Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the identity under a given binary operation I have two problems quite similar. The first:
In $\mathbb{Z}_8$ find the identity of the following commutative operation:
$$\overline{a}\cdot\overline{c}=\overline{a}+\overline{c}+2\overline{a}\overline{c}$$
I say:
$$\overline{a}\cdot\overline{i}=\overline{a}+\overli... | So since you want $$(\overline{a},\overline{b})=(\overline{a}+\overline{e_1},\overline{8}\overline{b}\overline{e_2})$$
You must have $\overline{a}=\overline{a}+\overline{e_1}$, meaning $\overline{e_1}=\overline{0}$, and from the second entry we see that
$\overline{b}=\overline{b}(\overline{8}\overline{e_2})$; since th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/118339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Trigonometric identity, possible error I need to prove the following trigonometric identity:
$$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$
I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it b... | Since all the trig values are squared, it seems as though the exercise is simply playing with shifts by odd or even multiples of $\pi/2$.
Loosely,
*
*Shifting by "$\frac{\pi}{2} \cdot \text{odd}$" switches "sin" and "cos" (and possibly affects the sign)
*Shifting by "$\frac{\pi}{2} \cdot \text{even}$" ($=$ "$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/119481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$ Compute
$$\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$$
I did
$$\lim_{x\to\infty} (\frac{x-2}{x+2})^x = \lim_{x\to\infty} \exp(x\cdot \ln(\frac{x-2}{x+2})) = \exp( \lim_{x\to\infty} x\cdot \ln(\frac{x-2}{x+2}))$$
But how do I continue? The hint is to use L Hop... | A nitpick: $\infty-\infty$ is not 0! It's undefined. Your limit is of the form $0/0$ though.
You can apply L'H'ôpital from the start if you like: $\lim\limits_{x\rightarrow\infty}{x-2\over x+2} =1$, and $\ln 1=0$.
So
$$
\lim_{x\rightarrow\infty} \Bigl(x \ln{x-2\over x+2} \Bigr)
=\lim_{x\rightarrow\infty} {\ln{x-2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
If z is one of the fifth roots of unity, not 1... If z is one of the fifth roots of unity, not 1, show that:
$1+z+z^2+z^3+z^4=0$
Which wasn't too bad, but the next part is killing me: show that:
$z-z^2+z^3-z^4=2i(sin(2\pi/5)-sin(\pi/5))$
Can anyone help? Thanks!
| For the first part you should note that by the formula for the geometric series,
$$1 + z + z^2 + z^3 + z^4 = \frac{z^5 - 1}{z- 1}.$$
So if you put $z = e^{2\pi i/5}$ into $z^5 - 1$ you get $(e^{2\pi i/5})^5 =1$ so that $1 + z + \ldots z^4 = 0$. You can see also that if you put $z = (e^{2\pi i/5})^n$ for $n=2,3,4$ you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$? I'm trying to evaluate the integral $\displaystyle\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$.
My book explains that to evaluate integrals of the form $\displaystyle\int_0^\infty x^\alpha R(x)dx$, with real $\alpha\in(0,1)$ and $R(x)$ a rational function, one first sta... | Here is the simplest I could get without introducing other variables or making strange substitutions. Everything is in complex on the main branch ($0 \le arg(z) < 2\pi$).
First we calculate a bound change :
$$\begin{eqnarray*}
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz &=& \int_{-\infty}^0 \frac{z^{1/3}}{1+z^2} dz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
How does one solve the differential equation $y'=\frac {y} {3x-y} $ ? I need to find a solution to the differential following equation:
$y'=\frac {y} {3x-y} $.
I tried to use use some kind of substitution, but I didn't manage to solve it.
Any suggestion\help?
Thanks a lot!
| $$ y' = \frac{y}{3x - y} = \frac{1}{\frac{3x}{y} - 1} $$
$$ \frac{y}{x} = v \Rightarrow y = vx \Rightarrow \frac{dy}{dx} = x \frac{dv}{dx} + v $$
$$ x \frac{dv}{dx} + v = \frac{1}{\frac{3}{v} - 1} = \frac{v}{3 - v} $$
$$ x \frac{dv}{dx} = \frac{v}{3-v} - v = \frac{v(v-2)}{3-v} $$
$$ \frac{(3-v)dv}{v(v-2)} = \frac{dx}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Exponential Equation with mistaken result I'm on my math book studying exponential equations, and I got stuck on this Problem:
What is sum of the roots of the equation:
$$\frac{16^x + 64}{5} = 4^x + 4$$
I decided to changed: $4^x$ by $m$, so I got: $$\frac{m^2 + 64}{5} = m + 4$$
working on it I've got: $m^2 - 5m + 44 ... | $$
\begin{align}
\frac{16^x+64}{5}&=4^x+4\\
16^x+64&=5\cdot 4^x+20\\
16^x-5\cdot 4^x&=-44\\
y^2-5y+44&=0 \quad \text{for } y=4^x\\
y&=\frac{1}{2}\left(5\pm i\sqrt{151}\right)\\
4^x&=\frac{1}{2}\left(5\pm i\sqrt{151}\right)\\
x&=\log_{4}\left( \frac{1}{2}\left(5\pm i\sqrt{151}\right)\right)\\
\sum(x)&=\log_{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find the limit $\lim\limits _{ x\to \infty } \left( \sqrt { x^2+3x } -\sqrt { x^2+x } \right) $? When I try to do this type of indeterminations I reach to this point:
$\lim\limits_{ x\to \infty } \dfrac { 2x }{ \sqrt { x^ 2 +3x } +\sqrt { x^2 +x } } $
but I don't know how to continue. Thanks.
| $\lim\limits_{ x\to \infty }{ \dfrac { 2x }{ \sqrt { { x }^{ 2 }+3x } +\sqrt { { x }^{ 2 }+x } } } $
$\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt \frac{x^2+3x}{x^2}+\sqrt \frac{x^2+x}{x^2}}}=\lim\limits_{x\to \infty }\frac {2}{\large{\sqrt{ 1+\frac{3x}{x^2}}+\sqrt{ 1+\frac{x}{x^2}}}}=\lim\limits_{x\to \infty ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to expand $\cos nx$ with $\cos x$? Multiple Angle Identities:
How to expand $\cos nx$ with $\cos x$, such as
$$\cos10x=512(\cos x)^{10}-1280(\cos x)^8+1120(\cos x)^6-400(\cos x)^4+50(\cos x)^2-1$$
See a list of trigonometric identities in english/ chinese
| \begin{equation}
\text{You can use De Moivre's identity:}
\end{equation}
\begin{equation}
\text{Let's Call:}\\\\
\end{equation}
\begin{equation}
\mathrm{z=\cos x+i \sin x}\\
\mathrm{\frac{1}{z}=\cos x-i \sin x}\\
\end{equation}
\begin{equation}
\text{Now, addind both equations together, we get:}\\
\end{equation}
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
} |
Number of positive solutions? What is the number of positive solutions to
$$ (x^{1000} + 1)(1 + x^2 + x^4 + \cdots + x^{998}) = 1000x^{999}? $$
I tried to solve it.
First I used by using sum of Geometric Progression.
Then the equation becomes too complicated and is in the power of 1998.
How can I get the number of po... | Consider$$(x^{m} + 1)(1 + x^2 + x^4 + \cdots + x^{m-2}) = mx^{m-1}?$$
by multiplying $x^2-1$ in tow sides of equation,we have
$$(x^{m} + 1)(x^m-1)=mx^{m-1}(x^2-1)$$,
Last we have
$$x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1 = 0 \tag{#}$$
Factorization
$$\begin{align*}
&x^{2m} - m(x^{m + 1} - x^{m - 1}) - 1\\
&=(x^{2m}-1)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/126146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Euclidean algorithm as a linear combination substitution and simplifying This problem is from my discrete mathematics textbook.
I'm trying to find $\gcd(420,66)$
I compute
$$\begin{align*}
420 &= 6 \times 66 + 24\\
66 &= 2\times 24 + 18\\
24 &= 1 \times 18 + 6\\
18 &= 3 \times 6 + 0
\end{align*}$$
then I rewrite t... | Distribute, reorder, associate:
$$\begin{align*}
24 - 1(66-2\times 24) &= 24 -1(66) -1(-2\times 24)\\
&= 24 - 66 +2\times 24\\
&= 24 + 2\times 24 - 66\\
&= (1+2)24 - 66\\
&= 3\times 24 - 66.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do you explain the concept of logarithm to a five year old? Okay, I understand that it cannot be explained to a 5 year old. But how do you explain the logarithm to primary school students?
| A logarithm is how much bigger or smaller a number is than another
number when you combine numbers using multiplication instead
of addition.
For instance, to see how much bigger $3$ is than $2$ in this way:
$$
\begin{array}{|cccccccccccccccc|}
\hline
2^1 & \cdot & 2^2 & \color{red}{2^3} & \cdot & 2^4 & \cdot & 2^5 & 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/129013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "94",
"answer_count": 21,
"answer_id": 16
} |
Help finding integral: $\int \frac{dx}{x\sqrt{1 + x + x^2}}$ Could someone help me with finding this integral
$$\int \frac{dx}{x\sqrt{1 + x + x^2}}$$
or give a hint on how to solve it.
Thanks in advance
| $\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{x\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}dx$.$\;\;$ Now let $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$, $dx=\frac{\sqrt{3}}{2}\sec^{2}\theta d\theta$
to get $\displaystyle\int\frac{1}{(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2})(\frac{\sqrt{3}}{2}\sec\theta)}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/129821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$ Evaluate $$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011}$$
So ...
$$(\frac{1}{-\sqrt{2}+\sqrt{2}i})^{2011} = (-\sqrt{2}+\sqrt{2}i)^{-2011}$$
$$\theta=\pi - \arctan(\frac{\sqrt{2}}{\sqrt{2}}) = \frac{3\pi}{4}$$
$$-\sqrt{2}+\sqrt{2}i=\cos{\theta} + i \sin{\theta}$$
$$(-\sqrt{2}... | Since everyone has been answering about the blue part, I will answer about the red part:
\begin{eqnarray*} 2^{-2011} e^{-(\color{red}{1508\pi i} + \frac{\pi i }{4})}
&=& 2^{-2011}e^{-(\color{red}{1508\pi i})} \cdot e^{\left( \frac{\pi i }{4}\right)} \\
&=& 2^{-2011}{\left(e^{(\color{red}{2\pi i})}\right)}^{-\frac{1508... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Applying difference of cubes to cube roots I am stumped as to why this application of the difference of cubes is valid...
I am rationalizing the denominator. I don't understand the reasoning of why the difference of cubes formula is applicable to cubed roots, removing the root one gets an exponent of $a^{1/3}$ - I kn... | You use the the formula
$$(x-y)(x^2+xy+y^2) = x^3-y^3$$
with $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$. You already have one of the factors on the left hand side, so you multiply by the other factor (and cancel it out).
If you have
$$\frac{1}{x-y}$$
then you can transform it into
$$\frac{1}{x-y} = \frac{x^2+xy+y^2}{(x-y)(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/132842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
A highschool factoring problem $x+y+z=0$
$x^3+y^3+z^3=9$
$x^5+y^5+z^5=30$
$xy+yz+zx=?$
I solved this problem by setting $xy+yz+zx=k$ and using the cubic equation with roots $x,y,z$. But is there any other methods?
| We have the Newton-Girard identities $$x^3+y^3+z^3=(x+y+z)^3+3xyz-3(x+y+z)(xy+xz+yz)$$ and $$\begin{split}x^5+y^5+z^5=&(x+y+z)^5-5(x+y+z)^3 (xy+xz+yz)+\\5(x+y+z)&(xy+xz+yz)^2-5xyz(xy+xz+yz)+5xyz(x+y+z)^2\end{split}$$
Replacing all instances of $x+y+z$ with $0$, we have the simultaneous equations
$$\begin{align*}
3xyz&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/138102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \fra... | Just in case you may be interested. More generally you can get a similar expression for such sums of powers of inverses. They basically come from the factorization
$$a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \cdots + a + 1)$$
which can be proved by multiplying out the right hand side and then most of the terms cancel out a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Highest power of a prime $p$ dividing $N!$ How does one find the highest power of a prime $p$ that divides $N!$ and other related products?
Related question: How many zeros are there at the end of $N!$?
This is being done to reduce abstract duplicates. See
Coping with *abstract* duplicate questions. and List of Gener... | Largest power of a prime dividing $N!$
In general, the highest power of a prime $p$ dividing $N!$ is given by
$$s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots$$
The first term appears since you want to count the numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "70",
"answer_count": 9,
"answer_id": 1
} |
Trouble manipulating a log expression This question sort of follows on from question Functions with logarithmic integrals. The book presents an example of integrating a function whose integral is logarithmic:
$$\int \frac{1}{4-3x} dx = -\frac{1}{3}\ln{|4 - 3x|} + K$$
$$= -\frac{1}{3}\ln{A|4 - 3x|}$$
$$= \frac{1}{3}\ln{... | From $-\dfrac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) $ we don't
get $\dfrac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }$, because
$$\begin{equation*}
-\frac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) \neq \frac{1}{3}
\ln \frac{A}{\left\vert 4-3x\right\vert }.
\end{equation*}$$
However if we wr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Solving an integral I am having trouble with this integral:
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx.$$
One obvious thing would be to complete the square: $x^2+2x+25=(x+1)^2+24$. But then, I don't know which substitution to use. Can anyone help? Thank you.
| You will have, by substitution of $y=\frac{x+1}{\sqrt{24}}$,
$$\int_0^4\frac{x}{\sqrt{x^2+2x+25}}dx=\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy$$
that becomes
$$\int_\frac{1}{\sqrt{24}}^1\frac{(\sqrt{24}y-1)}{\sqrt{y^2+1}}dy=\sqrt{24}\int_\frac{1}{\sqrt{24}}^1\frac{y}{\sqrt{y^2+1}}dy-\int_\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$. In this (btw, nice) answer to Twin primes of form $2^n+3$ and $2^n+5$, it was said that:
If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$?
I'm not familiar with these kind of calculations, so I'd like to see, if my answer is correct:
*
*Let $n=3k+2$ so then
$2^{3k+2}+3\equiv 2^{... | $$\begin{align*}
2^{3k+2}+3&\equiv 2^{3k+2}-4\\
&\equiv 4\left(2^{3k}-1\right)\\
&\equiv 4\left((2^3)^k-1\right)\\
&\equiv 4\left(1^k-1\right)\\
&\equiv 4\cdot0\\
&\equiv 0 \pmod 7\;,
\end{align*}$$
since $8\equiv1\pmod7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/148418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 1... | If $(n+1)((n+1)^2+8)=(n+1)(n^2+2n+9)$ then if $(n+1)= 0\ mod\ 3$, we're done. If not, than $(n+1)=1\mod 3$ or $(n+1)= 2\mod 3$. So if $n+1=1\mod 3 $ then $n=0\mod 3$ so $3|(n^2+2n+9)$, and if $n+1=2\mod 3$ then $n^2=n=1\mod 3$ now let $n^2=3k+1$ and $n=3l+1$ so we have $(n^2+2n+9)=(3k+1+6l+2+9)=3(k+2l+1+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/150425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 6
} |
Show $\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh}(t)}{t}dt=\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du$ How would I show (or explain) that
$$\int_\frac{1}{3}^\frac{1}{2}\frac{\operatorname{artanh} t}{t}dt,$$
$$\int_{\ln 2}^{\ln 3}\frac{u}{2\sinh u}du,$$
and
$$-\int_\frac{1}{3}^\frac{1}{2}\frac{\ln v}{1-v^2}dv$... | I'd like to give a suggestion for the equality
$\displaystyle \int_{1/3}^{1/2} \frac{\text{arctanh } t}{t}\text{ d}t=-\int_{1/3}^{1/2} \frac{\log v}{1-v^2}\text{ d}v$
The idea is to rewrite $\displaystyle \text{arctanh } t=\int_0^t \frac{1}{1-s^2}\text{ d}s$, so that we have
$\displaystyle \int_{1/3}^{1/2} \frac{\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/151390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Ratio and number theory The question goes as follows:
Let $K$ be a three digit number such that the ratio of the number to the sum
of its digit is least. What is the difference between the hundreds and
the tens digits of $K$?
Now I was able to do this question by trial and error, assuming hundredth digit place t... | Doing it in a single function, just for fun:
For $x \in \mathbb{Z}$ we want to minimize the ratio $\frac{100a+10b+c}{a+b+c}$ where $ a,b,c\in\mathbb{Z}$ and $100a+10b+c=x$ We can rewrite the ratio as: $$ f(x)=\frac{100\left\lfloor\frac{x}{100}\right\rfloor+10\left\lfloor\frac{x-100\left\lfloor\frac{x}{100}\right\rflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/153342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Long division in integration by partial fractions I am trying to figure out what my book did, I can't make sense of the example.
"Since the degree of the numberator is greater than the degree of the denominator, we first perform the long division. This enables us to write
$$\int \frac{x^3 + x}{x -1} dx = \int \left(x^2... | You did not do the long division correctly.
x^2 + x + 2
_________________________
x - 1 | x^3 + x
- x^3 + x^2
-----------
+ x^2 + x
- x^2 + x
--------
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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$3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
| As David Mitra commented, one may factor using polynomial division:
$\frac {x^3 - 8} {x-2} = (x^2+2x+4)$
To solve this, we can quickly complete the square as follows:
Notice that we are setting $x^2 + 2x+4 = 0$
$(x+1)^2 + 3 = 0$
$(x+1)^2 = -3$
$x+1 = \pm i \sqrt{3}$
$x = -1 \pm i \sqrt{3}$ and we have our answer. I fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$... | I agree with the answer given by Gerry Myerson. The only detail left out is, is to complete the square on the terms: $x^2-2ax = (x-a)^2 - a^2$ and the same for: $y^2-2by = (y-b)^2 - b^2$
adding these 2 equations and equal to zero then gives the equation in Cartesian form as shown above by Gerry Myerson. I am just addin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/154550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Given $x^2 + 2y^2 - 6x + 4y + 7 = 0$, find center, foci, vertex/vertices So the equation is:
$$ x^2 + 2y^2 - 6x + 4y + 7 = 0 $$
Find the coordinates of the center, the foci, and the
vertex or vertices.
What I did was put the equation in the form:
$$
\frac{(x-3)^2}{4}+ \frac{(y+1)^2}{4} = 1
$$
Now based on that, I s... | The equation should be $$\frac{(x-3)^2}{4}+\frac{(y+1)^2}{2}=1.$$ You've correctly identified the center and vertices. The focal length should be $\sqrt{a^2-b^2}$, not $\sqrt{a^2+b^2}$. Ellipses don't have asymptotes, you're thinking of hyperbolae.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/156663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Partial fractions for geometric probability-generating function wrong Let $X\sim \text{Geo}(1/4), Y\sim \text{Geo}(1/2)$ be given. First I have to compute $\mathbb{E}[z^{X+Y}]$:
$$\mathbb{E}[z^{X+Y}]=\mathbb{E}[z^{X}]\cdot\mathbb{E}[z^{Y}]=\frac{\frac{1}{4}z}{1-\left(1-\frac{1}{4}\right)z}\cdot\frac{\frac{1}{2}z}{1-\le... | Due to anons hint (thanks!) I found the mistake.
$$\frac{z^2}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}=1+\frac{\frac{10}{3}z-\frac{8}{3}}{8\left(1-\frac{1}{2}z\right)\left(1-\frac{3}{4}z\right)}$$
This leads to $\alpha=1/3,\;\beta=-1,\;\gamma=2/3$ which is correct.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx$ So, from here $$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)} dx$$
I divided by cos(x) and I got
$$\int \frac{\tan(x)}{2\cos^2(x)+1} dx$$
But I'm stuck here.
I tried to substitute $t=\cos(x)$
$$\int \frac{-1}{t\cdot(2t^2+1)} dt$$
Any help w... | $$
\begin{aligned}
& \int \frac{\sin x}{3 \cos ^{3} x+\sin ^{2} x \cos x} d x \\
=& \int \frac{\sin x d x}{\cos x\left(3 \cos ^{2} x+\sin ^{2} x\right)} \\
=& \int \frac{d(\cos x)}{\cos x\left(2 \cos ^{2} x+1\right)} \\
=& \int\left(\frac{1}{\cos x}-\frac{2 \cos x}{2 \cos ^{2} x+1}\right) d(\cos x) \\
=& \ln (\cos x)-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/157895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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What is the result of sum $\sum\limits_{i=0}^n 2^i$
Possible Duplicate:
the sum of powers of $2$ between $2^0$ and $2^n$
What is the result of
$$2^0 + 2^1 + 2^2 + \cdots + 2^{n-1} + 2^n\ ?$$
Is there a formula on this? and how to prove the formula?
(It is actually to compute the time complexity of a Fibonacci recur... | I thought I might post a little more elaborate version of Henning's hint (see his comment).
$$\begin{align}
1&=2^0\\
10&=2^1\\
100&=2^2\\
1000&=2^3\\
\vdots&=\vdots\\
10\dots0&=2^n\\
\hline
11\dots1&=2^0+2^1+\dots+2^n\\
1&=1\\
\hline
100\dots0&=2^0+2^1+\dots+2^n+1=2^{n+1}
\end{align}$$
Hence $2^0+2^1+\dots+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/158758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac... | The best answer is Byron's, but it's possible to show by induction that the characteristic polynomial of $M$ is $\big(\lambda - \frac{n+1}2\!\big)\big(\lambda - \frac 1 2\!\big)^{n-1}$, which means that the $M$ is (orthogonally) conjugate to a diagonal matrix with these (positive) eigenvalues along the diagonal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 6
} |
A simple quadratic inequality For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$
| Essentially, you want to prove that (multiplying throughout by $4$) $$c^2 -2cn -4c +n^2 + 8n \leq n^2 -2n +5$$
i.e. $$c^2 - 2cn -4c + 10n \leq 5$$
Since $n \geq c \geq 5$, we have that $(10-2c)n \leq (10-2c)c$ (since $10-2c \leq 0$). Hence, $$c^2 -4c + (10-2c)n \leq c^2 - 4c +(10-2c)c = -c^2 + 6c = c(6-c) = 9 - (c-3)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/159997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Integration of $\int\frac{1}{x^{4}+1}\mathrm dx$ I don't know how to integrate $\displaystyle \int\frac{1}{x^{4}+1}\mathrm dx$. Do I have to use trigonometric substitution?
| Directly by Sophie Germain's Identity or:
$$x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt2x)^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1)$$
After splitting the initial fraction we get:
$$ \int \frac{1}{x^4 +1} \ dx = \int \frac{\frac{x}{2\sqrt2}+\frac{1}{2}}{x^2+\sqrt2x+1} \ dx+\int \frac{\frac{-x}{2\sqrt2}+\frac{1}{2}}{x^2-\sqrt2x+1} \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/160157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 20,
"answer_id": 9
} |
Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
| $$
\left.\frac{(x+2)^2}{5}\right|_0^1 = \frac{(1+2)^2}{5} - \frac{(0+2)^2}{5} = \frac{9}{5} - \frac{4}{5} = \frac{5}{5} = 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/160542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Evaluating $\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}}+\frac{1}{\sqrt{16}+\sqrt{20}+\sqrt{25}}=?$ It's a question (not hw) I bumped into few years back. Couldn't make any real progress with. Maybe you can help?
$$\frac{1}{\sqrt{4}+\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}+\sqrt{16}... | The case of cube roots is probably more interesting than square roots; namely, simplifying
$$\frac{1}{\sqrt[3]{4}+\sqrt[3]{6}+\sqrt[3]{9}}+\frac{1}{\sqrt[3]{9}+\sqrt[3]{12}+\sqrt[3]{16}}+\frac{1}{\sqrt[3]{16}+\sqrt[3]{20}+\sqrt[3]{25}}. \tag{$\circ$}$$
To evaluate this, as sos440 did in the comments, one notes the st... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference ... | According to Maple, your integral is $\dfrac{\pi^2}{8} - \dfrac{10}{9}$, so your inequality becomes $\pi < \sqrt{89}/{3}$. In fact, an antiderivative is
$$F(x) = \dfrac{x^3 \ln(x)}{3} - \dfrac{x^3}{9} + x \ln(x) - x - \dfrac{\ln(x) \ln(x+1)}{2} - \dfrac{\text{dilog}(x)+\text{dilog}(x+1)}{2}$$
More generally, for $p > ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Is this proof that $\sqrt 2$ is irrational correct? Suppose $\sqrt 2$ were rational. Then we would have integers $a$ and $b$ with $\sqrt 2 = \frac ab$ and $a$ and $b$ relatively prime.
Since $\gcd(a,b)=1$, we have $\gcd(a^2, b^2)=1$, and the fraction $\frac{a^2}{b^2}$ is also in lowest terms.
Squaring both sides,
$2 =... | Hhmmm ... Here is a short one.
We all know that $\sqrt{2}$ is a root of $x^2-2 = 0$
Apply Rational Root Theorem and you're done
It goes something like this : if $\sqrt{2} = \dfrac{p}{q}$, then $q=1$ and $p$ is a factor of $2$. Hence the only possible rational root of $x^2-2=0$ would be $-2,-1,1,-2$.
A quick check shows... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Integral - using Euler Substitution I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot ... | You may use substitution x=1/t, or Euler substitution \sqrt{x^2+x+1}=tx-1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Solving a complex integral I need help solving an integral from John Conway book.
Lets $\alpha$ complex number different from 1 find integral $$\int\frac{dx}{1-2\alpha\cos{x}+{\alpha}^2}$$ from 0 to $2\pi$ in unit circle $$(z-\alpha)^{-1}(z-\frac{1}{\alpha})^{-1}$$
| Substitute: $z = e^{i x}$ then:
$$\cos x = \frac{1}{2} \left( z + \frac{1}{z} \right), \; dx = \frac{1}{i} \cdot \frac{dz}{z}$$
and we can rewrite the integral as:
$$ i \int_{|z|=1} \frac{dz}{a (z-a)(z - \frac{1}{a})}$$
There are two cases:
*
*$|a| < 1$ then only $z=a$ is a pole inside a circle and the residue is:
$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as which of the following? ${10 \choose 4}+{11 \choose 4}+{12 \choose 4}+\cdots+{20 \choose 4}$ can be simplified as ?
A. ${21 \choose 5}$
B. ${20 \choose 5}-{11 \choose 4}$
C. ${21 \choose 5}-{10 \choose 5}$
D. ${20 \choose 4}$
Plea... | The same thing, combinatorially. We want to choose $5$ positive integers from the first $21$. This can be done in $\binom{21}{5}$ ways.
We count the same thing in a different way. If the biggest chosen number is $21$, the rest can be chosen in $\binom{20}{4}$ ways. If the biggest is $20$, the rest can be chosen in $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/165044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Power (Laurent) Series of $\coth(x)$ I need some help to prove that the power series of $\coth x$ is:
$$\frac{1}{x} + \frac{x}{3} - \frac{x^3}{45} + O(x^5) \ \ \ \ \ $$
I don't know how to derive this, should I divide the expansion of $\cosh(x)$ by the expansion of $\sinh(x)$? (I've tried but without good results :( )
... | $$ \begin{eqnarray}
\coth(x) &=& \frac{\cosh(x)}{\sinh(x)} = \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{x + \frac{x^3}{6} + \frac{x^5}{120} + \mathcal{o}\left(x^5\right)} = \frac{1}{x} \frac{1 + \frac{x^2}{2} + \frac{x^4}{24} + \mathcal{o}\left(x^5\right)}{1 + \frac{x^2}{6} + \frac{x^4}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/167149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there a good way to solve for the inverse of $(u^2-u+4)$? I'm having trouble calculating the inverse of a polynomial. Consider the polynomial $f(x)=x^3+3x-2$, which is irreducible over $\mathbb{Q}$, as it has no rational roots. So $\mathbb{Q}[x]/(f(x))\simeq \mathbb{Q}[u]$ is a field.
How would I calculate $(u^2-u+... | Let's calculate the $\gcd$ of the two polynomials $x^3+3x-2$ and $x^2-x+4$ in $\mathbb{Q}[x]$. Using the Euclidean algorithm,
$$x^3+3x-2=x(x^2-x+4)+(x^2-x-2)$$
$$x^2-x+4=(x^2-x-2)+6$$
Because $2$ is a unit in $\mathbb{Q}[x]$, we see that the polynomials are relatively prime. Now, building back up, we see that
$$6=(x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/169335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $p$ divides $a^{p - 2} + a^{p - 3} b + a^{p - 4} b^2 + \cdots + b^{p - 2}$ when $p$ is prime, $a, b \in \mathbb{Z}$ and $a,b \lt p$? If $p$ is a prime number and $a, b \in \mathbb{Z}$ such that $a,b \lt p$, then how could we prove that $p$ divides
$\left(a^{p - 2} + a^{p - 3} b + a^{p - 4} b^2 + \cdots +... | We have
$$(a-b)(a^{p-2} + a^{p-3}b + \ldots + ab^{p-3} + b^{p-2}) = a^{p-1} - b^{p-1}.$$
If $p\nmid a$ and $p\nmid b$ then the right side is congruent to $1-1 = 0$ by Fermat's little theorem, i.e $p$ divides the product on the left. If furthermore $a \not\equiv b \pmod p$ then $p$ must already divide $a^{p-2} + a^{p-3}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{... | Completion of the square yields the following.
$$4x^2 - 6x + 9 = 4(x^2 - 3/2 x) + 9 = 4(x^2-3/2x + 9/16) + 27/4 = 4(x-3/4)^2 + 27/4.$$
This definitively shows your residual quadratic can have no real roots, since its graph never goes below the line $y = 27/4.$ This representation will allow you to find the complex one... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x... | Hint 1: For all $x$, $\sin^2x+\cos^2x=1$
Hint 2: For $at^2+bt+c=0$ the solutions are $t_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Finding remainders of large number This question was asked in a competitive exam.
Find the remainder of dividing $6^{83}+ 8^{83}$ by $49$
Are there any theorems/rules to compute the answer ?
| Using Euler's Totient theorem,
As $(6,49)=1$ and $\phi(49)=42$
$6^{42}\equiv 1\pmod {49}\implies 6^{42.k}\equiv1\pmod {49}$ where is any integer.
$\implies 6^{84}\equiv 1\pmod{49}\implies 6^{83}\equiv 6^{-1}\pmod{49}$
By observation, $6\times 8=48\equiv -1\pmod{49}\implies 6\times (49-8)\equiv 1\pmod{49}\implies 41\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent. I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.
I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor osci... | Here is another approach: If the sequence converges, then the difference between two successive terms $|a_{n+1}-a_n|$ must go to zero. So, if you can show that there are pairs of successive terms that do not go to zero, then the sequence diverges.
My suggestion is to consider pairs $a_{1+4k}, a_{2+4k}$, where $k$ is a ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Computing a sum of binomial coefficients: $\sum_{i=0}^m \binom{N-i}{m-i}$ Does anyone know a better expression than the current one for this sum?
$$
\sum_{i=0}^m \binom{N-i}{m-i}, \quad 0 \le m \le N.
$$
It would help me compute a lot of things and make equations a lot cleaner in the context where it appears (as some... | First note that $\dbinom{n-i}{m-i} = \dbinom{n-i}{n-m}$. Hence, you have
$$\dbinom{n-m}{n-m}+\dbinom{n-m+1}{n-m}+\cdots+\dbinom{n-1}{n-m}+\dbinom{n}{n-m}$$
Let $n-m = k$. Hence, we want to find an expression for
$$\dbinom{k}k+\dbinom{k+1}{k}+\cdots+\dbinom{n-1}{k}+\dbinom{n}{k}$$
From the Pascal's triangle, we have
$\h... | {
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"answer_count": 4,
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} |
Find coefficient of $x^ny^m$ how can I find the coefficient of $x^ny^m$ of following series
$$\frac{\log\big(\frac{1}{1-xy}\big)}{(1-x)(1-y)(1-xy)}$$
and
$$ \frac{\log(\frac{1}{1-x})}{(1-x)(1-y)(1-xy)}$$
where $1 \leq m \leq n$
| Use $\log\left(\frac{1}{1-x y}\right) = -\log(1- xy) = - \sum_{n=1}^\infty \frac{1}{n} x^n y^n$, and
$$
-\frac{\log(1-x y)}{1-x y} = \sum_{n=1}^\infty x^n y^n \sum_{m=1}^n \frac{1}{n} = \sum_{n=1}^\infty x^n y^n H_n
$$
Then use
$$
\frac{ \sum_{n \geqslant 1, m\geqslant 1} c_{n,m} x^n y^m}{(1-x)(1-y)} =
\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/173905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A (probably trivial) induction problem: $\sum_2^nk^{-2}\lt1$ So I'm a bit stuck on the following problem I'm attempting to solve. Essentially, I'm required to prove that $\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2} < 1$ for all $n$. I've been toiling with some algebraic gymnastics for a while now, but I can't seem... | Yet another approach :
Let us first analyze the sum till infinity. Let $$ S= \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+ \cdots \infty$$ $$\Rightarrow S=(\frac{1}{2^2}+\frac{1}{4^2}+ \cdots\infty) +(\frac{1}{3^2}+\frac{1}{5^2}+ \cdots\infty ) $$$$\Rightarrow S= \frac{1}{4}(1+\frac{1}{2^2}+\frac{1}{3^2}+\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/174828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 1
} |
Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$
I am trying to find all solutions to
(1) $y^3 = x^2 + x + 1$, where $x,y$ are integers $> 1$
I have attempted to do this using...I think they are called 'quadratic integers'. It would be great if someone could verify the steps and sugge... | I have not verified your work. This is just to say that $y^3=x^2+x+1$ is the equation of an elliptic curve (with $j$-invariant equal to $0$). Using Sage or Magma, you can find that this elliptic curve has rank $1$ and trivial torsion subgroup. The group of rational points is generated by the point $P=(0,1)$, and the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Power series expansion I recently had a problem. I know how to evaluate power series but I cannot seem to find an expansion for $\sqrt{x+1}$.
I've tried differentiating it, in order to bring it in reciprocal form but that didn't help. Due to the presence of square root, I cannot change it in the form of $1/(x+1)$.
Ki... | Here's one way. Start with the expansion you want, using $a_0$, $a_1$, $a_2$, $a_3$, etc. for the unknown coefficients:
$$\sqrt{x+1}\;=\;a_{0}\;+\;a_{1}x\;+\;a_{2}x^2\;+\;a_{3}x^3\;+\;a_{4}x^4\;+\;a_{5}x^5\;+ ...$$
Finding $a_0$: Plugging in $x=0$ on both sides leads to $a_{0}=1$.
Finding $a_1$: Differentiate both side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/178704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Commutative matrix
*
*If two matrices $A$ and $B$ are commutative then all rules for real numbers $a$ and $b$ apply for the matrices?
For example, if $AB=BA$ then:
$(A+B)^2=A^2 + 2AB + B^2$
$A^3 - B^3 = (A-B)(A^2+AB+B^2)$
and so on...
*If the matrix $A$ is invertible then is $A^m A^n = A^{(m+n)}$, where $m,n$ are ... | *
*"Do all rules for real numbers apply to the matrix?"
If by all rules for real numbers, you mean finite factorization laws like in your two examples, then yes. How might we prove such a thing? Let's consider $(A + B)^2$ and $(A+B)^3$.
$(A + B)^2 = A^2 + AB + BA + B^2$, and as $AB = BA$ we can write this as $A^2 + 2A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/181513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$ Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$
Prove the inequality
$$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$
Thanks :)
| Let $S=a^2bc+b^2cd+c^2da+d^2ab$. We can easily find that:
$$S-(ac+bd)(ab+cd)=-bd(a-c)(b-d);$$
$$S-(bc+ad)(bd+ac)=ac(a-c)(b-d)$$
which implies $$S\le \max\{(ac+bd)(ab+cd),(bc+ad)(bd+ac)\}.$$
By AG mean inequality:
\begin{align*}
(ac+bd)(ab+cd)&\le \left(\frac{(ac+bd)+(ab+cd)}{2}\right)^2\\
{}&=\frac{(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
I want to find out the angle for the expression $a^3 + b^3 = c^3$. like in pythagorean theorem angle comes 90 degree for the expression $a^2 + b^2 = c^2$, however I know that no integer solution is possible.
| Expanding on Brian M. Scott's answer, since
$\cos\theta=\frac{a^2+b^2-c^2}{2ab}$
and $c^3 = a^3+b^3$,
$\cos\theta=\frac{a^2+b^2-(a^3+b^3)^{2/3}}{2ab}
= \frac{1+(b/a)^2-(1+(b/a)^3)^{2/3}}{2}
=\frac{1+r^2-(1+r^3)^{2/3}}{2}
$
where $r = b/a$.
The derivative of the numerator
is $2r-(2/3)(3r^2)(1+r^3)^{-1/3} = 2r - 2r^2(1+r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/184352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Finding the sum of this alternating series with factorial denominator. What is the sum of this series?
$$ 1 - \frac{2}{1!} + \frac{3}{2!} - \frac{4}{3!} + \frac{5}{4!} - \frac{6}{5!} + \dots $$
| Hint: We have
$$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots.$$
Multiply both sides by $x$ and differentiate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.11 Let $a,b,c \gt 0$. Prove that
$$\sqrt{\frac{2a}{b+c}}... | Using cauchy Schwarz or AM-QM we have that $$LHS \leq \sqrt{3\sum_{cyc}\frac{2a}{b+c}}$$ It suffices to prove $$\sum_{cyc}\frac{2a}{b+c}\leq \sum_{cyc}\frac ab$$ By homogeneity we may suppose $a+b+c=1$.
Clearing out denominators this reduces to show $$2\sum_{cyc}a(a+b)(a+c)abc\leq \sum_{cyc}a^2c(a+b)(b+c)(c+a)$$ which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Deriving the addition formula for the lemniscate functions from a total differential equation The lemniscate of Bernoulli $C$ is a plane curve defined as follows.
Let $a > 0$ be a real number.
Let $F_1 = (a, 0)$ and $F_2 = (-a, 0)$ be two points of $\mathbb{R}^2$.
Let $C = \{P \in \mathbb{R}^2; PF_1\cdot PF_2 = a^2\}$.... | We follow the method of my answer to this question.
Let $u = \int_{0}^{x}\frac{dx}{\sqrt{1 - x^4}}$.
Then $x = s(u)$.
Let $v = \int_{0}^{y}\frac{dy}{\sqrt{1 - y^4}}$.
Then $y = s(v)$.
Let $c$ be a constant.
$u + v = c$ is a solution of the equation:
$$\frac{dx}{\sqrt{1 - x^4}} + \frac{dy}{\sqrt{1 - y^4}} = 0$$
It suffi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/189806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $ \lim\limits_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $ How would you evaluate the following series?
$$\lim_{n\to\infty} \sum_{k=1}^{n^2} \frac{n}{n^2+k^2} $$
Thanks.
| Here's another approach.
First, note that
$$\begin{eqnarray*}
\sum_{k=n^2+1}^\infty \frac{n}{n^2+k^2}
&<& \sum_{k=n^2+1}^\infty \frac{n}{k^2} \\
&\le& n\int_{n^2}^\infty \frac{dx}{x^2} \\
&=& \frac{1}{n}.
\end{eqnarray*}$$
We also need the partial fraction expansion of $\coth x$,
$$\begin{eqnarray*}
\coth x &=& \lim_{N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/190966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 2
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Prove that there is no value of the integers $x,y,z$ satisfied the equation: $19^x + 5^y + 1980z = 1975^{4^{30}} + 2010$ Prove that there is no value of the integers $x,y,z$ satisfying the equation: $$ 19^x + 5^y + 1980z = 1975^{4^{30}}+ 2010 $$
The equation $1975^{4^{30}}$ is like a double exponent :(
thanks again, it... | $ 19^x + 5^y =- 1980z + 1975^{4^{30}}+ 2010 $ an integer.
If, at least one of $x,y$ is $<0$,the LHS=$(19^x + 5^y)$ is a fraction.
If $x,y≥0$,
$ 19^x + 5^y + 1980z = 1975^{4^{30}}+ 2010 $
$\implies 19^x=1975^{4^{30}}+ 2010-1980z-5^y$
Observe that $(10a+5)^n$ leaves remainder $5$ when divided by $10$ ,where $n$ is posit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A binary quadratic form and an ideal of an order of a quadratic number field Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$.
We say $D = b^2 - 4ac$ is the discriminant of $F$.
If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $ax^2 + bxy + cy^2$ is primitive.
Is the following propo... | Proof that I is an ideal of $R$
By this question, $R = \mathbb{Z} + \mathbb{Z}\frac{(D + \sqrt D)}{2}$.
Hence it suffices to show that $a\frac{(D + \sqrt D)}{2} \in I$ and $\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} \in I$.
$a\frac{(D + \sqrt D)}{2} = \frac{(aD + a\sqrt D)}{2} = \frac{(aD + ab + a(-b + \sqrt D))}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<
| There are no real solutions, nor any complex solutions if you use the principal branch of the square root. Squaring both sides and simplifying gives you $20 \sqrt{x-4} = -92$.
EDIT: More generally, for any $a, b \ge 0$, $\sqrt{a + b} \le \sqrt{a} + \sqrt{b}$. Since
$(x+4) - (x-4) = 8$, the most $\sqrt{x+4} - \sqrt{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Homework Calculus line with circle Find an expression for the function whose graph consist of the line segment from the point $(1,-3)$ to the point $(5,7)$ together with the top of the circle with the center at $(8,7)$.
I don't understand what it means by together with a circle.
$$y = -\frac{5}{2}x + \frac{39}{2}$$
| I will assume that the circle has radius equal to $3$. If that is the case, then the circle is defined by
$C := \{ (x,y) \in \mathbb{R}^2 \mid (x-8)^2 + (y-7)^2 = 3^2\}$.
The equation that defines the circle can be rewritten in the form
$(y-7)^2 = 9 - (x-8)^2$
and, taking the square root of both sides, we obtain
$y = 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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A question with the sequence $e_{n}=\left(1+\frac{1}{n}\right)^{n}$ Prove that
$a$) the following sequence is increasing
$$e_{n}=\left(1+\frac{1}{n}\right)^{n},\quad n\ge1;$$
$b$) the inequality below holds
$$e_{n} \leq3,\quad n\ge1.$$
| In order to prove that the given sequence is strictly increasing, we are to demonstrate $e_{n+1} > e_n$:
\[
\bigg(1+ \dfrac{1}{n+1}\bigg)^{n+1} > \bigg(1 + \dfrac{1}{n} \bigg)^n.
\]
Let's rewrite the inequality above as:
\[ \bigg( \dfrac{1 + \dfrac{1}{n+1}}{ 1 + \dfrac{1}{n}} \bigg)^n > \dfrac{1}{1 + \dfrac{1}{n+1}}.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/193266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$... | There is a standard method for solving equations of the form:
$$
A \sin x + B \cos x = C
$$
Divide both sides by $\sqrt{A^2 + B^2}$:
$$
\frac{A}{\sqrt{A^2 + B^2}} \sin x + \frac{B}{\sqrt{A^2 + B^2}} \cos x = \frac{C}{\sqrt{A^2 + B^2}}
$$
Find $\theta \in [0, 2\pi)$ so that:
$$
\sin \theta = \frac{B}{\sqrt{A^2 + B^2}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/201399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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How to solve and see resolution of $13^{53} \pmod 7$ using Fermat little Theorem? How to solve and see resolution of $13^{53} \pmod 7$ using Fermat little Theorem? Using Fermat's Little Theorem, I know it gives me 6 as an answer to this problem..., but why? How is the resolution?
Thanks,
| Fermat's Little Theorem says $x^7\equiv x\pmod{7}$. When $x\not\equiv0\pmod{7}$, we can divide by $x$ to get
$$
x^6\equiv1\pmod{7}
$$
In the case of $13^{53}$, we get that $13^{53}=13^{6\cdot8+5}=\left(13^6\right)^8\cdot13^5\equiv1^8\cdot(-1)^5\equiv-1\pmod{7}$ since $13^6\equiv1\pmod{7}$ and $13\equiv-1\pmod{7}$.
Of c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
| Substition
$$x=4\tan y \Rightarrow dx=\frac{4}{\cos^2 y} dy$$
$$\tan y=\frac{x}{4} \Rightarrow y=\arctan\frac{x}{4}$$
$$$\int\frac{1}{(16+x^2)^2}dx=\int\frac{1}{(4^2+x^2)^2}dx$=$\int\frac{\frac{4}{\cos^2 y}}{(16+(4\tan y)^2)^2}dy$=$\int\frac{\frac{4}{\cos^2 y}}{(16+16\tan^2 y)^2}dy$$
$$= \int\frac{\frac{4}{\cos^2 y}}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/204961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find an ellipse whose length is the same as the outer rim of the monkey saddle
Given the monkey saddle $z=x^3-3xy^2$ over the unit circle $x^2+y^2
\leq 1$, find an ellipse whose length is the same as the length of the
outer edge of the monkey saddle.
I've already found a parameterization for the monkey saddle in cy... | The outer edge of the monkey saddle is at $r=1$, and thus $(x,y,z)=(\cos\theta,\sin\theta,\cos3\theta)$. The total arc length is
$$
\begin{align}
\int\mathrm ds
&=
\int\sqrt{\mathrm dx^2+\mathrm dy^2+\mathrm dz^2}
\\
&=
\int_0^{2\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm d\theta}\right)^2+\left(\frac{\mathrm dy}{\mathrm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/205149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving an equality of matrices I am asked to prove that $$\begin{pmatrix} \\ A & B\\ C &D\end{pmatrix}^{-1}=\begin{pmatrix} M & -MBD^{-1} \\ -D^{-1}CM & D^{-1}+D^{-1}CMBD^{-1} \end{pmatrix}$$
Where $M=(A-BD^{-1}C)^{-1}$.
Unfortunately, I have no idea what to do about it.
| I guess rather than verifying the inverse stated in the assignment, you should derive it. Let $$
\begin{pmatrix} A & B \cr C & D \end{pmatrix}^{-1} = \begin{pmatrix} U & V \cr W & X \end{pmatrix}
$$
We have:
$$
\begin{pmatrix} 1 & 0 \cr 0 & 1 \end{pmatrix} = \begin{pmatrix} A & B \cr C & D \end{pmatrix} \cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Triangle problem I have got one simple task to prove:
We have got a triangle $\triangle XYZ$.
Then we create points $A,B,C$ on $XY, YZ, ZX$ respectively, such that $XA = AB = BZ$ and $CZ = AY = AC$.
How to prove that $XY = \frac{XZ + YZ}{2}$?
| Choosing coordinates
Based on the variables $a$, $b$ and $c$ we define coordinates for your points like this, without loss of generality:
\begin{align*}
X &= \begin{pmatrix}-1\\0\end{pmatrix} &
Y &= \begin{pmatrix}1\\0\end{pmatrix} &
Z &= \begin{pmatrix}b\\c\end{pmatrix} \\
A &= \begin{pmatrix}a\\0\end{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/208824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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Sum of a nearly classic series Assuming we know that : $$\sum_{n=1}^{+\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}$$
How do you find the sum of a series where all terms are in this one ?
For instance, how do you prove that ?$$\sum_{n=1}^{+\infty}{\frac{1}{(2n-1)^2}} = \frac{\pi^2}{8}$$
| Observe that:
$$\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{6}$$
and
$$\sum_{n=1}^{\infty} \frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{24} $$
therefore
\begin{eqnarray*}
\sum_{n=1}^{\infty} \frac{1}{(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Analytic Geometry: Circle Given is a line with parametric equation:
$ x = 2 \lambda $
$ y = 1-\lambda $
Find out for which values of $\lambda$ the line is inside the circle of $x^2+4x+y^2-6x+5=0$
My attempt at solving this:
$x^2+4x+y^2-6x+5=0$
$x^2-2x+y^2+5=0$
$ (x-1)^2 -1 + y^2+5=0$
$ (x-1)^2 + y^2 =-4$
And that's ... | I will assume that we have a typo, and $x^2+4x+y^2-6x+5=0$ is intended to be $x^2+4x+y^2-6y+5=0$. Two reasons for this: (i) it would be strange to have two "$x$" terms and (ii) the given equation is not satisfied by any (real) pair $(x,y)$.
The modified equation can be rewritten as $(x+2)^2+(y-3)^2=8$, a circle with ce... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/213704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Help! Integrate using substitution method. I need help integrating the following function:
$$\int\frac{2x+5}{\sqrt{16-6x-x^2}}dx$$
| If you let $u=16-6x-x^2$, then $du=(-6-2x)dx=-(2x+6)~dx$. Now you can rewrite your integral:
$$\begin{align*}
\int\frac{2x+5}{\sqrt{16-6x-x^2}}dx&=\int\frac{(2x+6)-1}{\sqrt{16-6x-x^2}}dx\\
&=-\int\frac{du}{\sqrt{u}}-\int\frac1{\sqrt{16-6x-x^2}}dx\;.
\end{align*}$$
The first of these two integrals is an easy power rule ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/216692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
| The following argument is based on the suggestions by vesszabo. (I do not restrict the functions to the interval $[0, \, 0.1]$ as vesszabo did. That would erroneous because we are looking for the limit at $0$.) It is a rigorous argument and it does avoid using Taylor's Theorem ... but it is still hokey. The choice t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/217081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 4
} |
SVD and linear least squares problem Edit:
I've actually found an error:
Instead of full SVD I had to use, "economy size" SVD, where $U$ has only first $n$ columns, and $\Sigma$ becomes a square matrix. I also forgot to take the transpose of $V$, that's why I was getting wrong numbers. SO, the primary question is solv... | Problem definition
$$
\begin{align}
\mathbf{A} x &= b\\
\left(
\begin{array}{rr}
1 & 1 \\
0 & 1 \\
-1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{c}
x_{1} \\
x_{2}
\end{array}
\right)
&=
\left(
\begin{array}{c}
1 \\
1 \\
1 \\
\end{array}
\right)
\end{align}
$$
Normal equations
$$
\begin{align}
\mathbf{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/218333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the imaginary roots of polynomials I'm looking for a simple way to calculate roots in the complex numbers.
I'm having the polynomial $2x^2-x+2$
I know that the result is $1/4-$($\sqrt{15}/4)i$.
However I'm not aware of an easy way to get there.
| $$2x^2 - x +2 = 2\left(x^2 - \dfrac{x}2 + 1 \right) = 2\left(\underbrace{x^2 - 2 \times x \times \dfrac14 + \left(\dfrac14 \right)^2}_{a^2 - 2ab + b^2 = (a-b)^2} - \left(\dfrac14 \right)^2 + 1 \right) = 2 \left( \left(x- \dfrac14 \right)^2 + \dfrac{15}{16}\right)$$
Setting the above to zero, we get that
$$\left(x- \df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/225373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help with Trigonometry homework - prove an identity I need to prove the following identity:
$\sin^2 2\alpha-\sin^2 \alpha = \sin 3\alpha \sin \alpha$
What I have tried, is to work on each side of the identity. I have started with the left side:
\begin{align}
\sin^2 2\alpha-\sin^2 \alpha &= (\sin 2\alpha-\sin \alpha)(\s... | You made an error in each side of the identity. Both sides simplify to
$$\sin^2 \alpha(4\cos^2 \alpha-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/225499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The positive integer solutions for $2^a+3^b=5^c$ What are the positive integer solutions to the equation
$$2^a + 3^b = 5^c$$
Of course $(1,\space 1, \space 1)$ is a solution.
|
case : $b>2$ and $c>2$
A calculus with computer, modulo $N=2372625=5^3\times 19 \times 37 \times 3^3$, in $H=\mathbb Z/N\mathbb Z $ give
$\text{card}(<2>)=900$, $\text{card}(3^3\times <3>)=900$, $\text{card}(5^3\times<5>)=36$.
And $(<2>+(3^3\times <3>)) \cap 5^3\times <5>=\emptyset$
with $1\in <g>$ the subset of $H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/226415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Can $11^{13}-1$ be divided exactly by 6? Can $11^{13}-1$ be divided exactly by 6?
My solution:
$$11^2 \equiv 1 \pmod 6$$
$$11^{12} \equiv 1 \pmod 6$$
$$11^{13} \equiv 5 \pmod 6 $$
Hence, $(11^{13}-\mathbf{5})$ can be divided exactly by 6. However, according to the solution on my book, （$11^{13}-\mathbf{1}$）can be divid... | It is already not divisible by $ 3 $; notice that
\begin{align}
11^{13} - 1 &\equiv (-1)^{13} - 1 \\
&\equiv -1 - 1 \\
&\equiv -2 \\
&\equiv 1 \\
&\not\equiv 0 \, (\text{mod} \, 3).
\end{align}
Note $ 11 \equiv 2 \equiv -1 \, (\text{mod} \, 3) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/226874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Mixing things- ratios Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \fr... | When you mix up a m/n ratio of A/B to get C, then $m+n$ units of C will contain m units of A and n units of B, or $(m+n)C = mA+nB$. I prefer to express this as $C = \frac{m}{m+n}A + \frac{n}{m+n} B$. (Note that the sum of the fractions is exactly 1, so I can work with just the fraction of A, knowing that the fraction o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/229818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find the sum of the first $n$ terms of $\sum^n_{k=1}k^3$ The question:
Find the sum of the first $n$ terms of $$\sum^n_{k=1}k^3$$
[Hint: consider $(k+1)^4-k^4$]
[Answer: $\frac{1}{4}n^2(n+1)^2$]
My solution:
$$\begin{align}
\sum^n_{k=1}k^3&=1^3+2^3+3^3+4^3+\cdots+(n-1)^3+n^3\\
&=\frac{n}{2}[\text{first term} + \text{la... | Let's take the suggested hint, and consider $$(k+1)^{4}-k^{4}=4k^{3}+6k^{2}+4k+1$$
Summing up both sides from $k=1$ to $n$. Notice that $$\sum_{k=1}^{n}[(k+1)^{4}-k^{4}]=[2^{4}-1^{4}]+[3^{4}-2^{4}]+\ldots+[n^{4}-(n-1)^{4}]+[(n+1)^{4}-n^{4}]$$
Cancelling, we get $(n+1)^{4}-1$. So altogether,
$$(n+1)^{4}-1=4\sum_{k=1}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Why isn't it a Hilbert space Let $X$ be the vector space of all the continuous complex-valued functions on $[0,1]$. Then $X$ has an inner product $$(f,g) = \int_0^1 f(t)\overline{g(t)} dt$$ to make it an inner product space. But this is not a Hilbert space.
Why isn't is complete? Which Cauchy sequence in it is not conv... | Consider the sequence of continuous functions $f_n$ defined by $$
f_n(x) = \left\{ \begin{array}{rl}
0 &\mbox{ if $x\in \Big[0,\frac{1}{2}-\frac{1}{n}\Big]$} \\
\frac{nx}{2}-\frac{n}{4}+\frac{1}{2} &\mbox{ if $x\in\Big[\frac{1}{2}-\frac{1}{n},\frac{1}{2}+\frac{1}{n}\Big]$}\\
1 &\mbox{ if $x\in\Big[\frac{1}{2}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Finding sum of a series: difference of cubes I am trying to find sum of the infinite series:
$$1^{3}-2^{3}+3^{3}-4^{3}+5^{3}-6^{3} + \ldots$$
I tried to solve it by subtracting sum of even cubes from odd, but that solves only half of the numbers.
Any input is appreciated.
Thank you all (especially Marvis, Manzonni... | Using the identities
$$
\sum_{k=0}^\infty\binom{k}{n}x^k=\frac{x^n}{(1-x)^{n+1}}
$$
and
$$
k^3=\binom{k}{1}+6\binom{k}{2}+6\binom{k}{3}
$$
yields
$$
\sum_{k=1}^\infty k^3x^{k-1}=\frac{1}{(1-x)^2}+6\frac{x}{(1-x)^3}+6\frac{x^2}{(1-x)^4}
$$
Letting $x\to-1$, gives
$$
\begin{align}
\sum_{k=0}^\infty(-1)^{k-1}k^3
&=\frac14... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Function expansion help I want to write a function, $f(k,a,b)$, I made, in terms of combinations of the fractional part function, $$ j\left\{\frac{c \ }{d}k\right\},$$ where $c,d,$ and $j$ are any integers.
The function is as follows: $f(k,a,b)=1$ if $k\equiv b$ mod a
and $f(k,a,b)=0$, if it is not
I need a general m... | Recall that the fractional part function $\{x\}$ is defined as the unique real number $r$ with $0\le r<1$ such that $x-r$ is an integer. Recall also that the floor function $\lfloor x\rfloor$ is the unique integer $m$ such that $m\le x<m+1$. Finally, recall that these two functions are related by the identity $x=\lfloo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/244648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find an angle in a given triangle $\triangle ABC$ has sides $AC = BC$ and $\angle ACB = 96^\circ$. $D$ is a point in $\triangle ABC$ such that $\angle DAB = 18^\circ$ and $\angle DBA = 30^\circ$. What is the measure (in degrees) of $\angle ACD$?
| In $\triangle ADB,\angle ADB=(180-18-30)^\circ=132^\circ$
Applying sine law in $\triangle ADB,$ $$\frac{AB}{\sin 132^\circ}=\frac{AD}{\sin30^\circ}\implies AD=\frac{AB}{2\sin48^\circ}$$ as $\sin132^\circ=\sin(180-132)^\circ=\sin48^\circ$
$\angle ABC=\angle BAC=\frac{180^\circ-96^\circ}2=42^\circ$
Applying sine law in $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Does $x_{n+2} = (x_{n+1} + x_{n})/2$ converge? Let $0 < a < b < \infty$. Define $x_1=a$, $x_2=b$, and $x_{n+2} = \frac{x_n + x_{n+1}}{2}$ for $n \geq 1$. Does $\{x_n\}$ converge? If so, to what limit?
| $$x_3 = \dfrac{a+b}2$$ $$x_4 = \dfrac{a+3b}4$$ $$x_5 = \dfrac{3a+5b}8$$ $$x_6 = \dfrac{5a+11b}{16}$$ $$x_7 = \dfrac{11a+21b}{32}$$
In general, (by induction) $$x_n = \dfrac{J_{n-2} a + J_{n-1}b}{2^{n-2}}$$ where $J_n$'s are Jacobsthal sequence i.e. they satisfy the recurrence
$$J_{n+1} = J_n + 2J_{n-1}$$ with $J_0 = 0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factorization of polynomial I was just asked to factor $x^3+1$. I came to $(x^2-x+1)(x+1)$ after a while, and I was wondering, whether there is a good method to quicky factor such of polynomials.
| You are usually asked to remember $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^3+a^3=(x+a)(x^2-ax+a^2)$. In general,
$$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1})$$
If $$y=x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1}$$
$$xy=x^n+ax^{n-1}+...+a^{n-2}x^2+a^{n-1}x$$
$$-ay=-ax^{n-1}-a^2x^{n-2}-...-a^{n-1}x-a^n$$
Add $xy$ and $-ay$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integral by partial fractions
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)
factor out the 5, and use partial fraction
$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$
Solve for $A$ and $B$.
$A\left... | Distribute:
$$
\frac{5}{x-5}((x-5)\ln|x-5|-x)=5\left(\frac{x-5}{x-5}\ln|x-5|-\frac{x}{x-5}\right).
$$
Then
$$
\frac{x-5}{x-5}=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/247178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to simplify $\frac{4 + 2\sqrt6}{\sqrt{5 + 2\sqrt{6}}}$? I was tackling through an olympiad practice book when I saw one of these problems:
If $x = 5 + 2\sqrt6$, evaluate $\Large{x \ - \ 1 \over\sqrt{x}}$?
The answer written is $2\sqrt2$, but I can't figure my way out through the manipulations. I just know that ... | The square of $\sqrt{x} - {1 \over \sqrt{x}}$ is $x - 2 + {1 \over x}$. In this case $x = 5 + 2\sqrt{6}$, whose reciprocal is seen to be $5 - 2\sqrt{6}$ by rationalizing the denominator.
So
$$x - 2 + {1 \over x} = (5 + 2\sqrt{6}) - 2 + (5 - 2\sqrt{6})$$
$$= 8$$
So your answer is $\sqrt{8} = 2\sqrt{2}$. (You take the p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/249993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
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Problem when integrating $e^x / x$. I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong?
So, I'm trying... | There is no elementary antiderivative for you this function.
You can take a look at here: http://en.wikipedia.org/wiki/Exponential_integral
What you have calculated here:
$$\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( 1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) $$
is even some... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/251795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 3,
"answer_id": 2
} |
$f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$ How many ordered triples of rational numbers $(a,b,c)$ are there such that the cubic polynomial $f(x)=x^3+ax^2+bx+c$ has roots $a,b$ and $c$?
The polynomial is allowed to have repeated roots.
| [There was a mistake in the first version of this answer that caused one solution to go missing.]
The only monic polynomial with roots $a,b,c$ is $(x-a)(x-b)(x-c)$, so we must have
$$
\begin{align}
a&=-a-b-c\;,\\
b&=ab+bc+ca\;,\\
c&=-abc\;.
\end{align}
$$
If $c=0$, this becomes
$$
\begin{align}
a&=-a-b\;,\\
b&=ab\;,\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/252256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Evaluate $\prod_{x=2}^\infty\frac{x^4-1}{x^4+1}$ Difficult question from some test somewhere (I forget).
$$\prod_{x=2}^\infty\frac{x^4-1}{x^4+1}$$
$x$ is, of course, an integer.
| Write
$$\frac{x^4 - 1}{x^4+1} = \frac{(x-a_1)\ldots(x-a_4)}{(x-b_1)\ldots(x-b_4)}$$
where $a_j$ are the roots of $x^4-1$ and $b_j$ are the roots of $x^4+1$.
Then the partial product
$$ \prod_{x=2}^n \frac{x^4 - 1}{x^4+1} = \frac{\Gamma(n+1-a_1)\ldots \Gamma(n+1-a_4) \Gamma(2-b_1) \ldots \Gamma(2-b_n)}{\Gamma(2-a_1) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
| We will do this using only divisibility.
Implying that you have this knowledge.
$(a+b,a^2-ab+b^2)=\{1\;\text{and}\;3\}$ if $(a,b)=1$
Calling $d$ the $(a+b,a^2-ab+b^2)$
$$(a+b,a^2-ab+b^2)=d\Longrightarrow d\mid a+b\;\;\text{and}\;\;d\mid a^2-ab+b^2$$$$a^2-ab+b^2=a^2-ab+b^2\underbrace{+3ab-3ab}_{=0})=a^2+2ab+b^2-3ab=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/257392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Series of positive terms I want to show that $\displaystyle 1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+\cdots$ converges. I know that by using D'Alembert ratio test I easily show that this series converges but I am doing in this way:
\begin{align*}
s_{n}&=1+\frac{1}{2!}+\frac{1}{4!}+\cdots+\frac{1}{2(n-1)!}\\
&<1+\f... | Maybe you can compare it to exp(1). You would consider the term of your series to be
a2n = 1/(2n!) ,
a2n+1 = 0.
Obviously for each k, ak<=1/(k!) and therefore your series converges and its sum is less than exp(1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find $1/x^3 + 1/y^3$? If I am given, $x + y = a$ and $xy = b$, how would I find the value of $\dfrac1{x^3} + \dfrac1{y^3}$?
| \begin{align}
\dfrac1{x^3} + \dfrac1{y^3} & = \dfrac{x^3+y^3}{(xy)^3} = \dfrac{\left(x+y \right)\left(x^2+y^2-xy \right)}{(xy)^3}\\
& = \dfrac{\left(x+y \right)\left(\left(x+y \right)^2-3xy \right)}{(xy)^3} = \dfrac{a\left(a^2-3b \right)}{b^3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Variance formulae I know that the variance formula is
$$\sigma^2 = \frac{
\left( x_1 - \bar{x} \right) ^2 +
\left( x_2 - \bar{x} \right) ^2 +
\dots +
\left( x_n - \bar{x} \right) ^2
}{n}$$
Where $\sigma^2$ is the variance; $x_1,\ x_2,\ \dots,\ x_n$ are the statistical data, and $n$ is the number of data.... | Each of the $(x_i - \bar x)^2$ terms expands into $x_i^2 - 2x_i\bar x + \bar x^2$.
Since the sum of all $x_i$ is also equal to $n \bar x$ (by definition), we get that the sum of all the $2x_i\bar x$ is actually $2n\bar x^2$.
Then, the equation becomes: $\displaystyle \sigma^2 = \frac{x_1^2 + x_2^2 + \cdots + x_n^2 - 2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How can I resolve rational indefinite integral? $$\int\left( {20.56\over x^2-1.27}+x^{55}\right) dx$$
I came to something like this
$$20.56 \int {1\over (x-1)^2 - \frac{27}{100} }~dx + {x^{56}\over56} + \text{Constant}$$
Please can you help me to resolve this?
| Your integral is of the form
$$I = \int \left(\dfrac{a}{x^2 - b^2} + x^n \right)dx$$
Hence,
\begin{align}
I & = \int \left(\dfrac{a}{2b}\dfrac1{x - b} - \dfrac{a}{2b}\dfrac1{x + b} + x^n \right)dx \\
& = \dfrac{a}{2b} \log(\vert x-b \vert) - \dfrac{a}{2b} \log(\vert x+b \vert) + \dfrac{x^{n+1}}{n+1} + \text{constant}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Prove that $\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$ for positive $a,b,c$ Prove the following inequality: for
$a,b,c>0$
$$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\geq\frac{1}{2}(a+b+c)$$
What I tried is using substitution:
$p=a+b+c$
$q=ab+bc+ca$
$r=abc$
But I cannot reduce $a^2(b+c)... | By AM-GM inequality,
$$\frac{a^2}{a+b} + \frac{a+b}{4} \ge a$$
Add up the similar inequalities obtained by cyclic substitution, you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/264931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Distinct natural numbers such that $ab=cd=a+b+c+d-3$ Find the distinct natural numbers $a,b,c,d$ who satisfying $ab=cd=a+b+c+d-3$.
| Find natural numbers $(a,b)$ such that:
$c$ = $\dfrac{1}{2}(a b-\sqrt{(a b-a-b+3)^2-4 a b)}-a-b+3 )$
and $d$ = $\dfrac{1}{2}$ $(a b+\sqrt{(a b-a-b+3)^2-4 a b)}-a-b+3)$
are natural numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.