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Nonnegative Integer solutions of $x+y-xy=0$ I would like to see other methods, besides the one I use here to find all the nonnegative integer solutions of an equation like $$x+y-xy=0$$.
This is the one I used:
First we note that for $x=1$ we get $1+y-y=0$, but $1+y-y=1$ so $x\neq 1$.
Then as $x\neq 1$ we have $y=\frac{x}{x-1}=1+\frac{1}{x-1}$. Furthermore $y(0)=0$ and $y(2)=2$ so $x=y=0$ and $x=y=2$ are solutions. Then if $x>2$, $x-1>1>0$ so as $x-1>1$ we have $\frac{1}{x-1}<1$ and as $x-1>0$ we have $\frac{1}{x-1}>0$ so $0<\frac{1}{x-1}<1$ for $x>2$, and so $1<1+\frac{1}{x-1}<2$ for $x>2$. As $y=1+\frac{1}{x-1}$, we get that for $x>2$, $1<y<2$. As there is no integer between $1$ and $2$, for $x>2$, $y$ can't be a integer. So the only nonnegative solutions of the equations are $x=y=0$ and $x=y=2$.
How else can I prove this, using less theory (For example, I think it can be proven using only Peano and a little more)?
| If
$x+y-xy = 0$,
then
$1
=xy-x-y+1
=(x-1)(y-1)
$.
Therefore
$x-1$ and $y-1$
must be integers
that divide $1$.
The only solutions are
$x=y=2$
and
$x=y=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $a^2+b^2+c^2$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
If $a, b, c$ are distinct numbers such that $a^2 - bc = 2014$, $b^2 + ac = 2014$, $c^2 + ab = 2014$. Then compute $a^2 + b^2 + c^2$
(A)$4030$ (B)$4028$ (C)$4026$ (D)$4000$ (E)$2014$
Adding these three equations together
$$a^2 + b^2 + c^2 + ab + ac - bc = 3\times2014 \quad(1)$$
And also found that
\begin{align}
(a-b-c)^2 &= a^2 + b^2 + c^2 - 2ab - 2ac + 2bc\\
(a-b-c)^2 &= a^2 + b^2 + c^2 - 2(ab + ac - bc)\\
\end{align}
I don't know how to continue to reduce $ ab + ac - bc$, or am I using the wrong way to reducing it?
I'm very appreciate for those who have helped me to hint/explain me on how to do all these questions (I'm currently 10th grade (in US grade system), so I don't understand these much, all of these are outside my syllabus)
| Note that
$$b^2+ac=c^2+ab\iff b^2-c^2+ac-ab=0\iff (b-c)(b+c)-a(b-c)=0$$
$$\iff (b-c)(b+c-a)=0\iff a=b+c.$$
Now $$a^2+b^2+c^2=(b+c)^2+b^2+c^2=2(b^2+c^2+bc)$$
$$=2((b+c)^2-bc)=2(a^2-bc)=2\times 2014$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
From 1D gaussian to 2D gaussian I read this:
The Gaussian kernel for dimensions higher than one, say N, can be described as a regular product of N one-dimensional kernels. Example:
g2D(x,y,$\sigma_1^2 + \sigma_2^2$) = g1D(x,$\sigma_1^2$)g2D(y,$\sigma_2^2$)
saying that the product of two 1 dimensional gaussian functions with variances $\sigma_1^2$ and $\sigma_2^2$ is equal to a two dimensional gaussian function with the sum of the two variances as its new variance.
I tried to deduce this by using:
g1D(x,$\sigma1^2$)g2D(y,$\sigma2^2$) = $\frac{1}{\sqrt{2\pi}\sigma_1}e^{\frac{-x^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi}\sigma_2}e^{\frac{-y^2}{2\sigma_2^2}}$ = $\frac{1}{2\pi\sigma_1\sigma_2}e^{-(\frac{x^2}{2\sigma_1^2}+\frac{y^2}{2\sigma_2^2})}$
but I fail to obtain
$\frac{1}{2\pi(\sigma_1^2 + \sigma_2^2)}e^{\frac{-(x^2+y^2)}{2\sigma_1^2 + 2\sigma_2^2}}$
which is equal to g2D(x,y,$\sigma_1^2 + \sigma_2^2$).
Someone know how to get there?
| Hint: You see in the exponential that we have
$$ \exp \left ( - \frac{x^2 + y^2}{2( \sigma_1^2 + \sigma_2^2 ) } \right )=\exp \left ( \frac{-x^2 }{2( \sigma_1^2 + \sigma_2^2 ) } \right )\exp \left ( \frac{- y^2}{2( \sigma_1^2 + \sigma_2^2 ) } \right ) $$
The normalization hanging around should be
$$\frac{1}{2 \pi ( \sigma_1^2 + \sigma_2^2 ) } $$
as you've seen...but under the change of variables
$$ z_1 = \frac{ \sigma_1x}{\sqrt{\sigma_1^2 + \sigma_2^2}} \quad \& \quad z_2 = \frac{ \sigma_2y}{\sqrt{\sigma_1^2 + \sigma_2^2}} $$
What do you obtain? Also note that
$$ 2 \pi = \sqrt{2\pi} \sqrt{2 \pi} $$
Edit: We see that
$$\frac{1}{2 \pi ( \sigma_1^2 + \sigma_2^2 ) } \int_{\mathbb{R}^2} \exp \left ( - \frac{x^2 + y^2}{2( \sigma_1^2 + \sigma_2^2 ) } \right ) dxdy = $$
$$= \left[ \frac{1}{\sqrt{2 \pi ( \sigma_1^2 + \sigma_2^2 )} } \int_{\mathbb{R}} \exp \left ( - \frac{x^2 }{2( \sigma_1^2 + \sigma_2^2 ) } \right ) dx \right] \left [ \frac{1}{\sqrt{2 \pi ( \sigma_1^2 + \sigma_2^2 )} } \int_{\mathbb{R}} \exp \left ( - \frac{y^2 }{2( \sigma_1^2 + \sigma_2^2 ) } \right )dy \right]$$
$$ =\left[ \frac{1}{\sqrt{2 \pi }\sigma_1 } \int_{\mathbb{R}} \exp \left ( - \frac{z_1^2 }{2 \sigma_1^2 } \right ) dz_1 \right] \left [ \frac{1}{\sqrt{2 \pi }\sigma_2 } \int_{\mathbb{R}} \exp \left ( - \frac{z_2^2 }{2\sigma_2^2 } \right )dz_2 \right] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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How to prove the limit exists for function of two variables? Problem: Evaluate the indicated limit or explain why it does not exist: \begin{align*} \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2 + y^4} \end{align*}
The definition of limit my calculus textbook gives is:
We say that $\lim_{(x,y) \to (a,b)} f(x,y) = L$, provided that:
1) Every neighbourhood of $(a,b)$ contains points of the domain of $f$
different from $(a,b)$, and
2) For every positive number $\epsilon$ there exists a positive number
$\delta = \delta (\epsilon)$ such that $|f(x,y) - L| < \epsilon$ holds
whenever $(x,y)$ is in the domain of $f$ and satisfies $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$.
So in this specific problem I first checked the limiting behaviour when $(x,y)$ approaches $(0,0)$ from different directions:
When $y =0$ or $x = 0$ then $\lim_{(x,y) \to (0,0)} f(x,y) = 0$. When $y = x^2$, we also have that $\lim_{(x,y) \to (0,0)} f(x,x^2) = 0$.
So I believe the limit exists and is zero, and I want to prove it now by using the definition. So let $\epsilon > 0$. Then we need to find a $\delta > 0$ such that if $0 < \sqrt{x^2 + y^2} < \delta$, then $| \frac{x^2 y^2}{x^2 + y^4} - 0 | < \epsilon$.
Now, since $x^2 \leq x^2 + y^4$, it follows that $\frac{x^2}{x^2 + y^4} \leq 1$. Let $\delta = \sqrt{\epsilon}$. If $0 \leq \sqrt{x^2 + y^2} < \delta$, then it follows that $x^2 + y^2 < \epsilon$. Thus we also have that $y^2 < \epsilon$.
Since we already had that $\frac{x^2}{x^2 + y^4} \leq 1$ we can multiply this with $y^2$ and get $\frac{x^2 y^2}{x^2 + y^4} \leq y^2$, from which it follows that $\frac{x^2 y^2}{x^2 + y^4} < \epsilon$.
Can someone tell me if my reasoning is correct? Also, is this the right method to proof the existence of limits of functions of two variables? I mean, if you suspect that the limit exists, you have to use the delta-epsilon notation to prove it?
Also, I found an alternative solution:
Since $0 \leq \frac{x^2 y^2}{x^2 + y^4} \leq \frac{x^2 y^2}{x^2}$, and since $\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2} = 0$, we have also $\lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2 + y^4} = 0$ by the Squeeze Theorem.
| Let's to use polar coordinates $x=\rho\cos\phi$, $y=\rho\sin\phi$, and your limit is
$$
\lim_{\rho\to0} \frac{\rho^4 \cos^2\phi \sin^2\phi}{\rho^2\cos^2\phi + \rho^4\sin^4\phi} = \lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi}
$$
But $\lim_{\rho\to0}\rho^2 \cos^2\phi \sin^2\phi = 0$. If $\cos\phi\ne0$, then
$$
\lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi} = \frac{\lim_{\rho\to0}(\rho^2 \cos^2\phi \sin^2\phi)}{\lim_{\rho\to0}(\cos^2\phi + \rho^2\sin^4\phi)} = \frac{0}{\cos^2\phi} = 0.
$$
If $\cos\phi=0$,
$$
\lim_{\rho\to0} \frac{\rho^2 \cos^2\phi \sin^2\phi}{\cos^2\phi + \rho^2\sin^4\phi} = \lim_{\rho\to0} \frac{0}{\rho^2} = 0
$$
In any case, limit equal to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$? How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$ ?
Please let me know the fastest method you know of for such type of problems. Is there any way other than finding the left hand and right hand derivative using the concept of limits (That makes it huge)? Can intuition be used in any way?
| Observe that $(f(x))^2$ is given by
$$(x + 2\sqrt{2x - 4}) + 2 \sqrt{x + 2\sqrt{2x - 4}}\sqrt{x - 2\sqrt{2x - 4}} + (x - 2\sqrt{2x - 4})$$
$$= 2x + 2\sqrt{x^2 - 4(2x - 4)}$$
$$= 2x + 2\sqrt{(x - 4)^2}$$
$$= 2x + 2|x - 4|$$
Since $f(x) \geq 0$, for $x$ near $4$ we therefore have
$$ f(x) = \sqrt{2x + 2|x - 4|}$$
For $x > 4$ this is $\sqrt{4x - 8}$ and for $x < 4$ this is just $\sqrt{8}$. Hence the left and right derivatives do not match at $x = 4$ and the function is not differentiable there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Congruence using extended GCD $$\eqalign{ x &\equiv 5 \mod 15\cr
x &\equiv 8 \mod 21\cr}$$
The extended Euclidean algorithm gives $x≡50 \bmod 105$.
I understand now that if we combine the two it implies $15a-21b = 3$ but I don't understand how to use the extended GCD to go from there to finding $x$ and the corresponding modulus.
This is what I am using for the extended gcd computations:
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
From https://en.wikibooks.org/wiki/Algorithm_Implementation/Mathematics/Extended_Euclidean_algorithm#Python
| Since the GCD of the moduli is $(15,21)=3$, it is necessary that $x$ be the same thing in both equations mod $3$. That is,
$$
x\equiv5\pmod{15}\implies x\equiv2\pmod{3}
$$
and
$$
x\equiv8\pmod{21}\implies x\equiv2\pmod{3}
$$
If we didn't get that $x\equiv2\pmod{3}$ from both equations, a solution would not be possible.
This prompts us to look at $\frac{x-2}3\pmod{\frac{15}3}$ and $\frac{x-2}3\pmod{\frac{21}3}$. That is,
$$
\frac{x-2}3\equiv1\pmod{5}\tag{1}
$$
and
$$
\frac{x-2}3\equiv2\pmod{7}\tag{2}
$$
Using the Extended Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&1&2&2\\\hline
1&0&1&-2&5\\
0&1&-1&3&-7\\
7&5&2&1&0
\end{array}\tag{3}
$$
we get that
$$
\underbrace{5(3)}_{\large\color{#C00000}{15}}+\underbrace{\!7(-2)\!}_{\large\color{#00A000}{-14}}=1\tag{4}
$$
We can use $(4)$ to see that
$$
\begin{align}
\color{#00A000}{-14}&\equiv\color{#0000F0}{1}\pmod{5}\\
\color{#00A000}{-14}&\equiv\color{#0000F0}{0}\pmod{7}
\end{align}\tag{5}
$$
and that
$$
\begin{align}
\color{#C00000}{15}&\equiv\color{#0000F0}{0}\pmod{5}\\
\color{#C00000}{15}&\equiv\color{#0000F0}{1}\pmod{7}
\end{align}\tag{6}
$$
To solve $(1)$ and $(2)$ we can add $1$ times $(5)$ to $2$ times $(6)$ to get
$$
\begin{align}
16&\equiv\color{#0000F0}{1}\pmod{5}\\
16&\equiv\color{#0000F0}{2}\pmod{7}
\end{align}\tag{7}
$$
Equations $(7)$ tell us that $\frac{x-2}3\equiv16\pmod{35}$ or that
$$
\bbox[5px,border:2px solid #C0A000]{x\equiv50\pmod{105}}\tag{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differential equation: $Ay'' + By' + Cy = h(x)$ I'm stuck solving the equation $y'' - 3y' + 2y = 2x^3-30$.
The auxiliary equation is $k^2 - 3k + 2 = 0$ where $k_1 = 1, k_2=3$. Thus the general solution is: $$y_g = C_1e^x + C_2e^{3x}$$
Then, I tried to find the particular solution taking into consideration that $h(x) = Q_n(x) \cdot e^{\alpha x}$ where $\alpha$ is zero and doesn't equal one of auxiliary equation's roots, and $n$ is the order of $h(x)$ and equals 3. So, I get the equations as:
$$ y_p = Ax^3 + Bx^2 +Cx + D, \\
y'_p = 3Ax^2 + 2Bx + C, \\
y''_p = 6Ax + 2B$$
Having substituted they in the initial equation I get the system:
$$ \begin{cases}
A = 2, \\
B + 3A = 0, \\
C + 2B + 6A = 0, \\
D + C + 2B = -30
\end{cases} \implies
\begin{cases}
A = 2, \\
B = -6, \\
C = 0, \\
D = -18
\end{cases}
$$
Thus my particular solution is $y_p = 2x^3 - 6x^2 - 18$ but the correct answer is: $$y_p = x^3 + \frac{9}{2}x^2 + \frac{21}{2}x - \frac{15}{4}$$
Where was I wrong?
| First of all your factorization is wrong: $k_1=1, k_2=2$,so general solution will be,
$$y_g = C_1e^x + C_2e^{2x}$$
Next, $$y_p = Ax^3 + Bx^2 +Cx + D, \\
y'_p = 3Ax^2 + 2Bx + C, \\
y''_p = 6Ax + 2B$$and using all this you will get,
$$\begin{cases}
2A = 2, \\
-9A+2B = 0, \\
6A-6B+2C = 0, \\
2B-3C+2D = -30
\end{cases} \implies
\begin{cases}
A = 1, \\
B = 9/2, \\
C = 21/2, \\
D = -15/4
\end{cases}$$ Using these your $$y_p = x^3 + \frac{9}{2}x^2 + \frac{21}{2}x - \frac{15}{4}$$
Sol: $y=y_g+y_p$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\frac{1}{4/3} - \frac{1}{3/2}$ not the same as $\bigl(\frac{4}{3} - \frac{3}{2}\bigr)^{-1}$ If you have the problem:$$\frac{1}{\frac{4}{3}} - \frac{1}{\frac{3}{2}} =?$$
Why can't you change the problem to $(\frac{4}{3} - \frac{3}{2})^{-1}$ and get the same answer?
In the first scenario, the answer is $1/12$
In the second scenario, the answer is $-6$
Why doesn't this work?
| The original problem is $$\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}$$
for specific values of $x,y$. The modified problem is $$\frac{1}{x-y}$$
In order for them to be equal, $$\frac{y-x}{xy}=\frac{1}{x-y}$$
cross-multiplying, we get $$-(x-y)^2=xy$$
Expanding, we get $$-x^2+2xy-y^2=xy$$
or $$x^2-xy+y^2=0$$
The only real solutions this has are $x=0, y=0$, which are excluded since we are dividing by $x,y$. Hence this proposed simplification is never correct, even accidentally.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Determinant of a Certain 3 by 3 Block Matrix with Scaled Identity Blocks What is the determinant or/and eigenvalues of the following 3 by 3 block matrix:
$$\left[\begin{array}{ccc} \frac{3}{4}I & \frac{1}{4}I & \frac{1}{4}I \\ \frac{1}{4}I & \frac{3}{4}I & -\frac{1}{4}I \\ \frac{1}{4}I & -\frac{1}{4}I & \frac{3}{4}I \end{array}\right]$$
Where $I$ is the identity matrix of fixed size.
| The block matrix (let us denote by $M$) can be expressed as the Kronecker product of matrices $A$ and $I$ (the fixed size identity matrix, of dimension $n$) as follows:-
$$M=A\otimes I$$
where $A$ is the $3\times3$ matrix:-
$$A=\left[\begin{array}{ccc} \frac{3}{4} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{3}{4} & -\frac{1}{4} \\ \frac{1}{4} & -\frac{1}{4} & \frac{3}{4} \end{array}\right]$$
From the linked Wikipedia entry, for square matrices $A$ of size $m$ and $B$ of size $n$ the determinant of their Kronecker product is given by
$$|A\otimes B|=|A|^n|B|^m$$
Applying this to the matrix in question, results in
$$|M|=|A|^n|I|^3=\left(\frac{1}{4}\right)^n\times1=\frac{1}{4^n}$$
The eigenvalues of $A$ are
$\lambda_1=\frac{1}{4}, \lambda_2=1, \lambda_3=1$, while the eigenvalues of $I$ are $\mu_j=1$ for $j\in\{1,2,..,n\}$, leading to the eigenvalues of $M$ being $\lambda_i\mu_j$ for $i\in\{1,2,3\}$ and $j\in\{1,2,..,n\}$.
Thus, $M$ will have $n$ eigenvalues of value $\frac{1}{4}$ and $2n$ eigenvalues of value $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$ Please help me to evaluate this integral in a closed form:
$$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$
Using integration by parts I found that it could be expressed through integrals of elementary functions:
$$I_1=\int_0^\infty\log\left(1-e^{-\pi x}\right)\log\left(1+x^2\right)dx$$
$$I_2=\int_0^\infty x\log\left(1-e^{-\pi x}\right)\arctan x\,dx$$
| The following is another way to show that $$\sum_{n=1}^{\infty} \frac{(-1)^{k} \operatorname{Si}(k \pi )}{k^{3}}= - \frac{\pi^{3}}{18}.$$
(I'm surprised that replacing $\operatorname{Si}(k \pi)$ with its Maclaurin series and then switching the order of summation leads to the correct result because Fubini's theorem is not satisfied and $\sum_{k=1}^{\infty} (-1)^{k}k^{2n-4} = -\eta(4-2n)$ only when $n < 2$.)
The Fourier series of the piecewise continous function $$f(x) = \begin{cases} x &\text{if} - \pi < x< \pi\\ f(x+ 2\pi) \, &\text{otherwise} \end{cases}$$ is $$f(x) = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sin(kx).$$
Term-by-term integration gives $$\frac{x^{2}}{2} - \frac{\pi^{2}}{6} = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k} \cos(nx)}{k^{2}}, \quad - \pi \le x \le \pi.$$
And term-by-term integration a second time gives $$\frac{x^{3}}{6} - \frac{\pi^{2}x}{6} = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k} \sin(kx)}{k^{3}}, \quad - \pi \le x \le \pi. $$
Therefore, for $- \pi \le x \le \pi$, we have $$ \begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(kx)}{k^{3}} &= \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\int_{0}^{kx} \frac{\sin(t)}{t} \, \mathrm dt \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\int_{0}^{x} \frac{\sin(ku)}{u} \, \mathrm du \\ &= \int_{0}^{x} \frac{1}{u} \sum_{k=1}^{\infty} \frac{(-1)^{k} \sin(ku)}{k^{3}} \\ &= \int_{0}^{x} \frac{1}{u} \left(\frac{u^{3}}{12}- \frac{\pi^{2}u}{12} \right) \, \mathrm du \\ &= \frac{1}{12} \int_{0}^{x} \left(u^{2} - \pi^{2} \right) \, \mathrm du \\ &=\frac{1}{12} \left( \frac{x^{3}}{3}- \pi^{2}x \right). \end{align} $$
Letting $x= \pi$, we get $$\sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(k\pi)}{k^{3}} = \frac{1}{12} \left( \frac{\pi^{3}}{3}- \pi^{3} \right)= - \frac{\pi^{3}}{18}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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"answer_id": 3
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New idea to solve this equation I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rfloor+2x=1$$
1st: $$x=n+p \\n \in\mathbb{Z} , 0 \leq p<1 \to \left \lfloor x \right \rfloor=n ,p=x-n\\ $$
and
$$ \left \lfloor x\right \rfloor+2x=1\\n+2(n+p)=1 \to 3n+2p=1 \\3n=0 ,\pm3 ,\pm 6,\pm9,... $$in this case $$3n=0 \to 2p=1 \\n=0 , p=\frac{1}{2} \to x=n+p=0+\frac{1}{2} $$
2nd:
$$\left \lfloor x\right \rfloor+2x=1 \to \left \lfloor x\right \rfloor=1-2x$$ like $f(x)=g(x)$ by drawing both of them obtain the answer
With respect to the picture ,it suffice to solve $0=1-2x$s o the answer is $x=\frac{1}{2}$
3rd :
we know $$\left \lfloor x\right \rfloor =k \in \mathbb{Z} \to k \leq x <k+1 $$ so $$\left \lfloor x\right \rfloor=1-2x \to 1-2x=k \in \mathbb{Z}\\x=\frac{1-k}{2} \to \left \lfloor \frac{1-k}{2}\right \rfloor =k$$ so we have
$$k \leq \frac{1-k}{2} <k+1 \to \\\left\{\begin{matrix}
k\leq \frac{1-k}{2} \to & 2k \leq 1-k \to & k \leq \frac{1}{3} \to k=\left \{ 0,-1,-2,-3,... \right \}\\
\frac{1-k}{2}<k \to & 1-k<2k \to &k> -\frac{1}{3} \to k=\left \{ 0,1,2,3,... \right \}
\end{matrix}\right.\\
\left \{ ...,-3,-2,-1,0 \right \}\bigcap \left \{ 0,1,2,3,... \right \}=\left \{ 0 \right \}\rightarrow k=0 \\\rightarrow x=\frac{1-k}{2}=\frac{1}{2}$$
| Note that the function $f(x) =\lfloor x \rfloor + 2x$ is strictly increasing.
It is always between $3x-1$ and $3x$.
Solving $3x-1=1$ and $3x=1$ gives that $x$ is between $\frac13$ and $\frac23$, because $f$ is strictly increasing.
So $\lfloor x \rfloor = 0$, since $\frac13 < x < \frac23$. So we solve $2x=1$ so $x=\frac12$.
| {
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"url": "https://math.stackexchange.com/questions/1382103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x\rightarrow \infty $ I want to calculate limit $x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2})$ as $x \rightarrow \infty $ without L'Hôpital's rule.
I found this task on the Internet. The answer given by the author is $2$.
$$\lim_{x\rightarrow\infty} x \cdot \sin(\sqrt{x^2+3}-\sqrt{x^2+2}) = \\
= \lim_{x\rightarrow\infty} x \cdot \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot(\sqrt{x^2+3}-\sqrt{x^2+2}) = (*)\\
$$
at this point I took care of a sine interior
$$ \lim_{x\rightarrow\infty} \sqrt{x^2+3}-\sqrt{x^2+2} = \\
=\lim_{x\rightarrow\infty} \frac{(\sqrt{x^2+3}-\sqrt{x^2+2}) \cdot(\sqrt{x^2+3} + \sqrt{x^2+2})}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \\
= \lim_{x\rightarrow\infty} \frac{1}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \left[\frac{1}{\infty}\right] = 0 \\
$$
and whole sin function
$$ \lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}} = \left[\frac{\sin(y\rightarrow0)}{y\rightarrow0}\right] = 1 $$
finally
$$(*) =\lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot \frac{x}{(\sqrt{x^2+3}+\sqrt{x^2+2})} = \\ = \lim_{x\rightarrow\infty} \frac{\sin(\sqrt{x^2+3}-\sqrt{x^2+2})}{\sqrt{x^2+3}-\sqrt{x^2+2}}\cdot \frac{x}{x \left(\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{1}{x^2}}\right)} = \frac{1}{2} $$
Unfortunately $\frac{1}{2} \neq 2$ so who is wrong? (and if I'm wrong, then why?)
| You did nothing wrong, so you are correct. Good job! The answer is $\frac 12$.
| {
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"url": "https://math.stackexchange.com/questions/1383240",
"timestamp": "2023-03-29T00:00:00",
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Substitution to solve an initial value problem By using the substitution $y(x) = v(x)x$, how can I solve the initial value problem
$$
\frac{dy}{dx} = \frac{x^2+y^2}{xy - x^2},\quad y(1)=1
$$
And also keep my answer in the form $g(x,y)= 4e^{-1} xe^\frac{y}{x}$
| Here's an attempt:
$$ y = vx \implies \ \frac{dy}{dx} = v + x \frac{dv}{dx}$$
$$ \frac{dy}{dx} = \frac{x^2 + y^2}{xy -x^2} $$
$$ v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{x^2v - x^2}$$
$$ x\frac{dv}{dx} = \frac{v+1}{v-1}$$
Using separation of variables,
$$ \frac{v-1}{v+1}dv = \frac{v+1-2}{v+1} = \frac{1}{x}dx$$
$$ \left(1-\frac{2}{v+1}\right)dv = \frac{1}{x}dx $$
$$ v-2\ln |1+v| = \ln |x| +c $$ where c is a constant of integration
using the initial values $y(1) = 1$ gives $v =1 $:
$$ c = 1 -\ln 4 $$
$$ v - 2\ln |1+v| = \ln |x| +1 -\ln 4$$
substituting back $v(x) = \frac{y(x)}{x}$ gives:
$$ \frac{y}{x} = \ln \left(\frac{(x+y)^2}{4x}\right) + \ln e$$
where $1 = \ln e$
$$ \frac{e(x+y)^2}{4x} = e^\frac{y}{x} $$
$$ 4e^{-1} xe^\frac{y}{x} = (x+y)^2 $$
where $g(x, y) = (x+y)^2$
as required.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the approximate square root of $(a^4-a^{-4})/(a^2-a^{-2})$ Find the approximate square root of
$$\dfrac{\left( \dfrac{12}{5}\right)^4 - \left( \dfrac{5}{12}\right)^4 }{\left( \dfrac{12}{5}\right)^2 - \left( \dfrac{5}{12}\right)^2}$$
I tried using the formula for $(a^4-b^4)$ and $(a^2-b^2)$. Then cancelled the common $(a-b)$, substituted the values and simplified. Didn't work.
Answer is 13
Can someone tell me how to solve it with steps of possible?
It will be a great help
| The fraction itself has a value of around $6$, so the square root of the fraction will be somewhere between $2$ and $3$, and not $13$. How did you get $13$?
The most I was able to simplify the value of the whole expression is that it is equal to $$13^2\left(\frac1{5^2} + \frac1{12^2}\right) - 2$$
Which, I guess, you could say is close to $$\frac{13^2}{5^2} - 2 = \frac{12^2-5^2}{5^2}$$ which is "close" to $\frac{12^2}{5^2}$ and the root of this is $\frac{12}5=2.4$ Which is actually pretty close to the actual answer!
| {
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"url": "https://math.stackexchange.com/questions/1384196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
explicit formula for recurrence relation $a_{n+1}=2a_n+\frac{1}{a_n}$ For $n\in\mathbb N$,
$$a_{n+1}=2a_n+\frac{1}{a_n},\quad a_1=1.
$$
Can any one give an explicit formula for all $a_n$? If such an explicit general formula doesn't exist, please explain it. I've tried to figure out the $n$-iterated function $f^{(n)}$ where $f(x)=2x+1/x$ or even $f(\tan(t))$. But in either cases, I failed.Since the recurrence isn't linear nor homogeneous,the generating function method doesn't apply here.
| We can transform this equation in to $$\dfrac{b_{n+1}}{2}=b_n+\dfrac1{b_n}$$ by substituting $b_n=\sqrt2a_n.$
Note that $$b_{n+1}-4=\dfrac{2}{b_n}(b_n-1)^2$$ and $$b_{n+1}+4=\dfrac{2}{b_n}(b_n+1)^2$$ Now $$\dfrac{b_{n+1}/4-1}{b_{n+1}/4+1}=\left(\dfrac{b_n-1}{b_n+1}\right)^2$$ Continuing this processes we can obtain
$$\dfrac{b_{n+1}/4-1}{b_{n+1}/4+1}=\left(\dfrac{b_n-1}{b_n+1}\right)^2=\left(\dfrac{4b_{n-1}-1}{4b_{n-1}+1}\right)^{2^2}=\left(\dfrac{4^2b_{n-2}-1}{4^2b_{n-2}+1}\right)^{2^3}=\cdots=\left(\dfrac{4^{n-1}b_{1}-1}{4^{n-1}b_{1}+1}\right)^{2^n}$$ You can obtain $b_{n+1}$ and hence $a_n$ from here.
$$b_{n+1}=4\left(\dfrac{1+\left(\dfrac{4^{n-1}b_{1}-1}{4^{n-1}b_{1}+1}\right)^{2^n}}{1-\left(\dfrac{4^{n-1}b_{1}-1}{4^{n-1}b_{1}+1}\right)^{2^n}}\right).$$ Therefore
$$a_n=2\sqrt2\left(\dfrac{1+\left(\dfrac{4^{n-2}\sqrt2-1}{4^{n-2}\sqrt2+1}\right)^{2^{n-1}}}{1-\left(\dfrac{4^{n-2}\sqrt2-1}{4^{n-2}\sqrt2+1}\right)^{2^{n-1}}}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $\tan^2 \theta − 2 \sec \theta = 2$ Solve the given equation. Find all solutions of the equation (express your answer in terms of k, where k is any integer).
$$\tan^2 \theta − 2 \sec \theta = 2$$
What do I do to solve for $\theta$?
update: I got
$cos\theta$ $= \frac {1}{3}$ and $cos\theta$ $-1$
and then:
$1.231+2\pi k,\space5.052+2\pi k,\space\pi +2\pi k$
Thank you!
| Notice, we have $$\tan^2\theta-2\sec\theta=2$$ $$\sec^2\theta-1-2\sec\theta=2$$ $$\sec^2\theta-2\sec\theta-3=0$$ $$\sec^2\theta-3\sec\theta+\sec\theta-3=0$$
$$\underbrace{\sec^2\theta-3\sec\theta}\underbrace{+\sec\theta-3}=0$$ $$\sec\theta(\sec \theta-3)+(\sec\theta-3)=0$$ $$(\sec\theta-3)(\sec\theta+1)=0$$ $$\implies \sec\theta-3=0\iff \cos \theta=\frac{1}{3}\iff \color{blue}{\theta=2k\pi\pm\cos^{-1}\left(\frac{1}{3}\right)}$$
$$\implies \sec\theta+1=0\iff \cos \theta=-1\iff \color{blue}{\theta=2k\pi+\pi}$$
Where, $k$ is any integer & $\cos^{-1}\left(\frac{1}{3}\right)=1.231$
Your answers are obtained as follows, $$\color{red}{\theta}=2k\pi+\cos^{-1}\left(\frac{1}{3}\right)=\color{red}{2k\pi+1.231}$$
$$\color{red}{\theta}=2k\pi+2\pi-\cos^{-1}\left(\frac{1}{3}\right)=\color{red}{2k\pi+5.052}\ \ \ \text{(since, }\ \cos\theta=\cos(2\pi-\theta))$$ & $$\color{red}{\theta=2k\pi+\pi}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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IO is inclined to BC at an angle $\tan^{-1}(\frac{\cos B+\cos C-1}{\sin B-\sin C})$ Show that the line joining the inscribed center to the circumscribed center of a $\triangle ABC$ is inclined to BC at an angle $\tan^{-1}(\frac{\cos B+\cos C-1}{\sin B-\sin C})$
I tried taking the coordinates of $A(0,y_1),B(x_2,0),C(x_3,0)$.So the coordinates of incenter $I$ becomes $\left(\frac{bx_2+cx_3}{a+b+c},\frac{ay_1}{a+b+c}\right)$ and the circumcenter becomes $\left(\frac{x_2+x_3}{2},\frac{y^2_1+x_2x_3}{2y_1}\right)$
But answer is not appearing to come,calculations are messy and clumsy.Please help how should i prove it?
| Consider a $\triangle ABC$ with side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will be at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$ Hence, the inscribed center will be $$I\equiv \left(\frac{a\frac{a\tan C}{\tan B+\tan C}+b(0)+c(a)}{a+b+c}, \frac{a\frac{a\tan B\tan C}{\tan B+\tan C}+b(0)+c(0)}{a+b+c} \right)$$ From Sine rule, substitute $a=k\sin A$, $b=k\sin B$ & $c=k\sin C$, we get $$I\equiv \left(\frac{k(\sin A\sin C\cos B+\sin A\sin C)}{\sin A+\sin B+\sin C}, \frac{k\sin A\sin B\sin C}{\sin A+\sin B+\sin C} \right)$$
& the circumscribed center say $D$ can be calculated as $$D\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$ From Sine rule, substitute $a=k\sin A$, we get
$$D\equiv\left(\frac{k\sin A}{2}, \frac{k\cos A\cos B\cos C}{2} \right)$$
Hence, the slope of the line ID joining $I$ & $D$ with the BC (x-axis) is given as $$m=\frac{y_2-y_1}{x_2-x_1}$$
$$=\frac{\frac{k\cos A\cos B\cos C}{2}-\frac{k\sin A\sin B\sin C}{\sin A+\sin B+\sin C}}{\frac{k\sin A}{2}-\frac{k(\sin A\sin C\cos B+\sin A\sin C)}{\sin A+\sin B+\sin C}}$$
By simplifying we get
$$m=\frac{\cos B+\cos C-1}{\sin B-\sin C}$$ Hence the angle of the line joining I & D with side BC is given as $$\tan \theta=|m|$$ $$\implies \theta=\tan^{-1}\left|\frac{\cos B+\cos C-1}{\sin B-\sin C}\right|$$
| {
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Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $\cot\angle MAB + \cot\angle MBC + \cot\angle MCA \geq 3\sqrt{3}$ Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $$\cot\measuredangle MAB + \cot\measuredangle MBC + \cot\measuredangle MCA \geq 3\sqrt{3}$$
|
Let $A_1,B_1,C_1$ be the intersection points of $AM,BM,CM$ with $BC,CA,AB$ respectively, and let $D,E,F$ be the feet of perpendiculars from $M$ to $BC,CA,AB$ respectively.Then we have
\begin{align*}
\cot \angle MAB+\cot \angle MBC+\cot \angle MCA&=\dfrac {FA} {FM}+\dfrac {BD} {MD}+\dfrac {CE} {ME} \\
&=\sqrt{\dfrac {MA^2} {FM^2}-1}+\sqrt{\dfrac {MB^2} {MD^2}-1}+\sqrt{\dfrac {MC^2} {ME^2}-1}\\
&\ge \sqrt{\dfrac {MA^2} {MC_1^2}-1}+\sqrt{\dfrac {MB^2} {MA_1^2}-1}+\sqrt{\dfrac {MC^2} {MB_1^2}-1}
\end{align*}
Using Van Aubel theorem we have
$$\dfrac {MA} {MA_1}=\dfrac {AC_1} {C_1B}+\dfrac {AB_1} {B_1C}$$
$$\dfrac {MB} {MB_1}=\dfrac {BA_1} {A_1C}+\dfrac {BC_1} {C_1A}$$
$$\dfrac {MC} {MC_1}=\dfrac {CB_1} {B_1A}+\dfrac {CA_1} {A_1B}$$
Multiplying these and applying AmGm inequality we get
$$AM\cdot BM\cdot CM\ge 8MA_1 \cdot MB_1\cdot MC_1$$
Let $\dfrac {MA} {MC_1}=a,\dfrac {MB} {MA_1}=b,\dfrac {MC} {MB_1}=c $ so ti is sufficies to prove
$$\sqrt{a^2-1}+\sqrt{b^2-1}+\sqrt{c^2-1}\ge 3\sqrt3$$ where
$$a\ge1,b\ge1,c\ge1,abc\ge8$$
This one is an immediate consequence of Jensen inequality
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Factor $k^{4}+4k^{3}+8k^{2}+8k+4=0$ over $\mathbb C$ Any idea how to factor the polynomial $k^{4}+4k^{3}+8k^{2}+8k+4$ over $\mathbb C$?
Candidates for rational roots are $\pm1, \pm2, \pm4$ but none of them satisfies $k^{4}+4k^{3}+8k^{2}+8k+4=0$.
| Note that
\begin{align*}
k^4 + 4k^3 + 8k^2 + 8k + 4
&= (k^4 + 4k^3 + 6k^2 + 4k + 1) + (2k^2 + 4k + 2) + 1 \\
&= (k + 1)^4 + 2(k+1)^2 + 1 \\
&\ge 1 \text{ for all } k,
\end{align*}
so the polynomial has no real roots.
However, it does factor into two quadratics.
From the above,
\begin{align*}
k^4 + 4k^3 + 8k^2 + 8k + 4
&= (k + 1)^4 + 2(k+1)^2 + 1 \\
&= \big[(k+1)^2 + 1\big]^2 \\
&= (k^2 + 2k + 2)^2.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$
I showed that in
$$ \forall x \in [0,\frac{\pi}{4}] \quad \int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx \le \int_0^\frac{\pi}{4} \frac{1-\cos(\frac{\pi}{4})}{\frac{\pi}{4}} \,dx = 1 -\frac{\sqrt2}{2} $$
but it's not good enough
| Another way forward is to recall the Taylor series for the cosine
$$\cos x =\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \tag 1$$
Noting that for $0\le x\le \pi/4$ the terms of the series in $(1)$ alternate signs and are decreasing monotonically. Therefore, we have
$$\cos x \ge 1-\frac12 x^2$$
so that
$$\frac{1-\cos x}{x}\le \frac12 x$$
Finally, we see that
$$\begin{align}
\int_0^{\pi/4}\frac{1-\cos x}{x}\,dx &\le \int_0^{\pi/4} \frac12 x\, dx\\\\
&=\frac{\pi^2}{64}
\end{align}$$
as was to be shown!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all integer sided triangles whose area equals their perimeter. Find all integer sided triangles whose area equals their perimeter?
Taking sides to be $e,f,g$ we have,
$$\sqrt{(e+f+g)(e+f-g)(e+g-f)(g+f-e)}=4(e+f+g)$$
how to proceed next ?
| Let $a,b,c$ be the lengths of the sides and $s=(a+b+c)/2$.
By Heron's formula, We have
$$\sqrt{s(s-a)(s-b)(s-c)}=2s,$$
i.e.
$$(-a+b+c)(a-b+c)(a+b-c)=16(a+b+c)\tag1$$
Here, let $x=-a+b+c,y=a-b+c,z=a+b-c$, then we have
$$a=\frac{y+z}{2},\quad b=\frac{z+x}{2},\quad c=\frac{x+y}{2}\tag2$$
So, $(1)$ can be written as
$$xyz=16(x+y+z)\tag3$$
Here, note that $x,y,z$ are positive integers and all even from $(2)(3)$. So, let $x=2X,y=2Y,z=2Z$, then $(3)$ can be written as
$$XYZ=4(X+Y+Z).$$
We may suppose that $X\le Y\le Z$. So,
$$XYZ\le 4\cdot 3Z=12Z\Rightarrow XY\le 12.$$
$$X^2\le XY\le 12\Rightarrow X^2\le 12\Rightarrow X\le 3.$$
*
*For $X=1$, we have $(Y-4)(Z-4)=20\Rightarrow (Y,Z)=(5,24),(6,14),(8,9)$.
*For $X=2$, we have $(Y-2)(Z-2)=8\Rightarrow (Y,Z)=(3,10),(4,6)$.
*For $X=3$, we have $3YZ\le 4\cdot 3Z\Rightarrow Y\le 4$. If $Y=3$, then $9Z=4(6+Z)$. If $Y=4$, then $12Z=4(7+Z)$. Hence, there is no solution.
Thus, the answer is the followings :
$$(6,25,29),(7,15,20),(9,10,17),(5,12,13),(6,8,10).$$
| {
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Determining irreducible polynomial of $\zeta_n$ In a homework I did I had to determine the irreducible polynomials of some $\zeta_n$ functions over $\mathbb{Q}$. In $\zeta_6$, I set $\zeta_6=x$ and I know that $x^6=1 \rightarrow x^6-1=0$. So,when I factor it out I get $x^6-1=(x^3-1)(x^3+1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$ From there I know the expressions $(x^2-x+1)$ and $(x^2+x+1)$ are irreducible over $\mathbb{Q}$ but I don't know which one of those is the right answer. The solutions to the homework say its $(x^2+x+1)$ but I don't see why. Any hints? Thanks!
| It is sometimes helpful to write primitive roots of unity in their trigonometric form, and use De Moivre's theorem.
In this case, we may take $\zeta_6 = \cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})$, although the complex-conjugate would work as well (this is also the multiplicative inverse).
Then it is clear to see that $(\zeta_6)^3 = \cos(\pi) + i\sin(\pi) = -1 + i0 = -1$, that is:
$\zeta_6^3 + 1 = 0$.
So $\zeta_6$ is a root of $x^3 + 1$, not a root of $x^3 -1$.
We can also compute $f(\zeta_6)$ directly, where $f(x) = x^2 - x + 1$.
$\zeta_6^2 - \zeta_6 + 1 = (\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))^2 - \cos(\frac{\pi}{3}) - i\sin(\frac{\pi}{3}) + 1$
$= \cos(\frac{2\pi}{3}) - \cos(\frac{\pi}{3}) + 1 + i(\sin(\frac{2\pi}{3}) - \sin(\frac{\pi}{3}))$
$= -\frac{1}{2} - \frac{1}{2} + 1 + i(\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2})$
$= 0 + i0 = 0$.
In fact:
$(x - \zeta_6)(x - \zeta_6^{\ast}) = x^2 - 2\mathfrak{Re}(\zeta_6)x + \zeta_6(\zeta_6)^{\ast}$
$= x^2 - 2(\frac{1}{2})x + |\zeta_6|^2 = x^2 - x + 1$.
The above factorization shows that the splitting field of $x^3 + 1$ (indeed, even of $x^6 - 1$ since any cube root of unity is the square of a sixth root of unity) is indeed $\Bbb Q(\zeta_6)$, which therefore has degree $2$ over the rationals.
$2 = \phi(6)$, so this is not surprising. The sole non-trivial automorphism of $\Bbb Q(\zeta_6)$ which fixes $\Bbb Q$ is $\zeta_6 \mapsto \zeta_6^5 = (\zeta_6)^{-1} = \zeta_6^{\ast}$, that is, it is complex-conjugation restricted to $\Bbb Q(\zeta_6)$, and thus the Galois group is cyclic, of order $2$.
| {
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If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$, prove that $ x^2+y^2=1$. If $ \arcsin x+ \arcsin y=\frac{\pi}{2}$,
prove that $$ x^2+y^2=1$$
I tried taking sine of both the sides, I only come to this result: $$x^2 + y^2 -2x^2y^2 + 2xy\sqrt{(1-y^2)(1+x^2)}=1.$$
| Using the identity $arcsiny + arccosy = \frac{\pi}{2}$
$$arcsinx + arcsiny=\frac{\pi}{2}$$
$$arcsinx + \frac{\pi}{2} -arcccosy=\frac{\pi}{2}$$
$$arcsinx =arccosy$$
Let $arcsinx = \theta =arccosy \iff cos\theta=y \land sin(\theta) =x$.
Using the pythagorean identity to show that $cos^2\theta +sin^2\theta=1 \iff y^2 + x^2 =1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 6
} |
Factoring a polynomial with complex coefficients
Given $$3z^2+6z+3i=0$$ Find the complex roots and write in the form $a+bi$.
I want to see how to factor it when there is an $i$ being multiplied by the constant.
| $$3z^2+6z+3i=0\Longleftrightarrow$$
$$3(z^2+2z+i)=0\Longleftrightarrow$$
$$z^2+2z+i=0\Longleftrightarrow$$
$$z=\frac{-2\pm\sqrt{2^2-4\cdot 1 \cdot i}}{2\cdot 1}\Longleftrightarrow$$
$$z=\frac{-2\pm\sqrt{4-4\cdot i}}{2}\Longleftrightarrow$$
$$z=\frac{-2\pm\sqrt{4-4i}}{2}\Longleftrightarrow$$
$$z=\frac{-2+\sqrt{4-4i}}{2} \vee z=\frac{-2-\sqrt{4-4i}}{2}\Longleftrightarrow$$
$$z=-1+\frac{1}{2}\sqrt{4-4i} \vee z=-1-\frac{1}{2}\sqrt{4-4i}\Longleftrightarrow$$
$$z=-1+\sqrt{1-i} \vee z=-1-\sqrt{1-i}\Longleftrightarrow$$
$$z=-1+\frac{\sqrt{2+\sqrt{2}}}{2^{\frac{3}{4}}}-\frac{\sqrt{2-\sqrt{2}}}{2^{\frac{3}{4}}}i \vee z=-1-\frac{\sqrt{2+\sqrt{2}}}{2^{\frac{3}{4}}}+\frac{\sqrt{2-\sqrt{2}}}{2^{\frac{3}{4}}}i $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$n^{th}$ derivative of $y=x^2\cos x$ I am stuck with Leibniz formula
$$D^{n}y = \sum_{k=0}^{n} \binom{n}{k} \, x^{(2k)}\cos^{(n-k)}x$$
Could someone show how to do it?
| Write $$f^{(n)}(x) = a_n(x)\cos x +b_n(x)\sin x$$
The proceed by induction to detrmine recurrences for polynomials $a_n(x), b_n(x)$.
Specifically, if:
$$f^{(n)}(x)=(A_nx^2+B_nx+C_n)\cos x + (D_nx^2+E_nx+F_n)\sin x$$
You get the recurrence:
$$\begin{pmatrix}A_{n+1}\\B_{n+1}\\C_{n+1}\\D_{n+1}\\E_{n+1}\\F_{n+1}\end{pmatrix} = \begin{pmatrix}0&0&0&1&0&0\\
2&0&0&0&1&0\\
0&1&0&0&0&1\\
-1&0&0&0&0&0\\
0&-1&0&2&0&0\\
0&0&-1&0&1&0
\end{pmatrix}
\begin{pmatrix}A_{n}\\B_{n}\\C_{n}\\D_{n}\\E_{n}\\F_{n}\end{pmatrix} $$
So, by induction:
$$\begin{pmatrix}A_{n}\\B_{n}\\C_{n}\\D_{n}\\E_{n}\\F_{n}\end{pmatrix}= \begin{pmatrix}0&0&0&1&0&0\\
2&0&0&0&1&0\\
0&1&0&0&0&1\\
-1&0&0&0&0&0\\
0&-1&0&2&0&0\\
0&0&-1&0&1&0
\end{pmatrix}^n\begin{pmatrix}1\\0\\0\\0\\0\\0\end{pmatrix}$$
So now you need to know how to do exponentiation of a matrix. Turns out, the characteristic polynomial of this matrix is a very simple $(1+x^2)^3$, so the roots are $i$ and $-i$ and that means we know we can write:
$$A_n = p_A(n)i^n + q_A(n)(-i)^n$$
where $p_A$ and $q_A$ are polynomials of degree $2$ or less. So we can solve for $p_A$ and $p_B$ by using the first six values of $A_n$.
Similarly, there are polynomials $p_B,q_B,p_C,q_C$, etc.
| {
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"url": "https://math.stackexchange.com/questions/1390978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the value of $m$ if $\frac{m-a^{2}}{b^{2}+c^{2}}+\frac{m-b^{2}}{a^{2}+c^{2}}+\frac{m-c^{2}}{b^{2}+a^{2}}=3$
If $\dfrac{m-a^{2}}{b^{2}+c^{2}}+\dfrac{m-b^{2}}{a^{2}+c^{2}}+\dfrac{m-c^{2}}{b^{2}+a^{2}}=3;\ \ m,a,b,c \in\mathbb{R}$
Then the value of $m$ is...
Options
$\boldsymbol{1.)}\ a^{2}-b^{2}-c^{2} \quad \quad
\boldsymbol{2.)}\ a^{2}+b^{2}-c^{2}\\
\boldsymbol{3.)}\ a^{2}+b^{2} \quad \quad \quad \quad
\boldsymbol{4.)}\ a^{2}+b^{2}+c^{2}$
By observation if all the three value
$\dfrac{m-a^{2}}{b^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=\dfrac{m-b^{2}}{a^{2}+c^{2}}=1$
Then $m=a^{2}+b^{2}+c^{2}$
I want to know if their is any other simple and short method other than the stated observation .
I have studied maths up to $12$th grade.
| Let $a^2=x\;,b^2=y\;,c^2=z\;,$ Then equation convert into $\displaystyle \frac{m-x}{y+z}+\frac{m-y}{z+x}+\frac{m-z}{x+y}=1+1+1$
So $\displaystyle \left(\frac{m-x}{y+z}-1\right)+\left(\frac{m-y}{z+x}-1\right)+\left(\frac{m-z}{x+y}-1\right)=0$
So $\displaystyle (m-x-y-z)\left\{\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right\}=0$
So either $m=x+y+z$ or $\displaystyle \frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}=0$
So $m=x+y+z = a^2+b^2+c^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to solve the trigonometric equation $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$ the question is:
Find the solutions of the equation: $\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x=0$.
Let $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarly solving the above equation it comes out to be:
$$\sqrt2 u^2-u-\sqrt2=0=(u-\sqrt2)(\sqrt2u+1)=0 \implies \sin x+\cos x=\sqrt2 \quad(1)$$
and
$$\sin x+\cos x= \frac{-1} {\sqrt2}\quad (2)$$
so solving the results differently i got the answers:
$$x=2n\pi + \frac{\pi}4, 2n\pi +\frac{7\pi}{12}, 2n\pi-\frac{\pi}{12}$$
but the answers are:
$$x=2n\pi + \frac{\pi}4, 2n\pi -\frac{5\pi}{12}, 2n\pi+\frac{11\pi}{12}$$
I divided the eq(2) by $\sqrt2$
| This could be helpful:
\begin{align}
\sin x-\cos x-2(2)^{\frac 1 2}\sin x\cos x&=-\sqrt2\Big(\sin(x-\frac{\pi}{4})+\sin(2x)\Big)\\
&=-\frac{\sqrt2}{2}\cos(\frac{\pi}{8}-\frac{3}{2}x)\sin(\frac{\pi}{8}+\frac{1}{2}x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Given a condition on $n$ reals Let $a_1,a_2 \cdots a_n$ be reals such that $$\sqrt{a_1}+\sqrt{a_2-1}+ \cdots +\sqrt{a_n-(n-1)}=\frac{1}{2}(a_1+a_2+\cdots +a_n)-\frac{n(n-3)}{4}$$ Find the sum of the first $100$ terms of the sequence. I am just unable to think what to do. Pleae give some hints and ideas. Thanks.
| `Slightly' longer answer :)
I'll skip the base part of the MI. So, it is true for some $k$.
$$
\sum \limits_{i = 1}^k \sqrt{a_i - (i - 1)} = \frac 12 \sum \limits_{i = 1}^k a_i - \frac {k(k - 3)}4
$$
Now, let's observe what happens if we take $k + 1$ terms
$$
\sum \limits_{i = 1}^{k+1} \sqrt{a_i - (i - 1)} = \frac 12 \sum \limits_{i = 1}^{k+1} a_i - \frac {(k+1)(k - 2)}4
$$
Now, use the equality for the $k$ terms in LHS
$$
\frac 12 \sum \limits_{i = 1}^k a_i - \frac {k(k - 3)}4 + \sqrt{a_{k+1}-k} = \frac 12 \sum \limits_{i = 1}^{k+1} a_i - \frac {(k+1)(k - 2)}4 \implies \\
\sqrt{a_{k+1} - k} = \frac {a_{k+1}}2 - \frac 14\left [{(k+1)(k-2) - k(k-3)} \right ] = \frac {a_{k+1}}2 - \frac 12(k-1) \implies \\
4 (a_{k+1} - k) = a_{k+1}^2 - 2a_{k+1}(k-1) + (k-1)^2 \implies\\
a_{k+1}^2 - 2a_{k+1}(k-1+2) + (k-1)^2 + 4k = a_{k+1}^2 - 2a_{k+1}(k+1) + (k+1)^2 = 0 \implies \\
a_{k+1} = k+1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Proving limit of difference is $0$: $\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$ I am trying to show that
$$\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$$
and I know intuitively this must be so since the "+1" contribution in the first term becomes negligible as $n$ becomes sufficiently large. I am wondering whether showing the ratio between the two terms approaching $1$ is sufficient to show the difference is $0$; if so, how can this be formalized? If not, how can it more directly be shown that the difference approaches $0$?
| $$
a^7-b^7 = (a-b)(a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6)
$$
So for $a=\sqrt [ 7 ]{ n+1 } $ and $b = \sqrt [ 7 ]{n } $, we get:
$$
\lim _{ n\rightarrow +\infty }{ \sqrt [ 7 ]{ n+1 } -\sqrt [ 7 ]{ n } \quad } \quad =\quad \lim _{ n\rightarrow +\infty }{ \frac { n+1-n }{ (\sqrt [ 7 ]{ n+1 } )^{ 6 }+((\sqrt [ 7 ]{ n+1 } ))^{ 5 }\left( \sqrt [ 7 ]{ n } \right) +(\sqrt [ 7 ]{ n+1 } )^{ 4 }\left( \sqrt [ 7 ]{ n } \right) ^{ 2 }+(\sqrt [ 7 ]{ n+1 } )^{ 3 }\left( \sqrt [ 7 ]{ n } \right) ^{ 3 }+(\sqrt [ 7 ]{ n+1 } )^{ 2 }\left( \sqrt [ 7 ]{ n } \right) ^{ 4 }+(\sqrt [ 7 ]{ n+1 } )\left( \sqrt [ 7 ]{ n } \right) ^{ 5 }+\left( \sqrt [ 7 ]{ n } \right) ^{ 6 } } \quad } \\ \qquad \qquad \qquad \qquad \qquad =\quad \lim _{ n\rightarrow +\infty }{ \frac { 1 }{ (\sqrt [ 7 ]{ n+1 } )^{ 6 }+((\sqrt [ 7 ]{ n+1 } ))^{ 5 }\left( \sqrt [ 7 ]{ n } \right) +(\sqrt [ 7 ]{ n+1 } )^{ 4 }\left( \sqrt [ 7 ]{ n } \right) ^{ 2 }+(\sqrt [ 7 ]{ n+1 } )^{ 3 }\left( \sqrt [ 7 ]{ n } \right) ^{ 3 }+(\sqrt [ 7 ]{ n+1 } )^{ 2 }\left( \sqrt [ 7 ]{ n } \right) ^{ 4 }+(\sqrt [ 7 ]{ n+1 } )\left( \sqrt [ 7 ]{ n } \right) ^{ 5 }+\left( \sqrt [ 7 ]{ n } \right) ^{ 6 } } \quad } \\ \qquad \qquad \qquad \qquad \qquad =\quad 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
Gaussian Integration verification I have the following problem:
For the formula
$$\int_0^1 f(x) dx\approx w_1f(0)+w_2f(x_2)$$
determine the weights $w_1, w_2$ and the node $x_2$ so that the formula is exact for all polynomials of as degree as possible. What is the degree of precision of the formula?
This is my attempt. We require $3$ unknowns, namely: $w_1,w_2$ and $x_2$. We then use: $f(x)=1,f(x)=x$ and $f(x) = x^2$.Therefore I have the following system of equations
$$\int_0^1 1dx =1\Rightarrow1= w_1+w_2$$
$$\int_0^1 xdx =\frac{1}{2}\Rightarrow \frac{1}{2} = w_1x_1+w_2x_2$$
$$\int_0^1 x^2dx =\frac{1}{3}\Rightarrow \frac{1}{3} = w_1x_1^2+w_2x_2^2$$
Now, since $x_1=0$, we have:
$$1= w_1+w_2$$
$$ \frac{1}{2} = w_2x_2$$
$$\frac{1}{3} =w_2x_2^2$$
The solution of this system is given by $w_1 = \frac{1}{4}, w_2 =\frac{3}{4}$ and $x_2 = \frac{2}{3}$, so that
$$\int_0^1 f(x) dx\approx -2f(0)+3f(\frac{2}{3}).$$
The degree of precision is $2$, since for $f(x) = x^3$, we have:
$$\int_0^1 x^3 dx - [\frac{1}{4}f(0)+\frac{3}{4}f(\frac{2}{3})] = \frac{1}{36}\neq 0$$
| Considering the system of equations you wrote $$1= w_1+w_2$$ $$\frac{1}{2} = w_2x_2$$ $$\frac{1}{3} =w_2x_2^2$$ the solution should be $$\left\{{w_1}= \frac{1}{4},{w_2}=\frac{3}{4},{x_2}=
\frac{2}{3}\right\}$$ as A.G. commented.
The solution is immediate if you divide the third equation by the second : it gives $x_2= \frac{2}{3}$ and the other variables are immediately obtained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Need help with an inequality $1<\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{3001}<1\frac{1}{3}$
The first part is trivial with $AM-HM $ inequality. Having problem with the second part.
| For $0 \leqslant k \leqslant 1000$
$$\frac{1}{2001-k} + \frac{1}{2001+k} = \frac{4002}{2001^2 - k^2} \leqslant \frac{4002}{2001^2 - 1000^2} = \frac{4002}{1001 \cdot 3001} < \frac{1}{1000 \frac{1}{2}} \cdot \frac{4}{3}.$$
Adding up the inequalities for $k = 1, 2, \ldots, 1000$ and the inequality divided by $2$ for $k = 0$, we obtain
$$\frac{1}{1001} + \frac{1}{1002} + \ldots + \frac{1}{3001} < \frac{1000 \frac{1}{2}}{1000 \frac{1}{2}} \cdot \frac{4}{3} = \frac{4}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
sum $1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...$ find the sum if the following series converges:
$$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + ...$$
my attempt:
The series can be grouped into difference of a series of odd terms and a series of even terms.. but how to find sum?
| Let $L=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\dots \tag{1}$
$(1)$ converges by Alternating series test, so $L$ is finite
Go through this link to convince yourself that $L=\ln(2).$
Now multiply $L$ by $\frac{1}{2}$, we get $$\frac L2=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}\dots \tag 2$$
Now adding $(1)$ and $(2)$, we get
$$\frac{3}2L=1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}\dots \tag 3$$
Your question ends here as you have found the required sum as $\frac{3}2(\ln(2)) \approx 1.0397$
Now notice that $(3)$ is just a rearrangment of $(1)$, in a way, we list the first two +ve terms followed by first -ve term from $(1)$ to make up first three terms of $(3)$ , and then next two +ve terms followed by second -ve term, and so on.
Thus we see that rearrangments of a series are not necessarily to converge on the same limit. And thus, an interesting question pops up- "when does rearrangements converge to same limit?"
Answer-
When original series converges absolutely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In what system(s) of numeration is 11111 a perfect square? From Charles Trigg's "Mathematical Quickies: 270 Stimulating Problems with Solutions":
In what system(s) of numeration is 11111 a perfect square?
I have found one base that works: 3. I am not sure whether this is the only solution or not. So far, I have shown that for any solution, the base is not one less than a prime.
Let $S = \{ n^2 | n \in Z^+\}$ be the set of all positive perfect squares.
Let $y$ be the number written as $11111$ in some base, $x$.
For base 2, $y = \dfrac{2^5-1}{2-1} = 31 \notin S$
For base 3, $y = \dfrac{3^5-1}{3-1} = \dfrac{243-1}{2} = 11^2 \in S$
Assume a base $x|x\ge2,x\in\mathbb{Z^+}$.
Write $$y=1+x+x^2+x^3+x^4=n^2 \tag{1}$$
Rearrange (1) to
$$\begin{align}
1+x+x^3+x^4&=n^2-x^2\\
(1+x)(1+x^3)&=(n+x)(n-x)\\
(1+x)^2(1-x+x^2)&=(n+x)(n-x) \tag{2}
\end{align}$$
We can show $(1+x)^2\nmid(n+x)\text{ and }(1+x)^2\nmid(n-x)$ as follows:
$$\begin{align}
(1+x)^2\mid(n+x) &\implies n+x \ge x^2+2x+1 \\
&\implies n-x \ge x^2+1 \\
&\implies (n+x)(n-x) \ge (x^2+2x+1)(x^2+1) \\
&\implies n^2 - x^2 \ge x^4+2x^3+2x^2+2x+1 \\
&\implies n^2 - x^2 > x^4+x^3+x^2+x+1
\end{align}$$
which contradicts (2). Similarly, $(1+x)^2\mid(n-x)$ contradicts (2).
So $\boxed{(1+x)^2\nmid(n+x)}$ and $\boxed{(1+x)^2\nmid(n-x)}$
Assume then that $1+x$ divides both $n+x$ and $n-x$. If so:
$$(1+x)\mid 2x \implies x=1$$
which is disallowed.
So the assumption was false, and $\boxed{(1+x)\nmid(n+x)\text{ or }(1+x)\nmid(n-x)}$.
Therefore, $\boxed{(x+1)\text{ is not prime}}$.
Now let $p$ be a prime factor of $1+x$. Then:
*
*$p\nmid x$
*$p\mid\big((x^2+x)-2(x+1)\big) \implies p\mid(x^2-x-2) \implies p=3 \lor p\nmid(x^2-x+1)$
If $p\mid n+x$ and $p\mid n-x$, we have $p\mid2x \implies p=2\quad(\text{since }p\nmid x)$
Since $(1+x)^2\mid\big((n+x)(n-x)\big)$, if any prime $p_o>2$ divides $1+x$, then ${p_o}^2$ divides either $n+x$ or $n-x$, and $p_o$ does not divide the other.
| Let the base be $b$. Since
$$b^4 < b^4 + b^3 + b^2 + b + 1 < (b^2 + b)^2,$$
if it is a square, the square root must be $b^2 + c$ with $0 < c < b$. But
$$(b^2 + c)^2 = b^4 + 2c b^2 + c^2.$$
Since $c^2 < b^2$, it follows that we must have
$$2c = b+1 = c^2,$$
whence $c = 2$ and $b = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$
(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
| I'd like to add one more answer "by color-coding".
First, observe that
$$
\sqrt{n+\sqrt{k}}-\sqrt{n-\sqrt{k}} = \sqrt{2}\sqrt{n-\sqrt{n^2-k}}.
$$
Now the problem.
Designate
$$
x = \frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}},
$$
and calculate
$$
x-1 =
\frac{\overbrace{\sqrt{10+\sqrt{1}}-\sqrt{10-\sqrt{1}}}^{\color{red}{\sqrt{2}\sqrt{10-\sqrt{99}}}}
+ \cdots +
\overbrace{\sqrt{10+\sqrt{99}}-\sqrt{10-\sqrt{99}}}^{\color{blue}{\sqrt{2}\sqrt{10-\sqrt{1}}}} }{\color{blue}{\sqrt{10-\sqrt{1}}}+\cdots+\color{red}{\sqrt{10-\sqrt{99}}}}.
$$
Everything simplifies to $\sqrt{2}$, thus
$$
x = 1 + \sqrt{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
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Let $a,b,c>0$ so that $a+b+c=1$...
Let $a,b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6\geq 2\sqrt{2}\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )$$
What I think is we should replacing $1-a ,1-b ,1-c$ with $b+c,c+a, a+b$ respectively on the right hand side, I have no idea if this is true, any help will be appraciated.
| We have $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}=\sum \frac{b+c}{a}=\sum \frac{1-a}{a}$$ so $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6- 2\sqrt{2}\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )=\sum \left ( \frac{1-a}{a}-2\sqrt{2}\sqrt{\frac{1-a}{a}}+2 \right )=\sum \left ( \sqrt{\frac{1-a}{a}}-\sqrt{2} \right )^{2} \geq 0$$
Equality occurs when $$\sqrt{\frac{1-a}{a}}=\sqrt{\frac{1-b}{b}} = \sqrt{\frac{1-c}{c}}$$
Therefore $a=b=c=\frac{1}{3}$
AND WE'RE DONE
| {
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If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand it now.
| Notice, $$3x^2-2x+7=0$$ $$3x^2-2x+\frac{1}{3}+7-\frac{1}{3}=0$$ $$3\left(x^2-\frac{2x}{3}+\frac{1}{9}\right)+7-\frac{1}{3}=0$$ $$3\left(x-\frac{1}{3}\right)^2+\frac{21-1}{3}=0$$
$$3\left(x-\frac{1}{3}\right)^2=-\frac{20}{3}$$
$$\left(x-\frac{1}{3}\right)^2=-\frac{20}{9}$$
Hence, we get
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\left(x-\frac{1}{3}\right)^2=\color{blue}{-\frac{20}{9}}}}$$
| {
"language": "en",
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Find $\lim_{n\to\infty}\frac{a^n}{n!}$ First I tried to use integration:
$$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$
$$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$
But I could not express it as a riemann integral. Now I am thinking about sandwich theorem.
$$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$
Since $\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}$
$$\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$
since $\frac{a}{t+1}<1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$
Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$
And by using sandwich theorem, $y=0$. Is this correct?
| You can prove it as follows:
for every $\varepsilon >0$ and $m+1>\left| a \right| $ and if $n$ is big enough then
$$0<\left| \frac {a^n}{ n! } \right| =\frac { \left| a \right| }{ 1 } \cdot \frac { \left| a \right| }{ 2 } \cdots \frac { \left| a \right| }{ m } \cdot \frac { \left| a \right| }{ m+1 } \cdots \frac { \left| a \right| }{ n } <\frac { { \left| a \right| }^m }{ m! } { \left( \frac { \left| a \right| }{ m+1 } \right) }^{ n-m }<\varepsilon $$
| {
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Prove by induction: for $n \ge 0$, $\frac{(2n)!}{n!2^n}$ is an integer Another prove by induction question:
for $n \ge 0$, $$\frac{(2n)!}{n!2^n}$$ is an integer
Base step:
$$n = 0$$
$$\frac{(2 \times 0)!}{0! \times 2^0} = \frac{0!}{1 \times 1} = 1$$
Induction step: please help
| Assume $\frac{(2n)!}{n!2^n}$ is an integer. $\frac{2(n+1)!}{(n+1)!2^{n+1}}=\frac{(2n+2)\cdot(2n+1)\cdot(2n)!}{(n+1)\cdot n!\cdot 2^n\cdot 2}=\frac{(2n)!}{n!2^n}\cdot \frac{(2n+2)(2n+1)}{2(n+1)}$. Notice that $\frac{(2n)!}{n!2^n}$ is an integer by the inductive assumption, and that $\frac{(2n+2)(2n+1)}{2(n+1)}=\frac{2(n+1)(2n+1)}{2(n+1)}=2n+1$.
| {
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What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$?
Find a closed form expression for $$\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}$$
where $n$ is an even positive integer.
I tried using binomial identities, but since the binomial coefficient is in the denominator, I can't do much. Also, I'm seeking a non-calculus, elementary solution, if possible.
Thanks.
| We seek to find a closed form of
$$\sum_{r=0}^n (-1)^r {n\choose r}^{-1}.$$
Recall from MSE
4316307 the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We get for our sum
$$1+ \sum_{r=0}^n (-1)^r r [z^n] \log\frac{1}{1-z} (z-1)^{n-r}
\\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^n
[v^n] \frac{1}{1-v}
\sum_{r\ge 0} (-1)^r r (z-1)^{-r} v^r
\\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^n
[v^n] \frac{1}{v-1} \frac{v/(z-1)}{(1+v/(z-1))^2}
\\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^{n+1}
[v^{n-1}] \frac{1}{v-1} \frac{1}{(z-1+v)^2}.$$
The contribution from $v$ is
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{v-1} \frac{1}{(z-1+v)^2}.$$
The residue at infinity is zero so we may compute this by taking minus
the residue at one and minus the residue at $v=1-z.$ (Residues sum to
zero.) We get for the former
$$- [z^{n+2}] \log\frac{1}{1-z} (z-1)^{n+1}
= - {n+2\choose 1}^{-1} = -\frac{1}{n+2}$$
and the latter
$$- \left.\left( \frac{1}{v^n} \frac{1}{v-1} \right)'\right|_{v=1-z}
= - \left.\left( -\frac{n}{v^{n+1}} \frac{1}{v-1}
- \frac{1}{v^n} \frac{1}{(v-1)^2} \right)\right|_{v=1-z}
\\ = - \frac{n}{z (1-z)^{n+1}} + \frac{1}{(1-z)^n z^2}.$$
We get for the first term
$$- (-1)^{n+1} n [z^{n+1}] \log\frac{1}{1-z}
= (-1)^n \frac{n}{n+1}.$$
and for the second
$$(-1)^n [z^{n+2}] \log\frac{1}{1-z} (z-1)
= (-1)^n \frac{1}{n+1} - (-1)^n \frac{1}{n+2}.$$
Collecting everything we find
$$1+(-1)^n - \frac{1}{n+2} (1+(-1)^n).$$
This will produce zero when $n$ is odd. For $n$ even we find
$$\bbox[5px,border:2px solid #00A000]{
2 \frac{n+1}{n+2}.}$$
| {
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Factorization of a polynomials in complex number. Factorize this expression:
$$a^2+b^2+c^2-ab-bc-ca.$$
The result is
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$
How I can get $\Omega$ here?What's the approach?
| we have
$$(a+b\Omega+c\Omega^2)(a+b\Omega^2+c\Omega)$$$$=a^2+ab\Omega^2+ac\Omega+ab\Omega+b^2\Omega^3+bc\Omega^2+ac\Omega^2+bc\Omega^4+c^2\Omega^3$$
$$=a^2+\Omega^3b^2+\Omega^3c^2+(\Omega^2+\Omega)ab+(\Omega^2+\Omega^4)bc+(\Omega+\Omega^2)ac$$
$$=a^2+b^2+c^2-ab-bc-ac$$
It follows that $$\Omega^3=1$$
and
$$\Omega^2+\Omega=\Omega^2+\Omega^4=-1.$$
Solving $$\Omega^2+\Omega=-1$$ will give you two solutions that work (check that they satisfy all of the above equations). These happen to be the primitive third roots of unity as $$\Omega^2+\Omega+1$$ is the third cyclotomic polynomial.
| {
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} |
Express roots in polynomials of equation $x^3+x^2-2x-1=0$
If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.
I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-power terms, etc).
In the comment in this link, someone suggest that If $A$ is a root of $x^6−2x^5+3x^3−2x−1=0$, then so is $−A^5+2A^4−3A$, without further explanation (or maybe it's obvious to math experts, not to me) but I'm more interested in the underlying theory, preferably elementary, and techniques to solve problems of this kind.
Thanks!
| If $a$ is one root, then we get
$$
\frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1}
$$
Using the quadratic equation to solve this yields
$$
\frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2}
$$
for the other two roots.
After the Question Change
Changing the constant term only changes $(1)$ slightly:
$$
\frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3}
$$
and we still have $(2)$.
Since the discriminant is $9-2a-3a^2$, we know that if one root is between $\frac{-1-2\sqrt{7}}3$ and $\frac{-1+2\sqrt{7}}3$, then all three roots are. If any root is outside that interval, the other two roots will not be real.
After considering a comment by KCd, I see that
$$
(x-a)(x-a^2+2)(x+a^2+a-1)\equiv x^3+x^2-2x-1\pmod{a^3+a^2-2a-1}
$$
Therefore, if $a$ is a root, then $a^2-2$ and $-a^2-a+1$ are also roots.
| {
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How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$? How to prove that $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$ using the fact that $e^x=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n$ ?
So,
$$e^x=\lim_{n\to\infty }\sum_{k=0}^n \binom{n}{k}\left(\frac{x}{n}\right)^k=\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{(n-k)!n^k}\frac{x^k}{k!}.$$
I think that I have to prove that $\frac{n!}{(n-k)!n^k}=1$ but I didn't have success.
| 1. Solution using $\,e^{\,x} = \dfrac{d}{dx}\,e^{\,x}\,$ property of exponent
Assume that the exponent function can be represented as a series with unknown coefficients:
$$
e^{\,x} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots = \sum_{n=0}^{\infty} a_n x^n
$$
Recall the fundamental property of exponent $\, \dfrac{d}{dx} \big(e^{\,x} \big) = e^{\,x}$.
Applying this property to the series for of exponent, we get
\begin{align}
\dfrac{d}{dx} \,e^{\,x} = e^{\,x}
& \implies
\dfrac{d}{dx} \left(\sum_{n=0}^{\infty} a_n x^n \right) = \sum_{n=0}^{\infty} a_n x^n
\\ & \implies
\dfrac{d}{dx} \big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
\\ & \implies
0 + a_1 + 2 \, a_2\, x + 3 \, a_3\, x^2 + \ldots = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
\\ & \implies
\sum_{n=1}^{\infty}n \,a_n\,x^{n-1} = \sum_{n=0}^{\infty} a_n x^n
\iff
\sum_{n=0}^{\infty}\left(n+1\right) a_{n+1}\,x^{n} = \sum_{n=0}^{\infty} a_n x^n
\\ & \implies
\left(n+1\right)a_{n+1} = a_n
\\ &
\implies
a_{n+1} = \frac{a_n}{n+1}
\end{align}
The last equation can be rewritten as $\,a_{n} = \dfrac{1}{n}a_{n-1}, \,$ so that
$$
a_{n} = \frac{1}{n}\,a_{n-1} = \frac{1}{n}\,\frac{1}{n-1} \,a_{n-2}=
\frac{1}{n}\,\frac{1}{n-1}\,\frac{1}{n-2}\,a_{n-3} = \ldots =
\frac{1}{ n!}\,a_0\tag{1.1}
$$
Observer that $\,e^0 = 1,\,$ so we can write
$$
e^{\,x}\Big\rvert_{x=0} =
\big( a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots\big)\Big\rvert_{x=0}
= a_0 = 1
$$
This fact combined with equation $(1.1)$ gives us the explicit expression for coefficient $\,a_n = \dfrac{1}{n!}.\,$
Therefore we finally write
$$
\bbox[4pt, border:2.5pt solid #FF0000]{\ \ e^{\,x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}\ \,}
$$
Q.E.D.
EDIT: As requested in comments, here I provide solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ expression.
2. Solution using $\,\displaystyle e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ property of exponent
I do not believe that it is possible (at least within reasonable timespan) express the exponentsas $\,e^{\,x}=\sum_{k=0}^\infty \frac{x^k}{k!}\,$ using only algebraic operations and the expression formula $\, e^{\,x}=\lim_{n\to\infty }\left(1+\frac{x}{n}\right)^n\,$ as starting point.
However, it is possible to show the equivalence of these two definitions of $\,e^{\,x}\,$ as $\,n\to \infty.\,$
Indeed, for any $x\ge 0$ let us define
$$
S_n = \sum_{k=0}^n \frac{x^k}{k!}, \qquad
L_n = \left(1+\frac{x}{n}\right)^n.
$$
Then, by Newton's binomial
$$
\begin{aligned}
L_n & = \sum_{k=0}^n {n\choose k} \,\frac{x^k}{n^k}
= 1 + x + \sum_{k=2}^{n}
\frac{n\cdot \left(n-1\right)\cdot \left(n-2\right)\cdot \ldots\cdot \left(n-(k-1)\right)}{k! \,n^k}=
\\ & =
1 + x
+ \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right)
+ \frac{x^3}{3!}\,\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right)
+ \ldots
+ \frac{x^n}{n!}\,\left(1 - \frac{1}{n} \right) \cdots
\left(1 - \frac{n-1}{n} \right) \leq S_n
\end{aligned}
$$
Therefore
$$
\limsup_{n\to\infty}L_n \leq \limsup_{n\to\infty}S_n = e^{\,x}.\tag{2.1}
$$
On the other hand, for any positive integer $\, m\,$ such that $\,2\le m \le n\,$ we have
$$
1 + x
+ \frac{x^2}{2!}\,\left(1 - \frac{1}{n} \right)
+ \ldots
+ \frac{x^m}{m!}\,\left(1 - \frac{1}{n} \right)\left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)
\le L_n
$$
If we fix $\,m\,$ and let $\,n\to\infty,\,$ then we get
$$
S_m = 1 + x + \frac{x^2}{2!} + \ldots + \frac{x^m}{m!}
\leq \liminf_{n\to\infty}L_n\tag{2.2}
$$
Letting $\,m\to\infty\,$ in inequality $(2.2)$ and combining it with inequality $(2.1)$, we get
$$
e^{\,x} = \limsup_{n\to\infty}L_n\leq \lim_{n\to\infty} S_n \leq \liminf_{n\to\infty}L_n = e^{\,x}
$$
and thus
$$
\bbox[5pt, border:2.5pt solid #FF0000]{\lim_{n\to \infty}S_n = \sum_{n=0}^\infty \frac{x^n}{n!}=e^{\,x}}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
} |
Summing n times binomial(n,k) I'm trying to do $\sum_{n=a}^b \left( \begin{array}{rl} n \\ a \end{array} \right) n $ .
Is there a formula, that anybody knows?
| Use generating functions. First:
$\begin{align}
\sum_{n \ge 0} \binom{n}{a} z^n
&= \frac{z^a}{(1 - z)^{a + 1}} \\
z \frac{\mathrm{d}}{\mathrm{d} z} \sum_{n \ge 0} \binom{n}{a} z^n
&= \sum_{n \ge 0} n \binom{n}{a} z^n \\
&= \frac{z^a (z + a)}{(1 - z)^{a + 2}}
\end{align}$
But also, to get partial sums you divide by $1 - z$:
$$
\sum_{k \ge 0} \left( \sum_{0 \le n \le k} n \binom{n}{k}\right) z^k
= \frac{z^a (z + a)}{(1 - z)^{a + 3}}
$$
Now you are interested in the coefficient of $z^b$:
$\begin{align}
\sum_{0 \le n \le b} n \binom{n}{a}
&= [z^b] \frac{z^a (z + a)}{(1 - z)^{a + 3}} \\
&= [z^b] (z^{a + 1} + a z^a) \sum_{k \ge 0} (-1)^k \binom{-a - 3}{k} z^k \\
&= [z^b] (z^{a + 1} + a z^a)
\sum_{k \ge 0} \binom{a + 3 + k - 1}{a + 3 - 1} z^k \\
&= [z^{b - a - 1}] \sum_{k \ge 0} \binom{a + 3 + k - 1}{a + 3 - 1} z^k
+ a [z^{b - a}]\sum_{k \ge 0} \binom{a + 3 + k - 1}{a + 3 - 1} z^k \\
&= \binom{a + 2 + b - a - 1}{a + 2} + a \binom{a + 2 + b - a}{a + 2} \\
&= \binom{b + 1}{a + 2} + a \binom{b + 2}{a + 2}
\end{align}$
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Differentiation using Chain Rule Find $\frac{dy}{dx}$ if $y=7+5^{x^2+2x-1}$.
So far I have done $\frac{dy}{dx}=(5^{x^2+2x-1})'$. Now, the RHS can be found by $(e^{\ln 5\cdot (x^2+2x-1)})'=e^{\ln 5\cdot (x^2+2x-1)}(x^2+2x-1)'\ln 5=5^{x^2+2x-1}(2x+2)\ln 5$.
However, my textbook says the answer is $(2x^3+6x^2+2x-2)(5^{x^2+2x-1})$. Where did I go wrong? Thanks.
| Your work is correct
$$\frac { d\left( 7+5^{ x^{ 2 }+2x-1 } \right) }{ dx } =\frac { d\left( 7 \right) }{ dx } +\frac { d\left( 5^{ x^{ 2 }+2x-1 } \right) }{ dx } =0+\left( 5^{ x^{ 2 }+2x-1 } \right) \ln { 5 } \frac { d\left( x^{ 2 }+2x-1 \right) }{ dx } =\\ =\left( 5^{ x^{ 2 }+2x-1 } \right) \ln { 5 } \left( 2x+2 \right) \\ $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Error estimation for the Wallis product From the Wallis product we know $$\prod_{k=1}^{\infty} \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}$$
Let $a_n = \prod_{k=1}^{n} \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right)$ the n-th partial product. Is there an error estimate known for the relative error $\frac{a_n}{\pi/2}$?
| We have
$$\frac{a_n}{\pi/2} = \prod_{k = n+1}^\infty \frac{(2k-1)(2k+1)}{(2k)^2} = \prod_{k = n+1}^\infty \biggl(1 - \frac{1}{4k^2}\biggr).$$
To estimate products, it is often convenient to take logarithms. Here we can get the easy upper bound
$$\log \prod_{k = n+1}^\infty \biggl(1 - \frac{1}{4k^2}\biggr) = \sum_{k = n+1}^\infty \log \biggl( 1 - \frac{1}{4k^2}\biggr) < - \frac{1}{4}\sum_{k = n+1}^\infty \frac{1}{k^2} < -\frac{1}{4(n+1)}$$
and the lower bound
\begin{align}
\log \prod_{k = n+1}^\infty \biggl(1 - \frac{1}{4k^2}\biggr) &= - \log \prod_{k = n+1}^\infty \biggl(1 + \frac{1}{4k^2-1}\biggr)\\
&= - \sum_{k = n+1}^\infty \log \biggl(1 + \frac{1}{4k^2-1}\biggr)\\
&>-\sum_{k = n+1}^\infty \frac{1}{4k^2-1}\\
&= - \frac{1}{2}\sum_{k = n+1}^\infty \biggl( \frac{1}{2k-1} - \frac{1}{2k+1}\biggr)\\
&= -\frac{1}{4n+2}.
\end{align}
Thus
$$\exp \biggl(-\frac{1}{4n+2}\biggr) < \frac{a_n}{\pi/2} < \exp \biggl(-\frac{1}{4n+4}\biggr).$$
For the little work needed, these bounds are already decent. For better bounds, you can use better approximations of the logarithms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\lim\limits_{n\to\infty}\frac1{n}\sum\limits_{k=1}^{\infty}\left\lfloor\frac{n}{3^k}\right\rfloor=\frac{1}{2}$
Show that
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{3^k}\right\rfloor=\frac{1}{2}$$
I can do right hand.
$$\sum_{k=1}^{\infty}\left\lfloor\dfrac{n}{3^k}\right\rfloor\le \sum_{k=1}^{\infty}\dfrac{n}{3^k}=\dfrac{n}{2}$$
But how to solve left hand?
| $$0\leq \frac{n}{3^k}-\left\lfloor\frac{n}{3^k}\right\rfloor\leq 1$$
and the number of non-zero terms of the sum is bounded by $1+\log_3(n)$, hence:
$$\begin{eqnarray*} \sum_{k=1}^{+\infty}\left\lfloor\frac{n}{3^k}\right\rfloor=\sum_{k=1}^{\left\lceil\log_3(n)\right\rceil}\left\lfloor\frac{n}{3^k}\right\rfloor&\geq& -(1+\log_3(n))+\sum_{k=1}^{\left\lceil\log_3(n)\right\rceil}\frac{n}{3^k}\\&\geq&\frac{n}{2}-(1+\log_3(n))-\sum_{k>\left\lceil \log_3(n)\right\rceil}\frac{n}{3^k}\\&\geq&\frac{n}{2}-2\log(n)\end{eqnarray*}$$
for any $n$ big enough.
| {
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"url": "https://math.stackexchange.com/questions/1412638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by induction that $\big(1+\frac{1}{n}\big)^n\le3-\frac{1}{n}$, as in
$\hspace{.4 in}$ A simple proof that $\bigl(1+\frac1n\bigr)^n\leq3-\frac1n$?
| As suggested by Clement C., let:
$$ f(x)=\left(1+\frac{1}{x}\right)^{x}+\frac{1}{x}.\tag{1}$$
Then:
$$ f'(x) = \left(1+\frac{1}{x}\right)^{x}\left(\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right)-\frac{1}{x^2}\tag{2} $$
but, due to convexity:
$$\log\left(1+\frac{1}{x}\right)-\frac{1}{x+1}=-\frac{1}{x+1}+\int_{x}^{x+1}\frac{dt}{t}\geq \frac{1}{2(x+1)^2}\tag{3}$$
hence for any $x\geq 8$:
$$ f'(x)\geq \frac{\left(1+\frac{1}{8}\right)^8}{2(x+1)^2}-\frac{1}{x^2}>0.\tag{4} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Proof check: $(4n)!$ is divisible by $2^{3n}3^{n}$ Question: Show that $(4n)!$ is a multiple of $2^{3n}3^{n}$ for all $n$.
Proof: It's easy (involves kinda messy calculation tho) to show by induction that $(4n)!$ is a multiple of $2^{3n}$. Now, since every third number in the sequence $1,2,...,4n$ is divisible by $3$, there are at least $\dfrac{4n}{3}>n$ many 3's. This proves that $(4n)!$ is also a multiple of $3^n$ and the claim follows.
Is this proof valid?
| First, show that this is true for $n=3$:
$(4\cdot3)!=3\cdot11550\cdot(2^{3\cdot3}\cdot3^{3\cdot1})$
Second, assume that this is true for $n$:
$(4n)!=3\cdot{k}\cdot(2^{3n}\cdot3^{n})$
Third, prove that this is true for $n+1$:
$(4n+4)!=$
$\color\red{(4n)!}\cdot(4n+1)\cdot(4n+2)\cdot(4n+3)\cdot(4n+4)=$
$\color\red{3\cdot{k}\cdot(2^{3n}\cdot3^{n})}\cdot(4n+1)\cdot(4n+2)\cdot(4n+3)\cdot(4n+4)=$
$3\cdot{k}\cdot(2^{3n}\cdot3^{n})\cdot8\cdot(4n+3)\cdot(4n+1)\cdot(2n+1)\cdot(n+1)=$
$k\cdot(2^{3(n+1)}\cdot3^{n+1})\cdot(4n+3)\cdot(4n+1)\cdot(2n+1)\cdot(n+1)$
Please note that the assumption is used only in the part marked red.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculating in closed form $\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$ It's not hard to see that for powers like $1,2$, we have a nice closed form. What can be said about
the cubic version, that is
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx \ ?$$
What are your ideas on it? Differentiation under the integral sign? Other ways?
Mathematica 9 says that
$$\int_0^{\pi/2} \arctan\left(\sin ^3(x)\right) \, dx=\frac{2}{3} \, _5F_4\left(\frac{1}{2},\frac{2}{3},1,1,\frac{4}{3};\frac{5}{6},\frac{7}{6},\frac{3}{2},\frac{3}{2};-1\right).$$
| Just exploiting the arctangent Taylor series,
$$\begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \arctan(\sin^3 x)\,dx &=& \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\int_{0}^{\frac{\pi}{2}}\sin^{6m+3}(x)\,dx\\ &=& \frac{3\sqrt{\pi}}{4}\sum_{m\geq 0}\frac{(-1)^m\,\Gamma\left(3m+2\right)}{(3m+\frac{3}{2})\Gamma\left(3m+\frac{5}{2}\right)}\end{eqnarray*} $$
so, at least in principle, we may compute the RHS by applying a discrete Fourier transform to the power series:
$$ \sum_{m\geq 0}\frac{x^m\, \Gamma(m+2)}{\left(m+\frac{3}{2}\right)\Gamma\left(m+\frac{5}{2}\right)}$$
that is just a $\phantom{}_{3} F_2$ hypergeometric function, namely $\frac{8}{9\sqrt{\pi}}\;\phantom{}_3 F_2\!\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x\right).$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Differentiating $y=x^{-3/2}$ I'd like to differentiate $y=x^{\large\frac{-3}{2}}$. So,
$$y + dy = (x + dx)^{\large\frac{-3}{2}} = x^{\large\frac{-3}{2}}\Big(1 + \frac{dx}{x}\Big)^{\large\frac{-3}{2}}$$
is at least a start. How do I calculate the parentheses?
| Try squaring both sides and then eliminating less significant ${dy}^2$, ${dx}^2$, $dxdy$ and ${dx}^3$ etc. kind of terms.
Then you will get it.$$(y+dy)^2 = \frac{1}{(x+dx)^3}$$ $$\implies y^2+{dy}^2+2y\cdot dy = \frac{1}{x^3+{dx}^3+3x^2\cdot dx+3x\cdot {dx}^2}$$
Eliminating less significant higher degrees of $dx$ and $dy$
$$(y^2+2y\cdot dy)(x^3+3x^2\cdot dx) = 1$$$$\implies x^3y^2+2x^3y\cdot dy+3x^2y^2\cdot dx = 1$$
Using $y = x^{-3/2}$ here, it becomes
$$ 1+2x^{3/2}\cdot dy+3x^{-1}\cdot dx=1$$
$$\implies \frac{dy}{dx}=-\frac{3}{2}{x^{-5/2}}$$
Hope that helps.
| {
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Solving $(y'+1)\ln{\frac{x+y}{x+3}} = \frac{x+y}{x+3}$ Hello everyone :) This is another task I'm trying to solve and can't seem to get the same result as Wolfram Alpha.
Solve the following differential equation: $$(y'+1)\ln{\frac{x+y}{x+3}} = \frac{x+y}{x+3}$$
My attempt:
Use substitution $$e^z = \frac{x+y}{x+3}$$
and
$$y=e^z(x+3)-x$$
$$y=xe^z+3e^z-x$$
and
$$y'=e^z+z'xe^z+3z'e^z-1$$
Hence:
$$
(e^z+z'xe^z+3z'e^z)z = e^z
$$
Deviding by $e^z$
$$
z+zz'x+3zz' = 1$$
$$
1+z'x+3z' = \frac{1}{z}
$$
$$
z'(x+3)= \frac{1-z}{z}
$$
$$
\int \frac{z}{1-z}dz = \int \frac{1}{x+3} dx
$$
$$
-\ln{\frac{x+y}{x+3}} - \ln{\bigg|1-\ln{\frac{x+y}{x+3}} \bigg|=\ln{(x+3)}}
$$
EDIT:
After fixing the solution, I got the above solution. But still, this solution does not correspond to that on Wolfram Alpha nor can i find a way to transform them to be equal.
| I managed to solve this differential equation. Following from the last part of the question:
$$
-\ln{\frac{x+y}{x+3}} - \ln{\bigg|1-\ln{\frac{x+y}{x+3}} \bigg|=\ln{(x+3)}} +{C_1}
$$
Next is:
$$
-\bigg(\ln{\frac{x+y}{x+3}}+\ln{\bigg| 1- \ln{\frac{x+y}{x+3}}\bigg|}\bigg) = \ln{{(x+3)}}+\ln{C_2}
$$
$$
-\ln{\frac{x+y}{x+3}\bigg| 1-\ln{\frac{x+y}{x+3}} \bigg| } = \ln{(x+3)} + \ln{C_2}= \ln{(x+3)C_2}
$$
$$
\ln{\frac{x+y}{x+3}\bigg| 1-\ln{\frac{x+y}{x+3}} \bigg| } = \ln{\frac{1}{x+3}}
$$
$$
\frac{x+y}{x+3}\bigg( 1-\ln{\frac{x+y}{x+3}} \bigg) = \frac{C_3}{x+3}
$$
$$
{(x+y)}\bigg( 1-\ln{\frac{x+y}{x+3}} \bigg) = {C_3}
$$
$$
{(x+y)}\bigg( 1-\ln{\frac{x+y}{x+3}} \bigg) = {C_3}
$$
$$
x-x\ln{\frac{x+y}{x+3}}+y-y\ln{\frac{x+y}{x+3}} = -C
$$
$$
x\ln{\frac{x+y}{x+3}}+y \bigg( \ln{\frac{x+y}{x+3}} - 1 \bigg) - x = C
$$
| {
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"url": "https://math.stackexchange.com/questions/1428650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
I tried to solve this question but no luck.
My try:
$$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx=\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx\\
\int x^4(x^8+x^4+1)x^2(2+3x^{-4}+6x^{-8})^{1/4}dx$$
Now i got stuck,please help me reach the answer.Answer is $$\frac{x^5}{30}(2x^8+3x^4+6)^{\frac54}+C$$
| $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$=$\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx$
$\int (x^{11}+x^7+x^3)(2x^{12}+3x^{8}+6x^{4})^{1/4}dx$
Now just put $(2x^{12}+3x^{8}+6x^{4})$=$t$ so that $dt$=24$(x^{11}+x^7+x^3)$$dx$ and hence your integral finally simplifies to
(1/24)$\int(t^{1/4})dx$ which can be simplified to (1/30)$t^{5/4}$+c
Now put back t and you get : $(1/30)$$(2x^{12}+3x^{8}+6x^{4})^{5/4}$+c=
$(1/30)x^5$$(2x^{8}+3x^{4}+6)^{5/4}$+cHope this helps..
| {
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"timestamp": "2023-03-29T00:00:00",
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Check convergence of $\sum_{n=1} \frac{n+1}{n^3+4n}$ I have to check convergence of $\sum_{n=1} \frac{n+1}{n^3+4n}$.
$$\sum_{n=1} \frac{n+1}{n^3+4n}\le \sum_{n=1} \frac{n+1}{n^3}=\sum_{n=1} \frac{1}{n^3}+\sum_{n=1} \frac{1}{n^2}$$
(both series on the right side are obviously convergent)
| Still another way to check convergence:
$$\sum_{n\ge 1}\frac{n+1}{n^3+4n}\le \sum_{n\ge 1}\frac{2}{n^2+4}< 2\sum_{n\ge 1}\frac{1}{n^2}=\frac{\pi^2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find equation for circle related to a triangle write the equation of the circle of the triangle
with vertices $$A = (5 ,\ -4 ),\ B = (6 ,\ -1 ),\ C = ( 2,\ 3)$$
examine the relative position of this district and its image in axial symmetry about the line
$$3x + 4y + 26 = 0$$
I think about this: (drawing paper)
| Substitute $(A,B,C)$ into $(x-a)^2 + (y-b)^2 = R^2$ :
$$
(5-a)^2 + (-4-b)^2 = R^2 \\
(6-a)^2 + (-1-b)^2 = R^2 \\
(2-a)^2 + (3-b)^2 = R^2
$$ $$
a^2 - 10 a + 25 + b^2 + 8 b + 16 = R^2 \\
a^2 - 12 a + 36 + b^2 + 2 b + 1 = R^2 \\
a^2 - 4 a + 4 + b^2 - 6 b + 9 = R^2
$$
Subtract the second equation from the first one and the third equation from the second one:
$$
2 a - 11 + 6 b + 15 = 0 \\
- 8 a + 32 + 6 b - 3 = 0
$$ $$
2 a + 6 b = - 4 \\
- 8 a + 8 b = - 24
$$
Add the second equation to $4 \times$ the first one .. and do the rest:
$$
32 b = - 40 \quad \Longrightarrow \quad b = - 5/4 \\
2\cdot 2 a = - 8 + 15 = 7 \quad \Longrightarrow \quad a = 7/4 \\
R^2 = (2-a)^2 + (3-b)^2 = 145/8
$$
So the circle is:
$$
(x-7/4)^2 + (y+5/4)^2 = \left(\sqrt{145/8}\right)^2
$$
Still don't know what you mean by "district" ..
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
How do I prove the above expression using mathematical induction? So far I have only been proving simpler stuff. The base case P(1) is easy enough but I am lost as to where I should even start with my inductive step. I really don't know what the steps for P(k + 1) should be, and so help would be greatly appreciated.
| If $n$ is even,
$$
\begin{align}
\sum_{k=1}^n(-1)^{k-1}k^2
&=\sum_{k=1}^{n/2}\left[(2k-1)^2-(2k)^2\right]\\
&=\sum_{k=1}^{n/2}\left[1-4k\right]\\
&=-\frac{n^2+n}2
\end{align}
$$
If $n$ is odd, $n-1$ is even.
$$
\begin{align}
\sum_{k=1}^n(-1)^{k-1}k^2
&=n^2+\sum_{k=1}^{n-1}(-1)^{k-1}k^2\\
&=n^2-\frac{n^2-n}2\\
&=\frac{n^2+n}2
\end{align}
$$
Therefore,
$$
\sum_{k=1}^n(-1)^{k-1}k^2=(-1)^{n+1}\frac{n^2+n}2
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)}$? How can I evaluate this?
$$\sum_{n=1}^\infty \frac{1}{n(n+2)(n+4)} = \frac{1}{1\cdot3\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{3\cdot5\cdot7}+ \frac{1}{4\cdot6\cdot8}+\cdots$$
I have tried:
$$\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{2\cdot4\cdot6}+ \frac{1}{4\cdot6\cdot8}+\cdots
= \frac{1}{3\cdot5}\left(1+\frac{1}{7}\right)+\frac{1}{4\cdot6}\left(\frac{1}{2}+\frac{1}{8}\right)+\cdots$$
and so on... Been stuck for a while. Result should be $\dfrac{11}{96}$
| $$ \frac{1}{n(n+2)(n+4)}=\frac{1}{8}(\frac{1}{n}-\frac{1}{n+2})-\frac{1}{8}(\frac{1}{(n+2}-\frac{1}{n+4})$$
we know
$$\sum_{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+\sum_{n=1}^{\infty }\frac{1}{n+2}$$
$$\sum_{n=1}^{\infty }\frac{1}{n+2}=\frac{1}{3}+\frac{1}{4}+\sum_{n=1}^{\infty }\frac{1}{n+4}$$
$$\frac{1}{8}(\frac{1}{1}+\frac{1}{2})-\frac{1}{8}(\frac{1}{3}+\frac{1}{4})=\frac{11}{96}$$
| {
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"url": "https://math.stackexchange.com/questions/1433583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
| Hint The radical expressions that occur, $\sqrt{x}$ and $\sqrt[3]{x}$, are both integer powers of $$u := \sqrt[6]{x},$$ and so we can write the expression in the limit as a rational expression in $u$:
$$\frac{\sqrt[3]{x} - 1}{2(\sqrt{x} - 1)} = \frac{(\sqrt[6]{x})^2 - 1}{2 ((\sqrt[6]{x})^3 - 1)} = \frac{u^2 - 1}{2(u^3 - 1)}.$$ (We've thus made a rationalizing substitution.) Can you simplify this?
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Here is what I have:
$a+b =b+a$ since $R$ is commutative.
Now,
$$(b+a) \cdot b^{-1} = 1+ab^{-1} \\
a^{-1} \cdot (1+ab^{-1}) = a^{-1} + b^{-1}
$$
Thus, $a^{-1} + b^{-1} =a^{-1} \cdot (a+b) \cdot b^{-1}$
Therefore, $(a^{-1} + b^{-1})^{-1} = b \cdot (a+b)^{-1} \cdot a$
And thus, $a^{-1} + b^{-1}$ is a unit in $R$ as well.
Does my answer sound logical. Or are there errors in it?
| Your answer works perfectly fine. In fact, your answer works in non-commutative rings as well.
When we say that a ring is commutative, we mean $ab = ba$. We will have $a + b = b + a$ is any ring.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find all pairs of prime numbers $p , q$ for which: $p^2 | q^3 + 1$ and $q^2 | p^6 − 1$.
Find all pairs of prime numbers $p , q$ for which:
$$p^2 \mid q^3 + 1 \tag{A}$$
and
$$q^2 \mid p^6 − 1 \tag{B}$$
The question is from the Bulgaria National Olympiad 2014.
I'm looking for any solution I may have missed, and generally any alternative method that might reduce the case work (could combine cases 2.1 and 3.1, I suppose).
I will split the work into three cases:
*
*$p=q\ge2$.
*$p>q\ge2$.
*$q>p\ge2$.
Case 1
It is clear that neither (A) nor (B) are met.
Case 2
First consider two subcases:
*
*$p>q$ and $q\in\{2,3\}$.
*$p>q\ge5$.
Case 2.1
$$\begin{align}
q=2 &\implies p^2\mid9 &\implies p^2=9 &\implies p=3 \\
q=3 &\implies p^2\mid28 &\implies p^2=4 &\implies\text{ no solution} \\
\end{align}$$
So $\boxed{(p,q)=(3,2)}$ is the only solution for this case.
Case 2.2
(A) factorises as $p^2 \mid (q+1)(q^2-q+1)$. Now
$$q^2-q+1=(q+1)(q-2)+3 \implies \gcd(q+1,q^2-q+1)=
\begin{cases} 3,\quad\text{if }3\mid q+1\\1,\quad\text{otherwise}\end{cases}$$
Since $p>5$ is prime, we must have either $p^2\mid q+1$ or $p^2\mid q^2-q+1$. But this is impossible because $p>q\ge5 \implies p^2>q+1\text{ and }p^2>q^2>q^2-q+1$. So there are no solutions here.
Case 3
Consider two subcases:
*
*$q>p$ and $p\in\{2,3\}$.
*$q>p\ge5$.
Case 3.1
$$\begin{align}
p=2 &\implies q^2\mid63 &\implies q^2=9 &\implies q=3 \\
p=3 &\implies q^2\mid728 &\implies q^2=4 &\implies\text{ no solution} \\
\end{align}$$
So $\boxed{(p,q)=(2,3)}$ is the only solution for this case.
Case 3.2
(B) factorises as $q^2 \mid (p^3+1)(p^3-1)$. Now
$$p^3+1=(p^3-1)+2 \implies \gcd(p^3+1,p^3-1)=
\begin{cases} 2,\quad\text{if }p\text{ is odd}\\1,\quad\text{otherwise}\end{cases}$$
Since $q>5$ is prime, we must have either $q^2\mid p^3+1$ or $q^2\mid p^3-1$. If $q^2\mid p^3+1$, the the same arguments as in case 2.2 can be applied to show that $q^2\mid p+1$ or $q^2\mid p^2-p+1$ neither of which is possible when $q>p$.
So the only remaining possibility is $q^2\mid p^3-1$. This factorises as $q^2\mid (p-1)(p^2+p+1)$. Now
$$p^2+p+1=(p-1)(p+2)+3 \implies \gcd(p-1,p^2+p+1)=
\begin{cases} 3,\quad\text{if }3\mid p-1\\1,\quad\text{otherwise}\end{cases}$$
Since $q>5$ is prime, we must have either $q^2\mid p-1$ or $q^2\mid p^2+p+1$. But this is impossible because $q>p\ge5 \implies q^2>p-1\text{ and }q^2\ge (p+1)^2=p^2+2p+1>p^2+p+1$. So there are no solutions here either.
| The insight to simplify is that $p$ and $q$ must have different parities; they can't both be odd. At least one of them must be odd, because there are not two even primes. Then the other one must be even, because the addition or subtraction of $1$ makes one of the compound expressions even. Hence one of them must be 2. If we choose $p=2$, then $p^6-1=63$ and $q=3$. This checks out because $4\mid28$. If $q=2$, then $q^3+1=9$ and $p=3$. This checks out because $4\mid728$. $(p,q)=(2,3)$ in either order.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluation of $ \lim_{n\rightarrow \infty}\left[\prod^{n}_{r=1}\left(1+\frac{n}{r}\right)^{\frac{r}{n}}\right]^{\frac{1}{n}}$ Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\left[(1+n)^{\frac{1}{n}}\cdot \left(1+\frac{n}{2}\right)^{\frac{2}{n}}\cdot \left(1+\frac{n}{3}\right)^{\frac{3}{n}}.........2\right]^{\frac{1}{n}}$
$\bf{My\; Try::}$ Let $$ y = \lim_{n\rightarrow \infty}\left[(1+n)^{\frac{1}{n}}\cdot \left(1+\frac{n}{2}\right)^{\frac{2}{n}}\cdot \left(1+\frac{n}{3}\right)^{\frac{3}{n}}.........2\right]^{\frac{1}{n}}$$
Now taking $$ \ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\cdot \ln\left[(1+n)^{\frac{1}{n}}\cdot \left(1+\frac{n}{2}\right)^{\frac{2}{n}}\cdot \left(1+\frac{n}{3}\right)^{\frac{3}{n}}.........2\right]$$
So we get $$\ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\left[\frac{1}{n}\cdot \ln(1+n)+\frac{2}{n}\ln\left(1+\frac{n}{2}\right)+........+\frac{n}{n}\ln\left(1+\frac{n}{n}\right)\right]$$
So $$\ln y = \lim_{n\rightarrow \infty}\frac{1}{n}\sum^{n}_{r=1}\frac{r}{n}\ln\left(1+\frac{n}{r}\right)$$
Now Convertinto Reinmann Sum of Integral
So put $\displaystyle \frac{r}{n} = x\;,$ Then $\displaystyle \frac{1}{n}=dx$ and Calculate limit
So we get $$\ln y = \int_{0}^{1}x\cdot \ln\left(1+\frac{1}{x}\right)dx=\int_{0}^{1}x\cdot \left[\ln(1+x)-\ln x\right]dx$$
So we get $$\ln y = \int_{0}^{1}x\cdot \ln(x+1)dx-\int_{0}^{1}x\ln xdx$$
Now after Integrate we wil get $$\ln y = \frac{1}{2}\ln 2+\frac{1}{4}-\frac{1}{2}\ln 2+\frac{1}{4} = \frac{1}{2}$$
So we get $$\ln y = \frac{1}{2}\Rightarrow y = e^{\frac{1}{2}}=\sqrt{e}$$
Can we solve it any Shorter way, If yes then plz explain here
Thanks
| We could use Stolz theorem, which states:
$$\lim_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\ell\implies
\lim_{n \to \infty} \frac{a_n}{b_n}=\ell$$
Now we define $a_n$ and $b_n$ as follows:
$$a_n={\sum^{n}_{r=1}r\ln\left(1+\frac{n}{r}\right)} \quad b_n=n^2$$
So that:
$$
\ln y = \lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^{n}_{r=1} r\ln\left(1+\frac{n}{r}\right)\\~\\={\lim_{n\rightarrow \infty}\frac{1}{2n+1}\left[\sum^{n+1}_{r=1} r\ln\left(1+\frac{n+1}{r}\right)- \sum^{n}_{r=1} r\ln\left(1+\frac{n}{r}\right)\right]}\\~\\ \geq \lim_{n\rightarrow \infty}\frac{1}{2n+1}\left[ (n+1)\ln\left(1+\frac{n}{n+1}\right)\right]\\~\\={\lim_{n\rightarrow \infty}\frac{n}{2n+1}\ln\left(1+\frac{n}{n+1}\right)^{\frac{n+1}{n}}}\\~\\=\frac{1}{2}*ln(e)=\frac {1}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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An interview question: $2.1^{3.1}$ vs $3.1^{2.1}$,$ 2.1^{4.1}$ vs $4.1^{2.1}$, which is larger? While Mathematica told me that $2.1^{3.1} - 3.1^{2.1} = -0.786932$ and $2.1^{4.1} - 4.1^{2.1} = 1.58855$, I wonder how to compare them quickly, by hand.
I see $2^3 < 3^2$, so perhaps we have $2.1^{3.1} < 3.1^{2.1}$, but I'm not sure.
As for $2^4 = 4^2$, I guess $2.1^{4.1}$ is larger because the exponent is bigger, but I'm not too sure.
| Let's consider
$$(a+\epsilon)^{b+\epsilon} = e^{(b+\epsilon) \log{(a+\epsilon)}}$$
$$\begin{align}(b+\epsilon) \log{(a+\epsilon)} &= (b+\epsilon) \left [\log{a} + \log{\left (1+\frac{\epsilon}{a} \right )} \right ] \\ &= (b+\epsilon) \left [\log{a} +\frac{\epsilon}{a} - \frac12 \frac{\epsilon^2}{a^2} +O\left (\epsilon^3 \right ) \right ]\\ &= b \log{a} + \left (\frac{b}{a}+\log{a} \right )\epsilon + \left (\frac1{a}-\frac{b}{2 a^2} \right )\epsilon^2 + O\left (\epsilon^3 \right )\end{align}$$
Exponentiating, we get that
$$(a+\epsilon)^{b+\epsilon} = a^b \left [1+ \left (\frac{b}{a}+\log{a} \right )\epsilon + \left (\frac1{a}-\frac{b}{2 a^2}+ \frac12 \left ( \frac{b}{a}+\log{a} \right )^2 \right ) \epsilon^2 \right ]+ O\left (\epsilon^3 \right )$$
For the purposes of the interview, we only need look at the $O(\epsilon)$ term. Thus, when $a=2, b=3$ and $\epsilon=0.1$, we get that $2.1^{3.1} \approx 8 + 8 (1.5+\log{2}) (0.1) \approx 9.8$. In contrast, when $a=3, b=2$ and $\epsilon=0.1$, we get that $3.1^{2.1} \approx 9 + 9 (0.7+\log{3}) (0.1) \approx 10.6$. The difference is sufficiently greater than $a^b \epsilon^2$ that we don't need to consider the $O(\epsilon^2)$ term. Thus, we can say that $3.2^{2.1} \gt 2.1^{3.1}$.
The OP should be able to continue the analysis for the other case, which seems more interesting.
| {
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"timestamp": "2023-03-29T00:00:00",
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Stronger than Nesbitt inequality
For $x,y,z >0$, prove that
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \sqrt{\frac94+\frac32 \cdot \frac{(y-z)^2}{xy+yz+zx}}$$
Observation:
*
*This inequality is stronger than the famous Nesbitt's Inequality
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32 $$ for positive $x,y,z$
*We have three variables but the symmetry holds only for two variables $y,z$, resulting in a very difficult inequality. Brute force and Largrange Multiplier are too complicated.
*The constant $\frac32$ is closed to the best constant. Thus, this inequality is very sharp, simple AM-GM estimation did not work.
Update: As point out by Michael Rozenberg, this inequality is still unsolved
| Abstract :
We suggests a way to show the OP inequality so some part are left to the reader because there are easy . We spilt the problem in two cases , first case $2c-2b\le a$ and the second $2c-2b\ge a$ . As the problem is homogenous we suppose in addition $1\ge a\ge c\ge b$.Following that we use convexity and derivatives.
First case : $1\ge a\ge c\ge b$ and $2c-2b\le a$
we have :
$$\sum_{cyc}\frac{a}{b+c}\geq 1.5+0.5\frac{(a-b)^2}{ab+bc+ca}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}}$$
Or:
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)-\frac{\left(a-b\right)^{2}}{2}\geq 0\quad (C)$$
SubCase : $a\ge 2c\ge c\ge b$
$$\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)\geq \frac{2(a^2+b^2+c^2-ab-bc-ca)}{3}\geq\frac{\left(a-b\right)^{2}}{2}\quad $$
Wich is trivial using the $uvw's$ method and the assumptions on $a,b,c>0$
Subcase: $2c\geq a \geq c\geq b$ and $c+b\leq a $ and $a\geq 2c-2b$ remarking that :
$$f(c)=\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5\right)\left(ab+bc+ca\right)$$
Is a decreasing function always with the assumptions above .Remains to show the cases $a=c+b$ or $a=2c-2b$ wich is a one variable inequality since all the equalities/inequalities are homogenous.I shall prove that the function is decreasing later .
To prove that the function is decreasing we differentiate twice we have :
$$f''(c)=-\frac{2a^2b}{(a+c)^3}-\frac{2ab^2}{(b+c)^3 }+2$$
With the assumptions above the function $f(c)$ is convex so the derivative is increasing .Remains to show the inequality at the equality case wich is not hard .
Second case: $1\ge a\geq c\geq b>0$ and $2c-2b\ge a$
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge 1.5+\frac{\left(\sqrt{\frac{15}{4}}-1.5\right)\left(\left|a-b\right|\right)^{1.75}}{\left(ab+bc+ca\right)^{\frac{1.75}{2}}}\geq \sqrt{\frac{9}{4}+1.5\frac{(a-b)^2}{ab+bc+ca}} $$
To prove it we remark :
$$f(c)=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-1.5$$
Is increasing with $a\leq 2c-2b$ as $f(c)$ is a convex function we deduce that the first derivative is increasing . Remains to replace by the constraint $2c-2b=a$
$$g(c)=\left(ab+bc+ca\right)^{\frac{1.75}{2}}$$
Is increasing .
So the product of two positives increasing functions is also an increasing function always with constraints above .
It remains to show the case $a=2c-2b$ . As it's homogenous we get an inequality with one variable or a long polynomial .The RHS is trivial .
Last Edit 09/03/2022 :
We have the inequality :
Let $0\leq x\leq 1$ then we have :
$$\sqrt{\frac{9}{4}+1.5x}\leq g(x)=\left(2-a\right)\frac{1}{2}\sqrt{\frac{3}{5}}\left(ax-a\right)+\sqrt{\frac{9}{4}+1.5}+\frac{-\sqrt{\frac{3}{5}}}{20}\left(a^{-1}x-a^{-1}\right)^{2}$$
Where $a\geq 0$ is chosen as :
$$g(0)=1.5$$
We can do really better as it seems we have for $0\leq x\leq 1$ :
$$\left(2-a^{1.0735}\right)\frac{1}{2}\sqrt{\frac{3}{5}}\left(ax-a\right)+\sqrt{\frac{9}{4}+1.5}+\frac{-\sqrt{\frac{3}{5}}}{20}\left(a^{-1}x-a^{-1}\right)^{2}\geq \sqrt{\frac{9}{4}+1.5x}$$
Where $a\simeq 0.7748951014$ is chosen as :
$$g(0)=1.5$$
| {
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"url": "https://math.stackexchange.com/questions/1444352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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Is there another way to solve this quadratic equation? $$\frac { 4 }{ x^{ 2 }-2x+1 } +\frac { 7 }{ x^{ 2 }-2x+4 } =2$$
Steps I took:
$$\frac { 4(x^{ 2 }-2x+4) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } +\frac { 7(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } =\frac { 2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1) }{ (x^{ 2 }-2x+4)(x^{ 2 }-2x+1) } $$
$$4(x^{ 2 }-2x+4)+7(x^{ 2 }-2x+1)=2(x^{ 2 }-2x+4)(x^{ 2 }-2x+1)$$
$$11x^{ 2 }-22x+23=2x^{ 4 }-8x^{ 3 }+18x^{ 2 }-20x+8$$
I can keep going with all the steps I took, but is there a more elegant way to arrive at the solution for this equation? It seems as if I keep going the way I am, I will hit a dead end. No actual solution, please. Hints are much better appreciated.
| By inspection, the two denominators differ by $3$, just as the numerators do, hinting that to establish
$$\frac44+\frac77=2,$$ we may set
$$x^2-2x+1=(x-1)^2=4,$$ i.e. $$\color{green}{x=-1\lor x=3}.$$
The other two roots are a little more elusive.
But we can observe that when expanding
$$\frac4z+\frac7{z+3}-2=0,$$ we will get terms $-2z^2$ and $12$, so that the product of the $z$ roots is $-6$, and
$$(x-1)^2=-\frac32,$$ i.e. $$\color{green}{x=1-i\sqrt{\frac32}\lor x=1+i\sqrt{\frac32}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I solve $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1})$ indeterminate limit without the L'hospital rule? I've been trying to solve this limit without L'Hospital's rule as homework. So I tried rationalizing the denominator and numerator but it didn't work.
My best was: $\lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{x^3-1}) = \lim_{x \to 1} (\frac{1}{x-1}-\frac{2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{(x-1)(x+2)}{(x-1)(x^2+x+1)}) = \lim_{x \to 1} (\frac{x+2}{(x^2+x+1)}) = 3 ???$
| Picking up from your second step...
$$\lim_{x\to1}\frac{x^2+x+1-2}{(x-1)(x^2+x+1)}$$
$$=\lim_{x\to1}\frac{x^2+x-1}{(x^3-1)}$$
Let $x\to1$ and find the limit is unbounded or does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Solving without L'Hopital
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
That's
$$\lim_{x\to-\infty}\left(\sqrt{4x^2-6}\right)-\lim_{x\to-\infty}\left(\sqrt{4x^2+x}\right)$$
I have been taught to get the highest exponents, so...
$$\sqrt{4x^2}-\sqrt{4x^2}$$
It's the same for both sides, no?
$$2x-2x$$
$$-\infty+\infty$$
Which is wrong. The correct answer is
$$\frac{1}{4}$$
Why? I always just grab the highest exponent ($\sqrt{4x^2}$) and work with it. But this time it didn't go well.
| Notice, $$\lim_{x\to -\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$$
$$=\lim_{x\to \infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)$$
$$=\lim_{x\to \infty}\frac{\left(\sqrt{4x^2-6}-\sqrt{4x^2-x}\right)\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$
$$=\lim_{x\to \infty}\frac{4x^2-6-4x^2+x}{\left(\sqrt{4x^2-6}+\sqrt{4x^2-x}\right)}$$
$$=\lim_{x\to \infty}\frac{x-6}{x\left(\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}\right)}$$
$$=\lim_{x\to \infty}\frac{1-\frac{6}{x}}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4-\frac{1}{x}}}$$
$$=\frac{1-0}{\sqrt{4-0}+\sqrt{4-0}}=\frac{1}{2+2}=\color{red}{\frac{1}{4}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
| If we divide both sides by $x$ we get $$\frac{x^2-3x+1}{x} = x-3+\frac1x = 0\implies x+\frac1x = 3$$Squaring both sides$$\left(x+\frac1x\right)^2 = x^2+2+\frac1{x^2} = 9\implies x^2+\frac1{x^2}=7$$Squaring again$$\left(x+\frac1x\right)^4 = \left(x^2+\frac1{x^2}\right)^2 = x^4+2+\frac1{x^4} = 49\implies x^4+\frac{1}{x^4}=\boxed{47}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate radius of the surface of water at different points in time when it is poured into a spherical container? So I have a sphere with radius $1\ cm$, and I'm pouring in water at $0.5\ cm^3/s$.
How would I find a the radius of the surface of the water at any given time? So in the end this would look like a quadratic equation relating time and radius, where the maxima would be where radius $= 1\ cm$. How would I go about using differentiation in solving for this equation?
| Notice, let $r$ be the radius of the circular surface of water at any time $t$ then the volume of water poured (in a sphere of radius $R=1\ cm$ ) at the same time $t$ is given by using geometry $$\color{red}{V=\frac{\pi}{3}\left(2R^3-(2R^2+r^2)\sqrt{R^2-r^2}\right)}$$
Differentiating w.r.t. time $t$, we get
$$\frac{dV}{dt}=\frac{\pi}{3}\left(\frac{(2R^2+r^2)r}{\sqrt{R^2-r^2}}-2r\sqrt{R^2-r^2}\right)\frac{dr}{dt}$$
$$\frac{dV}{dt}=\frac{\pi r^3}{\sqrt{R^2-r^2}}\frac{dr}{dt}$$$$\implies \frac{dr}{dt}=\frac{\sqrt{R^2-r^2}}{\pi r^3}\frac{dV}{dt}$$
Now, setting the values $R=1\ cm$ & constant rate of pouring water $\frac{dV}{dt}=0.5\ cm^3/s$, we get
$$\frac{dr}{dt}=\frac{\sqrt{1-r^2}}{\pi r^3}(0.5)$$
$$\frac{2\pi r^3\ dr}{\sqrt{1-r^2}}=dt$$$$\implies \int \frac{2\pi r^3\ dr}{\sqrt{1-r^2}}=\int dt$$
$$\frac{-2\pi(2+r^2)\sqrt{1-r^2}}{3}=t+C$$
$\color{blue}{\text{Condition}}$: At initial time $t=0$, the radius of water surface $r=0$, hence, we get
$$\frac{-2\pi(2+0)\sqrt{1-0}}{3}=0+C\implies C=\frac{-4\pi}{3}$$
Now, setting the value of $C$, we get
$$\frac{-2\pi(2+r^2)\sqrt{1-r^2}}{3}=t-\frac{4\pi}{3}$$
$$\frac{2\pi(2+r^2)\sqrt{1-r^2}}{3}=\frac{4\pi}{3}-t$$
$$\color{red}{(1-r^2)(2+r^2)^2=\left(\frac{4\pi-3t}{2\pi}\right)^2}$$
Comment :
Setting maximum value of radius, $r=1\ cm$, we get
$$t=\frac{4\pi}{3}$$
We conclude that the maximum radius $\color{red}{r=1\ cm}$ of water surface will occur at the time $\color{red}{t=\frac{4\pi}{3} \ sec}$
| {
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Prove that the probability that an article chosen at random from $C$ is defective,is $\frac{8}{25}.$ Lot $A$ consists of 3 good and 2 defective articles.Lot $B$ consists of 4 good and 1 defective article.A new lot $C$ is formed by taking 3 articles from $A$ and $2$ from $B$.Prove that the probability that an article chosen at random from $C$ is defective,is $\frac{8}{25}.$
Required probabiliy$=$probability that either $A$ is chosen and then defective item is chosen from it or $B$ is chosen and then defective item is chosen from it.
$=\frac{1}{2}\times \frac{2}{5}+\frac{1}{2}\times \frac{1}{5}=\frac{3}{10}$
but my answer is wrong,means my logic is also wrong.What should be the correct logic to solve this problem.Please help me.Thanks.
| Let $G$ denote the event that the item is good.
Let $A$ denote the event that the item came from lot A, and likewise for $B$.
Then $P(G) = P(G | A)P(A) + P(G | B)P(B)$.
Of course $P(A) = \frac{3}{5}, P(B) = \frac{2}{5}$ by the way lot C was formed.
$P(G|A)$ is the probability that an item is good, given that it came from $A$. Well, we could have taken 3 good items, which would have happened with probability ${{3 \choose 3}{2 \choose 0} \over {5 \choose 3}} = \frac{3}{5} \frac{2}{4} \frac{1}{3} = \frac{1}{10}$, and then all of these would have been OK. Or we could have taken 2 good items, with probability ${{3 \choose 2}{2 \choose 1}} \over { 5 \choose 3}$, and then we'd have a $\frac{2}{3}$ chance in that case. Or we could have picked 1 good one, with probability ${{3 \choose 1}{2 \choose 2}} \over { 5 \choose 3}$ and then we'd have had a chance of $\frac{1}{3}$.
So $$P(G | A) = \frac{1}{10} \cdot 1 + {{{3 \choose 2}{2 \choose 1}} \over { 5 \choose 3}} \cdot \frac{2}{3} + {{{3 \choose 1}{2 \choose 2}} \over { 5 \choose 3}} \cdot \frac{1}{3}$$
We can compute $P(G | B)$ in a similar (somewhat simpler) way. Then compute $P(G)$ by the first formula.
| {
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Find all primes $p$ and $q$ such that $p^2-2q^2=1.$
Find all primes $p$ and $q$ such that $p^2-2q^2=1.$
My idea so far was to observe that since $2q^2$ is even, then $q^2$ must be odd or even. If $q^2$ is even, then $q$ is even and the only even prime is $2.$ Thus one pair of primes $(p,q)=(3,2).$
But, if $q^2$ is odd, then $q$ is also odd and $q=2d+1$ for some integer $d.$ Then,
$p^2-2q^2=1$ turns into $p^2=4(2d^2+2d)+3.$
From here I do not know how to proceed.
Any ideas or suggestions would be greatly appreciated.
| $p^2-2q^2\equiv p^2+q^2\equiv 1\pmod{3}$.
But $a^2\equiv \{0,1\}\pmod{3}$ for all $a\in\Bbb Z$, so exactly one of $p,q$ is equal to $3$.
$q=3$ gives $p^2=19$, impossible. $p=3$ gives $(p,q)=(3,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1461789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
} |
Find inverse $z$-transform of $\frac{5}{z^{2}-z-6}$ How can I find inverse z transform of $$X(z)=\frac{5}{z^{2}-z-6}$$
What I did:
first i factored denominator and i got (z+2)(z-3), now we get A(-2^{n}) + b(3^{n}). To get A and B i used Partial Fraction Decomposition and got A=-1 and B=1.
But Wolframalpha gave me another answer! $$=\frac{1}{6}(3(-2^{n}+2*3^{n}))$$ what i did wrong?
| Doing partial fractions as per usual,
$$\frac{5}{(z+2)(z-3)}= \frac{A}{z+2} + \frac{B}{z-3} \implies 5 = A(z-3)+ B(z+2).$$
Equating coefficients, we see that
$$A+B=0 \mbox{ and } 2B-3A=5.$$
This linear system has the unique solution $A=-1$, $B=1$. Therefore, we see that
$$\frac{5}{(z+2)(z-3)}=\frac{1}{z-3} - \frac{1}{z+2}.$$
So the poles of $X(z)$ are encircled, choose $C$ to be $\{z\in\mathbb{C}:|z|=4\}$ with standard orientation.
Therefore, by definition, the inverse $Z$-transform of $X(z)$ is
\begin{align*}
\frac{1}{2\pi j} \int_C X(z)z^{n-1} dz &= \frac{1}{2\pi j} \int_C \left(\frac{1}{z-3} - \frac{1}{z+2}\right)z^{n-1} dz \\
&= \frac{1}{2\pi j} \left(\int_C \frac{z^{n-1}}{z-3}dz - \int_C\frac{z^{n-1}}{z+2} dz\right). \\
\end{align*}
We then apply Cauchy's Integral Formula to get
$$\frac{1}{2\pi j} \left(\int_C \frac{z^{n-1}}{z-3}dz - \int_C\frac{z^{n-1}}{z+2} dz\right)=3^{n-1} - (-2)^{n-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Issue with trigonometry identity related to condition number of matrix So, in attempting to compute the condition number for the 2-norm of a matrix, I have stumbled upon a problem i can't resolve.
I have the formula
$$
\frac{1-\cos\left(\frac{n}{n+1} \pi\right)}{1-\cos\left(\frac{1}{n+1} \pi\right)}.
$$
I arrived at this because the eigenvalues for my matrix are $2\left(1-\cos\left(\frac{p}{n+1} \pi\right)\right)$ for $p=1,..,n$.
The problem comes from the fact that I need to show that this formula somehow ends up being equal to
$$\frac{1}{\tan^2\left(\frac{1}{2(n+1)}\pi\right)}
$$
I've arrived at
$$
\frac{1-\cos\left(\pi\frac{n}{n+1}\right)}{2\sin^2\left(\frac{\pi}{2(n+1)}\right)}
$$
but can't seem to get any further, mostly due to the $n$ in the numerator of the cosine.
Any help would be appreciated.
| You have the right idea in exploiting the identity
$$1 - \cos\theta \;=\; 2 \sin^2 \frac{\theta}{2}$$
to re-write the denominator. Using that on the numerator gives
$$1 - \cos\left( \pi \frac{n}{n+1} \right) = 2\sin^2\frac{\pi n}{2(n+1)} \tag{$\star$}$$
Now, simply observe that the arguments of the sines combine very conveniently:
$$\frac{\pi n}{2(n+1)} + \frac{\pi}{2(n+1)}= \frac{\pi(n+1)}{2(n+1)} = \frac{\pi}{2}$$
so that $(\star)$ becomes
$$2\sin^2\left( \frac{\pi}{2} - \frac{\pi}{2(n+1)} \right) = 2\cos^2\left(\frac{\pi}{2(n+1)}\right)$$
since $\sin(\pi/2-\theta) = \cos\theta$. The result follows. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}$ Can you tell me if it is possible to solve $$\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}\ ?$$
wolfram doesn't say it is not computable, but says there are no real solutions, whereas we know the solution is $x=1$
Any hints on how to proceed?
Thanks
| Let $\arctan\sqrt{\dfrac{10-x}x}=y\ \ \ \ (1)$
$\implies 0\le y\le\dfrac\pi2$ as $\sqrt{\dfrac{10-x}x}\ge0$
$\dfrac{10-x}x=\tan^2y\iff x=10\cos^2y$
$\implies\sqrt{\dfrac{10-x}{10}}=+\sin y$ as $0\le y\le\dfrac\pi2$
$\implies\arccos\sqrt{\dfrac{10-x}{10}}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)$
$\implies\arccos\sqrt{\dfrac{10-x}{10}}=\dfrac\pi2-y\ \ \ \ (2)$ as $0\le y\le\dfrac\pi2$
Use $(1),(2)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
if $\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$ find $w$ range let $w$ is postive integer,if there exist $a,b,c(\pi\le a<b<c\le 2\pi)$such
$$\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$$
Find the $w$ range.
My attempt：
since
$$\sin{(wa)}\le 1,\sin{(wb)}\le 1,\sin{(wc)}\le 1$$
then $$\sin{(wa)}=\sin{(wb)}=\sin{(wc)}=1$$
$$aw=\dfrac{\pi}{2}+2k_{1}\pi,k_{1}\in Z$$
$$bw=\dfrac{\pi}{2}+2k_{2}\pi,k_{2}\in Z$$
$$cw=\dfrac{\pi}{2}+2k_{3}\pi,k_{3}\in Z$$
what approaches do you think, I could take to solving the next step?
| We have
$$\begin{align}&\sin(aw)+\sin(bw)+\sin(cw)=3\\\\&\iff \sin(aw)=\sin(bw)=\sin(cw)=1\\\\&\small\iff\text{There exist three integers $k,l,m$ such that $aw=\frac{\pi}{2}+2k\pi,bw=\frac{\pi}{2}+2l\pi,cw=\frac{\pi}{2}+2m\pi$}\\\\&\small\iff\text{There exist three integers $k,l,m$ such that $\pi\le\frac{\pi+4k\pi}{2w}\lt\frac{\pi+4l\pi}{2w}\lt\frac{\pi+4m\pi}{2w}\le 2\pi$}\\\\&\small\iff \text{There exist three integers $k,l,m$ such that $2\le\frac{1+4k}{w}\lt\frac{1+4l}{w}\lt\frac{1+4m}{w}\le 4$}\end{align}$$
So, let us find a condition for $w$ such that there exist at least three distinct positive integers $n$ satisfying $2\le\frac{1+4n}{w}\le 4$, i.e. $\frac{2w-1}{4}\le n\le \frac{4w-1}{4}$.
For $w=2N$ where $N$ is a positive integer, the condition is
$$\left\lfloor\frac{4w-1}{4}\right\rfloor-\left\lceil\frac{2w-1}{4}\right\rceil+1\ge 3\iff (2N-1)-N+1\ge 3\iff N\ge 3.$$
For $w=2N-1$ where $N$ is a positive integer, the condition is
$$\left\lfloor\frac{4w-1}{4}\right\rfloor-\left\lceil\frac{2w-1}{4}\right\rceil+1\ge 3\iff (2N-2)-N+1\ge 3\iff N\ge 4.$$
Hence, the answer is $\color{red}{w\ge 6}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Limit in complex space How to prove that the limit in the complex space:
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=1$$
Please help me
| $$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{(x,y) \to (0,0)} \frac{\sqrt{(x+1)^2+y^2}-1}{x+iy}$$
if $x=y$
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{x\to 0} \frac{\sqrt{(x+1)^2+x^2}-1}{x+ix}\times \frac{\sqrt{(x+1)^2+x^2}+1}{\sqrt{(x+1)^2+x^2}+1}=\lim\limits_{x\to 0} \frac{{(x+1)^2+x^2}-1}{{(x+ix)(\sqrt{(x+1)^2+x^2}+1)}}=\lim\limits_{x\to 0} \frac{x(2x+2)}{x(1+i)\sqrt{(x+1)^2+x^2}+1}=\frac{2}{(1+i)\sqrt{(0+1)^2+0^2}+1}=\frac{1}{(1+i)} $$
similarly for $y=-x$
$$\lim\limits_{(x,y) \to (0,0)} \frac{\sqrt{(x+1)^2+y^2}-1}{x+iy} =\frac{1}{(1-i)}$$
hence the limit doesn't exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help showing $\lim\limits _{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right) = \frac{5}{4}$ I am stuck solving the following limit. I know the answer is 5/4, I just can't get it. This is the steps I have done so far.
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)$
Multiply by Conjugate
$\lim _ \limits{x\to -\infty \:}\left(\sqrt{4\cdot \:x^2-5\cdot \:x}+2\cdot \:x\right)\cdot \frac{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Multiply Out
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-5\cdot \:\:\:x-4\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Combine Like Terms
$\lim\limits_{x\to -\infty \:}\cdot \frac{\left(4\cdot \:\:\:x^2-9\cdot \:\:x\right)}{\left(\sqrt{4\cdot \:\:x^2-5\cdot \:\:x}-2\cdot \:\:x\right)}$
Factor out x
$\lim\limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{\left(\sqrt{x^2\left(4-\frac{5}{x}\right)}-2\cdot \:\:x\right)}$
Pull out x of sqrt and factor again
$\lim \limits_{x\to -\infty \:}\cdot \frac{x\left(4\cdot \:x-9\right)}{x\left(\sqrt{\left(4-\frac{5}{x}\right)}-2\right)}$
Now cancel x terms
$\lim \limits_{x\to -\infty \:}\frac{4\cdot \:\:x-9}{\sqrt{\left(4-\frac{5}{x}\right)}-2}$
Now I don't know what to do next.
If I plug in I get
$\frac{4\cdot \:\:\:-\infty \:-9}{\sqrt{\left(4-0\right)}-2}=\:\frac{-\infty \:}{0}$ Which doesn't equal $\frac{5}{4}$?
| \begin{align*}
\lim_{x\to-\infty}\sqrt{4x^2-5x}+2x&=\lim_{x\to-\infty}-2x\sqrt{1-\frac{5}{4x}}+2x\\
&=\lim_{x\to-\infty}2x\left(1-\sqrt{1-\frac{5}{4x}}\right)\\
&=\lim_{x\to-\infty}2x\cdot \frac{\frac5{4x}}{1+\sqrt{1-\frac{5}{4x}}}\\
&=5/4.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
How to solve $\lim _{x\to-\infty} \sqrt{x^2 + x - 1} + x$ I've tried everything with this question, once I multiply by the conjugate I can't simplify very well
$(x-1)/(\sqrt{x^2 + x - 1} - x)$
Then going to this line seems to mess with the limit, although it seems like I've done it right.
$(x-1)/(x(\sqrt{1 + 1/x - 1/x^2} - 1))$
I could also multiply by $(1/x)/(1/x)$ but again I get a similar problem
Once I cancel the $x$'s
$1/(\sqrt{1 + 1/x - 1/x^2} - 1) - 1/(\sqrt{x^2 + x - 1} - x)$
The first fraction becomes undefined
If someone could drop a stepping stone for me it would be very helpful
| $$\begin{align*}
\lim_{x\to-\infty}\left(\sqrt{x^2+x-1}+x\right)&=\lim_{x\to-\infty}\left(\sqrt{x^2+x-1}+x\right)\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{(x^2+x-1)-x^2}{\sqrt{x^2+x-1}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{x-1}{\sqrt{x^2}\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{x-1}{|x|\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}-x}\\[1ex]
&=\lim_{x\to-\infty}\frac{x-1}{-x\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}-x}\\[1ex]
&=-\lim_{x\to-\infty}\frac{1-\frac{1}{x}}{\sqrt{1+\frac{1}{x}-\frac{1}{x^2}}+1}\\[1ex]
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Base Number Addition Suppose that $p$ is prime and $1007_p+306_p+113_p+125_p+6_p=142_p+271_p+360_p$. How many possible values of $p$ are there?
I suppose I solve this by putting in $p^2$ and $p^1$ for the hundreds and tens digits?
I did that so I got $p^3$ + 24 = $p^2$ + 14p
| We must have that $p^3+7+3p^2+6+p^2+p+3+p^2+2p+5+6=p^2+4p+2+2p^2+7p+1+3p^2+6p$, meaning that $p^3-p^2-14p+24=0$. But the only prime solutions of this can be factors of $24$, i.e. $2$ and $3$. But $7$ is not a digit in base $2$ or $3$, so there are $\boxed{0}$ possible $p$!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Quadratic Sieve Can anyone explain how Quadratic Sieve (factorization algorithm) works?
I tried reading relevant articles but they didn't include clear explanation / implementation of it.
| This is an outline, an actual implementation is more complicated.
You want to factor $m$ where $m = pq$ , $p$ and $q$ are primes.
if $x^2 - y^2 = m$ then $(x-y)(x+y) = m$ so $(x-y)$ and $(x+y)$ are factors of $m$.
if $x^2 - y^2 = km$ then $(x-y)(x+y) = km$ so $(x-y)$ and $(x+y)$ could contain a single factor of $m$ each. One of them could contain all of $m$ which is useless.There are many more ways to find a solution to this problem.
The quadratic sieve tries to construct $x$ and $y$ so that $x^2 - y^2 = km$ then check if $(x-y)$ or $(x+y)$ has one common factor with $m$.
It generates many $x_n$ by $x_n = \lfloor \sqrt{m} \rfloor + n$ , $n = 1\dots$
$x_n^2 - km = (\lfloor \sqrt{m} \rfloor + n)^2 -km = p_1p_2\dots p_a \tag 1$
The right side is factored into primes.
By selecting some of the equations to match the primes so that there is an even power of each of them and multiplying the equations together it gets:
$$(x_1x_2\dots x_b)^2 - p_1^{2k_1}p_2^{2k_2}\dots p_a^{2k_a} = Km$$
$(x_1x_2\dots x_b)^2 - (p_1^{k_1}p_2^{k_2}\dots p_a^{k_a})^2 = Km \tag 2$
Then it tests if $(x_1x_2\dots x_b - p_1^{k_1}p_2^{k_2}\dots p_a^{k_a})$ or
$(x_1x_2\dots x_b + p_1^{k_1}p_2^{k_2}\dots p_a^{k_a})$ has one common factor with $m$. The greatest common divisor (gcd) algorithm is used.
In practice a factor base is used. This limits the set of primes $p_1\dots p_{max}$ that are used.If the right side of $(1)$ contains factors not in the factor base then that equation is rejected. Sieving methods ensure that only small primes are needed in the factor base i.e. smoothness.
Example: Let $m = 253 =11*23$, $\lfloor \sqrt{m} \rfloor = 15$
\begin{array}{|c|c|c|c|c|c|}
\hline
n & x_n & x_n^2-km & factors & odd\ power & comment\\
\hline
1 & 16 & 3 & 3 & 3 & \\
\hline
2 & 17 & 36 & 2,2,3,3 & & is\ square\\
\hline
3 & 18 & 71 & 71 & 71 & too\ large\\
\hline
4 & 19 & 108 & 2,2,3,3,3 & 3 & \\
\hline
5 & 20 & 147 & 3,7,7 & 3 & \\
\hline
\end{array}
Row 2 is square but for the sake of the demonstration lets ignore it.
The odd power column contains primes that have an odd power.
Multiplying rows 1 and 4 will give $3^2$ evening up the powers.
$x = x_1x_4 = 16.19 = 304$ , $y = \sqrt{3.108} = 2.3^2 = 18$
$gcd(x-y,m) = gcd(304-18,253) = 11$
$gcd(x+y,m) = gcd(304+18,253) = 23$
| {
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"url": "https://math.stackexchange.com/questions/1481589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Form a Committee Of 10 Senators The problem I'm working on reads: "How many different committees of 10 Senators can be formed if the two Senators from the same state, (50 States in All) are considered identical?"
This is the answer I got but I'm almost sure it's wrong:
$e_{1-50}=(1+x+x^2)$
$g(x)=(1+x+x^2)^{50}$
$(1+x+x^2)^{50}=\frac{(1-x^3)^{50}}{(1-x)^{50}}$
$=(1-x^3)^{50}*\frac{1}{(1-x)^{50}}$
$(1-x^3)^{50}=(1+(-1)(x^3 ))^{50}$
$=(1-C(50,1) x^3+C(50,2) x^6-C(50,3) x^9+⋯)$
$\frac{1}{(1-x)^{50}} =(1+C(50,1)x+C(51,2) x^2+⋯+C(59,10) x^{10}+⋯)$
$g(x)=(1-C(50,1) x^3+C(50,2) x^6-C(50,3) x^9+⋯)
*(1+C(50,1)x+C(51,2) x^2+⋯+C(59,10) x^{10}+⋯)$
$a_{r}=C(59,10)-C(50,1)*C(56,7)+C(50,2)*C(53,4)-C(50,3)*C(50,1)=51,590,216,930$
I'm wondering where I went wrong. Any help would be greatly appreciated!
| i think i might have a solution. you have at least $5$ states and at most $10$ with the $2$ senators in each state considered identical. if it is $5$ states there are none left to choose otherwise there are $2,4,6,8,10$ states selected that have $1$ senator chosen.
${5\choose 0}{50\choose 5}+{6\choose 2}{50\choose 6}+{7\choose 4}{50\choose 7}+{8\choose 6}{50\choose 8}+{9\choose 8}{50\choose 9}+{10\choose 10}{50\choose 10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How many five-digit numbers do not have three consecutive digits the same? How many five-digit numbers do not have three consecutive digits the same? Also, the initial digits might be $0$, but I'm not sure how that changes the answer.
This is the formula I've come up with for solving this problem.
Total number of numbers - $A - B - C + A \cap B + B \cap C + A \cap C - A \cap B \cap C$
$$10^5 - (10^3 \cdot 3) + (10^2 \cdot 2) - 10$$
So I have
set $A$, positions 1 2 3 are filled;
set $B$, positions 2 3 4 are filled;
set $C$, positions 3 4 5 are filled;
so each set will have $10 \cdot 10 \cdot 10$ subsets of numbers that have $3$ consecutive numbers.
I know that their should be double counting because having consecutive numbers in positions 1 2 3 4 and 2 3 4 5 should be added back, and consecutive numbers 1 2 3 4 5 will need to be subtracted.
So, I get
$$10^5 - (10^3 \cdot 3) + (10^2 \cdot 2) - 10 = 100000 - 3000 + 200 - 10
= 97190$$
However, this is not the correct answer. What is the correct procedure to solve this problem? Or what way am I to look at counting up the sets?
Thanks
| This problem can be broken down into two cases-
*
*When no consecutive digits are same
*When two consecutive digits are same
The answer would be the sum of the number of ways of the two indvidual cases.
If f(n) be the answer to the above problem having n number of digits. Then,
f(n) = 9*(f(n-1)+f(n-2))
The equation can be represented in matrix from as,
\begin{equation*}
\begin{bmatrix}
f(n) \\
f(n-1) \\
\end{bmatrix} =
\begin{bmatrix}
9 & 9 \\
1 & 0 \\
\end{bmatrix}*\begin{bmatrix}
f(n-1) \\
f(n-2) \\
\end{bmatrix}
\end{equation*}
Solving the recurrence, we get,
\begin{equation*}
\begin{bmatrix}
f(n) \\
f(n-1) \\
\end{bmatrix} =
\begin{bmatrix}
9 & 9 \\
1 & 0 \\
\end{bmatrix}^{n-2}*\begin{bmatrix}
f(2) \\
f(1) \\
\end{bmatrix}
\end{equation*}
Clearly, the base cases are f(1) = 10 and f(2) = 100.
This can now be solved using matrix exponentiation having computation complexity of (k3log(n)) where k=2 (size of the square matrix).
Now, considering the above asked question, plugging n=5 into the recurrence gives,
\begin{equation*}
\begin{bmatrix}
f(5) \\
f(4) \\
\end{bmatrix} =
\begin{bmatrix}
9 & 9 \\
1 & 0 \\
\end{bmatrix}^{3}*\begin{bmatrix}
100 \\
10 \\
\end{bmatrix}
\end{equation*}
i.e.,
\begin{equation*}
\begin{bmatrix}
f(5) \\
f(4) \\
\end{bmatrix} =
\begin{bmatrix}
891 & 810 \\
90 & 81 \\
\end{bmatrix}*\begin{bmatrix}
100 \\
10 \\
\end{bmatrix}
\end{equation*}
Therefore,
\begin{equation*} f(5)=891*100+810*10=97200\end{equation*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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Show that $\frac{1}{2}(0.5)+\frac{(1+\frac{1}{2})}{3}(0.5)^2+\frac{(1+\frac{1}{2}+\frac{1}{3})}{4}(0.5)^3+\cdots=(\log 2)^2$ Show that $$\frac{1}{2}(0.5)+\frac{(1+\frac{1}{2})}{3}(0.5)^2+\frac{(1+\frac{1}{2}+\frac{1}{3})}{4}(0.5)^3+\cdots=(\log 2)^2$$
I tried to prove it by using the Taylor series, But I've found it not useful as shown below
any help, thanks
| Note that your series is $f(\frac{1}{2})$, where
$$f(x) = \sum_{n=1}^\infty \frac{H_n}{n+1} x^n$$
($H_n$ being the hamonic series), which is a power series of radius of convergence $1$. Now, $x f(x)=\sum_{n=1}^\infty \frac{H_n}{n+1} x^{n+1}$ looks like (and is) the antiderivative of
$$F(x) = \sum_{n=1}^\infty H_n x^n$$
on $(-1,1)$, by properties of power series. So one can first try to find a closed form for $F$. Either because of a hunch ($H_n$ as coefficient looks like the coefficient of the Cauchy product between series of coefficients $1$ and $\frac{1}{n}$ respectively, which are known), or looking at the output of Mathematica (or both), we can guess what to do. (More details follow, with a full derivation.)
We will use the fact that, for $\lvert x\rvert < 1$,
$$\begin{align}
-\ln(1-x) &= \sum_{n=1}^\infty \frac{x^n}{n} \\
\frac{1}{1-x} &= \sum_{n=0}^\infty x^n
\end{align}$$
so that
$$\begin{align}
\frac{-\ln(1-x)}{1-x} &= \sum_{n=1}^\infty \left(\sum_{k=1}^n 1\cdot \frac{1}{k}\right) x^n = \sum_{n=1}^\infty H_n x^n
\end{align}$$
as wished (where, in the middle, we used the fact that $\left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{n=0}^\infty b_n x^n \right) = \sum_{n=0}^\infty c_n x^n$ with $c_n = \sum_{k=0}^n a_k b_{n-k}$). Now, computing the antiderivative:
$$
F(x) - F(0) = F(x) = \int_0^x \frac{-\ln(1-t)}{1-t}dt = \frac{1}{2}\ln^2(1-x)
$$
(recognizing an integral of the form $\int_0^x u^\prime u = \frac{1}{2}[u^2]^x_0$, for $u(x) = -\ln(1-x)$).
This implies that, for $\lvert x\rvert < 1$,
$$xf(x) = \frac{1}{2}\ln^2(1-x)$$
so that, for $x\in(0,1)$,
$$f(x) = \frac{\ln^2(1-x)}{2x}.$$
Plugging in $x=\frac{1}{2}$ leads to $$
f\!\left(\frac{1}{2}\right) = \ln^2 \frac{1}{2} = \ln^2 2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Integral $\int \sqrt{x+\sqrt{x^2+2}}dx$ $$\int \sqrt{x+\sqrt{x^2+2}}\ dx$$
I tried various solving methods but I am not coming forward. I unformed the term also to ${x+\sqrt{x^2+2} \over \sqrt{x+\sqrt{x^2+2}}}$ and even multiplied with ${\sqrt{x-\sqrt{x^2+2}} \over \sqrt{x-\sqrt{x^2+2}}}$.
Trigonometric and hyperbolic substitution didn't help either.
| Let $$\displaystyle I = \int \sqrt{x+\sqrt{x^2+2}}dx\;$$
Now Put $$\displaystyle (x+\sqrt{x^2+2}) = e^{2t}\;$$ Then $$\displaystyle \left(1+\frac{x}{\sqrt{x^2+2}}\right)dx = 2e^{2t}dt$$
So we get $$\displaystyle \left(\frac{e^{2t}}{\sqrt{x^2+2}}\right)dx = 2e^{2t}dt\Rightarrow dx = 2\sqrt{x^2+2}dt$$
Now Using $$\displaystyle \bullet\; \left(\sqrt{x^2+2}+x\right)\cdot \left(\sqrt{x^2+2}-x\right) = 2$$
So we get $$\displaystyle \left(\sqrt{x^2+2}-x\right) = \frac{2}{e^{2t}}$$
Now $$\displaystyle \sqrt{x^2+2} = \frac{1}{2}\cdot \left(e^{2t}+\frac{2}{e^{2t}}\right)$$
So Integral $$\displaystyle I = \int e^{t}\cdot \left(e^{2t}+\frac{2}{e^{2t}}\right)dt = \int e^{3t}dt+2\int e^{-t}dt$$
So we get $$\displaystyle I = \frac{1}{3}e^{3t}-2e^{-t}+\mathcal{C} = \frac{1}{3}\left(x+\sqrt{x^2+2}\right)^{\frac{3}{2}}-2\cdot \left(x+\sqrt{x^2+2}\right)^{-\frac{1}{2}}+\mathcal{C}$$
| {
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Number of cubes from sphere Reference : 2nd question -
.
From a spherical ball of 10cm radius, how many complete cubes of 5cm
side can be extracted out, if sphere cannot be molten
What I have tried.
*
*Volume of sphere = $\frac43 \pi 10^3$
*volume of cube = $5^3$
So find how many $5^3$ can appear in $\frac43 \pi 10^3$.
But I guess this is wrong and the statement sphere cannot be molten. makes this answer different. How this statement affect the answer. Please give pointers on what is the approach of solving this.
| Your computation yields only an upper bound of $33.51\ldots$ (so actually of $33$), but that is by sheer volume (so "with melting").
As the suggested answer show, this bound seems to be way too big.
Arranging $8$ cubes in a $2\times 2\times 2$ pattern produces a larger cube of side length $10$ and diameter $10\sqrt 3\approx17.3$, which therefore fits easily into a ball of diameter $20$.
Hence from the suggested answers, the largest seems to be appropriate.
However, could one not possibly extract $9$ or more cubes (a non-available answer)?
At least the straighforward, but larger step to $12$ cubes is not possible: An arrangement of $2\times 2\times 3$ cubes has a diameter of $5\cdot\sqrt 17\approx 20.6>20$. But $9$ cubes appear are possible: Take the $2\times 2\times 2$ arrangement and at one cube on the top, centered. The diameter of this object is $\sqrt{7.5^2+7.5^2+15^2}=\frac{15}2\sqrt{6}\approx18.4<20 $.
A sufficiently small diameter alone is not sufficient: for example a triangle of diameter $1$ does not fit into a circle of diameter $1$.
However, we can actually position our $2\times 2\times 2$ thing so that it touches the ball boundary at the "bottom"; then the lower vertices are $\sqrt{10^2-5^2-5^2}$ below the ball center; similarly, a single cube pushed to to "top" has its higher vertices $\sqrt{10^2-2.5^2-2.5^2}$ above the ball center.
From
$$ \sqrt{10^2-5^2-5^2}+\sqrt{10^2-2.5^2-2.5^2}\approx 16.4>15$$
we see that everything fits, even with a safety gap of $\approx 1.4$.
Even a block of two cubes, pushed as high up as possible ends up at $\sqrt{10^2-5^2-2.5^2}$ and then
$$ \sqrt{10^2-5^2-5^2}+\sqrt{10^2-5^2-2.5^2}\approx 15.4>15$$
shows that we can place $10$ cubes.
| {
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How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.
Thanks in advance.
| You got the right expression. $x^7+1=0$
The roots of the equation will be like $$x= \cos (\frac{2k\pi}{7})+ i\sin (\frac{2k\pi}{7}) , 0\le k\le6$$
Note: Stress on be like, this formula will not give you the exact roots. You have to change it a bit to account for the roots of negative unity.
EDIT: Complete solution follows:
$$x^7+1=0$$ or,
$$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0$$
This implies your equation has exactly 1 real root and 3 pairs of complex conjugate roots.
Hence now I can write
$$x^7+1=0$$ or,
$$x^7=-1$$ or,
$$x^7=\cos \pi + i\sin \pi = \cos (2k+1)\pi + i\sin (2k+1)\pi , 0 \le k\le 6 $$ or,
$$x=[\cos (2k+1)\pi + i\sin (2k+1)\pi]^{\frac{1}{7}} , 0 \le k\le 6$$ or,
$$x=\cos \frac{(2k+1)\pi}{7}+ i\sin \frac{(2k+1)\pi}{7} , 0 \le k\le 6$$
From the comment by Macavity: However,you have a spurious root included - $k=3$ . By multiplying by $x+1$ you introduced this root which is not a root of the original polynomial. So there are no real roots for the polynomial, only complex ones. Hence the final solutions are as follows: $$x=\cos \frac{(2k+1)\pi}{7}+ i\sin \frac{(2k+1)\pi}{7} , 0 \le k\le 6 \,\ \text{and} \,\ k \not = 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the sequence $a_n=1+\frac1{\sqrt{2}} +\frac1{\sqrt{3}} + \dots + \frac1{\sqrt{n}}$ is not Cauchy
Show directly (from the definition) that if
$$a_n=1+\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}}\;,$$
then $(a_n)$ is not a Cauchy sequence.
Attempt
Using $a_n$ is not Cauchy if
$$\exists\, \varepsilon > 0 : \forall N\, \exists\, n, m > N | a_{n} - a_{m}| \geq \varepsilon$$
taking $n>m$.
$$\begin{align}|a_n - b_m| = \left|\sum_{k=1}^{n} \frac{1}{\sqrt{k}} - \sum_{l=1}^{m} \frac{1}{\sqrt{l}}\right| &= \left|\sum_{k=m+1}^{n} \frac{1}{\sqrt{k}}\right| \ge \left|\sum_{k=m+1}^{n} \frac{1}{n}\right| = \frac{n-m}{n}\\ \end{align} $$
Letting $n = 2m$
$$\frac{n-m}{n} = \frac{2m-m}{2m} = \frac{m}{2m} = \frac{1}{2}$$
Thus
$$\exists\, \varepsilon = \frac{1}{2} : \forall N\, \exists\, n=2m > m > N : \left|\sum_{k=m+1}^{n} \frac{1}{\sqrt{k}}\right| \geq \frac{1}{2}$$
so $a_n$ is not Cauchy.
| Another proof:
If
$a_n
=\sum_{k=1}^n \frac1{\sqrt{k}}
$,
$a_{n^2}-a_n
=\sum_{k=n+1}^{n^2} \frac1{\sqrt{k}}
>\sum_{k=n+1}^{n^2} \frac1{\sqrt{n^2}}
=(n^2-n)\frac1{n}
=n-1
$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that $64$ divides $3^{2n+2}+56n+55$ by induction Let $n ≥ 0$ be an integer. Prove by induction: 64 divides $3^{2n+2} + 56n + 55$
general expression: $3^{2n+2} + 56n + 55 = 64m$
1st I substitute $P(0)$ and it gives me true:
$9+55 = 64$ (if m = 1 the condition is true)
2nd I assume $n = k$ and I substitute saying $3^{2k+2} + 56k + 55$
and then I don't know what to do to prove the claim.
| Assume that
$$3^{2k+2}+56k+55=64m.$$
Then, use
$$3^{2(k+1)+2}=3^{2k+4}=3^2\cdot 3^{2k+2}$$
and
$$3^{2k+2}=64m-56k-55.$$
$\begin{align}3^{2(k+1)+2}+56(k+1)+55&=9\cdot 3^{2k+2}+56k+56+55\\&=9(64m-56k-55)+56k+56+55\\&=9\cdot 64m-9\cdot 56k-9\cdot 55+56k+56+55\\&=9\cdot 64m-8\cdot 7\cdot 8k-6\cdot 64\\&=64(9m-7k-6)\end{align}$
| {
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Limits of sums in which the sum index is a function of the variable involved in the limit I came across this problem when trying to prove that the harmonic series diverges. I need to prove that:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} = 0
\end{equation}
Is he following methodology correct?
Firstly:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n}
= \lim_{N \rightarrow \infty} \sum_{n=1}^{2N+1} \frac{1}{N + n}
\end{equation}
Consider
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{m=1}^{M} \frac{1}{N+m} \qquad \textrm{for some } M > 1, M \in \mathbb{N}
\end{equation}
but
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{m=1}^{M} \frac{1}{N+m}
= \lim_{N \rightarrow \infty} \frac{1}{N+1} + \lim_{N \rightarrow \infty} \frac{1}{N+2} + \space ... + \lim_{N \rightarrow \infty} \frac{1}{N+M} = 0 \space + 0 \space + \space ... + \space 0 = 0
\end{equation}
Hence is true that:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{m=1}^{M} \frac{1}{N+m}
= 0 \qquad \forall \space M : M > 1, M \in \mathbb{N}
\end{equation}
Set $M = 2N + 1$
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{m=1}^{2N+1} \frac{1}{N+m}
= 0 \qquad N \in \mathbb{N}
\end{equation}
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} = 0
\end{equation}
Update:
The problem was misspecified. It was required to prove
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n}
\end{equation}
under the (false) assumption that the harmonic series has a finite limit $L$:
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=1}^{N} \frac{1}{n} = L
\end{equation}
In this case (relying on the fact that the sum is monotonically increasing):
\begin{equation}
\lim_{N \rightarrow \infty} \sum_{n=1}^{N} \frac{1}{n} = L
\Rightarrow \nexists \space N : \sum_{n=1}^{N} \frac{1}{n} > L
\Rightarrow \nexists \space N : \sum_{n=1}^{3N+1} \frac{1}{n} > L
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=1}^{3N+1} \frac{1}{n} \le L
\end{equation}
but
\begin{equation}
\sum_{n=1}^{3N+1} \frac{1}{n} = \sum_{n=1}^{N} \frac{1}{n} +\sum_{n=N+1}^{3N+1} \frac{1}{n}
\end{equation}
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=1}^{3N+1} \frac{1}{n} = L + \lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} \le L
\end{equation}
\begin{equation}
\Rightarrow \lim_{N \rightarrow \infty} \sum_{n=N+1}^{3N+1} \frac{1}{n} = 0
\end{equation}
Does that make sense?
| The sum can be written as $H_{3N}-H_N$, where $H_n$ is the $n^{th}$ harmonic number. At the
same time, we also have $H_N\simeq\ln N$, and $\ln a-\ln b=\ln\dfrac ab$. Combining the three, we
conclude that the limit is $\ln3$. Your reasoning is incorrect, because, according to it, we
could also write $~1~=~\lim\limits_{N\to\infty}\dfrac NN~=~\displaystyle\lim_{N\to\infty}\sum_{n=1}^N\dfrac1N~=~\lim_{N\to\infty}\dfrac1N~+~\lim_{N\to\infty}\dfrac1N~+~\cdots~+$
$+~\lim_{N\to\infty}\dfrac1N~=~0~+~0~+~\cdots~+~0~=~0,~$ which is absurd.
| {
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Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:
Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .
Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$
Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.
Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
So degree of $p(x)$ is atleast $4$.
But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .
So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$
But I can't get the step how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$
Kindly help with this.
| That fact comes from Galois Theory. Under the assumption that $K/F$ is Galois, the (monic) minimal polynomial $f$ of an element $\alpha \in K$ (over $F$) is of the form
$$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$
This implies that $\sigma \alpha$ are all roots of the minimal polynomial for $\alpha$.
In your case, let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ and $F = \mathbb{Q}$. Then $K/F$ is Galois and there are 4 automorphisms in $\text{Gal}(K/F)$: namely
*
*identity on $K$
*the one that sends $\sqrt{2} \mapsto -\sqrt{2}$ keeping $\sqrt{3}$ fixed: this automorphism sends $\sqrt{2} + \sqrt{3}$ to $-\sqrt{2} + \sqrt{3}$; hence $-\sqrt{2} + \sqrt{3}$ must be a root of $p(x)$.
*the one that keeps $\sqrt{2}$ fixed and sends $\sqrt{3} \mapsto -\sqrt{3}$: like above, we deduce $\sqrt{2} - \sqrt{3}$ is root of minimal polynomial $p(x)$
*the one that negate both roots: finally $- \sqrt{2} - \sqrt{3}$ must also be a root.
| {
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Find the solution of the differential equation that satisfies ${{dP} \over {dt}} = 8\sqrt {Pt} ,\,P(1) = 5$ Please help. My homework is grading my answer as incorrect, but I can't tell what I did wrong. The second photo is the work of the problem done correctly but with dp/dt=2sqrt(Pt). I based my work off of this, but my answer is incorrect. Thank you!
The link to the question is:
Q3
and here is my answer:
C
| $$\frac{dP}{dt}=8\sqrt{Pt}$$
$$\frac{dP}{\sqrt{P}}=8\sqrt{t}dt$$
$$\int \frac{dP}{\sqrt{P}}=\int 8\sqrt{t}dt$$
$$2\sqrt{P}=8\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}+c=\frac{16}{3}t^{\frac{3}{2}}+c$$
By initial conditions, $P(1)=5$
$$2\sqrt{5}=\frac{16}{3}+c$$
$$c=2\sqrt{5}-\frac{16}{3}$$
The answer should be $$2\sqrt{P}=\frac{16}{3}t^{\frac{3}{2}}+2\sqrt{5}-\frac{16}{3}$$
$$\sqrt{P}=\frac{8}{3}t^{\frac{3}{2}}+\sqrt{5}-\frac{8}{3}$$
$$P=\left(\frac{8}{3}t^{\frac{3}{2}}+\sqrt{5}-\frac{8}{3}\right)^2$$
| {
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Let $f(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$ and let $g(x)=f^{-1}(x)$.Find $g'''(0)$ Let $f(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$ and let $g(x)=f^{-1}(x)$.Find $g'''(0)$.
My attempt:As $g(x)=f^{-1}(x)\Rightarrow fog(x)=x\Rightarrow f'(g(x)).g'(x)=1$
$g'(x)=\frac{1}{f'(g(x))}$ .Similarly i found,$g''(x)=-\frac{f''(g(x))}{(f'(g(x)))^3}$ and Similarly i found $g'''(x)=-\frac{f'(g(x))f'''(g(x)).g'(x)-3(f''(g(x)))^2g'(x)}{(f'(g(x)))^4}$
So $g'''(0)=-\frac{f'(g(0))f'''(g(0)).g'(0)-3(f''(g(0)))^2g'(0)}{(f'(g(0)))^4}$
Now i am stuck here,i cannot find values of $g'(0),f'(g(0)),f''(g(0)),f'''(g(0))$.Please help me.
| Say $f(y) = y+y^2/2+y^3/3+y^4/4+y^5/5 = x$, then $f^{-1}(x) = y = g(x)$.
So $g(x)+g(x)^2/2+g(x)^3/3+g(x)^4/4+g(x)^5/5 = x$
First note that $g(0) = 0$.
Now doing implicit differentiation with chain rule, you would be able to obtain
$g'(x)(1+4g(x)) = 1.$
Plug in $0$ and you can get $g'(0) = \frac{1}{1+4g(0)} = 1 $.
Continue with implicit differential and you would be able to find $g''(0)$ then $g'''(0)$
$g'(x)(4g'(x)) + g''(x) (1+4g(x)) = 0$
$4(g'(x))^2 + g''(x) + 4g(x)g''(x) = 0$
so $4+g''(0)+4g''(0)=0$ and $g''(0) = -4/5$
Differentiate once more,
$8(g'(x))(g''(x))+ g'''(x) + 4g(x)g'''(x)+ 4g'(x)g'''(x) = 0$
$8(-4/5) + g'''(0) + 4g'''(0) + 4(-4/5)g'''(0) = 0$
so $g'''(0) = 32/9$
| {
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3 contestants choosing a smallest number to win a car Each of 3 contestants chooses a positive integer. The contestant who chooses the unique smallest positive integer number wins a car. If all of them chose the same number, then no body wins car. What is the optimal strategy?
I do not know how this is done, so I can only guess.
The players should play the same strategy, by the nature of the game.
I started off assuming that it makes no sense choosing 4 or above (so I assumed one should choose 1,2 or 3).
After some algebra bashing, I arrived at the result.
maximise $x(1-x)^2 + y(x^2+(1-x-y)^2)+(1-x-y)(x^2+y^2)$ subject to the condition $x+y\leq 1$
Somewhat surprisingly, this expression in symmetric in x and y. I used the Language's multiplier and arrive at $x=y=4/3-\sqrt{40}/6$, which is about 0.28.
This suggest the contestants should choose 1 or 2 with probability 0.28 and choose 3 with probability 0.44. The chance of winning is about 0.29.
However, it is clear that this is not the optimal strategy. If players A and B uses such strategy, then C can choose 1 or 2 with probability 1/2, then he has about > 0.5 chance of winning.
I suspected that the optimal strategy is all players choose 1 or 2 with probability 1/2. I seem to find is no way of deviating from this strategy which would improve each player's pay off. Is this right? how do we prove this?
| If there is a Nash equilibrium, and the other two are playing it, then it doesn't matter what number the first player plays. So if the others play with probability distribution p(1)=a,p(2)=b,p(3)=c,
First player wins playing 1 with probability $(1-a)^2$
First player wins playing 2 with probability $a^2+c^2$
First player wins playing 3 with probability $a^2+b^2$
So $(1-a)^2=a^2+c^2=a^2+b^2,a+b+c=1$
which has solution $a=2\sqrt{3}-3,b=2-\sqrt{3},c=2-\sqrt{3}$ and
first player wins with probability $28-16\sqrt{3}=0.287$
If they use four numbers, then $(1-a)^2=a^2+(1-a-b)^2=a^2+b^2+d^2=a^2+b^2+c^2$
and Wolfram says the chance of a win is 0.294
When more numbers are used, it seems that $a/b,b/c,c/d,...$ all approach the same ratio. That gives this solution for infinitely many:
$$(a_1,a_2,...)=a_1(1,x,x^2,x^3,x^4,...)\\a_1=1-x$$
Then, using $$(1-a_1)^2=a_1^2+(1-a_1-a_2)^2\\x^2=(1-x)^2+x^4\\x^3+x^2+x=1$$
we get the value of $x=0.5437$ for the Nash equilibrium, with a win probability of $(1-a_1)^2=0.2956$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1495543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Greatest common divisor of 2n+1 and 9n+4 Calculate
$GCD(2n+1,9n+4)$ and $GCD(2n-1,9n+4)$
$$GCD(2n+1,9n+4)=GCD(2n+1,9n+4-4 \cdot (2n+1))=GCD(2n+1,n)= GCD(n,1)=1$$
How to calculate $GCD(2n-1,9n+4)$
| Eliminate $n$
If integer $d$ divides both, $d$ must divide $2(9n+4)-9(2n-1)=17$
As the positive divisors of $17$ are $1,17$
The gcd will be $17$ iff $17$ divides both $9n+4,2n-1$
If $2n-1\equiv0\pmod{17}\iff2n\equiv1\equiv17+1\iff n\equiv9\pmod{17}$
and then $9n+4\equiv85\equiv0\pmod{17}$
For the first case,
if $D$ divides both $9n+4,2n+1,D$ must divide $9(2n+1)-2(9n+4)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Is $\arctan n$ always equal to $\arccos\sqrt{\frac{1}{n^2+1}}$? $$\arccos\sqrt\frac{1}{2}=\arctan 1$$
$$\arccos\sqrt\frac{1}{5}=\arctan 2$$
$$\arccos\sqrt\frac{1}{10}=\arctan 3$$
$$\arccos\sqrt\frac{1}{17}=\arctan 4$$
$$\arccos\sqrt\frac{1}{26}=\arctan 5$$
$$\arccos\sqrt\frac{1}{37}=\arctan 6$$
$$\arccos\sqrt\frac{1}{50}=\arctan 7$$
The answer is a sequence $n^2+1$ for the slope which is in the inverse of tangent. Digits that are whole numbers. Is there any explanation as to why this is true? Is it a well-known problem?
| $\arccos \sqrt \frac 1 n = x$ means $cos x = \frac 1 n$ which means $\sin x = \sqrt{1 - {\sqrt \frac 1 n}^2} = \sqrt{1 - \frac 1 n}$ so $\tan x = \frac{ \sqrt{1 - \frac 1 n}}{\sqrt \frac 1 n} = \sqrt{n - 1}$ so $\arctan{n -1} = x = \arccos \sqrt \frac 1 n$. So this is known.
====
Another way of looking at it is:
$\tan = \frac {\sin}{\cos} $ and $\sin^2 + \cos^2 = 1$, so $\tan = \frac {\sqrt{1 - \cos^2}}{\cos} = \sqrt {\frac 1{\cos^2} - 1}$ so if $n = \cos x$ then $\tan =\sqrt {\frac 1{n^2} - 1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Solving A Linear Recurrence Relation With Complex Roots Question:
For the given linear homogeneous difference equation, find the general solution:
$$y_{n+2} + y_{n+1} + y_n = 0$$
With the initial conditions of:
$$y(0)=\sqrt3, y(1) = 0$$
Attempted Answer:
I approached the problem normally as one would except by solving the auxiliary equation which yields:
$$m^2 + m + 1 = 0$$
$$\implies m = \frac{-1}{2} \pm \frac{i\sqrt3}{2}$$
Now the general solution is of form:
$$y_n = A \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)^n
+ B \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right)^n $$
Here is the part where I get stuck when I substitute the initial conditions to form a system of equations:
$$A + B = \sqrt{3}$$
$$A \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)
+ B \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right) = 0$$
Now to make my life easier (which it really didn't), I decided to take a 'shortcut' and rewrite the 2nd equation as the sum of an imaginary and real part:
$$\frac{-1}{2}(A + B) + i\frac{\sqrt3}{2}(-A + B) = 0 + 0i$$
Since the real part on the LHS must equal the real part on the RHS, and the same applies for the imaginary part, we obtain:
$$\frac{-1}{2}(A + B) = 0$$
$$\frac{\sqrt3}{2}(-A + B) = 0$$
So the above implies that $A = B = -B$ which can only be true if $A = B = 0$. However if that is the case, then the first equation in the system of equations ($A + B = \sqrt{3}$) implies that 0 = $\sqrt{3}$.
I know that there is a mistake somewhere in the logic of my reasoning (I think perhaps when I equated each real and imaginary part of the LHS to the RHS) but I don't know why and where exactly. Please point out my mistake.
| What happens
when the roots of the
characteristic polynomial are complex
is that the solutions
have a periodic component.
Your case is
$y_n
= a \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)^n
+ b \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right)^n
=a r^n + bs^n
$.
Note that
$r+s = -1$
and
$r-s = -i\sqrt{3}$.
If
$y_0 = u$
and
$y_1 = v$,
then
$a+b = u$
and
$ar+bs = v$.
Since
$b = u-a$,
$v
= ar+(u-a)s
=a(r-s)+us
$,
so
$a
=\frac{v-us}{r-s}
$
and
$b
=u-a
=u-\frac{v-us}{r-s}
=\frac{u(r-s)-v+us}{r-s}
=\frac{ur-v}{r-s}
$.
In your case,
$u=\sqrt{3},
v = 0,
r=\frac{-1}{2} - \frac{i\sqrt3}{2},
s=\frac{-1}{2} + \frac{i\sqrt3}{2},
r-s=-i\sqrt{3}
$,
so
$a
=\frac{v-us}{r-s}
=\frac{-\sqrt{3}(\frac{-1}{2} + \frac{i\sqrt3}{2})}{-i\sqrt{3}}
=-i(\frac{-1}{2} + \frac{i\sqrt3}{2})
=\frac{i}{2} + \frac{\sqrt3}{2}
$
and
$b
=u-a
=\frac{\sqrt3}{2}-\frac{i}{2}
$.
(We are getting close.)
We have
$y_n
=ar^n+bs^n
$.
By the magic of
assigned problems,
$|r| = |s| = 1$.
$r = e^{-2i\pi/3}
$,
so
$r^n
=e^{-2ni\pi/3}
=\cos(-2n\pi/3)+i\sin(-2n\pi/3)
=\cos(2n\pi/3)-i\sin(2n\pi/3)
$.
Similarly,
$s = e^{2i\pi/3}
$,
so
$s^n
=e^{2in\pi/3}
=\cos(2n\pi/3)+i\sin(2n\pi/3)
$.
Also
$a
=e^{i\pi/6}
$
and
$b
=e^{-i\pi/6}
$.
Therefore
(finally!)
$\begin{align*}
y_n
&=ar^n+bs^n\\
&=e^{i\pi/6}e^{-2in\pi/3}+e^{-i\pi/6}e^{2in\pi/3}\\
&=e^{\pi i(1/6-2n/3)}+e^{\pi i(-1/6+2n/3)}\\
&=\cos(\pi (1/6-2n/3))+i\sin(\pi (1/6-2n/3))
+\cos(\pi (-1/6+2n/3))+i\sin(\pi (-1/6+2n/3))\\
&=\cos(\pi (1/6-2n/3))+\cos(\pi (-1/6+2n/3))
+i(\sin(\pi (1/6-2n/3))+\sin(\pi (-1/6+2n/3)))\\
&=2\cos(\pi (1/6-2n/3))
\qquad\text{since }
\cos(-x)=\cos(x) \text{ and }\sin(x) = -\sin(-x)\\
\end{align*}
$
You can get explicit values for
values of $n$ mod 6,
but this shows how the result is periodic.
A check is that
the result is real,
with no imaginary part.
This has to hold,
since the initial values
and the recurrence coefficients
are all real.
| {
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"url": "https://math.stackexchange.com/questions/1499090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sum_0^{\infty}\frac{z^{2^n}}{z^{2^{n+1}}-1}$ converges to $\frac{z}{z-1}$ or $\frac{-1}{1-z}$ How do we show that the sum below converges to $\frac{z}{z-1}$ if $|z|<1$ and $\frac{-1}{1-z}$ if $|z|>1$?
$$\sum_0^{\infty}\frac{z^{2^n}}{z^{2^{n+1}}-1}=\frac{z}{z^2-1}+\frac{z^2}{z^4-1}+\frac{z^4}{z^8-1}+...$$
Any help is appreciated, thanks!
| Note
\begin{align}\frac{z^{2^n}}{z^{2^{n+1}}-1} &= \frac{z^{2^n}(z^{2^n}-1)}{(z^{2^n}-1)(z^{2^{n+1}}-1)} = \frac{z^{2^{n+1}}-z^{2^n}}{(z^{2^n}-1)(z^{2^{n+1}}-1)} \\
&= \frac{(z^{2^{n+1}}-1) - (z^{2^n}-1)}{(z^{2^n}-1)(z^{2^{n+1}}-1)}\\
& = \frac{1}{z^{2^n} - 1} - \frac{1}{z^{2^{n+1}}-1}.
\end{align}
So for every positive integer $N$,
$$\sum_{n = 0}^N \frac{z^{2^n}}{z^{2^{n+1}}-1} = \frac{1}{z-1} - \frac{1}{z^{2^{N+1}}-1}.\tag{*}$$
by telescoping. Now evaluate the limit as $N\to \infty$ of the right-hand side (*) depending on the cases $|z| < 1$ and $|z| > 1$.
| {
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"url": "https://math.stackexchange.com/questions/1502741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as $x$ varies over the reals?
How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as x
varies over the reals ?
Suggestions please!
| $$f(x)-1=3^x + 5^x - 9^x + 15^x - 25^x-1=-\frac{1}{2}\left( (3^x-1)^2+(5^x-3^x)^2+(1-5^x)^2\right) \leq 0$$
Second solution By C-S inequality
$$(3^x+5^x+15^x)^2 \leq (1 +5^x+3^x)( 3^x+1+5^x) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Find the indefinite integral $\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $ Find the indefinite integral
$$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$
Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..
| Let $x=\sin\theta\implies dx=\cos\theta \ d\theta$, $$\int \frac{dx}{(1+x^2)\sqrt{1-x^2}}$$ $$=\int \frac{\cos\theta\ d\theta}{(1+\sin^2\theta)\cos\theta}$$
$$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta(1+\sin^2\theta)}$$
$$=\int \frac{\sec^2\theta\ d\theta}{\sec^2\theta+\tan^2\theta}$$
$$=\int \frac{\sec^2\theta\ d\theta}{1+2\tan^2\theta}$$
$$=\int \frac{ d(\tan\theta)}{1+(\sqrt2 \tan\theta)^2}$$
$$=\frac{1}{\sqrt 2}\tan^{-1}\left(\sqrt 2\tan\theta\right)+C$$
$$=\color{red}{\frac{1}{\sqrt 2}\tan^{-1}\left(\frac{x\sqrt 2}{\sqrt{1-x^2}}\right)+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$ are real
If $a,b,c,d\in \mathbb{R}$ and roots of the quadratic equation $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0$
are real.Then prove that roots are equal.
$\bf{My\; Try::}$ Given $(a^4+b^4)x^2+4abcdx+(c^4+d^4)=0\;$ Then we can write it as
$$\left[(a^2x)^2+c^4-2a^2c^2x+(b^2x)^2+d^4-2b^2d^2x+2a^2c^2x+2b^2d^2x+4abcdx\right]=0$$
So $$(a^2x-c^2)^2+(b^2x-d^2)^2+2x(ac+bd)^2=0$$
Now I did not understand How can I proceed further.
Although I have a knowledge of $\bf{Discriminant\; Method.}$
So plz explain me above method which i am trying above.
Thanks.
| Solving the equation requires you to compute square roots of the discriminant$$\Delta=(4abcd)^2-4(a^4+b^4)(c^4+d^4).$$If the roots are real, then $\Delta$ has to be non-negative. But$$0\leq (a^2-b^2)=a^4+b^4-2a^2b^2,$$so that$$0\leq 2a^2b^2\leq a^4+b^4$$and similarly$$0\leq 2c^2d^2\leq c^4+d^4.$$Taking the product of these inequalities yields$$4a^2b^2c^2d^2\leq (a^4+b^4)(c^4+d^4),$$so $\Delta \leq 0$. Thus $\Delta=0$, and the roots must both be equal to$$\frac{-4abcd\pm\sqrt{0}}{2(a^4+b^4)}=\frac{-2abcd}{a^4+b^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Divisibility of a number $N$ by $4,7,9$
Let $$N=\binom{20}{7}-\binom{20}{8}+\binom{20}{9}-\binom{20}{10}+\binom{20}{11}-\binom{20}{12}+.......-\binom{20}{20}.$$
Then $N$ is divisible by
$\bf{Options::}\;\;:: (a)\; 4\;\;\;\;\;\; (b)\;\; 7\;\;\;\;\;\; (c)\;\; 9\;\;\;\;\;\; (d)\;\; None\; of\; these.$
$\bf{My\; Try::}$ Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+..........+\binom{n}{n}x^n.$$
Now put $x=-1\;,n=20\;,$ We get $$0=\binom{20}{0}-\binom{20}{1}+\binom{20}{2}+.......+\binom{20}{20}=S$$
So we get $$-N=-\binom{20}{7}+\binom{20}{8}-\binom{20}{9}+\binom{20}{10}-\binom{20}{11}+\binom{20}{12}+.......+\binom{20}{20}$$
So $$-N = S-\binom{20}{0}+\binom{20}{1}-\binom{20}{2}+\binom{20}{3}-\binom{20}{4}+\binom{20}{5}-\binom{20}{6}$$
So we get $$N=\binom{20}{0}-\binom{20}{1}+\binom{20}{2}-\binom{20}{3}+\binom{20}{4}-\binom{20}{5}+\binom{20}{6}$$
So we get $$N=1-20+\frac{20\cdot 19}{2}-\frac{20\cdot 19\cdot 18}{3\cdot 2}+\frac{20\cdot 19\cdot 18 \cdot 17 }{4\cdot 3\cdot 2}-\frac{20\cdot 19\cdot 18\cdot 17\cdot 16}{ 5\cdot 4\cdot 3\cdot 2}+\frac{20\cdot 19\cdot 18\cdot 17\cdot 16\cdot 15}{ 6\cdot 5\cdot 4\cdot 3\cdot 2}$$
Now How can I proceed after that.
or If there is any short method then plz explain here.
Thanks
| You began as I would have. If you have a calculator, you can just find the value of $N$ directly. If you don't have a calculator or you want to avoid such calculations, you can proceed using only modular arithmetic, which is easy enough to do without a calculator.
Each fraction in your last formula for $N$ is an integer, of course. You can strike out common factors from the numerators and denominators, then simplify the calculation by using only modular arithmetic. To check the divisibility by $4$, use modulo $4$.
$$\begin{align}
N\ &\equiv (1) - (20) + (10\cdot 19) - (20\cdot 19\cdot 3) + (5\cdot 19\cdot 3\cdot 17) - (19\cdot 3\cdot 17\cdot 16) + (19\cdot 17\cdot 8\cdot 15)\\
&\equiv (1)-(0)+(2\cdot -1)-(0\cdot 19\cdot 3)+(1\cdot -1\cdot -1\cdot 1)-(19\cdot 3\cdot 17\cdot 0)+(19\cdot 17\cdot 0\cdot 15) \\
&\equiv 1-0+-2-0+1-0+0 \\
&\equiv 0 \pmod 4
\end{align}$$
We conclude that $N$ is divisible by $4$.
Redo that modulo $7$ and $9$ to get your final answer. I get that $N$ is also divisible by $7$ but not by $9$. It seems your final answer should be both choices $(a)$ and $(b)$. Can your test/homework handle that?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to find the range of $1 / (1+x^2)^{1/2}$? How to find the range of $$\frac{1}{\sqrt{1+x^2}}$$?
Ok. I've revised the (easy theory). I would like to complete the exercise finding the derivative of f(x) and setting equal to zero. I do it correctly, $$f'(x) = - x / (1+x^2)^{3/2}$$ but I don't have clear how to solve this equation once set equal to zero. I simplify the denominator but I don't know how to isolate the x.
| We try to rewrite $x$ in terms of $y$:
$$
0 < \frac{1}{\sqrt{1+x^2}} = y \le 1 \quad (x \in \mathbb{R})\Rightarrow \\
0 < \frac{1}{1+x^2} = y^2 \le 1 \Rightarrow \\
1 \le \frac{1}{y^2} = 1 + x^2 < \infty \Rightarrow \\
0 \le \frac{1-y^2}{y^2} = x^2 < \infty
$$
and get
$$
x =
\pm \frac{\sqrt{1-y^2}}{y} \quad (y \in (0,1])
$$
$y$ is not a function anymore, for every $y \in (0,1)$
there are two choices of $x$.
Here is an image:
Addendum:
$$
0 \le x^2 \quad (x \in \mathbb{R}) \Rightarrow \\
1 \le 1 + x^2 \quad (x \in \mathbb{R}) \Rightarrow \\
\frac{1}{1+x^2} \le 1
$$
because $1+x^2 \ge 1$ thus $1+x^2 \ne 0$ we can divide both sides of the inequality by it, because $1+x^2 > 0$ the direction of the comparison operator did not change.
The RHS is $y^2$, so
$$
y^2 = \frac{1}{1+x^2} \le 1 \Rightarrow \\
y = \frac{1}{\sqrt{1+x^2}} \le \sqrt{1} = 1
$$
This one holds because $\sqrt{.}$ is monotone:
$$
x_1 \le x_2 \Rightarrow \sqrt{x_1} \le \sqrt{x_2}
$$
Addendum:
A criterion for local extrema is the vanishing of the first derivative:
$$
y = \frac{1}{\sqrt{1+x^2}} \Rightarrow \\
y' = -\frac{1}{2\left(1+x^2\right)^{3/2}}(2x) = -\frac{x}{\left(1+x^2\right)^{3/2}}
$$
This means
$$
0 = y' = -\frac{x}{\left(1+x^2\right)^{3/2}} \Rightarrow x = 0
$$
Thus $y(0) = 1$ might be a local extremum. To know more we look at
the second derivative:
$$
y''
=
-\frac{\left(1+x^2\right)^{3/2}
- x \frac{3}{2}\left(1+x^2\right)^{1/2}(2x)}
{\left(1+x^2\right)^3}
=
-\frac{\left(1 - 2 x^2 \right) \left(1 + x^2\right)^{1/2}}
{\left(1+x^2\right)^3}
= \frac{2x^2 - 1}
{\left(1+x^2\right)^{5/2}}
$$
We see $y''(0) = -1 \ne$ so it is a local maximum.
| {
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"timestamp": "2023-03-29T00:00:00",
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"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.