Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
The $n-th$ term of a sequence is the LCM of the integers from $1$ to $n$
The $n-th$ term of a sequence is the least common multiple (LCM) of the integers from $1$ to $n$. Which term of the sequence is the first one that is divisible by $100$?
How I'll solve this?
Note: This is a problem from BDMO - $2012$
| For some prime $p$, the greatest $p^n$ that divides the LCM of the integers $1$ through $n$ is the greatest $p^n$ that is less than $n$. For example, for the LCM of the integers $1$ through $50$, the greatest $7^n$ that divides the LCM is $49$ since $7^2=49<50$, but $7^3=343>50$. $7^3$ can not divide the LCM of the int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding segment in a right triangle. Here is the picture of the question:
*
*$ABC$ is a right triangle.
*$m(CBA)=90^\circ$.
*$m(BAD)=2m(DAC)=2\alpha$.
*$D$ is a midpoint of $[BC]$.
*$E$ is a point on $[AD]$.
*$m(BED)=90^\circ$.
*$|DE|=3$.
*What is $|AB|=x$?
Tried lots of things which gi... | This is one approach. I don't pretend it's the nicest.
Let $u = \tan\alpha$. Then
$$
\tan2\alpha = \frac{2u}{1-u^2}
$$
$$
\tan3\alpha = \frac{u(3-u^2)}{1-3u^2}
$$
Since $\tan3\alpha = 2\tan2\alpha$, we have
$$
\frac{4u}{1-u^2} = \frac{u(3-u^2)}{1-3u^2}
$$
$$
(1-u^2)(3-u^2) = 4(1-3u^2)
$$
$$
3-4u^2+u^4 = 4-12u^2
$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculating determinant of 100x100 matrix I was trying to calculate the determinant of 100x100 matrix:
$$A=\begin{bmatrix}5&5&5&...&5&5&-1\\5&5&5&...&5&-1&5\\5&5&5&...&-1&5&5\\ \\5&5&-1&...&5&5&5\\5&-1&5&...&5&5&5\\-1&5&5&...&5&5&5\end{bmatrix}$$
I thought that using LU decomposition would quickly show a pattern, but t... | Let $D_n$ be the determinant of such a matrix that is $n\times n$.
In any one column except the last column, the bottom $n-1$ entries sum to $5n-11$. So add $\frac{-5}{5n-11}$ times each of row 2, row 3, etc. to row 1, and you have
$$\begin{bmatrix}0&0&0&...&0&0&-6\frac{5n-6}{5n-11}\\5&5&5&...&5&-1&5\\5&5&5&...&-1&5&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A Quadrilateral's area given four sides and a diagonal Assume there exists a quadrilateral called ABCD and AB=5cm,BC=13cm,CD=16cm, DA=20cm and diagonal AC=12cm. The exercise now states that I should calculate the area of a quadrilateral. Thank you for your time!!! :)
| By the Law of Cosines:
$$AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos \angle ABC$$
$$\iff \cos \angle ABC=\frac{5^2+13^2-12^2}{2\cdot 5\cdot 13}=\frac{5}{13}$$
Analogously: $$\cos \angle CDA=\frac{16^2+20^2-12^2}{2\cdot 16\cdot 20}=\frac{4}{5}$$
Now, we know that $\angle ABC$ and $\angle CDA$ are both acute, because $\an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
A sequence to achieve $\frac{1}{a_{2016}}$ It is given that $a_ka_{k-1} + a_{k-1}a_{k-2} = 2a_k a_{k-2}$ , $k\geq3$ and $a_1=1$.
We have $S_q= \sum_{k=1}^{q} \frac{1}{a_k} $ and given that $\frac{S_{2q}}{S_{q}}$ is independent of q then $\frac{1}{a_{2016}}$ is = ?
I think no information is given to find $a_2$. How ... | You get some information, I just don't think you get enough (unless I'm misreading the problem).
Given $a_1 = 1, a_2$, we can determine $a_3 = \frac{a_2}{2-a_2}$ and $a_4 = \frac{a_2^2}{3a_2-2a_2^2}$. Based on the conditions imposed on $S_q$, we also have
$$
\frac{S_2}{S_1} = \frac{S_4}{S_2}
$$
$$
1+\frac{1}{a_2} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Laurent series expansion of $f$ Find the Laurent series expansion of $f(z)=\dfrac{1}{2z^2-13z+15}$ about the annulus $\dfrac{3}{2}<|z|<5$.
I did like this :
$f(z)=\dfrac{2}{7}(\dfrac{3}{3-2z}-\dfrac{1}{z-5})$
Then I took $\dfrac{3}{3-2z}=\dfrac{1}{1-\dfrac{2z}{3}}=(1-\dfrac{2z}{3})^{-1}=1+\dfrac{2z}{3}+(\dfrac{2z}{3})... | There is a small error in your partial fraction decomposition, it should be
$$
f(z) = \frac{1}{7}(\dfrac{2}{3-2z}+\dfrac{1}{z-5})
$$
(However, this does not change the result for $a_1/a_2$, see below.)
The main problem is here:
$$
\frac{3}{3-2z}=\frac{1}{1-\frac{2z}{3}}=(1-\frac{2z}{3})^{-1}=1+\frac{2z}{3}+(\frac{2z}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a+b=8$ and $ab+c+d = 23$ and $ad+bc=28$ and $cd=12\;,$ Then $abcd$
If $a+b=8$ and $ab+c+d = 23$ and $ad+bc=28$ and $cd=12\;,$ Then value of
$(1)\;\; a+b+c+d=$
$(2)\;\; ab+bc+cd+da = $
$(3)\;\; abcd=$
My attempt: Let $x=a\;,b$ be the roots of $(x-a)(x-b)=0$ and $x=c\;,d$ be the roots
of $(x-c)(x-d)=0.$
So, $(x-a)(... | This problem actually checks your observational skills of algebraic products.
Now take this:
$(x^2+ax+c)(x^2+bx+d) $
$=x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd $
$=x^4+8x^3+23x^2+28x+12=f(x) $
$12$ is the product of roots so by putting $ x=−1 $ we find $f(−1)=0 $
$⟹(x+1) $ is factor of $ f(x) $
Similarly $(x+2) $ and $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$f(x) - f'(x) = x^3 + 3x^2 + 3x +1; f(9) =?$ Given the following $f(x) - f'(x) = x^3 + 3x^2 + 3x +1$
Calculate $f(9) = ?$
I have tried to play with different number of derivatives. Also tried to solve it by equations.
Maybe there is some geometric meaning that could shade the light ?
I feel it is no complex problem at... | It is most convenient to search $f$ among the third degree polynomials with the leading coefficients $1$. Let $f_0 (x) = x^3 + a x^2 + b x + c$. Then the equation becomes $$(f_0 - f'_0) (x) := x^3 + (a - 3) x^2 + (b - 2a) x + (c - b) = x^3 + 3 x^2 + 3 x + 1.$$ Hence, we have $a = 6$, $b = 15$ and $c = 16$. Then $$f_0 (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
$ \int\frac{\sin(nx) \sin x}{1-\cos x} \,dx$ by elementary methods What is an elementary way to show that for positive integer $n$
$$
\int\frac{\sin(nx) \sin x}{1-\cos x} \,dx= x + \frac{\sin (nx)}{n} + 2 \sum_{k=1}^{n-1}\frac{\sin(kx)}{k}
$$
This cropped up when trying to answer Proving $\int_0^{\pi } f(x) \, \mathrm{... | To prove
$$\sin(nx) \sin(x)= (1-\cos(x)) \left(1 + \cos (nx) + 2 \sum_{k=1}^{n-1}\cos(kx) \right)(**)$$
by induction, first note that the basis ($N=1)$ is easy: For $n=1$ the sum is vacuously zero, and
$$\sin(1\cdot x) \sin(x)= \sin^2 x = 1-\cos^2 x = (1-\cos(x)) \left(1 + \cos (1\cdot x) \right)$$
Now assume that fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that the equation of the tangent at P is $ \frac {xx_1}{a^2} - \frac {yy_1}{b^2} = 1 $ (Hyperbolas)
Question: Point P ($x_1 , y_1$) is on the hyperbola $\frac {x^2}{a^2}$ - $\frac {y^2}{b^2}$ = 1
Prove that the equation of the tangent at P is
$$ \frac {xx_1}{a^2} - \frac {yy_1}{b^2} = 1 $$
What I have attem... | Since $(x_1,y_1)$ lie on hyperbola, they satisfy the equation $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1$
$$ \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2} = 1$$
$$\implies\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the polynomial $P$ of smallest degree with rational coefficients and leading coefficient $1$ such that $ P(49^{1/3}+7^{1/3})=4 $
Find the polynomial $P$ of smallest degree with rational coefficients
and leading coefficient $1$ such that $$ P(49^{1/3}+7^{1/3})=4 $$
(Source:NYSML)
My attempt
Let $$ 49^{1/3}+7^... | A fast way to get the polynomial is to observe that
$$(x+y)^3=x^3+3x^2y+3xy^3+y^3=x^3+y^3+3xy(x+y) (**)$$
Now with your $a=49^{1/3}+7^{1/3}$ setting $x=49^{1/3}, y=7^{1/3}$ you have
$$x^3=49, y^3=7, xy=7$$
Replace those in $(**)$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
constrained stars and bars problem I want to know number of solutions for following equation, where $r_k$'s are non-negative integers, and there is a constraint on $r_k$'s such that $r_1 \geq r_2 \geq \cdots \geq r_K$
\begin{equation}
r_1+r_2+\cdots+r_K=N
\end{equation}
| EDIT: This response answers a slightly different question, where $K$ (the number of summands) is allowed to vary. I will leave it here for posterity, but it doesn't deserve any upvotes.
These are the integer partitions of $N$. There is no closed formula for $p(N)$, the number of such partitions, although there is a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
solution of differential equation $\left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$
The solution of differential equation $\displaystyle \left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}+y=0$
$\bf{My\; Try::}$ Let $\displaystyle \frac{dy}{dx} = t\;,$ Then Diferential equation convert into $t^2-xt+y=0$
So Its solution is gi... | Here's another approach. Differentiating gives $y''\left(2y'-x\right)=0$, so $y''=0$ or $y'=\frac{x}{2}$. The former option gives $y=ax+b$ so $a^2-ax+ax+b=0$ and $b=-a^2$. The latter option gives $y=\tfrac{x^2}{4}+c$ so $\frac{x^2}{4}-\frac{x^2}{2}+\tfrac{x^2}{2}+c=0$, which work iff $c=0$. The solution is $y=ax-a^2$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove that $a^ab^bc^c\geq (\frac{a+b}{2})^{\frac{a+b}{2}} (\frac{b+c}{2})^{\frac{b+c}{2}}(\frac{c+a}{2})^{\frac{c+a}{2}}$
Prove that $$a^ab^bc^c\geq \left(\frac{a+b}{2}\right)^{\frac{a+b}{2}} \left(\frac{b+c}{2}\right)^{\frac{b+c}{2}}\left(\frac{c+a}{2}\right)^{\frac{c+a}{2}}\geq \left(\frac{a+b+c}{3}\right)^{a+b+c}$$... | For $x > 0 \Rightarrow f(x) = x\ln x \Rightarrow f'(x) = 1 + \ln x \Rightarrow f''(x) = \dfrac{1}{x} > 0\Rightarrow f(a)+ f(b) \geq 2f\left(\dfrac{a+b}{2}\right)\Rightarrow a\ln a + b\ln b \geq 2\left(\dfrac{a+b}{2}\right)\ln\left(\dfrac{a+b}{2}\right)\Rightarrow \ln(a^ab^b) \geq \ln\left(\left(\dfrac{a+b}{2}\right)^{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{7}{12}<\ln 2<\frac{5}{6}$ using real analysis I studying in Real Analysis 2, but I have no idea how to solve this problem. My guess is to use Mean Value Theorem or a similar theorem? Could any one help me?
Thanks.
| Consider the following inequalities from Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$
as suggested by @labbhattacharjee
$$\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx>0$$
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx>0$$
The integrals are positive because the integrands are positive for $0<x<1$.
Let us evaluate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Is there any way to solve integral of $\sqrt{8-x^{2}}$ without using $\sin$ or $\cos$ formulas? I was thinking about the following integral if I could solve it without using trigonometric formulas. If there is no other way to solve it, could you please explain me why do we replace $x$ with $2\sqrt 2 \sin(t)$? I'm reall... | Let
$$I=\int\sqrt{8-x^2}\ dx\tag 1$$
using integration by parts,
$$I=\sqrt{8-x^2}\int 1\ dx-\int \left(\frac{-2x}{2\sqrt{8-x^2}}\right)\cdot x\ dx$$
$$I=\sqrt{8-x^2}(x)-\int \frac{(8-x^2)-8}{\sqrt{8-x^2}} \ dx$$
$$I=x\sqrt{8-x^2}-\int \left(\sqrt{8-x^2}-\frac{8}{\sqrt{8-x^2}} \right)\ dx$$
$$I=x\sqrt{8-x^2}-\int\sqrt{8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
finding real roots by way of complex I was given
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i... | From $$x^4 + 1 = (x^2+i)(x^2-i)$$ we need to find the roots of each factor. Remember that $i = e^{i\pi/2}$, so $x=e^{i\pi/4}$ is a root of $x^2-i$. Hence, so is its conjugate $e^{-i\pi/4}$. By Euler's formula,
$$
e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{\sqrt{2}}{2} + i\frac{\sqrt 2}{2}.
$$
I will leave you to d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+... | Another way: For $|x|<1$, we have:
$$(x^3+x^4+x^5+...)^3=x^9(1+x+x^2+...)^3=x^9(1-x)^{-1}.$$
Now $(1-x)^{-3}$ is half of the second derivative of $(1-x)^{-1}.$ The second derivative of $(1-x)^{-1}=(1+x+x^2+...)$ is $(1.2+2.3 x+3.4 x^2+4.5 x^3+...). $ Half the co-efficient of $x^3$ in this, which is $(1/2).4.5=10,$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
How many numbers between $1$ and $9999$ have sum of their digits equal to $8$? $16$? How many numbers between $1$ and $9999$ have sum of their digits equal to $8$? $16$?
Can someone tell me if I got the right answers? I solved both cases and I've got $148$ for $8$ and $633$ for $16$. I solved this problem using $x_1+x_... | We treat a number with fewer than four digits as a number with leading zeros. For instance, we regard the number $235$ as $0235$ and the number $8$ as $0008$. Then the number of positive integers between $1$ and $9999$ with digit sum $8$ is equal to the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 8 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Multi-variable Limit Question: Does the following limit exist, if so what does it equal?
$$\lim_{(x, y) \to (0,0)} \frac{x^2 y^5}{2x^4 +3y^{10}}$$
-
Solution 1: The limit DOES NOT exist.
Let $ x=y^{5/2}$
$$\lim_{y \to 0} \frac{(y^{5/2})^2 y^5}{2(y^{5/2})^4 +3y^{10}} = \lim_{y \to 0} \frac{y^{10}}{2y^{10} +3y^{10}} = \... | The first solution is correct, the limit does not exists. The second one is wrong because the "limit"
$$ \frac{0}{0 + 2 \cos^4\theta}$$
is not well-defined if $\theta = \frac\pi 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
the maximum value of $(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$
If $a, b, c, d \in [\frac12, 2]$ and $abcd =1$, find the maximum value of
$$(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a)$$
Thought: $$\begin{split}(a+\frac1b)(b+\frac1c)(c+\frac1d)(d+\frac1a) &= \frac{(ab+1)(bc+1)(cd+1)(da+1)}{abcd} \\ & \stackrel... | I think it can help:
http://www.artofproblemsolving.com/community/c6h621917
See an elim's proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $1280000401$ is Composite I tried to prove $N=1280000401$ as composite using complex cube roots of unity:
we can write $$N=1+400+(128*10^{7})$$ which gives
$$N=1+20^2+20^{7}$$
now if $F(x)=1+x^2+x^7$, $w$ and $w^2$ are roots of $F(x)=0$ where $w=\frac{-1+i\sqrt{3}}{2}$ and $w^2$ its conjugate.
Hence $x^2+x+1... | You can say $1+x^2+x^7=1+x+x^2+(x^7-x)=1+x+x^2+x(x^3-1)(x^3+1)=(1+x+x^2)(1+x(x-1)(x^3+1))$
but your approach is a good one to find this. People have been very clever in finding things to try.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of
$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$
the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but ca... | @user230452 asks for an enlightened way of doing this.
The expression
$$
E = E(a, b, c) = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3
$$
is symmetric in $a, b, c$, and homogeneous of total degree $3$, so it must be an integral combination of the elementary symmetric polynomials in $a, b, c$, of the form
$$
E = x \sigma_{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How many distinct combinations of five marbles could be drawn? A bag contains ten marbles of the same size: 3 are identical green marbles, 2 are identical red marbles, and the other 5 are five distinct colors. If 5 marbles are selected at random, how many distinct combinations of five marbles could be drawn?
(A) 41
(B)... | Let $k$ be the number of green and red marbles chosen, and let $n$ be the number of other marbles chosen.
Then $k+n=5$, and if $g(k)$ is the number of ways to choose the $k$ marbles,
then $g(0)=g(5)=1,\; \;g(1)=g(4)=2,\;\;g(2)=g(3)=3$.
Since there are $\dbinom{5}{n}$ ways to choose the other $n$ marbles,
there are $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof that the sum of the first $n$ odd numbers is $n^2$. Here is what I have so far:
The $n$th odd number is $2n-1$.
So we prove that $1+3+...+(2n-3)+(2n-1)= n^2$.
Separate the last term and you get: $[1+3+...+(2n-3)]+(2n-1)$
$[1+3+...+(2n-3)]$ is the sum of the first $(n-1)$ odd numbers.
Here is where I get stuck. Th... | Another way of looking at is ...
$$\begin{align}&1 + 3 + 5 + {\dots} + 2n - 1 \\
&= (n - (n - 1)) + {\dots} + (n - 4) + (n - 2) + n + (n + 2) + (n + 4) + {\dots} + (n + n - 1)\\
&= n + n +n + {\dots} +n (n times)\\
&= n^{2}\end{align}$$
It can also be proved with the sum of arithmetic progression. The sum of an $A.P.$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 1
} |
How to find Ratio And proportion If $\frac{4x+3y}{4x-3y}=\frac{7}{5}$.Find the Value of the $\frac{2x^2-11y^2}{2x^2+11y^2}$.
Okay I just want hint how to solve this problem.
| Just notice that
$$\frac{4x+3y}{4x-3y}=\frac{7}{5}\qquad\implies\qquad 5(4x+3y)=7(4x-3y)\qquad\implies\qquad 36y=8x$$
So $y=\frac{2}{9}x$, then
$$\frac{2x^2-11y^2}{2x^2+11y^2}=\frac{2x^2-\frac{44}{81}x^2}{2x^2+\frac{44}{81}x^2}=\frac{2\cdot 81-44}{2\cdot 81+44}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \right... | Expand the square root in a taylor series to three terms, to get $1+k/2n^2 -k^2/8 n^4+ k^3/16 n^6,$ then sum.
The sum of the $1$s gives you $n$. The sum of $k/n^2$ gives $n(n+1)/4n^2,$ the sum of $k^2/n^4$ gives you $(1+n)(2+n)/48 n^3,$ and the next term is $O(1/n^2).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$ How do you go about factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$
Which factoring method would be the best?
| Notice that $t^4/4 = (t^2/2)^2$ so let $u = t^2/2$
Then $({16+4t^2+\frac{t^4}{4}})^{1/4} = (16 + 8u + u^2)^{1/4}$
$16 + 8u + u^2 = (4+u)^2$ and $(a^2)^{1/4} = a^{1/2}$
So $({16+4t^2+\frac{t^4}{4}})^{1/4} = (16 + 8u + u^2)^{1/4} = (4 +u)^{1/2} = (4+t^2/2)^{1/2}$
Is that simplified enough?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Primes of the form $x^3+y^3+z^3 - 3xyz$ Do quadruplets $(x,y,z,p)$ of positive integers exist for which $p$ is a prime number and $$x^3+y^3+z^3 = 3xyz + p?$$
I've tried looking for solutions in mathematica for $x,y,z<1000$, without finding any. Unfortunately, looking at the equation $\mod 3$ or $\mod 9$ has yielded no... | We use the identity
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left( \frac{ (x-y)^2 + (y-z)^2 + (z-x)^2 }{2}\right)$$
to try to find some solutions using linear and quadratic equations with the following system:
$$\begin{cases}x+y+z=1\\(x-y)^2+(y-z)^2+(z-x)^2=2p\end{cases}$$
Putting $z=1-x-y$ one gets
$$(x-y)^2 + (y-z)^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.
Prove that for positive integer $n$ we have $169| 3^{3n+3}-26n-27$.
It is easy to show that if we take the expression modulo $13$ we have $3^{n+3}-26n-27 \equiv 1^{n+1}-1 \equiv 1-1 \mod 13 = 0 \mod 13$. How do I prove it is divisible by $13^2$?
| $A(n) = 3^{3n+3}-26n-27$
$A(n+1) = 3^{3n+6}-26(n+1)-27$
$A(n+1) - A(n) = 3^{3n+3}(3^3-1) - 26(n+1)-27 + 26n + 27 = 3^{3n+3} \cdot 26 - 26 = (3^{3n+3} - 1) \cdot 26 = (27^{n+1} - 1) \cdot 26$
From here, it's quite obvious now, since 13 divides both multipliers.
$13 / (27^{n+1} - 1) $ and
$13 / 26 $
He... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Proof for $\forall x\in \mathbb{Z}^+, \exists t \in \mathbb{Z}, 5 \nmid x \to ((x^2= 5t + 1) \vee (x^2 = 5t – 1))$ I am trying to write a proof for the following: If $x$ is a positive integer that is not divisible by $5$, then $x^2$ can be written either as $x^2 = 5t+1$ or $x^2 = 5t−1$ for some integer $t$. So far, I h... | If $5 \not \mid x$, then
*
*$x \equiv 1\mod 5$,
*$x \equiv 2 \mod 5$,
*$x \equiv 3 \mod 5$ or
*$x \equiv 4 \mod 5$.
Now $1^2 \equiv 4^2 \equiv 1\mod 5$ and $2^2 \equiv 3^2 \equiv -1\mod 5$ and therefore $x^2 \equiv 1 \mod 5$ or $x^2 \equiv -1 \mod 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Strengthen inequality It is known that for positive integers $ a, b, c, d $ and $n $, the inequalities $ a + c <n $ and $ \frac ab + \frac cd <1 $ hold. Prove that
$$ \frac ab + \frac cd <1 - \frac {1} {n ^ 3} $$
My work so far:
$ a, b, c, d, n -$ positive integers $\Rightarrow a+c \le n-1$
and $ad+bc \le bd-1.$
| For convenience, let us to think that $a\le c$ and $t=d-c\in\mathbb{N}$.
1. Not hard to see that if $\frac{a}{b}+\frac{c}{c+t}<1$ then
\begin{equation}
\frac{a}{b}+\frac{c}{c+t}\le\frac{a}{\left\lfloor \frac{ac}{t}\right\rfloor+a+1} + \frac{c}{c+t} < 1.
\end{equation}
2. Now, let us fix only $a\le c$ and let $t\ge1$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$ What is the maximum of $\quad 2(a+b)-ab\ $ if we have: $a^2+b^2=8 \ (a,b\ real) $
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow $$
$$ab\le4 $$
On the other hand , from $\ a^2+b^2... | Using differentiation: Let $f(t)=2\sqrt{8+2t}-t$, then $f'(t)=\frac{2}{\sqrt{8+2t}}-1$. Then $f(t)$ increases when $-4<t<-2$ and decreases when $-2<t<4$. Thus $f(t)$ has a maximum at $t=-2$ and $f(-2)=2\sqrt{8-4}+2=6$. Since $|ab|\le 4$, $2(a+b)-ab$ has a maximum $6$.
Alternative method: Let $a=2\sqrt{2}\cos\theta$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides a, b and c. This is an A-level trigonometric problem.
Using the usual notation for a triangle write $\sin^2A$ in terms of the sides $a$, $b$ and $c$.
Answer: $$\frac{(a+b-c)(a-b+c)(a+b+c)(-a+b+c)}{4b^2c^2}$$
The last step of the solution requ... | $$\begin{align}&-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2a^2c^2\\&=-a^4+(2b^2+2c^2)a^2-b^4+2b^2c^2-c^4\\&=-a^4+2(b^2+c^2)a^2-(b^2-c^2)^2\\&=\color{red}{-a^4+2(b^2+c^2)a^2-(b^2+c^2)^2}+(b^2+c^2)^2-(b^2-c^2)^2\\&=\color{red}{-(a^2-(b^2+c^2))^2}+4b^2c^2\\&=(2bc)^2-(a^2-b^2-c^2)^2\end{align}$$
Now use $A^2-B^2=(A+B)(A-B)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find seventh term of the evaluation of $(2a+b^2)^8$ I need to find the seventh term of
$(2a+b^2)^8$
But, at this moment, the only way to solve this I can make up is to distribute the brackets..
Is there any other way to do this?
| See here.
Using the binomial theorem, we can expand $(a+b)^8$ to be
${8 \choose 0}a^8 + {8 \choose 1}a^7b + {8 \choose 2}a^6b^2 + {8 \choose 3}a^5b^3 + {8 \choose 4}a^4b^4 + {8 \choose 5}a^3b^5 + {8 \choose 6}a^2b^6+{8 \choose 7}ab^7 + {8 \choose 8}b^8$
From this we get the seventh term to be ${8 \choose 6}a^2b^6$ or $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Inverse Fourier Transform I've got a problem where I need to find the IFT of
$$F(\omega) = \frac{1 + i\omega}{6-\omega^2+5i\omega}$$
I've been trying to solve it through partial fractions, but that gives us
$$\frac{1 + i\omega}{6-\omega^2+5i\omega} = \frac{1 + i\omega}{(-\omega + 3i)(\omega-2i)} = \frac{i(\omega - i)}... | By using partial fractions we have
\begin{align}
\frac{i\omega +1}{6-\omega^2+5i\omega}&=\frac{i\omega+1}{(i\omega+2)(i\omega+3)}\\
&=\frac{A}{i\omega+3}+\frac{B}{i\omega+2}
\end{align}
Where $A$ and $B$ are constants such that
\begin{align}
A(i\omega+2)+B(i\omega+3)&=i\omega+1\\
(A+B)i\omega+2A+3B&=i\omega+1
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$
Find the value of $$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\pi}{14}$$
$$\sin\frac{\pi}{7}\sin\frac{2\pi}{7}\sin\frac{3\pi}{7}\... | It can be seen as an instance of the well-known identity:
$$ \prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{2n}{2^n} \tag{1} $$
that for $n=14$ gives:
$$ \left[\prod_{k=1}^{6}\sin\left(\frac{k\pi}{14}\right)\right]^2 = \frac{14}{2^{13}} \tag{2}$$
from which:
$$ \prod_{k=1}^{6}\sin\left(\frac{k\pi}{14}\right) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Extracting the divergent part of an integral I want to evaluate the integral
$$ \int_0^1 \frac{2x(x-2)(1-x)}{(1-x)^2 + ax} \, \mathrm{d}x$$
in the limit of small $a$. For $a = 0$ this integral is divergent due to the $1/(1-x)$ pole. The exact expression for this integral is, according to Mathematica, a horrible combina... | The integral can be rewritten explicitely as
$$
-2\int_0^1 \frac{x(x^2-3x+2)}{x^2-(2-a)x+1}\mathrm{d}x=-2 \int_0^1 x\left(1-\frac{(a+1)x-1}{x^2-(2-a)x+1}\right)\mathrm{d}x,
$$
hence it is equal to
$$
-1+2\int_0^1 \frac{(a+1)x^2-x}{x^2-(2-a)x+1}\mathrm{d}x=-1+2(a+1)+2\int_0^1 \frac{(-a^2+a+1)x-(a+1)}{x^2-(2-a)x+1}\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve the non-homogeneous linear recurrence $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$? The problem: $ a_{n+1} - a_n = 2n+3 $, given that $a_0=1$
First I solved the associated homogeneous recurrence and got $a_n = A(1)^n = A$, where A is a constant, but I got stuck solving the rest. My final answer was $a_n=2n... | You got in trouble when you set up $p_n$. Since the forcing term is linear, we expect $p_n$ to be quadratic in $n$; $p_n=d_0n^2+d_1n+d_2$. Now we have
$$\begin{align*}
2n+3&=p_{n+1}-p_n\\
&=d_0(n+1)^2+d_1(n+1)+d_2-d_0n^2-d_1n-d_2\\
&=d_0(2n+1)+d_1\\
&=2d_0n+(d_0+d_1)\;.
\end{align*}$$
This has to hold for all $n$, so w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Suppose $n$ divides $3^n + 4^n$. Show that $7$ divides $n$. Suppose $n \geq 2$ and $n$ is a divisor of $3^n + 4^n$. Prove that $7$ is a divisor of $n$.
My work so far:
I had a hypothesis that if $n| 3^n + 4^n$, then $n = 7^k$ for some $k\in\mathbb{N}$. But this is not necessarily so. Take $n = 7⋅379$, where $3^7 + 4^7 ... | Assume $\enspace n\mid 3^n+4^n$, for some $n\in\mathbb{N}$, $n\geq 2$.
Hence, $$3^n+4^n=nm, \quad for \enspace m\in\mathbb{N}.$$
Since $3^n+4^n$ is odd $\enspace (odd\cdot odd=odd, \enspace even\cdot even=even, \enspace odd+even=odd)$, and since $\enspace n \mid 3^n+4^n$, we know that $n$ is also odd. Thus, we can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\small 2\sin{(2m+1)x} -\sin{(2m-1)x}=\sin{x}\left(1+2\cos{(2mx)}+2\sum\limits_{k=1}^{m}\cos{(2kx)}\right)$? By chance I found:
$$
2\sin(2m+1)x - \sin(2m-1)x=\sin(x)\left(1+2\cos(2mx)+2\sum_{k=1}^{m}\cos(2kx)\right)
$$
Any idea how to prove it?
| it's simple if u know this identity:
$$\sum_{i=0}^n cos(nb) = \frac{sin(\frac{(n+1)b}{2}) \times cos(\frac{nb}{2})}{sin(\frac{b}{2})}$$
I'll proof this identity if u want but first we use this to solve the above problem:
substituting $b = 2x$ and $n=m$ we get:
$$1+2\sum_{k=1}^{m}\cos(2kx) =-1+ 2 \sum_{k=0}^{m}\cos(2kx)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Multiple Nested Radicals $\sqrt{9-2\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}}$
I have no idea how to unnest radicals, can anyone help?
| Observe that
$$(\sqrt{2}-1)^2=2-2\sqrt{2}+1=3-2\sqrt{2}\qquad\implies\qquad\color{blue}{\sqrt{3-2\sqrt{2}}=\sqrt{2}-1}$$
Then,
$$\color{blue}{\sqrt{10+4\sqrt{3-2\sqrt{2}}}}=\sqrt{10+4(\sqrt{2}-1)}=\sqrt{6+4\sqrt{2}}=\sqrt{4+4\sqrt{2}+2}=\color{blue}{\sqrt{(2+\sqrt{2})^2}}$$
So
\begin{align}
\sqrt{23-6\sqrt{10+4\sqrt{3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 1
} |
How to prove that $\binom{n}{1}\binom{n}{2}^2\binom{n}{3}^3\cdots \binom{n}{n}^n \leq \left(\frac{2^n}{n+1}\right)^{\binom{n+1}{2}}$?
How can we prove that $$\binom{n}{1}\binom{n}{2}^2\binom{n}{3}^3\cdots \binom{n}{n}^n \leq \left(\frac{2^n}{n+1}\right)^{\binom{n+1}{2}}$$
$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M\;,}$... | Hint: You are close to the answer. Here is the right way: Apply the AM-GM:
$(a_1+2a_2+\cdots + na_n)^{\binom{n+1}{2}} \geq \binom{n+1}{2}^{\binom{n+1}{2}}a_1a_2^2\cdots a_n^n$, with $a_k = \binom{n}{k}$, and the left side is a popular sum that can be calculated by several methods one of which is using derivative of $(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding the order of $2 \times 2 $ matrix. Finding the order of $2 \times 2 $ matrix.
$$\begin{pmatrix}1&-1\\1&0\end{pmatrix}$$
Is there an easier way to find the order of this matrix?
I have been multiplying the matrix for five times or so, and I still haven't found the order yet.
| Not that long:
$$A^2=\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}\begin{pmatrix}1&\!-1\\1&0\end{pmatrix}=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}$$
$$A^4=\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}\begin{pmatrix}0&\!-1\\1&\!-1\end{pmatrix}=\begin{pmatrix}-1&1\\\!-1&0\end{pmatrix}$$
$$A^6=\begin{pmatrix}0&\!-1\\1&\!-1\end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$ The question is to evaluate the given limit :
$$\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}$$
When I'm trying to evaluate this I'm getting $1$ as my answer but the answer given in the text is $0$. I'm wondering how can this limit be... | We also have
\begin{align*}
\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}&=\lim_{x \to 3} \frac{x - 3}{\sqrt {x - 2} - \sqrt {4 - x}}\cdot\frac{\sqrt{x-2}+\sqrt{4-x}}{\sqrt{x-2}+\sqrt{4-x}}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(x-2)-(4-x)}\\
&=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Induction Proof $k^2 \times 2^k$ I need help on this proof. I am not able to do it after setting m=m+1.
Prove by induction on n that sum of $k^2 \times 2^k$ from $k=1$ to $n$ is equal to $(n^2-2n+3) \times 2^{n+1}-6$
Base case:
Let $k=1$ so L.H.S side is $2$
Let $n=1$ so R.H.S side is $2$
Inductive hypothesis:
Let $n=m... | For the inductive step the computations would be as follows:
$$
\begin{array}{rcl}
\sum_{k = 1}^{m+1} k^{2} 2^{k} & = & (m+1)^{2}2^{m+1} + \sum_{k = 1}^{m} k^{2}2^{k}\\
& = & (m+1)^{2}2^{m+1} + (m^2 - 2m + 3)2^{m+1} - 6\\
& = & 2^{m+1} \left((m+1)^2+m^2 - 2m +3\right) - 6\\
& = & 2^{m+1} (2m^2 + 4) - 6\\
& = & 2^{m+2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{(x,y) \to (0,0)} \frac{\sin(x^3y^2)}{(x^2+y^2)^2}$
Find $$\lim_{(x,y) \to (0,0)} \frac{\sin(x^3y^2)}{(x^2+y^2)^2}$$
I tried multiplying $x^3y^3$ the nominator and denominator, didn't work.
I tried the polar way too, $x = r \cos\phi$, $y=r \sin\phi$,$x^2 + y^2 = r^2$ too. which gives me $\frac{"0"}{"0"}$
... | Switching to polar is a good idea and gives
$$\lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^4}$$
You probably know that $\sin(a)/a$ tends to 1 for $a \to 0$; so:
$$\begin{array}{cc}
\displaystyle \lim_{r \to 0} \frac{\sin\left(r^5 \cos^3t\sin^2t \right)}{r^4} & \displaystyle
= \lim_{r \to 0} \left( \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Derive a polynomials formula by setting $n = 1$, $n = 2$, $n = 3$ to determine the coefficients. Derive a formula for the sum of squares $1^2 + 2^2 + 3^2 + … + n^2$.
Hint: assume the formula is a polynomial of degree 3, i.e. $an^3 + bn^2 + cn + d$, and use the cases of $n=0, n=1, n=2$, and $n=3$ to determine its coeff... | .I'll answer your question, but I sincerely encourage you to have an attempt at your question before you ask a question in this website.
So we assume that the sum of squares up till $n$ is given by a cubic polynomial $an^3+bn^2+cn+d$. What we do is put n=0,1,2,3 on both sides of the equation:
$$
\sum_{i=1}^n i^2 = an^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
differential equation of family of circles passing through origin How do I find the DE of all circles passing through origin? I tried something like this
The family of circles passing through the origin is given by
$$
(x- r \cos \theta )^2 + (y- r\sin \theta)^2 = r^2
$$Differentiating once, we get
$$
... | Let the the circle be of radius $r$. Then,
$$\tag1(x-r\cos\theta)^2+(y-r\sin\theta)^2=r^2$$
$$\tag22(x-r\cos\theta)+2(y-r\sin\theta)y'=0$$
$$2x-2r\cos\theta+2yy'-2r\sin\theta y'=0$$
$$1+yy''+(y')^2-r\sin\theta y''=0$$
$$\tag3r\sin\theta=\frac{1+yy''+(y')^2}{y''}$$
Substitute $(3)$ in $(1)$,
$$(x-r\cos\theta)^2+(y-r\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1694561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$? What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$, that is $1$ and $-1$, two at a time alternating?
| The sequence $a_0=1$, $a_1=1$, $a_2=-1$, $a_3=-1$, $a_4=1$ and so on satisfies the recursion
$$
a_0=1,\quad a_1=1,\qquad a_{n+2}=-a_n
$$
so its characteristic equation is $t^2+1=0$. Thus the general solution is of the form
$$
xi^n+y(-i)^n
$$
The initial conditions tell that $x+y=1$ and $xi-yi=1$, thus
$$
\begin{cases}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
} |
Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$? Is $A$ diagonalizable? Let $A$ be the $n\times n$ matrix with a $1$ in every entry. What are the eigenvalues of $A$ and a basis for each eigenspace? Is $A$ diagonalizable?
Having some trouble with this one. I tried using the f... | Here is a matrix $P$ that I made up some time ago. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal and therefore independent. The columns are eigenvectors for your matrix, in the 10 by 10 case. Multiply your matrix of all ones (10 by 10) on the left, this matrix on the right. What happens... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to calculate $\lim_{(x,y) \rightarrow (1,-1)} \frac{x^2 - y^2}{x^3 + y^3}$ $$\lim_{(x,y) \rightarrow (1,-1)} \frac{x^2 - y^2}{x^3 + y^3}$$
By approaching along x = 1 or y = -1 and using L'Hopitals rule, I can establish that if the limit does exist, then it is $\frac{2}{3}$.
This matches the result given by Wolfram ... | \begin{align}
\lim_{(x,y)\to (1,-1)}\frac{x^2-y^2}{x^3+y^3}&=\lim_{(x,y)\to(1,-1)}\frac{(x+y)(x-y)}{(x+y)(x^2-xy+y^2)}\\
&=\lim_{(x,y)\to(1,-1)}\frac{x-y}{x^2-xy+y^2}\\
&=\frac{1-(-1)}{1-1\cdot(-1)+1}\\
&=\frac{2}{3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$
Let $a$ be a constant. Find a formula for $f^{(n)}(x)$ where $f(x) = \dfrac{1}{x^2-a^2}$.
After computing a few derivatives, the derivatives seem to have factorials in them sometimes and other times not. For example, $\dfrac{d^5}{dx^5} \left (\dfrac{1... | $$
{1 \over {x^{\,2} - a^{\,2} }} = {1 \over {\left( {x + a} \right)\left( {x - a} \right)}} = {1 \over {2\,a}}\left( {{1 \over {\left( {x - a} \right)}} - {1 \over {\left( {x + a} \right)}}} \right)
$$
... then it goes easily ..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1700939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \s... | $7cos(2x)-sin(2x)+5=0$
$7cos(2x)/\sqrt{50}-sin(2x)/\sqrt{50}=-5/\sqrt{50}$
Let $\alpha=sin^{-1}(7/\sqrt{50})$
Then
$sin(\alpha)cos(2x)+cos(\alpha)sin(2x)=-5/\sqrt{50}$
$sin(2x+\alpha)=-5/\sqrt{50}$
$2x+\alpha=sin^{-1}(-1/\sqrt{2})$
So
$x=(sin^{-1}(-1/\sqrt{2})-\alpha)/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 8
} |
Geometric series $ar^n$ where $n \ne 1,2,3,4 \cdots$ The probability of rolling a seven on two dice is as follows
$$p=1/6+1/6(5/6)^2+1/6(5/6)^4 + \cdots$$
what is the probability of rolling a $7$?
Is there an advantage to rolling first?
My attempt at a solution; $n=1,2,3,4,5 \cdots$
$$\sum_{n=1}^\infty \frac 1 6 \left ... | You are complicating things, note that :
$$p=\frac{1}{6} [1+(\frac{5}{6})^2+(\frac{5}{6})^4+\cdots ]$$
This is a geometric progression, where $a=1, r=\frac{25}{36}$ The infinite sum of which is:
$$\frac{a}{6(1-r)}=\frac{1}{6}\cdot\frac{36}{36-25}=\color {blue}{\frac{6}{11}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $
Question: Prove $$ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
RHS: $$ \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$
$$ ⇔ \frac{\frac{1}{\sin^2(x... | HINT:
Writing $\cos x\sin x=u$
$$1-u-2u^2=(1-2u)(1+u)$$
$1-2u=(\sin x-\cos x)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively. Let $a$ and $b$ be the coefficient of $x^3$ in $(1+x+2x^2+3x^3)^4$ and $(1+x+2x^2+3x^3+4x^4)^4$ respectively.Find $(a-b).$
I tried to factorize $(1+x+2x^2+3x^3)$ and $(1+x+2x^2+3x^3+4x^4)$ into product of two ... | Here is a slightly different variation of the theme. If we introduce the polynomial
\begin{align*}
p(x)=1+x+2x^2+3x^3
\end{align*}
and write $[x^n]$ to denote the coefficient of $x^n$ of a polynomial,
we observe
\begin{align*}
&[x^3]\left(p(x)\right)^4-[x^3]\left(p(x)+4x^4\right)^4\\
&\quad=[x^3]\left(p(x)\right)^4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Relation between two sequences or summations Let us define two sequences
\begin{equation}
G_{n}=\sum_{i=1}^n a^{-i}t_{i-1}
\end{equation}
and
\begin{equation}
g_{n}=\sum_{i=1}^n t_{i-1}
\end{equation}
where $a$ is an integer and $t_n$ is an another sequence.
Can we find a relation between $G_n$ and $g_n$ ?
| Here are relationships of $G_n$ and $g_n$ in terms of the delta operator $\Delta$ and of generating functions which could be helpful.
We define the generating functions
\begin{align*}
G(z)=\sum_{n\geq 1}G_{n}z^n\qquad\text{and}\qquad
g(z)=\sum_{n\geq 1}g_{n}z^n
\end{align*}
and show
The following is valid
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem with showing convergence or divergence of $\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$ Determine if the following series is absolutely convergent, conditionally convergent or divergent.
$$\sum_{n=2}^\infty\frac{(-1)^{n+1}}{3n+4}$$
So I showed that$\sum_{n=2}^\infty\bigl|\frac{(-1)^{n+1}}{3n+4}\bigr|$ is divergent... | The alternating test is OK.
Now, there is no contradition with your second point, because
$$
\frac{-1}{3n+4} \le \frac{(-1)^{n+1}}{3n+4} \le \frac{1}{3n+4}
$$ yields
$$
-\infty \le \sum_{n\geq1}\frac{(-1)^{n+1}}{3n+4} \le +\infty
$$ which does not exclude convergence of the initial series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The number of possible options to get colored balls I have a bag with 5 white balls and 3 blue balls. We getting out 3 balls (without returning).
How many possible ways is to get at least 1 white ball?
My question is why this is not the answer:
$${{5}\choose{1}}*{7 \choose 2}$$
| The number of ways of selecting $k$ of the five white balls and $3 - k$ of the three blue balls is
$$\binom{5}{k}\binom{3}{3 - k}$$
Thus, the number of ways of selecting at least one white ball is
$$\binom{5}{1}\binom{3}{2} + \binom{5}{2}\binom{3}{1} + \binom{5}{3}\binom{3}{0}$$
Alternatively, we could find the same ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
determinant of 3x3 matrix algebra not matching CAS Find
$$\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}$$
we will expand along first row
$$ \begin{aligned}
\begin{vmatrix}
... | $$ \begin{aligned}
\begin{vmatrix}
i & 2 & -1
\\ 3 & 1+i & 2
\\ -2i & 1 & 4-i \end{vmatrix}
&=i \begin{vmatrix} 1+i & 2 \\ 1 & 4-i \end{vmatrix}
-2 \begin{vmatrix} 3 & 2 \\ -2i& 4-i \end{vmatrix}
-1 \begin{vmatrix} 3 & 1+i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$ Here is the expression:
$$2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$$
The exercise is to evaluate it.
In my text book the answer is $0$
I tried to factor the expression, but it got me nowhere.
| Consider $a^3+b^3=(a+b)^3-3ab(a+b)$ and $a^2+b^2=(a+b)^2-2ab$. For $a=\sin^2x$ and $b=\cos^2x$ we have
$$
2(\sin^6x+\cos^6x)-3(\sin^4x+\cos^4x)+1=
2(a+b)^3-6ab(a+b)-3(a+b)^2+6ab+1
$$
However, $a+b=\sin^2x+\cos^2x=1$, so the expression simplifies to
$$
2-6ab-3+6ab+1=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$
I tried by subtracting $I_1$ and $I_2$
$$I_1-I_2=\int_{0}^{1}\left(... | Factoring the integrand gives
$$
\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x
=\int_0^1\left(1-x^{50}\right)^{100}\,\mathrm{d}x
-\int_0^1\left(1-x^{50}\right)^{100}x^{50}\,\mathrm{d}x
$$
Integration by parts gives
$$
\int_0^1\left(1-x^{50}\right)^{101}\,\mathrm{d}x
=101\cdot50\int_0^1\left(1-x^{50}\right)^{100}x^{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Definition of modulo division Could anyone tell me why
$3/7\equiv 9 \mod12$? How is this operation defined and why sometimes modulo division is impossible, like in the case of $3/4 \mod 12$?
See here for more details, I'm refering to page 95.
I can't grasp why $3/7\equiv 9 \mod12$. I divide $3$ by $7$, what I get is $3... | Think of it like this:
$$\frac{3}{7} = 3 \cdot \frac{1}{7} = 3 \cdot 7^{-1}.$$
So then
\begin{align}
\frac{3}{7} \mod 12 &= (3 \cdot 7^{-1}) \mod 12\\
&= (3 \mod 12) \cdot (7^{-1} \mod 12).
\end{align}
$3 \mod 12$ is just 3. What's $7^{-1} \mod 12$? In other words, find $x$ so that $7x \equiv 1 \mod 12$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $ without L'hopital Please help me to solve this limit without using L'Hôpital's rule. I don't know what other method can't be used to solve this limit.
$$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $$
| Use Taylor series
$$\cos(x) \approx 1 - \frac{x^2}{2}$$ thence
$$\frac{1 - \cos(x)}{x\cos(x)} = \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x\cdot\left(1 - \frac{x^2}{2}\right)} = \frac{x^2}{2x - x^3} = \frac{x^2}{x^2\left(\frac{2}{x} - x\right)} = \frac{1}{\frac{2}{x}} = \frac{x}{2}$$
And since $x\to 0$ the limit is
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
} |
How to prove that: if $q= b+d$, then $p = a+c$? Let $a,b,c,d,p$, and $q$ be natural numbers such that $ad-bc = 1$ and $\frac{a}{b} > \frac{p}{q} > \frac{c}{d}$.
How to prove that: if $q= b+d$, then $p = a+c$? Is there a simple way?
| Quite an easy problem. while it is informed that all the variables in the problem is natural numbers so, $p = a+c$ is equivalent to $a+c-1 < p <a+c+1$, that is to say $\frac{a+c-1}{b+d}<\frac{c}{d}$ and $\frac{a+c+1}{b+d}>\frac{a}{b}$.in fact:
$$
ad - d<bc
$$
$$
ad+cd -d <bc+cd
$$
$$
\frac{a+c-1}{b+d}<\frac{c}{d}
$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Inequality $(x-1)(y-1)(z-1)\geq 8$ provided that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1$ How can I prove that if $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1,$$ then $(x-1)(y-1)(z-1) \geq 8$?
Edit:
$x,y,z \in \mathbb R_{>0} $
Thanks
| $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$$
$$xyz=xy+yz+zx$$
$$AM \ge GM \Rightarrow (x+y+z)\left (\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)\ge 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{\frac 1{xyz}}=9 \Rightarrow$$
$$\Rightarrow (x+y+z)\cdot 1 \ge 9$$
Now $$(x-1)(y-1)(z-1)=(xy-x-y+1)(z-1)=$$
$$=(xyz-xz-yz-xy)+(x+y+z)-1=0+(x+y+z)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can you find a series whose sum is $.... = (a) *\sqrt{\frac{1}{1-n}}$ I discovered here that $ \frac{1}{n-1}= (n-1)^-1$ is the sum om a geometric series:
$$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$$
can you tell me if $(\sqrt{1-n})^-1$ $$\sqrt{\frac{1}{1-n}}$$
can be considered the result/sum of any ... | $\sum_{n=m}^{\infty} (a_{n}-a_{n+1})
=a_m
$,
so
$\begin{array}\\
\frac1{\sqrt{m}}
&=\sum_{n=m}^{\infty} (\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}})\\
&=\sum_{n=m}^{\infty} (\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}})\frac{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}{\frac1{\sqrt{n}}+\frac1{\sqrt{n+1}}}\\
&=\sum_{n=m}^{\infty} \frac{\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$
My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac... | In the same spirit as Olivier Oloa's answer.
By Taylor $$\tan(\frac{\pi}{4}+x))=1+2 x+2 x^2+\frac{8 x^3}{3}+\frac{10 x^4}{3}+\frac{64 x^5}{15}+O\left(x^6\right)$$ $$\log\Big(\tan(\frac{\pi}{4}+x)) \Big)=2 x+\frac{4 x^3}{3}+\frac{4 x^5}{3}+O\left(x^6\right)$$ $$\frac 1x\log\Big(\tan(\frac{\pi}{4}+x)) \Big)=2+\frac{4 x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solution of the first order differential equation $y'=\frac{x(x^2+y^2)^2}{4y}$. I m stuck in finding the solution of the following differential equation
$$y'=\frac{x(x^2+y^2)^2}{4y}.$$
Please give me some hint.
| $\frac{4yy'}{(x^2+y^2)^2}=x$
then define $x^2+y^2=h^2 \Rightarrow 2hh'=2yy'+2x \Rightarrow y'=\frac{hh'-x}{y}$
then by substituting we get: $\frac{4(hh'-x)}{h^4}=x \Rightarrow4(hh'-x)=xh^4 \Rightarrow \frac{4hh'}{4+h^4}=x$
now define $g=h^2$ and solve the problem: $\Rightarrow g'=2hh'$
$$\frac{2g}{4+g^2}=x \Rightarro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the interval of convergence for $\sum_{n=1}^∞ sin(\frac{1}{n}) tan(\frac{1}{n})x^n$ Find the interval of convergence for:
$$\sum_{n=1}^{\infty} \sin(\frac{1}{n})\tan(\frac{1}{n})x^n$$
I have no idea how to do this question. Any help would be greatly appreciated
| Here $a_n=\sin(\frac{1}{n})\tan(\frac{1}{n})x^n$
So
\begin{align*}
\Big|\frac{a_{n+1}}{a_n}\Big|&=\Big|\frac{\sin\Big(\frac{1}{n+1}\Big)\tan\Big(\frac{1}{n+1}\Big)x^{n+1}}{\sin\Big(\frac{1}{n}\Big)\tan\Big(\frac{1}{n}\Big)x^n}\Big|\\
&=\Big|\frac{\sin\Big(\frac{1}{n+1}\Big)}{\frac{1}{n+1}}\frac{\tan\Big(\frac{1}{n+1}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1718200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find two different generalized inverse of the given matrix Definition:
For a given matrix $A_{m\times n}$, a matrix $G_{n\times m}$ is said to be a generalized inverse of $A$, if it satisfies
$$AGA=A.$$
Question:
Find two different generalized inverse of the given matrix
$$\begin{pmatrix}
1 & 0 &-1 & 2\\2 & 0 &-2 & 4... | The matrix and its Moore-Penrose pseudoinverse are
$$
\mathbf{A} =
\left[
\begin{array}{rrrr}
1 & 0 & -1 & 2 \\
2 & 0 & -2 & 4 \\
-1 & 1 & 1 & 3 \\
-2 & 2 & 2 & 6 \\
\end{array}
\right],
\qquad
\mathbf{A}^{\dagger} = \frac{1}{40}
\left[
\begin{array}{rrrr}
8 & 16 & -5 & -10 \\
-2 & -4 & 3 & 6 \\
-8 & -16 & 5 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors i... | Just manipulate the Wallis product for $\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Maximum value of $f(x) = \cos x \left( \sin x + \sqrt {\sin^2x +\sin^2a}\right)$ Can we find maximum value of $$f(x) = \cos x \left( \sin x + \sqrt
{\sin^2x +\sin^2a}\right)$$
where '$a$' is a given constant.
Using derivatives makes calculation too complicated.
| the first derivative is given by $$f'(x)=-\sin \left( x \right) \left( \sin \left( x \right) +\sqrt { \left(
\sin \left( x \right) \right) ^{2}+ \left( \sin \left( a \right)
\right) ^{2}} \right) +\cos \left( x \right) \left( \cos \left( x
\right) +{\frac {\sin \left( x \right) \cos \left( x \right) }{\sqrt
{ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1723636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Show that $||X||$ is a norm in $\mathbb{R}^2$ Let $a,b \in \mathbb{R}$ fixed numbers and the following norm in $\mathbb{R}^2$:
$$ ||x|| = \sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}} \qquad x=(x_1,x_2)\in\mathbb{R}^2$$
Prove that it is, in fact, a norm.
Taking the rules of a norm, I get the following:
*
*$||x||\geq0$:... | Notice that $x+y=(x_1+y_1,x_2+y_2)$ so we have that $$||x+y||=\sqrt{\frac{(x_1+y_1)^2}{a^2}+\frac{(x_2+y_2)^2}{b^2}}$$ is the correct first step. You then need to manipulate it to show that it is greater than $$||x||+||y||=\sqrt{\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}}+\sqrt{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the smallest value of $n$ such that the final digit of $13^n$ is one more than the digit adjacent to it?
What is the smallest value of $n$ such that the final digit of $13^n$ is one more than the digit adjacent to it?
If you have a computer, it is easy to check that the answer to this question is $14$, but I ... | I suppose you want to find $n$ such that $13^n$ is of the form $*(x)(x+1)$ for some $x\in\{0,\cdots,9\}$.
This is equivalent to finding $n$ such that $13^n\equiv 11x+1\pmod{100}.$
Then notice that $\phi(100)=\phi(4)\times\phi(25)=2\times20=40,$ so, by Fermat's little theorem, $13^{40}\equiv 1\pmod{100}.$ In fact, $$\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
All 3 digit numbers the sum of whose digits is not greater than 16 My working: Let the digits of the numbers be $x,y,z$ where
\begin{align}1\leq x&\leq 9\\0\leq y,z&\leq 9\\x+y+z &\leq 16 \end{align}
I tried to solve it by making different cases here and using counting but it got really complicated and i couldnt get ve... | Let $k\in\{1,2,...,16\}$ a fixed number. If you count the number of triplets $(x,y,z)$ such that $x+y+z=k$, say $S_k$, then you are searching for $S_0+S_1+...+S_{16}$,
Now, by generating functions method, we use the polinomyal $p(x)=(x+x^2+...+x^9)(1+x+x^2+...+x^9)^2$ and we want to find the coefficient of $x^k$.
But
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$
Find all integer solutions $(x, y)$ of the equation
$$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$
What have done is that:
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so,
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we... | Another answer that does not follow your approach is like this:
Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:
$3x+3y = 2xy$.
Thus $4xy - 6x - 6y = 0$ and hence $(2x-3)(2y-3) = 9$.
Now it remains to find all factorizations of $9$ as a product of two integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of $-1^2-2^2+3^2+4^2-5^2-6^2+\cdots$
Find the sum of $$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2$$
By expanding the given summation,
$$\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2=-1^2-2^2+3^2+4^2-5^2-6^2+\cdots+(4n-1)^2+(4n)^2$$
$$=(3^2-1^2)+(4^2-2^2)+(7^2-5^2)+(8^2-6^2)+\cdots+[(4n-1)^2-(4n-3)^2]+((4n)^2-... | \begin{equation*}
\begin{aligned}
\sum_{k=1}^{4n}(-1)^{\frac{k(k+1)}{2}}k^2 &= \sum_{k=1}^{n}[((4k - 1)^2 - (4k - 3)^2) + ((4k)^2 - (4k - 2)^2)]\\
&= \sum_{k=1}^{n}[2(8k - 4) + 2(8k - 2)] \\
&= 4\sum_{k=1}^{n}[(8k - 3)] \\
&= -12n + 32\sum_{k=1}^{n}k\\
&= -12n + \frac{32n(n+1)}{2} \\
&= 16n^2 + 4n. \\
\end{aligned}
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Calculating exponential limit I've been breaking my mind over this one.
Find the limit.
$\lim\limits_{n \to \infty} (\frac{n^2+3}{n^2+5 n-4})^{2n} $
I know it equals $\frac{1}{e^{10}} $ but can't figure out how to find it.
Help?
| Divide your fraction by $n^2$ and look at the asymptotics for $n\rightarrow \infty$
$$\frac{n^2+3}{n^2+5n-4}=\frac{1+\frac{3}{n^2}}{1+\frac{5}{n}-\frac{4}{n^2}}
= 1 - \frac{5}{n} + O(n^{-2}) \sim 1 - \frac{5}{n}$$
And now for the power:
$$\left(\frac{n^2+3}{n^2+5n-4}\right)^{2n}
\sim \left(1 - \frac{5}{n}\right)^{2n}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Applying root test to sequence $\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots$ The following is an example from Principles of Mathematics, by Rudin. I've been trying to understand the example but haven't quite grasped it because it seems I can solve it differently.
Given the following sequence: $\d... | As Abstraction says, the book uses $n$ for two different things.
$a_n$ alternates between $\frac1{2^m}$ and $\frac1{3^m}$. So the first, third, fifth... terms are powers of $1/2$, and the second,fourth,sixth... are powers of $1/3$.
For the root test, for example, $\frac1{3^6}$ is the twelfth term, so you look at $\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I determine the transformation matrix T of the coordinate transformation from the base E to the base B? In $\mathbb{R}^3$ the canonical basis $E=\left (\mathbf{e_1},\mathbf{e_2},\mathbf{e_3} \right )$ and $B=\left (\mathbf{b_1},\mathbf{b_2},\mathbf{b_3} \right )$ with
$\mathbf{b_1}=(1,2,4)^T$, $\mathbf{b_2}=(0... | Hint:
The matrix
$$ M=
\begin{bmatrix}
1 & 0 & 2\\
2 & -1 & 3\\
4 & 1 & 8
\end{bmatrix}
$$
represents the transformation:
$$
\mathbf{e_1}\to \mathbf{b_1} \qquad \mathbf{e_2}\to \mathbf{b_2} \qquad \mathbf{e_3}\to \mathbf{b_3}
$$
and its inverse:
$$M^{-1}
\begin{bmatrix}
-11 & 2 & 2\\
-4 & 0 & 1\\
6 & -1 & -1
\end{bmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How many ordered pairs $(x,y)$ are there such that ${1\over\sqrt x}+{1\over\sqrt y}={1\over\sqrt {20}}$, where both $x$ and $y$ are positive integers $${1\over \sqrt x}+{1\over \sqrt y}={1\over \sqrt {20}}$$
I could find one ordered pair that satisfies above equation, that is $(80,80)$. But the answer says that there a... | If you manipulate the expression, we get
$$
\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{20}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{20}}{\sqrt{20x}}.
$$
Therefore, $y=\frac{20x}{x-2\sqrt{20x}+20}.$ In order for there to be any hope of $y$ being an integer, $\sqrt{20x}=2\sqrt{5x}$ must be an integer. Therefore, the only option... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1747862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$
Write $4\cos x - 3 \sin x$ in the form $k\sin(x - \alpha)$ where $0 \le \alpha \le 360$.
$4\cos x - 3 \sin x = k\sin(x - \alpha)$
=> $k(\sin x \cos \alpha - \cos x \sin \alpha)$
=> $k\cos \alpha \sin x - \sin \alpha \cos x$
Equati... | $$4\cos x - 3 \sin x = k\sin(x - \alpha)$$
$$= k(\sin x \cos \alpha - \cos x \sin \alpha)$$
$$=k\cos \alpha \sin x - k\sin \alpha \cos x$$
Hence:
$$k\cos \alpha \sin x = - 3 \sin x$$
$$- k\sin \alpha \cos x = 4\cos x$$
so:
$$k\cos \alpha = -3$$
$$\color{red}{k\sin \alpha = - 4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of limit at infinity: $\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$ $$\lim_{x\to\infty} x^2 \sin(\ln(\cos(\frac{\pi}{x})^{1/2}))$$
What I tried was writing $1/x=t$ and making the limit tend to zero and writing the cos term in the form of sin
| Notice that
\begin{align*}
\ln\big(\cos(\pi/x)^{1/2}\big) &= \frac 1 2 \ln \cos \frac{\pi}{x} \\
&= \frac 1 2 \ln \left(1 - \frac 1 2\frac{\pi^2}{x^2} + O\left(\frac 1 {x^4}\right)\right) \\
&= -\frac 1 4 \frac{\pi^2}{x^2} + O \left( \frac 1 {x^4}\right)
\end{align*}
Now do you see how to show that the desired limit is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Considering the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Consider the power series $\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n+1)(2n-1)}$. Find a closed form expression for all x which converge and hence evaluate $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n+1)(2n-1)}$.
Attempt at the solu... | First, by using the standard power series,
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{t^{2n-1}}{2n-1}=\arctan t, \quad |t|<1,\tag1
$$ we multiply $(1)$ by $t$ then we are alowed to integrate termwise:
$$
\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{2n+1}}{(2n-1)(2n+1)}=\int_0^x t\arctan t\: dt, \quad |x|<1,\tag2
$$ then integrating by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$
$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we ... | You get a quadratic equation in terms of "$a$". Using the quadratic formula for $a$,
$$
a=\frac{2x^2+1\pm|2x-1|}{2}
$$
From this, we get $a=x^2+x$ or $a=x^2-x+1$, so
$$
x=\frac{-1\pm \sqrt{1+4a}}{2}\text{ or } x=\frac{1\pm\sqrt{-3+4a}}{2}
$$
Now we must get rid of possiblities:
*
*If $x=\frac{-1-\sqrt{1+4a}}{2}$, it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Why is there a difference of $\frac{-1}{6}$ when integrating $(1-2x)^2$ in two different ways? I integrated $(1-2x)^2$ by expanding the expression first, that is $1 - 4 x + 4 x^2$ and I get $$x - 2 x^2 + \frac{4 x^3}{3} +C$$
When I integrate by substitution I get $$\frac{-(1-2x)^3}{6} + C$$
Expanding $(-(1-2x)^3)/6 +C$... | in $x - 2 x^2 + \frac{4 x^3}{3} +C$ you did right and got a constant C.
in $-\frac{1}{6} + x - 2 x^2 + \frac{4 x^3}{3} +C$ what happened is that you got different C, you can call it D. and it is a constant so $-\frac{1}{6}$ is also a part of that same constant. your expression should be $x - 2 x^2 + \frac{4 x^3}{3} +E$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do I show that $\frac xy + \frac yz + \frac zx \ge 1 + \frac {z + x}{x + y} + \frac {x + y}{z + x}$? Show that
$$\frac xy + \frac yz + \frac zx \ge 1 + \frac {z + x}{x + y} + \frac {x + y}{z + x}$$
for $x, y, z \gt 0$.
I observed that this is a homogeneous inequality so normalization might work. I tried to set $x =... | Using CS inequality in slightly different way,
$$\frac{x}y + \frac{y}z + \frac{z}x + 1 \geqslant \frac{(x+y+z+x)^2}{xy+yz+zx+x^2} = \frac{(\color{red}{x+y}\:+\:\color{blue}{z+x})^2}{(\color{red}{x+y})\cdot(\color{blue}{z+x})} = \frac{x+y}{z+x}+2 + \frac{z+x}{x+y}$$
The cyclic counterparts of the inequality can be show... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all solutions of the equation $3 \cdot 2^{x+2}+5^x=8\cdot 3^x+5$
Find all solutions of the equation
$$3 \cdot 2^{x+2}+5^x=8\cdot 3^x+5$$
My work so far:
Let $f(x)=3 \cdot 2^{x+2}+5^x-8\cdot 3^x-5$
$f'(x)=12\cdot 2^x\ln2+5^x\ln5-8\cdot3^x\ln3$
| Your equation can be changed to $$5(5^{x-1}-1)=24(3^{x-1}-2^{x-1})$$
At least $x=0$, $x=1$ and $x=3$ are solutions.
If your question is a Diophantine problem:
With $n_1,n_2\in\mathbb{N}$ you have the conditions $$3^{x-1}-2^{x-1}=5n_1$$ $$5^{x-1}-1=24n_2$$
EDIT:
$x\in\mathbb{R}$ o.k.
But with the values 0,1,3 you can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I go about solving this? I have tried substitution, but it is not working for me.
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+1)}+\sin(x)+n\cos(x)}=\int_0^\pi \frac{n dx}{\sqrt{(n^2+1)}+n\sin(x)+\cos(x)}=2
$$
General form of this integral is
$$
\int_0^\pi \frac{dx}{\sqrt{(n^2+m^2)}+m\sin(x)+n\cos(x)}=\frac{2}{m}
$$
| This seems so easy in polar form. Let $1=r\cos\phi$, $n=r\sin\phi$, then $r=\sqrt{1+n^2}$, $\phi=\tan^{-1}n$ and then let $x+\phi=y+\frac{\pi}2$
$$\begin{align}\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}+\sin x+n\cos x}&=\int_0^{\pi}\frac{dx}{\sqrt{1+n^2}(1+\sin(x+\phi))}\\
&=\int_{-\frac{\pi}2+\phi}^{\frac{\pi}2+\phi}\frac{dy}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
For which primes $p\not=2$ is $5$ a square mod $p$? For which primes $p\not=2$ is $5$ a square mod $p$?
Using the Legendre symbol, $5$ is a square modulo $p$ if
$$\left(\frac{5}{p}\right)=5^{\dfrac{p-1}{2}} \equiv 1 \pmod{p}$$
Now we have
$$5^{\dfrac{p-1}{2}} =
(5^{p-1})^{1/2} \equiv 1\pmod{p}$$
So $5$ is a square fo... | You can't take the square root like this. By Fermat's little theorem, we know that for $p$ prime we have that $5^{p-1} \equiv 1 \mod p$. This means that $p \mid 5^{p-1} -1$, and hence $$p \mid \left(5^{\tfrac{p-1}{2}} -1\right)\left(5^{\tfrac{p-1}{2}} +1\right)$$
Therefore $5^{\tfrac{p-1}{2}} \equiv 1 \mod p$ or $5^{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$ If $a,A,b,B,c,C$ are non negative reals such that $a+A=b+B=c+C=k$ Prove that $aB+bC+cA \le k^2$
I substituted $B=k-b,C=k-c,A=k-a$ and plugged them to get a quadratic of $k$ which I had to show positive .SO, the discriminant ... | Think of a equilateral triangle $\triangle ABC$ where each side is $k$, then let $$\overline {AD}=A, \overline {DB}=a, \overline {BE}=B, \overline {EC}=b, \overline {CF}=C, \overline {FA}=c$$Where $D,E,F$ are points on $\overline {AB}, \overline {BC}, \overline {CA}$ respectively. Now, notice $$\frac{\sqrt 3}{4}Ac+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1763788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$ If we have
$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$
Then, what will be the set of $x$ for which this equation is true?
I tried to solve it by putting $x = \sin a$ or $\cos a$ but got no result.
I am totally stuck on how to do it.
| Consider the function
$$
f(x)=\arcsin(2x\sqrt{1-x^2})-2\arcsin x
$$
which is defined in the domain where
$$
\begin{cases}
|2x\sqrt{1-x^2}|\le 1\\[4px]
|x|\le 1
\end{cases}
$$
The first inequality becomes, writing $t=|x^2|$,
$$
4t-4t^2\le 1
$$
that's satisfied for every $t$. So our function $f$ is defined over $[-1,1]$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Definite integral of a positive continuous function equals zero? Let's calculate $$\int_0^{\frac\pi 2} \frac {dx}{\sin^6x + \cos^6x}$$
We have
$$\int \frac {dx}{\sin^6x + \cos^6x} = \int \frac {dx}{1 - \frac 34 \sin^2{2x}}$$
now we substitute $u = \tan 2x$, and get
$$\int \frac {dx}{1 - \frac 34 \sin^2{2x}} = \frac{1}{... | Hint
$$\sin^6(x)+\cos^6(x)=(\sin^2(x)+\cos^2(x))((\sin^2(x)+\cos^2(x))^2-\sin^2(x) \cos^2(x))= \\ = 1-\frac{\sin^2(2x)}{4}$$
note I have reduced $a^3+b^3$ and then it's easy
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
How many different values does the expression $1 \pm 2 \pm 3 \pm \cdots \pm n$ take?
I was wondering about this problem and didn't think it was immediately obvious. The answer can't be $2^{n-1}$ since the combinations all might not be unique. ... | Let us use generating functions
$$G_n(X)=X\prod_{k=2}^n(X^{-k}+X^k)=\dfrac{1}{X^p}\prod_{k=2}^n(1+X^{2k}) \ \ \ \text{with} \ \ \ p=\frac{n(n+1)}{2}-2$$
(very close to the proposal of @Jack D'Aurizio). In this way, we transfer the issue into counting how many combinations as positive or negative exponents of indetermin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
Which irrational number represents the infinite simple continued fraction [0;7]? Which irrational number represents the infinite simple continued fraction [0;7]?
-So from my current understanding [o;7] can be represented as the following:
$ = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$
So, x would be $x = \frac{1}{7 ... | Let $x$ be the value of the continued fraction $[0;7]$. We have that
$$ x = \frac{1}{7 + \frac{1}{7 + \frac{1}{7 +...}}}$$
Now we notice that $x$ appears in this continued fraction, so we can write it as:
$$x = \frac{1}{7 + x}$$
This simplifies to $x^2+7x-1=0$, and solving for $x$ gives $x=\frac{-7}{2}\pm \frac{\sqrt{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to properly find supremum of a function $f(x,y,z)$ on a cube $[0,1]^3$? Solving an applied problem I was faced with the need to find supremum of the following function
$$f(x,y,z)=\frac{(x-xyz)(y-xyz)(z-xyz)}{(1-xyz)^3}$$
where $f\colon\ [0,1]^3\backslash\{(1,1,1)\} \to\mathbb{R}$.
I used Wolfram Mathematica and it... | tl; dr: The supremum of $f$ can be found by restricting to the diagonal of the cube and approaching $(1, 1, 1)$.
Partition the cube $C$ into level sets of $xyz$. If $0 < c < 1$, the level set
$$
L_{c} = \{(x, y, z) \in C : f(x, y, z) = c\}
$$
is compact, so $f$ achieves absolute extrema in $L_{c}$. (In fact, $L_{c}$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve by factoring, then perform synthetic division. $${x^4 - 3x^3 + 3x^2 - x = 0}$$
$${x(x^3 - 3x^2 + 3x - 1)}$$
Now, we have to perform synthetic division, if I'm correct we get the 1 to divide by from the X outside of the parentheses from the second equation, which is above this paragraph, and is also the factored f... | Synthetic division is not the greatest tool, in my opinion, because it makes the division process very mysterious. It's a shorthand for long division, and I would recommend writing out the steps of long division instead.
Anyway, to use synthetic division here, you can notice by plugging in a few values of $x$ that $x=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I make sense of this basic algebraic manipulation backwards? $$\frac{1}{n^2-1} = \frac12\left(\frac1{n-1}-\frac1{n+1}\right)$$
It is easy for me to understand this algebraic from right to left ←, but I struggle to find a reasonable way to work from left to right.
Regarding expected difficulty, the level of alg... | Write $$R = \frac{1}{n^2 - 1} = \frac{1}{(n - 1)(n + 1)}$$
Now, we wish to decompose $R$ into something with denominator $(n - 1)$ and denominator $(n + 1)$
The standard way to do this is by a method knows as "partial fractions". We assign $R$ to
$$R = \frac{A}{n - 1} + \frac{B}{n + 1} = \frac{1}{(n - 1)(n + 1)}$$
Cr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.