Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}... | The definition of $x_n$ can be directly used to evaluate the limit. To simplify typing let $s_n=\sum_{i=0}^{n}1/i!$. Then we have $$x_n=\frac{\log s_n-n\log(1+n^{-1})}{\log(1+n^{-1})}$$ Since $n\log(1+(1/n))\to 1$ the desired limit is equal to the limit of $$n\log s_n-n^2\log(1+(1/n))$$ and now we need to use Taylor se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
| By Jensen's inequality with $\phi\left(t\right)=\frac{1}{3t^2+1}$ we have
$$\phi\left(\frac13\sum_{i=1}^{3}x_i\right)\le \frac13\sum_{i=1}^{3}\phi\left(x_i\right) $$
since $x+y+z= 6$ we get
$$\frac{1}{13}=\phi\left(2\right)\le \frac13\left(\frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3z^2+1}\right) $$
hence $$\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of Gamma functions I need to prove:
$$\lim_{p \to\infty}\frac{\Gamma \left ({\dfrac{p+1}{2}} \right )}{\Gamma(p/2)\sqrt{p\pi}}= \frac{1}{\sqrt{2\pi}}.$$
Where $\Gamma (x)$ represents the gamma function.
To provide some context, I need to find the limiting distribution of a student's t distribution , which should... | As $p\to +\infty$ the Gamma function has the asymptotic approximation:
$$\Gamma\left(\frac{p+1}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$
and
$$\Gamma\left(\frac{p}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration with multiple constants Question:
$$\int \frac{ax^2-b}{x\sqrt{{c^2x^2-(ax^2+b)^2}}}\ \text dx$$
My approach:
I can't understand whether I should integrate it normally or use a trigonometric function.
| Let $$I =\int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx = \int\frac{ax^2-b}{x\cdot x \sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$
So $$I = \int\frac{a-\frac{b}{x^2}}{\sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$
Put $\displaystyle \left(ax+\frac{b}{x}\right) = t\;,$ Then $\displaystyle \left(a-\frac{b}{x^2}\right)dx = dt$
$$I = \int\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $f(x)=\sin^2 x-x^2\cos x, \forall x\in [0,\frac{\pi}{2}]$ is monotonic increasing I would like to show that $f(x)=\sin^2 x-x^2\cos x, \forall x\in [0,\frac{\pi}{2}]$ is monotonic increasing.
If we can show that $f'(x)>0$, $\forall x\in [0,\frac{\pi}{2}]$, then $f(x)$ is increasing there.
We have
$f'... | One has \begin{align}2\sin x\cos x-2x\cos x+x^2\sin x &=2\cos x(\sin x-x)+x^2\sin x \\ &\ge x^2\sin x-\frac{x^3\cos x}{3} \\&=x^2\cos x\left(\tan x-\frac{x}3\right)\\&\ge \frac23x^3\cos x\ge0,\end{align}where we have used the Taylor expansion of $\sin x$ and the fact that $\tan x\ge x$ on $\left[0,\dfrac\pi2\right]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x... | You need an $(x+1)^2$ in the denominator of at least one of the partial fractions, or when you sum them they would not have an $(x+1)^2$ term in the denominator of the sum.
But you might find it easier to solve:
$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{Bx + C}{(x+1)^2}$
And that is completely valid.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number ... | Just a quick correction to your transition matrix...
$$P =\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given odd positive integer, $n=2m-1$, $n \equiv 1 \pmod 4 \implies m \equiv 1 \pmod 2$ $n=2m-1\implies n+1=2m$. So, $n\equiv 1\pmod 4, 2m=n+1\equiv 2\pmod 4\implies 2m=n+1\equiv 0\pmod 2$ $\implies n\equiv -1\pmod 2\implies n\equiv 1\pmod 2$. But, how to find $m$ from this last line of equivalence relations is not clea... | $$n = 2m-1\equiv 1\pmod{4}\quad\Rightarrow\quad
2m \equiv 2\pmod{4}.$$
Now, note that in general, if $k\ne 0$ and $ak\equiv bk\mod{ck}$, this means that $ak$ is $bk$ plus a multiple of $ck$, say $rck$:
$$ak = bk + rck.$$
Dividing through by $k$ gives $a = b+cr$, so that $a\equiv b\mod c$. Applying this to $2m\equiv 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$
Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$.
Here is my solution I am looking to see if it is correct or if there is another way to do it, thanks!
The only way to form an $x^{25}$ term is to gather two $x^8$ and three $x^3$ . Since there are ${... | A formal way:
$$(1 + x^3 + x^8)^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{3k}(1+x^5)^k=\sum_{k=0}^{10}\sum_{j=0}^{k}\binom{10}{k}\binom{k}{j}x^{3k+5j}.$$
Now $25=3k+5j$ for $ 0\leq j\leq k\leq 10$ is solved iff $(k,j)=(5,2)$ and
$$[x^{25}](1 + x^3 + x^8)^{10}=\binom{10}{5}\binom{5}{2}=2520.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{... | You can end it by the following way:
$$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)=$$
$$=\left(\frac{1}{2}+\cos9^{\circ}\right)\left(\frac{1}{2}+\sin9^{\circ}\right)\left(\frac{1}{2}+\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1... | The standard way is to consider
$$(1-\frac{5}{n^4})^{(2018n+1)^4}=e^{(2018n+1)^4 \log{(1-\frac{5}{n^4}) }}\to e^{-5\cdot2018^4}$$
indeed by standard limits
$${(2018n+1)^4 \log{(1-\frac{5}{n^4}) }}=(2018n+1)^4(\frac{5}{n^4})
\frac{\log {(1-\frac{5}{n^4})}}{\frac{5}{n^4}}\to -5\cdot2018^4$$
Or as an alternative
$$(1-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4... | Let $z=\cos t+i\sin t$. Then $|z^2+\bar{z}^2|=1$ becomes
$$2|\cos(2t)|=1 $$
or
$$ \cos(2t)=\pm\frac{1}{2}$$
So
$$ \cos t=\pm\sqrt{\frac{1+\cos(2t)}{2}}=\pm\frac{\sqrt3}{2}\text{ or }\pm\frac{1}{2}$$
and
$$ \sin t=\pm\sqrt{\frac{1-\cos(2t)}{2}}=\pm\frac12\text{ or }\pm\frac{\sqrt3}{2}. $$
Thus
$$ z=\cos t+i\sin t=\pm\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2... | You nearly had the answer yourself: In the line
$$\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
& =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
& =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\
\end{align}$$
For large $n$ $$\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}} \appro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power Series of $\ \frac x3 \ln(1+x^2).$ I have a question regarding the power series representation of the function $$\ \frac x3 \ln(1+x^2).$$
I got the answer $$\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+1}}}{3n} $$
by using the Maclaurin series expansion for $\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{n}}}{n}$ an... | So continuing from where your teacher left off, we have $\frac{2x}3 \int\frac{xdx}{1+x^2}$. We start with the series expansion for $\frac{1}{1+y}$:
$$ \frac1{1+y} = \sum_{n=0}^\infty (-1)^ny^n$$
We substitute $y = x^2$:
$$ \frac1{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n}$$
We multiply by $x$:
$$ \frac x{1+x^2} = \sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the value of $\sin^2 (\frac{\pi}{10}) \sin^2 (\frac{3\pi}{10})$?
PROBLEM
$$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$
My Try
$$
\left(1 + \cos \frac{\pi}{10}\right)
\left(1 + \cos \frac{9\pi}{10}\right)
\left(1 + \cos \frac{7\pi}{10}\right)
\left(1 + \cos \frac{3\pi}{10}\right)... | $$\sin^2 {\frac {\pi}{10}} \sin^2 \frac {3\pi}{10}=\left(\frac {4\sin {\frac {\pi}{5}}. \cos {\frac {2\pi}{10}}\cos {\frac {4\pi}{10}}}{4\sin \frac {\pi}{5}}\right)^2=\left(\frac {\sin \frac {4\pi}{5}}{4\sin \frac {\pi}{5}}\right)^2=\frac {1}{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\fr... | Note that by standard limits since
$$\frac{a^x-1}{x}\to \log a$$
we have that
$$\frac{3^x+2^x-2}{4^x+2^x-2}=\frac{\frac{3^x-1}x+\frac{2^x-1}x}{\frac{4^x-1}x+\frac{2^x-1}x}\to\frac{\log 3+\log2}{\log 4+\log 2}=\frac{\log 6}{\log 8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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This formula gives $8^{-1}$ (mod $n$). Is there a deeper pattern lurking here? Pick $n\equiv 1$ (mod $4$) which is not a multiple of $3$ and such that $n>5$. Consider the sum
$$S(n):=2\cdot\frac{n-1}2+3\cdot\frac{n-3}2+\ldots+m(m+2)+m+1,$$
where $m:=\frac{n-1}4$. For example, for $n=17$ we are considering
$$2\cdot 8+3\... | This is a partial answer.
Using that
$$\sum_{i=0}^{N}1=N+1,\qquad \sum_{i=0}^{N}i=\frac{N(N+1)}{2},\qquad\sum_{i=0}^{N}i^2=\frac{N(N+1)(2N+1)}{6}$$
we have
$$\small\begin{align}8S(n)&=8\sum_{i=0}^{m-2}(2+i)\left(\frac{n-1}{2}-i\right)+8m+8\\\\&=-8\left(\sum_{i=0}^{m-2}i^2\right)+(4n-20)\left(\sum_{i=0}^{m-2}i\right)+(8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Eigenvalue and Eigenvector of $\small\pmatrix{0 & 0 \\ 0 & -7}$ I need help working out the eigenvectors for this matrix.
$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $
The original matrix is $ \begin {pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix} $ , eigenvalues are 5,-2,
but I am not sure how to about the eigenv... | From first equation you deduce whatever is x and y the equation holds
$$0x+0y=0$$
From second equation you deduce that $y=0$
$$0x-7y=0 \implies -7y=0 \implies y=0$$
So
$$(x,y)=(x,0)=x(1,0)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}... | A hint, as requested:
What you've done looks pretty good; I haven't checked every bit of the algebra, but the symmetry of the result makes me think that you've probably done it right.
What you have not done is use the important fact given at the start: that $A, B,$ and $C$ are the angles of a triangle (hence they sum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$
Find the Cartesian equation of the locus described by $|z+2-7i| =
2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.
This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If an... | Consider z= (x+iy),Now put the value of Z in the given question.
after applying rules of modulus
we get `x^2+Y^2-28x+10y+121=0
Now we can arrange in square form like
(x-14)^2+(y+5)^2=45
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2... | Given $n-1, n, n+1$ and $n+2$ then $2$ divides two of them and $4$ divides one of them and $8$ divides the product $(n-1)n(n+1)(n+2)$. If $n$ is odd then we know the two that $2$ and $4$ divides must be $n-1$ and $n+1$ and that $8$ divides the product $(n-1)(n+1)$.
Given $n-1, n , n+1$ then $3$ divides exactly one of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\m... | Presuming $s_1 > 0$, by AM-GM inequality,
$$s_{n} = \frac{s_{n-1}+\frac{3}{s_{n-1}}}{2} \ge \sqrt{3}$$ for $n \ge 2$.
Also,
$$s_{n+1}-s_{n} = \frac{1}{2}\left(\frac{3}{s_n}-s_n\right)\le 0 \iff s_{n} \ge \sqrt{3}$$
for $n \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A step in finding the slant asymptote I'm curious why this textbook shows an extra step when finding the slant asymptote in this function: $$f(x) = \frac {x^3}{x^2 + 1}.$$
So long division gives
$$f(x) = \frac{x^3}{x^2+1} = x - \frac{x}{x^2+1},$$
so
$$f(x) - x = - \frac{x}{x^2 + 1} = -\frac{\frac{1}{x}}{1 + \frac{1}{x^... | Because $-\frac{x}{x^2+1}$ is in the indeterminate form $\frac{\infty}{\infty}$ and thus the additional step is to factor out an $x^2$ term before to take the limit in order to have a form $\frac{0}{1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the mistake in this? What is wrong in this proof?
In $\triangle ABC$ with right angle at $B$, $BD$ is drawn as an altitude on $AC$. Let $AB=a$ and $BC=b$. So by similarity, $AD=\frac{a^2}{\sqrt{a^2+b^2}}$ and similarly $CD=\frac{b^2}{\sqrt{a^2+b^2}}$. Now, $\triangle ABD,BDC$ are also right angled at $D$. So a... | Your problem is actually not even really a misuse of a rule. It's a typo.
$$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{a^2}{\sqrt{a^2+b^2}}$$
This should be:
$$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{b^2}{\sqrt{a^2+b^2}}$$
Reducing that down, we get:
$$(a^2)\sqrt{a^2+b^2}-a^2=(b^2)\sqrt{a^2+b^2}-b^2$$
Factor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $B$ is non singular and that $AB^{-1}A=A$
$$A_{n\times n}=\begin{bmatrix}a & b & b & b &. &.&.&&b\\b & a
&b&b&.&.&.&&b\\b & b &a&b&.&.&.&&b\\b & . &.&.&.&.&.&&b\\b & .
&.&.&.&.&.&&b\\b & b &b&b&.&.&.&&a\end{bmatrix}\text{ where }
a+(n-1)b =0$$
Define $l^t=\begin{bmatrix}1&1&1&1&1&....1\end{bmatrix}$ Where ... | Let $\mathbf{1}_n$ denote the $n\times n$ matrix with all entries equal to one and $E_n$ denote the $n\times n$ identity matrix. Then
$$
A_n = b\mathbf 1_n + (a-b)E_n = b\mathbf 1_n -nbE_n
$$
and
$$
B_n = A_n + \frac{1}{n}\mathbf 1_n = \left(b+\frac{1}{n}\right)\mathbf 1_n - nbE_n.
$$
In general, for a matrix $X_n=\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2... | Ok you already proved it if $x \in (-\infty ,0]\cup [1,+\infty) $ , it suffice to prove it for $x\in (0,1)$ to conclude : you can write it as ,
$(x^{2}-\frac{x}{2})^{2}+\frac{3x^{2}}{4}-x+\frac{1}{2}=(x^{2}-\frac{x}{2})^{2}+\frac{x^{2}}{4}+\frac{1}{2} (x-1)^{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
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Prove $\Delta=\left|\begin{smallmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\-z+x&-z+y&3z\end{smallmatrix}\right|=3(x+y+z)(xy+yz+zx)$
Prove that$$
\Delta=\begin{vmatrix}
3x&-x+y&-x+z\\
-y+x&3y&-y+z\\
-z+x&-z+y&3z
\end{vmatrix}=3(x+y+z)(xy+yz+zx)
$$
using factor theorem and polynomials.
My Attempt
$$
\begin{matrix}
\color{r... | As you have mentioned, $x+y+z$ is a factor. As the determinant is symmetric, the factor other than $x+y+z$ must also be symmetric. As it is quadratic, it must be in the form $a(x^2+y^2+x^2)+b(xy+yz+zx)$. (We don't have to guess. $a(x^2+y^2+x^2)+b(xy+yz+zx)$ is the general form of symmetric quadratic polynomial in $x$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
| $$\begin{align}\sin(n+1)x&=\sin nx\cos x+\cos nx\sin x\\
\sin(n-1)x&=\sin nx\cos x-\cos nx\sin x\\
\hline\sin(n+1)x+\sin(n-1)x&=2\sin nx\cos x\end{align}$$
Now that we know that
$$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$$
We can go forward to
$$\begin{align}\sin 0x&=0\\
\sin1x&=\sin x\\
\sin2x&=2\sin x\cos x-\sin0x=2\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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number of distinct solution $x\in[0,\pi]$ of the equation satisfy $8\cos x\cos 4x\cos 5x=1$
The number of distinct solution $x\in[0,\pi]$ of the equation which satisfy $8\cos x\cos 4x\cos 5x=1$
Try: $$4\bigg[\cos(6x)+\cos(4x)\bigg]\cos 4x=1$$
$$2\bigg[\cos(10x)+\cos(2x)+1+\cos (8x)\bigg]=1$$
So $$\bigg[\cos(10x)+\cos... | Let $\cos{x}=t$.
Thus, we need to solve
$$8t(8t^4-8t^2+1)(16t^5-20t^3+5t)=1$$ or
$$(8t^3+4t^2-4t-1)(8t^3-4t^2-4t+1)(16t^4-16t^2+1)=0,$$
which gives $10$ roots on $[-1,1]$.
We can just solve this equation.
An interesting root: $t=\cos\frac{2\pi}{7}$ or $t=\cos15^{\circ}.$
Indeed, $$16t^4-16t^2+1=0$$ gives
$$1-16\cos^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Definite integration that results in inverse trigonometric functions I try to evaluate this integrat
$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx$$
It seems simple.
$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx=\arctan(\sqrt{3})-\arctan(1)$$
My question is what exact number it is?
Should it be $\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12... | $$
\arctan\sqrt3 - \arctan 1 = \frac \pi 3 - \frac \pi 4 = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4-3)\pi}{12} = \frac \pi {12}.
$$
One of a number of ways to see that this need not involve any of the "nonprincipal" values of the arctangent is this:
$$
\text{If } 1 \le x \le \sqrt 3 \text{ then } \frac 1 2 \ge \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Powers for the sum of three squares Say $n$ can be represented as a sum of three non-zero squares. (i.e. $n = a^2 +b^2+ c^2$, for some $n,a,b,c \in \mathbb{N}$)
Is it possible that every natural power of $n$ is also a sum of three non-zero
squares? (i.e. $n^k = x^2+y^2+z^2$ for $x,y,z,k \in \mathbb{N}$)
| Yes, this is possible. Take $n=3$. Clearly $3=1^2+1^2+1^2$. Now any power of $3$ is also the sum of three squares by the Three-Squares Theorem, because $3^k$ is not of the form $4^n(8m+7)$. This may include zero summands, though. Otherwise one has to give a direct solution. If $n=3^k$ with $k=2m+1$, then $3^k=(3^m)^2+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Prove that $3^{2n-1} + 2^{n+1}$ is always a multiple of $7$. I'm trying to prove the following statement:
$P(n) = 3^{2n-1} + 2^{n+1}$ is always a multiple of $7$ $\forall n\geq1$.
I want to use induction, so the base case is $P(1) = 7$ so that's okay.
Now I need to prove that if $P(n)$ is true then $P(n+1)$ is true.
S... | If
$$2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot 18d$$
Then
$$2 \cdot 3^{2n+1} + 2 \cdot 2^{n+2} = 7\cdot 18d - 7 \cdot 2^{n+2}$$
And we conclude
$$3^{2n+1} + 2^{n+2} = \frac{7(18d - 2^{n+2})}{2} = 7(9d - 2^{n+1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
| $$
(2k-1)!! = \frac{(2k-1)!}{2^{k-1}(k-1)!}
$$
$$
(2k)!! = 2^k k!
$$
$$
\frac{(2k-1)!!}{(2k)!!} = \frac{1}{2^{2k-1}}\frac{(2k-1)!}{k!(k-1)!}=\frac{1}{2^{2k-1}}\frac12\frac{2k(2k-1)!}{k!k(k-1)!}=\frac{1}{2^{2k}}\frac{(2k)!}{k!k!}
= \frac{1}{2^{2k}}\binom{2k}{k}
$$
We can express that central binomial coefficient in term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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In this approximation of $\pi$, do you need to know $\pi$ make these calculations? $\pi = 2n\dfrac{\cos (x)}{\sin (x)+1}$
where $x = 90°\dfrac{n-2}{n}$
and $n \to \infty$
A high school student came up with the idea for this approximation of $\pi$, and I helped develop it. It is based on an inscribed polygon. Is this a ... | With some trigonometric identities we can rewrite this as $\pi=\lim_{n\to\infty}n\tan\frac{\pi}{n}$, which doesn't require knowledge of $\pi$ provided we consider values of $n$ that are powers of $2$. The insight is that $\tan 2x =\frac{2\tan x}{1-\tan^2 x}$ implies $$\tan 2x\tan^2 x +2\tan x - \tan 2x=0,\,\tan x =\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Approximate $\sqrt{(1.02)^3+(1.97)^3}$ using differential So $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$
So the differential will be $$\sqrt{1^3+2^3}+\frac{3(1)^2}{2\sqrt{1^3+2^3}}\Delta x+\frac{3(2)^2}{2\sqrt{1^3+2^3}}\Delta y=3+\frac{1}{2}\Delta x+2\Delta y$$
is $\Delta x=0.02 \text{ and } \Delta y=-0.0... | For $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$
The correct choice is $$\Delta x=0.02 \text{ and } \Delta y=-0.03$$
Note that you have changed your $x$ from $1$ to $1.02$, which means your change in $x$ is
$ \Delta x=0.02$
Similarly for change in $y$ from $2$ to $1.97$ we have $\Delta y=-0.03$
| {
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"url": "https://math.stackexchange.com/questions/2686554",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
| The equation $2^x=m 3^y7^z+1$ can have infinitely many solutions;
We use Fermat little theorem; for primes p and q we may write:
$2^{p-1} ≡ 1 \ mod p$ ⇒ $2^{ k p}≡ 1 \mod p$
$2^{q-1} ≡ 1 \ mod q $⇒ $2^{k q}≡ 1 \mod q$
If $p=7 $ we have:
$2^{7-1}=2^6 ≡ 1 \ mod 7$ ⇒ $2^{6 k_1}≡ 1 \mod 7$
For $p=3$ we have:
$2^{3-1}=2^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2688972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Solving $x^4-15x^2-10x+ 24 = 0$ using Ferrari’s method
Ferrari’s method for solving a quartic equation
$$x^4-15x^2-10x+ 24 = 0$$
begins by writing:$$x^4= 15x^2+ 10x-24$$and then adding a term of
the form:$$-2bx^2+b^2$$to both sides.
(a) Explain why this is good idea and what it accomplishes.
(b) Use $b= 7$ to find t... | $$(x^2-7)^2 = (x+5)^2$$
$$(x^2-7)^2-(x+5)^2=0$$
$$(x^2-7-x-5)(x^2-7+x+5)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Studying the extrema of $f(x,y) = x^4 + y^4 -2(x-y)^2$
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x,y) = x^4 + y^4 - 2(x-y)^2$. Study its extrema.
So here was my approach.
We have $$\frac{\partial f}{\partial x}(x,y) = 4(x^3 -x + y),\frac{\partial f}{\partial y}(x,y)= 4(y^3 -y + x) $$
I have to find $(x... | The matrix $H(0,0)$ is negative semidefinite since
$$\left\langle \begin{bmatrix}-4 & 4 \\ 4 & -4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}, \begin{bmatrix}x \\ y\end{bmatrix}\right\rangle = -4(x-y)^2 \le 0$$
with equality iff $x = y$.
This is a necessary condition for a local maximum, but not sufficient. Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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How to evaluate this anti-derivative? How to evaluate
$$ \int \frac{1}{ \ln x} \ \mathrm{d} x, $$
where $\ln x$ denotes the natural logarithm of $x$?
My effort:
We note that
$$ \int \frac{1}{ \ln x} \ \mathrm{d} x = \int \frac{x}{x \ln x} \ \mathrm{d} x = \int x \frac{ \mathrm{d} }{ \mathrm{d} x } \left( \ln \ln... | I = $\large\int\frac{1}{ln(x)}dx$
let ln(x) = u
$\,e^u = x$
$\,dx = e^udu$
I = $\,\int \frac{e^u}{u}du$
expanding e$^u$,
I=$\,\int\frac{1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\frac{u^5}{5!}..............}{u}du$
I = $\,\int\frac{1}{u}+1+\frac{u}{2!}+\frac{u^2}{3!}+\frac{u^3}{4!}+\frac{u^4}{5!}...........du$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Algebra precalculus factorization I must factor this out:$(a+1)(a+3)(a+5)(a+7)+15$
I know that it might have an artifice or technique to transform those factors, but i cannot find any idea on how to do it.
PS.The majority of my questions are about factoring polynomials. Is there an app or site on internet where I can ... | take $$ b = a+4 $$
$$ (b-3)(b-1)(b+1)(b+3) + 15 $$
$$ (b-1)(b+1)(b-3)(b+3) + 15 $$
$$ (b^2-1)(b^2-9) + 15 $$
$$ b^4 - 10 b^2 + 9 + 15 $$
$$ b^4 - 10 b^2 + 24 $$
take $u = b^2$
$$ u^2 - 10 u + 24 $$
$$ (u - 4)(u - 6) $$
$$ (b^2 - 4)(b^2 - 6) $$
$$ (a^2 + 8 a + 16 - 4) (a^2 + 8 a + 16 - 6) $$
$$ (a^2 + 8 a + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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To compute the sum using 2011 th roots of unity
Question:
Let $x$ be a complex number such that $x^{2011}=1$ and $x\neq1$ then, compute the sum
$$S=\dfrac{x^2}{x-1}+\dfrac{x^4}{x^2-1}+\dfrac{x^6}{x^3-1}+\dots+\dfrac{x^{4020}}{x^{2010}-1}$$
My attempt:
$$x=(1)^{\frac{1}{2011}}\implies x=1,e^{\frac{2\pi i}{2011}},e^{\f... | Supposing that $\zeta$ is a primitive $n$th root of unity we consider
$$f(z) = \frac{z^2}{z-1} \frac{n z^{n-1}}{z^n-1}
= \frac{1}{z-1} \frac{n z^{n+1}}{z^n-1}
= \frac{n}{z-1} \left(z + \frac{z}{z^n-1}\right)
\\ = n \left(1 + \frac{1}{z-1} \right)
\left(1 + \frac{1}{z^n-1}\right).$$
We seek $$S_n = \sum_{k=1}^{n-1} \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2702334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding integer values of $n$
Finding integer values of $n$ for which the equation
$$x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0$$
has at least one integer solution.
Try: let $\alpha,\beta,\gamma$ be the roots of the equation. Then
$\alpha+\beta+\gamma=-(n+1)$ and $\alpha\beta+\beta\gamma+\gamma\alpha=1-2n$ and $\alpha\bet... | Suppose $x,n$ are integers such that
$$x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\tag{eq1}$$
Suppose first that $n$ is odd.
Then, reducing mod $2$, we can replace $n$ by $1$, so
\begin{align*}
&x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\\[4pt]
\implies\;&x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\;(\text{mod}\;2)\\[4pt]
\implies\;&x^3-2x^2-x -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Validity of proving identities by showing LHS-RHS =0 or using reversible steps? When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:
Method 1:
Show... | This question works both from LHS $\rightarrow$ RHS and RHS $\rightarrow$ LHS to prove it.
| {
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"url": "https://math.stackexchange.com/questions/2705645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Differentiate $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}
Find $\dfrac{\mathrm dy}{\mathrm dx}$ if
$y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
I can solve it as follows:
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\frac{d}{dx}\Big(2x\sqrt{1-x^2}\... | The idea is very good. However, the notation $\sin^{-1}t$ usually doesn't mean “the set of all angles $\varphi$ such that $\sin\varphi=t$”, but rather
$\sin^{-1}t$ denotes the unique angle $\phi\in[-\pi/2,\pi/2]$ such that $\sin\varphi=t$.
The fact that $\sin^{-1}$ is not the inverse function of the sine is the reaso... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Minimum of product of three trigonometric function
If $x,y,z$ are real numbers with $x\geq y\geq z \geq 15^\circ$ and $x+y+z=90^\circ$, then find the range of $\cos x\cos y\cos z$.
I tried Jensen inequality and AM-GM inequality:
$$3(\cos x\cos y\cos z)^{\frac{1}{3}}\leq \cos x+\cos y+\cos z\leq 3 \cos\bigg(\frac{x+y+... | Note that $x\geqslant y\geqslant z\geqslant15°$ and $x + y + z = 90°$ implies $2y + z\leqslant90°$ and $15°\leqslant z\leqslant30°$, then\begin{align*}
\cos x\cos y\cos z &= \frac12(\cos(x + y) + \cos(x - y))\cos z\\
&= \frac12(\cos(90° - z) + \cos(90° - 2y - z))\cos z\\
&= \frac12(\sin z + \sin(2y + z))\cos z \geqslan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Graph of the equation The graph of the equation $$ x^2y^3=(2x+3y)^5$$
is same as $$x=-y$$ Why is this so?
I understand it satisfies the equation, but is there a way to derive it? Don't the other roots count?
The graph
Edit:
Something I just realized.
$$ dy/dx=y/x$$ for the equations $$x^my^n=(x+y)^{m+n}$$
we have the s... | $$
\begin{aligned}
(2x + 3y)^5 - x^2y^3 &= (2(x+y)+y)^5 - x^2y^3\\
&= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 40(x+y)y^3 + 10y^4) \\
&\qquad - y^3(x^2 - y^2)\\
&= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 40(x+y)y^3 + 10y^4) \\
&\qquad - y^3(x+y)(x+y-2y)\\
&= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 39(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction decomposition with a 3rd degree numerator I have the following function to decompose using PFD:
$H(x) = \frac{x^3 + 4x^2 - 11x - 48}{x^3 + 6x^2 + 3x - 10}$
The poles are $1$, $-2$ and $-5$ so I tried to do it this way: finding $a$, $b$ and $c$ such as:
$H(x) = \frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{... | Since the degree of the numerator is $\geq$ to the degree of the denominator you should first divide them. Hence
$$\frac{x^3 + 4x^2 - 11x - 48}{x^3 + 6x^2 + 3x - 10}=1-\frac{2x^2 + 14x + 38}{x^3 + 6x^2 + 3x - 10}=1+\frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{x+5}$$
for some real constants $a$, $b$, $c$ (the same that you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2711885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s
| The real root $a$ of $x^3+x+1$ belongs to the interval $\left(-1,-\frac{1}{2}\right)$ and in order to check that
$$ \sqrt[3]{(3a^2-2a+2)(3a^2+2a)}=1-a^2 \tag{1}$$
holds it is enough to check that
$$ (3a^2-2a+2)(3a^2+2a)=1-3a^2+3a^4+a^6 \tag{2}$$
holds, or that
$$ (1+a+a^3)(-1+5a+a^3) = 0 \tag{3} $$
holds, but that is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Why $\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$? Why does an integral $$\int \frac{dx}{1-x^2}$$ with the limitless (undefined) interval equal to $$\frac 12\ln\frac{1+x}{1-x},$$ yet an integral $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}$$ ... | $$\int \frac{dx}{1-x^2} = $$
$$\frac {1}{2} \int \bigg(\frac{1}{1+x} + \frac{1}{1-x} \bigg) dx =$$
$$\frac 12\ln\frac{1+x}{1-x}$$Upon Evaluation at the upper and lower limits, for $$a=\sin (5\pi /{12})$$
Note that $$-a=\sin (-5\pi /{12})$$
Thus $$\frac 12\ln\frac{1+x}{1-x}\bigg|_{-a}^a = $$
$$\frac 12\bigg(\ln\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How to evaluate $\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$? Evaluate the definite integral:
$$\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$$
I think the series involving $x$ converges because $x\in[0,1]$, but I cannot form an expression for the series. If I let
$$... | Claim 1: For $k\geq 1$, we have that
\begin{align}
\int^1_0 \frac{1-x}{\log x}x^{2^k}\ dx = -\log \frac{2^k+2}{2^k+1}.
\end{align}
Claim 2: We have
\begin{align}
\prod^\infty_{k=0}\left( 1+\frac{1}{2^k+1}\right) = 3
\end{align}
Using the claims, we have the series
\begin{align}
-\sum^\infty_{k=0} \log \left(\frac{2^k+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)... | Let $\sin a=x\implies a=\sin^{-1}x$ and $\sin b=\sqrt{x}\implies b=\sin^{-1}\sqrt{x}$
$$
y=\sin^{-1}\Big[ \sin a|\cos b|-\sin b.|\cos a| \Big]\\
$$
$0<x<1\implies 0<\sin^{-1}x=a<\frac{\pi}{2}\implies |\cos a|=\cos a$ and
$0<x<1\implies 0<\sqrt{x}<1\implies0<\sin^{-1}\sqrt{x}=b<\frac{\pi}{2}\implies|\cos b|=\cos b$
$$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integration by Substitution of Fraction involving e
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$
The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e... | You have made a mistake:
Note that $$\int\frac{1}{u^2-4u}du \ne \int u^{-2}-\frac{1}{4}u^{-1}du$$
By the way
$$\int\frac{2dx}{e^{2x}+4} = \int\frac{2e^{2x}dx}{e^{2x}(e^{2x}+4)} = $$
$$\int\frac{du}{u(u+4)}$$
Where $$u= e^{2x}$$
is a short way to go.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the greatest common divisor of $f(x)=2x^3+2x^2+x+4$ and $g(x)=x^4+3x^3+4x^2+3x$ Find the greatest common divisor of $f(x)=2x^3+2x^2+x+4$ and $g(x)=x^4+3x^3+4x^2+3x$ in $\mathbb{Z}_{11}[x]$
| We have to perform the Euclidean algorithm modulo $11$:
$$\begin{align}
h_1(x)&:=g(x)+(5x-1)f(x)\equiv 7x^2+7,\\
h_2(x)&:=f(x)+(6x+6)h_1(x)\equiv -x+2,\\
h_3(x)&:=h_1(x)+(7x+3)h_2(x)\equiv 2.\\
\end{align}$$ Hence, in $\mathbb{Z}_{11}[x]$,
$$\gcd(g(x),f(x))=\gcd(h_1(x),f(x))=\gcd(h_1(x),h_2(x))=\gcd(h_3(x),h_2(x))=1.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4... | Alt. hint - by AM-GM:
$$3600 = (2x)^2 + (3y)^2 \ge 2 \sqrt{(2x)^2 \cdot (3y)^2} = 12 \,|x|\,|y| = 3\,A$$
Equality holds iff $\,2x=3y\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Evaluate $\int_0^\infty \frac{x^2}{x^4 + 6x^2 + 13}dx$ In the context of the residue theorem, I have this integral to evaluate. The function is even, and $|\int_0^\pi\frac{R^2e^{2i\theta}iRe^{i\theta}}{R^4e^{4i\theta}+6R^2e^{2i\theta} + 13}d\theta| \leq \int_0^\pi2\frac{R^3}{R^4}d\theta \to 0$, so the problem is to fin... | You need both square-roots in the upper half-plane. The difference between those two square-roots is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$ $I=\displaystyle\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$
Attempt:
$\dfrac{ A_0 }{ x }+\dfrac{ A_1 }{ x +1 }+\dfrac{ A_2 }{ x + 2 }...+\dfrac{ A_m }{ x +m } =\dfrac{1}{x(x+1)(x+2)(x+3)...(x+m)}$
But things got very messy.
I also thought that 1)a... | $f(x) = \frac {1}{x(x+1)(x+2)\cdots(x+m)} = \frac {A_0}{x} + \cdots + \frac {A_n}{x+m}\\
\lim_\limits{x\to -n} (x-n)f(x) = A_n\\
A_n = \prod_\limits {i\ne n} \frac 1{i-n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
The difference between the radii of the largest and smallest circles having centres on the circumference of $x^2+2x+y^2+4y=4$ Owing to restriction of 150 characters in the title section I include the latter part of the problem here below in bold and italics
Also given that both the circles(largest and smallest) pass th... | Draw the line passing through $(a,b)$ and the centre of the circle, meeting the circle at $P$ and $Q$. $P$ and $Q$ are the nearest and the farthest points on the circle from $(a,b)$. So they are the centres of the smallest and the largest circles. If $d$ is the distance between $(a,b)$ and the centre of the given circl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Pre-calc optimization of a rational function A student I am tutoring is in a pre-calc class and they just had a test on rational functions. One of the questions my student said they got was to minimize the area of the outer rectangle, given that the area of the inner rectangle is a fixed 48, and the distance to the out... | Explanation for the surface area:
\begin{align}
\text{Surface area of outer rectangle} &= (x+2)(y+3) \\
&= \left(\frac{48}{y} + 2 \right)(y+3) \\
&= 48 + 2y + \frac{144}{y} + 6 \\
&= \color{red}{54} + 2y + \frac{144}{y}
\end{align}
Though OP wrote $56$ instead, it didn't affect the minimiser (the value of $y$ which min... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Is there a closed form of this sequence? The sequence is $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{2}{1}$ , $\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ , $\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ , ...
and I need to find n th..
Here's my approach..
I bound them with who has same s... | You can break the sequence in groups of $N$ terms with $N=1,2,3,\dots$ of the form
$$\frac{k}{N+1-k}\quad \text {for $k=1,2,\dots ,N$}.$$
So the first $N-1$ groups have $\sum_{i=1}^{N-1}i=\binom{N}{2}$ terms.
Now, given $n$, there is a unique $N$ such that $\binom{N}{2}<n\leq \binom{N+1}{2}$, which means that the $n$-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2727395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the power series representation and interval of convergence for the function Find the power series representation and interval of convergence for the function $ \ f(x)=\frac{3}{2+x} \ $
Answer:
$f(x)=\frac{3}{x+2}=\frac{3}{2} (1+\frac{x}{2})^{-1}=\frac{3}{2} (1-\frac{x}{2}+(\frac{x}{2})^2-(\frac{x}{2})^3+............ | Rewrite the series as $\; \sum_{n\ge 0}(-1)^n 16^n x^{4n+5}= x^5\sum_{n\ge 0}(-1)^n 16^n(x^4)^n$.
The radius of convergence of the power series (setting $x^4=u$) is given by
$$\frac1R=\limsup_n\,\bigl(16^n\bigr)^{\tfrac1n}=16,$$
so the given power series converges if
$$ x^4<\frac1{16}\iff |x|<\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2729915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that
$$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$
My attempt:
$a$, $b$, $c$ are in arithmetic ... | Another way to look at it is just compute the tangents in terms of the sides of the triangle $\triangle ABC$.
From Law of Cosines we have that $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, then follows
$$2\sin^2\dfrac{A}{2}=1-\cos A=\dfrac{a^2-(b-c)^2}{2bc}=\dfrac{(a+b-c)(a-b+c)}{2bc}=\dfrac{2(p-c)(p-b)}{bc}$$
where $2p=a+b+c$. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Asymptotic expansion of complete elliptic integral of third kind Is there a way to compute the expansion of the complete elliptic integral of third kind
$\Pi(n,k)=\int_0^{\pi/2} \frac{d\varphi}{(1-n\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}}$
for
$\Pi(1+\epsilon,1-\epsilon)\ , \qquad \epsilon\to 0$,
and if so, what is it?... | Assuming $\epsilon > 0$, I get
\begin{equation}
\Pi (1 + \epsilon, 1- \epsilon ) = - \frac{1}{\epsilon \sqrt{3}} \left( \textrm{arcsinh} ( \tfrac{1}{\sqrt{2}} ) + \frac{i \pi}{2} \right) - \frac{1}{4} \log \epsilon + \frac{1}{36} \left( 6 - 8 \sqrt{3} \: \textrm{arcsinh} ( \tfrac{1}{\sqrt{2}} ) + 27 \log 2 - 4 i \pi \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with r... | A very simple way:
$$ \begin{eqnarray*}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}\right)&=&\sum_{n\geq 1}\int_{0}^{1}\left( 2x^{2n-1}-x^{2n-2}-x^{2n}\right)\,dx\\&=&\int_{0}^{1}\sum_{n\geq 1}x^{2n-2}(2x-1-x^2)\,dx\\&=&\int_{0}^{1}\frac{2x-1-x^2}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x-1}{x+1}\,dx=\left[x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Matrix in canonical form of an orthogonal transformation
Let:
$$A = \frac 12 \begin{pmatrix} 1 & -1 & -1 &-1 \\ 1 & 1 & 1 &-1 \\ 1 & -1 & 1 & 1\\ 1 &1 &-1 & 1\end{pmatrix} $$
Prove there exists an orthogonal transformation $\phi$ of the
Euclidean Vector Space $\mathbb{R^4}$ such that $A=M_{Bs}(\phi)$ Then
find... | Turns out you just need to get the eigenvalues, which will be complex and come in conjugate pairs. For each eigenvalue, find an eigenvector, make real vectors of the real and imaginary parts, if necessary use Gram-Schmidt to make that pair of vectors orthonormal. For this one, it was only necessary to adjust the length... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Number of non-negative integral solutions This is a problem I've seen a couple times around here, but I couldn't find one quite like this.
Say we have ten variables, $a, b,$ and $c_1, c_2, c_3,\dots, c_8$. How many non-negative integral solutions are there to the following problem such that $a\leq 5$ and $b\geq 5$:
$$a... | We wish to solve the equation
$$a + b + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 + c_7 + c_8 = 100 \tag{1}$$
in the nonnegative integers subject to the constraints $a \leq 5$ and $b \geq 5$.
Since $b \geq 5$, $b' = b - 5$ is a nonnegative integer. Substituting $b' + 5$ for $b$ in equation 1 yields
\begin{align*}
a + b' + 5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in ad... | Hint - telescope:
$$
\begin{align}
&\quad \frac{\big(\color{red}{(6-5)\cdot}(6+5)\big)\;(6^2+5^2)(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{\color{red}{(6-5)\cdot}3^{1024}} \\
&= \frac{\big((6^2-5^2)(6^2+5^2)\big)\;(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\
&= \frac{\big((6^4-5^4)(6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Given $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest? The graph below shows a portion of the curve defined by the quartic polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest?
$(A)$ $P(-1)$
$(B)$ The product of the zeros of $P$
$(C)$ The product of the non... | We have, since the non-real root are conjugates $$P(x)=(x-(m+ni))(x-(m-ni))(x-a)(x-b)$$ or
$$P(x)=x^4-(a+b-2m)x^3+(ab+2m(a+b)+m^2+n^2)x^2-(2abm+(a+b)(m^2+n^2))x+ab(m^2+n^2)$$ It is not hard to compare $m^2+n^2$ which corresponds to $(C)$ above with the other quantities $(A),(B),(D),(E)$.
For example, calculate a quar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the coefficients $p_0,p_1,p_2,q_1,q_2,q_3$ of Padé approximation Determine the Padé approximation of degree $5$ with $ n =2 $ and $ m= 3$ for $f(x) = e^{-x}$.
Suppose $r$ is a rational function of degree $N$.$$
r(x) = \frac{p(x)}{q(x)} = \frac{p_0 +p_1x + \cdots + p_nx^n}{q_0 +q_1x+ \cdots + q_mx^m}.
$$
The Pa... | $\newenvironment{gmatrix}{\left\lgroup\begin{matrix}}{\end{matrix}\right\rgroup}$In general, to find the Padé approximation $r(x) = \dfrac{\sum\limits_{k = 0}^n p_k x^k}{\sum\limits_{k = 0}^m q_k x^k}$ for $f(x) = \sum\limits_{k = 1}^\infty a_k x^k$, it reduces to find the solution to$$
\begin{gmatrix}
a_0 &&&\\
a_1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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An inequality involving three consecutive primes Can you provide a proof or a counterexample to the following claim :
Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .
I have tested this claim up to $10^{10}$ .
For $p>5$ we get $\pi... | Here is a stronger form of the above inequality. The prime number theorem implies that for every $\epsilon$, there exists a prime $p_{\epsilon}$ such that for all $p > p_{\epsilon}$, we have $1 - \epsilon < \frac{p}{q}, \frac{q}{r} < 1$. Hence for all positive real $a$ and all primes greater than some $p_{\epsilon_a}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Geometric intuition of composition of hyperbolic and inverse hyperbolic trig functions We can evaluate $\cos(\arcsin x)$ with simple geometric intuition. Write $y = \cos(\arcsin x)$. Letting $\theta = \arcsin x$, $y = \cos \theta$ corresponds to a right triangle with angle $\theta$, adjacent side length $y$, and hypote... | Solved it myself. We can mirror the argument above, but using the hyperbolic identity $\cosh^2 - \sinh^2 = 1$ in place of Pythagoras.
Write $y = \cosh(\mathrm{arcsinh}\ x) = \cosh(\theta)$. Then $y = \cosh \theta, x = \sinh \theta$, so $y^2 - x^2 = 1 \implies y = \sqrt{1 + x^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find variance of $X$ given its cumulative distribution function A rv $X$ has the cumulative distribution function
$$ F(x) = \begin{cases} 0 \; \; \; \; x<1 \\ \frac{x^2-2x+2}{2} \; \; \; \; 1 \leq x < 2 \\ 1 \; \; \; \; x \geq 2 \end{cases} $$
Calculate the variance of $X$
attempt
First since $F'(x) = f(x)$, then
$$ f... | The cdf 'jumps' at $x=1$, so $P(X=1)=0.5$.
$$\begin{align} Var(X) &= E(X^2) - E(X)^2 = \int\limits_1^2 (x^3 - x^2) dx + 0.5 - \left( \int\limits_1^2 (x^2-x )dx+0.5\right)^2 \\
& \approx 1.4167+0.5-(0.83333+0.5)^2 \\
& \approx 0,1389 \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding joint prob mass function
Suppose two balls are chosen from a box containing 3 white, 2 red and
5 blue balls. Let X = the number of white balls chosen and Y = the
number of blue balls chosen. Find the joint pmf of X and Y.
Attempt
We want $p_{XY}(x,y) = P(X = x \cap Y =y ) $
Let's count the number of ways to s... | Numerator should be white's, blue's and red's:
$$ p_{XY}(x,y) = \frac{ {3 \choose x}{5 \choose y } {10-5-3 \choose 2-x-y}}{{10 \choose 2 } }$$
Our answers are the same if
$${7 \choose 7-x} {5 \choose 5-y} = {10-5-3 \choose 2-x-y}$$
which is not the case (Note: I guess you meant ${5 \choose 5-y}$ and not ${5 \choose 5-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ ... | Write $x:=u+1$. Then
$$(x+1)^n=(u+2)^n=\sum_{k=0}^n{n\choose k}2^{n-k}u^k\ .$$
Dividing by $(x-1)^3=u^3$ we get the remainder
$$r={n\choose0}2^n+{n\choose 1}2^{n-1}u+{n\choose 2}2^{n-2}u^2=2^n+n2^{n-1}(x-1)+n(n-1)2^{n-3}(x-1)^2\ .$$
This can be written in the form
$$r=2^{n-3}\bigl((n^2-n)x^2+(-2n^2+6n)x+(n^2-5n+8)\bigr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$f(x) = \frac{x^3}{6}+\frac{1}{2x}$, $\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$ $$f(x) = \frac{x^3}{6}+\frac{1}{2x}$$
$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$
Let's start by deriving the function, we have
$$f'(x) = \dfrac{x^4-1}{2x^2}$$
Hence we get
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx =... | Note that
$$f(x) = \frac{x^3}{6}+\frac{1}{2x}\implies f'(x)=\frac {x^2}{2}-\frac1{2x^2}$$
$$ \implies I=\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx =\int^{3}_{1} \sqrt{(\frac {x^2}{2}+\frac 1{2x^2})^2}\, dx $$
$$\implies I=\left . \frac {x^3}6-\frac 1 {2x}\right |^3_1 = \frac {14}3$$
And that you can't write this
$$\int^{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Negative variance.
I've an urn with 5 red and 3 white balls. I'm randomly drawing a ball
from the urn each round (without replacement). Let $X$ be the number
of balls taken from the urn until the first white ball is chosen. I'm
asked for the pmf, expected value and variance of $X.$
Since we can at most take 5 r... | Continuing from comments:
Letting $X$ be the random variable counting the number of balls pulled until pulling the first white ball, we try to find the probability distribution.
To do this, temporarily label each of the balls uniquely. For the $k$'th ball to be the first white ball, that implies that the first $k-1$ b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2755035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\cos^4 20^{\circ}+\cos^4 40^{\circ}+\cos^4 60^{\circ}+\cos^4 80^{\circ}$ Compute $$\cos^4 20^{\circ}+\cos^4 40^{\circ}+\cos^4 60^{\circ}+\cos^4 80^{\circ}$$
Suppose that this is a scenario where calculator isn't allowed.
I want to say that this expression has something to do with this equation $\cos^2 20^{\c... | We can use the "inverse" rule : just like how we compute $\cos n \theta$ from powers of $\cos \theta$, we may compute powers of $\cos \theta$ using values of $\cos n\theta$, which is what we should desire here, since the sequence $20n$ has some regularity(remainder on division by $90$, values for which we know $\sin \b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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throwing a two-dice 7 with in total 6 dice We have been stuck on the following questions regarding the game Qwixx:
What is the probability that there is a combination of two dice which add up to exactly 7 when throwing with a total of 6 dice once?
Thank you in advance!
| Note: what follows has several cases each of which is somewhat error prone. The calculation should be checked carefully.
The pairs that add to $7$ are $(1,6), (2,5), (3,4)$. We'll say that a pair is "hit" if at least one member of the pair comes up in your six tosses. We'll work from the complement...that is we wil... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that a constraint is not convex
Is the following problem convex?
$$\begin{array}{ll} \text{minimize} & x+y^2\\ \text{subject to} & x + y \leq 2\\ & \frac{x+y^2}{x^2+2} \leq 3\\ & x^2 = 1\end{array}$$
The problem is, in fact, not convex, but I'm having problems showing that the constraint $$\frac{x+y^2}{x^2+2}... | $$\frac{x+y^2}{x^2+2}\le 3$$
$$y^2 \le 3x^2-x+6$$
Hence if $y$ is fixed, and $x$ is large enough in magnitude, then this construct will be satisfied.
Let $y=3$, then $(2,3)$ and $(-2,3)$ both satisfies the inequality, but $(0,3)$ doesn't.
Remark:
For this question, we know that $x=1$ or $x=-1$.
If $x=1$, $y \le 1$ an... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Almost done solving IVP using Laplace transform .. Need advice/guidance I have the following IVP:
$$\begin{cases}y''-2y'+5y = -8e^{-t},\\ \ \\ y(0) = 2\\ \ \\ y'(0) = 12\end{cases}$$
I will show my steps so far.
After taking the transform of each term I get:
$$s^2Y(s) - 2sY(s) + 5Y(s) - 2s - 8 = \frac {-8}{s+1}$$
I s... | You have $s^2-2s+5=(s-1)^2+4$. So the first term looks like
$$
\frac {3s+5}{(s-1)^2+4}
=\frac {3(s-1)+8}{(s-1)^2+4}
=\frac {3(s-1)}{(s-1)^2+4}
+\frac {8}{(s-1)^2+4}.
$$
Since $\frac{s}{s^2+4}$ is the transform of $\cos 2t$ and $\frac2{s^2+4}$ of $\sin 2t$, with the shifting rule you get
$$
\mathcal L^{-1}\left[\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2759716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Could someone elaborate on this algebraic transformation? I have a solution for an exercise and one part of it is not clear to me:
$
\frac{\frac{1}{\sqrt{x - 1}} - 1}{x - 2} = \frac{1 - \sqrt{x - 1}}{(x - 2)\sqrt{x - 1}}
$
Could anyone explain, please, how the result was obtained?
The full solution is this:
$
\frac{\fr... | You simply multiply the top and bottom by $\sqrt{x-1}$:
$$\frac{\frac{1}{\sqrt{x-1}}-1}{x-2}=\frac{\Bigl(\frac{1}{\sqrt{x-1}}-1\Bigr)\sqrt{x-1}}{(x-2)\sqrt{x-1}}=\frac{1-\sqrt{x-1}}{(x-2)\sqrt{x-1}}$$
Edit: This is valid as long as the thing you are multiplying top and bottom by is not $0$. In this case, $\sqrt{x-1} \n... | {
"language": "en",
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Minimum of the maximum Let $a, b,c,d,e,f,g$ seven nonnegative real numbers with the sum equal to 1. Let considered the sums $ a+b+c$, $b+c+d$, $c+d+e$, $d+e+f$, $e+f+g$. What is the minimum of the maximum of those sums when we consider all the possibilities?I have no idea. One hint, ple... | The language is obfuscating but:
Suppose $a=b=c=0$ then $d+e+g = 1$. $d+e+g$ and that is the maximum of the sums. But that's pretty high maximum. What if we don't have three elements equal to $0$. Then any sum of three numbers will not include a fourth possible none zero term. So the sum of any three can not be as... | {
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"source": "stackexchange",
"question_score": "1",
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Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$ I need to tell if a following series convergent and if so, find it's value:
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}
$$
I've noticed that
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _... | Yes, you are correct that the sum telescopes quite nicely. The partial sum formula is$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left[\left(1+\frac 12+\cdots+\frac 1m\right)-\left(\frac 14+\frac 15+\cdots+\frac 1{m+3}\right)\right]$$Notice how anything past $\frac 14$ in the first sum is automatically canceled from t... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Residue of $\frac{1}{z(e^{z}-1)}$ I was trying to find the residue of $\dfrac {1}{z (e^z - 1)}$. I have written the Taylor series for $e^z$ which is $1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!}..$ Thus, for $e^z - 1$ I have series of the form $z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}..$. But now I am stuck as I have a problem... |
We obtain
\begin{align*}
\color{blue}{\frac{1}{z\left(e^z-1\right)}}&=\frac{1}{z(z+\frac{1}{2}z^2+O(z^3))}\tag{1}\\
&=\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}\tag{2}\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{z^2}-\frac{1}{2z}+O(1)}
\end{align*}
and we conclude ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $\dfrac{a+b+c}{b-a}$
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
Find the minimum value of $\dfrac{a+b+c}{b-a}$
Attempt:
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the express... | Actually, you are on the right track! Now let's complete it with only elementary calculus.
$$\frac{a+b+\frac{b^2}{4a}}{b-a}=\frac{4a^2+4ab+b^2}{4a(b-a)}=\frac{(2a+b)^2}{4a(b-a)}$$
Knowing that $b>a$, let's set $b=ka, k>1$.
$$\frac{(2a+b)^2}{4a(b-a)}=\frac{(2+k)^2a^2}{4(k-1)a^2}=\frac{(2+k)^2}{4(k-1)}$$
Now you just nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How do you set up this Tricky u-sub? Tricky u-sub. Can you point me in the right direction?
$$\int{{x^3}\sqrt{5-2x^2}}dx$$
$u = 5-2x^2$
$ du = -4x dx$
Obviously, this does not match fully.
I tried breaking up the $x^3 = x*x^2$
and I continued with:
$u=5-2x^2$
$2x^2 = 5-u$
$x^2=\frac{5-u}{2}$
But, I can't see how to ... | Another approach:
We can write the integrand as
\begin{align*}
{x^3}\sqrt{5-2x^2}&=-\tfrac12x(5-2x^2)\sqrt{5-2x^2}+\tfrac52x\sqrt{5-2x^2}&&\text{then}\\[5pt]
\int{x^3}\sqrt{5-2x^2}dx&=-\frac12\int x\left(5-2x^2\right)^{3/2}dx+\frac52\int x\left(5-2x^2\right)^{1/2}dx\\[5pt]
&=\color{red}{\frac1{20}\left(5-2x^2\right)-\f... | {
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"url": "https://math.stackexchange.com/questions/2772173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20... | $$\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2 = \binom{2}{2} \underbrace{-\binom{3}{2} + \binom{4}{2}}_{\binom{3}{1}} \underbrace{-\binom{5}{2} + \binom{6}{2}}_{\binom{5}{1}}- \ldots \underbrace{-\binom{21}{2} + \binom{22}{2}}_{\binom{21}{1}}$$
$$ = 1 + 3 + 5 + \ldots + 21 = 121$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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How to prove that $(1 + \frac{1}{n})^{\sqrt{n(n+1)}} < e$? I've been trying to solve the following problem:
Show that $\ln{(k+1)} - \ln{k} = \ln{(1 + \frac{1}{k})} \leq \frac{1}{\sqrt{k(k+1)}}$
EDIT: the title was inaccurate, my bad.
So what we have to prove is that the upper limit of $(1 + \frac{1}{n})^{\sqrt{n(n+1)}... | More information...
As $n \to +\infty$, we have
\begin{align*}
\sqrt{n(n+1)} &= n + \frac{1}{2} - \frac{1}{8n} + O(n^{-2})
\\
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n} - \frac{1}{2n^2} +
\frac{1}{3n^3} + O(n^{-4})
\\
\sqrt{n(n+1)}\log\left(1+\frac{1}{n}\right) &=
1-\frac{1}{24n^2}+O(n^{-3})
\end{align*}
Therefore, ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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finding solution to simultaneous equations I have three equations
$$
\begin{align*}
c_1 & = \frac{a}{1+a+b} + \frac{a}{a+b}+\frac{a}{1+a}\\
c_2 & = \frac{b}{1+a+b} + \frac{b}{a+b}+\frac{b}{1+b}\\
c_3 & = \frac{1}{1+a+b} + \frac{1}{1+a}+\frac{1}{1+b}\\
\end{align*}$$
where $c_1,c_2,c_3$ are known constant.
I ... | Before undertaking calculations which are guaranteed to be tedious, sometimes it helps to graph the equations involved. In this case if we turn these into equations in $x$ and $y$ by letting $x=a$ and $y=b$. Using the graphing site desmos.com and letting $c_1,\,c_2,\,c_3$ be 'sliders' we see that the places where the t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ Compute the limit
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n
$$
How can this be done? The best I could do was rewrite the limit as
$$
\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \r... | Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781063",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Minimum value of $\frac{pr}{q^2}$ in quadratic equation If $y^4-2y^2+4+3\cos(py^2+qy+r)=0$ has $2$ solutions and $p,q,r\in(2,5)$. Then minimum of $\displaystyle \frac{pr}{q^2}$ is.
solution I try
$$-3\cos(py^2+qy+r) =(y^2-1)^2+3\geq 3$$
$$\cos(py^2+qy+r)\leq -1\implies \cos(py^2+qy+r)=-1$$
$$py^2+qy+r=(4n+1)\pi\implies... | The condition for
$$
y^4-2y^2+4+3\sigma(y)=0
$$
with $-1 \le \sigma(y) \le 1$
to have real roots is that $\sigma(y) = -1$ or $p y^2+q y+ r = \pi$ but then $y = y_0 = \pm 1$ hence
$$
\min \frac{pr}{q^2}\\
\mbox{subjected to}\\
p y^2_0+q y_0 + r = \pi \\
(p,q,r)\in (2,5)^3
$$
and the minimum value is $\frac{1}{2(2-\pi)}$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration using "method of judicious guessing" My calculus-book gives an example of integration using the method of judicious guessing. But I do not intuit the method very well.
QUESTION: How does the derivative of $f_{mn}(x)$ "suggest that we try" $I=Px^4\left(\log {x}\right)^2 +Qx^4\log{x}+Rx^4+C$? Where does this... | Note that$$
f_{m, n}'(x) = mx^{m - 1} (\ln x)^n + nx^{m - 1} (\ln x)^{n - 1} = P_{m, n}(x, \ln x),
$$
where $P_{m, n}(x, y) = mx^{m - 1} y^n + nx^{m - 1} y^{n - 1}$ is a polynomial of $x$ and $y$, and $\deg_x P_{m, n} = m - 1$, $\deg_y P_{m, n} = n$.
First, the primitive function of $f_{m_0, n_0}(x) = x^{m_0} (\ln x)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Take $\epsilon>0$, I want to find $\delta>0$ such that:
$$\lVert (x-1,y-2)\rVert <\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert... | On bounding $\frac{1}{|x+y|}$. If you take $\delta<1/2$, then
$$
|x-1|<\delta
$$
and
$$
|y-2|<\delta
$$
yields
$$
1/2<x
$$
and
$$
3/2<y
$$
and so
$$
|x+y|=x+y>2
$$
which is the part you are worried about.
The intuition behind this: Note that you are worried about $|x+y|$ getting very small, but of course it wont s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$... | $(a^2-1)(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)$
$=a^{2m}+a^{2m-2}+a^{2m-4}+\cdots+a^4+a^2-(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)$
$=a^{2m}+a^{2m-2}-a^{2m-2}+a^{2m-4}-a^{2m-4}+\cdots+a^4-a^4+a^2-a^2-1$
$=a^{2m}-1$
$\Rightarrow \dfrac{a^{2m}-1}{a+1}=\dfrac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the second derivative of y when y is given in terms of x. (Solved) y = $(2+1/x)^3$
Find y''.
Explanation for help:
The correct answer is y'' = $6/x^3(2+1/x)(2+2/x)$.
So far, I'm at y' = $-3/x^2(2+1/x)^2$.
I'm not sure how to get from y' to y'', though.
Could someone please show how to solve for y'' starting from y... | Starting with $y' = −3/x^2(2+1/x)^2$, we solve for $y''$,
$$
y'' = \frac{\mathrm{d}}{\mathrm{d}x}\Big[-3/x^2\Big](2+1/x)^2 + \frac{\mathrm{d}}{\mathrm{d}x}\Big[(2+1/x)^2\Big](-3/x^2).
$$
We now simplify and then factor.
\begin{array}{rl}
y'' = & 6/x^3\cdot(2+1/x)^2 + (-2/x^2)\cdot(2+1/x)\cdot(-3/x^2) \\
= & 6/x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown of... | we prove $\sqrt11$ is irrational by contradiction.
let $\sqrt11$ is rational,$\exists a,b$ such that $(a,b)=1$ and
$\sqrt{11}=\frac{a}{b}$
$\implies 11=\frac{a^2}{b^2}$
$a^2=11.b^2 \implies 11\mid a^2 $ and $11$ is prime
therefore, $11\mid a, \exists a_1$ such that $a=11a_1$
$$a^2=(11a_1)^2=121a_1^2=11b^2$$
$$\impli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Solving Recurrence Relation for Series Solution of an ODE I am trying to solve the below problem:
Assume $y = \sum_{n=0}^{\infty}a_nx^n$ is a solution to $(x-1)y''-(x-3)y'-y=0$. Find $a_n$.
I took both derivatives of $y$, plug them into the equation, modify the indices until each series has the same $x^n$, and take ter... | Following along up until the difference equation for the coefficients leads one to
$$a_{n+2} = \frac{(n+3) \, a_{n+1} - a_{n}}{n+2} \hspace{5mm} n \geq 0.$$
Now, for a few values of $n$ it is discovered that
\begin{align}
a_{2} &= \frac{3 a_{1} - a_{0}}{2!} \\
a_{3} &= \frac{4 a_{2} - a_{1}}{3} = \frac{10 a_{1} - 4 a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=... | We can eliminate the square root with an Euler substitution of
$$t = \sqrt{x^2-x} - x \implies x = -\frac{t^2}{1+2t} \implies dx = -\frac{2t(1+t)}{(1+2t)^2} \, dt$$
Then
$$\begin{align*}
I &= \int \sqrt{x^2-x} \, dx \\[1ex]
&= -2 \int \left(t-\frac{t^2}{1+2t}\right) \frac{t(1+t)}{(1+2t)^2} \, dt \\[1ex]
&= -2 \int \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2790831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Evaluation of $\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad $ without using L'Hospital rule My aim is to evaluate the following limit
$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad$
which evaluates to ... | We have
$$
\sin(\arcsin(x)-\arctan(x))=\frac{x(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}
$$
and
$$
\lim_{x\to 0}\left(\frac{\arcsin(x)-\arctan(x)}{x^3}\right)\equiv \lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)}\right)
$$
then
$$
\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)} = \frac{x^3(1-\sqrt{1-x^2})}{x^2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$ Problem
Evaluate $$\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$$
My solution
Notice that $$\lim_{n \to \infty}\frac{n+n^2+n^3+\cdots +n^n}{n^n}=\lim_{n \to \infty}\frac{n(n^n-1... | We shall prove that
$$
\frac{1^n+2^n+\cdots+n^n}{n^n}\to \frac{\mathrm{e}}{\mathrm{e}-1}\tag{$\star$}
$$
First of all, $\log (1-x)<-x$, for all $x\in(0,1)$ and hence
$$
\log\left(1-\frac{k}{n}\right)<-\frac{k}{n}\quad\Longrightarrow\quad
\left(1-\frac{k}{n}\right)^n<\mathrm{e}^{-k}, \quad \text{for all $n>k$}
$$
and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 0
} |
There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$ Find the other side. There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$. Find the equation of
other side.
I first let the sides are $OM$: $y = m_1x$ and $ON:$ $y = m_2x$. Where $m_1m_2= \frac{-2h}{b}... | Put $A(0,0),B(x,0),C(?,?)$. The coordinates of the vertex $C$ are intersection of the perpendiculars to the altitudes passing by $A$ and $B$, so we have after easy calculation $C(p,\frac{p(x-p)}{q})$.
Now one has $$(AC)^2=p^2+\left(\frac{p(x-p)}{q}\right)^2\\(BC)^2=(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2$$ The quotient... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792568",
"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.