Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$
| Let $\triangle$ denote the area.
$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$
Take $b+c-a=x$
$c+a-b=y$ and $a+b-c=z$.
$\dfrac{x}{2}=(s-a)$
$\dfrac{y}{2}=(s-b)$
$\dfrac{z}{2}=(s-c)$
$\dfrac{x+y+z}{2}=s$
Now use AM-GM inequality.
$\dfrac{x+y+z}{4} \ge ((xyz)(x+y+z))^\frac{1}{4}$
$\dfrac{2}{x+y+z} \le \dfrac{1}{((xyz)(x+y+z))^\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How can I determine limit: $\lim\limits_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$? How can I determine limit:
$$\lim_{x\to4}\frac{x-4}{x-\sqrt[]{x}-2}$$
Thanks
| If you are unfamiliar with L'Hopital's rule, this is what you can do.
$$ \lim_{x \rightarrow 4} \frac{ x - 4 } { x - \sqrt{x} - 2} = \lim_{x \rightarrow 4} \frac{ (\sqrt{x} - 2) ( \sqrt{x} + 2) } { (\sqrt{x} -2) ( \sqrt{x} +1) }= \lim_{x \rightarrow 4} \frac{ ( \sqrt{x} + 2) } { ( \sqrt{x} +1) } = \frac{ 4}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/365856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Strings and Substrings So here is one of the last homeworks we are doing in my Discrete math class. It seems like it should be simple but I am really stuck. Any help would be greatly appreciated.
*
*Find the ordinary generating series with respect to length
for the strings in $\{0,1,2\}^*$ having no "$22$" su... | A general way to solve such problems is to set up a system of recurrences. Call $a_n$, $b_n$ and $c_n$ the number of strings of length $n$ of interest that end in 0, 1, 2 respectively. Then:
$$
\begin{align*}
a_{n + 1} &= a_n + b_n + c_n \\
b_{n + 1} &= a_n + b_n + c_n \\
c_{n + 1} &= a_n + b_n
\end{align*}
$$
It is al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
expand $ \arctan\left(\frac{3x+2}{3x-2}\right)$ into pwer series, find radius of convergence (check solution) I would be grateful if someone could check what I've worked out:
$$ f(x)=\arctan\left(\frac{3x+2}{3x-2}\right)\implies f'(x)=\frac{1}{1+(\frac{3x+2}{3x-2})^2}\cdot \frac{3(3x-2)-3(3x+2)}{(3x-2)^2}$$
$$=\frac{(... | You missed $(-1)^k\left(\frac{3}{2}\right)^{k}:$ $$f'(x)=-\frac{3}{2} \sum_{k=0}^{\infty}\left(-\frac{3}{2}x^2\right)^k=\frac{3}{2} \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}x^2\right)^k= \sum_{k=0}^{\infty}(-1)^{k+1}\left(\frac{3}{2}\right)^{k+1} x^2.$$
Then
$$
f(x)=\int{f'(x)\,dx}= \sum_{k=0}^{\infty}(-1)^{k+1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/370087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$
Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$
No idea if I'm on the right track. May... | We have
$$I(a) = \int_{0}^{2 \pi} \dfrac{dx}{1+a^2-a \cos(x)} = \dfrac1{1+a^2} \int_0^{2\pi} \dfrac{dx}{1 - \dfrac{a}{1+a^2} \cos(x)}$$
Note that we always have $b = \dfrac{a}{1+a^2} < 1$. Hence, we can write
$$\dfrac1{1 - b \cos(x)} = \sum_{k=0}^{\infty}b^k \cos^k(x)$$
Now we have
$$\int_0^{2\pi} \dfrac{dx}{1 - b \cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/370711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show abelian groups of order 3240? Show how to get all abelian groups of order $2^3 \cdot 3^4 \cdot 5$.
I just started learning this and was wondering how you would do this?
Is this correct?
$2^3 \cdot 3^4 \cdot 5 = 3240$. Therefore the number of abelian groups of order $3240$ is $3 \cdot 4 = 12$.
Is this the entire pr... | You have the right idea, but there are 5 abelian groups of order $3^4$, not 4. You can have:
*
*$\def\zt{\Bbb Z_3}\Bbb Z_{81}$
*$ \Bbb Z_{27}\times\zt$
*$\Bbb Z_{9}\times\Bbb Z_{9}$
*$\Bbb Z_{9}\times\zt\times\zt$
*$\zt\times\zt\times\zt\times\zt$
These correspond to the 5 ways (not 4) that you can express 4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving this 3-degree polynomial I'm trying to factor the following polynomial by hand:
$-x^3 + 9x^2 - 24x + 20 = 0$
The simplest I could get is:
$-x^2(x-9) - 4(5x+5) = 0$
Any ideas on how I could go ahead and solve this by hand? This seems pretty tough.
| You have not a polynomial, rather an equation, whose LHS is the cubic polynomial $$-x^{3}+9x^{2}-24x+20.\tag{0}$$
To get rid of the negative coefficient of $x^3$ multiply $(0)$ by $-1$
$$
\begin{equation*}
-\left( -x^{3}+9x^{2}-24x+20\right) =x^{3}-9x^{2}+24x-20.\tag{1}
\end{equation*}
$$
By trial and error$^1$, inspec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Sum of the first $n$ triangular numbers - induction Question:
Prove by mathematical induction that $$(1)+(1+2)+(1+2+3)+\cdots+(1+2+3+\cdots+n)=\frac{1}{6}n(n+1)(n+2)$$ is true for all positive integers n.
Attempt:
I did the the induction steps and I got up to here:
$$RTP:\frac{1}{6}n(n+1)(n+2)+(1+2+3+\cdots+n+(n+1))=\f... | You can use two nested inductions.
In order to show the equality $\forall n, 1 + (1+2) + (1+2+3) + \ldots (1+2+\ldots+n) = \frac 1 6 n (n+1) (n+2)$,
you check that it is true for $n=1$, and then you are left to show the equality $\forall n, \frac 1 6 n (n+1) (n+2) + (1+2+ \ldots+n+(n+1)) = \frac 1 6 (n+1) (n+2) (n+3)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Complex analytic function and Cauchy-Riemann conditions question here again asking for some guidance
Let me state the problem and working on and then my question,
So I have two functions:
$$u(x, y) = \frac{x(1+x)+y^2}{(1+x)^2 + y^2}$$
and
$$v(x, y) = \frac{y}{(1+x)^2 + y^2}$$
and I'm being asked if they can serve as re... | The Cauchy-Riemann equations on a pair functions $u(x,y)$ and $v(x,y)$ are the two equations:
$\displaystyle \frac{du}{dx} = \frac{dv}{dy}$ and $\frac{du}{dy} = -\frac{dv}{dx}$.
Checking these two equations for the given functions $u(x,y)$ and $v(x,y)$ gives that
$\displaystyle \frac{du}{dx} = \frac{1+ 2x + x^2 - y^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Differentiating $\;y = x a^x$ My attempt:
$$\eqalign{
y &= x{a^x} \cr
\ln y &= \ln x + \ln {a^x} \cr
\ln y &= \ln x + x\ln a \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr
{1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\... | Proceeding with your method in approaching the problem, but recalling that we can take $a$ as representing some constant: that is, $\;$ '$a$' $\;$ is a value that is NOT a function of $x$, and so does not get differentiated with respect to $x$. Similarly, so we can take $\ln a$ to be a constant.
So in your third line,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Is the sum of two singular matrices also singular? If $A$ and $B$ are $n \times n$ singular matrices, is $A+B$ also singular?
|
The sum of two singular $n × n$ matrices may be non-singular.
For example, consider two $2 × 2$ matrix $A=\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}$ and $B=\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}$
Here $\det A = 1-1=0$ and $\det B=1-1=0$
So both the matrix $A \quad \text{and} \quad B$ are singular matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/376923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Continued fraction for $\sqrt{14}$ I'm referencing this page: An Introduction to the Continued Fraction, where they explain the algebraic method of solving the square root of $14$.
$$\sqrt{14} = 3 + \frac1x$$
So, $x_0 = 3$, Solving for $x$, we get
$$x = \frac{\sqrt{14} + 3}5$$
However, in the next step, how do we get t... | It's really easy to compute $\left\lfloor \frac{\sqrt a + b}{c}\right\rfloor$ for integers $a,b,c$. Just use the fact that
$$\left\lfloor \frac{r + b}{c}\right\rfloor = \left\lfloor \frac{\lfloor r \rfloor + b}{c}\right\rfloor$$
for real $r$ and integers $b,c$.
Here, $\lfloor\sqrt{14}\rfloor = 3$, so $x = \left\lfloor\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
help in solving a recursion EDIT: Turns out that the given solution has an error in it.
I have the following recursion question and following that is my answer. The problem is it doesn't seem to agree with the marking scheme. Can you help me by pointing out where i have gone wrong?
Question:
Use substitution to solve t... | So u have the the the difference between the nth term and (n-1)th term is n^2
And the zeroth term is 7,
Thus f(0) = 7, f(1) = 8, f(2) = 12, f(3) = 21
Now we know the difference is quadratic in the term number so the function itself will be (via integration) cubic. Thus it has form
F(n) = an^3 + bn^2 + cn + d
We know d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A series: $1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\cdots$ Denote $$b_1=1,b_{n}=b_{n-1}-\dfrac{S(b_{n-1})}{n},(n>1 )\tag1$$
where $S(x)=1$ if $x>0,S(x)=-1$ if $x<0$, and $S(0)=0.$
So $b_n=1,\dfrac{1}{2},\dfrac{1}{6},-\dfrac{1}{12},\dfrac{7}{60},-\dfrac{1}{20},\dfrac{13}{140},-\dfrac{9}... | I can't answer question 1, but I do have an answer for question 2. In particular, the fact that
$$
\left|\sum_{i=4}^n S(b_i)\right| \leq 1
$$
is an immediate consequence of the following theorem.
Theorem. For all $n\geq 2$, the terms $b_{2n}$ and $b_{2n+1}$ have opposite signs.
Proof: It is easy to check that the theo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
To find the x and y-intercepts of the line $ax+by+c=0$ Please check if I've solved the problem in the correct way:
The problem is as follows:
Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.
My solution:
We have to find the x and y-intercepts of the line.... | Your work is exemplary, Samama. You know your definitions well, and your answers are entirely correct.
A nice "shortcut" is to take advantage of what you already know:
*
*the $x$ intercept is the value of $x$ when $y = 0$, and
*the $y$ intercept is the value of $y$ when $x = 0$.
*$ax + by + c = 0 \iff ax + by = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/388873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving that $x - \frac{x^3}{3!} < \sin x < x$ for all $x>0$
Prove that $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$
This should be fairly straightforward but the proof seems to be alluding me.
I want to show $x - \frac{x^3}{3!} < \sin(x) < x$ for all $x>0$. I recognize this shouldn't be too difficult but perh... | A possible solution without use a Taylor series.Observe that:
$\\ \\ \displaystyle \sin(3\gamma)=\sin(2\gamma)\cos(\gamma)+\sin(\gamma)\cos(2\gamma)=2\sin(\gamma)\cos^2(\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=2\sin(\gamma)(1-\sin^2\gamma)+\sin(\gamma)(1-2\sin^2(\gamma))=3\sin(\gamma)-4\sin^3(\gamma)\Rightarrow \sin^3(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Find max of $x^7+y^7+z^7$
Find max of $x^7+y^7+z^7$ where $x+y+z=0$ and $x^2+y^2+z^2=1$
I tried to use the inequality:$$\sqrt[8]{\frac {x^8+y^8+z^8} 3}\ge\sqrt[7]{\frac {x^7+y^7+z^7} 3}$$ but stuck
| Here is an elementary way to proceed, which may not be wholly convenient, but which illustrates some interesting techniques.
If $x,y,z$ are the roots of a cubic $t^3-s_1t^2+s_2t-s_3=0$, then we have$$s_1=x+y+z=0$$$$2s_2=2xy+2yz+2zx=(x+y+z)^2-(x^2+y^2+z^2)=-1$$
Thus we have (for some $u$) $$2t^3-t-u=0$$
This gives also ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/391454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
finding nth term Let
3,8,17,32,57 . . . . .
be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.
| If you want the smallest degree polynomial that gives these numbers (which is only one of the infinitely many possible explanations for them), we can use the calculus of finite differences:
$$\fbox{3}\quad 8\quad 17\quad 32\quad 57 \\
\fbox{5}\quad 9\quad 15\quad 25\\
\fbox{4}\quad 6\quad 10\\
\fbox{2} \quad 4\\
\fbox{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Partial fraction decomposition with a nonrepeated irreducible quadratic factor I'm trying to do a partial fraction decomposition on the following rational eqn with a nonrepeated irreducible quadratic factor:
$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)}$$
I've broken it down into an identiy: $-28x^2 -92 = A((x-4)(x^2-1))+B(x^2+1... | This is what you have done...
$$\dfrac{-28x^2-92}{(x-4)^2(x^2+1)}
= \dfrac{A}{(x-4)} + \dfrac{B}{(x-4)^2} + \dfrac{Cx+D}{x^2+1}$$
$$-28x^2 -92 = A(x-4)(x^2+1) + B(x^2+1) + (Cx+D)(x-4)^2$$
Letting $x=4$, we get
$$-540 =17B$$
$$B = -\dfrac{540}{17}$$
What I would do now...
Let $B = -\dfrac{540}{17}$ and simplify.
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Circular motion trig We have $x_P = -2 + 4 \cos (-\pi t)$ and $y_P = 1 + 4 \sin ( - \pi t)$ with $t$ in seconds.
We have to find the coordinates of the intersection with the y-axis. So I use trig and I eventually end up with:
$ t = -\dfrac{1}{3} + k \cdot 2$ and $ t = \dfrac{1}{3} + k \cdot 2$
The correction model do... | $$\text{If }t=2k+ \frac13,$$
$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi+\frac\pi3\right)=-\sin\left(\frac\pi3\right)=-\frac{\sqrt3}2$$
$$\text{If }t=2k- \frac13,$$
$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi-\frac\pi3\right)=-\sin\left(-\frac\pi3\right)=\sin\left(\frac\pi3\right)=+\frac{\sqrt3}2$$
So, the two coordi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem?
$$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$
It has two poles at $\pm i$ and branch point of $-1$ while the integral is to b... | If you're still interested in approaches that use contour integration, consider the function $$f(z) = \frac{\log(1+z) \log(-z)}{1+z^{2}}.$$
Using the principal branch of the logarithm, there is a branch cut along $[0,\infty)$ and a branch cut along $(-\infty, -1]$.
Then integrating counterclockwise around a keyhole co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 0
} |
How can I show this field extension equality? How can I show this field extension equality $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?
| The easier direction is $\mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$ Can you tell us why this is true?
Now we have $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$
Since $\sqrt{2}$ is a root of the degree 2 polynomial $X^2-2$ which is i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What am I doing wrong in these quartic formula calculations? I was a bit surprised that there is a general formula for the roots of a quartic equation, so I decided to test Wikipedia's version of it myself. To my surprise, I have arrived at a correct answer only once in about five attempts, using only integer coefficie... | Everytime there is a complicated cube root or square root, there is a simplification that can occur.
For example, you can compute $(35+18i\sqrt 3)^{1/3}$ approximately in $\Bbb C$, or you can magically realize that $(-1-2i\sqrt 3)^3 = 35+18i\sqrt 3$ and so you can pick $Q = -1-2i\sqrt 3$ (don't try to find this cube ro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$
I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$
I then combined like terms $x^2 + 17x + 59 = 0$
I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$
However, the answer is 3
| When you square both sides you should get $9(x+13) = (x+9)^2$ which rearranges to
\[x^2+9x-36 = 0 \ ,\]
which has the solutions
\[x_{1,2} = -\frac 92 \pm \sqrt{\frac{81}{4}+36} = \frac{-9\pm 15}{2} \: ,\]
i.e. $\begin{cases}x_1 = 3 \\ x_2 = -12\end{cases}$. By putting these into the original equation $3\sqrt{x+13} = x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/398051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Plane geometry tough question $\triangle ABC$ is right angled at $A$. $AB=20, CA= 80/3, BC=100/3$ units. $D$ is a point between $B$ and $C$ such that the $\triangle ADB$ and $\triangle ADC$ have equal perimeters. Determine the length of $AD$.
|
Add up the the sides of $\triangle ADB$ and $\triangle ADC$ and you get their perimeters and then equate them:
$x+\frac{100}{3}-y+\frac{80}{3}=x+y+20$
Solve for $y$ to get $y=20$
and then for $x$ using the cosine rule we get
$x=\sqrt{20^2+20^2-2(20)(20)\cos\left(\tan^{-1}\frac{80}{20\times3}\right)}$
$x=8\sqrt{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/400079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$ I tried to solve this inequality:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$
with
$$a+b+c+d+e=1$$
I am stuck at this. I don't want the full solution, a hint would be enough.
| Hint:
You have $1-a=b+c+d+e$ and
$a=1-(b+c+d+e)$
$\dfrac{1-(b+c+d+e)}{b+c+d+e}=\dfrac{1}{b+c+d+e}-1$ and $b+c+d+e=1-a$
Using AM $\ge$ HM
$(\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}+\dfrac{1}{1-e}) \ge \dfrac{5^2}{1+1+1+1+1-(a+b+c+d+e)} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving that either $2^n-1 $ or $ 2^n+1$ is not prime Not sure what approach to take with this:
Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$
| As $n>2,2^n-1>3$
If $n$ is even, $=2m$(say),
$2^n-1=2^{2m}-1=4^m-1$ which is divisible by $4-1=3$ as $(a-b)$ divides $a^n-b^n$ for integer $n\ge 0$
If $n$ is odd, $=2m+1$(say),
$2^{2m+1}+1=2^{2m+1}+1^{2m+1}$ which is divisible by $2+1=3$ as $(a+b)$ divides $a^{2m+1}+b^{2m+1}$ for integer $m\ge 0$
Alternatively,
$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Solving summation $2n+2^2(n-1)+2^3(n-2)+....+2^n$ Can anyone help me with this summation? I tried to use the geometric series on this, but I can't use that.
$$2n+2^2(n-1)+2^3(n-2)+....+2^n$$
I am trying to do this for studying algorithms. Can we get a closed form for this ?
| This does have a nice closed form.
We can write your original sum, $2n+2^2(n-1)+2^3(n-2)+....+2^n$, as
$(2+2^2+2^3+...2^n)+(2+2^2+2^3+...2^{n-1})+2+2^2+2^3+...2^{n-2})+...+2$, because this sum will have $n$ $2$'s, $n-1$ $2^2$'s, and so on. By the geometric series formula, this sum simplifies to $(2^{n+1}-2) + (2^n-2) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/403384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?
I thought it would be 12 this as per pemdas rule:
$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$
Wanted to confirm the right answer from you guys. Thanks for your help.
| Multiplication takes priority over addition, so this
$1+1+1+1+1+1+1+1+1+1+1+1\times0+1$
becomes:
$1+1+1+1+1+1+1+1+1+1+1+0+1$
Now add everything which gives you 12.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/405543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Why dividing by zero still works Today, I was at a class. There was a question:
If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$.
What my teacher did was this:
$x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he... | The division was a division in the polynomial ring $\mathbb{C}[x]$
by a non-zero polynomial
For example: by dividing $$x^{2}+2x+1$$ by $$x+1$$ which is a non-zero
polynomial we get $$(x+1)(x+1)=x^{2}+2x+1$$ since the quotient is
$x+1$ and the reminder is $0$. The result is an equality
of polynomials, and it is valid if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 0
} |
$X$ is an odd number, $Y$ is a natural number more than 36. If $\frac{1}{X}+\frac{2}{Y}=\frac{1}{18}$, find the set $(X,Y)$? $X$ is an odd number, $Y$ is a natural number more than 36. If $\frac{1}{X}+\frac{2}{Y}=\frac{1}{18}$, find the set $(X,Y)$ ?
Re arranging the given equation, we have,
$\frac{2}{Y}=\frac{X-18}{18... | As you have $Y=\frac{36X}{X-18},$
So, $Y=\frac{36(X-18)+36\cdot 18}{X-18}=36+\frac{3^4\cdot2^3}{X-18}$
also as, $Y>0, X>18$
and as you have identified $X-18=3^r ,0\le r\le 4 $
Alternatively
HINT:
So, $Y=\frac{36X}{X-18}$ which is even $=2Z$(say)
So, we have $$\frac1 X+\frac1 Z=\frac1{18}\implies XZ-18(X+Z)=0$$
$$\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/408984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Quadratic Equation find the value of $\lambda$ when other roots are given in restriction Problem :
If $\lambda$ be an integer and $\alpha, \beta$ be the roots of $4x^2-16x+\lambda$=0 such that $ 1 < \alpha <2$ and $2 < \beta <3$, then find the possible values of $\lambda$
My approach :
The roots $\alpha, \beta = \fr... | Using Vieta's formulas, $\alpha+\beta=4$ and $\alpha \beta=\frac{\lambda}4\implies \lambda=4\alpha \beta $
If $1< \alpha<2 \iff 1< 4-\beta <2\iff -1>\beta-4>-2\iff 3>\beta>2$
So, one condition implies the other
Now, $\lambda=4\alpha \beta=4\alpha(4-\alpha)=16-(2\alpha-4)^2$
As $1< \alpha<2 \implies 0>2\alpha-4>-2 \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove $(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$ without calculations I read somewhere that I can prove this identity below with abstract algebra in a simpler and faster way without any calculations, is that true or am I wrong?
$$(a-b)^3 + (b-c)^3 + (c-a)^3 -3(a-b)(b-c)(c-a) = 0$$
Thanks
| An equivalent identity is that $x+y+z = 0$ implies $x^3+y^3+z^3-3xyz = 0$.
So suppose $x+y+z = 0$.
Then the determinant $ \begin{vmatrix} x & z & y\\ y & x & z\\ z & y & x \end{vmatrix}$ must be zero, because the sum of the elements of each column is zero.
Expanding the determinant, we have the required result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 12,
"answer_id": 4
} |
Find the length of the parametric curve Find the length of the parametric curve defined by:
$x=t+\dfrac{1}{t}$
and
$y=\ln{t^2}$
on the interval
$(1 \le t \le 4)$.
| $$\begin{align*}s&=\int_1^4\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\\&=\int_1^4\sqrt{\left(1-\frac1{t^2}\right)^2+\left(\frac2t\right)^2}dt\\&=\int_1^4\sqrt{1-\frac2{t^2}+\frac1{t^4}+\frac4{t^2}}dt\\&=\int_1^4\sqrt{\left(\frac1{t^2}\right)^2+2\left(\frac1{t^2}\right)+1}dt\\&=\int_1^4\sqrt{\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find $\int_0^{2\pi} \frac{dt}{1+2\cos(t)}$ The problem is
$$\int_0^{2\pi} \frac{dt}{1+2\cos(t)}.$$ I know it is equal to
$$\int\limits_{|z|=1}\frac{2dz}{i(1+z)^2}$$
but I don't know how I should calculate the last integral.
| If you write
$$
\begin{equation*}
z=e^{it}\qquad \left( 0\leq t\leq 2\pi \right),
\end{equation*}
$$
since
$$
\begin{equation*}
\cos t=\frac{z+z^{-1}}{2},\qquad dt=\frac{dz}{iz},
\end{equation*}
$$
the integral takes the form
$$
\begin{eqnarray*}
\int\limits_{|z|=1}\left( \frac{1}{1+2\frac{z+z^{-1}}{2}}
\right) \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions. This is how I tried to prove it. Is it correct? Thanks!!
$2^n = x^2+23$
$x^2$ must be odd, therefore $x^2 = 4k+1$, where $k \in \mathbb{N}$.
$2^n=4k+24$
$k=2(2^{n-3}-3)$
Since $x^2=4k+1$,$ \ \ \ \ $ $k_1 = \frac{x_1^2-1}{4}$
and $k_2=\frac... | This seems wrong to me.
You went from
$k_1 = \frac{x_1^2-1}{4}$
to $x_1^2 = 4 k_1^2+1$.
Somehow the $k_1$ got squared.
This is not a correct proof.
Also, having to download that big image is a pain.
Please learn how to enter math in $\LaTeX$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
The angle at which a circle and a hyperbola intersect?
$x^2 - 2y^2 = 4$
$ (x-3)^2 + y^2 = 25 $
How do you calculate the angle at which a circle and a hyperbola intersect?
If I express $y^2$ from the first equation and apply it to the second equation, I get the following:
$y^2 = -2 + \frac{x^2}{2}$
$(x-3)^2 + -2 + \... | HINT:
$(1):$ From the article $151$ of The elements of coordinate geometry (Loney), the gradient $(m_1)$ of $x^2+y^2+2gx+2fy+c=0$ is $=-\frac{x_1+g}{y_1+f}$ where $(x_1,y_1)$ is the given point on the circle
and from the Article $305,262$ the gradient $(m_2)$ of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{b^2\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Cyclotomic integers: Why do we have $x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$? Why do we have the factorization $$x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$$ for $\zeta$ a primitive $n^{\text{th}}$ root of unity where $n$ is an odd prime?
| If $y = 0$, the factorisation reads $x^n = \underbrace{x\,.x \dots x}_{n\ \text{times}}$.
Consider $y\neq 0$. If $\zeta \in \mathbb{C}$ is a primitive $n^{\text{th}}$ root of unity then $\zeta^n = 1$ and $\zeta^k \neq 1$ for $k = 1, \dots, n-1$. In particular, $\zeta, \zeta^2, \dots, \zeta^{n-1}, \zeta^n=1$ are all dis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/416310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
About Rayleigh's formula How to use $\displaystyle j_n(x)=(-1)^nx^n\left(\frac{1} {x} \frac{d} {dx}\right)^n \left(\frac{\sin x}{x}\right)$?
for example, to find $j_3(x)$
| The formula you provided uses a compact way of writing operators.
For example, at $n = 2$, the term $\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2$ should be understood to mean the operator
$$\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2=\frac{1}{x} \frac{d}{dx}\frac1x \frac d{dx}.$$
You can see the val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to determine the rank and determinant of $A$? let $A$ be $$A_{a} = \begin{pmatrix} a & 1 & 1 & 1 \\ 1 & a & 1 & 1\\ 1 & 1 & a & 1\\ 1 & 1 & 1 & a \end{pmatrix}$$
How can I calculate the rank of $A$ by the Gauss' methode and $\det A$?
| Add all the columns to the first and we find
$$ B=\begin{pmatrix} a+3 & 1 & 1 & 1 \\ a+3 & a & 1 & 1\\ a+3 & 1 & a & 1\\ a+3 & 1 & 1 & a \end{pmatrix}$$
then subtract the first row from the other we find
$$ C=\begin{pmatrix} a+3 & 1 & 1 & 1 \\ 0 & a-1 & 0 & 0\\ 0 & 0 & a-1 & 0\\ 0 & 0 & 0 & a-1 \end{pmatrix}$$
hence
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
proving $\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+$ Without Induction i proved that:
$$
\begin{align}
& {} \quad \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+\cdots+\frac{1}{(2n-1)\cdot 2n} \\[10pt]
& =\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\text{ for }n\in \math... | $$\text{As }\frac1{(2r-1)2r}=\frac{2r-(2r-1)}{(2r-1)2r}=\frac1{2r-1}-\frac1{2r},$$
$$\text{we can write }\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}+\frac{1}{5\cdot 6}+...+\frac{1}{(2n-1)\cdot 2n}$$
$$=\left(\frac11-\frac12\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1{2n-1}-\frac1{2n}\right)$$
$$=\left(\frac11+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine whether a multi-variable limit exists $\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$ I need to determine whether the next limit exists:
$$\lim_{(x,y)\to(0,0)}f(x,y)=\lim_{(x,y)\to(0,0)}\frac{\cos x-1-\frac{x^2}2}{x^4+y^4}$$
Looking at the numerator $(-1-\frac{x^2}2)$ it immediately reminds me of ... | I think you can assume the following limit as well to show that it doesn't exist:
$$\lim_{n\to\infty}f(x,y_n),~~y_n=\pi/n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/418208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\lim_{x\to1}\frac{\sqrt{x^2+3}-2}{x^2-1}$? I tried to calculate, but couldn't get out of this:
$$\lim_{x\to1}\frac{x^2+5}{x^2 (\sqrt{x^2 +3}+2)-\sqrt{x^2 +3}}$$
then multiply by the conjugate.
$$\lim_{x\to1}\frac{\sqrt{x^2 +3}-2}{x^2 -1}$$
Thanks!
| Use L'Hospital's Rule. Since plugging in $x=1$, gives you indeterminate form, take the derivative of the numerator and the derivative of the denominator, and try the limit again.
$\lim_{x\to 1}\frac{(x^2+3)^{\frac{1}{2}}-2}{x^2-1}\implies$ (Via L
Hospital's Rule...) $\lim_{x\to 1}\frac{\frac{1}{2}(x^2+3)^{-\frac{1}{2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Proving a formula for $\int_0^\infty \frac{\log(1+x^{4n})}{1+x^2}dx $ if $n=1,2,3,\cdots$ I came across the formula
$$\int_0^\infty \frac{\log \left(1+x^{4n} \right)}{1+x^2}dx = \pi \log \left\{2^n \prod_{k=1 ,\ k \text{ odd}}^{2n-1} \left(1+\sin \left( \frac{\pi k}{4n}\right) \right)\right\} $$
where $n=1,2,3,4,\cdot... | Evaluate
$J=\int_0^\infty\frac{\ln(1+x^{4n})}{1+x^2} \,dx $
with
$1+x^{4n} = \prod_{k=1}^{2n}(1+e^{i\pi\frac{2n+1-2k}{2n} }x^2)
$
and
$\int_0^\infty \frac{\ln(1+r x^2)}{1+x^2}dx= \pi\ln(1+r^{\frac12})
$
\begin{align}
J& =\int_0^\infty\frac{dx}{1+x^2} \sum_{k=1}^{2n}
\ln (1+e^{i\pi\frac{2n+1-2k}{2n} }x^2)
= \pi\sum_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
} |
How can evaluate $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.
$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x... | Using the identity
$$
\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)
$$
we get
$$
\begin{align}
\lim_{x\to0}\frac{\sin\left(x^2+\frac1x\right)-\sin\left(\frac1x\right)}{x}
&=\lim_{x\to0}\frac{2\sin\left(\frac{x^2}{2}\right)\cos\left(\frac{x^2}{2}+\frac1x\right)}{x}\\
&=\lim_{x\to0}x\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/423938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the natural number $n$ satisfy the condition Find the natural number $n$ satisfy the condition
$$\dfrac{1}{2}C_{2n}^1 - \dfrac{2}{3} C_{2n}^2 + \dfrac{3}{4} C_{2n}^3 - \dfrac{4}{5} C_{2n}^4 + \cdots - \dfrac{2n}{2n+1} C_{2n}^{2n} =\dfrac{1}{2013}.$$
| HINT:
$$\frac r{r+1}\binom {2n}r=\binom {2n}r-\frac{(2n)!}{r!(2n-r)!(r+1)}$$
$$=\binom {2n}r-\frac1{2n+1}\cdot \binom{2n+1}{r+1}$$
$$\text{So, }\sum_{1\le r\le 2n}\frac r{r+1}(-1)^{r-1}\binom {2n}r$$
$$=\sum_{1\le r\le 2n}(-1)^{r-1}\binom{2n}r-\frac1{2n+1}\cdot \sum_{1\le r\le 2n}(-1)^{r-1}\binom{2n+1}{r+1}$$
$$\text{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Compute this limit $\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$ using L'Hôpital's rule I have asked this problem before, but I can't understand the explanation, I couldn't understand how the sin multiply for cos, and too multiply for A + and - B: $$\sin(A)-\sin(B)=2\sin\left(\frac{A-B}{2}\right)\cos\le... | It's easier if you use the plain old addition formula:
$$\sin{\left(x^2+\frac{1}{x}\right)} = \cos{x^2}\, \sin{\frac{1}{x}} + \sin{x^2} \,\cos{\frac{1}{x}}$$
which behaves as
$$\left(1-\frac12 x^2\right) \sin{\frac{1}{x}} + x^2 \cos{\frac{1}{x}}$$
so that
$$\frac{\sin{\left(x^2+\frac{1}{x}\right)} - \sin{\frac{1}{x}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/424701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$a+b+c =0$; $a^2+b^2+c^2=1$. Prove that: $a^2 b^2 c^2 \le \frac{1}{54}$ If a,b,c are real numbers satisfying $a+b+c =0; a^2+b^2+c^2=1$.
Prove that $a^2 b^2 c^2 \le \frac{1}{54}$.
| Several good answers have been posted. Just feeling like adding a single variable solution with a geometric twist.
The set of points $(a,b,c)$ is the intersection of the unit sphere and the plane $T:x+y+z=0$ with normal vector $(1,1,1)$. IOW a circle of radius one in the plane $T$ centered at the origin. An occasionall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 1
} |
How to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$? I don't know how to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$,appreciate any help!
Is there any general rule for solving such problems?
| $$\text{As } e^x=\sum_{0\le r<\infty}\frac{x^r}{r!} $$
$$\text{and }\frac{(2n+1)}{n!}x^{2n+1}=2x^3 \cdot \frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \frac{(x^2)^n}{n!}$$
$$\text{So,} \sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$$
$$=2x^3 \cdot \sum_{n=0}^{\infty}\frac{(x^2)^{n-1}}{(n-1)!}+x\cdot \sum_{n=0}^{\infty}\frac{(x^2)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Ramanujan-Sums... How do they do? Prove that:
$\displaystyle \frac{1}{e^{2\pi}-1}+\frac{2}{e^{4\pi}-1}+\frac{3}{e^{6\pi}-1}+...=\frac{1}{24}-\frac{1}{8\pi}$
I would like to see different ways of solving this beautiful sum, whoever is encouraged? :)
Thanks to all.
| This problem is already solved in Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$ and has a nice solution there.
But then the simplest proof comes from Ramanujan. He uses the transformation formula for "eta" function and logarithmic differentiation. If we define $\eta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
If $a,b$ are roots for $x^2+3x+1=0$.Calculating $(\frac{a}{b+1})^2 +(\frac{b}{a+1})^2$ If $a,b$ are roots for the equation $x^2+3x+1=0$.How to calculate $$\left(\frac{a}{b+1}\right)^2 +\left(\frac{b}{a+1}\right)^2$$
| $$a+b=-3,ab=1$$
$$(a+b)^2=a^2+b^2+2ab\implies a^2+b^2=7$$
$$(a+b)^3=a^3+b^3+3ab(a+b)\implies a^3+b^3=-18$$
$$(a^2+b^2)^2=a^4+b^4+2(ab)^2\implies a^4+b^4=47$$
$${\left(\dfrac {a}{b+1}\right)}^2+{\left(\dfrac {b}{a+1}\right)}^2$$
$$\dfrac {a^2(a+1)^2+b^2(b+1)^2}{((b+1)(a+1))^2}$$
$$\dfrac {a^4+2a^3+a^2+b^4+2b^3+b^2}{(ab+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 1
} |
Find $S$ where $S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$, why am I getting an imaginary number? $\large S=\sqrt[3] {5+2 \sqrt {13}}+\sqrt[3]{5-2 \sqrt {13}}$
Multiplying by conjugate:
$\large S=\dfrac {-3}{\sqrt[3] {5+2 \sqrt {13}}-\sqrt[3]{5-2 \sqrt {13}}}$
From the original:
$\large S-2\sqrt[3]{5-2 \sqr... | You can find quickly $S$ if you note that $$5+2\sqrt{13}=\left(\frac{1+\sqrt{13}}{2}\right)^3$$ and similarly for $5-2\sqrt{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/431671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only I have found a proof using complex analysis techniques (contour integral, residue theorem, etc.) that shows $$\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$$ for $n\in \m... | In general, let $$y=\dfrac{1}{1+x^b}\quad\Rightarrow\quad x=\left(\dfrac{1-y}{y}\right)^{\frac1b}\quad\Rightarrow\quad dx=-\left(\dfrac{1-y}{y}\right)^{\frac1b-1}\ \dfrac{dy}{by^2}\ ,$$ then
\begin{align}
\int_0^\infty\dfrac{x^{a-1}}{1+x^b}\ dx&=\int_0^1 y\left(\dfrac{1-y}{y}\right)^{\large\frac{a-1}b}\left(\dfrac{1-y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 4,
"answer_id": 2
} |
Proving the inequality $\frac{a^3}{b^2-bc+c^2}+\frac{b^3}{a^2-ac+c^2}+\frac{c^3}{a^2-ab+b^2}\geq a+b+c$ I am trying to prove the following inequality
For all positive numbers $a$, $b$ and $c$ we have
$$\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{a^2-ac+c^2}+\dfrac{c^3}{a^2-ab+b^2}\geq a+b+c$$
I can probably solve this by reduc... | $$\Longleftrightarrow \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\ge a+b+c$$
by cauchy-Schwarz we have
$$\left[ \dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+ba^2}+\dfrac{c^4}{ca^2-abc+cb^2}\right][bc(b+c)+ac(a+c)+ab(a+b)-3abc]\ge (a^2+b^2+c^2)^2$$
$$\Longleftrightarrow (a^2+b^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Integration trig substitution $\int \frac{dx}{x\sqrt{x^2 + 16}}$ $$\int \frac{dx}{x\sqrt{x^2 + 16}}$$
With some magic I get down to $$\frac{1}{4} \int\frac{1}{\sin\theta} d\theta$$
Now is where I am lost. How do I do this? I tried integration by parts but it doesn't work.
| UPDATE 2 The first integral is not $\int \frac{dx}{x\sqrt{x^{2}-16}}$ but $\int
\frac{dx}{x\sqrt{x^{2}+16}}$ (as noticed by julien), because
\begin{eqnarray*}
I &=&\int \frac{dx}{x\sqrt{x^{2}+16}},\qquad x=\tan \theta ,dx=\sec
^{2}\theta d\theta \\
&=&\int \frac{\sec ^{2}\theta }{\left( \tan \theta \right) 4\sec \th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Is it possible to find the value of
$$\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
Does it help if I set it equal to $x$? Or I mean what can I possibly do?
$$x=\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$$
$$x^2=1+2\sq... | We can write
$$
\begin{align}
x & = \sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\cdots}}}}} \\
& = \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+\frac{4\sqrt{4}}{3}\sqrt{1+\frac{5\sqrt{5}}{4}\sqrt{1+\cdots}}}}} \\
&< \sqrt{1+2\sqrt{2}\sqrt{1+\frac{3\sqrt{3}}{2}\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "51",
"answer_count": 5,
"answer_id": 2
} |
How to solve these three equations? If α ,β ,γ are three numbers s.t.:
$\ α^ \ $ + $\ β \ $ + $ γ \ $ = −2
$\ α^2 \ $ + $\ β^2 \ $ + $ γ^2 \ $ = 6
$\ α^3 \ $ + $\ β^3 \ $ + $ γ^3 \ $ = −5,
then $\ α^4 \ $ + $\ β^4 \ $ + $ γ^4 \ $ is equal to ??
I tried out substituting the values of each equation to one other .... | HINT:
$$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$$
Now $$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$$
$$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$$ we can find $ab+bc+ca$ from here
$$a^3+b^3+c^3-3abc$$
$$=(a+b)^3-3ab(a+b)+c^3-3abc$$
$$=(a+b+c)\{(a+b)^2-(a+b)c+c^2\}-3ab(a+b+c)$$
$$=(a+b+c)\{(a+b)^2+c^2-3ab\}$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
show that the function $z = 2x^2 + y^2 +2xy -2x +2y +2$ is greater than $-3$ Show that the function
$$z = 2x^2 + y^2 +2xy -2x +2y +2$$
is greater than $-3$
I tried to factorize but couldn't get more than $(x-1)^2 + (x+y)^2 +(y-1)^2 - (y)^2$.
Is there any another way to factorize or another method??
| $$z=2x^2 + y^2 +2xy -2x +2y +2$$
$$\implies 2x^2+2x(y-1)+y^2+2y+2-z=0$$
As $x$ is real, the discriminant of the above Quadratic equation must be $\ge0$
$$\implies \{2(y-1)\}^2-4\cdot 2\cdot(y^2+2y+2-z)\ge0 $$
$$\implies (y-1)^2- 2\cdot(y^2+2y+2-z)\ge0 $$
$$\implies 2z\ge y^2+6y+3$$
$$\text{Now, }y^2+6y+3=(y+3)^2-6\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/436679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Factor Equation Help me with this,
Question: factor $x^3y-x^3z+y^3z-xy^3+xz^3-yz^3$.
Solution:
$$\begin{eqnarray}&=&x^3y-x^3z+y^3z-xy^3+xz^3-yz^3\\
&=&x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)\\
&=&x\left[(z-y)\left(z^2+zy+y^2\right)\right]+y\left[(x-z)\left(x^2+xz+z^2\right)\right]+z\left[(y-x)... | You were on the right track!
$x\left(z^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$
$=x\left(z^3-x^3 + x^3-y^3\right)+y\left(x^3-z^3\right)+z\left(y^3-x^3\right)$
$=\left(z^3-x^3\right)(x-y) + \left(x^3-y^3\right) (x - z)$
$=(z-x)(x-y)[z^2 + x^2 + zx -x^2 -y^2 -xy]$
$=(z-x)(x-y)[z^2 -y^2 + zx -xy]$
$=(z-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0 $ Let $a,\,b,\,c,\,d$ be distinct real numbers and $a$ and $b$ are the roots of quadratic equation $x^2 -2cx-5d=0$ and $c$ and $d$ are the roots of quadratic equation $x^2 -2ax-5b=0$. Then ... | Vieta's formulas $\Rightarrow \ \ $ $a+b=2c$, $c+d=2a$ $\ \ \Rightarrow \ \ $ $a+b+c+d=2(a+c)$ $\ \Rightarrow \ $ $a+c=b+d$.
Denote
$$m = \dfrac{a+c}{2}=\dfrac{b+d}{2}, \qquad p=\dfrac{c-a}{2}\color{gray}{=m-a=c-m}.$$
Then
$$
\left\{
\begin{array}{r}
a = m-p, \quad b=m+3p, \\
c = m+p, \quad d=m-3p.
\end{array}
\right.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Taylor series expansion for $f(x)=\sqrt{x}$ for $a=1$ I seem to be stuck defining an alternating sequence of terms in this series because $f^{(0)}(x)=f(x)$ is positive, as well as $f'(x)$, but then every other term starting with $f''(x)$ is negative. How can I define $f^{(n)}(x)$ given this?
\begin{array}{ll}
f(x)=... | Well you're not defining $f(x)$ to be negative, you're finding that certain derivatives of $f(x)$ are negative - if you think about the graph of this $f(x)$ it decreasingly increases; that is, it is increasing, but the rate at which it increases is decreasing, so it should seem to me that in the least, the second deriv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of $\frac{2}{x^3-x^2}$ How can I integrate $\dfrac2{x^3-x^2}$? Can someone please give me some hints? Thanks a lot!
| $$\int \frac2{x^3-x^2}\, dx =\frac2{x^2(x-1)}= \int \left(\frac{A}x+\frac{B}{x^2}+\frac{C}{x-1}\right)\,dx$$
$$Ax(x-1) + B(x-1) + Cx^2 = 2$$
We choose values for $x$ that "zero out" some of the terms, to simplifying the computation of the unknown values $A, B, C$:
$$x = 1 \implies A(1)(1 - 1) + B(1-1) + C(1^2) = 2 \iff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find five prime factors of $3^{140 }- 1$ I tried to simplify $3^{140}$ but I couldn't go past $81^{35}$, any help would be greatly appreciated.
| $$a^n - b^n = (a-b) ( a^{n-1} + a^{n-2} b + \cdots + b^{n-1})$$
Using $n = 140$, we get that $3 - 1$ is a factor, which gives $2$ as a factor.
Using $n = 35$, we get that $3^4 - 1$ is a factor, which gives $5$ as a factor.
Using $n = 28$, we get that $3^5 - 1$ is a factor, which gives $11$ as a factor.
Using $n = 20$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/443692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Prove that : $\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a} \le 3\sqrt[4]{2}.$ For $a,b,c\in\mathbb{R}^+$ and $a+b+c=3$ .
Prove that : $\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a} \le 3\sqrt[4]{2}.$
| Oops, give a wrong answer in previous edit. Here is the corrected version.
Notice both $\sqrt[4]{x}$ and $\sqrt[4]{x^2+1}$ are strictly increasing function in $x$.
The list of numbers $(\sqrt[4]{a},\sqrt[4]{b},\sqrt[4]{c})$ are in same sorted order as $(\sqrt[4]{a^2+1},\sqrt[4]{b^2+1},\sqrt[4]{c^2+1})$.
By Rearrangemen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/446448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
tough algebric problem? I wanted to know how can i prove that if
$xy+yz+zx=1$, then
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}
= \frac{2}{\sqrt{(1+x^2)(1+y^2)(1+z^2)}}$$
I did let $x=\tan A$, $y=\tan B$, $z=\tan C$
given $xy+yz+zx =1$ we have $\tan A \tan B+ \tan B \tan C+\tan C \tan A=1$
$\tan C(\tan... | HINT:
$$\frac x{1+x^2}=\frac {\tan A}{1+\tan^2A}=\frac{2\sin A\cos A}2=\frac{\sin2A}2$$
Now , $$\sin2A+\sin2B+\sin2C=2\sin(A+B)\cos(A-B)+2\sin C\cos C$$
$$=2\sin\left(\frac\pi2-C\right)\cos(A-B)+2\sin C\cos C$$
$$=2\cos C\{\cos(A-B)+\cos(A+B)\}$$ as $\sin C=\sin\{\frac\pi2-(A+B)\}=\cos(A+B)$
$$\implies \sin2A+\sin2B+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/448545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Prove that $2^n < \binom{2n}{n} < 2^{2n}$ Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction.
First part: $2^n < \binom{2n}{n}$. The base $n = 1$ is trivial. Assume inductively that some $k$ satisfies our statement. The indu... | For the first part, we don't need induction as $$\binom {2n}n=\prod_{0\le r\le n-1}\frac{2n-r}{n-r}=2\prod_{1\le r\le n-1}\frac{2n-r}{n-r}>2\cdot 2^{n-1}$$ and $2n-r>2(n-r)$ for $r>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/448861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 1
} |
On $x^3+y^3+z^3 = 1$ and a Pell equation Given,
$$(1-ac+bc)^3 + (a+c^2-ac^3)^3 + (ac^3-b-c^2)^3 = 1\tag{1}$$
where,
$$a,b,c,r = 12qrt,\;\; 3(q-r)(3q+r)t,\;\; 3s^2t^2,\;\; p-18qs^3t^3$$
then $(1)$ holds true if $p,q,s,t$ satisfies,
$$p^2-3(108s^6t^6-1)q^2=s\tag{2}$$
For $s=1$, this gives the Pell equation,
$$p^2-3(108... | Yes, there are other integers $s$. Just put $t=0$, so the equation becomes $p^2+3q^2=s$. Then pick any integers $p$ and $q$, and most likely $p^2+3q^2$ won't be either a square or $3$. For instance, if $p=2$ and $q=1$ then $s=7$.
If you require that $t$ be nonzero, then there are still solutions to (2) for other int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/449063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Matrix determinant using Laplace method I have the following matrix of order four for which I have calculated the determinant using Laplace's method.
$$
\begin{bmatrix}
2 & 1 & 3 & 1 \\
4 & 3 & 1 & 4 \\
-1 & 5 & -2 & 1 \\
1 & 3 & -2 & -1 \\
\end{bmatrix}
$$
Finding the determinant gives me $-726$. Now if I chec... | The Laplace development can be performed with respect to any row or column. Let's see when developing with respect to the first row:
\begin{align}
\det\begin{bmatrix}
2 & 1 & 3 & 1 \\
4 & 3 & 1 & 4 \\
-1 & 5 & -2 & 1 \\
1 & 3 & -2 & -1
\end{bmatrix}={}&
(-1)^{1+1}\cdot 2 \cdot
\det\begin{bmatrix}
3 & 1 & 4 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/450334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$n$ and $n^5$ have the same units digit? Studying GCD, I got a question that begs to show that $n$ and $n^5$ has the same units digit ...
What would be an idea to be able to initiate such a statement?
testing
$0$ and $0^5=0$
$1$ and $1^5=1$
$2$ and $2^5=32$
In my studies, I have not got "mod", please use other means, i... | We know if unit digits of two numbers are same, their difference is divisible by 10 and vice versa.
Method $1a:$
Using Fermat's Little Theorem $n^5-n\equiv0\pmod 5$
and $n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$ which is divisible by $n(n-1)$ which is always even
$\implies 2|(n^5-n)$ and we have $5|(n^5-n)$
$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/451927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing. How do you show the sequence $(1 - \frac{1}{n})^{-n}$ is decreasing?
I understand that the binomial theorem should be used here but I don't see how we can use it to prove that $a_{n+1} < a_n$.
I will rewrite the sequence as,
\begin{align*}
(1 - \frac{1}{n})^{-n}... | Ultimately, you want to show that$^{(1)}$ for $n\geqslant 2$ $${\left( {1 + \frac{1}{{\left( {n - 1} \right)\left( {n + 1} \right)}}} \right)^n} > 1 + \frac{1}{n}$$
Using the Binomial Theorem, the left hand side is $${\left( {1 + \frac{1}{{\left( {n - 1} \right)\left( {n + 1} \right)}}} \right)^n} > 1 + \frac{n}{{\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/452841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Finding $\frac{a+b}{a-b}$ such that $a^2+b^2=6ab$ For $a,b > 0$ such that $a^2+b^2=6ab$ .How to find $\frac{a+b}{a-b}$
| $$(a-b)^2=4 a b = (a+b)^2-(a-b)^2$$
which means that
$$1 = \left ( \frac{a+b}{a-b}\right)^2 - 1$$
or
$$\frac{a+b}{a-b} =\pm \sqrt{2}$$
depending whether $a > b$ or not.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/453044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
show that $\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$ show that
$$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$
using different ways
thanks for all
| In this answer, the more general integral
$$
\int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x
$$
is calculated.
Your integral is that integral for $n=3$.
A Different Way
In a fashion similar to this answer, we will use the equation
$$
\frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\sin^3(kx)}{k^3}
=\frac{9\sin(3kx)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 0
} |
Fractional part of rational power arbitrary small I think that $\{a^n\}$ (where $\{x\}$ is $x \pmod 1$), where $a$ is fixed rational greater than 1 and $n$ is positive integer, is dense in $[0,1]$ is unsolved. However what about $\{a^n\}$ is arbitrary small for some $n$ ($a$ is fixed rational as well).
| I will show that
if $a = 1+\sqrt{2}$
then the limit points of
$\{a^n\}$
are $0$ and $1$.
I know that this doesn't tell anything
about rational $a$,
but this might be of use.
Note that this method can show this
for $a$ and $b$ roots of
$x^2-2ux-v = 0$
where $u$ and $v$ are
positive integers such that
$v < 2u+1$.
This ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$ $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $, $\p... | Hint: A linear transformation is completely determined by how it acts on a basis for your vector space (a linearly independent and spanning set), and a linear transformation can send basis vectors to wherever it wants (you can send them anywhere). Show these fact if you haven't yet!!
For $\phi_1$, note that $\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/453988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the sum : $\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$ Problem :
Find the sum of :
$$\sin^{-1}\frac{1}{\sqrt{2}}+\sin^{-1}\frac{\sqrt{2}-1}{\sqrt{6}}+\sin^{-1}\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}+\cdots$$
My approach :
Here the $n$'th term... | Hint: Use the fact that $\sin^{-1}a+\sin^{-1}b = \sin^{-1}(a\sqrt{1-b^2}+b\sqrt{1-a^2})$. The first two terms then give you $\sin^{-1}(\sqrt{2/3})$. Then applying the same identity with this term and the third term gives you $\sin^{-1}(\sqrt{3/4})$, and...
Edit: Induction step gives some headache, so let me write it fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/454523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
How can prove this $\sum_{k=0}^{p}\binom{p}{k}(\pm i\sqrt{3})^k\equiv 1\pm (i\sqrt{3})^p-p\sum_{k=1}^{p-1}\frac{(\mp i\sqrt{3})^k}{k}\pmod{p^2}?$
For any prime $p>3$ show that
$$p\sum_{j=0}^{p-1}\dfrac{(-3)^j}{2j+1}\equiv \left(\dfrac{p}{3}\right)\pmod{p^2}$$
where $\left(\dfrac{p}{3}\right)$ denotes the Legendre... | Now,I have konw why
$$S_{0}\equiv 1-2^{p-1}\pmod {p^2},S_{1}\equiv 2^{p-1}\left(\frac{p}{3}\right)-(-3)^{(p-1)/2}\pmod {p^2}$$
Because use
$$(1\pm i\sqrt{3})^p=2^{p-1}(1\pm i\left(\frac{p}{3}\right)\sqrt{3})$$
and other hand we have
$$(1\pm i\sqrt{3})^p\equiv 1\pm i\sqrt{3}(-3)^{(p-1)/2}-S_{0}\pm i\sqrt{3}S_{1}\pmod{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes :
... | Hint $$\begin{align}{\sin x}&=2^1\sin\frac x 2 \cos\frac x2\\{}\\\sin x& =2^2\sin \frac x4\cos\frac x 4\cos \frac x 2\\{}\\\sin x& =2^3\sin \frac x8\cos \frac x8\cos\frac x 4\cos \frac x 2\\{}\\\cdots\;&=\hspace{2cm }\cdots\end{align} $$
One further hint
$$\sin x = {2^n}\sin \frac{x}{{{2^n}}}\prod\limits_{k = 1}^n {\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/455995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 3,
"answer_id": 2
} |
Prove: $\frac{1}{1^2} +\frac{1}{2^2} + \cdots + \frac{1}{n^2} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2} < 2$ While I don't doubt that this question is covered somewhere else I can't seem to find it, or anything close enough to which I can springboard. I however am trying to prove $$\frac{1}{1^2} +\frac{1}{2^2} + \cdo... | One way is to look at the lower Riemann sum with width $1$ under the graph of $f(x) = {1 \over x^2}$, from $x = 1$ to $x = \infty$. Since ${1 \over x^2}$ is decreasing, the lower
Riemann sum will be $\sum_{n=2}^{\infty} f(n) = \sum_{n=2}^{\infty} {1 \over n^2}$, which must be less than the integral $\int_1^{\infty} {1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/456595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Splitting Stacks Nim A game is played with two players and an initial stack of $n$ pennies ($n\geq 3$). The players take turns choosing one of the stacks of pennies on the table and splitting it into two stacks. When a player makes a move that causes all the stacks to be of height $1$ or $2$ at the end of his or her tu... | Here for reference is a table of nim-values for every nontrivial position with less than 12 pennies. Piles of size 1 have been omitted.
A nim-value of 0 indicates a position that is a win for the player who moves to it; any other value indicates a position that is a win for the next player to move.
$$
\begin{array}{lr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/457360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
$2+2 = 5$? error in proof $$\begin{align} 2+2 &= 4 - \frac92 +\frac92\\
&= \sqrt{\left(4-\frac92\right)^2} +\frac92\\
&= \sqrt{16 -2\times4\times\frac92 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{16 -36 + \left(\frac92\right)^2} +\frac92\\
&= \sqrt {-20 +\left(\frac92\right)^2} + \frac92\\
&= \sqrt{25-45 +\left(\fra... | $$\sqrt{a^2}=|a|\not=a$$
Watch the signs.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/457490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 9,
"answer_id": 6
} |
Nested Radical of Ramanujan I think I have sort of a proof of the following nested radical expression due to Ramanujan for $x\ge 0$.
$$\large x+1=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\cdots}}}}$$ for $ x\ge -1$
I just want to know if my proof is okay or there is a flaw, and if there is one I request to give som... | Variant of your proof, for some clarity.
Define $a_1(x)=\sqrt{1+x}$ and $a_{n+1}(x)=\sqrt{1+xa_n(x+1)}$.
Define $b_n(x)=1+x-a_n(x)$. By your proof, we know that $b_n(x)$ is decreasing and bounded below by zero. You want to show that $b_n(x)\to 0$, and then you are done.
Now (this is pretty much exactly your proof, but ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
what are real and imaginary part of this expression I have $M:=\sqrt{\frac{a\cdot(b+ic)}{de}}$ and all variables $a,b,c,d,e$ are real. Now I am looking for the real and imaginary part of this, but this square root makes it kind of hard.
| $$\sqrt{\frac{a(b+ic)}{de}}=\sqrt{\frac{a}{de}}\cdot\sqrt{b+ic}$$
Let $$\sqrt{b+ic}=x+iy$$
$$\implies b+ic=(x+iy)^2=x^2-y^2+2xyi$$
Equating the real & the imaginary parts, $b=x^2-y^2, c=2xy$
So, $b^2+c^2=(x^2-y^2)^2+(2xy)^2=(x^2+y^2)^2\implies x^2+y^2=\sqrt{b^2+c^2}$
We have $$x^2-y^2=b$$
$$\implies 2x^2=\sqrt{b^2+c^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/458947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What would have been our number system if humans had more than 10 fingers? Try to solve this puzzle. Try to solve this puzzle:
The first expedition to Mars found only the ruins of a civilization.
From the artifacts and pictures, the explorers deduced that the
creatures who produced this civilization were four-legg... | Many people believe that since humans have $10$ fingers, we use base $10$. Let's assume that the Martians have $b$ fingers and thus use a base $b$ numbering system, where $b \neq 10$ (note that we can't have $b=10$, since in base $10$, $x=8$ shouldn't be a solution). Then since the $50$ and $125$ in the equation are a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "74",
"answer_count": 13,
"answer_id": 1
} |
How Many License Numbers Consist of $4$ digits and $4$ letters, where one letter or digit is repeated? If license numbers consist of four letters and four numbers how many different licenses can be created having at least one letter or digit repeated?
$$26 \cdot 26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 \cdot 10 ... | Assuming the licences are of the form $\text{LLLLNNNN}$ where $\text{L}$ and $\text{N}$ denote letters and numbers respectively:
Number of licences with no letters or numbers repeated:
$26\cdot 25\cdot 24\cdot 23\cdot 10\cdot 9\cdot 8\cdot 7$.
So, number of licences with at least one letter or number repeated:
$\text{T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/460805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
show that if $ n-1$ and $n+1$ are both primes and $n>4$, then $\phi(n) \leq n/3$ Show that if $n-1$ and $n+1$ are both primes and $n>4$, then $\phi(n)$ is less than or equal to $n/3$
I tried a few cases
If $n=6$, $n-1=5$, and $n+1=7$ then $~\phi(6)=2=n/3$
If $n=12$, $n-1=11$, and $n+1=13$ then $~\phi(12)=4=n/3$
$\phi... | You need to show that $$\phi(n)\leq \frac{n}{3}$$
If $n>4$, with $n-1$ and $n+1$ both being primes,
Now since $n-1$ is a prime $p =(n-1)> 3$
And therefore odd, it follows that, $$(n-1)+1=n, \text{ is even}$$
Likewise we know that at least one of $3$ successive integers, must be divisible by $3$.
Now looking at the int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/461379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using the definition of the derivative prove that if $f(x)=x^\frac{4}{3}$ then $f'(x)=\frac{4x^\frac{1}{3}}{3}$ So I have that $f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
and know that applying
$f(x)=x^\frac{4}{3}=f\frac{(x+h)^\frac{4}{3} -x^\frac{4}{3}}{h}$
but am at a loss when trying to expand
$(x+h)^\frac{4}{3... | You can use the generalized binomial coeficients to solve this problem, which gives you
$(x + h)^\frac{4}{3} = x^\frac{4}{3}(1 + \frac{4h}{3x} + \frac{4h^2}{18x^2} + \ldots)$
It will go on forever, but since you divide by $h$, most will tend to zero, and you'll be left with $x^\frac{4}{3} \frac{4}{3x} = \frac{4}{3}x^\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/463656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Easy trigonometry question How is
$$\frac{2 \sin x}{(1+ \cos x)^2}= \tan\left(\frac{x}{2}\right)\sec^2\left(\frac{x}{2}\right)\;?$$
It should be easy.
But somehow I don't get it.
Can you help me with this?
| $$\frac{2 \sin x}{(1+ \cos x)^2}=\frac{4 \sin \frac{x}{2}\cos \frac{x}{2}}{(1+ 2\cos^2 \frac{ x}{2}-1)^2}=\frac{ \sin \frac{x}{2}\cos \frac{x}{2}}{( \cos^4 \frac{ x}{2})}=\tan\frac{x}{2}\sec^2\frac{x}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/465014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Solving recurrence equation using generating function method I've been trying to solve the following equation intuitively (I only know the method if there are minuses in the equation - $a_{n-1}, a_{n-2}...$).
$$a_{n+2}=4a_{n+1}-4a_{n}$$
$$a_{0}=3$$
$$a_{1}=8$$
$$
\begin{align}
A(x)&=\sum\limits_{n>=0}a_{n}x^{n} \\
&= \... | Not sure where you went wrong, so I'll start from scratch with a different method:
Let $A(x)=\sum_{n=0}^\infty a_nx^n$. Note that $a_n$ satisfies $a_{n+2}-4a_{n+1}+4a_n=0$. Let $p(x)=1-4x+4x^2$. We then define
$$
B(x) = \sum_{n=0}^\infty b_nx^n=p(x)A(x)
$$
Note that for $n\geq2$, we have $b_n=a_{n+2}-4a_{n+1}+4a_n=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/468385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $\int x^2 \cos \frac{x}{2} \mathrm{d}x$ I am trying to compute the following integral:
$$\int x^2 \cos \frac{x}{2} \mathrm{d}x$$
I know this requires integration by parts multiple times but I am having trouble figuring out what to do once you have integrated twice. This is what I have done:
Let $u = \cos \frac{... | let S be the integrate... That is, S x"2cosx/2 dx........let u=x"2 and dv=cosx/2dx. so that, du=2x and v=2sinx/2 dx. Therefore, S x"2cosx/2 dx= x"2(2sinx/2)-S 2sinx/2(2x) =2x"2sinx/2-S 2x.2sinx/2 dx = 2x"2sinx/2-4 S xcosx/2 dx. Apply another integration. S xcosx/2 dx. let u=x and dv=cosx/2dx. so that du=(1)dx=dx and v=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/469344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$. Prove that $\sum_{n=1}^{\infty} \frac{1}{2^n - 1}$ is convergent and $\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < 2$.
I'm not sure, but I suppose that $$\sum_{n=1}^{\infty} \frac{1}{2^n - 1} < \sum_{n=0}^{\inft... | This is a formalization of your proof
$$
\sum\limits_{n=1}^\infty\frac{1}{2^n-1}
=1+\sum\limits_{n=2}^\infty\frac{1}{2^n-1}
=1+\sum\limits_{n=1}^\infty\frac{1}{2^{n+1}-1}
<1+\sum\limits_{n=1}^\infty\frac{1}{2^n}=2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How prove that $10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$ let $a,b,c\ge 0$, such that $a+b+c=1$, prove that
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\le\dfrac{9}{4}$$
This problem is simple as 2005, china west competition problem
$$10(a^3+b^3+c^3)-9(a^5+b^5+c^5)\ge 1$$
see:(http://www.artofproblemsolving.com/Forum/viewtopi... | We need to prove that
$$40(a^3+b^3+c^3)(a+b+c)^2-36(a^5+b^5+c^5)\leq9(a+b+c)^5$$ and since the last inequality is homogeneous already,
it's enough to prove this inequality for all non-negatives $a$, $b$ and $c$.
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, since our inequality is fifth degree, we need to p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the value of $\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right)$ What is the value of $$\textrm{cosec}^2\left(\frac\pi7\right) +\textrm{cosec}^2\left(\frac{2\pi}7\right)+\textrm{cosec}^2\left(\frac{4\pi}7\right) \qquad\qquad ? $$
I tried to ... | $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
On the Pell-like $Ax^2-By^2 = 1$ This is connected to the post, Mere coincidence? (prime factors). I was looking at NeuroFuzzy's dataset and noticed the line,
{{{1, {4, 2}}, {1, 4, 2, 4, 2}, 23762}}
It seems this could be generalized. Is it true that given the equation,
$$(n+1)x^2-ny^2 = 1\tag{1}$$
then its solutions a... | Here
$$
\left( \begin{array}{c}
x \\
y
\end{array}
\right)
$$
will be a solution of $$ (n+1) x^2 - n y^2 = 1. $$
We have an "automorph" or generator of the automorphism group or isometry group or orthogonal group of the indefinite binary quadratic form depicted; the form is $ (n+1) x^2 - n y^2. $
$$ A =
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/470734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove by induction $\frac{n^n}{3^n}$\dfrac{n^n}{3^n}<n!<\dfrac{n^n}{2^n}$
The case $n!<\dfrac{n^n}{2^n}$ is easier.
| Another way this might be viewed is by rearranging the inequalities as
$$2^n \ < \ \frac{n^n}{n!} \ \ \text{and} \ \ \frac{n^n}{n!} \ < \ 3^n , $$
the entirety of which holds for $ \ n \ = \ 6 \ , $ as already mentioned by user84413 . The "induction step" for the ratio is then
$$\frac{(n+1)^{n+1}}{(n+1)!} \ = \ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the value of the the term The sequence $a_1,a_2,a_3,\ldots$ satisfies $a_1=1$, $a_2=2$, and $$a_{n+2}=\frac2{a_{n+1}}+a_n\;;$$ find the value of
$$\frac{a_{2012}2^{2009}}{2011}$$
| As lab bhattacharjee shows, $a_{n+2}a_{n+1}=2n+2$, and thus,
$$
\begin{align}
\frac{a_{n+2}}{a_n}
&=\frac{a_{n+2}}{a_{n+1}}\frac{a_{n+1}}{a_n}\\
&=\frac{n+1}{n}
\end{align}
$$
Therefore,
$$
\begin{align}
\frac{a_{2n}}{a_2}
&=\frac{2n-1}{2n-2}\frac{2n-3}{2n-4}\cdots\frac32\\
&=\color{#C00000}{\frac1{4^{n-1}}\frac{(2n-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/472989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ Find the value
$$\sum\limits_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}.$$
This problem is from this and I am interested in this problem, but I can't solve it.
Here is ... | You can take that last expression and turn it into an integral that gets you a result that agrees with both Mathematica and the answer linked to (and not really quite explained).
Your last sum may be rewritten as
$$\begin{align}\sum_{n=0}^{\infty} (-1)^n \int_0^{-1} dx \frac{x^n}{1-x} \, \int_0^{-1} dy \frac{y^n}{1-y}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/473996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
How many factors does 6N have? Given a number $2N$ having 28 factors another number $3N$ having 30 factors, then find out the number of factors of $6N$.
| Let $p_i$, be primes in increasing order. Consider the prime factorization of $n$.
$$ n = 2^{a_1} \times 3^{a_2} \times \prod p_i^{a_i}, $$
where $a_i$ are non-negative integers. For this question, let the indexing run over $i\geq 3$.
You are given that
$$ (a_1 + 2) \times (a_2 +1 ) \times \prod (a_i + 1) = 28 $$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/475002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Differentiate $x \sqrt{1+y}+y \sqrt{1+x}=0$
If $x \sqrt{1+y}+y \sqrt{1+x}=0$, prove that $(1+x^2)\frac{dy}{dx}+1=0.$
The answer I got is $$\frac{dy}{dx}= -\frac{2 \sqrt{1+x} \sqrt{1+y}+y}{x+2 \sqrt{1+x}\sqrt{1+y}}$$ but I cannot simplify it further.
Please provide your assistance.
| $$x\sqrt{1+y} = -y\sqrt{1+x}$$
squaring both sides
$$ x^2(1+y) = y^2(1+x)$$
simplifying
$$x^2 - y^2 = xy(y - x)$$
$$x+y = -xy$$
$$y =\frac{-x}{1+x}$$
further take derivatives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/475824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can we determine the number of terms which we have to take in a series to get a particular accurate? As I remember , two days ago , there was a question ( here ) asks for calculating this limit
$\displaystyle \lim \limits_{x\rightarrow \infty } \frac{x^3}{e^x}$ and the question was answered .
of course this is an... | It took me a while to figure out what the question was asking. It seems the question is why
$$
\begin{align}
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}}&=\infty\tag{1}\\
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}}&=6\tag{2}\\
\lim_{x\to\infty}\frac{x^3}{1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/476875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\ta... | $$A+B=180^\circ-C\\
\tan(A+B)=\tan(180^\circ-C)\\
\frac{\tan A+\tan B}{1-\tan A\tan B} =-\tan C\\
\tan A +\tan B +\tan C =\tan A \tan B \tan C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 10
} |
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.