy} \big)$? Suppose that $x,y$ are positive real numbers and that
$$ (x-y)^2=2\big( (x+y)-2\sqrt{xy} \big). \tag{*}$$
Then Mathematica claims that one of the following $3$ options holds:
$$1. \, \, \, x=y.$$
$$2. \, \, \, x = y - 2 \sqrt 2 \sqrt y + 2.$$
$$3. \, \, \, x = y + 2 \sqrt 2 \sqrt y + 2.$$
(Option $3$ is in fact impossible over the reals.)
Also, if I interpret the results correctly, the last two options $(2,3)$ are only possible when $xy \le \frac{1}{4}$. When $xy \ge \frac{1}{4}$ only option $(1)$ is possible.
Is there a way to prove this analytically (without using a computer)?
Edit:
We assume $x \ge 0,y \ge 0$.
In the answer below, we rewrite equation $(*)$ as
$(x - y)^2 = 2 ( \sqrt{x} - \sqrt{y})^2$, which gives
$$ (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})=x - y =\pm \sqrt{2} ( \sqrt{x} - \sqrt{y}).$$
Thus, either $x=y$ or $\sqrt{x} + \sqrt{y}=\pm \sqrt{2}$.
If $x,y$ are non-negative reals, then $\sqrt{x} + \sqrt{y}=- \sqrt{2}$ is ruled out.
So, we are left with
$$\sqrt{x} + \sqrt{y}= \sqrt{2} \Rightarrow \sqrt{x} = \sqrt{2}-\sqrt{y} \Rightarrow x=y - 2 \sqrt 2 \sqrt y + 2.$$
The third option actually comes from the branch where $\sqrt{x} + \sqrt{y}=- \sqrt{2}$, and then $x,y$ are complex numbers.
Next, we show that $\sqrt{x} + \sqrt{y}= \sqrt{2}$ is possible if and only if $xy \le \frac{1}{4}$.
By the AM-GM inequality $\frac{1}{\sqrt 2}=\frac{\sqrt{x} + \sqrt{y}}{2}\ge \sqrt{\sqrt{xy}}$.
On the other hand, suppose that $\sqrt{xy} =s \le \frac{1}{2}$. Writing $a=\sqrt x,b=\sqrt y$, we are looking for $a,b \ge 0$ such that $ab=s,a+b=\sqrt{2}$. This is a quadratic equation, and since the AM-GM holds it has real solutions, which must be positive.
(Indeed, since $s=ab$ is positive, then $a,b$ have the same sign, and $a+b=\sqrt{2}>0$ implies they are both positive.
The quadratic is
$$ t^2-\sqrt 2 t+s=0.$$
| Factorise to $(x - y)^2 = 2 ( \sqrt{x} - \sqrt{y})^2$. Which in turn gives $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = \pm \sqrt{2} (\sqrt{x} - \sqrt{y})$. So $x=y$ or $\sqrt{x} = \pm \sqrt{2} - \sqrt{y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on both sides:
$$1 = \frac{p^2}{h^2}+\frac{b^2}{h^2}$$
since $\sin x = \frac ph$ and $\cos x = \frac bh$,
$$1 = \sin^2x+\cos^2x$$
Are there any more innovative ways of proving this common identity?
| Here's a couple of different methods:
Proof using the cosine angel sum formula:$$
1=\cos(0) =\cos(x + (-x)) =\cos(x)\cos(-x) - \sin(x)\sin(-x) = \cos^2(x) + \sin^2(x)
$$
Proof using differential equations:
Recall the differential equation definitions of $\sin$ and $\cos$: They are the solutions to $f'' = - f$ with the appropriate initial conditions $f(0) = 0$, $f'(0) = 1$ for $\sin(x)$ and $f(0) = 1$, $f'(0) = 0$ for $\cos(x)$. Since solutions to differential equations are unique given the initial conditions, we immediately get $\sin'(x) = \cos(x)$ under this definition. Then:\begin{eqnarray}
\frac{d}{dx} \left(\sin^2(x) + \cos^2(x)\right) &=& \frac{d}{dx} \left(\sin^2(x) + \left(\sin'(x)\right)^2\right)\\
&=& 2\sin(x)\sin'(x) + 2\sin'(x)\sin''(x)\\
&=& 2\sin(x)\sin'(x) +2\sin'(x)(-\sin(x)) = 0
\end{eqnarray}
hence $\sin^2(x) + \cos^2(x)$ is constant, since its derivative is 0. Plugging in $x=0$ we see that it must equal 1.
Proof using Euler's formula:
$$
\sin^2(x) + \cos^2(x) = \left(\cos x + i \sin x\right)\left(\cos x - i \sin x\right) =\left(\cos x + i \sin x\right)\left(\cos (-x) + i \sin (-x)\right)= e^{i x} e^{-i x} = 1
$$
Proof using Taylor series:
Using the Taylor series:\begin{eqnarray}
\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\\
\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}
\end{eqnarray}
we have $$
\sin^2(x) = \left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\right)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{2n + 2m + 2}}{(2n+1)!(2m+1)!} = \sum_{k=0}^\infty x^{2k+2}(-1)^k \sum_{j=0}^k \frac{1}{(2j+1)!(2(k-j)+1)!}= \sum_{k=0}^\infty \frac{x^{2k+2}(-1)^k}{(2k+2)!}\sum_{j=0}^k \binom{2k+2}{2j+1} = -\sum_{k=1}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^{k-1} \binom{2k}{2j+1}
$$
and $$
\cos^2(x) =\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\right)^2 = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m} x^{2n + 2m}}{(2n)!(2m)!} = \sum_{k=0}^\infty x^{2k}(-1)^k \sum_{j=0}^k \frac{1}{(2j)!(2(k-j))!} = \sum_{k=0}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^k \binom{2k}{2j} = 1 + \sum_{k=1}^\infty \frac{x^{2k}(-1)^k}{(2k)!}\sum_{j=0}^k \binom{2k}{2j}
$$
Adding these together:$$
\sin^2(x) + \cos^2(x) = 1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^k}{(2k)!}\left(\sum_{j=0}^k \binom{2k}{2j}-\sum_{j=0}^{k-1} \binom{2k}{2j+1}\right)=1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^k}{(2k)!}\left(\sum_{j=0}^{2k} (-1)^j\binom{2k}{j}\right) = 1+\sum_{k=1}^\infty\frac{x^{2k}(-1)^l}{(2k)!} (1 + (-1))^{2k} = 1
$$
We use the binomial theorem on the second last step.
Carlson's Theorem: Major overkill to use this one, but it still works. Notice $\cos(x)$, and $\sin(x)$ have exponential type 1. Thus $$
f(x) = \cos^2(\frac{\pi}{4}x) + \sin^2(\frac{\pi}{4}x) - 1
$$
has exponential type at most $\frac{\pi}{2} < \pi$. By inspection, you can see that $f(x) = 0$ for $x\in \mathbb{N}$. Hence Carlson's theorem implies $f(x) = 0$ identically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$.
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.
| Note
$$\alpha (\alpha^2+2\alpha -4)\beta =-1$$
Thus,
$$\beta =- \frac1{\alpha (\alpha^2+2\alpha -4)}
=- \frac1{\alpha^3+2\alpha^2-4\alpha}
=- \frac1{5\alpha -1 -4\alpha} =\frac1{1-\alpha}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find (ab) if $ \int \frac{15x^2dx}{\sqrt{2x-1}}=\left(ax^2\:+bx\:+b\right)\sqrt{2x-1}+C $ $$
\int \frac{15x^2dx}{\sqrt{2x-1}}=\left(ax^2\:+bx\:+b\right)\sqrt{2x-1}+C
$$
What is the answer written as integer
$$
225\sqrt{2x-1}+C
$$
(excuse my English)
| \begin{eqnarray*}
\int{\frac{15 x^{2}}{\sqrt{2 x-1}} dx} &=& \left(2+2 x+ 3 x^{2} \right) \sqrt{2 x -1} + C
\end{eqnarray*}
which means, $a=3, b=2$ and hence $ab=6$. Is this that you are seeking here?
Note: $y^{2} = 2 x-1$, which means $2 y d y = 2 dx $ or $ \frac{dx}{y}=dy$. The problem then translates to,
\begin{eqnarray*}
\int{\frac{15 x^{2}}{\sqrt{2 x-1}} dx} &=& \int{15 \left(\frac{y^{2}+1}{2} \right)^{2} dy} \\
&=& \frac{y}{4} \left(15+10 y^{2}+ 3 y^{4}\right) +C\\
&=& \frac{\sqrt{2 x -1} }{4} \left(8+8 x+ 12 x^{2} \right) + C \\
&=&\left(2+2 x+ 3 x^{2} \right) \sqrt{2 x -1} + C
\end{eqnarray*}
| {
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"source": "stackexchange",
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"answer_id": 0
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Showing if all eigenvalues of $A$ have negative real parts then our system has a strong Lyapunov function of the form $x^TSx$. Can I please have help solving the problem? I am having a tough time working out the details for $S$ and how to do it without assuming diagonability. Thank you!
Show that if all eigenvalues of $A$ have negative real parts then our system has a strong Lyapunov function of the form $x^TSx$. I want to show that $$S = \int_0^\infty e^{\tau A^T} e^{\tau A} d\tau,$$ satisfies $A^TS + SA = -I.$ So I know we can left-multiply by $e^{\tau A^T}$ and right-multiply by $e^{\tau A}$ and show that the result is a total derivative. Then we can use this to compute $S$ for the matrix $$A = \begin{pmatrix} -2 & 1 \\ 0 & -2 \end{pmatrix}.$$
$\textbf{Solution:}$ Assume $\lambda$ is an eigenvalue of $A$. Then the real parts of $\lambda$ are less than $0$.
Let us consider $L = x^TSx, S^T = S$ and $x'= Ax$. By differentiating, we arrive at $$\frac{dL}{dt} = x^T(A^TS + SA)x.$$ From the hint, $$\frac{dL}{dt} = x^T(-I)x = -||x|| < 0.$$ Thus, $L$ is a strong Lyapunov function.
Next, let $$S = \int_0^\infty e^{\tau A} e^{\tau A} d\tau.$$ Moreover, $$A^TS + SA = \int_0^\infty A^Te^{\tau A^T} + e^{\tau A} A d\tau$$ $$= \int_0^\infty \frac{d}{d\tau} (e^{\tau A^T} e^{\tau A}) d\tau$$ and $$A^Te^{\tau A^T} +e^{\tau A} A = \frac{d}{d\tau}(e^{\tau A^T} + e^{\tau A}).
Now, without loss of generality, assume $A$ is diagonalizable such that $$A = PDP^{-1} \implies A^T = (P^{-1})^TDP^T$$ implies $$A^TS + SA = [e^{\tau A^T} e^{\tau A}]_0^\infty$$ $$= [Pe^{\tau D} P^{-1}(P^{-1})^Te^{\tau D} P^T]_0^\infty$$ $$= [Pe^{-\infty}P^{-1}(P^{-1})^Te^{-\infty} P^T] - [(PIP^{-1})(P^{-1})^T I P^T] = - I.$$
| Assume all eigenvalues of $A$ have negative real parts. Consider $x'=Ax,$ $$L=x^TSx \text{ where } S=S^T \text{ and } x\ne 0.$$ We must show that $\frac{dL}{dt} < 0$ so $$\frac{dL}{dt} = (x^T)'Sx + x^TSx' = (x')^TSx + x^TSx' = x^TA^TSx + x^TSAx$$ $$=x^T(A^TS+SA)x.$$ Since we know that $A^TS+SA = -I$ so $$\frac{dL}{dt} = -x^T(I)x = -||x||^2 < 0 \text{ since } x\ne 0.$$ So $L$ is a strong Lyapunov function.
Next, let $$S= \int_0^\infty e^{\tau A^T}e^{\tau A} d\tau.$$ We need $A^TS + SA=-I.$ So $$A^TS+SA = \int_0^\infty A^Te^{\tau A^T}+e^{\tau A} A d\tau$$ $$= \int_0^\infty \frac{d}{d\tau} (e^{\tau A^T}e^{\tau A})=[e^{\tau A^T}e^{\tau A}]_0^\infty. \hspace{8pt} (1)$$ Since $A$ has negative eigenvalues, $A^T$ does to. Assume $A$ diagonalizable, so $A=PDP^{-1}$, $A^T = (P^{-1})^TDP^T.$ $P$ consists of eigenvectors corresponding to eigenvalues of $A$, $D$ has eigenvalues of $A$ in the diagonal. So, from (1) we have $$(1) = \left[(P^{-1})^Te^{\tau D}P^T(Pe^{\tau D}P^{-1}) \right ]_0^\infty = 0 - \left[((P^{-1})^TIP^T)(PIP^{-1})\right] = -I \text{ and implies } A^TS+SA=-I.$$
Take, $$A= \begin{pmatrix} -2 & 1 \\ 0 & -2\end{pmatrix}$$ $$A^T = \begin{pmatrix} -2 & 0 \\ 1 & -2 \end{pmatrix}$$ $$A=-2I + N, \text{ where } N = \begin{pmatrix} 0&1 \\0&0 \end{pmatrix}$$ $$A^T = -2I + M, \text{ where } M = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}$$ $$N^2 = M^2 = \begin{pmatrix} 0 & 0 \\ 0&0 \end{pmatrix} \text{ and } -2IN = N(-2I), -2IM = M(-2I)$$ $$\implies e^{\tau A} = e^{\tau(-2I + N)}= e^{-2\tau}Ie^{\tau N} = e^{-2\tau}(I)(I+N) = e^{-2\tau}I + e^{-2\tau}N$$ $$e^{\tau A^T} = e^{-2\tau}I e^{-2\tau}M=e^{-2\tau}I+e^{-2\tau}M$$ $$\implies S = \int_0^\infty (e^{-2\tau}I+e^{-2\tau}N)(e^{-2\tau}I+e^{-2\tau}M)d\tau$$ $$= \int_0^\infty (e^{-4\tau}I+e^{-4\tau}R_1+e^{-4\tau}R_2) d\tau$$ where $R_1 = \begin{pmatrix} 0&1\\1&0\end{pmatrix}$ and $R_2 = \begin{pmatrix} 1&0\\0&0\end{pmatrix}.$
So $$S= I\int_0^\infty e^{-4\tau}d\tau + R_1\int_0^\infty e^{-4\tau}d\tau + R_2\int_0^\infty e^{-4\tau}d\tau$$ $$=I\left[\frac{e^{-4\tau}}{-4} \right]_0^\infty + R_1\left[\frac{e^{-4\tau}}{-4} \right]_0^\infty + R_2\left[\frac{e^{-4\tau}}{-4} \right]_0^\infty$$ $$=I(\frac{1}{4}) + R_1(\frac{1}{4}) + R_2(\frac{1}{4})$$ $$=\begin{pmatrix} \frac{1}{4} & 0 \\ 0 & \frac{1}{4} \end{pmatrix} +\begin{pmatrix} 0&\frac{1}{4} \\ \frac{1}{4} &0 \end{pmatrix} + \begin{pmatrix} \frac{1}{4} & 0 \\ 0 & 0 \end{pmatrix}$$ $$=\begin{pmatrix} \frac{1}{2} & \frac{1}{4} \\ \frac{1}{4} & \frac{1}{4} \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647221",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Piecewise Laplace transformation The piecewise function is:
$$
\begin{array}{cc}
&
\begin{array}{cc}
t & 0\leq t< 1 \\
2-t & 1\leq x\leq 2 \\
0 & t>2
\end{array}
\end{array}
$$
Now putting the piecewise into LaPlace form:
$$\int_0^1te^{-st}dt +2\int_1^2e^{-st}dt-\int_1^2te^{-st}dt+0$$
The first and third integral are the same integration by parts: $u=t,du=dt,dv=e^{-st},v=-\frac{1}{s}e^{-st}$.
Then the expression becomes:
$$-\frac{t}{s}e^{-st}+\int_0^1\frac{1}{s}e^{-st}dt+2\int_1^2e^{-st}dt-\frac{t}{s}e^{-st}+\int_1^2\frac{1}{s}e^{-st}dt$$
Which after integrating and evaluating equals:
$$-\frac{1}{s}e^{-s}-\frac{1}{s^2}e^{-s}+\frac{1}{s^2}-\frac{2}{s}e^{-2s}+\frac{2}{s}e^{-s}-\frac{2}{s}e^{-s}-\frac{2}{s}e^{-2s}-\frac{1}{s^2}e^{-2s}+\frac{1}{s}e^{-s}+\frac{1}{s^2}e^{-s}$$
Are my calculations correct or did I mess up somewhere?
| I believe you've made a few small sign slips. Your starting point is correct,
$$\tilde f(s) = \int_0^1 t e^{-st} + \int_1^2 (2-t) e^{-st} \; dt$$
If we're not careful, we'll get far too many terms (like your final answer, which is difficult to figure out whether it's correct), so let's evaluate
\begin{align}
I(t) =\int t e^{-st} \; dt &= -\frac{t}{s} e^{-st} + \frac{1}{s} \int e^{-st} \; dt \\
&= -e^{-st} \left[\frac{t}{s} + \frac{1}{s^2} \right]
\end{align}
where we're dropping the constant of integration since we're only considering definite integrals in $\tilde f(s)$.
Then
\begin{align}
\tilde f(s) &= I(1) - I(0) -\frac{2}{s} \left( e^{-2s} - e^{-s} \right) - I(2) + I(1) \\
&= 2 I(1) - I(0) - I(2) - \frac{2}{s} \left( e^{-2s} - e^{-s} \right) \\
&= -2e^{-s} \left[ \frac{1}{s} + \frac{1}{s^2} \right] + \frac{1}{s^2} +e^{-2s} \left[ \frac{2}{s} + \frac{1}{s^2} \right] - \frac{2}{s} \left(e^{-2s} - e^{-s} \right) \\
&= e^{-s} \left[ -\frac{2}{s} - \frac{2}{s^2} + \frac{2}{s} \right] + \frac{1}{s^2} e^{-2s} + \frac{1}{s^2} \\
&= -\frac{2}{s^2} e^{-s} + \frac{1}{s^2} \left(e^{-2s} + 1 \right) \\
&= \frac{e^{-2s} - 2e^{-s} + 1}{s^2} = \left(\frac{1- e^{-s}}{s} \right)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the particular integral? The question is to find the particular integral of the expression
$$ (D^3+1)y = cos(2x-1) $$
I’m not sure how to go about it. How do I find the PI for this expression?
| $\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$This is an interesting differential equation because it yields some really nice graphs when some initial $y(0),y'(0),y''(0)$ are sampled.
As for the solution, first solve the homogenous problem:
$$y''' + y = 0 \xRightarrow{y = e^\lambda} \lambda^3 e^\lambda + e^\lambda = 0 \Rightarrow \lambda = \begin{cases} -1, \\ \dfrac{1}{2} + \dfrac{i\sqrt{3}}{2}, \\ \dfrac{1}{2} - \dfrac{i\sqrt{3}}{2}.\end{cases}$$
Thus, the general solution is:
\begin{align*}
y_g(x) &= c_1e^{-x} + c_2 e^{x\left(\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)} + c_3 e^{x\left(\frac{1}{2} - \frac{i\sqrt{3}}{2}\right)} \\ &= c_1 e^{-x} + c_2 e^{x/2}\sin\left(\frac{\sqrt{3}}{2}x\right) + c_3e^{x/2} \cos\left(\frac{\sqrt{3}}{2}x\right).
\end{align*}
For the particular solution (aka your particular integral), based on the expression of the given differential, as Ninad Munshi mentioned in the comments, one can "guess" the expression:
$$y_p(x) = PI = A\cos(2x-1) + B\sin(2x-1).$$
Substitute then in the initial expression and calculate the values of $A$ and $B$.
Finally, the interesting part that I promised:
$\qquad \qquad \qquad \qquad$
$\qquad \qquad \qquad \qquad$
| {
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"url": "https://math.stackexchange.com/questions/3654986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\frac{1}{\sqrt{2}}\left(\sqrt{46-6\sqrt{5}}-2\sqrt{6-2\sqrt{5}}\right)=$$
$$=\frac{1}{\sqrt{2}}\left(\sqrt{(3\sqrt{5}-1)^2}-2\sqrt{(\sqrt5-1)^2}\right)=\frac{1}{\sqrt{2}}\left(3\sqrt{5}-1-2(\sqrt{5}-1)\right)=$$
$$=\frac{1}{\sqrt{2}}(\sqrt{5}+1)=\frac{1}{\sqrt{2}}\sqrt{6+2\sqrt{5}}=\sqrt{3+\sqrt{5}}.$$
But this way does not work for the starting identity.
How to prove the starting identity without squaring?
Thank you!
| It is already said everything about the solution, but i was curious to see if there is any "straightforward" way to proceed in this and similar cases by working in the appropriate quadratic field, here in $K=\Bbb Q(s)$, with $s=\sqrt {17}$ (with class number one) and asking the computer, here sage, for factorizations of the numbers under the radicals. It gave the decompositions:
$$
\begin{aligned}
23 -s &= (4+s)^3\cdot\frac{s+3}2\cdot\left(\frac{s-3}2\right)^8\ ,\\
7 -s &= (4+s)\cdot\frac{s+3}2\cdot\left(\frac{s-3}2\right)^4\ ,\\
71 -17s &= (4+s)\cdot\frac{s+3}2\cdot\left(\frac{s-3}2\right)^6\ ,
\end{aligned}
$$
and each R.H.S. above is of the shape a common factor
$f=(4+s)\cdot\left(\frac{s+3}2\right)\cdot\left(\frac{s-3}2\right)^4$, taken
times the one or the other square element in $K$. The square root of the common factor $f$ takes us out of $K$, but after factorizing it, the posted equality lives in $K$ and is:
$$
(4+s)\cdot\left(\frac{s-3}2\right)^2
-
2
=
\left(\frac{s-3}2\right)
\ .
$$
Easily checked.
Code used to reproduce the above:
K.<s> = QuadraticField(17)
print(f'K has class number {K.class_number()}.')
for r in (23 - s, 7 - s, 71 - 17*s):
print(f'{r} = {r.factor()}')
Results:
K has class number 1.
-s + 23 = (-65*s - 268) * (-1/2*s - 3/2) * (-1/2*s + 3/2)^8
-s + 7 = (-s - 4) * (-1/2*s - 3/2) * (-1/2*s + 3/2)^4
-17*s + 71 = (-s - 4) * (-1/2*s - 3/2) * (-1/2*s + 3/2)^6
The factor $(65s-268)$ is an integral unit, and thus a power of the generator $(4+s)$, which is easily found. The final check:
sage: (4 + s)*( (s-3)/2 )^2 - 2 == (s-3)/2
True
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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$\frac{x^2 -4}{x-4} \rightarrow -5$ as $x \rightarrow 3$, proving the limit from first principles. ($\delta - \epsilon$). This is a follow up to a previous post. I am sorry of this is considered spamming but I have made significant progress of the question since then. Hopefully I'm on the right track, an tips or advice on my methods would be much appreciated!
I have been asked to prove the limit of $\frac{x^2 -4}{x-4} \rightarrow -5$ as $x \rightarrow 3$. First I fix $\epsilon >0$ and find $\delta>0$ such that.
$$0<|x-3| <\delta \implies \bigg|\frac{x^2 -4}{x-4} + 5\bigg| < \epsilon$$
$$\impliedby \bigg|\frac{x^2+5x -24}{x-4}\bigg| < \epsilon$$
$$\impliedby \bigg|\frac{(x+8)(x-3)}{x-4}\bigg| < \epsilon$$
$$\impliedby \bigg|(x-3)\frac{x+8}{x-4}\bigg| < \epsilon$$
If $|x-3| \leq \frac{1}{2}$, then $[2\frac{1}{2},3\frac{1}{2}]$. Then $\frac{x+8}{x+4} \geq \frac{1}{2}$. Therefore $\frac{|x-3|}{2} < \epsilon \iff |x-3|<2\epsilon $.
Thus any $\delta \leq$ min$(\frac{1}{2}, 2\epsilon)$ has the required property.
Thanks for your time!
| Your choice for $\delta$ does not work.
Let $\varepsilon = \frac{1}{4}$.
Then your formula gives $\delta = \frac{1}{2}$.
If $x = 3.4$ then
$$\quad \bigg|(x-3) \; \frac{x+8}{x-4}\bigg| = 7.6$$
If $x \in [2\frac{1}{2},3\frac{1}{2}]$ then
$$\quad |x+8| \le 12 $$
If $x \in [2\frac{1}{2},3\frac{1}{2}]$ then
$$\quad |x-4| \ge \frac{1}{2} $$
It follows that for every $x \in [2\frac{1}{2},3\frac{1}{2}]$ that
$$\quad \bigg|\frac{x+8}{x-4}\bigg| = \frac{|x+8|}{|x-4|} \le 24$$
To solve the OP's problem, for any $\varepsilon \gt 0$ challenge let
$$ \delta = \text{min}\bigg(\frac{1}{2}, \frac{\varepsilon}{24}\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to compute $\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$ I have first written $x^4-2x^2+2=(x^2-1)^2+1$ then I choose $t=x^2-1$. From this I got $$\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$$
From this how to proceed?
| HINT:
$$\frac{\text{d}}{\text{dn}}\left(\int_0^\infty\frac{\ln\left(1+\text{n}x^2\right)}{\sqrt{x}\left(1+x\right)}\space\text{d}x\right)=\int_0^\infty\frac{x^\frac{3}{2}}{\left(1+\text{n}x^2\right)\left(1+x\right)}\space\text{d}x=$$
$$\frac{\pi}{2\left(1+\text{n}\right)}\left(2-\frac{\sqrt{2}\left(\sqrt{\text{n}}-1\right)}{\text{n}^\frac{3}{4}}\right)\tag1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$ Show that if $0 \leq x \leq a$ and $n \in \mathbb{N}$, then $$1 + \frac{x}{1!} + \dots + \frac{x^n}{n!} \leq e^x \leq 1 + \frac{x}{1!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{e^{a}x^n}{n!}$$
Let $$e^x = 1 + x + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!} + \frac{x^n}{n!} + \frac{x^{n+1}}{(n+1)!}e^x$$
where $e^x$ is expanded by Taylor's theorem, but I am not sure if I am correctly applying Taylor's theorem because it isn't clear to me where the right side of the inequality comes from.
How does Taylor's theorem get used to complete the inequality? Is that even the right approach to this problem?
| By Taylor's theorem,
$$
e^x = 1 + x + \frac{{x^2 }}{{2!}} + \cdots + \frac{{x^{n - 1} }}{{(n - 1)!}} + \frac{{x^n }}{{n!}}e^c
$$
for all $x\geq 0$ with a suitable $0\leq c\leq x$. Now just note that if $0\leq x\leq a$ then
$$
1 \leq e^c \leq e^x \leq e^a .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Vieta's formula for $f(\frac{1}{x})$
Let $f(x)$ be a monic cubic polynomial. Let the solutions of the equation $f(\frac{1}{x})=0$ be $\alpha, \beta$ and $\gamma$. If $\alpha+\beta+\gamma = 10$ and $\alpha\beta\gamma=15$, what is $\left \lceil{f(10)-f(-10)}\right \rceil$?
So since $f(x)$ is a monic cubic we can write it as $f(x) = x^3+bx^2+cx+d$ and from here $f(\frac{1}{x}) = \frac{1}{x^3} + b \frac{1}{x^2} + c \frac{1}{x}+d$.
The question stated that $\alpha, \beta$ and $\gamma$ were solutions for $f(\frac{1}{x})$ and I see that this would imply to use Vieta's here, but how do we go about it with something like $f(\frac{1}{x})$?
| First notice that:
\begin{gather}
f(10)-f(-10) = 10^3+10^2b+10c+d - (-10)^3 - (-10)^2b-(-10)c -d =\\
=2000+20c
\end{gather}
So if we find the value of $c$ we are done.
As you noticed $f\left(\frac{1}{x}\right) = \frac{1}{x^3}(dx^3+cx^2+bx+1) = \frac{1}{dx^3}\left(x^3+\frac{c}{d}x^2+\frac{b}{d}x+\frac{1}{d}\right)$ where $d$ is different to zero because by hypotesis we have three solutions.
Thanks to Vieta's formula we have
\begin{gather}
\frac{c}{d} = -(\alpha + \beta + \gamma) =-10\\
\frac{1}{d} = -(\alpha\beta\gamma) -15
\end{gather}
Hence $c=\frac{2}{3}$ and the result is:
$$
f(10)-f(-10) =2000+20c = \frac{6040}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\right]\left[ \frac{d^r}{dx^r}(1-x)^{-\frac{1}{2}} \right].$$
Since we know that $ \frac{d^{2}}{dx^2}(1+x) = 0 $, we can simplify the above sum to the following:
$$ 0+\ldots+\binom{100}{98}\left[\frac{d^{2}}{dx^2}(1+x)\right] \left[ \frac{d^{98}}{dx^{98}}(1-x)^{-\frac{1}{2}} \right]
+\binom{100}{99}\left[\frac{d}{dx}(1+x)\right] \left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] +\binom{100}{100}(1+x) \left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right], $$
which then gives us,
$$ 100\left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] + (1+x)\left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right].$$
This is where I am stucked. I'm not sure resolve those differentials.
| $$y=\dfrac{1+x}{\sqrt{1-x}}=\dfrac{2-(1-x)}{\sqrt{1-x}}=2(1-x)^{-1/2}-(1-x)^{+1/2}$$
Now use this by writing $$(1-x)^m=(-1)^m(x-1)^m$$
$$\dfrac{d^n(x-1)^m}{dx^n}=m^n(x-1)^{m-n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3669653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.