Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$ Find the limit following:
$$L=\lim_{ _{\Large {n\to \infty}}}\:\sqrt{\frac{1}{2}+\sqrt[\Large 3]{\frac{1}{3}+\cdots+\sqrt[\Large n]{\frac{1}{n}}}}$$
P.S
I tried to find the value of $\:L$, but I found myself stuc... | Actually, there is another way which gives better upper boundary. I'm posting it as a separate answer because of the size.
First we notice that:
$$\lim_{n \to \infty} \left( \frac{1}{n} \right)^{\frac{1}{n}}=\lim_{n \to \infty} \left(1- \left(1- \frac{1}{n} \right) \right)^{\frac{1}{n}}=1$$
This is not a proof, but t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 4,
"answer_id": 3
} |
Math Induction Proof: $(1+\frac1n)^n < n$ So I have to prove:
For each natural number greater than or equal to 3,
$$(1+\frac1n)^n<n$$
My work:
Basis step: $n=3$
$$\left(1+\frac13\right)^3<3$$
$$\left(\frac43\right)^3<3$$
$$\left(\frac{64}{27}\right)<3$$
which is true.
Now the inductive step, assume $P(k)=\left(1+\... | Two approaches
Bernoulli-like Inductive Step
$$
\begin{align}
\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n}
&=\overbrace{\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}}^{\lt1}\frac{n+1}n\\
&\lt\frac{n+1}n
\end{align}
$$
This shows that if $\left(1+\frac1n\right)^n\lt n$, then $\left(1+\frac1{n+1}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/582459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Showing $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$
I want to show that $x^4-24x^2+4$ is irreducible over $\mathbb{Q}$.
I have tried Eisenstein's criterion using $p=2$ and substituting $x+1$ and $x-1$ but it won't work. Then what I tried was using the rational root test. I got as my possible rational roots $\pm ... | $x^4-24x^2+4=t^2-24t+4=0 \Rightarrow t=12\pm 2\sqrt{35}\Rightarrow x^2-(12\pm 2\sqrt{35})=0$
I was a bit excited (Not careless ) and concluded that given polynomial is irreducible.
But It is not proper to conlcude at this stage but rather find all other roots and then check for possible combinations.
$x^2=12+2\sqrt{35}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $x$ such that $\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$ Find $x$ such that $$\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$$
I tried many ways: $$\sqrt{x+\sqrt{x+7}}=n$$
$$\sqrt{x+\sqrt{x+7}}^2=n^2$$
$$x+\sqrt{x+7}=n^2$$
then solve for $x$ but didn't do with success.
I think this is the most difficult problem in my lifetime
Also $x... | To see what's going on in this problem, it helps to generalize it slightly. (This should also help put martini's wonderful answer in context.) Let's look for integers $x$ that make an expression of the form
$$\sqrt{x+\sqrt{x+A}}\in\mathbb{N}$$
for an arbitrary $A\in\mathbb{N}$.
As other answers have shown, setting th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/592266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 8,
"answer_id": 3
} |
Find $x^4+y^4$ and $x^3+y^3$ if $x+y=2$ and $x^2+y^2=8$ Find $x^4+y^4$ if $x+y=2$ and $x^2+y^2=8$
So i started the problem by nothing that $x^2+y^2=(x+y)^2 - 2xy$ but that doesn't help!
I also seen that $x+y=2^1$ and $x^2+y^2=2^3$ so maybe $x^3+y^3=2^5$ and $x^4+y^4=2^7$ but i think this is just coincidence
So how can ... | Solve the given equations simultaneously to obtain $x=1-\sqrt 3, y=1+\sqrt 3$ or vice versa. Then just compute $x^4+y^4$ directly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/592963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Series $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges or diverges.
I tried to use the Ratio test:
$|\frac{a_{n+1}}{a_n}|= \frac{n+2}{2^{n+1}} \frac{2^n}{n+1} = \frac{1}{2} \frac{n+1+1}{n+1} = \frac{1}{2} (1+ \frac{1}{n+1})$
$\lim\limits_{n \rig... | You can also break it down into a sum of two series $1/2^n $ and $n/2^n$ , each of which converges; first is a geometric series (with $r<1$), the second converges by , e.g., the zeta test; $n/2^n < 1/n^2$ for all $n>N$ ; $N$ a finite value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/595981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$ I'm new here, but I need your help so much to solve an equation:
$$2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$$
I tried a lot like making $2[(\sin^2 x)^3 + (\cos^2 x)^3$
| Hint: $x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$. Then look for squares.
Also, it is not equal to zero, but to $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/598045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$... | Your proof is correct!
Here is a different proof.
First note that the sequence, let's call it $(a_n)_{n\in\mathbb N}$, can be defined recursively as $a_1=\sqrt{2}$ and $a_{n+1}=\sqrt{2a_n}$. Then you can show that it is increasing and bounded above by 2, therefore it converges.
Its limit, $\ell$, must satisfy $\ell=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/601045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
How many Positive integer solutions does the equation $x + y + z + w = 15$ have? How many Positive integer solutions does the equation $x + y + z + w = 15$ have?
Attempt:
Let $x = m + 1, y = n + 1, z = o + 1, w = p + 1 $
Then, $ m + 1 + n + 1 + o + 1 + p + 1 = 15$
$ m + n + o + p = 11 $
| Hint:
Find the coefficient of $x^{15}$ of the following function:
$$f(x)=(x+x^2+x^3+\cdots+x^{15})^4$$
Why this gives you the number of integer solution of your equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/602547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Calculate the determinant of $3\times 3$ matrix with $\sin x$ and powers of $\cos x$ How to calculate the determinant of this matrix
$A=\begin{bmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x & 1 & 1
\end{bmatrix}$
$$\left[A\right]=\begin{vmatrix}
\sin x & \cos^2x & 1 \\
\sin x & \cos x & 0 \\
\sin x ... | $$\left[A\right]=\begin{vmatrix} \sin x & \cos^2x & 1\\ \sin x & \cos x & 0 \\ \sin x & 1 & 1 \end{vmatrix}\begin{vmatrix} \sin x & \cos^2x \\ \sin x & \cos x \\ \sin x & 1 \end{vmatrix}=\sin x\cos x+\sin x-\sin x\cos x -\sin x\cos^2x=\\=\sin x\left(1-\cos^2 x\right)=\\=\sin x\cdot \sin^2x=\sin^3 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/602731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How do I solve this limit: $\lim _{x \to 0} \left(\frac{ \sin x}{x}\right)^{1/x}$? $$\lim _{x \rightarrow 0} \left(\frac{
\sin x}{x}\right)^{1/x}$$
I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Than... | Hint : Take its logarithm, use the fact that $\ln a^b=b\ln a=\dfrac{\ln a}{1/b}$, and apply l'Hopital $3$ or $4$ times.
$$\ln L=\lim_{x\to0}\frac1x\cdot\ln\frac{\sin x}x=\lim_{x\to0}\frac{\ln\sin x-\ln x}x=\lim_{x\to0}\frac{\dfrac{\cos x}{\sin x}-\dfrac1x}1=\lim_{x\to0}\frac{x\cdot\cos x-\sin x}{x\cdot\sin x}=$$
$$=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/605202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
If $abc=1$ and $a,b,c$ are positive real numbers, prove that ${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1$.
If $abc=1$ and $a,b,c$ are positive real numbers, prove that $${1 \over a+b+1} + {1 \over b+c+1} + {1 \over c+a+1} \le 1\,.$$
The whole problem is in the title. If you wanna hear what I've tried, ... | let $$a=x^3,b=y^3,c=z^3\Longrightarrow xyz=1$$
since
$$y^3+z^3\ge y^2z+yz^2$$
so
$$\dfrac{1}{1+b+c}=\dfrac{xyz}{xyz+y^3+z^3}\le\dfrac{xyz}{xyz+y^2z+yz^2}=\dfrac{x}{x+y+z}$$
so
$$\sum_{cyc}\dfrac{1}{1+b+c}\le\sum_{cyc}\dfrac{x}{x+y+z}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/606380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
The roots of the equation $z^n=(1+z)^n$... The complex roots of the equation
$$z^{n}=(1+z)^{n}$$
$A.$ are vertices of a regular polygon
$B.$ lie on a circle
$C.$ are collinear
$D.$ none of these
Don't know where to start.. Please help.!
NOTE: $z$ is a complex number.
| Clearly, $z\ne0,$
So dividing either sides by $z^n,$ we have $$\left(1+\frac1z\right)^n=1$$
Using de Moivre's formula & Euler's formula, $$1+\frac1z=e^{\frac{2m\pi i}n}=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n$$ where $m$ is any integer
$$\implies \frac1z=\cos\frac{2m\pi}n+i\sin\frac{2m\pi }n-1=i2\sin\frac{m\pi}n\cos\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/607487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find equation of a tangent on $y= \sin2x$ Find equation of a tangent on $y= \sin2x$ in intersections with $y=\frac{1}{2}$
What I calculated:
Intersections: $$\sin2x= \frac{1}{2}$$
...
$$0=tg^2x-4tgx-1$$
$$tgx_{1}=2+\sqrt{3}$$
$$x_{1}=75+k\pi$$
$$tgx_{2}=2-\sqrt{3}$$
$$x_{2}=15+ k\pi$$
$$T_{1}(75+k\pi,\frac{1}{2})$$ a... | \begin{align*}
\ \sin{2x} &= \frac{1}{2} \\
\ \arcsin{\big(\sin{2x}}\big) &= \arcsin{\frac{1}{2}} \\
\ 2x &= \frac{\pi}{6} \\
\ x &= \frac{\pi}{12} \qquad\text{to be more precise you can say:}\\
\ x &= \frac{\pi}{12} + \pi k\quad ,\quad k\in\mathbb Z\\
\end{align*}
So we know $x=\dfrac{\pi}{12}$ when they intersec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/610616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can I show that $\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$ for any $x\in{\bf R}$?
Show that
$$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right|\leq\frac{\pi}{2}$$
It is true for $x=0$. But I don't see how it is true for any $x\in{\bf R}$. The identity $\frac{\pi}{2}=\int_0^\infty\frac{1}... | We know that $$\left|\sum_{n=1}^\infty\frac{x}{n^2+x^2}\right| < \int_0^\infty{\frac{x}{n^2+x^2}dn}$$ for any $x\in{\Bbb R}$
Evaluating the integral we get $$\begin{align}
\int_0^\infty{\frac{x}{n^2+x^2}dn} &= tan^{-1}\bigg(\frac{n}{x}\bigg)\bigg|_0^\infty \\
&= \lim_{t\to\infty} tan^{-1}\bigg(\frac{t}{x}\bigg) \\
&= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/611186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Prove Trig Identity
For any three angles $\alpha,\beta,\gamma$, show that $$\sin(\alpha-\beta)+\sin(\alpha-\gamma)+\sin(\beta-\gamma)=4\cos\frac{\alpha-\beta}2\sin\frac{\alpha-\gamma}2\cos\frac{\beta-\gamma}2$$
This is what I've tried:
$$2\sin\overbrace{\left(\frac{A-C}2\right)}^x\cos\overbrace{\left(\frac{A-B}2\righ... | $$
\begin{align}
4\cos(\phi)\cos(\theta)\sin(\phi+\theta)
&=2\big[\cos(\phi-\theta)+\cos(\phi+\theta)\big]\sin(\phi+\theta)\tag{1}\\
&=\sin(2\phi)+\sin(2\theta)+\sin(2\phi+2\theta)\tag{2}
\end{align}
$$
Justification:
$(1)$: $2\cos(\phi)\cos(\theta)=\cos(\phi-\theta)+\cos(\phi+\theta)$
$(2)$: $\begin{array}{}2\cos(\phi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/611324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integral $-\int_{0}^{a}{at^{3}\,{\rm d}t\over\,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,}$ This problem's taking me a lot longer than I like (probably doing things the hard way...). I have 9 different terms to integrate, some of which are messier than others. This one, though, is... | Using the substitution $u=a^2-t^2$ as suggested by André Nicolas, we get:
$\dfrac{\mathrm du}{\mathrm dt}=-2t$. So,
$\begin{align}&\int-\frac{at^3}{\sqrt{(a^2-t^2)(b^2t^2+a^4-a^2t^2)}}\mathrm dt\\
&=\frac{a}2\int\frac{t^2(-2t)}{\sqrt{(a^2-t^2)((b^2-a^2)t^2+a^4)}}\mathrm dt\\
&=\frac{a}2\int\frac{(a^2-u)}{\sqrt{u((b^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/614425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Function : Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$ Problem :
Find the range of $f(x) = \sin^{-1}x +\tan^{-1}x +\cos^{-1}x$
Solution : Since, $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$
Since range of $\tan^{-1}x$ is $ (\frac{-\pi}{2}, \frac{\pi}{2})$
$\therefore, \frac{-\pi}{2} \leq \tan^{-1}x \le... | This is not correct because you need to restrict $\tan^{-1}$ to the same domain as the other two functions: $[-1,1]$. The extreme vales that $\tan^{-1}(x)$ takes on $[-1,1]$ are $\pm\frac{\pi}{4}$. So the range of your $f$ is $\left[\frac{\pi}{2}-\frac{\pi}{4},\frac{\pi}{2}+\frac{\pi}{4}\right]=\left[\frac{\pi}{4},\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Cubic equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots. How can I find their value in terms of $a,b,c$? If the equation $ax^3+3bx^2+3cx+d = 0$ has $2$ equal roots, then equal root must
be equal to $\displaystyle \frac{bc-ad}{2(ac-b)^2}.$
My Try:: Let $x=\alpha,\alpha,\beta$ be the roots of given equation. Then using... | Since $\beta=-2\alpha-\frac{3b}{a}$, we have
$${\alpha}^2+2\alpha \left(-2\alpha-\frac{3b}{a}\right)=\frac{3c}{a}\Rightarrow a{\alpha}^2+6b\alpha+c=0\Rightarrow 2a{\alpha}^3+12b{\alpha}^2+2c\alpha=0,$$
$${\alpha}^2\left(-2\alpha-\frac{3b}{a}\right)=-\frac da\Rightarrow 2a{\alpha}^3+3b{\alpha}^2-d=0.$$
Hence, we have
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont... | Write $$\frac{3}{x+1}-\frac{2}{2x+5}=\frac{4x+13}{\left(x+1\right)\left(2x+5\right)}\leq0$$ There are
'special' values $-1,-2.5$ and $-3.25$, found if one of the factors in numerator or denominator is asked to equalize $0$.
Ask yourself the question:
What is the sign of this fraction if I fill in a value $x>-1$ or
$-2.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/618661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
$\omega^2+\omega+1$divides a polynomial The question is
Show that $f(n)=n^5+n^4+1$ is not prime for $n>4$.
The solution is given as
Let $\omega$ be the third root of unity. Then $\omega^2+\omega+1=0$. Since $\omega^5+\omega^4+1=\omega^2+\omega+1$, we see that $\omega^2+\omega+1$ is a *factor of the polynomial. So *$n^... | This is rather poor phrasing (in the highlighted solution). What it is saying is that that $\omega,$ which is a root of the irreducible polynomial $x^2+x+1$ is also a root of $x^5+x^4+1,$ and so the gcd of the two polynomials is not $1,$ and since the first polynomial is irreducible, it must divide the second. Now, tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/619783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Help with Evaluating triple integral Evaluate $\int \int \int_B xyz^2 dV$, where B is a cuboid bounded by the regions $ 0 \le x \le 1 $, $ -1 \le y \le 2 $, $ 0 \le z \le 3 $.
I keep getting $ \frac{27}{4}$ as my answer but apparently it's incorrect...Any help would be appreciated.
| This is equivalent to
\begin{align*}
\int_0^1 x dx \int_{-1}^2 y dy \int_0^3 z^2 dz &= \left(\frac{1}{2} x^2 \Big|_0^1\right) \left(\frac 1 2 y^2\Big|_{-1}^2\right) \left(\frac 1 3 z^3 \Big|_0^3 \right) \\
&= \frac{1}{2} \cdot \left(\frac{4}{2} - \frac{1}{2}\right)\cdot \frac{3^3}{3} \\
&= \frac{1}{2} \cdot \frac{3}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the pos... | This is called the "Chinese Remainder Theorem". You are given two congruent condition and you have to find the answer. I will show an example using slightly bigger numbers.
Suppose we are given
$$ n \equiv 3 \mod 7, ~~\hbox{and}~~ n\equiv 5\mod 11$$
From the first we can write
$$ n = 7 k + 3$$
Substituting in the seco... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/620994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find the exact value of $\tan\left ( \sin^{-1} \left ( \sqrt 2/2 \right )\right )$
Find the exact value of $\tan\left ( \sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )\right )$ without using a calculator.
I started by finding $\sin^{-1} \left ( \dfrac{\sqrt{2}}{2} \right )=\dfrac{\pi}{4}$
So, $\tan\left ( \sin^{-1} \... | Hint: $$\tan\left(\frac{\pi}{4}\right) = \frac{\sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How find this ODE:$y''-y'+y=x^2e^x\cos{x}$ Find this follow ODe solution
$$y''-y'+y=x^2e^x\cos{x}$$
I konw solve this follow three case
$$y''-y'+y=x^2\cos{x}$$
$$y''-y'+y=e^x\cos{x}$$
$$y''-y'+y=x^2e^x$$
But for $f(x)=x^2e^x\cos{x}$, I can't.
Thank you very much!
| One more way to solve it is using the Laplace transformation $\mathcal{L}.$
Let $\mathcal{L}(y)=Y(p).$ Then
$$
\mathcal{L}(y')=p Y(p)-y(0),\\
\mathcal{L}(y'')=p^2 Y(p)-y'(0)-p y(0),\\ \text{and, using the standard rules for L.t.}\\
\mathcal{L}({x}^{2}{{\rm e}^{x}}\cos \left( x \right))=\frac{1}{\left( p-1-i \right) ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/624715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Roots of unity and a system of equations by Ramanujan Is it immediately apparent that the solution to the system of equations,
$$\begin{aligned}
x_1^2 &= x_2+2\\
x_2^2 &= x_3+2\\
x_3^2 &= x_4+2\\
&\vdots\\
x_n^2 &= x_1+2\\
\end{aligned}\tag{1}$$
can be given by the roots of unity? Specifically,
$$x =\frac{y_k^2+1}{y_k... | Let $f(x)=x^2-2$. Then $$f(2\cos u)=4\cos^2u-2=2(2\cos^2u-1)=2\cos2u$$ so $$f^n(2\cos u)=2\cos2^nu$$ Then $f^n(x)=x$ becomes $$\cos u=\cos2^nu$$ for which solutions are given by $$(2^n-1)u=2k\pi,k=0,\pm1,\pm2,\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\binom{2n}{n}\frac{1}{n+1} = \prod \limits_{i = 2}^n \frac{2i-1}{i+1} $? How to prove this closed form involving Catalan numbers? $$\binom{2n}{n}\frac{1}{n+1} = \prod \limits_{i = 2}^n \frac{2 \times (2i-1)}{i+1} $$
I have seen this being used here. Not sure how to derive it. Any ideas?
| Note that $\binom{2n}{n}=\frac{(2n)!}{n!n!}$. But
$$(2n)!=\left(1\cdot 3\cdot 5\cdots \cdot(2n-1)\right)\left(2\cdot 4\cdot 6\cdots\cdot(2n)\right)$$
(we separated the odd numbers and the even numbers).
The product of the even numbers $2\cdot 4\cdot 6\cdots(2n)$ is $2^n n!$.
From this it follows that
$$\frac{1}{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/627026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How would you explain to a 9th grader the negative exponent rule? Let us assume that the students haven't been exposed to these two rules: $a^{x+y} = a^{x}a^{y}$ and $\frac{a^x}{a^y} = a^{x-y}$. They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^... | Multiplication of exponents:
$4^2 \times 4^3 = 4^5$
The powers are added: $2 + 3 = 5$
Division of exponents:
$\dfrac{4^2}{4^3}\ = 4^{-1}$
The powers are subtracted: $2 - 3 = -1$
Now tell them to try and find the value of $4^{-1}$
Since $4^2 = 16$, and $4^3 = 64$
$\dfrac{16}{64}\ = 1/4 = 4^{-1}$
This is the way I would... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/629740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 24,
"answer_id": 12
} |
How to prove the inequality: $\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$ Prove that:
$$\frac{(1+x)^2}{2x^2+(1-x)^2}+\frac{(1+y)^2}{2y^2+(1-y)^2}+\frac{(1+z)^2}{2z^2+(1-z)^2}\leq 8$$
subject to the constraints:
$$x,y,z >0$$
and
$$x+y+z=1.$$
| if $a+b+c=1,a,b,c>0$
$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2+2a+1}{3a^2-2a+1}\le 8$$
since
$$\dfrac{a^2+2a+1}{3a^2-2a+1}\le 4a+\dfrac{4}{3}$$
this is true becasuse
$$\Longleftrightarrow 3(4a+1)(a-\dfrac{1}{3})^2\ge 0$$
so
$$\sum_{cyc}\dfrac{(a+1)^2}{2a^2+(1-a)^2}\le 4(a+b+c)+4=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/629811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the integral: $ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x\;?$ My problem, is how to find the integral of $$ \int \frac{2x}{\sqrt {4x-1}}\, \mathrm{d}x$$
Can I do this?
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
$$ \frac... | You've got the right idea, but there are two places you erred:
$$ \int \frac{\frac{1}{2}\cdot4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x = \int \frac {2x - 1 + 1}{\sqrt{4x - 1}}\,dx \neq \frac{1}{2}\int \frac{4x-1+1}{\sqrt {4x-1}}\, \mathrm{d}x$$
You need to have: $$\int \dfrac{2x}{\sqrt{4x-1}}\,dx = \int \frac{\frac{1}2(4x-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/630684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to calculate limit without L'Hopital How can I calculate limit without using L'Hopital rule?
$\displaystyle\lim_{x\to 0 }\frac{e^{\arctan(x)}-e^{\arcsin(x)}}{1-\cos^3(x)}$.
| Let $\arctan x = t$ so that $x = \tan t$ so that $\sin t = x/\sqrt{1 + x^{2}}$ or $\arctan x = t = \arcsin (x/\sqrt{1 + x^{2}})$.
Now we have
$\displaystyle\begin{aligned}\arctan x - \arcsin x &= \arcsin (x/\sqrt{1 + x^{2}}) - \arcsin x\\
&= \arcsin\left(x\sqrt{\frac{1 - x^{2}}{1 + x^{2}}} - \frac{x}{\sqrt{1 + x^{2}}}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/632993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Limit of two sequences
*
*$ \lim_{n\to \infty} \sqrt[n]{3^n+4^n} $ . I think the limit is $4$. I did :
$ \sqrt[n]{3^n+4^n} = 4 \sqrt[n]{(\frac{3}{4}) ^n+1}$ .Am I right?
*$ \lim_{n\to \infty} \frac{1}{1\cdot 4 } + \frac{1}{4\cdot 7} +...+\frac{1}{(3n-2)(3n+1)} $. I know that for each $k$ , this sequence is the sum ... | Hint:
For the second one use this fact that :
$$\frac{1}{(3n+1)(3n-2)}=\frac{-1}{3(3n+1)}+\frac{1}{3(3n-2)}$$ So the sum can be reduced to this one: $$S_n=\frac{1}3\left(\frac{1}{1}+\frac{1}{(3n+1)(3n-2)}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/633522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y$ is a given positive odd integer Let $k$ be a fixed positive odd integer. Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y=k$
My approaches:
*
*Since $k$ is odd, $x$ and $y$ have different parity.... | Assume wlog that $x < y$ (they cannot be equal since $k$ is odd). Let's compare $x^2 + y^2$ to $(x + 1)^2 + (y-1)^2$.
$$
(x + 1)^2 + (y-1)^2 - (x^2 + y^2) = 2x -2y+ 2
$$
Since $x < y$, the result is non-positive, and as long as $x \neq y-1$, it is strictly negative, mening the net value has decreased. So as long as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/633675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Are there real numbers such that ... $x^2 + xy =3$ and $x - y^2 = 2$?
I graphed it and there are no intersections so obviously there is no real number solutions, but is there a "mathier" (read algebraic) way to prove this?
| From the second equation, $x = 2 + y^2$. Substituting this into the first equation gives
$$3 = x^2 + xy = (2 + y^2)^2 + (2 + y^2) y = (2 + y^2)(2 + y^2 + y) $$
Now $2 + y^2 \geq 2$ and $$y^2 + y + 2 = \left(y + \frac 1 2\right)^2 +\frac7 4 \geq \frac 74$$
So the right hand side of the above equation is at least
$$2 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How to prove that a $3 \times 3$ Magic Square must have $5$ in its middle cell? A Magic Square of order $n$ is an arrangement of $n^2$ numbers, usually distinct integers, in a square, such that the $n$ numbers in all rows, all columns, and both diagonals sum to the same constant.
How to prove that a normal $3\times 3$... | Let
\begin{array}{|c|c|c|}
\hline
a & b & c\\
\hline
d & e & f \\
\hline
g & h & i \\
\hline
\end{array}
be a magic square of order three with sum row, column, and diagonal sums equal to S. then
\begin{align}
a+e+i &= S\\
b+e+h &= S\\
c+e+g &= S\\
d+e+f &= S \\
\hline
(a+b+c)+(d+e+f)+(g+h+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 6,
"answer_id": 1
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Integration method for $\int_0^\infty\frac{x}{(e^x-1)(x^2+(2\pi)^2)^2}dx=\frac{1}{96} - \frac{3}{32\pi^2}.$ The following definite integral is obtained directly from Hermite's integral representation of the Hurwitz zeta function. But is it possible to obtain the same result via the residue calculus or another technique... | This integral may be done via the residue theorem, by considering the integral
$$\oint_C dz \frac{z \log{z}}{(e^z-1)(z^2+4 \pi^2)^2}$$
where $C$ is a keyhole contour about the positive real axis, having an outer circle of radius $R$ and inner circle of radius $\epsilon$, each centered about the origin. One may show th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/636896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{... | $\cos^4x=(\cos^2x)^2=(1-\sin^2x)^2=1+ \sin^4x-2\sin^2x$
Now,
$\dfrac{\sin^4x}{a} +\dfrac{cos^4x}{b}=\dfrac{1}{a+b}$
$\Rightarrow \dfrac{\sin^4x}{a} + \dfrac{(1+\sin^4x-2\sin^2x)}{b} = \dfrac{1}{a+b}$
$\dfrac{[b\cdot\sin^4x + a(\sin^4x-2\sin^2x+1)]}{ab} = \dfrac{1}{a+b}$
$\Rightarrow \dfrac{[(a+b)\sin^4x-2a \sin^2x+a]}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Show that $\forall p\in \mathbb{R}_3[X]$ we have $p(a) + p(b) + 4p(\frac{a + b}{2}) = \frac{6}{b - a}\int_{a}^{b} p(t)dt$ Let $a, b$ and $c$ three reals distinct.
*
*Let $P(X) = (X-a)(X-b)(X-c)$
determine a necessery and suffecient condition for $$\int_{a}^{b} p(t)dt = 0$$
*Show that $\forall p\in \mathbb{R}_3[X]$ ... | *
*Using 2. for this particular $p(X)=(X-a)(X-b)(X-c)$, as $p(a)=p(b)=0$, a necessary and sufficient condition for the integral to vanish would be
$$4p\left(\frac{a+b}2\right)=0$$
that is, $\displaystyle\frac{a+b}2$ would be also a root of $p$, but that means exactly $c=\displaystyle\frac{a+b}2$.
*
The matrix of $\Ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Hints on evaluating this $\frac{3}{2\pi}\int_0^{2\pi}\frac{e^{-ikx}}{5 - 4\cos(x)} \mathrm dx$? I have the following integral I'm trying to solve:
$$\frac{3}{2\pi}\int_0^{2\pi}\frac{e^{-ikx}}{5 - 4\cos(x)} \mathrm dx, \quad k \in \mathbb{Z}.$$
I've tried writing the exponential in terms of sines and cosines and then us... | For $k\ge 1$ put $z=e^{ix}$ to obtain
$$\frac{3}{2\pi} \int_{|z|=1} \frac{1/z^k}{5-2z-2/z} \frac{1}{iz} dz
= 3\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{5z-2z^2-2} dz
\\= 3\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{(1-2z)(z-2)} dz
= -\frac{3}{2}\times \frac{1}{2\pi i} \int_{|z|=1} \frac{1/z^k}{(z-1/2)(z-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/643040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding amplitude of oscillation So I know how to find the amplitude of oscillation. It's just the coefficient of the trig function. In general it would be $$ x = A\sin\left(\omega t +\phi_0\right) $$ and $A$ would be the amplitude. But my problem is asking for the amplitude of $$ x=-\cos(t)+3\sin\left(t-\frac{\pi}{6}... | Expanding the explanations given in the comments by Artem and André Nicolas to the question.
We can always find constants $A,B$ and $C$ such that the equation
\begin{equation*}
x(t)=A\cos \omega t+B\sin \omega t
\end{equation*}
is identical to the equation
\begin{equation*}
x(t)=C\sin \left( \omega t+\phi _{0}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/645693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be app... | Here is a different approach.
The sum of each pair of terms is
$$
\begin{align}
a_{2n-1}+a_{2n}
&=\frac{-(2n-1)^2+(2n-1)}{(2n-1)^3+1}+\frac{(2n)^2+2n}{(2n)^3+1}\\
&=\frac{4n^3-6n^2+5n-1}{n(4n^2-2n+1)(4n^2-6n+3)}\\
&=\frac1{4n^2}\frac{1-\frac3{2n}+\frac5{4n^2}-\frac1{4n^3}}{\left(1-\frac1{2n}+\frac1{4n^2}\right)\left(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/646610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Double integral -- tricky?
If $f(x,y) = x^2+y^2$ and $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\geq1, x^2+y^2-2x\leq0 \text{ and } y\geq0\}$, find $\displaystyle\int\displaystyle\int_D f$.
$D$ looks like the intersection between two circumferences, as I draw below:
Polar coordinates seems a obvious choice, but using them I'd... | There is no need to split this integral into pieces. The inequality $x^2 + y^2 \geq 1$ is the same as $r \geq 1$, and the inequality $x^2 + y^2 - 2x \leq 0$ is the same as $r^2 - 2 r \cos \theta \leq 0$ or just $r \leq 2\cos \theta$. The circles $r = 1$ and $r = 2 \cos \theta$ intersect at $\theta = \frac{\pi}{3}$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/648004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\lef... | If
$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)\ \mathrm{d}x,$$ then
\begin{align}
I'(a)&=\int_{0}^{\pi}\frac{2a-2\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x,\\
I'(a) & = \frac{1}{a}\int_{0}^{\pi}\frac{2a^2-2a\cos x}{1-2a\cos x+a^2} \ \mathrm{d}x \\
& = \frac{1}{a}\int_{0}^{\pi}\frac{1-1+a^2+a^2-2a\cos x}{1-2a\cos x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/650513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 5
} |
Equality of triangle inequality in complex numbers $z$ and $w$ be nonzero complex numbers. How do I show that $|z+w|=|z|+|w|$ if and only if $z=sw$ for some real positive number $s$.
I approached this by letting $z=a+ib$, and $w=c+id$, and kinda play around with it. I also tried to square both sides when proving forwa... | First try to prove it in one direction.
Let $z = a + bi$ and $w = sa + sbi$. Then we have:
$$\mid z + w \mid = \mid a + bi + sa + sbi \mid = \mid |a(1+s) + (1+s)bi\mid = \sqrt{(1+s)^2a^2 + (1+s)^2b^2} = (1+s)\sqrt{a^2 + b^2} = \sqrt{a^2 + b^2} + \sqrt{s^2a^2 + s^2b^2} = \mid z \mid + \mid w \mid$$
Now for the other dir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Partial Fraction Solution? This is the original fraction : $$\dfrac{8s^2}{(2s-3)^3}$$
From this I get :
$$4As^2 - 12As + 9A + 2Bs - 3B + C = 8s^2$$
On the solution it says $4A = 8, A=2$, okay I understand this, $4As^2 = 8s^2$, and the rest is cancelled in the equation since there is only one $s^2$, but how $B=12$ and ... | As alternative to expanding polynomial expression and equating coefficients as you have done, you can exploit the fact that two polynomials are equal iff they have the same values for all possible values of the variable. So starting with $8s^2 = A(2s-3)^2 + B(2s-3) + C$, we can choose values for $s$ and equate both sid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/653362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I calculate the maximum and minimum of this function?
This is incorrect and I have a feeling I am not doing this correctly.
| I understand your unease. The origin is a saddle point, as the Hessian can say:
$$H(0,0) = \begin{vmatrix} f_{xx} (0,0) & f_{xy} (0,0) \\ f_{xy} (0,0) & f_{yy} (0,0) \end{vmatrix} = \begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1.$$
As for nonzero points, I have continued from your system the following way: we can wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/655549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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is there a nicer way to $\int e^{2x} \sin x\, dx$? I have to solve this:
$\int e^{2x} \sin x\, dx$
I managed to do it like this:
let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$
$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$
If I substitute these values in... | we have
$$
\int e^{2x}\sin x \mathrm d x = (A\cos x + B\sin x)e^{2x}
$$
Equation is the same at both ends of the derivative $x$
$$
e^{2x}\sin x = 2e^{2x}(A\cos x +B \sin x) + (B\cos x -A\sin x)e^{2x}
$$
Finishing
$$
e^{2x}\sin x =（2B-A)\sin x\,e^{2x} +(2A+B)\cos x\,e^{2x}
$$
Compare coefficient，we have
$$
2A+B = 0\\
2B... | {
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Proof sought for a sum involving binomials that simplifies to 1/2 A proof of:
$$\begin{align*}(1/2)^{2m+1} \sum_{k=0}^{m} \binom{m}{k} \sum_{j=0}^{k} \binom{m+1}{j} = \frac{1}{2} \end{align*} $$
Conjecture based on the following Maple code:
Q := (1/2)^(2*m+1) * sum( binomial(m, k) * sum(binomial(m+1, j), j = 0 .. k), k... | Suppose we seek to verify that
$$\sum_{k=0}^m {m\choose k}
\sum_{j=0}^k {m+1\choose j} = 2^{2m}.$$
We use the integral
$${m+1\choose j} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z^{j+1}} \; dz.$$
This yields for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{m+1}}{z}
\sum_{j=0}^k \fr... | {
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How to maximize $\left({a+b \choose a} 2^{-a-b}\right)$? How can you maximize $\left({a+b \choose a} 2^{-a-b}\right)$ assuming, $a,b \geq 0$ and $0< (a+b) \leq n$, where all the variables are non-negative integers? Is the maximum when $a=b=n/2$, assuming $n$ is even?
| The expression can be viewed as the probaility of having exactly $a$ heads in $a+b$ fair coin tosses, so intuitively, this should be maximized if $a=b$. Moreover, the more coin tosses we make, the less likely it will become to obtain exactly (rounded) half of them heads.
Formally, with $f(a,b)={a+b\choose a}2^{-a-b}$ w... | {
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"source": "stackexchange",
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radius of a sphere containing a circle Consider the circle $C$ which is the intersection of the sphere $x^2+y^2+z^2-x-y-z=0$ and the plane $x+y+z=1$. The radius of the sphere with centre at the origin, containing the circle $C$ is
a. $1$
b. $2$
c. $3$
d. $4$
| When we complete the squares in the equation for the sphere, we find
$$ (x^2 - x + \frac{1}{4}) + (y^2 - y + \frac{1}{4}) + (z^2 - z + \frac{1}{4}) \ = \frac{3}{4} $$
$$ \Rightarrow \ (x - \frac{1}{2})^2 \ + \ (y - \frac{1}{2})^2 \ + \ (z - \frac{1}{2})^2 \ = \frac{3}{4} \ \ . $$
An immediate way to find the circle wo... | {
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Pre-Calculus: Fractions with Exponents If $x = \frac{2}{3}$ and $y = \frac{1}{9}$ find the value of $\dfrac{x^3y^2}{xy^5}$
I've tried working it out multiple ways, but my answer is wayyyy off. I'm not sure how to deal with this problem.
Regards,
| We could simply, but to avoid this we can also just do the math. Since $x=\frac{2}{3}$ and $y=\frac{1}{9}$, we have
$$
x^3=\left(\frac{2}{3}\right)^3=\frac{8}{27}\text{ and }y^2=\left(\frac{1}{9}\right)^2=\frac{1}{81} \text{ and }y^5=\left(\frac{1}{9}\right)^5=\frac{1}{59049}
$$
then we have
$$
\frac{x^3y^2}{xy^5}=\fra... | {
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Find a non-diagonizable matrix that commutes with $\begin{bmatrix}0 & 0& -1\\1 & 1 &1\\0 & 0 & 1\end{bmatrix}$ I'm really lost on this one, with no clue on where to begin. Any help would be appreciated.
| Not a general approach, but one solution is easy to see. You have for example
$$\begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}\cdot
\begin{pmatrix}0&0&-1\\1&1&1\\0&0&1\end{pmatrix} = \begin{pmatrix} 0&0&0 \\ 0&0&1\\0&0&0\end{pmatrix}=\begin{pmatrix}0&0&-1\\1&1&1\\0&0&1\end{pmatrix}\cdot\begin{pmatrix} 0&0&0 \\ 0&0&... | {
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"source": "stackexchange",
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$\lim\limits_{(x,y)\rightarrow(0,0)} \frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}$ Does the following limit exist?
$$\lim_{(x,y)\rightarrow(0,0)} \frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}$$
I've shown that it exist but i'm unsure about how to find the value of the limit.
| Note that
$$\frac{7 \sin(2 x) x^2 y}{2 x^3 + 2 x y^2}=7\cdot\frac{\sin(2x)}{2x}\cdot\frac{x^2y}{x^2+y^2}.$$
When $(x,y)\to (0,0)$, we have $x\to 0$ and $r=\sqrt{x^2+y^2}\to 0$, therefore, we have
$$\left|\frac{\sin(2x)}{2x}\right|\to 1\mbox{ as }x\to 0$$
and
$$\left|\frac{x^2y}{x^2+y^2}\right|=\left|\frac{r^3\cos^2\t... | {
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how can I show that the sequence $x_{n}=\frac{n}{n^{2}+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n},n=1,2,...$ converges? How can I check if the sequence $x_{n}=\frac{n}{n^{2}+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n},n=1,2,...$ converges?Is there a theorem that I could use?
| Notice the following:
Every two integers $n,k$ such that $1\leq k \leq n$ satisfy
$$\frac {1}{n+1} = \frac{n}{n^2+n} \leq \frac{n}{n^2+k} \leq \frac{n}{n^2}=\frac{1}{n}$$
So we get
$$ \frac{1}{1+\frac{1}{n}} = n \cdot \frac {1}{n+1} \leq x_n \leq n \cdot \frac{1}{n} =1$$
Thus by the Sandwich Theorem we get $x_n\righ... | {
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How to resolve this integration $\int\frac{dx}{1+x^2+\sin^2x}$? I have tried Trigonometric Substitution, but I can´t get an already known function to be easy for integrate:
$$\int\frac{dx}{1+x^2+\sin^2x}$$
I entered this on Wolfram and it gave me the same function. I'm not asking for the exact solution, just a good way... | Hint:
$\because0\leq\dfrac{\sin^2x}{x^2+1}<1$ $\forall x\in\mathbb{R}$
$\therefore\int\dfrac{dx}{1+x^2+\sin^2x}$
$=\int\dfrac{dx}{(x^2+1)\left(1+\dfrac{\sin^2x}{x^2+1}\right)}$
$=\int\dfrac{1}{x^2+1}\sum\limits_{m=0}^\infty\dfrac{(-1)^m\sin^{2m}x}{(x^2+1)^m}dx$
$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfra... | {
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| By C-S
$$\frac{b}{1+b}+\frac{c}{c+1}=\frac{b^2}{b^2+b}+\frac{c^2}{c^2+c}\geq\frac{(b+c)^2}{b^2+c^2+b+c}\geq\frac{(b+c)^2}{(b+c)^2+b+c}=$$
$$=\frac{b+c}{b+c+1}=1-\frac{1}{b+c+1}>1-\frac{1}{a+1}=\frac{a}{a+1}.$$
Done!
| {
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A coordinate geometry question. On pg. 46 of "Coordinate geometry" by S.L. Loney, the following question has been posed:
Show that the equations to the straight lines passing through the point $(3,-2)$ and inclined at $60^\circ$ to the line $\sqrt{3}x+y=1$ are $y+2=0$ and $y-\sqrt{3}x+2+3\sqrt{3}=0$.
I am getting di... | Let $\theta$ be the angle between two lines whose slopes are $m_1,m_2$. Then we have:
$$\tan \theta = |\frac{m_1-m_2}{1+m_1m_2}|$$
As we know $m_1$, we can just plug in the value and get $m_2$:
$$\sqrt{3} = |\frac{-\sqrt{3}-m_2}{1-\sqrt{3}{m_2}}|$$
$$\sqrt{3}-3m_2 = -\sqrt{3}-m_2 \implies m_2 = \sqrt{3}$$ or
$$-\sqrt{... | {
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How to evaluate the integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$ Evaluating this integral: $I=\int_0^3\frac{x\sqrt{x+1}dx}{x^2+x+1}$
I've tried:
Set : $\sqrt{x+1}=t\mapsto dx=2t\,dt \Longrightarrow I=2\int_1^2\frac{t^4-t^2}{t^4-t^2+1}dt=2\int_1^2(1-\frac{1}{t^4-t^2+1})dt$
And then, I set $(t^2-\frac{1}{2})=\frac... | HINT:
As $\displaystyle t^4-t^2+1=(t^2+1)^2-3t^2=(t^2+\sqrt3t+1)(t^2-\sqrt3t+1)$
$$\frac1{t^4-t^2+1}=\frac1{2\sqrt3t}\left(\frac1{t^2-\sqrt3t+1}-\frac1{t^2+\sqrt3t+1}\right)$$
Using Trigonometric substitution ,
$$t^2-\sqrt3t+1=\frac{4t^2-4\sqrt3t+4}4=\frac{(2t-\sqrt3)^2+1}4$$
So set, $2t-\sqrt3=\tan\phi$ etc.
| {
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Proving $\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)= -\frac{\log^33}{6}+\frac{\pi^2}{6}\log 3-\frac{13\zeta(3)}{6}$? Ramanujan gave the following identities for the Dilogarithm function:
$$
\begin{align*}
\operatorname{Li}_2\left(\frac{1}{3}\right)-\frac{1}{6}\operatorname{Li}_2\left(\fr... | By "anywhere in literature" did you include L. Lewin, Polylogarithms and Assoicated Functions ?
In the (fortcoming) solution for Monthly problem 11654, you will see that very equation. For the proof, our solver (Richard Stong) used these identities:
$$
\mathrm{Li}_3\left(\frac{1-z}{1+z}\right) - \mathrm{Li}_3\left(\fr... | {
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Prove that for every integer greater than 1. $\dbinom{n}{1}-2\dbinom{n}{2}+3\dbinom{n}{3}+.....+(-1)^{n-1}n\dbinom{n}{n}=0$ How do you show the following.
Prove that for every integer greater than 1.
$\dbinom{n}{1}-2\dbinom{n}{2}+3\dbinom{n}{3}+.....+(-1)^{n-1}n\dbinom{n}{n}=0$
My idea is that is that the whole sequenc... | Hint: Expand $(1-x)^n$ using the Binomial Theorem and differentiate term by term.
Alternately, note that $k\binom{n}{k}=n\binom{n-1}{k-1}$.
Details: For the first hint, let $n\gt 1$, and let $f(x)=(1-x)^n$. Then $f'(x)=n(1-x)^{n-1}$ and therefore $f'(1)=0$.
We now compute $f'(x)$ another way, by first expanding $f(x... | {
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Let $\theta=\frac{2\pi}{5}$. Show $2\cos(2\theta)+2\cos(\theta)+1=0$. I have been at this for a while. Any ideas?
| If $\displaystyle5\theta=2\pi, 3\theta=2\pi-2\theta$
$\displaystyle\implies\cos3\theta=\cos(2\pi-2\theta)=\cos2\theta$
Using Triple and Double angle formulas,
$$4c^3-3c=2c^2-1\iff 4c^3-2c^2-3c+1=0\ \ \ \ (1)$$
where $ c=\cos\theta$
Again, $\displaystyle\cos3\theta=\cos2\theta$
$\displaystyle\implies3\theta=2n\pi\pm2\t... | {
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"answer_id": 7
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Is there an easy way to prove that $3a^2+2=b^2$ does not have any rational solutions? I am a physicist who needs to prove (for his research) that there are no $a,b\in\mathbb{Q}$ such that $3a^2+2=b^2$. Is there an easy way to do this?
| If there is a rational solution $(a,b)$, it can be expressed as $a=\frac{A}{z}$, $b=\frac{B}{z}$ where $A$ and $B$ are integers, and $z$ is a non-zero integer. Clearing denominators, we get
$$3A^2+2z^2=B^2.$$
If the above equation has solutions with $B\ne 0$, there is a triple of integers, not all $0$, such that $B$ i... | {
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2014 AMC 12 B problem 25 What is the sum of all positive real solutions $x$ to the following equation? $$2\cos(2x)\left( \cos(2x) - \cos{\left(\frac{2014\pi^2}{x^2}\right)} \right) = \cos(4x) - 1 $$
| Using $\cos2A=2\cos^2A-1,$
$$2\cos^22x-2\cos2x\cos\left(\frac{2014\pi^2}{x^2}\right)=2\cos^22x-1-1$$
$$\implies\cos2x\cos\left(\frac{2014\pi^2}{x^2}\right)=1$$
As for real $y,-1\le\cos y\le1;$ We need
$$\cos2x=\cos\left(\frac{2014\pi^2}{x^2}\right)=\pm1$$
Taking the '+' sign,
$\displaystyle\cos2x=1\implies 2x=2n\pi\i... | {
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Minimum value of the function $\sqrt{(1+1/m)(1+1/n)}$ If $m, n$ are positive real variables whose sum is a constant $k$, then what is the minimum value of $$\sqrt{\bigg(1 + \frac{1}{m}\bigg)\bigg(1 + \frac{1}{n}\bigg)}$$
| $\sqrt{1+\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{mn}}=\sqrt{1+\dfrac{m+n}{mn}+\dfrac{1}{mn}}=\sqrt{1+\dfrac{k+1}{mn}}\ge\sqrt{1+\dfrac{k+1}{\bigg(\dfrac{m+n}{2}\bigg)^2}}$
$1+\dfrac{4(k+1)}{k^2}=\dfrac{k^2+4k+4}{k^2}=\bigg(\dfrac{k+2}{k}\bigg)^2$
So the least value of your desired expression is $\dfrac{k+2}{k}$
| {
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Checking multivariable function's differentiability at $(0,0)$ I have a function: $$f(x,y) = \begin{cases} (x^2+y^2)\sin (x^2+y^2)^{-1} \ \ \ &\text{, if } x^2+y^2 \neq 0 \\ 0 &\text{, if } x^2+y^2=0 \end{cases}$$
I calculated the partial derivatives:
$$f_x(x,y)=2x\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2... | Hint: You can use Taylor series as
$$ \sin(h_1^2+h_2^2) = (h_1^2+h_2^2)-\frac{(h_1^2+h_2^2)^3}{3!}+\dots\,. $$
| {
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"question_score": "2",
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How to show whether a $3\times 4$ matrix has no solution, a unique solution or infinitely many solutions? The system is :
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
After Gaussian elimination, I found that
$$
\begin{array}{cccc|cc}
1 & -4 & 6 & a & ... | Constraints on $a$
In general, we have existence of a solution when the data vector is in the column space of the target matrix $\mathbf{A}$. That is, if the data vector can be written as a linear combination of the fundamental columns of $\mathbf{A}$.
Additionally, if the nullspace $\mathcal{N} \left( \mathbf{A}^{*} \... | {
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Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt... | You have a very elegant demonstration given by Sami Ben Romdhane.
If, instead of summing to $\infty$, you sum to $m$, you will get the following formulas for the different sums
$$\sqrt{1}-\frac{1}{\sqrt{m+1}}$$
$$\sqrt{1}+\sqrt{2}+H_m^{\left(\frac{1}{2}\right)}+H_{m+1}^{\left(\frac{1}{2}\right)}-2
H_{m+2}^{\left(\... | {
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Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false... | You have too many terms, $x^{n-n}y^{n-1}$ is the last, which in this case is your second term $x^{n-2}y$. The third term would in your case be $x^{n-3}y^{n-2}$ which we don't want since $n=2$ and the exponent of $x$ would become negative.
To more easily see this, write the factorization in this form:
$$x^n - y^n = (x-y... | {
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How to find a matrix $C$ such that $C^{-1}AC$ is in Jordan block form. $A:=\begin{bmatrix}
6 & -1\\
4 & 2
\end{bmatrix}$
Now, just to show I've done some working, at least to find $A$'s eigenvalues and deduced that it's not diagonalisable:
Any hints/advice?
Thanks
| For a $2 \times 2$ (say complex) matrix $A$ which is not diagonalizable, there is a method to determine the conjugating matrix $C$ such that $C^{-1}AC$ is in Jordan form. The only way that $A$ can fail to be diagonalizable is if the characteristic polynomial of $A$ has a repeated root $s,$ but $A \neq sI.$ Then the onl... | {
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How to find multiplicative orders of all elements in field $\Bbb F$ (say $\Bbb F_{13}$)? I am working on some finite fields and I was referring to some online class material. Is there any way to find the multiplicative orders of all elements in a field $\Bbb F$?
| Just compute the powers:
\begin{align}
2^2&\equiv_{13}4 &2^3&\equiv_{13}8 &
2^4=4^2&\equiv_{13}16\equiv_{13}3&
2^5&\equiv_{13}6&2^6&\equiv_{13}12\equiv_{13}-1\\
3^2&\equiv_{13}9\equiv_{13}-4 &3^3&\equiv_{13}-12\equiv_{13}1\\
5^2&\equiv_{13}-1\\
6^2&\equiv_{13}-3&6^3&\equiv_{13}-5&6^4&\equiv_{13}-4&6^5&\equiv_{13}2&6^6... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration of $1/(1+\sin x)$ I solved it using $t=\tan(\frac{x}{2})$ substitution and got $-2/(1+\tan(x/2))+C$, but in my math book solution is $\tan(x/2-\pi/4)+C$. Are those the same expressions and if they are, how do I transform from one to another, or are one(or both) solutions incorrect ?
| $\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{1+\sin x}dx = \int\frac{1}{1+\cos \left(\frac{\pi}{2}-x\right)}dx = \int\frac{1}{2\cos^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)}dx$
So $\displaystyle I = \frac{1}{2}\int \sec^2 \left(\frac{\pi}{4}-\frac{x}{2}\right)dx = -\frac{1}{2}\tan \left(\frac{\pi}{4}-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/704163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Integral $\int_0^a \ln \left( \frac{b-\sqrt{a^2-x^2}}{b+\sqrt{a^2-x^2}} \right)dx$ Hi I am trying to calculate,
$$
\int_0^a \ln \left( \frac{b-\sqrt{a^2-x^2}}{b+\sqrt{a^2-x^2}} \right)dx
$$
where $a,b$ are positive real constants. I Know $\ln(xy)=\ln x +\ln y$, but I do not know how to evaluate this integral then.
... | This is a possible way (I assume $a\geq 0$, $b\geq 0$):
1) substitute $x= a \sin t $.
2) integrate the resulting integral by parts to get rid of $\ln$
3) you should end up with $$-\int_0^{\pi/2}\frac{2 a^2 b \sin^2 t\,dt }{b^2 -a^2 \cos^2 t} .$$
4) it is possible to integrate the last integral by elementary means or u... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| it is easily shown that as $z$ approaches zero
$$
\log(1+z) = z + O(z^2)
$$
so we may write
$$
\log \sqrt{(x+a)(x+b)} =\log x + \log \sqrt{(1+\frac{a}x)(1+\frac{b}x)} \\= \log x + \frac{a+b}{2x} + O(x^{-2})
$$
taking exponentials
$$
\sqrt{(x+a)(x+b)} = x\bigg(1+O(x^{-2})\bigg) \exp\frac{a+b}{2x} \\
= (x+O(x^{-1})\big... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
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"Identity" of volume of a solid of revolution (Limited Weierstrass function) The question is: Is this correct?
This is pretty much the first thing I've tried to come up with by myself. I wanted to see what would happen if I tried to calculate volume of a solid of revolution of the Weierstrass-like function. I'm pretty ... |
please do not up- or downvote this answer. It's just impossible to do this by pure commenting. I copy pasted your calculus and added some remarks.
$V_{w(x)_n} =\pi\displaystyle\int_{\alpha}^{\beta} (w(x)_n)^2dx = \pi\displaystyle\int_{\alpha}^{\beta} (\displaystyle\sum_{k= 0}^{n} a^k\cos(b^k\pi x) )^2dx =$
$\pi\di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$ The question statement from my homework booklet goes:
Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$
is true for all integers $n \geq 2$.
I've managed to come up with this for the induction step (the base case is trivial), but I am not sure what to do ... | For $k$ we have
$\large k^{k+1}>k(k+1)^{k-1}$
which we hold to be true.
For $k+1$ we have
$\large(k+1)^{k+2}=(k+1)^{k+1}(k+1)=k^{k+1}(1+\frac{1}{k})^{k+1}(k+1)
\\\large>k(1+k)^k(1+\frac{1}{k})^{k+1}$
as $k^{k+1}>k(k+1)^{k-1}$
Now
$\large k(k+1)^k(1+\frac{1}{k})^{k+1}=k(k+1)^k(1+\frac{1}{k})^{k+1}(\frac{k+2}{k+2})^k
\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Need help solving this equation with complex numbers $$(z^3 + 1)^3 = 1$$
where $z$ is an element of the complex number system.
Can someone show me the most efficient way of finding all the solutions for $z$ here and also if possible please demonstrate how both sides of the equation can be converted into polar coordinat... | As $\displaystyle(z^3+1)^3=1=\cos0+i\sin 0,$
using this, $\displaystyle z^3+1=\cos\dfrac{(2k\pi+0)}3+i\sin\dfrac{(2k\pi+0)}3$ where $k=0,1,2$
$\displaystyle\implies z^3=\cos\dfrac{(2k\pi+0)}3+i\sin\dfrac{(2k\pi+0)}3-1$
Using Double-Angle Formulas,
$\displaystyle z^3=2i\sin\frac{k\pi}3\cos\frac{k\pi}3-2\sin^2\frac{k\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/713728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove: $99^{100}+100^{101}+101^{99}+1$ is a prime number
Prove or disprove: $$99^{100}+100^{101}+101^{99}+1$$ is a prime number.
My idea: let $100^{101}=x^{x+1}$,then
$$99^{100}+100^{101}+101^{99}+1=(x-1)^{x}+x^{x+1}+(x+1)^{x-1}+1$$ is prime number？
I have solved following problem before:
Show that $$5^{... | Similar Mathematica code is FactorInteger[99^100 + 100^101 + 101^99 + 1, 2] which returns $825277$ and $\frac{99^{100} + 100^{101} + 101^{99} + 1}{825277}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/715561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
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If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$ If $x,y,z \in \mathbb{R^+}$ such that $x+y+z=3$. Prove the inequality $\sqrt x+\sqrt y+\sqrt z\ge xy+yz+zx$.
My work: We have
$$3(x+y+z)=x^2+y^2+z^2+2(xy+yz+zx) \implies (xy+yz+zx)=\dfrac12(3x-x^2+3y-y^2+3z-z^... | Hint: Now, one way to proceed: show $x^2-3x+2\sqrt x>0$. Let $x=w^2$ and factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/715823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to find continued fraction of pi I have always been amazed by the continued fractions for $\pi$. For example some continued fractions for pi are:
$\pi=[3:7,15,1,292,.....]$ and many others given here.
Similarly some nice continued fractions for $e$ and it's derivatives are given here.I have tried to prove that the ... | A neat method to construct a continued fraction for $\pi$ is to use the addition formula for $\arctan$:
$$\arctan(x) + \arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which can also be written
$$\arctan\left(\frac{1}{x}\right) + \arctan\left(\frac{1}{y}\right) = -\arctan\left(\frac{x+y}{1-xy}\right) = \arctan\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$ Ok, so I know that since the numerator has a higher power that long division is needed. So after doing that, the main fraction is $\frac{-6x-36}{x^2 + 12x + 36}$. I think that's right. But my problem is that after you factor the denominator they're both equal to $(x+6)$. ... | After the observation that the denominator is equal to $(x+6)^2$ (which was mentioned in other answer), this problem can be solved relatively easy using the substitution $y=x+6$.
$$
\begin{align*}
\frac{x^3}{(x+6)^2}
&=\frac{(y-6)^3}{y^2}=\\
&=\frac{y^3-3\cdot 6 y^2+3\cdot 6^2 y - 6^3}{y^2}=\\
&=\frac{y^3-18y^2+108y-21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/722890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Resolving a Limit I'm studying for college exams and I don't know how to solve this type of limit:
$$\begin{align}
\lim_{x\to -3} \sqrt{\frac{x^2-9}{2x^2+7x+3}}
\end{align}$$
Any help?
Update: I know that the solution is:
$$\begin{align}
\frac{1}{5} \sqrt{30}
\end{align}$$
| We have by the L'Hôpital theorem:
$$\begin{align}
\lim_{x\to -3} \frac{x^2-9}{2x^2+7x+3}=\lim_{x\to -3}\frac{2x}{4x+7}=\frac{6}{5}\end{align}$$
Notice that
$$\lim_{x\to -3}{\sqrt{\frac{x^2-9}{2x^2+7x+3}}}=\sqrt{\lim_{x\to -3}\frac{x^2-9}{2x^2+7x+3}}\quad\text{Why?}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/723209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find complicated Taylor Series According to some software, the power series of the expression,
$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$
around $x=0$ is
$$\sqrt{x}-x^{3/2}+\mathcal{O}(x^{5/2}).$$
When I try to do it I find that I can't calculate Taylor because there are divisions by zero. Also I do not understand how Tayl... |
$$f(x) =\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$
The argument of the square root is 0, hence $f$ has no derivative in 0. That is why you get non integer powers.
First, get the equivalent:
$$\sqrt{1+8 x}-1 \sim 4x\\
f(x) = \frac 12\sqrt{-1+\sqrt{1+8 x}}\sim \frac 12\sqrt{4x} = \sqrt{x}
$$
and then:
$$
f(x) -\sqrt{x}
= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723296",
"timestamp": "2023-03-29T00:00:00",
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Help with surface integral question Find the surface area of the plane
$$x+\frac{1}{\sqrt{2}}y+\frac 14 z=1$$
limited by the coordinate system planes
My findings :
I suppose we should express the scalar $z=f(x,y) \rightarrow z=4-4x-2\sqrt{2}y$ and then $\mathbf{r}(s,t)$ is
$$\mathbf{r}(s,t)=s\mathbf{i}+t\mathbf{j}+(4-... | The area of a surface is $$\iint\limits_{\Sigma} dS,$$ where $\Sigma$ is the surface and $dS$ is the surface area element. It is computed by finding the length of the normal vector to the surface. In your case we have the graph of a function, therefore
$$dS = \sqrt{ 1 + \left( \frac{\partial f}{\partial s} \right)^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/723386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to prove the formula of altitude from this following triangle? Given: Right triangle $\triangle ABC$ with $A$ as right angle.
If $t_A$ is altitude that drawn from point $A$ to $\overline{BC}$, called $\overline{AD}$.
Prove that $t_A = \sqrt{2}\cdot\dfrac{bc}{b+c}$
Thanks
| By the picture that you drew, we have $\frac{AD}{c}=\frac{b}{\sqrt{b^2+c^2}}$, which gives $AD=\frac{bc}{\sqrt{b^2+c^2}}$.
This is hardly ever equal to $\sqrt{2}\frac{bc}{b+c}$. For if we had equality, we would have $\frac{\sqrt{2}}{b+c}=\frac{1}{\sqrt{b^2+c^2}}$, or equivalently $2(b^2+c^2)=(b+c)^2$. This simplifies ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integers that can be expressed as $a^3+b^3+c^3-3abc$ $$S=\{a^3+b^3+c^3-3abc|a,b,c\in\Bbb Z\}$$
Can we decide $S$? that is, we want to find all integers of the form $a^3+b^3+c^3-3abc$.
obviously,
*
*if $m,n\in S$, then $mn\in S$, so we only need to consider primes;
*if $n\in S$, then $-n\in S$.
Let $f(a,b,c)=a^3... | Note that $$(a\pm1)^3+a^3+a^3-3(a\pm1)a^2=3a^3\pm3a^2+3a\pm1-3a^3\mp3a^2=3a\pm1$$
and
$$(a+1)^3+a^3+(a-1)^3-3(a+1)a(a-1)=3a^3+6a-3a^3+3a=9a$$
Suppose $n=a^3+b^3+c^3-3abc$ where $n\in\mathbb Z$.
So we can get all $n$ such that $3\nmid n$ or $9\mid n$.
Note that $(3p+k)^3\equiv k^3\equiv k\ \ (\text{mod 9})$ for $k\in\{-... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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Minimum area of a triangle In triangle inscribed circle with radius $r = 1$ and one of it sides $a=3$. Find the minimum area of triangle? Ans = 5.4
My reasonings:
$BC = a$, $AC = b$, $AB = c$
$AD=AF=x$
$FC=CE=y$
$BD=BE=z$
$a=z+y$, $b=x+y$, $c=x+z$
The radius of the incircle is $$r =\frac{A_{ABC}}{s}$$ where $s = \frac... | Trigonometric approach:
In your notation,
$AD=AF=x$, $FC=CE=y$, $BD=BE=z$,
denote in addition
$\alpha=\angle OAF=\angle OAD$,
$~~~\beta=\angle OCF=\angle OCE$,
$~~~\gamma=\angle OBE=\angle OBD$.
So, if $r=1$, then
$x=\dfrac{1}{\tan\alpha}$,
$~~~y=\dfrac{1}{\tan\beta}$,
$~~~z=\dfrac{1}{\tan\gamma}$.
Then, as you said... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$ In solving the definite integral $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$$I tried to do $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx=\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx\Longrightarrow$$ If $x=\frac{2}{\sqrt{3}}\sec(u)\Longrightarrow dx=\frac2{\sqrt3... | If $\tan t=\frac{\sqrt{3x^2-2}}{2}$ and $\cos t=\frac{2}{\sqrt{3}x} $, then $x=\frac{2}{\sqrt{3}}\sec t \Rightarrow dx=\frac{2}{\sqrt{3}}\sec t \tan t dt$ and so
$$\int \frac{x-1}{x\sqrt{3x^2-4}}dx=\int \frac{(\frac{2}{\sqrt{3}}\sec t-1)}{\frac{2}{\sqrt{3}}\sec t\cdot 2\tan t} \cdot \frac{2}{\sqrt{3}}\sec t\tan t dt=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/725168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help finishing proof via induction for a summation So I have to prove the following equation using induction for n >= 2:
$$
\sum\limits_{i=1}^n 4/5^i < 1
$$
However the question asks me to prove something stronger such as this:
$$
\sum\limits_{i=1}^n 4/5^i <= 1 - \frac{1}{5^n}
$$
first to imply the first equation is t... | $$\sum_{i=1}^{3}1/5^i=(1/5+1/5^2+1/5^3)*(1-1/5)/(1-1/5)=(1-1/5^3)/4 $$
By induction it is easy to see that
$$\sum_{i=1}^{n}1/5^i = (1-1/5^n)/4 $$
And so
$$\sum_{i=1}^{n}4/5^i = (1-1/5^n) $$
Which is strictly less than one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the largest circle that fits in $\sin(x)?$ Imagine dropping a circle into the trough of $\sin(x)$. Would it reach the bottom or get wedged between two points on the curve? Depends on the size of the circle.
So, what is the radius of the largest circle that will reach the bottom of the curve $y=\sin(x)$?
This p... | Nice question.
If we are dropping the circle so that its center is right above $\frac{3\pi}{2}$ then the coordinates of the center of the dropped circle will be $(x_0,y_0) = (3\pi/2,r-1)$ where $r$ is the radius of the circle. Now, the distance from this point to any point $(x,y)$ on the curve of $\sin$ is
$$
\sqrt{(... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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show that $a^3+4a=b^2$ then exist $M\in\mathbb{N}$ s.t $a=2M^2$ I would appreciate if somebody could help me with the following problem:
Q: show that
If $a,b$ satisfy equation
$$a^3+4a=b^2(a,b\in\mathbb{N})$$ then there exist $M\in\mathbb{N}$ s.t $a=2M^2$
| Let $d = \gcd(a,b)$. So $a = xd$, and $b = yd$. Thus: $x^3d^3 + 4xd = y^2d^2$. So $d^2x^3 + 4x = dy^2$. This gives: $x$ divides $dy^2$. But $\gcd(x,y) = 1$ so $\gcd(x,y^2) = 1$. Thus $x$ divides $d$. So $d = kx$. So: $k^2x^5 + 4x = kxy^2$. Thus: $k^2x^4 + 4 = ky^2$. Hence: $4 = k(y^2 - kx^4)$. This means $k$ divides $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/727424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A question on multinomial theorem using binomial theorem $(3x^2+2x+c)^{12}=\sum A_r x^r$
and
$\frac{A_{19}}{A_5}=\frac 1 {2^7}$
Find $c$. I really have no idea what to do with this. This was on a test. I have studied only binomial theorem. So, please use it only(combinatorics approach is always welcome).
| By comparing coefficients
$$
A_{19} = \frac{12!}{9! \cdot 1! \cdot 2!} \cdot 3^9 \cdot 2 \cdot c^2 + \frac{12!}{8! \cdot 3! \cdot 1!} 3^8 \cdot 2^3 \cdot c + \frac{12!}{7! \cdot 5! \cdot 0!} 3^7 \cdot 2^5 = 3^7\left(\frac{12!}{9! \cdot 1! \cdot 2!}18c^2 + \frac{12!}{8! \cdot 3! \cdot 1!}24c + \frac{12!}{7! \cdot 5! \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/731335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$
Suppose $P(x)$ is a polynomial of degree $n \geq 1$ such that $\int_{0}^{1}x^{k}P(x)\,dx = 0$ for $k = 1, 2, \ldots, n$. Show that $$\int_{0}^{1}\{P(x)\}^{2}\,dx = (n + 1)^{2}\left(\int_{0}^{1}P(x)\,dx\right)^{2}$$
If we assum... | Since, $\displaystyle \int_0^1 x^kP(x)\,dx=\frac{a_0}{n+k+1}+\frac{a_1}{n+k}+\ldots+\frac{a_n}{k+1}=0$, for each $k=1,2,\ldots,n$
Then $f(x)=\dfrac{a_0}{n+x+1}+\dfrac{a_1}{n+x}+\ldots+\dfrac{a_n}{x+1}=\dfrac{Q(x)}{(n+x+1)\ldots(x+1)}$ (where, Q is a polynomial of degree atmost n), has $n$ zeros $x=1,2,\ldots,n$.
Thus, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/732132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$
| You could factorise it, in the manner of $(x-2)^4=0$. I saw those factors immediately.
One process is to note that $16$ has divisors, and one can try various combinations of this such that the sum gives eight.
Possibilities include $2, 2, 2, 2$ and $4, 4, 1, -1$. However, one can not produce the second set to give +1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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The derivative of $2^{x+1}$ in the point $-3$ They ask me to compute the derivative of $2^{x+1}$ in the point $-3$. Since the function is continous in that point, all I have to do is to compute the lateral derivatives. The left one first:
$$\lim_{x \to -3^-} \frac{2^{x+1} - \frac{1}{4}}{x+3}$$
I've tried to solve this ... | $$\lim_{x \to -3^-} \frac{2^{x+1} - \frac{1}{4}}{x+3} = \lim_{x \to -3^-} \frac{2^{x+1}\frac{2^2}{2^2} - \frac{1}{4}}{x+3} = \lim_{x \to -3^-} \frac{\frac{2^{x+3}}{4} - \frac{1}{4}}{x+3} =$$
$$= \frac{1}{4}\lim_{x \to -3^-} \frac{2^{x+3}- 1}{x+3} = \frac{1}{4}\ln{2}$$
| {
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"url": "https://math.stackexchange.com/questions/738128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to integrate $\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx$?
How to integrate $\displaystyle\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx$ ?
If $\theta\in(0,\frac{1}{2})$
$$
\begin{align}
& \int_0^{\theta}\frac{x}{(1-x^2)(1-x)}\,dx \\[8pt]
= & \int_0^{\theta}\frac{x}{(1-x)(1+x)(1-x)}\,dx \\[8pt]
= & \int_0^{\theta}\frac{... | How about partial fraction decomposition?
$$\begin{align} \displaystyle\int_0^{\theta}\frac{x}{(1-x^2)(1-x)}dx & = \int_0^\theta \frac x{(1-x)(1+x)(1 - x)} \,dx \\ \\ & = \int_0^\theta \frac x{(1+x)(1-x)^2}dx \\ \\ & = \int_0^\theta \left[ \frac A{1+x} + \frac B{1-x} + \frac C{(1-x)^2}\right]dx\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/738424",
"timestamp": "2023-03-29T00:00:00",
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coefficient of $x^2$, in $(1+x+x^2)^{10}$
How to find coefficient of $x^2$, in $(1+x+x^2)^{10}$, without actually expanding it?
I think the fact $\dfrac{1-x^3}{1-x}=1+x+x^2$ may help.
But can't use it!
| Method $\#1:$
Following your process, $$(1+x+x^2)^{10}=\left(\frac{1-x^3}{1-x}\right)^{10} =(1-x^3)^{10}(1-x)^{-10}$$
$$=\left[1-\binom{10}1x^3+\binom{10}2x^6-\cdots\right]\left[1-10(-x)+\frac{-10(-10-1)}2(-x)^2+\cdots\right]$$
Clearly the coefficient of $x^2$ is $\displaystyle \frac{-10(-10-1)}2=55$
Method $\#2:$
For ... | {
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"timestamp": "2023-03-29T00:00:00",
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)... | Notice that for all real $a,b,c$,
$$(a - 1)^2 + (b-1)^2 + (c - 1)^2 \ge 0.$$
$$a^2 + b^2 + c^2 - 2a - 2b - 2c + 3 \ge 0.$$
$$a^2 + b^2 + c^2 \ge -3 + (a + b + c) + (a + b + c).$$
But by AM-GM, $a + b + c \ge 3\sqrt[3]{abc} = 3$. So,
$$a^2 + b^2 + c^2 \ge -3 + 3 + (a + b + c).$$
$$a^2 + b^2 + c^2 \ge a + b + c.$$
Equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Find domains of convergence of the series I tried to solve this question but the final solution which I obtain is not the as same as in the text book
Find domains of convergence of the series
$$\sum_{n=1}^ \infty \frac{1.3.5...(2n-1)}{n!}(\frac{1-z}{z})^n$$
The book answer is $|z-\frac{4}{5}|\lt \frac{2}{3}$
But my an... | @DonAntonio Your solution is correct :
$$\left|\frac{1-z}z\right|<\frac12\iff|2-2z|<|z|$$
Now put $\;z=x+iy\;,\;\;x,y\in\Bbb R\implies 2-2z=(2-2x)-2yi\;$ , so::
$$|2-2z|^2<|z|^2\iff(2-2x)^2+4y^2<x^2+y^2\iff$$
$$4-8x+4x^2+4y^2<x^2+y^2\iff3x^2-8x+3y^2+4<0\iff$$
$$3\left(x-\frac43\right)^2-\frac{16}3+3y^2+4<0\iff 3\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Use the integration formula $\frac{1}{a}\arctan\frac{x}{a}$ to solve $\frac{1}{2} \int_{-1}^1 \mathrm{ \frac{dx}{1+\sqrt{2}x+x^2} }\, $ As question states, I am trying to figure out how to use the integration formula to solve the integral. My issue is that the integral isn't of the form $\frac{dx}{a^2+x^2}$
| If I amy interfer in this discussion, you have been shown by various answers and comments that through the appropriate change of variable dictated by completing the square
$$\frac{1}{2} \int_{} \frac{dx}{1+\sqrt{2}x+x^2} =\frac{\tan ^{-1}\left(x\sqrt{2} +1\right)}{\sqrt{2}}$$ So, putting the bounds $$\frac{1}{2} \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
solve the equation in Z Solve the equation over $\textbf{Z}$ :
2$x^2$ - 2$xy$ - 5$x$ - $y$ + 19 = 0
I tried to obtain some $(A+B)^2$ terms, but I didn't make it.
Thanks for your time!
| $2x^2 - 2xy - 5x - y + 19 = 0 \to y = \dfrac{2x^2 - 5x + 19}{2x + 1} = x - 3 + \dfrac{22}{2x + 1}$. So $2x + 1 | 22$, and $2x + 1 = 1, -1 , 11, -11$. we can then find $y$ for each value of $x$ listed.
| {
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"url": "https://math.stackexchange.com/questions/743126",
"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.