Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Showing that $x^x/(2x)!=0$ as $x$ approaches infinity How would I show
$$\lim_{x\to\infty} \frac{x^x}{(2x)!}=0$$
I know $x^x$ grows faster than $(2x)!$
So then would I do
$$\frac{x^x}{2x(2x-1)(2x-2)(2x-3)\cdots(2x-(2x-1))}$$
But how do I proceed.
| \begin{align}
\frac{x^x}{(2x)!} & = \frac{\overbrace{x\cdots x}^{x\text{ factors}}}{\underbrace{1\cdot2\cdot3\cdots x}_{x\text{ factors}} \cdot \underbrace{(x+1) \cdot (x+2) \cdots (2x)}_{x\text{ factors}}} \\[10pt]
& = \frac 1 {x!} \cdot \underbrace{\frac x {x+1} \cdot \frac x {x+2} \cdot \frac x {x+3} \cdots \frac x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$. Show that there is no rational number $x$ satisfying the equation $x^2-[x]=4$. Find the roots of the equation $x^2-[x]=4$ where $[x]$ is the greatest integer less than or equal to $x$.
I tried to solve this equatio... | Using $\displaystyle \bullet \lfloor x \rfloor \leq x$
So here Given $x^2-4 = \lfloor x \rfloor $
So we get $x^2-4\leq x\Rightarrow x^2-x-4\leq 0$
So we get $\displaystyle \frac{1-\sqrt{17}}{2}\leq x\leq \frac{1+\sqrt{17}}{2}$
So we get Approx $-1.6\leq x \leq 2.55$
So we get $\lfloor x \rfloor =-2,-1,0,1,2$
Now we wil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1446094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Modulo polynomials, excluding selected solution combinations Given
$$
(x-a)(x-b)(x-c) \equiv 0\ \ \pmod {p}
$$
$$
(x-d)(x-e)(x-f)\ \equiv 0\ \ \pmod {q}
$$
x: unknown variable. p,q : known primes. a,b,c,d,e,f : known values.
Are there one or more modulo equations that will exclude the combination
$$
(x \equiv a \pmod ... | Let
$\left(\frac{a}{p}\right)=\left(\frac{a}{q}\right)=1$ ,
$\left(\frac{b}{p}\right)=\left(\frac{b}{q}\right)=-1$
$$x^{\frac{p-1}{2}} \equiv \pm1\ (mod\ p)\ \ , \ x^{\frac{q-1}{2}} \equiv \pm1\ (mod\ q)$$
$$x^{\frac{p-1}{2}} = \pm1\ + k_1 p\ \ ,\ x^{\frac{q-1}{2}} = \pm1\ + k_2 q$$
$$qx^{\frac{p-1}{2}} = \pm q\ + k_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving $\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$ I'm trying to resolve the $$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}}$$
First answer is $\frac{0}{0}$
Applying formula:
$$\lim_{x\to0}\frac{x}{\sqrt{3+x}-\sqrt{3-x}} = \lim_{x\to0}\frac{x(\sqrt{3+x}+\sqrt{3-x})}{(\sqrt{3+x}-\sqrt{3-x})(\sqrt{3+x}+\sqrt{3-x})}$$
A... | Let $f(x) =\sqrt {3+x} - \sqrt {3-x}.$ Note $f(0)= 0.$ Our expression has the form
$$\frac{x-0}{f(x) - f(0)}.$$
By the definition of a derivative, the above $\to 1/f'(0)$ as $x\to 0.$ This is an easy computation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove that using induction that $\binom22+\dots+\binom n2 = \binom{n+1}2$ so I have this math problem where I have to prove this using induction. $$\begin{pmatrix}2\\2\end{pmatrix}+\begin{pmatrix}3\\2\end{pmatrix}+\begin{pmatrix}4\\2\end{pmatrix}+\cdots+\begin{pmatrix}n\\2\end{pmatrix}=\begin{pmatrix}n+1\\3\end{pmatrix... | You said, $${k+1\choose3} = {k\choose2}+{k\choose3}$$
$$\implies {k+2\choose3} = {k+1\choose2}+{k+1\choose3}$$
Now use you induction hypothesis, i.e. $${2\choose2}+{3\choose2}+\dots+{k\choose2} = {k+1\choose3}$$
Substituting the value of ${k+1\choose3}$ in the above equation you get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inequality involving an exponent I wish to prove the following inequality
$$x^{\frac{3}{x-1}} > 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3}, \quad x > 1.$$
Graphically the above inequality appears to be true since if one plots
$$g(x) = x^{\frac{3}{x-1}} - \left (1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \rig... | We want to prove that
$$ \forall x\in(0,1),\qquad x^{\frac{3x}{x-1}}>1+x+x^2+x^3\tag{1}$$
but it is enough to prove that:
$$ \forall x\in(0,1),\qquad 3x\log(x)<(x-1)(x+x^2+x^3)=x^4-x\tag{2} $$
or:
$$ \forall x\in(0,1),\qquad 1+3\log(x) < x^3\tag{3} $$
or (by replacing $x$ with $z^{1/3}$):
$$ \forall z\in(0,1),\qquad 1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the Taylor series about $x = 1$ for $f(x) = \dfrac{1}{(x − 2)^2}$ . Find the Taylor series about $x = 1$ for $f(x) = \dfrac{1}{(x − 2)^2}$ . Express your answer in sigma notation, simplified as much as possible.
This is a practice question that I am having trouble with. I know how to get it about $x=0$, but it is ... | $f(x)
= \dfrac{1}{(x − 2)^2}
$
Let $y = x-1$,
or $x= y+1$.
We want to find the expansion
about $y = 0$.
$\begin{array}\\
g(y)
&=f(y+1)\\
&= \dfrac{1}{(y-1)^2}\\
&= \dfrac{1}{(1-y)^2}\\
&=\sum_{n=0}^{\infty} (n+1)y^n\\
&=\sum_{n=0}^{\infty} (n+1)(x-1)^n\\
&\text{and we could stop here, but, for the fun of it,...}\\
&=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Formula for $A^n$ where $n \in \{1, \ 2, \ \cdots \ \}$ for the matrix $A = \begin{bmatrix} 1 && b \\ 0 && 1 \end{bmatrix}$ Formula for $A^n$ where $n \in \{1, \ 2, \ \cdots \ \}$ for the matrix $A = \begin{bmatrix} 1 && b \\ 0 && 1 \end{bmatrix}.$
Please help with the question if you can, it is for my Linear Algebra ... | $$A^n=\begin{pmatrix}1&nb\\0&1\end{pmatrix}$$
Suppose this is true until $n$. Then
$$A^{n+1}=A.A^n$$
Computing the right side almost completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
how to show cubic function is one to one using assumption Okay I know the function $f(x)= x^3 +2x+3$ is one to one but how do I show that using the assumption that $f(a)=f(b)$ where $a=b$?
So if I do the process I have
$ a^3+2a+3 = b^3+2b+3$ my 3's cancel and i get
$a^3+2a= b^3+2b$
I know I can factor out an $a$ and ... | Notice that
$$a^3+2a= b^3+2b\iff a^3-b^3+2(a-b)=0\iff (a-b)(a^2+ab+b^2+2)=0...(1)$$
Since $$a^2+ab+b^2+2=\left(a+\frac{b}{2}\right)^2+\frac{b^2}{4}+2\ge 2$$
it follows that the only real solution of ($1$) is $a=b$. Then $f$ is one-to-one.
Another approach:
Since $f'(x)=3x^2+2>0$ for any $x\in\mathbb{R}$ it follows $f$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$? Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$
then obviously :
$$ 1 - \frac13 +\frac15 - \fra... | To prove the sum $\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dotsc$ I don't see how it could be useful to refer to
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dotsc=\frac{\pi}{4}$$
It might, however, be convenient to recall the famous sum of Euler:
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dotsc=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1454960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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To Substitute or Not ... $2\sin 2x = 3\tan x$, in interval $0\le x<360^\circ$ The Question
Solve $2\sin 2x = 3\tan x$, in interval $0\le x<360^\circ$
I thought this straightforward enough, used the identity $\sin^2 x + \cos^2 x \equiv 1$, shown below in the working, which clearly is not right. My question is what have ... | Your original solution is correct as well. It's just that $x=45^\circ$ doesn't satisfy $\sin^2 x = \frac14$. If $\sin^2 x = \frac14$, then $\sin x = \pm\frac12$, while $\sin 45^\circ = \frac{1}{\sqrt{2}}$. The solutions for $\sin^2 x = \frac14$ with $0\leq x<360^\circ$ are $30^\circ, 150^\circ, 210^\circ$ and $330^\cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Doing $\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$ without L'Hopital Without L'Hopital,
$$\lim_{x\to0}\frac{\sin x - \tan x}{x^2\cdot\sin 2x}$$
This is
$$\frac{\sin x -\frac{\sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\frac{\sin x \cdot \cos x - \sin x}{\cos x}}{x^2\cdot\sin 2x} = \frac{\sin x\cdot \cos x - \sin... | $$\frac{\sin x-\tan x}{x^{2}\sin 2x}=\frac{\sin x}{\sin 2x}\frac{\cos x -1}{x^{2} \cos x}=\frac{\sin x}{\sin 2x}\frac{-2\sin^{2}\frac{x}{2}}{x^{2}\cos x}$$
Now, $$\frac{\sin x}{\sin 2x}\to \frac{1}{2}$$
$$\frac{\sin^{2}\frac{x}{2}}{x^{2}}=1/4\frac{\sin^{2}\frac{x}{2}}{(x/2)^{2}}\to 1/4$$
And $\cos x \to 1$
So the limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the maximum of $\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$ for nonnegative $a$, $b$, $c$.
Show that the maximum of
$$\frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}$$
is $4$ for nonnegative $a$, $b$, $c$.
An elegant elementary solution is preferred.
Generally, is there an easy way to show that
$$
\frac{
\left( a_1 + \... | Not a very elegant solution, I will admit.
We want to show that
$$ \frac{(a+b+c)^3-27abc}{a^3+b^3+c^3-3abc}\leq 4$$
which is equivalent to (since the denominator is non-negative by AM-GM)
$$(a+b+c)^3-27abc \leq 4(a^3+b^3+c^3) - 12abc$$
or
$$a^3 + b^3 + c^3 + 3a^2(b+c) + 3b^2(c+a)+3c^2(a+b) + 6abc - 27abc \leq 4(a^3+b^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Computing $\int (1 - \frac{3}{x^4})\exp(-\frac{x^{2}}{2}) dx$ How does one compute
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
Mathematica gives $(x^{-3} - x^{-1})\exp(-x^2/2)$ which indeed the correct answer, but how does one get there?
Integration by parts gives $(y + y^{-3})e^{-y^2/... | i tired integration by parts to show to OP but i think it doesn't work
Let's compute this integral :
$$\int \left(1 - \frac{3}{x^4}\right)\exp\left(-\frac{x^{2}}{2}\right) dx?$$
indeed,
$$ u=e^{\dfrac{-x^{2}}{2}},dv = (1 - 3/x^4)\qquad du=-e^{-\dfrac{x^2}{2}}x,\, v=\frac{1}{x^3}+x $$
\begin{align}
\int \left(1 - \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate Integral $\int\frac{1+x}{1+x^2}\ dx$ $\int\frac{1+x}{1+x^2}\ dx$
Let $u=1+x^2$
Then $du = 2x\ dx$
Here is my work.
Split integral $\int\frac{1}{1+x^2}\ dx$ + $\int\frac{x}{1+x^2}\ dx$
Integrate first integral term:
$\int\frac{1}{1+x^2}\ dx=tan^{-1}x$
$\int\frac{x}{1+x^2}\cdot\frac{du}{2x} $
I am stuck when it... | $\frac{x}{1+x^2}$ is something of the form $\frac{1}{2}\cdot\frac{f'}{f}$, hence:
$$ \int\frac{1+x}{1+x^2}\,dx = \arctan(x)+\frac{1}{2}\log(1+x^2)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1464192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the expansion? I encounter the following formula in some textbook. However, I can not understand what the expansion in this formula is. Is there anyone giving some tips?
$$
\begin{align}
\alpha &= -\frac{iU}{2 \epsilon} \pm \sqrt{\frac{i\omega}{\epsilon} - \frac{U^{2}}{4 \epsilon^{2}}} \\
&= \frac{iU}{2 \epsilo... | Hint:
They factored out $\;-\dfrac{U^2}{4\varepsilon^2}=\biggl(\dfrac{iU}{2\varepsilon}\biggr)^2$ under the radical:
$$\sqrt{\frac{i\omega}{\epsilon} - \frac{U^{2}}{4 \epsilon^{2}}}=\sqrt{\frac{-U^2}{4\varepsilon^2}\biggl(1-\frac{4i\omega\varepsilon}{U^2}\biggr)}=\dfrac{iU}{2\varepsilon}\sqrt{1-\frac{4i\omega\varepsilo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that if $a \neq b$ and a and b are positive then $\frac{a}{b}+\frac{b}{a}$ is never an integer
Some observations I made is for $\frac{a}{b}+\frac{b}{a}$, is either:
*
*the denominator has to be one,
*the numerator has to be a multiple of the denominator or
*the numerator and denominator have to be the same.... | (I assume that $a$ and $b$ are both integers...)
Let $x=\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{a^2+b^2}{ab}$. We can assume that $a$ and $b$ are relatively prime, since substituting $a/g$ and $b/g$ instead of $a$ and $b$ doesn't change the value of $x$.
If $x$ is integer, than $ab$ should divide $a^2+b^2$, so
$$a|(a^2+b^2) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1468189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Maximum value of a variable from a system of 2 equations From the following $2$ equations, find the maximum value of $d$.
$a + b + c + d = 8$ and $ab + ac + ad + bc + bd + cd = 12$
How to go about with this problem?
Please help.
Thankyou.
| Without loss of generality we can assume
$a = p + q$, $b = p - q$, with $q > 0$.
The first equation is simplified as
(1)
$$
c + d = 8 - 2 \, p.
$$
Similarly, for the second equation, we have
\begin{align}
12 & = c d + (a + b)(c+ d) + a b \\
& = c d + 2\,p (c + d) + p^2 - q^2, \\
& = c d + 2\,p (8 - 2p) + p^2 - q^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1469579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx$
If $\displaystyle I = \int_{0}^{1}\frac{2x-(1+x^2)^2\cdot \cot^{-1}(x)}{(1+x^2)[1-(1+x^2)\cot^{-1}(x)]}dx\;,$ Then value of $100(I-\ln 2) =$
$\bf{My\; Try::}$ Let $\cot^{-1}(x)=t\;,$ Then $\displaystyle \frac{1}{1... | You can start by rewriting your integrand as
$$
\begin{aligned}
1&+\frac{2x-(1+x^2)}{(1+x^2)\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]}\\
&=1+\frac{2x\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]+(1+x^2)\bigl[2x\,\text{arccot}\,x-1\bigr]}{(1+x^2)\bigl[1-(1+x^2)\,\text{arccot}\,x\bigr]}\\
&=1+\frac{2x}{1+x^2}+\frac{1-2x\,\text{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$? The question states:
If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, then what is $\frac{dy}{dx}$?
The options are:
*
*$$\sqrt{\frac{1-x^2}{1-y^2}}$$
*$$\sqrt{\frac{1-y^2}{1-x^2}}$$
*$$\frac{1-x^2}{1-y^2}$$
*$$\frac{1-y^2}{1-x^2}$$
I trie... | Your approach is absolutely fine and the result that you have obtained is correct. If you handed me a homework like this I would be satisfied, since it is clear that you have understood the procedure of implicit differentiation.
Still, the problem is somewhat pedantic and asks you to do a final superfluous simplificati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1472086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=$ If $n=12m$, where $m\in\Bbb{N}$, prove that
$$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=(-1)^m\left(\frac{2\sqr... | Hint: First try to prove that the sum can be simplified to
$$\binom{n}{0}-\frac{1}{(2+\sqrt3)^2}\binom{n}{2}+\frac{1}{(2+\sqrt3)^4}\binom{n}{4}-\frac{1}{(2+\sqrt3)^6}\binom{n}{6}+.....=\operatorname{Re}\left(\frac{1}{2+\sqrt3}+i\right)^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A problem on complex no. Q.If $z$ and $w$ be two complex no, such that $|z|=|w|=1$ and$|z+iw|=|z-iw|=2$.Then find z
I tried to break $z$ and $w$ into $a+ib$ form separately and carry on linear equation by substituting the values.But,It leads to me nowhere.Any hints?
| Let z = a + bi
Let w = c + di
iw = -d + ci
z + iw = (a-d) + i(b+c)
z - iw = (a+d) +i(b-c)
|z| = |w| = 1 => $a^2 + b^2 = c^2 + d^2 = 1$
|z + iw| =2 => $(a - d)^2 + (b + c)^2 = 4$ so $a^2 + b^2 + c^2 + d^2 - 2ad + 2bc =4$ so $2 + 2bc - 2ad =4$ so $bc - ad = 1$
Likewise |z - iw| =2 => $(a + d)^2 + (b - c)^2 = 4$ so $a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solve the recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n $
Solve the recurrence relation
$$a_n = 4a_{n-1} - 3a_{n-2} + 2^n $$
With initial conditions:
$a_1 = 1$
$a_2 = 11$
I have done similar recurrence relation problems to this, but none that were a non-homogeneous recurrence relation such as this one.
So far ... | Start by finding the general solution to the homogeneous recurrence relationship:
$$a_n = 4a_{n-1} - 3a_{n-2}$$
This has auxiliary equation $\lambda^2=4 \lambda-3$
$\lambda^2-4\lambda+3=0$
$\lambda_1=1, \lambda_2=3$
$$a_n = A(1)^n +B(3)^n$$
You want a particular solution to the non-homogeneous relationship.
Try $a_n=k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1475469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Proof of trigonometric relationship of angle bisector I have to prove the following equation:
$$CD=\frac{2ab\cos\frac{C}{2}}{a+b}$$ where $CD$ is the angle bisector of $C$ in $\Delta ABC $.
My attempt:
I've started by rearranging the equation in terms of $\cos \frac{C}{2}$ ,then by squaring and subtracting $\sin^2 \fr... | Drop perpendiculars $AF$ and $BE$ onto $CD$.
By the definition of cosine,
$$
\left|CF\right|=b\cos(C/2)\quad\text{and}\quad\left|CE\right|=a\cos(C/2)\tag{1}
$$
Since $\triangle AFC\sim\triangle BEC$ and $\triangle AFD\sim\triangle BED$, we get
$$
\frac{\left|CD\right|-\left|CE\right|}{\left|CF\right|-\left|CD\right|}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1479557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I poste... | $(1 - x)(1 + x + x^2 + ..... )=$
$(1 + x + x^2 .... )(-x - x^2 - x^3 -......) = 1 + (x -x) + (x^2 - x^2)... = 1$ so
$1 + x + x^2 + .... = \frac{1}{1 - x}$
Let x = -1.
$ 1 - 1 + 1 - 1 + 1 - 1 .... = \frac{1}{1 -(-1)} = \frac 12$
but clearly $1 - 1 + 1 - 1 +... = (1-1) + (1-1) +... = 0$.
So $0 = \frac 12$ (and also 1, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 8
} |
show this diophantine equation has at least is $3n+3\lfloor \frac{n+1}{3}\rfloor+1$ postive integer solution For any postive integer $n\ge 4$, let $s(n)$ denote the number of ordered pairs $(x,y,z)$ of positive integers for which
$$\color{red}{xy+yz+xz=n(x+y+z)}$$
show that
$$s(n)\ge 3n+3\lfloor \dfrac{n+1}{3}\rfloor+1... | Here's an approach to construct the necessary quantity of solutions which uses the ideas suggested by individ:
So we fix some $x=k$ where $1\le k \le 10$ and obtain the equation
$$(y-n+k)(z-n+k)=n^2-kn+k^2$$
Now, we are searching for a factorization of the RHS.
There is one obvious factorization which works for all $k$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given known and and unknown line equation and maximising area of triangle condition.
Q) If from a point $P\equiv(4,4)$ perpendiculars to the straight lines
$-3x+4y+5=0$ and $y=mx+7$ meet at $Q$ and $R$ and area of triangle
$PQR$ is maximum then $m=__ $ ?
First I drew the known line $-3x+4y+5=0$ and the unknown li... | First, I found the equation of lines which form the triangle.
*
*$y-4=\frac{-1}{m}(x-4)$ which is the line perpendicular to $y=mx+7$ and passing through $P(4,4)$
*$y-4=\frac{4}{3}(x-4)$ represents the line perpendicular to $3x+4y+5=0$ and passing through $P(4,4)$
Next, I need information about vertices of the tria... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find solutions of $2^m\cdot p^2+1=q^5$ $2^m\cdot p^2+1=q^5$
$p$ and $q$ are prime numbers
find $p$ and $q$
I think it will be useful to transfer $1$ to the other side of the equation
$2^m\cdot p^2=(q-1)(q^4+q^3+q^2+q+1)$
and we know $gcd(q-1,q^4+q^3+q^2+q+1)=1$ or $5$
we know if
I)$q-1|2^m \implies p^2|q^4+q^3+q^2+q+1... | You can write the equation as:
$$2^m\cdot p^2=q^5-1$$
From here you can note that the prime number $q>3$.
Noe you obtain:
$$2^m\cdot p^2=(q-1)(q^4+q^3+q^2+q+1)$$
therefore you can note that $q^4+q^3+q^2+q+1$ is odd; this yields $$2^m=q-1$$
and $$p^2=q^4+q^3+q^2+q+1$$
but $p^2=q^4+q^3+q^2+q+1$ has solutions only for $p=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Does $c\cdot\sum\limits_{n=k}^{\infty}a_{n}= \sum\limits_{n=k}^{\infty}c\cdot a_{n}$ provided that the series converge? I am struggling to find what is wrong about this reasoning when calculating a series that does not start at $n=0$.
For instance, let $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n$.
Then $\... | If the others' help was not sufficient, let me explain it to you like this:
You tried to calculate $S = \sum\limits_{n=2}^{\infty}\left(\frac{1}{2}\right)^n.$ The first terms of this sum are $$S=\frac14 + \frac18 + \frac1{16}+\dots.$$You tried to do this by premultiplying by $1/4,$ probably because you wanted the first... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.
2.
$$
\begin{align*}
(-8)^{4/3} &= (-8)^{1+1/3} \\
&= -8\times(-8)^{1/3} \\
&= -8\times (-1)^{1/3}\times 8^{1/3} \\
&= -2\times 8\times (-1)^{1/3} \\
... | Let $a$ be any real number. We can show that the equation
$$
x^3 = a
$$
has one and only one real root. It is called the cube root of $a$ and
is denoted $a^{1/3}$ or $\sqrt[3]{a}$.
From $(-1)^3 = -1$, we conclude that $(-1)^{1/3} = -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
The limit of $(1-\cos^{1/3} x)/(1-\cos x^{1/3})$ as $x\to 0$ I have a problem to find answer for this limit.
$$\lim_{x\to 0}\frac{1-{\cos^{1/3} x}}{1-\cos(x^{1/3})}$$
I did it like this $$\dfrac{\dfrac{\sin x}{3×\sqrt[3]{\cos x²}}}{\sin \sqrt[3]x×\dfrac 1{3×\sqrt[3]{x²}}}$$
| We need two standard limits $$\lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \frac{1}{2},\,\,\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{1 - \cos^{1/3}x}{1 - \cos x^{1/3}}\notag\\
&= \lim_{x \to 0}\frac{1 - \cos^{1/3}x}{1 - \cos x}\cdot\frac{1 - \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find all five solutions of the equation $z^5+z^4+z^3+z^2+z+1 = 0$ $z^5+z^4+z^3+z^2+z+1 = 0$
I can't figure this out can someone offer any suggestions?
Factoring it into $(z+1)(z^4+z^2+1)$ didn't do anything but show -1 is one solution.
I solved for all roots of $z^4 = -4$ but the structure for this example was more si... | Note that $(z-1)(1+z+z^2+z^3+z^4+z^5) = z^6-1$. Solve $z^6-1=0$ and discard the $z=1$ solution (which comes from the $z-1$ factor).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1486464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 2
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What is the best way of resolving an expression with square roots in denominator? I was resolving the following question from my textbook:
Write each of the following expressions as a single fraction, simplifying your answer where possible:
$4 - \frac{1}{\sqrt{12}} + \frac{10}{\sqrt{3}}$
I have resolved it this way:... | Your answer of $$\frac{24 - \sqrt{3} +
20\sqrt{3}}{6}=\frac{24 +
19\sqrt{3}}{6}$$
matches the answer given by the book within a factor of $\sqrt 3\over \sqrt 3$:
$$\frac{24 +
19\sqrt{3}}{6}=\frac{8\sqrt3+19}{2\sqrt 3}$$
Your method was effective in solving the problem, although putting $\sqrt{12}\sqrt 3$ as the deno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Evaluating a logarithmic integral with square roots We need to evaluate $$I = \int x\ln\left(x^2 + a^2 + \sqrt{x^2 - a^2}\right)\, \mathrm{d}x$$ so we using $u = \ln f(x)$ and $\mathrm{d}v = x$ so that $\mathrm{d}u = \frac{f'(x)}{f(x)}$ and $v = \frac{x^2}{2}$ (where $f(x) =$ argument of the logarithm yielding:
$$I = ... | Using substitution will be more successful that 'by parts'.
Let $t=x^2-a^2$ so $dt=2xdx$
Then the integral becomes:
$$ I=\frac{1}{2}\int \ln(t + \sqrt{t}+2a^2)dt$$
Next using 'by parts':
Let $u=\ln(t+\sqrt{t}+2a^2)$ and let $dv=1$
We get: $v=t$ and
$$du = \frac{1+\frac{1}{2}t^{-\frac{1}{2}}}{t+\sqrt{t}+2a^2}dt$$
$$I=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to Compute $\frac{d}{dx}\left(\left(1+x^2\right)^x\right)$? This is what I worked out:
Let $y = (1 + x^2)^x$ and let $a = 1 + x^2$
Then, by the chain rule of differentiation:
$\frac{dy}{dx} = \frac{dy}{da}\cdot\frac{da}{dx} = a^x\cdot ln(a) \cdot 2x$
$\frac{dy}{dx} = (1 + x^2)^x \cdot ln(1 + x^2) \cdot 2x $
But whe... | In order to calculate $\frac{d}{dx}\left(\left(1+x^2\right)^x\right)$ then this expression can be placed into the form
\begin{align}
\frac{d}{dx} \left(\left(1+x^2\right)^x\right) &= \frac{d}{dx}\left[ e^{x \, \ln(1+x^2)} \right] \\
&= e^{x \, \ln(1+x^2)} \, \left[ x \, \frac{d}{dx} \ln(1+x^2) + \ln(1+x^2) \right] \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1492320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$ without L'Hopital's rule
Find the following limit
$$\lim_{x\to0}\frac1{x^3}\left(\left(\frac{2+\cos x}{3}\right)^x-1\right)$$
without using L'Hopital's rule.
I tried to solve this using fundamental limits such as $\lim_{x\to0}\left(1+x... | Consider first $$A=\left(\frac{2+\cos (x)}{3}\right)^x$$ $$\log(A)=x\log\left(\frac{2+\cos (x)}{3}\right)$$ Now, Taylor series for $\cos(x)$ and then for $\log(1+y)$$$\log(A)=x\log\left(\frac{2+\cos (x)}{3}\right)=x\log\left(\frac{2+1-\frac{x^2}{2}+\cdots}{3}\right)$$ $$\log(A)=x\log\left(\frac{3-\frac{x^2}{2}+\cdots}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1496192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Two similar integration about continued fractions Prove that \begin{align*}
\int_0^{+\infty} \cfrac{\sin nx}{x + \cfrac{1}{x + \cfrac{2}{x + \cfrac{3}{x + \cdots}}}} \, dx &= \cfrac{\sqrt {\cfrac{\pi }{2}} }{n + \cfrac{1}{n + \cfrac{2}{n + \cfrac{3}{n + \cdots}}}}\\
\int_0^{+\infty} \cfrac{\sin \cfrac{n\pi x}{2}}{x + \... | Hint
Note
$$4\int_0^{+\infty}\cos(4t\ln x)\cdot e^{-2n\pi t} \, dt = \frac{2n\pi}{n^2\pi^2 + 4\ln^2 x}$$
then you only find
$$\int_0^{+\infty}\left(\int_0^1 \frac{\cos(4t\ln x)}{1+x^2} \, dx\right) e^{-2n\pi t} \, dt$$
you only let
$x=e^{-u}$, then
$$\int_0^1 \frac{\cos(4t\ln x)}{1+x^2} \, dx = \int_0^{+\infty} \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
prove trig equivalence $$\sin x + \sin y = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
I want to use the the $e^{ix}$ identities but I'm not sure if it can be done that way, let alone how to do it.
Any tips would be appreciated.
| \begin{align}
\sin u \cos v + \cos u \sin v & = \sin(u+v) \\
\sin u \cos v - \cos u \sin v & = \sin(u-v)
\end{align}
Apply the above in the case where $u = \dfrac{x+y} 2$ and $v=\dfrac{x-y} 2$:
\begin{align}
\sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + \cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1500472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
We have $12$ balls numbered with $1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 9, 9$ in a bin.
We have $12$ balls in a bin, numbered from $1$ to $8$ and the other $4$ numbered with the number $9$. We take one of them, write down the number and then put back into the bin again. We do this $3$ times. Which is the probability of obtain... | I think your methodology is off. For instance, $\binom{12}{1}\cdot\binom{12}{1}\cdot\binom{12}{1} = 12^3$ (not $36$), is the number of ways to choose any three balls, not to choose two $3$'s and a $1$.
Think about them as $12$ balls: if we pick three of them with replacement, we have $12^3$ possible sequences of ball... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find limit of the expression $$\lim_{x \to 0}\frac{\cos(x) - 8x\sin(x/2) - \cos(3x)}{x^4}$$
I think I should replace by equivalent, such as $\sin(x)$ ~ $x$, but got nothing.
Thank you for answers, but what about solving without using L'Hôpital's rule?
| $\cos(x)-\cos(3x) = 2\sin(x)\sin(2x) = 4\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\sin(2x)$, hence we want to compute:
$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)-2x}{x^3}=2\cdot\lim_{x\to 0}\frac{\sin(2x)-2x}{x^3}=16\cdot\lim_{z\to 0}\frac{\sin(z)-z}{z^3} $$
that equals $-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find probability of having a female in every group 10 male students and 5 female students are split into 5 groups (every grup consists of 3 students). What is the probability that there is a female student in every group?
I should solve this with combinatorics. Any idea on how to appraoch the problem?
| total possible: ${15\choose 3}{12\choose 3}{9\choose 3}{6\choose 3}{3\choose 3}$
one female in every group: ${10\choose 2}{5\choose 1}{8\choose 2}{4\choose 1}{6\choose 2}{3\choose 1}{4\choose 2}{2\choose 1}{2\choose 2}{1\choose 1}$
solution: ${{10\choose 2}{5\choose 1}{8\choose 2}{4\choose 1}{6\choose 2}{3\choose 1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1503926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find : $\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$ in its algebraic form. Find : $$\sqrt[6]{\frac{\sqrt{2}+(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^7}{(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)^{11}}}$$ in its algebraic form.
Now, I kinda think i... | The binomial quantities being raised to powers are easy to simplify: $$\sqrt[6]{\frac{\sqrt{2}-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i}{-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}}$$
Now it's quite easy to continue. $$\sqrt[6]{\frac{\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i}{-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}}=\sqrt[6]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$ Let $m$ be a positive integer and let $\alpha$ be a real number. Show that
$$\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\ri... | This is known as Hermite's identity. A nice proof is due to Matsuoka "On a Proof of Hermite's Identity", AMM 71:10 (1964), pp. 1115, DOI: 10.2307/2311413.
Let:
$\begin{align}
f(x)
= \lfloor m x \rfloor
- \lfloor x \rfloor
- \left\lfloor x + \frac{1}{m} \right\rfloor
- \left\lfloor x + \frac{2}{m} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$ Write out the Form of the Partial Fraction Decomposition: $\int\frac{x^6}{x^2-4}$
The book says use long division my answer was $x^3+\frac{4x^3}{x^2-4}$
The answer manual is $\frac{x^6}{x^2-4}=x^4+4x^2+16+\frac{64}{(x+2)(x-2)}$
| Notice, the formula $a^3-b^3=(a-b)(a^2+b^2+ab)$,
Now, re-arrange the numerator as follows $$\frac{x^6}{x^2-4}=\frac{x^6-4^3+4^3}{x^2-4}$$
$$=\frac{((x^2)^3-4^3)+4^3}{x^2-4}$$
$$=\frac{(x^2)^3-4^3}{x^2-4}+\frac{64}{x^2-4}$$ $$=\frac{(x^2-4)((x^2)^2+4^2+4x^2)}{x^2-4}+\frac{64}{x^2-4}$$
$$=\frac{(x^2-4)(x^4+4x^2+16)}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1510978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find all possible remainders.
Suppose $b\in\mathbb Z$. Find all possible remainders of $b^3$ divided by $7$.
I know $b^6=1\bmod7$ which means $(b^2)^3=1\bmod7$ so all square numbers^3 leave $1$ as a remainder, but how to continue?
| Notice that this depends only on the value of $b$ mod $7$, since $(b+7k)^3 = b^3 + (\text{multiple of 7})$. So we can work wlog with $b = 0, 1, \dots, 6$.
Your method is also faulty: just because $b^6 \equiv 1 \pmod{7}$, that doesn't mean $b^2 \equiv 1 \pmod{7}$. For example, $3^2 \equiv 2 \pmod{7}$.
There's a really ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Am I correctly finding the standard matrix?
Question:
Let $F: \Bbb R^3 \to \Bbb R^3$ be the linear transformation satisfying
$F(1,0,1)=(-3,-3,1)$,$F(0,1,0)=(0,1,1)$, and $F(0,1,1)=(2,-2,1)$. Find
the standard matrix $A$ of $F$.
My Approach:
I used a method that I haven't been taught, so I'm not sure if I am corre... | Let's check your solution,
So you got $T(1,0,0)=(3,-1,-2)$, $~T(0,1,0)=(0,0,1)$ and $~T(0,0,1)=(2,-1,-1)$
Now $T(1,0,1)=(3,-3,1)$
also $~T(1,0,1)=T(e_1+0.e_2+e_3)=Te_1+Te_3=(3,-1,-2)+(2,-1,-1)=(5,-2,-3) \not=(3,-3,1).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1515520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Help with $\lim \frac {{1}^{p} + 3^p + ... + (2n+1)^p}{n^{p+1}}$. I'm using Stolz–Cesàro theorem to show that this limit = $\frac {2^p}{P+1}$. I took $(X_{n})$ = $\sum_{i=0}^{n}{2i+1}^p$ and $(Y_n)$ = $n^{p+1}$. Then, by the theorem, $\lim = \frac {X_{n+1}-X_n}{Y_{n+1}-Y_n}$.
In the numerator I get $(2n+3)^{p}$. I've... | Try to link the summation to a Riemann sum:
\begin{align}
& \frac{1^p + 3^p + \cdots + (2n + 1)^p}{(n + 1)^{p + 1}} \\
= & \frac{1}{2}\sum_{i = 1}^{n + 1}\left(\frac{2i - 1}{n + 1}\right)^p\times\frac{2}{n + 1} \\
\to & \frac{1}{2} \int_0^2 x^p dx = \frac{2^p}{p + 1},
\end{align}
during which we partitioned interval $[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1516954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to find the Minimum of the Function. Minimize $xy$ on the ellipse $b^2x^2+a^2y^2=a^2b^2$
So what I did first was take the gradient of $f$ and $g$
$∇ f = (y,x)$ $\qquad$ $∇ g= (2x^2b^2,a^22y)$
Then we do the lagrange multiplier
$y= 2x^2b^2λ$
$x=2ya^2λ$
Then we equate the functions. By multiplying.
$... | $$L(x,y,z)=xy+\lambda(b^2x^2+a^2y^2-a^2b^2)$$
$$\nabla_{x,y,\lambda} L(x,y,z)=\langle y+2\lambda b^2 x, x+2\lambda a^2 y , b^2x^2+a^2y^2-a^2b^2 \rangle=\langle 0,0,0 \rangle$$
$$y=-2 \lambda b^2 x$$
$$x=-2 \lambda a^2 y$$
assuming $\lambda \ne0$:
$$\frac{y}{x}=-2 \lambda b^2$$
$$\frac{y}{x}=\frac{1}{-2 \lambda a^2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving an equation by telling the value of $x^2+y^2$. I have a problem solving an equation. The equation is:
$xy+x+y=44$ and $x^2y+xy^2=448$
and we have to tell the value of $x^2+y^2$
First I tried solving this by doing the following:
$xy+x+y=44~\to~x+y=44-xy~\to~x^2+2xy+y^2=44^2-88xy+x^2y^2~\Rightarrow$
$\Rightarrow~... | $$(x+y)=44-xy..............1$$
$$xy(x+y)=448........................2$$
substitute 1 in 2
$$xy(44-xy)=448$$
$$x^2y^2-44xy+448=0$$
use the quadratic formula
$$xy=22\pm6$$
or
$$x^2y^2=(22\pm6)^2$$
now square the equ.1
$$x^2+2xy+y^2=44^2-88xy+x^2y^2$$
$$x^2+y^2=44^2-90xy+x^2y^2$$
hence
$$x^2+y^2=44^2-90(22\pm6)+(22\pm6)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Which natural numbers satisfy $2^n > n^2$?
Which natural numbers satisfy $2^n > n^2$ ?
My work. Step 1: $n = 1 $, $2^1 > 1^2$. True.
For $n = k$, $2^k > k^2$. For $n = k+1$,
$$ 2^{(k+1)} > (k+1)^2 \\
2\cdot 2^k>k^2+2k+1 \\
2^k+2^k > k^2+2k+1$$
*
*$2^k > k^2 \text{ - from step 1}$
*$2^k > k^2+2k+1$
How I can find... | (A rather informal approach, just for fun!)
As $n$ increases by one, the expression $2^n$ doubles.
As $n$ increases by one, the expression $n^2$ increases by the $(n+1)$st odd.
This is because $n^2$ becomes $(n+1)^2 = n^2 + (2n+1)$ where $2n+1$ is the $(n+1)$st odd.
Note that the expressions are equal when $n = 4$: $2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the determinent of a $4 \times 4$ matrix with the letter $a$ in it Any idea how to compute the determinant of $4 \times 4$ matrix $A$ when
\begin{equation}
A = \begin{bmatrix} 1 & 4 & 8 & 1\\ 0 & 30 & 1 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 2 & 9 & a \end{bmatrix}.
\end{equation}
The $a$ in the $A_{44}$ location is really c... | Just expand across the third row, as @Surb suggested:
\begin{equation}
|A| = -2 \left|
\begin{pmatrix}
1 & 8 & 1\\
0 & 1 & 0 \\
1 & 9 & a
\end{pmatrix} \right|
=-2 \left|
\begin{pmatrix}
1 & 1 \\ 1 & a
\end{pmatrix} \right| = -2(a - 1) = 2 - 2a.
\end{equation}
Another option is to expand down the fourth column:
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1520819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove the identity $\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$
Let $n,m \in \mathbb{N}$. Prove the identity $$\sum^{n}_{k=0}\binom{m+k}{k} = \binom{n+m+1}{n}$$
This seems very similar to Vandermonde identity, which states that for nonnegative integers we have $\sum^{m}_{k=0}\binom{m}{k}\binom{n}{r-k} = \binom{m... | We can write $\displaystyle \sum^{n}_{k=0}\binom{m+k}{k} = \sum^{n}_{k=0}\binom{m+k}{m} = \binom{m+0}{m}+\binom{m+1}{m}+........+\binom{m+n}{m}.$
Now Using Coefficient of $x^r$ in $(1+x)^{t} $ is $\displaystyle = \binom{t}{r}.$
So we can write above series as...
Coefficient of $x^m$ in $$\displaystyle \left[(1+x)^m+(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Finding the vertex and focus of a rotated parabola So I begun with the following equation : $x^2+2xy+y^2+2\sqrt{2}x-2\sqrt{2}y+4=0$
I transformed it in the following : $y'=\frac{x'^2}{2}+1$
I had to do a rotation of $\frac{\pi}{4}$ of the xy axis. (counter clock wise) My question is how do you find the focus and vertex... | I'm going to assume the OP wanted the vertex and focus of the original tilted parabola since they already had rotated it to a standard form where those values are easier to find.
Rather than rotating, its convenient to do everything in place since we're looking for the focus and by definition: for any point on the par... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Convergence of $\sum _{n=2}^{\infty }\:\frac{1}{n\ln\left(n\right)}$ Is this convergent or not?
$$\sum _{n=2}^{\infty }\:\frac{1}{n\ln\left(n\right)}$$
I tried using the ratio test but the limit is giving me 1, which doesn't help me. I don't think I'm supposed to use the integral test, since we haven't studied it.
| If you are familiar with Cauchy Condensation Test, the condens series becomes
$$\sum_{n} 2^n \frac{1}{2^n \ln (2^n)}$$
If not, you can use the idea of the test:
$$\frac{1}{2 \ln2 } \geq \frac{1}{2 \ln 2 } \\
\frac{1}{3 \ln 3 } \geq \frac{1}{4 \ln 2^2 } \\
\frac{1}{4 \ln 4 } \geq \frac{1}{4 \ln 2^2 } \\
\frac{1}{5 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Centre of gravity of eighth of a sphere Could I ask for help to show where I'm going wrong?
Question
Find the position of the center of gravity of that part of a thin spherical shell x^2 + y^2 + z^2 = a^2 which exists in the first octant.
My Answer
Working in spherical coordinates:
The $2_{nd}$ moment of area of a smal... | edit: Never mind this answer. I mistakenly though you were talking about a solid eight of a sphere.
Why use spherical coordinates at all?
Let $A$ be the one-eight sphere located in the positive octant. The total volume of $A$ is just $\frac{a^3\pi}{6}$. So to calculate the $z$ component of the center of mass, and by s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to compute eigenvalues of big $5×5$ matrix (symmetric matrix) . How to compute eigenvalues of big $5×5$ matrix (symmetric matrix) .
$\left(\begin{matrix}
0 & 0 & 0 & 0 & 1\\
0 & 0 & 1 & 1 & 0\\
0 & 1 & 0 & 1 & 0\\
0 & 1 & 1 & 0 & 0\\
1 & 0 & 0 & 0 & 0\\
\end{matrix}\right)
$
By characteristics equation w... | There is a lot of $0$... you can compute it as usual...
$$\left|\begin{matrix}-t&0&0&0&1\\0&-t&1&1&0\\ 0&1&-t&1&0\\0&1&1&-t&0\\1&0&0&0&-t\end{matrix}\right|=
-t\left|\begin{matrix}
-t&1&1&0\\ 1&-t&1&0\\1&1&-t&0\\0&0&0&-t
\end{matrix}\right|+
\left|\begin{matrix}
0&0&0&1\\-t&1&1&0\\ 1&-t&1&0\\1&1&-t&0
\end{matrix}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\lfloor{(2+\sqrt{3})^n}\rfloor$ is an odd number? How can we prove that $\lfloor{(2+\sqrt{3})^n}\rfloor$ is an odd number? My teacher did something like this:
$$\lfloor{(2+\sqrt{3})^n}\rfloor=(2+\sqrt{3})^n+(2-\sqrt{3})^n-1$$
And she said a few more words, but I understood nothing. Could you please tell me ... | $$(2+\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}+ \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+\binom{n}{n}2^{0}\sqrt{3}^{n}\\
(2-\sqrt{3})^n=\binom{n}{0}2^{n}\sqrt{3}^{0}- \binom{n}{1}2^{n-1}\sqrt{3}^{1}+\binom{n}{2}2^{n-2}\sqrt{3}^{2}+...+(-1)^n\binom{n}{n}2^{0}\sqrt{3}^{n}$$ note that for sum of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of the sequence $\left( \sqrt {2n^{2}+n}-\sqrt {2n^{2}+2n}\right) _{n\in N}$ My answer is as follows, but I'm not sure with this:
$\lim _{n\rightarrow \infty }\dfrac {\sqrt {2n^{2}+n}}{\sqrt {2n^{2}+2n}}=\lim _{n\rightarrow \infty }\left( \dfrac {2n^{2}+n}{2n^{2}+2n}\right) ^{\dfrac {1}{2}}$
$\lim _{n\ri... | Multiplying it by
$$\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}$$
gives
$$\begin{align}&\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\\\\&=\lim_{n\to\infty}(\sqrt{2n^2+n}-\sqrt{2n^2+2n})\cdot\frac{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}{\sqrt{2n^2+n}+\sqrt{2n^2+2n}}\\\\&=\lim_{n\to\infty}\frac{(2n^2+n)-(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the coefficient of partial expansion $\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}$ I want to decompose the equation:
$$\frac{2x+3}{(1-x)(1+0.5x+0.5x^2)}=\frac{A}{1-x}+\frac{B}{1+0.5x+0.5x^2}$$
I found $A$ by multiple both side with $1-x$ and plug $x=1$. However, it is so difficult to find $B$. Could you help me to find $B$
| \begin{equation*}
f(x)=\frac{x+\frac{3}{2}}{(x-1)(2+x+x^{2})}=\frac{A}{x-1}+\frac{Bx+C}{%
x^{2}+x+2}
\end{equation*}
cover-up method leads to
\begin{equation*}
A=\left. f(x)\left( x-1\right) \right\vert _{x=1}=\left. \frac{x+\frac{3}{2}%
}{(2+x+x^{2})}\right\vert _{x=1}=\frac{\frac{5}{2}}{4}=\frac{5}{8}.
\end{equation*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle of $\frac{\pi}{3}$ Find the equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at an angle of $\frac{\pi}{3}$.
Let the direction rat... | Hint:
Write the equation of the line in parametric form, so that an arbitrary point is expressed in terms of just one letter. Then you can use the dot product to form a quadratic equation in the parameter giving you two solutions.
So we have$$\frac{x-3}{2}=\frac{y-3}{1}=\frac z1=\lambda$$
$$\Rightarrow \underline{r}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What is $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2} $ Find the following limit
$$ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3-y^3}{x^2-y^2}$$
This question has been bugging me for some time. Couldn't find it anywhere on the Internet
| You can cancel a common factor:
$$\frac{x^3-y^3}{x^2-y^2}=\frac{(x-y)(x^2+xy+y^2)}{(x-y)(x+y)}=\frac{x^2+xy+y^2}{x+y} $$
(This also continuously extends the function to most of the line $x=y$).
Along any straight line $(x,y)=(at,bt)$, $t\to 0$, this is $\frac{a^2 +ab+b^2}{a+b}\cdot t$ and tends to $0$. But this of cour... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove $\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$ using the half angle identities. Given is the sequence $x_1=0,\; x_{n+1}=\sqrt{2+x_n}$. Prove: $$\lim_{n\rightarrow \infty} 2^n \sqrt{2-x_n}=\pi$$
Hint:
Use the following formulas:
$$\cos\left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$$
$$\sin\left(\frac{x}{2}\... | Note that as $x_1 \le 2$, $x_n \le 2 \forall n$. This can be shown inductively.
Thus, we may write $x_n = 2 \cos \theta_n$ for some $\theta_n \in [0, \pi/2]$, with $\theta_1 = \pi/2$
Now, $2 \cos \theta_{n+1} = \sqrt{2(1+ \cos \theta_n)} = 2 \cos \frac{\theta_n}2 \implies \theta_{n+1} = \frac{\theta_n}{2} = \frac{\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim_{x \to \infty} x^{3/2}(\sqrt[3]{x^3 + 1} - \sqrt[3]{x^3 - 1})$ Calculate $$\lim_{x \to \infty} x^{3/2}(\sqrt[3]{x^3 + 1} - \sqrt[3]{x^3 - 1})$$
I found out that we can simplify it by multiplying:
$$\lim_{x \to \infty} x^{3/2}\frac {{x^3 + 1} - (x^3 - 1)} {\sqrt[3]{(x^3 + 1)^2} + \sqrt[3]{(x^3 + 1)(x^3 -... | Dividing both the numerator and the denominator by $x^{3/2}$ gives
$$\frac{2}{(x^{3/2}+2x^{-3/2}+x^{-9/2})^{1/3}+(x^{3/2}-x^{-9/2})^{1/3}+(x^{3/2}-2x^{-3/2}+x^{-9/2})^{1/3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
String in an kleene star alphabet Let Σ = {a, b}.
How many strings of length 10 are in the language (bb + aab)*?
If this a matter of writing them out or is there a formula to it?
| We could also associate a geometric series with $$(bb+aab)^\star=\varepsilon+(bb+abb)+(bb+abb)^2+\ldots$$ Since each word in this language is uniquely identified by the occurrences of $bb$ and $aab$ parts and since we are only interested in the length of the substrings $bb$ and $aab$, we consider the series
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show $p$ prime s.t. $p \not\equiv 1 \mod 3$ is represented by the binary quadratic equation. I am working on the following question:
Let $p>3$ be a prime such that $p \not\equiv 1 \mod 3$. Show that $p$ is not represented by the binary quadratic equation $f(x, y) = x^2 + xy + y^2$.
I would appreciate any help.
| Remember that you can just add up mods.
If $x \equiv 1 \bmod 3$ and $y \equiv 1 \bmod 3$, then $x^2 + xy + y^2 \equiv 0 \bmod 3$.
If $x \equiv 1 \bmod 3$ and $y \equiv 2 \bmod 3$, then $x^2 + xy + y^2 \equiv 1 \bmod 3$.
If $x \equiv 1 \bmod 3$ and $y \equiv 0 \bmod 3$, then $x^2 + xy + y^2 \equiv 1 \bmod 3$.
If $x \equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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A new approach to find value of $x^2+\frac{1}{x^2}$ When I was teaching in a college class, I write this question on board.
If we now $\bf{x+\dfrac{1}{x}=4}$ show the value of $\bf{x^2+\dfrac{1}{x^2}=14}$
Some student asks me for a multi idea to show or prove that.
I tried these : $$ $$1 :$$\left(x+\frac{1}{x}\right)... | Here's a trigonometric approach:
Let $x=\tan\theta=\sin\theta/\cos\theta$. Then
$$4=x+{1\over x}={\sin\theta\over\cos\theta}+{\cos\theta\over\sin\theta}={1\over\sin\theta\cos\theta}={2\over\sin2\theta}$$
so $\sin2\theta=1/2$. Now
$$x^2+{1\over x^2}={\sin^2\theta\over\cos^2\theta}+{\cos^2\theta\over\sin^2\theta}={1\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Find the limit $\lim\limits_{x\to\infty} f(x)= \lim\limits_{x \to \infty} \left(\frac{x}{x+1}\right)^x$ I need to find the following:
$$\lim_{x\to\infty} f(x)= \lim_{x \to \infty}\left (\frac{x}{x+1} \right )^x$$
I know that this limit = $\frac{1}{e}$ from plugging it into a calculator, but I have to prove it without u... | First you set the entire equation equal to y:
$$
(\frac {x}{(x+1)})^x = y
$$
We can then insert both sides into $ln(x)$:
$$
ln((\frac x{(x+1)})^x) = ln(y)
$$
Then pull out the x power:
$$
xln(\frac x{(x+1)}) = ln(y)
$$
Split the natural log:
$$
xln(x) - xln(x+1) = ln(y)
$$
Shift the left term into the bottom of a fract... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Power series summation Trying to find the sum of the following infinite series:
$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$
Any ideas on how to find this sum?
| Note that for suitable $t$,
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+\cdots.$$
Integrate term by term from $0$ to $x$. We get
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Divide by $x$. We get
$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots.$$
Finally, let $x=\frac{1}{\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Is $a\sqrt[3]{2} + b\sqrt[3]{4}$ irrational? I need to prove that $$ a\sqrt[3]{2} + b\sqrt[3]{4}$$ is irrational, while $a$,$b $ are non zero rationals.
I know that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational and I also know how to prove it, but I can't think of any reasonable implication that would state: $a\sqrt[3]{2... | If $b\ne0$, then $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational iff $\sqrt[3]{2}$ is a root of a quadratic polynomial over $\mathbb Q$.
$\sqrt[3]{2}$ is a root of $X^3-2$, which is irreducible over $\mathbb Q$.
So, every polynomial having $\sqrt[3]{2}$ as root must be a multiple of $X^3-2$. In particular, $\sqrt[3]{2}$ is n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| Well assume that $\sqrt{a+\sqrt{b}}$ can be written as sum of 2 square roots
$$\sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y}\\a+\sqrt{b}=x+y+\sqrt{4xy}\\a=x+y\\b=4xy\\x=a-y\\b=4(a-y)y\\b=4ay-4y^2\\4y^2-4ay+b=0\\y_{1,2}=\frac{4a\pm\sqrt{16a^2-16b}}{8}\\y_{1,2}=\frac{a\pm\sqrt{a^2-b}}{2}\\x_{1,2}=\frac{a\mp\sqrt{a^2-b}}{2}$$
Now i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 0
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Show this function of two variables $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at (0,0) Is the function $f(x,y) = \frac{2x^2y^3}{x^4+y^6}$ continuous at $(0, 0)$?
Trying to use the epsilon-delta definition of continuity to prove this but can't figure it out.
$\sqrt{x^2+y^2} < \delta$ implies $\left|\frac{2x^2y^3}{x... | Note $f(x,y)$ is continuous at $(0,0)$ if $\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$, which is equivalent to for all $(x_n,y_n)$ such that $x_n\to 0,y_n\to 0$, we have $\lim f(x_n,y_n)$ exists and equal to $f(0,0)$.
In your example, the function is not continuous at $(0,0)$. To show this, we show $\lim_{(x,y)\to (0,0)}f(x,y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the expression $( \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 ) / ( \cos 5x + 5 \cos 3x + 10 \cos x ) $ equal to? $$\frac{ \cos 6x + 6 \cos 4x + 15 \cos 2x + 10 }{ \cos 5x + 5 \cos 3x + 10 \cos x }$$
My approach so far : Tried to represent the denominator as a factor of numerator by manipulating numerator's $\cos 6x =... | $$
2^6 \cos^6 x = (e^{ix}+e^{-ix})^6=e^{6ix}+6e^{4ix}+15e^{2ix} +20+15e^{-2ix}+6e^{-4x}+e^{-6ix} \\
= 2(\cos 6x + 6 \cos 4x +15 \cos 2x +10)
$$
and so forth
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1545398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find $A^x$ where A is a $2\times 2$ matrix So, I need to calculate $A^x$ where
$A=\pmatrix{4&-3\\3&-2}$
First of all, $A^x = e^{x\ln A}$, so I get that
$\ln A = \pmatrix{2&0\\-1&1}$
Then I use that
$f(A)= T^{-1}f(J)T$
Here $J$ is the Jordan Canonical form. I got that
$J=\pmatrix{1&1\\0&1}$
and
$T=\pmatrix{3&1\\2... | The result is this one
\begin{equation}
A^x=\left(\begin{array}{c}
\begin{array}{ccccc}
3x+1 & -3x\\
3x & -3x+1\\
\end{array}\end{array}\right)
\end{equation}
Now I have to exit, but as soon as I'm back I will explain it if needed...
Edit>>:
There are many ways to reach the same answer. The more plain one is that:
\beg... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Improper integral $\int_{-\infty}^{+\infty}\frac{1}{a x^2 + bx + c} dx$, $a > 0,a x^2 + bx + c>0$ I have a question about improper integral. If you can help me , I appreciate that.
If a > 0 and the graph of $y=a x^2 + bx + c$ lies entirely above the x-axis, show that
$$
\int_{-\infty}^{+\infty} \frac{dx}{a x^2 + bx + c... | Note that if $ax^2+bx+c$ has real roots then the integral doesn't exists, since we will end up with an integral of the form $$\dfrac1{a}\int_{-\infty}^{\infty} \dfrac{dx}{(x-\alpha)(x-\beta)}$$where $\alpha, \beta \in \mathbb{R}$, which clearly diverges. Hence, we can assume that the discriminant is strictly negative, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Let $a$, $b$, and $c$ be positive real numbers with $aSource: AoPS
My attempt:
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\implies144=50+2(ab+bc+ca) $$
$$\implies ab+bc+ca=47$$
and
$$a^3+b^3+c^3-3abc=(a^2+b^2+c^2-(ab+bc+ca))(a+b+c)\implies216-3abc=12(50-47)$$$$\implies abc=(216-36)/3=60$$
So, $ab=\frac{60}{c}$
Now, $$a^3+b^... | Note that you have found the three elementary symmetric polynomials in the variables $a,b,c$, which determine a cubic polynomial whose roots are $a,b,c$. So $a,b,c$ are the solutions to
\begin{equation*}
\begin{aligned}
&\mathrel{\phantom{=}} x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc \\
&= x^3 - 12x^2 + 47x - 60 \\
&= (x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
| Let the desired limit be $L$. Then we have
\begin{align}
\log L &= \log\left(\lim_{x \to 0}\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\right)\notag\\
&= \lim_{x \to 0}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integral of $\frac{1}{x^2+4}$ Different approach underneath is a brief method of partial fractions integration on the problem given in the title
Using a standard trigonometric result it is known that:
$$ \int \frac{1}{x^2+4}dx=\frac{1}{2}\tan^{-1}(\frac{x}{2})+C$$
But also:
$$\frac{1}{x^2+4}=\frac{A}{x+2i}+\frac{B}{x-2... | Also
$$ tan^{-1}(\frac 2x )= cot^{-1}(\frac x2 ) = \frac \pi 2 -tan^{-1}(\frac x2 ) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $\min \left((x-2)^2+y^2\right)$ $s.t. \>x^2\leq ky^2+1$, $x\geq 0$ Consider the problem $$\min \left((x-2)^2+y^2\right)$$ $$s.t. \>x^2\leq ky^2+1$$
$$x\geq 0$$
where $k \in \mathbb{R}$ is a parameter of the problem.
Determine the status of the point $(1,0)$ for each value of $k$. For what values of $k$ is the poin... | You are looking for the closest point to the point $(2,0)$ in an area limited by the hyperbole $x^2-ky^2=1$ on the right side and the y axis on the left side.
The solutions of the minimisation problem are clearly on the hyperbole.
So $x^2-ky^2=1$, $\lambda_2=0$
If $y \neq 0$ then $k=\frac{1}{\lambda_1}$ and $x(1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove this binomial identity using the following equality: Use the equation $\frac{(1-x^2)^n}{(1-x)^n} = (1+x)^n$ to prove the following identity:
$\displaystyle \sum_{k=0}^\frac{m}{2}(-1)^k{n\choose k}{n+m-2k-1\choose n-1}={m\choose n}$, $m\leqslant n$ and $m$ even
I am really at a loss for how to do this. Where did ... | Let me just briefly present a different generating function.
Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor m/2\rfloor}
{n\choose k} (-1)^k {m-2k+n-1\choose n-1}$$
where $m\le n$ and introduce
$${m-2k+n-1\choose n-1} = {m-2k+n-1\choose m-2k}
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m-2k+1}} (1+z)^{m-2k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Inverse of an exponential function I am having difficulties forming the inverse of this
$f(x) = 3 \cdot2^{3x+1} \cdot 5^{3x-1}$.
What I have done so far:
$3 \cdot 2^{3y} \cdot 2^1 \cdot 5^{3y}\cdot5^{-1} \Leftrightarrow 3\cdot 2\cdot \frac{1}{5}\cdot (5\cdot 2)^{3y} \Leftrightarrow \frac{6}{5}\cdot 10^{3y} \Leftrightar... | $x = \frac{\log \left(\frac{y}{6}\right)+\log (5)}{3 (\log (2)+\log (5))}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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find all integers $a,b,c$ such that $a^2=bc+1,$ $b^2=ca+1.$ This was one of the problems in a math contest in India. This is how I tried it:
Subtracting the second equation from the first one gives $a^2-b^2=bc-ca$ or $(a-b)(a+b)=c(b-a)$. $a-b=-(b-a)$. Therefore a+b has to be -c. Thus all integers satisfying the conditi... | $c=-(a+b)$ is not a general solution: for example with $a=b=2$ and $c=-4$ would give $4=-8+1$, which is not correct.
Instead, in cases of $c=-(a+b)$, you need to solve $a^2=-ab-b^2+1$, i.e. $a =\frac{-b \pm\sqrt{4-3b^2}}{2}$ and that only has integer solutions when $b=-1,0,1$ in which case $a=0\text{ or }1, -1\text{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove the identity $\sum\limits_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$ for $0 \leq x \leq 1$ I'd like to prove this identity:
$$\sum_{k=0}^n\left(x-\frac{k}{n}\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=\frac{1}{n}x(1-x)$$
for $x\in[0,1]$ and $n\in\mathbb{N}$.
I've worked on this pr... | Brute force approach. There might be a more elegant approach.
Multiply by $n^2$, and you want:
$$\sum_{k=0}^n\left(nx-k\right)^2 \binom{n}{k}x^k(1-x)^{n-k}=nx(1-x)$$
First part:
$$\sum_{k=0}^n n^2x^2\binom{n}{k}x^k(1-x)^{n-k} = n^2x^2(x+(1-x))^n = n^2x^2.\tag{1}$$
Second part:
$$\begin{align}
\sum_{k=0}^n 2nkx \binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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find $ \cfrac{1}{r^2} +\cfrac{1}{s^2} +\cfrac{1}{t^2}$ given that $r,s,t$ are the roots of $x^3-6x^2+5x-7=0$
I am asked to find $$ \cfrac{1}{r^2} +\cfrac{1}{s^2} +\cfrac{1}{t^2}$$
given that $r,s,t$ are the roots of $x^3-6x^2+5x-7=0$ .
So what I did was to get the polynomial whose roots are the reciprocals of $r,s,... | Still another approach. We have that
$$ M=\begin{pmatrix}0 & 0 & 7 \\ 1 & 0 & -5 \\ 0 & 1 & 6\end{pmatrix}$$
is the companion matrix of your polynomial, and:
$$ \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}=\text{Tr}(M^{-2})=\text{Tr}\,\begin{pmatrix}0 & 7 & 42 \\ 0 & -5 & -23 \\ 1 & 6 & 31\end{pmatrix}^{-1} $$
hence:
$$ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove that $\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)+\ldots+\cos\left(\frac{2(n-1)\pi}{n}\right)=-1$ May you help on how to start, or where to look for the following question?
By using the $n$-th roots of the unity, show that:
*
*$\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}\right)... | Consider $n$-th roots of the unity $z$, we have $1+z+z^2+\cdots+z^{n-1}=0$, i.e. $$1+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})^2+\cdots+(\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n})^{n-1}=0,$$ by De Moivre's formula, we have $$1+\cos\left(\frac{2\pi}{n}\right)+\cos\left(\frac{4\pi}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Generating function, finding coefficient (decomposing) I just started learning about generating functions, and there is a problem that I have the solution to, but I'm wondering if there is a better general method to solve problems of that kind.
If I want to find the coefficient of $x^8$ in
$$\left( 1+x+x^2+x^3+x^4 \ri... | You could try:
$\begin{align*}
(1 + z + z^2 + z^3 + z^4) \sum_{k \ge 0} (-1)^k \binom{-2}{k} z^k
&= \frac{1 - z^5}{1 - z} \cdot (1 - z)^{-2} \\
&= (1 - z^5) \cdot (1 - z)^{-3} \\
&= (1 - z^5) \cdot \sum_{k \ge 0} (-1)^k \binom{-3}{k} z^k \\
&= (1 - z^5) \cdot \sum_{k \ge 0} \binom{k + 2}{2} z^k
\end{align*}$
so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating limit by sandwich theorem: $\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$ $$\lim_{n\to\ \infty}\frac{1}{1+n^2} +\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}$$
For using Sandwich theorem I need two functions such that $g(x)<f(x)<h(x)$
$$\frac{1}{n+n^2} +\frac{2}{n+n^2}+\cdots+\frac{n}{... | HINT:
If you are familiar with the harmonic numbers, you can also proof:
$$\lim_{n\to\infty}\frac{1}{1+n^2}+\frac{2}{2+n^2}+\cdots+\frac{n}{n+n^2}=\lim_{n\to\infty}\sum_{m=1}^{n}\frac{m}{m+n^2}=$$
$$\lim_{n\to\infty}n\left(n\text{H}_{n^2}-n\text{H}_{n^2+n}+1\right)=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Go from A to D in three equal steps Given two parallel lines $r$ and $s$, line $p$, perpendicular to both, and points $A$ and $D$ on different sides of $p$ with respect to the parallel lines, how can I prove the existence of two points, $B$ and $C$, respectively on $r$ and $s$, such that segments $AB$, $BC$ and $CD$ ar... | Let $a,x,d,y$ the distances shown in the picture and $t=\overline{AB}=\overline{BC}=\overline{CD}$
Then we have
$$a^2+x^2=t^2\tag{1}$$
$$d^2+y^2=t^2\tag{2}$$
$$(x-y)^2+e^2=t^2\tag{3}$$
If we add $a^2d^2$ to $(3)$ and reorder the terms we get
$$x^2+a^2-2xy+y^2+b^2+e^2=t^2+a^2+d^2$$
an further
$$2xy=t^2+e^2-a^2-d^2\tag... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find $ \int \frac{1}{2\sin(x)-3\cos(x)}dx$.
Find $\displaystyle \int \dfrac{1}{2\sin(x)-3\cos(x)}dx$.
My book said to solve this by saying $u = \tan \left(\dfrac{x}{2} \right)$ since $\cos(x) = \dfrac{1-u^2}{1+u^2}$ and $\sin(x) = \dfrac{2u}{1+u^2}$. I don't see how this will help since $du = \dfrac{1}{\cos(x)+1}dx $... | Hint:
the suggested substitution works well because:
$$
\cos x+1=\frac{1-u^2}{1+u^2}+1=\frac{2}{1+u^2}
$$
so:
$$
du=\frac{1}{2\cos^2(x/2)}dx= \frac{dx}{\cos x+1} \Rightarrow dx=\frac{2\,du}{1+u^2}
$$
the integral becomes:
$$
2\int\frac{du}{3u^2+4u-3}
$$
that, completing the square, becomes:
$$
2\int\frac{du}{(\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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I want to calculate the limit of: $\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $ I want to calculate the limit of: $$\lim_{x \to 0} \left(\frac{2^x+8^x}{2} \right)^\frac{1}{x} $$
or prove that it does not exist. Now I know the result is $4$, but I am having trouble getting to it. Any ideas would be great... | Notice, $$\lim_{x\to 0}\left(\frac{2^x+8^x}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\left(\frac{2^x+2^{3x}}{2}\right)^{1/x}$$
$$=\lim_{x\to 0}\exp\frac{1}{x}\ln\left(\frac{2^x+2^{3x}}{2}\right)$$
Applying L'hospital's rule for $\frac 00$ form $$=\lim_{x\to 0}\exp\frac{\frac{d}{dx}\ln\left(\frac{2^x+ 2^{3x}}{2}\right)}{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1582106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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Find $n$ if the area between the curve of $y=x^n$ and the $y$ axis is $3$ times the area between the curve and the $x$-axis
For this question i tried to find the area for red and blue sections, and equate them by red= 3 blue.
However, it didnt work out and I got $b-a = 3(b^n - a^n)$ for the outcome.
In the book, it ... | The blue area is $$\int_a^b x^n dx = \frac{b^{n+1}-a^{n+1}}{n+1}.$$
The red (orange?) area is the area of the big rectangle with sides, $b$, $b^n$, minus the small (white) rectangle with sides $a$, $a^n$, minus the above integral. That is $$b^{n+1}-a^{n+1}-\frac{b^{n+1}-a^{n+1}}{n+1}.$$ Hence, we need $$ b^{n+1}-a^{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Involutory matrix $2 \times 2$ I want to find out how many $2 \times 2$ involutory matrices are there over $\mathbb Z_{26}$.
$ $
Is there any formula to calculate this?
$ $
Thanks for your help.
| Let $p$ be an odd prime, and $A=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\in M_2(\mathbb Z/p\mathbb Z)$.
Then $A^2=I$ is equivalent to $a^2+bc=1,$ $d^2+bc=1,$ $b(a+d)=0,$ $c(a+d)=0.$
If $a+d\ne0$, then $b=c=0$, so $a^2=d^2=1$, and there are $2$ solutions.
If $a+d=0$, then consider the following cases:
(i) $a=0... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\prod \frac {2k - 1} {2k}$ I came across to this: evaluate $$\prod_{k = 1}^{n} \frac {2k - 1} {2k}.$$
I brought it to a closed-form expression in asymptotics but I suspect my asymptotics is bad and I also want to see different solutions, hence the question.
My original solution: Write
$$\begin {eqnarray}
\pr... | $$\prod_{k=1}^{n}\frac{2k-1}{2k}=\frac{2-1}{2}\cdot\frac{4-1}{4}\cdot\frac{6-1}{6}\cdot\dots\cdot\frac{2n-1}{2n}=$$
$$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\dots\cdot\frac{2n-1}{2n}=\frac{1\cdot3\cdot5\dots\cdot(2n-1)}{2\cdot4\cdot6\dots\cdot2n}=\frac{\frac{2^n\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}}}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Derive quadratic formula I cannot understand how quadratic formula to solve for $x$ was derived.
On this website, it explains the steps
Following I understand
but I cannot understand how they got
$b/2a$
and why they are squaring it
$b^2/4a^2$
Really, I am baffled!
| $x^2+\frac{b}{a}x+K^2=(x+K)^2$
$x^2+\frac{b}{a}x+K^2=x^2+2xK+K^2$
$\frac{b}{a}x=2xK$
$k=\frac{b}{2a}$
Then we know that
$x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(x+\frac{b}{2a})^2$
$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=(x+\frac{b}{2a})^2$
This process is called completing the square.
Now we can factor a perfect square. Is this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $0 \leq ab^2-ba^2 \leq \frac{1}{4}$ with $0 \leq a \leq b \leq 1$.
Let $a$ and $b$ be real numbers such that $0 \leq a \leq b \leq 1$. Prove that $0 \leq ab^2-ba^2 \leq \dfrac{1}{4}$.
Attempt
We can see that $ab^2-ba^2 = ab(b-a)$, so it is obvious that it is greater than or equal to $0$. But how do I show ... | There are more generic ways of solving such problems.
Since $f(a,b) = ab^2 - ba^2$ is continuous and differentiable, and the domain $0 \le a \le b \le 1$ is a compact simplex, it attains its max and min at one of the following points:
*
*A point where $0 < a < b < 1$ and $\nabla f (a,b) = 0$, i.e. $f_x(a,b) = f_y(a,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Arithmetic: Prove that is multiple of 30 Prove that $n^{19}-n^7$ is multiple of $30$
I've seen $6$ can divide it because
$$n^{19}-n^7=n^7(n^{12}-1) = n^7(n^6+1)(n^6-1)=n^4(n^6+1)(n^3-1)n^3(n^3+1)$$
And there are three consecutive numbers, so, at least one is multiple of $3$ and up to two even numbers.
But, how to pr... | Notice, one should re-write & factorize as follows $$n^{19}-n^7=n^7(n^{12}-1)$$
$$=n\cdot \underbrace{n^6\color{blue}{(n^{6}-1)}\color{red}{ (n^{6}+1)}}_{\text{divisible by 5}}$$
$$=n^7\color{blue}{(n^{3}-1)(n^3+1)}\color{red}{(n^{2}+1)(n^4-n^2+1)}$$
$$=n^7\color{blue}{(n-1)(n^2+n+1) (n+1)(n^2-n+1)}\color{red}{(n^{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Subadditivity of square root function Any hint to prove
$$ \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 $$
| Another:
$x,y \geq 0$ then the product $\sqrt{x} \sqrt{y}$ have sence and now
$$\sqrt{x + y} \leq \sqrt{x + 2\sqrt{x}\sqrt{y} + y} = \sqrt{(\sqrt{x} + \sqrt{y})^2} = \sqrt{x} + \sqrt{y}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Problem related to non-coplanar unit vectors,equally inclined to one another at an angle $\theta$ I've lately been facing lot of trouble in solving vector equations.Like the one below :
Let $a,b,c$ be non-coplanar unit vectors,equally inclined to one
another at an angle $\theta$.If $a×b+b×c=pa+qb+rc$,find the scalar... | First of all, observe that we can safely assume $0 < \theta < \frac{2}{3} \pi$. Indeed, it makes sense to consider $\theta$ to be the smallest (positive) angle between, say, $a$ and $b$. Furthermore, clearly for $\theta = 0$ we have $a = b = c$, while for $\theta = \frac{2}{3} \pi$ we have $a,b,c$ coplanar.
Now, recall... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a combinatorial recurrence relation with three variables This question is from the generating functionology textbook,
Let $f(n, m,k)$ be the number of strings of n $0$’s and $1$’s that contain
exactly $m$ $1$’s, no $k$ of which are consecutive.
Find a recurrence formula for $f$. It should have $f(n, m,k)$
on th... | We can break down $f(n,m,k)$ into how many consecutive ones it ends with
$$
f(n,m,k)=\left\{\begin{array}{}
0&\text{if }m<0\text{ or }n<m&\text{impossible}\\[9pt]
[n<k]&\text{if }n=m&\text{only ones}\\
\displaystyle\sum_{j=0}^{k-1}f(n-j-1,m-j,k)&\text{if }n\gt m&\begin{array}{l}\text{ends with a zero}\\[-4pt]\text{and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$... | Yes, you can! By AM-GM we obtain:
$$\sum_{cyc}\frac{2}{a+b}\leq\sum_{cyc}\frac{1}{\sqrt{ab}}\leq\sum_{cyc}\frac{\frac{1}{a}+\frac{1}{b}}{2}=\sum_{cyc}\frac{1}{a}.$$
The right inequality we can prove also by AM-GM:
$$2\sum_{cyc}\frac{1}{a+b}=\frac{1}{a+b+c}\sum_{cyc}(a+b)\sum_{cyc}\frac{1}{a+b}\geq$$
$$\geq\frac{1}{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that ${x^7-1 \over x-1}=y^5-1$ has no integer solutions I want to show that $${x^7-1 \over x-1}=y^5-1$$
cannot have any integer solutions. The only observation I have made so far is that the left hand side is the $7$th cyclotomic polynomial
$$\Phi_7(x)= {x^7-1 \over x-1}=x^6+x^5+x^4+x^3+x^2+1$$
If I remember co... | A start to solving this equation is to first prove that if $x$ is an integer and $p$ is a prime divisor of the left hand side $\frac{x^{7}-1}{x-1}$ then either $p=7$ or $p \equiv 1(\mod{7})$.
Proof: First notice by Fermat's little theorem that $x^{p-1}-1$ is divisible by $p$. Also by our hypothesis $x^{7}-1$ is divis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
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.