Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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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=\fr... | 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",
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"answer_id": 2
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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)$... | 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",
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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 functi... | 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
$$\frac{1}{2 \pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the limit exists for function of two variables? Problem: Evaluate the indicated limit or explain why it does not exist: \begin{align*} \lim_{(x,y) \to (0,0)} \frac{x^2 y^2}{x^2 + y^4} \end{align*}
The definition of limit my calculus textbook gives is:
We say that $\lim_{(x,y) \to (a,b)} f(x,y) = L$, provi... | Let's to use polar coordinates $x=\rho\cos\phi$, $y=\rho\sin\phi$, and your limit is
$$
\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}
$$
But $\lim_{\rho\to0}\rho^2 \cos^2\phi \sin^2\phi = 0$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$? How to prove $f(x)=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}$ is not differentiable at $x=4$ ?
Please let me know the fastest method you know of for such type of problems. Is there any way other than finding the left hand ... | Observe that $(f(x))^2$ is given by
$$(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|}$$
For $x >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Congruence using extended GCD $$\eqalign{ x &\equiv 5 \mod 15\cr
x &\equiv 8 \mod 21\cr}$$
The extended Euclidean algorithm gives $x≡50 \bmod 105$.
I understand now that if we combine the two it implies $15a-21b = 3$ but I don't understand how to use the extended GCD to go from there to finding $x$ and th... | Since the GCD of the moduli is $(15,21)=3$, it is necessary that $x$ be the same thing in both equations mod $3$. That is,
$$
x\equiv5\pmod{15}\implies x\equiv2\pmod{3}
$$
and
$$
x\equiv8\pmod{21}\implies x\equiv2\pmod{3}
$$
If we didn't get that $x\equiv2\pmod{3}$ from both equations, a solution would not be possible.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential equation: $Ay'' + By' + Cy = h(x)$ I'm stuck solving the equation $y'' - 3y' + 2y = 2x^3-30$.
The auxiliary equation is $k^2 - 3k + 2 = 0$ where $k_1 = 1, k_2=3$. Thus the general solution is: $$y_g = C_1e^x + C_2e^{3x}$$
Then, I tried to find the particular solution taking into consideration that $h(x) = ... | First of all your factorization is wrong: $k_1=1, k_2=2$,so general solution will be,
$$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}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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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 $1/12$
I... | 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 so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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:
$$\left[\begin{array}{ccc} \frac{3}{4}I & \frac{1}{4}I & \frac{1}{4}I \\ \frac{1}{4}I & \frac{3}{4}I & -\frac{1}{4}I \\ \frac{1}{4}I & -\frac{1}{4}I & \frac{3}{4}I \e... | The block matrix (let us denote by $M$) can be expressed as the Kronecker product of matrices $A$ and $I$ (the fixed size identity matrix, of dimension $n$) as follows:-
$$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}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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:
$$I_1=\int_0^\inf... | The following is another way to show that $$\sum_{n=1}^{\infty} \frac{(-1)^{k} \operatorname{Si}(k \pi )}{k^{3}}= - \frac{\pi^{3}}{18}.$$
(I'm surprised that replacing $\operatorname{Si}(k \pi)$ with its Maclaurin series and then switching the order of summation leads to the correct result because Fubini's theorem is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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New idea to solve this equation I was teaching $\left \lfloor x \right \rfloor$ function properties and equation . I solved this equation in my class . My works are show below. Some students ask me for new Idea...,and now I am looking for various method to solve this (like this) equation .$$\left \lfloor x\right \rf... | 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=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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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 $2$.
$$\lim_{x\rightarrow\infty} x \cdot \sin(\sqrt{x^2+... | You did nothing wrong, so you are correct. Good job! The answer is $\frac 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Substitution to solve an initial value problem By using the substitution $y(x) = v(x)x$, how can I solve the initial value problem
$$
\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$$
$$ \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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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 com... | 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 co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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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 $f^{(n)}$ w... | 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
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to solve $\tan^2 \theta − 2 \sec \theta = 2$ Solve the given equation. Find all solutions of the equation (express your answer in terms of k, where k is any integer).
$$\tan^2 \theta − 2 \sec \theta = 2$$
What do I do to solve for $\theta$?
update: I got
$cos\theta$ $= \frac {1}{3}$ and $cos\theta$ $-1$
and then:
$... | Notice, we have $$\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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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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... | 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 $$I\equiv \left(\frac{a\frac{a\tan C}{\tan ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $\cot\angle MAB + \cot\angle MBC + \cot\angle MCA \geq 3\sqrt{3}$ Let $M$ be an arbitrary point located inside the triangle $ABC$. Prove that $$\cot\measuredangle MAB + \cot\measuredangle MBC + \cot\measuredangle MCA \geq 3\sqrt{3}$$
|
Let $A_1,B_1,C_1$ be the intersection points of $AM,BM,CM$ with $BC,CA,AB$ respectively, and let $D,E,F$ be the feet of perpendiculars from $M$ to $BC,CA,AB$ respectively.Then we have
\begin{align*}
\cot \angle MAB+\cot \angle MBC+\cot \angle MCA&=\dfrac {FA} {FM}+\dfrac {BD} {MD}+\dfrac {CE} {ME} \\
&=\sqrt{\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Factor $k^{4}+4k^{3}+8k^{2}+8k+4=0$ over $\mathbb C$ Any idea how to factor the polynomial $k^{4}+4k^{3}+8k^{2}+8k+4$ over $\mathbb C$?
Candidates for rational roots are $\pm1, \pm2, \pm4$ but none of them satisfies $k^{4}+4k^{3}+8k^{2}+8k+4=0$.
| Note that
\begin{align*}
k^4 + 4k^3 + 8k^2 + 8k + 4
&= (k^4 + 4k^3 + 6k^2 + 4k + 1) + (2k^2 + 4k + 2) + 1 \\
&= (k + 1)^4 + 2(k+1)^2 + 1 \\
&\ge 1 \text{ for all } k,
\end{align*}
so the polynomial has no real roots.
However, it does factor into two quadratics.
From the above,
\begin{align*}
k^4 + 4k^3 + 8k^2 + 8k + 4
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$ prove that $\int_0^\frac{\pi}{4} \frac{1-\cos x}{x} \,dx < \frac{\pi^2}{64}$
I showed that in
$$ \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{... | 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
$$\frac{1-\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find all integer sided triangles whose area equals their perimeter. Find all integer sided triangles whose area equals their perimeter?
Taking sides to be $e,f,g$ we have,
$$\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=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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 $x^6-1=(x^3-1)(x^3+1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)$... | 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 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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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^... | {
"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
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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.
| $$3z^2+6z+3i=0\Longleftrightarrow$$
$$3(z^2+2z+i)=0\Longleftrightarrow$$
$$z^2+2z+i=0\Longleftrightarrow$$
$$z=\frac{-2\pm\sqrt{2^2-4\cdot 1 \cdot i}}{2\cdot 1}\Longleftrightarrow$$
$$z=\frac{-2\pm\sqrt{4-4\cdot i}}{2}\Longleftrightarrow$$
$$z=\frac{-2\pm\sqrt{4-4i}}{2}\Longleftrightarrow$$
$$z=\frac{-2+\sqrt{4-4i}}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$n^{th}$ derivative of $y=x^2\cos x$ I am stuck with Leibniz formula
$$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{pmatri... | {
"language": "en",
"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... | 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}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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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 $\sin x+\cos x=u \text{ and } \sin x \cos x=v \implies \sin^2x+\cos^2x+2\sin x\cos x=u^2 \implies v=\frac {u^2-1} 2$
similarl... | 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
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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
$$
\sum \limits_{i = 1}^{k+1} \sqrt{a_i - (i - 1)} = \frac 12 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Proving limit of difference is $0$: $\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$ I am trying to show that
$$\lim_{n\rightarrow\infty} [(n+1)^{\frac{1}{7}} - (n)^{\frac{1}{7}}] = 0$$
and I know intuitively this must be so since the "+1" contribution in the first term becomes negligible as $n... | $$
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+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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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.... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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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.... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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... | 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",
"answer_count": 1,
"answer_id": 0
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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:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$... | I'd like to add one more answer "by color-coding".
First, observe that
$$
\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}}},
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 0
} |
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... | 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 $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+6- 2\sqrt{2}\left ( \sqrt{\frac{1-a}{a}}+\sqrt{\frac{1-b}{b}} + \sqrt{\frac{1-c}{c}}\right )=\sum \left ( \frac{1-a}{a}-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $3x^2 -2x+7=0$ then $\left(x-\frac{1}{3}\right)^2 =$? If $\ 3x^{2}-2x+7=0$ then $$\left(x-\frac{1}{3}\right)^2 =\text{?} $$
I am so confused. It is a self taught algebra book.
The answer is: $ \large -\frac{20}{9}$ but I don't know how it was derived.
Please explain.
Thanks for everyone who commented! I understand ... | Notice, $$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 #C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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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 s... | You can prove it as follows:
for every $\varepsilon >0$ and $m+1>\left| a \right| $ and if $n$ is big enough then
$$0<\left| \frac {a^n}{ n! } \right| =\frac { \left| a \right| }{ 1 } \cdot \frac { \left| a \right| }{ 2 } \cdots \frac { \left| a \right| }{ m } \cdot \frac { \left| a \right| }{ m+1 } \cdots \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove by induction: for $n \ge 0$, $\frac{(2n)!}{n!2^n}$ is an integer Another prove by induction question:
for $n \ge 0$, $$\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$$
Induction step: please help
| Assume $\frac{(2n)!}{n!2^n}$ is an integer. $\frac{2(n+1)!}{(n+1)!2^{n+1}}=\frac{(2n+2)\cdot(2n+1)\cdot(2n)!}{(n+1)\cdot n!\cdot 2^n\cdot 2}=\frac{(2n)!}{n!2^n}\cdot \frac{(2n+2)(2n+1)}{2(n+1)}$. Notice that $\frac{(2n)!}{n!2^n}$ is an integer by the inductive assumption, and that $\frac{(2n+2)(2n+1)}{2(n+1)}=\frac{2(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$?
Find a closed form expression for $$\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}$$
where $n$ is an even positive integer.
I tried using binomial identities, but since the binomial coefficient is in the denominator, I can't do much. Also, I'm seeking a non-calculus, el... | We seek to find a closed form of
$$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 1
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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
$$\Omega^2+\Ome... | {
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"url": "https://math.stackexchange.com/questions/1403505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Express roots in polynomials of equation $x^3+x^2-2x-1=0$
If $\alpha$ is a root of equation $x^3+x^2-2x-1=0$, then find the other two roots in polynomials of $\alpha$, with rational coefficients.
I've seen some other examples [1] that other roots were found for equations with certain properties (having only even-powe... | If $a$ is one root, then we get
$$
\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... | {
"language": "en",
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"question_score": "8",
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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,
$$e^x=\lim_{n\to\infty }\sum_{k=0}^n \binom{n}{k}\left(\frac{x}{n}\right)^k=\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{(n-k)!n^k}... | 1. Solution using $\,e^{\,x} = \dfrac{d}{dx}\,e^{\,x}\,$ property of exponent
Assume that the exponent function can be represented as a series with unknown coefficients:
$$
e^{\,x} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots = \sum_{n=0}^{\infty} a_n x^n
$$
Recall the fundamental property of exponent $\, \dfrac{d}{dx} \... | {
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"url": "https://math.stackexchange.com/questions/1407111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Summing n times binomial(n,k) I'm trying to do $\sum_{n=a}^b \left( \begin{array}{rl} n \\ a \end{array} \right) n $ .
Is there a formula, that anybody knows?
| Use generating functions. First:
$\begin{align}
\sum_{n \ge 0} \binom{n}{a} z^n
&= \frac{z^a}{(1 - z)^{a + 1}} \\
z \frac{\mathrm{d}}{\mathrm{d} z} \sum_{n \ge 0} \binom{n}{a} z^n
&= \sum_{n \ge 0} n \binom{n}{a} z^n \\
&= \frac{z^a (z + a)}{(1 - z)^{a + 2}}
\end{align}$
But also, to get partial sums you divide b... | {
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"url": "https://math.stackexchange.com/questions/1408954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Differentiation using Chain Rule Find $\frac{dy}{dx}$ if $y=7+5^{x^2+2x-1}$.
So far I have done $\frac{dy}{dx}=(5^{x^2+2x-1})'$. Now, the RHS can be found by $(e^{\ln 5\cdot (x^2+2x-1)})'=e^{\ln 5\cdot (x^2+2x-1)}(x^2+2x-1)'\ln 5=5^{x^2+2x-1}(2x+2)\ln 5$.
However, my textbook says the answer is $(2x^3+6x^2+2x-2)(5^{x^2... | Your work is correct
$$\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 ... | {
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"url": "https://math.stackexchange.com/questions/1409543",
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"source": "stackexchange",
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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 ... | 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
$$\log \prod_{k = n+1}^\infty \biggl(1 - \frac{1}{4k^2}\biggr) = \sum_{k = n... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim\limits_{n\to\infty}\frac1{n}\sum\limits_{k=1}^{\infty}\left\lfloor\frac{n}{3^k}\right\rfloor=\frac{1}{2}$
Show that
$$\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 ... | $$0\leq \frac{n}{3^k}-\left\lfloor\frac{n}{3^k}\right\rfloor\leq 1$$
and the number of non-zero terms of the sum is bounded by $1+\log_3(n)$, hence:
$$\begin{eqnarray*} \sum_{k=1}^{+\infty}\left\lfloor\frac{n}{3^k}\right\rfloor=\sum_{k=1}^{\left\lceil\log_3(n)\right\rceil}\left\lfloor\frac{n}{3^k}\right\rfloor&\geq& -(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1412638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that $(1+\frac{1}{n})^n+\frac{1}{n}$ is eventually increasing I would like to find a way to show that the sequence $a_n=\big(1+\frac{1}{n}\big)^n+\frac{1}{n}$ is eventually increasing.
$\hspace{.3 in}$(Numerical evidence suggests that $a_n<a_{n+1}$ for $n\ge6$.)
I was led to this problem by trying to prove by ind... | As suggested by Clement C., let:
$$ 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+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
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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 $3$, ... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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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 unde... | Just exploiting the arctangent Taylor series,
$$\begin{eqnarray*} \int_{0}^{\frac{\pi}{2}} \arctan(\sin^3 x)\,dx &=& \sum_{m\geq 0}\frac{(-1)^m}{2m+1}\int_{0}^{\frac{\pi}{2}}\sin^{6m+3}(x)\,dx\\ &=& \frac{3\sqrt{\pi}}{4}\sum_{m\geq 0}\frac{(-1)^m\,\Gamma\left(3m+2\right)}{(3m+\frac{3}{2})\Gamma\left(3m+\frac{5}{2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
Differentiating $y=x^{-3/2}$ I'd like to differentiate $y=x^{\large\frac{-3}{2}}$. So,
$$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}$$
Eliminating less significant higher degrees of $dx$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving $(y'+1)\ln{\frac{x+y}{x+3}} = \frac{x+y}{x+3}$ Hello everyone :) This is another task I'm trying to solve and can't seem to get the same result as Wolfram Alpha.
Solve the following differential equation: $$(y'+1)\ln{\frac{x+y}{x+3}} = \frac{x+y}{x+3}$$
My attempt:
Use substitution $$e^z = \frac{x+y}{x+3}$$
... | I managed to solve this differential equation. Following from the last part of the question:
$$
-\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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$ $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$
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 ... | $\int(x^{12}+x^8+x^4)(2x^8+3x^4+6)^{1/4}dx$=$\int x^4(x^8+x^4+1)(2x^8+3x^4+6)^{1/4}dx$
$\int (x^{11}+x^7+x^3)(2x^{12}+3x^{8}+6x^{4})^{1/4}dx$
Now just put $(2x^{12}+3x^{8}+6x^{4})$=$t$ so that $dt$=24$(x^{11}+x^7+x^3)$$dx$ and hence your integral finally simplifies to
(1/24)$\int(t^{1/4})dx$ which can be simplified to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Check convergence of $\sum_{n=1} \frac{n+1}{n^3+4n}$ I have to check convergence of $\sum_{n=1} \frac{n+1}{n^3+4n}$.
$$\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}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find equation for circle related to a triangle write the equation of the circle of the triangle
with vertices $$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
$$
Subtract the second equation from the first one and the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer
How do I prove the above expression using mathematical induction? So f... | If $n$ is even,
$$
\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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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:
$$\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+\frac{1}{2\cdot4\cdot6}+ \fr... | $$ \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}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
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
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
... | 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 $u$:
$$\frac{\sqrt[3]{x} - 1}{2(\sqrt{x} - 1)} = \frac{(\sqrt[6]{x})^2 - 1}{2 ((\sqrt[6]{x})^3 - 1)} = \frac{u^2 - 1}{2(u^3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1434528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Suppose that elements $a, b$ and $a+b$ are units in a commutative ring $R$. Show that $a^{-1} + b^{-1}$ is also a unit.
Here is what I have:
$a+b =b+a$ since $R$ is commutative.
Now,
$$(b+a) \cdot b... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
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 g... | 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. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1435718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
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]^{\fr... | 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:
$$
\ln y = \lim_{n\rightarrow \infty}\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1440972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
An interview question: $2.1^{3.1}$ vs $3.1^{2.1}$,$ 2.1^{4.1}$ vs $4.1^{2.1}$, which is larger? While Mathematica told me that $2.1^{3.1} - 3.1^{2.1} = -0.786932$ and $2.1^{4.1} - 4.1^{2.1} = 1.58855$, I wonder how to compare them quickly, by hand.
I see $2^3 < 3^2$, so perhaps we have $2.1^{3.1} < 3.1^{2.1}$, but I'm ... | Let's consider
$$(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^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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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
$$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \geqslant \frac32 ... | Abstract :
We suggests a way to show the OP inequality so some part are left to the reader because there are easy . We spilt the problem in two cases , first case $2c-2b\le a$ and the second $2c-2b\ge a$ . As the problem is homogenous we suppose in addition $1\ge a\ge c\ge b$.Following that we use convexity and deri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 2,
"answer_id": 1
} |
Is there another way to solve this quadratic equation? $$\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+... | 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 w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 1
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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... | 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.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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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...
$$... | Notice, $$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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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$$\left(x+\frac1x\right)^4 = \left(x^2+\frac1{x^2}\right)^2 = x^4+2+\frac1{x^4} = 49\implies x^4+\frac{1}{x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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How to calculate radius of the surface of water at different points in time when it is poured into a spherical container? So I have a sphere with radius $1\ cm$, and I'm pouring in water at $0.5\ cm^3/s$.
How would I find a the radius of the surface of the water at any given time? So in the end this would look like a ... | Notice, let $r$ be the radius of the circular surface of water at any time $t$ then the volume of water poured (in a sphere of radius $R=1\ cm$ ) at the same time $t$ is given by using geometry $$\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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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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 ch... | 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 fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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 od... | $p^2-2q^2\equiv p^2+q^2\equiv 1\pmod{3}$.
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",
"answer_count": 6,
"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 anot... | 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
$$\frac{5}{(z+2)(z-3)}=\frac{1}{z-3} - \frac{1}{z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Issue with trigonometry identity related to condition number of matrix So, in attempting to compute the condition number for the 2-norm of a matrix, I have stumbled upon a problem i can't resolve.
I have the formula
$$
\frac{1-\cos\left(\frac{n}{n+1} \pi\right)}{1-\cos\left(\frac{1}{n+1} \pi\right)}.
$$
I arrived at ... | You have the right idea in exploiting the identity
$$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{$\star$}$$
Now, simply observe that the arguments of the sines combine very c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}$ Can you tell me if it is possible to solve $$\arctan{\sqrt{\frac{10-x}{x}} + \arccos\sqrt{\frac{10-x}{10}}}= \frac\pi{2}\ ?$$
wolfram doesn't say it is not computable, but says there are no real solutions, whereas we know the sol... | Let $\arctan\sqrt{\dfrac{10-x}x}=y\ \ \ \ (1)$
$\implies 0\le y\le\dfrac\pi2$ as $\sqrt{\dfrac{10-x}x}\ge0$
$\dfrac{10-x}x=\tan^2y\iff x=10\cos^2y$
$\implies\sqrt{\dfrac{10-x}{10}}=+\sin y$ as $0\le y\le\dfrac\pi2$
$\implies\arccos\sqrt{\dfrac{10-x}{10}}=\arccos(\sin y)=\dfrac\pi2-\arcsin(\sin y)$
$\implies\arccos\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
if $\sin{(aw)}+\sin{(bw)}+\sin{(cw)}=3$ find $w$ range let $w$ is postive integer,if there exist $a,b,c(\pi\le a<b<c\le 2\pi)$such
$$\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}... | We have
$$\begin{align}&\sin(aw)+\sin(bw)+\sin(cw)=3\\\\&\iff \sin(aw)=\sin(bw)=\sin(cw)=1\\\\&\small\iff\text{There exist three integers $k,l,m$ such that $aw=\frac{\pi}{2}+2k\pi,bw=\frac{\pi}{2}+2l\pi,cw=\frac{\pi}{2}+2m\pi$}\\\\&\small\iff\text{There exist three integers $k,l,m$ such that $\pi\le\frac{\pi+4k\pi}{2w}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Limit in complex space How to prove that the limit in the complex space:
$$\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+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
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\r... | \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",
"answer_count": 4,
"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^... | $$\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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1477375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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",
"url": "https://math.stackexchange.com/questions/1480743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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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$ e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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... | 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}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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 num... | 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 equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 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$ 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$$
I tried to prove it by using the Taylor series, But I've found it no... | Note that your series is $f(\frac{1}{2})$, where
$$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$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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 hyperboli... | 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 $$\displa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Number of cubes from sphere Reference : 2nd question -
.
From a spherical ball of 10cm radius, how many complete cubes of 5cm
side can be extracted out, if sphere cannot be molten
What I have tried.
*
*Volume of sphere = $\frac43 \pi 10^3$
*volume of cube = $5^3$
So find how many $5^3$ can appe... | Your computation yields only an upper bound of $33.51\ldots$ (so actually of $33$), but that is by sheer volume (so "with melting").
As the suggested answer show, this bound seems to be way too big.
Arranging $8$ cubes in a $2\times 2\times 2$ pattern produces a larger cube of side length $10$ and diameter $10\sqrt 3\a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.
Thanks in advance.
| You got the right expression. $x^7+1=0$
The roots of the equation will be like $$x= \cos (\frac{2k\pi}{7})+ i\sin (\frac{2k\pi}{7}) , 0\le k\le6$$
Note: Stress on be like, this formula will not give you the exact roots. You have to change it a bit to account for the roots of negative unity.
EDIT: Complete solution fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that the sequence $a_n=1+\frac1{\sqrt{2}} +\frac1{\sqrt{3}} + \dots + \frac1{\sqrt{n}}$ is not Cauchy
Show directly (from the definition) that if
$$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\, ... | 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
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487746",
"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... | 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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1488976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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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 ... | 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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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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: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3}... | That fact comes from Galois Theory. Under the assumption that $K/F$ is Galois, the (monic) minimal polynomial $f$ of an element $\alpha \in K$ (over $F$) is of the form
$$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$
This implies that $\sigma \alpha$ are all roots of the minimal polynomial for $\alph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Find the solution of the differential equation that satisfies ${{dP} \over {dt}} = 8\sqrt {Pt} ,\,P(1) = 5$ Please help. My homework is grading my answer as incorrect, but I can't tell what I did wrong. The second photo is the work of the problem done correctly but with dp/dt=2sqrt(Pt). I based my work off of this, but... | $$\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{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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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... | 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 $.
Con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1494876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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 o... | 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 play... | {
"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 $D$ di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is $\arctan n$ always equal to $\arccos\sqrt{\frac{1}{n^2+1}}$? $$\arccos\sqrt\frac{1}{2}=\arctan 1$$
$$\arccos\sqrt\frac{1}{5}=\arctan 2$$
$$\arccos\sqrt\frac{1}{10}=\arctan 3$$
$$\arccos\sqrt\frac{1}{17}=\arctan 4$$
$$\arccos\sqrt\frac{1}{26}=\arctan 5$$
$$\arccos\sqrt\frac{1}{37}=\arctan 6$$
$$\arccos\sqrt\frac{1}{5... | $\arccos \sqrt \frac 1 n = x$ means $cos x = \frac 1 n$ which means $\sin x = \sqrt{1 - {\sqrt \frac 1 n}^2} = \sqrt{1 - \frac 1 n}$ so $\tan x = \frac{ \sqrt{1 - \frac 1 n}}{\sqrt \frac 1 n} = \sqrt{n - 1}$ so $\arctan{n -1} = x = \arccos \sqrt \frac 1 n$. So this is known.
====
Another way of looking at it is:
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1498718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving A Linear Recurrence Relation With Complex Roots Question:
For the given linear homogeneous difference equation, find the general solution:
$$y_{n+2} + y_{n+1} + y_n = 0$$
With the initial conditions of:
$$y(0)=\sqrt3, y(1) = 0$$
Attempted Answer:
I approached the problem normally as one would except by solvin... | What happens
when the roots of the
characteristic polynomial are complex
is that the solutions
have a periodic component.
Your case is
$y_n
= a \left(\frac{-1}{2} - \frac{i\sqrt3}{2}\right)^n
+ b \left(\frac{-1}{2} + \frac{i\sqrt3}{2}\right)^n
=a r^n + bs^n
$.
Note that
$r+s = -1$
and
$r-s = -i\sqrt{3}$.
If
$y_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1499090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Show that $\sum_0^{\infty}\frac{z^{2^n}}{z^{2^{n+1}}-1}$ converges to $\frac{z}{z-1}$ or $\frac{-1}{1-z}$ How do we show that the sum below converges to $\frac{z}{z-1}$ if $|z|<1$ and $\frac{-1}{1-z}$ if $|z|>1$?
$$\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}+...$$
An... | Note
\begin{align}\frac{z^{2^n}}{z^{2^{n+1}}-1} &= \frac{z^{2^n}(z^{2^n}-1)}{(z^{2^n}-1)(z^{2^{n+1}}-1)} = \frac{z^{2^{n+1}}-z^{2^n}}{(z^{2^n}-1)(z^{2^{n+1}}-1)} \\
&= \frac{(z^{2^{n+1}}-1) - (z^{2^n}-1)}{(z^{2^n}-1)(z^{2^{n+1}}-1)}\\
& = \frac{1}{z^{2^n} - 1} - \frac{1}{z^{2^{n+1}}-1}.
\end{align}
So for every positiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1502741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as $x$ varies over the reals?
How to find the maximum value of $3^x + 5^x - 9^x + 15^x - 25^x$ as x
varies over the reals ?
Suggestions please!
| $$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) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1504006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the indefinite integral $\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $ Find the indefinite integral
$$\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\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1505148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
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
$$\left[(a^2x)^2+c^4... | Solving the equation requires you to compute square roots of the discriminant$$\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.$$Taking the product of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1506409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Divisibility of a number $N$ by $4,7,9$
Let $$N=\binom{20}{7}-\binom{20}{8}+\binom{20}{9}-\binom{20}{10}+\binom{20}{11}-\binom{20}{12}+.......-\binom{20}{20}.$$
Then $N$ is divisible by
$\bf{Options::}\;\;:: (a)\; 4\;\;\;\;\;\; (b)\;\; 7\;\;\;\;\;\; (c)\;\; 9\;\;\;\;\;\; (d)\;\; None\; of\; these.$
$\bf{My\; Try::}$... | You began as I would have. If you have a calculator, you can just find the value of $N$ directly. If you don't have a calculator or you want to avoid such calculations, you can proceed using only modular arithmetic, which is easy enough to do without a calculator.
Each fraction in your last formula for $N$ is an intege... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1507415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to find the range of $1 / (1+x^2)^{1/2}$? How to find the range of $$\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}$$ but I don't have clear how to solve th... | We try to rewrite $x$ in terms of $y$:
$$
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... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1509833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.